CS CS 466 466 In Introduct ctio ion t to B Bio ioin - - PowerPoint PPT Presentation

CS CS 466 466 In Introduct ctio ion t to B Bio ioin informatics ics

Lecture 2 Part 2

Mohammed El-Kebir January 28, 2020

Outline

• 1. Edit distance recap
• 2. Global alignment
• 3. Fitting alignment
• 4. Local alignment
• 5. Gapped alignment

Reading:

• Jones and Pevzner. Chapters 6.6, 6.8 and 6.9
• Lecture notes

2

Weighted Edit Distance – Practice Problem

• Compute weighted edit distance between 𝐰 = AGT and 𝐱 = ATCT.

3

1 2 3 4 1 2 3

A T C G A G T V w

d[i, j] = min                0, if i = 0 and j = 0, d[i 1, j] + 1, if i > 0, d[i, j 1] + 1, if j > 0, d[i 1, j 1] + 2, if i > 0, j > 0 and vi 6= wj, d[i 1, j 1], if i > 0, j > 0 and vi = wj.

Weighted Edit Distance – Practice Problem

• Compute weighted edit distance between 𝐰 = AGT and 𝐱 = ATCT.

4

1 2 3 4 1 2 3 4 1 1 1 2 3 2 2 1 2 3 2 3 3 2 1 2 3

A T C G A G T V w

d[i, j] = min                0, if i = 0 and j = 0, d[i 1, j] + 1, if i > 0, d[i, j 1] + 1, if j > 0, d[i 1, j 1] + 2, if i > 0, j > 0 and vi 6= wj, d[i 1, j 1], if i > 0, j > 0 and vi = wj.

Edit Distance – Additional Insights

• An alignment corresponds to a series of elementary operations

5

Examples from http://profs.scienze.univr.it/~liptak/ACB/files/StringDistance_6up.pdf

Edit Distance – Additional Insights

• An alignment corresponds to a series of elementary operations
• But not every series of elementary operations corresponds to an alignment! Why?

6

Examples from http://profs.scienze.univr.it/~liptak/ACB/files/StringDistance_6up.pdf

Distance Function / Metric

7

A distance function (metric) on a set 𝑌 is a function 𝑒 ∶ 𝑌 × 𝑌 → ℝ s.t. for all 𝑦, 𝑧, 𝑨 ∈ 𝑌: i. 𝑒 𝑦, 𝑧 ≥ 0 [non-negativity]

• ii. 𝑒 𝑦, 𝑧 = 0 if and only if 𝑦 = 𝑧

[identity of indiscernibles]

• iii. 𝑒 𝑦, 𝑧 = 𝑒(𝑧, 𝑦)

[symmetry]

• iv. 𝑒 𝑦, 𝑧 ≤ 𝑒 𝑦, 𝑨 + 𝑒(𝑨, 𝑧)

[triangle inequality] Question: Is edit distance a distance function?

Edit Distance is a Distance Function

8

Edit distance 𝑒(𝐰, 𝐱) is the minimum number of elementary operations to transform 𝐰 ∈ Σ∗ into 𝐱 ∈ Σ∗. Claim: edit distance is a distance function. Proof: Let 𝐯, 𝐰, 𝐱 ∈ Σ∗. i. 𝑒 𝐰, 𝐱 ≥ 0 [non-negativity] Edit distance is defined by an alignment. This in turn uniquely determines a series of elementary operations, each with cost either 0 (match) or 1 (otherwise). Thus, 𝑒 𝐰, 𝐱 ≥ 0.

Edit Distance is a Distance Function

9

Edit distance 𝑒(𝐰, 𝐱) is the minimum number of elementary operations to transform 𝐰 ∈ Σ∗ into 𝐱 ∈ Σ∗.

