Elze de Groot & Anastas ia Berdnikova 1 Multiple Alignment by Multiple Alignment by profile HMM training profile HMM training and and Phylogenetic Trees Phylogenetic Trees Elze de Groot & Anastacia Berdnikova
Dec 30, 2015
Elze de Groot & Anastasia Berdnikova
1
Multiple Alignment by profile Multiple Alignment by profile HMM trainingHMM training
andandPhylogenetic TreesPhylogenetic Trees
Elze de Groot
&
Anastacia Berdnikova
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TopicsTopics
Multiple alignment with known HMMHMM training from unaligned sequencesAvoiding local maxima
– Simulated annealing– Noise injection– Stochastic sampling traceback algorithm
Model surgeryPhylogenetic trees
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Multiple alignment with known Multiple alignment with known profile HMMprofile HMM
Multiple alignment and model known -> align large number of other family members
Calculating Viterbi alignment for every sequence
Residues in same match state are aligned in columns
That´s a difference between profile HMM and traditional multiple alignment
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Example continuedExample continuedThe most probable paths and alignment
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Profile HMM training from Profile HMM training from unaligned sequencesunaligned sequences
Algorithm:
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Initial ModelInitial Model
Choose length of model
- M is number of match states
- set M to be the average length
Choose initial models carefullyRandomness in choice of initial model
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Parameter EstimationParameter 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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Baum-Welch re-estimation Baum-Welch re-estimation equationsequations
Expected emission counts from sequence x
)()()(
1)(
)()()(
1)(
|
|
ibifxP
aE
ibifxP
aE
k
i
kk
k
i
kk
Iaxi
II
Maxi
MM
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Baum-Welch re-estimation Baum-Welch re-estimation equationsequations
Expected transition counts from sequence x
)1()()(
1
)1()1()()(
1
)1()1()()(
1
111
1111
ibaifxP
A
ibxeaifxP
A
ibxeaifxP
A
kkkkkk
kkikkkk
kkkkkkk
Di
DXXDX
Ii
iIIXXIX
Mi
iMMXXMX
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Avoiding local maximaAvoiding 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 -> many options to get wrong solution
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Avoiding local maximaAvoiding local maxima
Use stochastic search algorithm
Commonly used: Simulated annealing
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Simulated annealingSimulated 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)Simulated annealing (2) ´temperature´ T Probability of ´state´ x is given by Gibbs
distribution
Partition function:
x usually multidimensional so impossible to calculate Z
E(x)
TZP(x)
1exp
1
dxxET
Z
)(
1exp
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Simulated annealing (3)Simulated annealing (3)
T0, 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 HMMSimulated annealing for HMM
Natural energy function negative log of likelihood –logP(data|)
Non-trivial, the two methods I´m going to mention are approximations
´´)|data(
)|data()|data(
1)|data(log
1exp
1/1
/1/1
dP
PP
ZP
TZ T
TT
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Noise injectionNoise 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 Simulated annealing Viterbi estimationestimation
If there are N sequences, there´s 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 Simulated annealing Viterbi estimationestimation
Denominator is Z, the partition function -> sum over all paths
Can be obtained by modified forward algorithm using exponentiated transmission and emission parameters
´
/1
/1
)|´,()|,(
)(Prob
T
T
xPxP
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Simulated annealing Viterbi Simulated annealing Viterbi estimationestimation
Exponentiated transmission parameter– âij = aij
1/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 Simulated annealing Viterbi estimationestimation
Algorithm: Stochastic sampling traceback algorithm for HMMs
k ikiiii iiiiâfâf ,,1,,11 /|Prob
11
Initialisation: πL+1 = End.Recursion: for L+1 ≥ i ≥ 1,
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Simulated annealing Viterbi vs Simulated annealing Viterbi vs ViterbiViterbi
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 SurgeryModel 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 SurgeryModel 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 surgeryModel surgery
If match state is used by less than ½ sequences -> 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 Model surgery – Example SAMSAM
I tried a sequence in SAM with and without model surgery
Same 7 sequences as in example beforeParameters <cutinsert 0.25> <cutmatch
0.5> -> 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 Model surgery – Example SAMSAM
Without model surgery>seq1FPHFD.....L...S.....-HGSAQ>seq2FESFG.....D...LstpdaVMGNPK>seq3FDRFKhlkteA...E.....MKASED>seq4FTQFA.....G...Kdles.IKGTAP>seq5FPKFK.....G...LttadqLKKSAD>seq6FSFLK.....GtseV.....PQNNPE>seq7FGFSG.....A...-.....--SDPG
With model surgery>seq1FPHF.DLS-..-..--HGSAQ>seq2FESF.GDLStpD..AVMGNPK>seq3FDRF.KHLK..TeaEMKASED>seq4FTQFaGKDL..E..SIKGTAP>seq5FPKF.KGLTtaD..QLKKSAD>seq6FSFL.KGTS..E..VPQNNPE>seq7FGFS.G---..-..--ASDPG
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OverviewOverview
The tree of life – descriptionBackground on trees
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Multiple alignment and treesMultiple 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 lifeThe 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 & Pauling’s 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 => we have to check the assumption carefully.
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Gene duplication and speciationGene 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 orthologuesA tree of orthologues: alpha haemoglobins HBA_ACCGE, : alpha haemoglobins HBA_ACCGE, HBA_AEGMO, HBA_AILFU, HBA_AILME, HBA_ALCAA, HBA_AEGMO, HBA_AILFU, HBA_AILME, HBA_ALCAA, HBA_ALLMI, HBA_AMBME, HBA_ANAPL (SWISS-PROT).HBA_ALLMI, HBA_AMBME, HBA_ANAPL (SWISS-PROT).
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A tree of paraloguesA tree of paralogues: HBAT_HUMAN, HBAZ_HUMAN, : HBAT_HUMAN, HBAZ_HUMAN, HBA_HUMAN, HBB_HUMAN, HBD_HUMAN, HBE_HUMAN, HBA_HUMAN, HBB_HUMAN, HBD_HUMAN, HBE_HUMAN, HBG_HUMAN, MYG_HUMAN (SWISS-PROT).HBG_HUMAN, MYG_HUMAN (SWISS-PROT).
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Background on treesBackground 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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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 labellingCounting and labelling
Rooted tree:– n leaves, plus (n-1) branch nodes in
addition to leaves -> we have 2n-1 nodes in all, and 2n-2 edges.
– leaves – 1..n, branch nodes – n+1 .. 2n-1, (2n-1)th node is root.
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Counting and labellingCounting and labelling
Unrooted tree:– n leaves, 2n-2 nodes and 2n-3 edges.– a root can be added at any of its
edges => we can get 2n-3 rooted trees.
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Number of rooted and unrooted Number of rooted and unrooted treestrees
A root can be added at any edge, producing 2n-3 rooted trees from unrooted tree => 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 3x5=15 unrooted trees with five leaves.
● The number of unrooted trees with n leaves is equal to 3*5*...*(2n-5) = (2n-5)!! So, we have (2n-3)!! rooted trees with n leaves.
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Building phylogenetic treesBuilding phylogenetic trees
Questions?
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Exercise 7.2Exercise 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 7.3Exercise 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 7.4Exercise 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 = (3m+2n-1)N m,n-1 + (n+1)N m-1,n+1
(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.)