1. Reconstruction on trees and Phylogeny 3
Elchanan Mossel,
U.C. Berkeley
Supported by Microsoft Research and the Miller Institute
1/18/2019

2. Phylogeny
“Phylogeny is the true evolutionary relationships between groups of living things”
Noah
Shem
Japheth
Ham
Cush
Kannan
Mizraim
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3. History of Phylogeny Prehistory: “animal kingdom” or “plant kingdom.”
Intuitively:
More scientifically: morphology, fossils, etc. Darwin …
But: Is a human more like a great ape or like a chimpanzee?
No brain,
Can’t move
Stupid
Walks
Stupid
Swims
Stupid
Flies
Too smart
Barely moves
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4. Molecular Phylogeny
Molecular Phylogeny: Based on DNA, RNA or protein sequences of organisms.
Rooted / Unrooted trees: Evolution from common ancestor modeled on a rooted tree. Usually reconstruct unrooted trees.
Mutation mechanisms:
Substitutions
Transpositions
Insertions, Deletions, etc.
Will only consider substitutions
and assume sequences are aligned.
Noah
acctga
Shem
Japheth
Ham
acctaa
acctga
acctga
Put
Cush
Kannan
Mizraim
acctga
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acctga
agctga
acctga

5. Genetic substitution models and trees
Assumption 1: Letters of sequences (“characters”) evolve independently and identically.
Assumption 2: Trees are binary -- All internal degrees are 3 (bifurcating speciation; results valid if degrees are ¸ 3).
Given a set of species (labeled vertices) X, an X-tree is a tree which has X as the set of leaves.
Two X-trees T1 and T2 are identical if there’s a graph isomorphism between T1 and T2 that is the identity map on X. u u
Me’ v
Me’ Me’’
Me’’ w w d a c b d a b c c a b d
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6. Substitution model – finite state space
Finite set A of information values (|A| = 4 for DNA).
Tree T=(V,E) rooted at r.
Vertex v 2 V, has information σv 2 A.
Edge e=(v, u), where v is the parent of u, has a mutation matrix Me of size |A| £ |A|:
Mi,j (v,u) = P[u = j | v = i]
Will focus on the CFN model:
A character is (v)v 2 T.
For each character , the data is T = (v)v 2 T,
where T is the boundary of the tree; |T| = n.
We are given k independent characters 1T,…, kT.
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7. A diagram
Length of sequence!
Interested to know k = #characters needed to reconstruct the tree with n = #leaves, given a range [max,min] for mutation rate .
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8. Phylogeny: Conjectures and results
Statistical physics
Phylogeny
Binary tree in ordered phase
conj
k = O(log n)
Binary tree unordered
conj
k = poly(n)
Percolation
critical  = 1/2
Random Cluster
M-Steel2003
CFN
M-2003
Ising model
critical : 22 = 1
Sub-critical
representation
High mutation
M-2003
Problems: How general?
What is the critical point? (extremality vs. spectral)
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9. Cavendar-Farris-Neyman model:
The CFN model
Cavendar-Farris-Neyman model:
2 data types: 1 and –1 (“purine-pyrimidine”)
Mutation along edge e: with probability (e) copy data from parent. Otherwise, choose 1/-1 with probability ½ independently of everything else
Thm[CFN] Suppose that for all e, 1 -  > (e) >  > 0.
Then given k characters of the process at n leaves,
It is possible to reconstruct the underlying topology with probability 1 - , if k = nO(-log ).
Steel 94: Trick to extend to general Me provided
that det(Me) [-1,-1+]  [- , ]  [1 - , 1],
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10. Phase transition for the CFN model
Th1[M2003]: Suppose that n=3 £ 2q and
T is a uniformly chosen (q+1)-level 3-regular X-tree.
For all e, (e) < , and 22 < 1.
Then in order to reconstruct the topology with probability  > 0.1, at least k = (n(-2log2() - 1)) characters are needed.
Proof: Information theoretic variant of the proof for random cluster model.
Same proof applies to any model for which the reconstruction problem is unsolvable.
more formally, for models for which I(,n) decays exp. fast in n.
