Multiple Alignment by profile HMM training and Phylogenetic Trees

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Multiple Alignment by profile HMM
trainingandPhylogenetic Trees

• Elze de Groot
• Anastacia Berdnikova

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Topics

• Multiple alignment with known HMM
• HMM training from unaligned sequences
• Avoiding local maxima
• Simulated annealing
• Noise injection
• Stochastic sampling traceback algorithm
• Model surgery
• Phylogenetic trees

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Multiple alignment with known profile HMM

• Multiple alignment and model known -gt align large
  number of other family members
• Calculating Viterbi alignment for every sequence
• Residues in same match state are aligned in
  columns
• Thats a difference between profile HMM and
  traditional multiple alignment

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Example

• Model estimated from an alignment

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Example continued

• The most probable paths and alignment

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Profile HMM training from unaligned sequences

• Algorithm

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Initial Model

• Choose length of model
• - M is number of match states
• - set M to be the average length
• Choose initial models carefully
• Randomness in choice of initial model

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Parameter Estimation

• Use forward and backward variables to re-estimate
  emission and transition probability parameters
• Baum-Welch re-estimation can be replaced by
  viterbi alternative

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Forward Algorithm
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Backward algorithm
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Baum-Welch re-estimation equations

• Expected emission counts from sequence x

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Baum-Welch re-estimation equations

• Expected transition counts from sequence x

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Avoiding local maxima

• Baum-Welch guaranteed to find local maxima
• Not guaranteed it is anywhere near global optimum
  or biologically reasonable solution
• Reason models are long -gt many options to get
  wrong solution

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Avoiding local maxima

• Use stochastic search algorithm
• Commonly used Simulated annealing

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Simulated annealing

• Some compounds only cristallise if they are
  slowly annealed from high to low temperature
• Optimisation problem minimise function energy
  E(x)
• Maximising function same as minimising negative
  value of function

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Simulated annealing (2)

• temperature T
• Probability of state x is given by Gibbs
  distribution
• Partition function
• x usually multidimensional so impossible to
  calculate Z

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Simulated annealing (3)

• T?0, all configurations except with lowest energy
  are prob 0 (system is frozen)
• T??, All configuration have same prob (system is
  molten)
• With crystallisation minimum can be found by
  sampling this distribution at high temperature
  first and then decreasing temperatures

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Simulated annealing for HMM

• Natural energy function negative log of
  likelihood logP(data?)
• Non-trivial, the two methods Im going to mention
  are approximations

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Noise injection

• Adding noise to counts estimated in
  forward-backward procedure and let size of noise
  decrease slowly
• In Krogh et al.1994 the noise was generated by
  a random walk in the initial model

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Simulated annealing Viterbi estimation

• If there are N sequences, theres an exact
  translation from the N paths ?1,, ?N to the
  parameters of the model
• Treat the paths as fundamental parameters in
  which to maximise the likelihood
• Simulated annealing done in these variables
  instead of the model parameters

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Simulated annealing Viterbi estimation

• Denominator is Z, the partition function -gt sum
  over all paths
• Can be obtained by modified forward algorithm
  using exponentiated transmission and emission
  parameters

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Simulated annealing Viterbi estimation

• Exponentiated transmission parameter
• âij aij1/T
• Exponentiated emission parameter
• êj(x) ej(x)1/T
• Used in place of unmodified probability
  parameters in forward algorithm
• Z is result of forward algorithm

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Simulated annealing Viterbi estimation

• Algorithm Stochastic sampling traceback
  algorithm for HMMs

Initialisation pL1 End. Recursion for L1
i 1,
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Simulated annealing Viterbi vs Viterbi

• Key difference
• Viterbi selects highest probable path for each
  sequence
• Simulated annealing samples each path according
  to the likelihood of the path

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Model Surgery

• During training a model two things can happen
• (a) some match states are redundant and should be
  absorbed in insert state
• (b) one or more insert states aborb too much
  sequence, in which case they should be expanded

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Model Surgery

• How much is a certain transition used by training
  sequences
• Usage of match state is sum of counts for all
  letters in state

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Model surgery

• If match state is used by less than ½ sequences
  -gt delete module
• If more than ½ of sequences use the transitions
  into an insert state, this is expanded to new
  modules

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Model surgery Example SAM

• I tried a sequence in SAM with and without model
  surgery
• Same 7 sequences as in example before
• Parameters ltcutinsert 0.25gt ltcutmatch 0.5gt -gt
  delete any match state used by fewer than half
  the sequences, and insert match states for any
  insert node used by greater than one quarter of
  the sequences

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Model surgery Example SAM

• Without model surgery
• gtseq1
• FPHFD.....L...S.....-HGSAQ
• gtseq2
• FESFG.....D...LstpdaVMGNPK
• gtseq3
• FDRFKhlkteA...E.....MKASED
• gtseq4
• FTQFA.....G...Kdles.IKGTAP
• gtseq5
• FPKFK.....G...LttadqLKKSAD
• gtseq6
• FSFLK.....GtseV.....PQNNPE
• gtseq7
• FGFSG.....A...-.....--SDPG

