From Branching processes to Phylogeny:

1
From Branching processes to Phylogeny A history
of reproduction
Elchanan Mossel, U.C. Berkeley
2
Trees in genetics

• Darwin Today - tree of species.
• Today tree of fathers.
• Today tree of mothers.
• Today Descendants for a few generations if there
  are no crossovers (graph has no cycles).

Noah
Shem
Ham
Japheth
Cush
Put
Mizraim
Kannan
3

No brain, Cant move
Information on the tree

• Tree of species characteristics. Wont discuss.
• Tree of mothers Mitochondria.
• Tree of fathers Y chromosome.
• No crossover Full DNA.

Stupid Flies
Stupid Walks
Stupid Swims
Too smart Barely moves
acctga
Noah
Shem
Ham
Japheth
acctga
acctaa
acctga
Put
Cush
Mizraim
Kannan
acctga
acctga
agctga
acctga
4
Genetic mutation models

• How does the D.N.A and other genetic data mutates
  along each edge of the tree?
• Actual mutation is complicated (for Y indels,
  snips, micro satellites, mini satellites).
• A simplified model Split sequence to
  subsequences. All subsequences have same mutation
  rate.
• Thats the standard model in theoretical
  phylogeny.

5
Formal model finite state space

• Finite set A of information values.
• Tree T(V,E) rooted at r.
• Vertex v in V, has information sv in A.
• Edge e(v, u), where v is the parent of u, has a
  mutation matrix Me of size A byA
• A sample of the process is (?v)v in T.
• For each sample, we know ?v, for v in B(T),
• where B(T) is the boundary of the tree B(T)
  n.
• We are given k independent samples ?1B(T),,
  ?kB(T).

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Formal model infinite state space

• Infinite set A assuming no mutations back
  Homoplasy-free models
• Defined on an un-rooted tree T(V,E).
• Edge e has (non-mutation) parameter ?(e).
• Sample Perform percolation edge e open with
  probability ?(e).
• All the vertices v in the same open-cluster have
  the same color ?v. Different clusters get
  different colors.
• We are given k independent samples ?1B(T),,
  ?kB(T).

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Reconstructing the topology

• Given ?1B(T),, ?kB(T), want to reconstruct the
  topology, i.e., the structure of the underlying
  tree on the set of labeled leaves.
• Formally, we want to find for all u and v in B(T)
  their graph-metric distance d(u, v).
• Assume all internal degrees are at least 3.
• Assume B(T) consists only of leaves vertices of
  degree 1.

u
u
Me
Me Me
v
Me
w
w
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Summary Conjectures and results
Statistical physics
Phylogeny
Binary tree in ordered phase
k O(log n)
conj
Binary tree unordered
k poly(n)
conj
Percolation
Homoplasy free
critical ? 1/2
Ising model
CFN
critical ? 2?2 1
Sub-critical random cluster
High mutation
Problems How general? What is the critical
point? (extremality vs. spectral)
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Homoplasy free models and percolation

• Th1M2002 Suppose that n32q and
• T is a uniformly chosen (q1)-level 3-regular
  tree.
• For all e, ?(e)lt ?, and ?lt 1/2.
• Then in order to reconstruct the topology with
  probability ? 0.1, at least k (?q) nO(-log
  ?) samples are needed.
• Th2M2002 Suppose that T has n leaves and
• For all e, ½ ? lt ?(e)lt 1 - ?.
• Then k O(log n log ?) samples suffice to
  reconstruct the topology with probability 1-?.

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The CFN model

• Cavendar-Farris-Neyman model
• 2 data types 1 and 1.
• Mutation along edge e with probability ?(e) copy
  data from parent. Otherwise, choose 1/-1 with
  probability ½ independently of everything else.
• ThmCFN Suppose that for all e, 1 - ? gt ?(e) gt ?
  gt 0.
• Then given k samples of the process at the n
  leaves,
• It is possible to reconstruct the underlying
  topology with probability 1 - ?, if k nO(-log
  ?).