Proof: Let 𝐯, 𝐰, 𝐱 ∈ Σ∗. ii. 𝑒 𝐰, 𝐱 = 0 if and only if 𝐰 = 𝐱 [identity of indiscernibles] (=>) By the premise, 𝑒 𝐰, 𝐱 = 0. By definition, the optimal alignment can only consist

• f operations with cost 0. That is, the alignment consist of only matches. Thus, 𝐰 = 𝐱.

(<=) By the premise, 𝐰 = 𝐱. Thus, there exists an alignment where every pair of columns is a match. This means that |𝐰| = |𝐱| and each letter 𝑤A equals 𝑥A (where 𝑗 ∈ [|𝐰|]). Moreover, only the match operations has cost 0, the other operations have cost

• 1. Hence, this is the optimal alignment with cost 𝑒 𝐰, 𝐱 = 0.

Claim: edit distance is a distance function.

Edit Distance is a Distance Function

10

Edit distance 𝑒(𝐰, 𝐱) is the minimum number of elementary operations to transform 𝐰 ∈ Σ∗ into 𝐱 ∈ Σ∗.

Proof: Let 𝐯, 𝐰, 𝐱 ∈ Σ∗. iii. 𝑒 𝐰, 𝐱 = 𝑒(𝐱, 𝐰) [symmetry] Let 𝐁 = [𝑏A,H] be the optimal alignment corresponding to 𝑒 𝐰, 𝐱 , i.e. 𝐁 is an 2 × 𝑙 matrix where 𝑙 ∈ {max( 𝐰 , 𝐱 ), … , 𝐰 + 𝐱 }. Define the function 𝑔 𝐁 = 𝐂 such that 𝐂 is obtained by interchanging the two rows of 𝐁. Since the cost of any insertion, deletion and mismatch is 1, we have that alignment 𝐂 has cost 𝑒 𝐰, 𝐱 . The existence

• f an alignment from 𝐱 to 𝐰 with cost less than 𝑒 𝐰, 𝐱 , yields a contradiction as it

implies that 𝐁 is not an optimal alignment from 𝐰 to 𝐱. Hence, 𝑒 𝐱, 𝐰 = 𝑒 𝐰, 𝐱 .

Claim: edit distance is a distance function.

Edit Distance is a Distance Function

11

Edit distance 𝑒(𝐰, 𝐱) is the minimum number of elementary operations to transform 𝐰 ∈ Σ∗ into 𝐱 ∈ Σ∗.

Proof: Let 𝐯, 𝐰, 𝐱 ∈ Σ∗. iv. 𝑒 𝐰, 𝐱 ≤ 𝑒 𝐰, 𝐯 + 𝑒(𝐯, 𝐱) [triangle inequality] Assume for a contradiction that 𝑒 𝐰, 𝐱 > 𝑒 𝐰, 𝐯 + 𝑒(𝐯, 𝐱). Let 𝑇 be the sequence

• f elementary operations for transforming 𝐰 into 𝐯. Let 𝑇′ be the sequence of

elementary operations for transforming 𝐯 into 𝐱. Note that 𝑒 𝐰, 𝐯 = |𝑇| and 𝑒 𝐯, 𝐱 = |𝑇′|. Concatenate 𝑇 and 𝑇′ and remove redundant operations, yielding sequence 𝑇′′. By definition, 𝑇VV ≤ 𝑇 + 𝑇V . We can obtain an alignment of 𝐰 and 𝐱 from 𝑇′′ with cost 𝑇VV ≤ 𝑒 𝐰, 𝐯 + 𝑒(𝐯, 𝐱). This yields a contradiction with 𝑒 𝐰, 𝐱 > 𝑒 𝐰, 𝐯 + 𝑒(𝐯, 𝐱) being the cost of the optimal alignment of 𝐰 and 𝐱.

Claim: edit distance is a distance function.