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11. CFN Logarithmic reconstruction
Th2[M2003]: If T is an X-tree on n leaves s.t.
For all e, min < (e)< max and 22min > 1, max < 1.
Then k = O(log n – log ) characters suffice to reconstruct the topology with probability 1- .
Need either a “balanced tree” – all leaves at the same distance from a root.
Or, “molecular clock” – (e) = e-t(e), where t(e) is the time interval between the two endpoints of the interval + all leaves are at the same time.
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12. Main Lemma [M2003]
Lemma: Suppose that 2  min2 > 1, then there exists an L, and  > 0 such that the CFN model on the binary tree of L levels with
(e)   min, for all e not adjacent to ∂T.
(e)    min , for all e adjacent to ∂T.
satisfies E[σr Maj(σ∂)]  .
Roughly, given boundary data of “quality  ”, we can reconstruct the root data with “quality  ”.
In phylogeny – can treat known pieces of the tree as vertices.
Main problem: how to reconstruct pieces of the tree?
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13. Metric spaces on trees Let D be a positive function on the edges E.
Define D(u,v) =  {D(e) : e 2 path(u,v)}.
Claim: Given D(v,u) for all v and u in T, it is possible to reconstruct the topology of T.
Proof: Suffices to find d(u, v) for all u, v 2 T where d is the graph metric distance.
d(u1,u2) = 2 iff for all w1 and w2 it holds that
D’(u1,u2,w1,w2) := D(u1,w1)+D(u2,w2) –D(u1,u2)–D(w1,w2) ¸ 0 (“Four point condition”). w1 u1 w1 u1 w2 u2 w2 u2
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14. Metric spaces on trees
Continue by replacing known sub-trees T on vertices (v1,…,vr) by a single vertex v.
The distance between (v1,…,vr) and (u1,…us) is defined as d(v1,u1).
D’(u1,u2,w1,w2) > 0 ) D’(u1,u2,w1,w2)  2 min_e D(e).
Suffices to have D with accuracy min_e D(e)/4.
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15. Metric spaces on trees Let T be a balanced tree.
The L-topology of T is
d¤(u,v) := min{d(u,v},2L}.
Claim: If T is balanced, then in order to recover the L-topology of T it suffices to have
For each leaf u of T a set U(u) containing all elements at distance · 2L+2 from u.
For all u and all w1,w2,w3,w4 2 U(u) the sign of D’(w1,w2,w3,w4).
“proof”: If d(u1,u2) > 2, then either
u2 is not in U(u1), or
Let v be a sister of u1 and v’ a cousin of v.
D’(u1,v,u2,v’) > 0.
We have a witness that u1 and u2 are not siblings. u2 v’ v u1
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16. Proof of CFN theorem Define D(e) = - log (e).
D(u,v) = -log(Cov(v,u)), where Cov(v, u) = E[vu].
Estimate Cov(v, u) by Cor(v, u) where
Need D with accuracy m = min D(e)/4 = c, or
Cor = (1  c)Cov.
Cor(v, u) is a sum of k i.i.d.  1 variables with expected value Cov(v, u).
Cov(v, u) may be a small as  2 depth(T) = n-O(-log ).
Given k = nΩ(-log ) characters, it is possible to estimate D and therefore reconstruct T with high probability.
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17. Reconstructing the topology [M2003]
The algorithm: Repeat the following:
Reconstruct the topology up to l levels from the boundary using 4-points method.
For each sample, reconstruct the data l levels from the boundary using majority algorithm. + - - +
Reconstruction near the boundary take O(log n) samples.
By main lemma quality stays above .
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18. Proving main Lemma
Need to estimate E[σr Maj(σ∂)]. Estimate has two parts:
Case 1: For all e adjacent to ∂T, (e) is small. Here we use a perturbation argument, i.e. estimate partial derivatives of E[σr Maj(σ∂)] with respect to various variables (using something like Russo formula).
Case 2: Some e adjacent to ∂T has large (e). Use percolation theory arguments.
Both cases uses isoperimetric estimates for the discrete cube.
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