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Phase transition for the CFN model

• Th1M2002 Suppose that n32q and
• T is a uniformly chosen (q1)-level 3-regular
  tree.
• For all e, ?(e)lt ?, and 2?2 lt 1.
• Then in order to reconstruct the topology with
  probability ? gt 0.1, at least k (2-q?-2q)
  nO(-log ?) samples are needed.
• Th2M2002 Suppose that T is a balanced tree on
  n leaves and
• For all e, ?min lt ?(e)lt ?max and 2?2min gt 1, ?max
  lt 1.
• Then k O(log n log ?) samples suffice to
  reconstruct the topology with probability 1-?.

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Metric spaces on trees

• Let D be a positive function on the edges E.
• Define
• Claim Given D(v,u) for all v and u in B(T), it
  is possible to reconstruct the topology of T.
• Proof Suffices to find d(u, v) for all u, v
  B(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) is non-negative (Four point
  condition).

w1
u1
w1
u1
w2
u2
w2
u2
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Metric spaces on trees

• Similarly, d(u1,u2) 4 iff for all w1 and w2
  s.t. for i1,2 and j1,2, d(ui,wj) 4, it holds
  that
• D(u1,u2,w1,w2) 0.
• Remark 1 If D(u1,u2,w1,w2) gt 0, then
• D(u1,u2,w1,w2) ? 2 min_e D(e). Therefore,
  it suffices to know D with accuracy min_e D(e)/4.
• Remark 2 For a balanced tree, in order to
  reconstruct the underlying topology up to l
  levels from the boundary, it suffices to know
  D(u,v) for all u and v s.t. d(u,v) ? 2 l 2
  (with accuracy min_e D(e)/4).

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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 nO(-log ?) samples, it is possible to
  estimate D and therefore reconstruct T with high
  probability.

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Extensions

• Steel 94 Trick to extend to general Me provided
  that that det(Me)? -1,-1? ? - ?, ? ? 1 - ?,
  1,
• using D(e) -log( det(Me) ) (more or less).
• ESSW If tree has small depth then k is smaller.
• For Balanced trees In order to reconstruct up to
  l levels from the boundary, suffices to have k
  T(log(n)).
• Proof Cov(?v, ?u). for u and v with d(u,v) ? 2 l
  2 is at
• least ?2l 2.

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Whats next?

• Its important to minimize the number of samples.
• Do we need k nO(1), or
• do we need k polylog(n)?
• Since the number of trees on n leaves is
  exponential in n log n, and each sample consists
  of n bits, we need at least O(log(n)) samples.

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The Ising model on the binary tree

• The (Free)-Ising-Gibbs measure on the binary
  tree
• Set sr, the root spin, to be /- with probability
  ½.
• For all pairs of (parent, child) (v, w), set sw
  sv, with probability ?, otherwise sw /-
  with probability ½.
• Different Perspective Topology is known and
  looking at a single sample.

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-



-
-


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The Ising model on the binary tree

• Studied in statistical physics Spitzer 75,
  Higuchi 77, Bleher-Ruiz-Zagrebnov 95,
  Evans-Kenyon-Peres-Schulman 2000, M 98
• Interesting phenomena double phase transition
  (different from Ising model in Zd).
• When 2 ? ? 1, unique Gibbs measure.
• When 2 ? 2 ? 1, free measure is extremal.
• In other words,

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The Ising model on the binary tree
From BRZ or EKPS
mutual information H(s?) H(sr)) - H(sr,s?)
Temp ? sr s? 1 Uniq I(sr,s?) Free measure
high lt 1/2 unbiased V ? 0 extremal
med. (1/2,1/v2) biased X ? 0 extremal
low gt 1/v2 biased X Inf gt 0 Non-ext
Remark 2 ? 2 1 phase transition also
transition for mixing time of Glauber dynamics
for Ising model on tree (with Kenyon and Peres)
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Lower bound on number of samples

• Th1M2002 Suppose that n32q and
• T is a uniformly chosen (q1)-level 3-regular
  tree.
• For all e, ?(e)lt ?, and 2?2 lt 1.
• Then in order to reconstruct the topology with
  probability ? gt 0.1, at least k (2-q?-2q)
  nO(-log ?) samples are needed.
• Proof of lower bound uses information
  inequalities.
• Lemma 1EKPS For the single sample process on
  the binary tree of q levels, I(sr,s?) ? (2 ? 2)
  q.

