Class 9: Multiple Sequence Alignment


Andrés Aravena

27 November 2020

Multiple sequence alignment


  • to study closely related genes or proteins
    • to find conserved domains
  • to find the evolutionary relationships
    • the base of phylogenetic trees
  • to identify shared patterns among related genes
    • conserved sequences can be binding sites

In the previous chapter…

We discussed pairwise alignment

  • Global, using Needleman & Wunsch algorithm
    • also for semi-global alignment
  • Local, using Smith & Waterman algorithm

Both methods use a dot-plot matrix

We build a matrix with \(m\) rows and \(n\) columns

The cost (in time and memory) is \[O(mn)\]

(\(m\) is the query length, \(n\) is the subject length)

Filling the dot-plot matrix

To find the optimal alignment, we look for diagonals in the matrix that maximize the total score

Every cell \(M_{ij}\) in the matrix has initially the value \[M_{ij}=Score_2(s_{1}(i),s_{2}(j))\] where \(s_{1}(i)\) is the letter in position \(i\) of sequence \(s_1\),
and \(s_{2}(j)\) is the letter in position \(j\) of \(s_2\)

Pairwise to three-wise alignment

To aligning two sequences, we build a dot-plot matrix.
That is, a rectangle.

To align three sequences, we need a three-dimensional array.
That is, a cube.

Each cell \(M_{ijk}\) has value \[M_{ijk}=Score_3(s_{1}(i),s_{2}(j), s_{3}(k))\]

Cost of three-wise alignment

If the three sequences have length \(m_1, m_2,\) and \(m_3,\) then building the cube has cost \[O(m_1\cdot m_2\cdot m_3)\]

To simplify, we assume that all sequences have length \(m\)

Then the cost of three-wise alignment is \[O(m^3)\]

Multiple alignment

Following the same idea…

To align \(N\) sequences, we need a dot-plot in \(N\) dimensions.


Therefore, if the average sequence length is \(m,\) then the cost is \[O(m^N)\]

How much is that?

To fix ideas, assume that \(m=1000\)

Therefore the cost is \[O(1000^N)=O(10^{3N})\]

How many seconds

Now assume that the computer can do one million comparisons each second

The number of seconds is then \[O(10^{3N-6})\]

How much is that?

Under these hypothesis we have this table

\(N\) Seconds
2 \(10^0\)
4 \(10^6\)
8 \(10^{18}\)
12 \(10^{30}\)


  • Translate these numbers to days, years, etc.

  • How do these numbers change if \(m\) changes?

  • What happens if the computers are 1000 times faster?

  • What is the largest multiple alignment that you can do in your life?

  • What is the largest number of sequences that can be aligned?

  • What can we do to align more sequences?

Write your answers in the Quiz


This is clearly too expensive, so we need heuristics

(i.e. solving a similar but simpler problem)

One common idea is to do a progressive alignment

  • start with a pairwise alignment,
  • and then align the rest one by one

Too many heuristics

There are several ways to simplify the original problem

Thus, there are many approximate solutions

The main differences are:

  • How to decide what to align first?
  • Can new sequences change the previous alignment?
    • Can \(s_k\) change the alignment of \((s_1,…,s_{k-1})\)?
  • How to use additional information (like 3D structure)?
  • What is the formula for \(Score_k(s_{1}(i_1),…,s_{k}(i_k))\)?

Some common Multiple-Alignment tools

  • Clustal
    • ClustalW, ClustalX, Clustal Omega
  • T-Coffee
    • 3D-Coffee


Clustal Omega

Clustal was the first popular multiple sequence aligner

Versions: Clustal 1, Clustal 2, Clustal 3, Clustal 4, Clustal V, Clustal W, Clustal X, Clustal Ω

Only the last is currently in use

Clustal W

This version is easier to explain

Scoring function

First, there should be a scoring function

Clustal uses a simple one. The sum of all v/s all \[\begin{aligned} Score_k(s_{1}(i_1),…,s_{k}(i_k))= & Score_2(s_{1}(i_1),s_{2}(i_2)) + \\ & Score_2(s_{1}(i_1),s_{3}(i_3)) + \cdots+ \\ & Score_2(s_{k-1,i_{k-1}},s_{k}(i_k)) \end{aligned} \] where \(Score_2(s_{a}(i),s_{b}(j))\) is PAM, BLOSUM, or a suitable substitution scoring matrix

Progressive alignment order

We start by comparing sequences all-to-all

That is, comparing all pairs of sequences

We store them in a distance matrix

How many pairs can be done with \(N\) sequences?