• With model surgery
• gtseq1
• FPHF.DLS-..-..--HGSAQ
• gtseq2
• FESF.GDLStpD..AVMGNPK
• gtseq3
• FDRF.KHLK..TeaEMKASED
• gtseq4
• FTQFaGKDL..E..SIKGTAP
• gtseq5
• FPKF.KGLTtaD..QLKKSAD
• gtseq6
• FSFL.KGTS..E..VPQNNPE
• gtseq7
• FGFS.G---..-..--ASDPG

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Building phylogenetic trees
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Overview

• The tree of life description
• Background on trees

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Multiple alignment and trees

• Alignment of sequences should take account of
  their evolutionary relationship. Sankoff, Morel
  Cedergren, 1973
• Several progressive alignment algorithms use a
  guide tree (to guide the clustering process).
• We begin to build trees.

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The tree of life

• The similarity of molecular mechanisms of the
  organisms that have been studied strongly
  suggests that all organisms on Earth had a common
  ancestor. Thus any sets of species is related,
  and this relationship is called a phylogeny.
• Usually the relationship can be represented by a
  phylogenetic tree.

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• Zuckerkandl Paulings paper 1962 showed that
  molecular sequences provide sets of morphological
  characters that can carry a large amount of
  information.
• An assumption the sequencies we want to analyze
  on the phylogeny matter have descended from some
  common ancestral gene in a common ancestral
  species.
• Gene duplication exists gt we have to check the
  assumption carefully.

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Gene duplication and speciation

• By another mechanism, gene duplication, two
  sequences can also be separated and diverge from
  the common ancestor.
• Genes which diverged because of speciation are
  called orthologues. Genes which diverged by gene
  duplication are called paralogues.

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A tree of orthologues alpha haemoglobins
HBA_ACCGE, HBA_AEGMO, HBA_AILFU, HBA_AILME,
HBA_ALCAA, HBA_ALLMI, HBA_AMBME, HBA_ANAPL
(SWISS-PROT).
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A tree of paralogues HBAT_HUMAN, HBAZ_HUMAN,
HBA_HUMAN, HBB_HUMAN, HBD_HUMAN, HBE_HUMAN,
HBG_HUMAN, MYG_HUMAN (SWISS-PROT).
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Background on trees

• All trees will be assumed to be binary (an edge
  that branches splits into two daughter edges).
• Each edge of the tree has a certain amount of
  evolutionary divergence associated to it. We
  adopt the general term length, which will be
  represented by lengthes of edges on figures.
• A true biological phylogeny has a root, or
  ultimate ancestor of all sequences.

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Rooted and unrooted tree
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• A tree with a given labelling will be called a
  labelled branching pattern.
• We refer to this as the tree topology and denote
  it by T.
• Lengths of the edges ti with a suitable
  numbering scheme for the is.

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Counting and labelling

• Rooted tree
• n leaves, plus (n-1) branch nodes in addition to
  leaves -gt we have 2n-1 nodes in all, and 2n-2
  edges.
• leaves 1..n, branch nodes n1 .. 2n-1,
  (2n-1)th node is root.

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Counting and labelling

• Unrooted tree
• n leaves, 2n-2 nodes and 2n-3 edges.
• a root can be added at any of its edges gt we can
  get 2n-3 rooted trees.

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Number of rooted and unrooted trees
A root can be added at any edge, producing 2n-3
rooted trees from unrooted tree gt there are
(2n-3) times as many rooted trees as unrooted
trees, for a given number n of leaves.
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Instead of the root, we can add an extra edge or
branch with a distinct label in its leaf.
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• There are three such trees with (2n-3)5 leaves
  they are distinct labelled branching patterns.
• There are then five ways of adding a further
  branch labelled with a distinct label (5),
  giving in all 3x515 unrooted trees with five
  leaves.
• The number of unrooted trees with n leaves is
  equal to 35...(2n-5) (2n-5)!! So, we have
  (2n-3)!! rooted trees with n leaves.

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Building phylogenetic trees

• Questions?

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Exercise 7.2

• The trees with three and four leaves in Figure
  7.3 all have the same unlabelled branching
  pattern. For both rooted and unrooted trees, how
  many leaves do there have to be to obtain more
  than one unlabelled branching pattern? Find a
  recurrence relation for the number of rooted
  trees. (Hint consider the trees formed by
  joining two trees at their root).

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Exercise 7.2
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Exercise 7.3

• All trees considered so far have been binary, but
  one can envisage ternary trees that, in their
  rooted form, have three branches descending from
  a branch node. If there are m branch nodes in an
  unrooted ternary tree, how many leaves are there
  and how many edges?

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Exercise 7.4

• Consider next a composite unrooted tree with m
  ternary branch nodes and n binary branch nodes.
  How many leaves are there, and how many edges?
  Let Nm,n denote the number of distinct labelled
  branching patterns of this tree. Extend the
  counting argument for binary trees to show that
• Nm,n (3m2n-1)N m,n-1 (n1)N m-1,n1
• (Hint the first term after the counts
  the number of ways that a new edge can be added
  to an existing edge, thereby creating an
  additional binary node the second term
  corresponds to edges added at binary nodes,
  thereby producing ternary nodes.)