.

Sequence Alignment IIILecture #4

This class has been edited from Nir Friedman’s lecture which is available at www.cs.huji.ac.il/~nir. Changes made by Dan Geiger and Shlomo Moran.

Background Readings: The second chapter (pages 12-45) in the text book, Biological Sequence Analysis, Durbin et al., 2001.Chapter 3 in Introduction to Computational Molecular Biology, Setubal and Meidanis, 1997.

2

Score(s[i],t[i])

Log Odd-Ratio Test for AlignmentTaking logarithm of Q yields

])[(])[(

])[],[(log

])[(])[(

])[],[(log

)|,Pr(

)|,Pr(log

itqisq

itisp

itqisq

itisp

Rts

Mts

ii

If log Q > 0, then s and t are more likely to be related.If log Q < 0, then they are more likely to be unrelated.

How can we relate this quantity to a score function ?

3

Estimating p(·,·) for proteinsGenerate a large diverse collection of accepted mutations. An accepted mutation is a mutation due to an alignment of closely related protein sequences. For example, Hemoglobin alpha chain in humans and other organisms (homologous proteins).

Let pa = na/n where na is the number of occurrences of letter a and n is the total number of letters in the collection, so n = ana.

Mutation counts be the number of mutations a b, be the total number of mutations that involve a, be the total number of amino acids involved in a mutation.

abb aba ff|

baab ff

Note that f is twice the number of mutations.

a aff

4

PAM-1 matricesDefine Mab to be the symmetric probability matrix for switching between a and b. We set, Maa = 1 – ma, so that ma is the probability that a is involved in a change.

aab maababaMa

ab

f

fchanged)Pr(changed)|Pr()Pr(

We define Mab, such that only 1% of amino acids change according to this matrix or 99% don’t. Hence the name, 1-Percent Accepted Mutation (PAM). In other words,

99.011 a aaaa aaaa a mpmpMp

5

PAM-1 matrices

fpK

fm

a

aa where K is a proportional constant.

We wish that ma will be proportional to the relative mutability of letter a compared to other letters.

So K=100 for PAM-1 matrices. Note that K=50 yields 2% change, etc.

We select K to satisfy the PAM-1 definition:

01.0 aa amp

fKp

fp

a

aa a KfK

fa a 1

6

Evolutionary distance

The choice that 1% of amino acids change (and that K =100) is quite arbitrary. It could fit specific set of proteins whose evolutionary distance is such that indeed 1% of the letters have mutated.

This is a unit of evolutionary change, not time because evolution acts differently on distinct sequence types.

What is the substitution matrix for k units of evolutionary time ?

7

Model of Evolution

We make some assumptions:

1. Each position changes independently of the rest

2. The probability of mutations is the same in each position

3. Evolution does not “remember”

Timet t+ t+2 t+3 t+4

A A C C GT T T C G

8

Model of Evolution

How do we model such a process? This process is called a Markov Chain

A chain is defined by the transition probability P(Xt+ =b|Xt=a) - the probability that the next

state is b given that the current state is a We often describe these probabilities by a matrix:

M[]ab = P(Xt+ =b|Xt=a)

9

Multi-Step Changes

Thus M[2] = M[]M[]

By induction (HMW exercise): M[n] = M[] n

c

ttttt

tt

aXcXPaXcXbXP

aXbXP

)|(),|(

)|(2

2

Based on Mab, we can compute the probabilities of changes over two time periods

c cbac

c

tttt

MM

aXcXPcXbXP )|()|( 2Using Conditional independence (No memory)

10

A Markov Model (chain)X1 X2 Xn-1 Xn

•Every variable xi has a domain. For example, suppose the domain are the letters {a, c, t, g}.•Every variable is associated with a local probability table

P(Xi = xi | Xi-1= xi-1 ) and P(X1 = x1 ).•The joint distribution is given by

n

iiin xpxxp

11 )|(),,( paIn short, we write:

n

iiiii

nnnnnn

PxXp

xXxXPxXxXPxXPxXxXp

1

1111221111

)|(

)|()|()(),,(

paa

where Pai are the parents of variable/node Xi ,namely, none or Xi-1.

