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Dynamic Programming: Sequence alignmentPowerPoint Presentation

Dynamic Programming: Sequence alignment

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Dynamic Programming: Sequence alignment. CS 498 SS Saurabh Sinha. DNA Sequence Comparison: First Success Story . Finding sequence similarities with genes of known function is a common approach to infer a newly sequenced gene’s function

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DNA Sequence Comparison: First Success Story

- Finding sequence similarities with genes of known function is a common approach to infer a newly sequenced gene’s function
- In 1984 Russell Doolittle and colleagues found similarities between cancer-causing gene and normal growth factor (PDGF) gene
- A normal growth gene switched on at the wrong time causes cancer !

Cystic fibrosis (CF) is a chronic and frequently fatal genetic disease of the body's mucus glands. CF primarily affects the respiratory systems in children.

If a high % of cystic fibrosis (CF) patients have a certain mutation in the gene and the normal patients don’t, then that could be an indicator of a mutation that is related to CF

A certain mutation was found in 70% of CF patients, convincing evidence that it is a predominant genetic diagnostics marker for CF

Cystic FibrosisBring in the Bioinformaticians

- Gene similarities between two genes with known and unknown function alert biologists to some possibilities
- Computing a similarity score between two genes tells how likely it is that they have similar functions
- Dynamic programming is a technique for revealing similarities between genes

Dynamic programming example:Manhattan Tourist Problem

Imagine seeking a path (from source to sink) to travel (only eastward and southward) with the most number of attractions (*) in the Manhattan grid

Source

*

*

*

*

*

*

*

*

*

*

*

*

Sink

Dynamic programming example:Manhattan Tourist Problem

Imagine seeking a path (from source to sink) to travel (only eastward and southward) with the most number of attractions (*) in the Manhattan grid

Source

*

*

*

*

*

*

*

*

*

*

*

*

Sink

Manhattan Tourist Problem: Formulation

Goal: Find the longest path in a weighted grid.

Input: A weighted grid G with two distinct vertices, one labeled “source” and the other labeled “sink”

Output: A longest path in Gfrom “source” to “sink”

MTP: An Example

0

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j coordinate

source

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2

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0

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9

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13

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15

19

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i coordinate

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0

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20

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2

sink

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23

4

MTP: Greedy Algorithm Is Not Optimal

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source

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promising start, but leads to bad choices!

5

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sink

18

MTP: Simple Recursive Program

MT(n,m)

ifn=0 or m=0

returnMT(n,m)

x MT(n-1,m)+

length of the edge from (n- 1,m) to (n,m)

y MT(n,m-1)+

length of the edge from (n,m-1) to (n,m)

returnmax{x,y}

What’s wrong with this approach?

Here’s what’s wrong

- M(n,m) needs M(n, m-1) and M(n-1, m)
- Both of these need M(n-1, m-1)
- So M(n-1, m-1) will be computed at least twice
- Dynamic programming: the same idea as this recursive algorithm, but keep all intermediate results in a table and reuse

j

0

1

source

1

0

1

S0,1= 1

i

5

1

5

S1,0= 5

- Calculate optimal path score for each vertex in the graph
- Each vertex’s score is the maximum of the prior vertices score plus the weight of the respective edge in between

MTP: Dynamic Programming (cont’d)

j

0

1

2

source

1

2

0

1

3

S0,2 = 3

i

5

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-5

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S1,1= 4

3

2

8

S2,0 = 8

MTP: Dynamic Programming (cont’d)

j

0

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source

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0

1

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S3,0 = 8

i

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-5

1

1

5

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13

S1,2 = 13

5

3

-5

2

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9

S2,1 = 9

0

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8

S3,0 = 8

MTP: Dynamic Programming (cont’d)

j

0

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source

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0

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i

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-5

-5

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-5

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S1,3 = 8

5

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-3

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-5

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12

S2,2 = 12

0

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S3,1 = 9

greedy alg. fails!

MTP: Dynamic Programming (cont’d)

j

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source

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-3

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-5

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15

S2,3 = 15

0

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-5

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9

S3,2 = 9

MTP: Dynamic Programming (cont’d)

j

0

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source

1

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1

3

8

Done!

i

5

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-5

-5

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-5

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13

8

(showing all back-traces)

5

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-3

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-5

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-5

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16

S3,3 = 16

si-1, j + weight of the edge between (i-1, j) and (i, j)

si, j-1 + weight of the edge between (i, j-1) and (i, j)

max

si, j =

MTP: RecurrenceComputing the score for a point (i,j) by the recurrence relation:

The running time is n x m for a n by mgrid

(n = # of rows, m = # of columns)

A2

A3

A1

B

sA1 + weight of the edge (A1, B)

sA2 + weight of the edge (A2, B)

sA3 + weight of the edge (A3, B)

max of

sB =

Manhattan Is Not A Perfect GridWhat about diagonals?

