Title: Multiple Alignment by profile HMM training and Phylogenetic Trees
1Multiple Alignment by profile HMM
trainingandPhylogenetic Trees
- Elze de Groot
-
- Anastacia Berdnikova
2Topics
- 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
3Multiple 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
4Example
- Model estimated from an alignment
5Example continued
- The most probable paths and alignment
6Profile HMM training from unaligned sequences
7Initial 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
8Parameter Estimation
- Use forward and backward variables to re-estimate
emission and transition probability parameters - Baum-Welch re-estimation can be replaced by
viterbi alternative
9Forward Algorithm
10Backward algorithm
11Baum-Welch re-estimation equations
- Expected emission counts from sequence x
12Baum-Welch re-estimation equations
- Expected transition counts from sequence x
13Avoiding 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
14Avoiding local maxima
- Use stochastic search algorithm
- Commonly used Simulated annealing
15Simulated 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
16Simulated annealing (2)
- temperature T
- Probability of state x is given by Gibbs
distribution - Partition function
- x usually multidimensional so impossible to
calculate Z
17Simulated 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
18Simulated annealing for HMM
- Natural energy function negative log of
likelihood logP(data?) - Non-trivial, the two methods Im going to mention
are approximations
19Noise 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
20Simulated 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
21Simulated 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
22Simulated 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
23Simulated annealing Viterbi estimation
- Algorithm Stochastic sampling traceback
algorithm for HMMs
Initialisation pL1 End. Recursion for L1
i 1,
24Simulated 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
25Model 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
26Model Surgery
- How much is a certain transition used by training
sequences - Usage of match state is sum of counts for all
letters in state
27Model 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
28Model 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
29Model 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
30Building phylogenetic trees
31Overview
- The tree of life description
- Background on trees
32Multiple 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.
33The 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.
34- 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.
35Gene 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.
36A tree of orthologues alpha haemoglobins
HBA_ACCGE, HBA_AEGMO, HBA_AILFU, HBA_AILME,
HBA_ALCAA, HBA_ALLMI, HBA_AMBME, HBA_ANAPL
(SWISS-PROT).
37A tree of paralogues HBAT_HUMAN, HBAZ_HUMAN,
HBA_HUMAN, HBB_HUMAN, HBD_HUMAN, HBE_HUMAN,
HBG_HUMAN, MYG_HUMAN (SWISS-PROT).
38Background 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.
39Rooted and unrooted tree
40- 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.
41Counting 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.
42Counting 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.
43Number 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.
44Instead of the root, we can add an extra edge or
branch with a distinct label in its leaf.
45- 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.
46Building phylogenetic trees
47Exercise 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).
48Exercise 7.2
49Exercise 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?
50Exercise 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.)