11

Markov Model of Evolution Revisited

In the evolution model we studied earlier we hadP(x1) = (pa, pc, pg, pt) which sum to 1 and called the prior probabilities, andP(xi|xi-1) = M[] which is a stationary transition probability table, not depending on the index i.

The quantity we computed earlier from this model was the joint probability table

nxx

nn Mxpxxp

1][)(),( 11

X1 X2 Xn-1 XnM M

12

Longer Term Changes

Estimate M[] = M (PAM-1 matrices) Use M[n] = Mn (PAM-n matrices) Define

Use this quantity to define the score for your application of interest.

abn

a Mpbap ),(

13

Comments regarding PAM

Historically researchers use PAM-250. (The only one published in the original paper.)

Original PAM matrices were based on small number of proteins (circa 1978). Later versions use many more examples.

Used to be the most popular scoring rule, but there are some problems with PAM matrices.

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Degrees of freedom in PAM definition

With K=100 the 1-PAM matrix is given by

fp

f

fpK

f

f

fm

f

fM

a

ab

a

a

a

aba

a

abab

100

With K=50 the basic matrix is different, namely:

fp

fM

a

abab

50

'

Use the 1-PAM matrix to the fourth power: M[4] = M[] 4

OrUse the K=50 matrix to the second power: M[4] = M[2] 2

Thus we have two different ways to estimate the matrix M[4] :

15

Problems in building distance matrices

How do we find pairs of aligned sequences? How far is the ancestor ?

earlier divergence low sequence similarity later divergence high sequence similarity

E.g., M[250] is known not reflect well long period changes.

Does one letter mutate to the other or are they both mutations of a third letter ?

16

BLOSUM Outline

Idea: use aligned ungapped regions of protein families.These are assumed to have a common ancestor. Similar ideas but better statistics and modeling.

Procedure: Cluster together sequences in a family whenever more

than L% identical residues are shared. Count number of substitutions across different clusters

(in the same family). Estimate frequencies using the counts.

Practice: Blosum50 and Blosum62 are wildly used. (See page 43-44 in the text book).

Considered state of the art nowadays.

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Multiple Sequence Alignment

S1=AGGTC

S2=GTTCG

S3=TGAACPossible alignment

A-T

GGG

G--

TTA

-TA

CCC

-G-

Possible alignment

AG-

GTT

GTG

T-A

--A

CCA

-GC

18

Multiple Sequence Alignment

Definition: Given strings S1, S2, …,Sk a multiple (global) alignment map them to strings S’1, S’2, …,S’k that may contain blanks, where:

1. |S’1|= |S’2|=…= |S’k|

2. The removal of spaces from S’i leaves Si

Aligning more than two sequences.

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Multiple alignmentsWe use a matrix to represent the alignment of k sequences,

K=(x1,...,xk). We assume no columns consists solely of blanks.

M Q _ I L L L

M L R - L L -

M K _ I L L L

M P P V L I L

The common scoring functions give a score to each column, and set: score(K)= ∑i score(column(i))

For k=10, a scoring function has 2k -1 > 1000 entries to specify. One need not have a general function. For example, the order of arguments need not matter: score(I,_,I,V) = score(_,I,I,V).

x1

x2

x3

x4

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SUM OF PAIRS

M Q _ I L L L

M L R - L L -

M K _ I L L L

M P P V L I L

A common scoring function is SP – sum of scores of the projected pairwise alignments: SPscore(K)=∑i<j score(xi,xj).

In order for this score to be written as ∑i score(column(i)),we must specify score(-,-)=0. Why ?

Because these entries appear in the sum of columns but not in the sum of projected pairwise alignments (lines).

Note that we need to specify the score(-,-) because a column may have several blanks (as long as not all entries are blanks).

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SUM OF PAIRS

M Q _ I L L L

M L R - L L -

M K _ I L L L

M P P V L I L

Definition: The sum-of-pairs (SP) value for a multiple global

alignment A of k strings is the sum of the values of all projected

pairwise alignments induced by A where the pairwise alignment

function score(xi,xj) is additive.

2

k

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Example

Consider the following alignment:

a c - c d b -

- c - a d b d

a - b c d a d

Using the edit distance and for ,

this alignment has a SP value of

0, xx 1, yx yx

33 +43 + 4 + 5 = 12

23

Multiple Sequence Alignment

Given k strings of length n, there is a natural generalization of the dynamic programming algorithm that finds an alignment that maximizes

SP-score(K) = ∑i<j score(xi,xj).