- The score at point B is given by:

of

sy + weight of vertex (y, x) where

y є Predecessors(x)

sx =

Manhattan Is Not A Perfect Grid (cont’d)Computing the score for point x is given by the recurrence relation:

- Predecessors (x) – set of vertices that have edges leading to x
- The running time for a graph G(V, E) (V is the set of all vertices and Eis the set of all edges) is O(E) since each edge is evaluated once

Traveling in the Grid

- By the time the vertex x is analyzed, the values sy for all its predecessors y should be computed – otherwise we are in trouble.
- We need to traverse the vertices in some order
- For a grid, can traverse vertices row by row, column by column, or diagonal by diagonal
- If we had, instead of a (directed) grid, an arbitrary DAG (directed acyclic graph) ?

Traversing the Manhattan Grid

a)

b)

- 3 different strategies:
- a) Column by column
- b) Row by row
- c) Along diagonals

c)

Topological Ordering

- A numbering of vertices of the graph is called topological ordering of the DAG if every edge of the DAG connects a vertex with a smaller label to a vertex with a larger label
- In other words, if vertices are positioned on a line in an increasing order of labels then all edges go from left to right.
- How to obtain a topological sort (???)

Longest Path in DAG Problem

- Goal: Find a longest path between two vertices in a weighted DAG
- Input: A weighted DAG G with source and sink vertices
- Output: A longest path in G from source to sink

sv=

Longest Path in DAG: Dynamic Programming- Suppose vertex v has indegree 3 and predecessors {u1, u2, u3}
- Longest path to v from source is:
In General:

sv = maxu (su + weight of edge from uto v)

su1 + weight of edge fromu1to v

su2 + weight of edge fromu2to v

su3+ weight of edge fromu3to v

mismatches

insertions

deletions

4

1

2

match

2

mismatch

deletion

indels

insertion

Aligning DNA SequencesAlignment : 2 x k matrix ( km, n )

n = 8

V = ATCTGATG

m = 7

W = TGCATAC

V

W

Longest Common Subsequence (LCS) – Alignment without Mismatches

- Given two sequences
- v = v1v2…vm and w = w1w2…wn
- The LCS of vand w is a sequence of positions in
- v: 1 < i1 < i2 < … < it< m
- and a sequence of positions in
- w: 1 < j1 < j2 < … < jt< n
- such that it -th letter of v equals to jt-letter of wand t is maximal

1

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0

0

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LCS: Examplei coords:

elements of v

A

T

--

C

--

T

G

A

T

C

elements of w

--

T

G

C

A

T

--

A

--

C

j coords:

(0,0)

(1,0)

(2,1)

(2,2)

(3,3)

(3,4)

(4,5)

(5,5)

(6,6)

(7,6)

(8,7)

positions in v:

2 < 3 < 4 < 6 < 8

Matches shown inred

positions in w:

1 < 3 < 5 < 6 < 7

Every common subsequence is a path in 2-D grid

si-1, j

si, j-1

si-1, j-1 + 1 if vi = wj

max

si, j =

Computing LCSLet vi = prefix of v of length i: v1 … vi

and wj = prefix of w of length j: w1 … wj

The length of LCS(vi,wj) is computed by:

LCS Problem as Manhattan Tourist Problem

A

T

C

T

G

A

T

C

j

0

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5

6

7

8

i

0

T

1

G

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C

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A

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C

7

Edit Graph for LCS Problem

A

T

C

T

G

A

T

C

j

0

1

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8

Every path is a common subsequence.

Every diagonal edge adds an extra element to common subsequence

LCS Problem: Find a path with maximum number of diagonal edges

i

0

T

1

G

2

C

3

A

4

T

5

A

6

C

7

Backtracking

- si,j allows us to compute the length of LCS for vi and wj
- sm,n gives us the length of the LCS for v and w
- How do we print the actual LCS ?
- At each step, we chose an optimal decision si,j = max (…)
- Record which of the choices was made in order to obtain this max

si-1, j

si, j-1

si-1, j-1 + 1 if vi = wj

max

si, j =

Computing LCSLet vi = prefix of v of length i: v1 … vi

and wj = prefix of w of length j: w1 … wj

The length of LCS(vi,wj) is computed by:

Printing LCS: Backtracking

- PrintLCS(b,v,i,j)
- if i = 0 or j = 0
- return
- ifbi,j = “ “
- PrintLCS(b,v,i-1,j-1)
- printvi
- else
- ifbi,j = “ “
- PrintLCS(b,v,i-1,j)
- else
- PrintLCS(b,v,i,j-1)

From LCS to Alignment

- The Longest Common Subsequence problem—the simplest form of sequence alignment – allows only insertions and deletions (no mismatches).
- In the LCS Problem, we scored 1 for matches and 0 for indels
- Consider penalizing indels and mismatches with negative scores
- Simplest scoring schema:
+1 : match premium

-μ : mismatch penalty

-σ : indel penalty

Simple Scoring

- When mismatches are penalized by –μ, indels are penalized by –σ,
and matches are rewarded with +1,

the resulting score is:

#matches – μ(#mismatches) – σ (#indels)

The Global Alignment Problem

Find the best alignment between two strings under a given scoring schema

Input : Strings v and w and a scoring schema

Output : Alignment of maximum score

→ = -σ

= 1 if match

= -µ if mismatch

si-1,j-1 +1 if vi = wj

si,j = max s i-1,j-1 -µ if vi ≠ wj

s i-1,j - σ

s i,j-1 - σ

m : mismatch penalty

σ : indel penalty

{

Scoring Matrices

To generalize scoring, consider a (4+1) x(4+1) scoring matrixδ.