Building a guide tree

Once we get all pairwise “distances”
(that is, scores)

We make a tree by hierarchical clustering

Hierarchical clustering

Start with one leaf node for each sequence, and no branches

  • Take the two sequences (\(A\) and \(B\)) that are more similar (highest score)
  • create a new node \(C\) connected to \(A\) and \(B\)
  • \(Score_2(X,C)\) between each old node \(X\) and \(C\) is the average of \(Score_2(X,A)\) and \(Score_2(X,B)\) \[Score_2(X,C)=\frac{Score_2(X,A)+Score_2(X,B)}{2}\]
  • forget about \(A\) and \(B\)

This is not a phylogenetic tree

The guide tree is built without seeing the big picture

So it is not safe to assign any meaning to it

We will talk more about trees and build phylogenetic trees later

Guide tree guides the alignment

Clustal aligns the sequences following the guide tree

First, it aligns the more similar sequences

Then it adds the nearest sequence, and so on

These are semi-global alignments

Uses \(Score_k()\) when there are \(k\) sequences


Search for CLUSTAL

Try with

Solving without aligning


Short patterns, usually without gaps

  • Binding sites in DNA
  • Active sites in proteins

Sometimes they can be found with experiments like ChIP-seq

Finding motifs characteristics

If we know several instances of binding sites, we can align them

After we align them, we can calculate the frequency of each symbol on each column

If we know these frequencies, we can build a position specific score matrix

These PSSM represent the motif

Finding new motif instances

If we know the PSSM, it is easy to find new sites matching the motif

We just use the PSSM to give a score to each symbol, and we get a total score.

High scores will be new motif instances

How to find motifs without alignment

It is easy to prepare PSSM in a multiple alignment

But we have seen that multiple alignment can be expensive

When we are looking for short patterns (a.k.a motifs) we can use other strategies

Nucleic Acids Research, Volume 34, Issue 11, 1 January 2006, Pages 3267–3278,


Given \(N\) sequences of length \(m\): \(s_1, s_2, ..., s_N\), and a fixed motif width \(k,\) we want

  • a subsequence of size \(k\) (a \(k\)-mer) that is over-represented among the \(N\) sequences
  • or a set of highly similar \(k\)-mers that are over-represented

Testing all sets of \(k\)-mers is computationally intensive, so several heuristics have been proposed

Gibbs Sampling Algorithm

Given \(N\) sequences of length \(m\) and desired motif width \(k\):

  • Step 1: Choose a position in each sequence at random: \[a_1 \in s_1, a_2 \in s_2, ..., a_N \in s_N\]

  • Step 2: Choose a sequence at random from the set (let’s say, \(s_j\)).

  • Step 3: Make a PSSM of width \(k\) from the sites in all sequences except \(s_j.\)

Gibbs Sampling Algorithm (cont)

  • Step 4: Assign a score to each position in \(s_j\) using the PSSM constructed in step 3: \[p_1, p_2, p_3, ..., p_{m-k+1}\]

  • Step 5: Choose a starting position in \(s_j\) (higher score have higher probability) and set \(a_j\) to this new position.

Gibbs Sampling Algorithm (cont)

  • Step 6: Choose a sequence at random from the set (say, \(s_{j'}\)).

  • Step 7: Make a PSSM of width \(k\) from the sites in all sequences except \(s_{j'}\), but including \(a_j\) in \(s_j\)

  • Step 8: Repeat until convergence (of positions or motif model)

Step 1

Choose a position \(a_i\) in each sequence at random

Step 2 and 3

Choose a sequence at random, let’s say \(s_1.\) Make a PSSM from the sites in all sequences except \(s_1\)

Step 4

Assign a score to each position in using the PSSM

Step 5

Choose a new position in \(s_1\) based on this score

Step 6

Build a new PSSM with the new sequence

Step 7

Choose randomly a new sequence from the set and repeat


We repeat the steps until the PSSM no longer changes

Alternative: we stop when the positions \(a_i\) do not change

The methods probably gives us a over-represented motif