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Sequence Alignment III
Lecture #4
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.
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.
.
Log Odd-Ratio Test for Alignment
Taking logarithm of Q yields
Score(s[i],t[i])
Pr( s, t | M )
p( s[i ], t[i ])
p( s[i ], t[i ])
log
 log 
  log
Pr( s, t | R)
q( s[i ]) q(t[i ])
i
i q ( s[i ]) q (t[i ])
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 ?
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Estimating p(·,·) for proteins
Generate 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
f ab  f ba
be the number of mutations a  b,
f a  b|b  a f ab be the total number of mutations that involve a,
f  a f a be the total number of amino acids involved in a mutation.
Note that f is twice the number of mutations.
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PAM-1 matrices
Define 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.
M ab
f ab
 Pr( a  b)  Pr( a  b | a changed)  Pr( a changed) 
ma
fa
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,

a
pa M aa  a pa 1  ma   1  a pa ma  0.99
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PAM-1 matrices
We wish that ma will be proportional to the relative mutability of
letter a compared to other letters.
fa
ma 
K  pa f
where K is a proportional constant.
We select K to satisfy the PAM-1 definition:
 fa
a pa ma  a pa  Kp f
 a




fa
1
 0.01

Kf
K
a
So K=100 for PAM-1 matrices. Note that K=50 yields 2% change, etc.
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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 ?
6
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”
T
A
T
A
T
C
C
C
t
t+
t+2
t+3
G
G
t+4
Time
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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)
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Multi-Step Changes
on Mab, we can compute the probabilities of
changes over two time periods
 Based
P( X t  2   b | X t  a) 
 c P ( X t  2   b | X t    c , X t  a ) P ( X t    c | X t  a )
Using Conditional independence (No memory)
 c P( X t  2   b | X t    c) P( X t    c | X t  a)
 c M ac M cb
 Thus
 By
M[2] = M[]M[]
induction (HMW exercise): M[n] = M[]
n
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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
p( X 1  x1 ,, X n  xn )  P( X 1  x1 ) P( X 2  x2 | X 1  x1 )  P( X n  xn | X n 1  xn 1 )
n
  p( X i  xi | Pai  pa i )
i 1
where Pai are the parents of variable/node Xi ,namely, none or Xi-1.
n
In short, we write: p( x1 ,, xn )   p( xi | pa i )
i 1
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Markov Model of Evolution Revisited
X1
M
X2
M
Xn-1
Xn
In the evolution model we studied earlier we had
P(x1) = (pa, pc, pg, pt)
which sum to 1 and called the prior probabilities, and
P(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
n

p( x1 , xn )  p( x1 )  M []

x1 xn
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Longer Term Changes
M[] = M (PAM-1 matrices)
 Use M[n] = Mn (PAM-n matrices)
 Define
 Estimate
 
p ( a, b)  pa M
n
ab
 Use
this quantity to define the score for your
application of interest.
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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
M ab
f ab
f ab  f a

ma 

fa
f a  K  pa f

f ab

 100  pa f
With K=50 the basic matrix is different, namely:
f ab
M 'ab 
50  pa f
Thus we have two different ways to estimate the matrix M[4] :
Use the 1-PAM matrix to the fourth power: M[4] = M[] 4
Or
Use the K=50 matrix to the second power: M[4] = M[2] 2
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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 ?
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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
Possible alignment
S3=TGAAC
AGG T - C - G - T T C G
TG - A A C -
Possible alignment
AGG T - C GTT - - C G
- TGA A A C
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Multiple Sequence Alignment
Aligning more than two sequences.
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
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Multiple alignments
We use a matrix to represent the alignment of k sequences,
K=(x1,...,xk). We assume no columns consists solely of blanks.
The common scoring functions give a score to each column,
and set: score(K)= ∑i score(column(i))
x1
M
Q
_
I
L
L
L
x2
M
L
R
-
L
L
-
x3
M
K
_
I
L
L
L
x4
M
P
P
V
L
I
L
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).
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SUM OF PAIRS
A common scoring function is SP – sum of scores of the
projected pairwise alignments: SPscore(K)=∑i<j score(xi,xj).
Note that we need to
specify the score(-,-)
because a column may
have several blanks (as
long as not all entries are
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
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).
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SUM OF PAIRS
Definition: The sum-of-pairs (SP) value for a multiple global
alignment A of k strings is the sum of the values of
k 
all  
2
projected
pairwise alignments induced by A where the pairwise alignment
function score(xi,xj) is additive.
M
Q
_
I
L
L
L
M
L
R
-
L
L
-
M
K
_
I
L
L
L
M
P
P
V
L
I
L
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Example
Consider the following alignment:
ac-cdb-c-adbd
a-bcdad
Using the edit distance  x, x  0 and  x, y   1 for x  y ,
this alignment has a SP value of
3+
+44 + 5 = 12
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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.
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The idea via K=2
Recall the notation:
V [i , j ]  d (s [1..i ],t [1.. j ])
and the following recurrence for V:
V [i , j ]   (s [i  1 ],t [ j  1 ]) 


V [i  1 , j  1 ]  max V [i , j  1 ]   (s [i  1 ],) 
V [i  1 , j ]   ( ,t [ j  1 ]) 


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).
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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):
s(i)  max{s(i  b)  score(column(i, b))}
b
Where
•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”).
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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.
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Time saving heuristics: Relevance tests
Heuristic: Avoid computing score(i) for irrelevant vectors.
x1
M
Q
_
I
L
L
L
x2
M
L
R
-
L
L
-
x3
M
K
_
I
L
L
L
x4
M
P
P
V
L
I
L
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.
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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
These computations
done in linear space.
Build such a table a x u x v (iu , iv )
for every two sequences
s=xu and t=xv, 1  u, v  k.
This entry encodes the
optimum through (iu,iv).
s
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Time saving heuristics:
Relevance test
Let L be a lower bound on SP-score( x1 ,.., x k ).
We say that i  (i1 ,.., ik ) is irrelevant if the
score of any alignment through i is less than L.
Definition: L(i)   axu xv (iu , iv )
1u  v  k
Claim 1: If L> L(i) then i  (i1 ,.., ik ) is irrelevant.
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)
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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.
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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
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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.
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