Instead of a 2-dimensional table, we now have a k-dimensional table to fill.

For each vector i =(i1,..,ik), compute an optimal multiple alignment for the k prefix sequences x1(1,..,i1),...,xk(1,..,ik).

The adjacent entries are those that differ in their index by one or zero. Each entry depends on 2k-1 adjacent entries.

24

The idea via K=2

])[,(],[

)],[(],[

])[],[(],[

max],[

1jtj1iV

1is1jiV

1jt1isjiV

1j1iV

])..[],..[(],[ j1ti1sdjiV

V[i,j] V[i+1,j]

V[i,j+1] V[i+1,j+1] Note that the new cell index (i+1,j+1) differs from previous indices by one of 2k-1 non-zero binary vectors (1,1), (1,0), (0,1).

Recall the notation:

and the following recurrence for V:

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The idea for arbitrary k

Order the vectors i=(i1,..,ik) by increasing order of the sum ∑ij. Set s(0,..,0)=0, and for i > (0,...,0):

( ) max{ ( ) ( ( , ))}s s score columnb

i i b i b

•The vector b ranges over all non-zero binary vectors. •The vector i-b is the non-negative difference of i and b. •The jth entry of column(i,b) equals cj= xj(ij) if bi=1, and cj= ‘-’ otherwise.(Reflecting that b is 1 at location j if that location changed in the “current comparison”).

Where

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Complexity of the DP approach

Number of cells nk.

Number of adjacent cells O(2k).Computation of SP score for each column(i,b) is o(k2)

Total run time is O(k22knk) which is utterly unacceptable !

Not much hope for a polynomial algorithm because the problem has been shown to be NP complete.

Need heuristic to reduce time.

27

Time saving heuristics: Relevance testsHeuristic: Avoid computing score(i) for irrelevant vectors.

M Q _ I L L L

M L R - L L -

M K _ I L L L

M P P V L I L

x1

x2

x3

x4

Let L be a lower bound on the optimal SP score of a multiple alignment of the k sequences. A lower bound L can be obtained from an arbitrary multiple alignment, computed in any way.

Main idea: Using L, compute lower bounds Luv for the optimal score for every two sequences s=xu and t=xv, 1 u < v k. When processing vector i=(..iu,..iv…), the relevant cells are such that in every projection on xu and xv, the optimal pairwise score is above Luv.

28

Recall the Linear Space algorithm V[i,j] = d(s[1..i],t[1..j]) B[i,j] = d(s[i+1..n],t[j+1..m]) F[i,j] + B[i,j] = score of best alignment through

(i,j)t

s

These computations done in linear space.

),( vuxxiia vuBuild such a table

for every two sequences s=xu and t=xv, 1 u, v k. This entry encodes the optimum through (iu,iv).

29

Time saving heuristics:Relevance test

1

1

1

Let be a lower bound on SP- ( ,.., ).We say that ( ,.., ) is irrelevant if the

score of any alignment through is less than .Definition: ( ) ( , )

Claim 1: If L> ( ) the

u v

k

k

u vx xu v k

L score x xi i

LL a i i

L

ii

i

i

1n ( ,.., ) is irrelevant.ki ii

But can we go over all cells determine if they are relevant or not ?No. Start with (0,…,0) and add to the list relevant entries until reaching (n1,…,nk)

30

Star Alignments

Rather then summing up all pairwise alignments, select a fixed sequence x0 as a center, and set

Star-score(K) = ∑j>0score(x0,xj).

The algorithm to find optimal alignment: at each step, add another sequence aligned with x0, keeping old gaps and possibly adding new ones.

31

Tree Alignments

Assume that there is a tree T=(V,E) whose leaves are the sequences. • Associate a sequence in each internal node.• Tree-score(K) = ∑(i,j) Escore(xi,xj).

Finding the optimal assignment of sequences to the internal nodes is NP Hard.

We will meet again this problem in the next topic:Phylogenetic trees

32

Multiple Sequence Alignment – Approximation Algorithm

In tutorial time you will see an O(k2n2) multiple alignment algorithm that errs by a factor of at most 2(1-1/k) < 2.