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Lower bound on number of samples

• Lemma 2Fanos inequality X ? Y
• Want to reconstruct the random variable X given
  the random variable Y,
• X takes m values.
• Let pe be the error probability. Then
• 1 pe log2(m) gt H(XY).
• Lemma 3Data Processing Lemma X ? Y ? Z
• If X and Z are independent given Y, then
• I(X, Z) ? min I(X, Y) , I(Y, Z) .

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Lower bound on number of samples

• Assume that topology of bottom q - l levels is
  known, i.e., we know d(u, v) if d(u, v) ? 2(q
  l).
• Then the the conditional distribution of the
  topology T of the first l levels is uniform.
• Let slt for t1,,k be the data at level l for
  sample t. Recall that we are given the Data
  (s?t).
• By Data Processing,
  
  

l
X
?
?
Y
-----
k
q - l
Known
Known
Z
k
-----------
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Lower bound on number of samples

• By independence,
  
• After some manipulations and by EKPS Lemma
• So
• Let
  Since T
• is chosen uniformly,
• Now
• By Fanos Lemma, the probability of error in
  reconstructing the topology, pe, satisfies
• In order to get pe lt 1 - ?, need
• 
  

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Upper bound on number of samples

• Th2M2002 Suppose that T is a balanced tree on
  n leaves and
• For all e, ?min lt ?(e)lt ?max and 2?2min gt 1, ?max
  lt 1.
• Then k O(log n log ?) samples suffice to
  reconstruct the topology with probability 1-?.
• Proves a conjecture of M. Steel.
• 
  

25
Algorithmic aspects of phase transition

• Higuchi 77 For Ising model, when 2 ? 2 gt 1,
  and all q level binary trees Esr Maj(s?) gt ? gt
  0, where Maj is the majority function (? is
  independent of q).
• Looks good for phylogeny because can apply Maj
  even when do not know the topology.
• But, doesnt work when ? is non-constant.
• All edges on blue area have ? 1
• All edges on black area have ? 2
• ? 1 lt ? 2 is close to 1.
• Maj(s?) is very close to Maj of black tree.
• Maj of black tree very close to sv .
• sv and sr are weakly correlated.

r
v
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Algorithmic aspects of phase transition

• Mossel 98 For Ising model, when 2 ? 2 gt 1,
  and all q level binary trees Esr Rec-Majl(s?) gt
  ? gt 0, where Rec-Majl is the recursive majority
  function of l levels (? is independent of q l
  depends on ?).

Rec-Majl for l1

• Looks bad for phylogeny, as we need to know the
  tree topology.
• But main lemma of Mossel98 is extendable to
  non-constant ?.

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Main Lemma M2001

• Lemma Suppose that 2 ? min2 gt 1, then there
  exists an l, and ? gt 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 Esr Maj(s?) ? ?.
• Roughly, given data of quality ? ?, we can
  reconstruct the root with quality ? ?.

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Reconstructing the topology M2001

• 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.
• Recall reconstruction near the boundary take
  O(log n) samples.
• By main lemma quality stays above ?.
• Remark The same algorithm gives (almost) tight
  upper bounds also when 2 ? min2 lt 1.

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Remarks and open problems

• For CFN model, algorithm is very nice
• Polynomial time.
• Adaptive (dont need to know ? min and ? max in
  advance).
• Nearly optimal.
• Main problem extending main lemma to
  non-balanced trees and other mutation models
  (reconstructing local topology still works).
• Secondary problem extending lower bounds to
  other models.

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Proving main Lemma

• Need to estimate Esr Maj(s?). 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 Esr Maj(s?)
  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.