In the case of an amino acid sequence alignment, the scoring matrix would be a (20+1)x(20+1) size. The addition of 1 is to include the score for comparison of a gap character “-”.

This will simplify the algorithm as follows:

si-1,j-1 + δ(vi, wj)

si,j = max s i-1,j + δ(vi, -)

s i,j-1 + δ(-, wj)

{

Making a Scoring Matrix

- Scoring matrices are created based on biological evidence.
- Alignments can be thought of as two sequences that differ due to mutations.
- Some of these mutations have little effect on the protein’s function, therefore some penalties, δ(vi , wj), will be less harsh than others.

KAAANK

-1 + (-1) + (-2) + 5 + 7 + 3 = 11

Scoring Matrix: Example- Notice that although R and K are different amino acids, they have a positive score.
- Why? They are both positively charged amino acids will not greatly change function of protein.

Conservation

- Amino acid changes that tend to preserve the physico-chemical properties of the original residue
- Polar to polar
- aspartate glutamate

- Nonpolar to nonpolar
- alanine valine

- Similarly behaving residues
- leucine to isoleucine

- Polar to polar

Local vs. Global Alignment

- The Global Alignment Problem tries to find the longest path between vertices (0,0) and (n,m) in the edit graph.
- The Local Alignment Problem tries to find the longest path among paths between arbitrary vertices (i,j) and (i’, j’) in the edit graph.
- In the edit graph with negatively-scored edges, Local Alignment may score higher than Global Alignment

Compute a “mini” Global Alignment to get Local Alignment

Local Alignment: ExampleLocal alignment

Global alignment

Local Alignments: Why?

- Two genes in different species may be similar over short conserved regions and dissimilar over remaining regions.
- Example:
- Homeobox genes have a short region called the homeodomain that is highly conserved between species.
- A global alignment would not find the homeodomain because it would try to align the ENTIRE sequence

The Local Alignment Problem

- Goal: Find the best local alignment between two strings
- Input : Strings v, w and scoring matrix δ
- Output : Alignment of substrings of v and w whose alignment score is maximum among all possible alignment of all possible substrings

The Problem Is …

- Long run time O(n4):
- In the grid of size n x n there are ~n2 vertices (i,j) that may serve as a source.

- For each such vertex computing alignments from (i,j) to (i’,j’) takes O(n2) time.

- This can be remedied by allowing every point to be the starting point

Notice there is only this change from the original recurrence of a Global Alignment

The Local Alignment Recurrence- The largest value of si,j over the whole edit graph is the score of the best local alignment.
- The recurrence:

0

si,j = max si-1,j-1 + δ(vi, wj)

s i-1,j + δ(vi, -)

s i,j-1 + δ(-, wj)

{

This is more likely. recurrence of a Global Alignment

This is less likely.

Affine Gap Penalties- In nature, a series of k indels often come as a single event rather than a series of k single nucleotide events:

ATA__GC

ATATTGC

ATAG_GC

AT_GTGC

Normal scoring would give the same score for both alignments

Accounting for Gaps recurrence of a Global Alignment

- Gaps- contiguous sequence of spaces in one of the rows
- Score for a gap of length x is:
-(ρ +σx)

where ρ >0 is the penalty for introducing a gap:

gap opening penalty

ρ will be large relative to σ:

gap extension penalty

because you do not want to add too much of a penalty for extending the gap.

Affine gap penalty in DP recurrence of a Global Alignment

- When computing si,j, need to look at si,j-1, si,j-2, si,j-3,…. and si-1,j, si-2,j, …
- Each cell needs O(n) time for update
- O(n2) cells
- Therefore, O(n3) algorithm
- We can still do this in O(n2) time

Affine Gap Penalty Recurrences recurrence of a Global Alignment

Continue Gap in w (deletion)

si,j = s i-1,j - σ

max s i-1,j –(ρ+σ)

si,j = s i,j-1 - σ

max s i,j-1 –(ρ+σ)

si,j = si-1,j-1 + δ(vi, wj)

max s i,j

s i,j

Start Gap in w (deletion): from middle

Continue Gap in v (insertion)

Start Gap in v (insertion):from middle

Match or Mismatch

End deletion: from top

End insertion: from bottom

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