Outline

• 1. Edit distance recap
• 2. Global alignment
• 3. Fitting alignment
• 4. Local alignment
• 5. Gapped alignment

Reading:

• Jones and Pevzner. Chapters 6.6, 6.8 and 6.9

12

Biological Sequence Alignment

• Weighted edit distance: find

alignment with minimum distance

• Shortest path in weighted

edit graph

• Sequence alignment: find

alignment with maximum similarity

• Longest path in weighted

edit graph

• Score function:

𝜀 ∶ Σ ∪ −

Z → ℝ

13

1 2 3 4 O O O O O 1 O O O O O 2 O O O O O 3 O O O O O 4 O O O O O

W A T C G A T G T V

match mismatch insertion deletion

• "

#

$%

• $%

"

#

$% "

#

𝜀(𝑤A, −) 𝜀(−, 𝑥H) 𝜀(𝑤A, 𝑥H)

Question: What is an example of 𝜀?

Scoring Matrices

14

Transitions: interchanges among purines (two rings) or pyrimidines (one ring)

• A <--> G
• C <--> T

Transversions: interchanges between purines (two rings) and pyrimidines (one ring)

• A <--> C, A <--> T
• G <--> C, G <--> T

Transitions more likely than transversions!

A C G T

Scoring Matrices

15

Transitions: interchanges among purines (two rings) or pyrimidines (one ring)

• A <--> G
• C <--> T

Transversions: interchanges between purines (two rings) and pyrimidines (one ring)

• A <--> C, A <--> T
• G <--> C, G <--> T

Transitions more likely than transversions!

𝜀 A T C G

• A

1

• 2
• 2
• 1
• 1

T

• 2

1

• 1
• 2
• 1

C

• 2
• 1

1

• 2
• 1

G

• 1
• 2
• 2

1

• 1
• 1
• 1
• 1
• 1

−∞

Global Alignment – Needleman-Wunsch Algorithm

• An alignment is a source-to-sink path in the edit graph
• An alignment 𝐁 = [𝑏A,H] is a 2 × 𝑙 matrix s.t. (i) 𝑙 = {max 𝑛, 𝑜 , … , 𝑛 + 𝑜},

(ii) 𝑏A,H ∈ Σ ∪ − and (iii) there is no 𝑘 ∈ [𝑙] where 𝑏_,H = 𝑏Z,H = −

16

Global Alignment problem: Given strings 𝐰 ∈ Σ` and 𝐱 ∈ Σa and scoring function 𝜀, find alignment with maximum score. s[i, j] = max          0, if i = 0 and j = 0, s[i − 1, j] + δ(vi, −), if i > 0, s[i, j − 1] + δ(−, wj), if j > 0, s[i − 1, j − 1] + δ(vi, wj), if i > 0 and j > 0.

deletion insertion match/ mismatch

Demonstration

• http://alfehrest.org/sub/nwa/index.html
• 𝐰 = ATGTTAT and 𝐱 = ATCGTAC.

17

𝜀 A T C G

• A

1

• 2
• 2
• 1
• 1

T

• 2

1

• 1
• 2
• 1

C

• 2
• 1

1

• 2
• 1

G

• 1
• 2
• 2

1

• 1
• 1
• 1
• 1
• 1

−∞

Outline

• 1. Edit distance recap
• 2. Global alignment
• 3. Fitting alignment
• 4. Local alignment
• 5. Gapped alignment

Reading:

• Jones and Pevzner. Chapters 6.6, 6.7 and 6.9
• Lecture notes

18

Next Generation Sequencing (NGS) Technology

19 November, 2017

Log Scale 1,000 10,000 100,000,000 10,000,000 1,000,000 100,000

NGS

Allow for inexact matches due to:

• Sequencing errors
• Polymorphisms/mutations in

reference genome

20

NGS Characterized by Short Reads

Genome Millions -billions nucleotides Next-generation DNA sequencing 10-100’s million short reads Short read: 100 nucleotides

… GGTAGTTAG … … TATAATTAG … … AGCCATTAG … … CGTACCTAG … … CATTCAGTAG … … GGTAAACTAG …

Allow for inexact matches due to:

• Sequencing errors
• Polymorphisms/mutations in

reference genome

21

NGS Characterized by Short Reads

Genome Millions -billions nucleotides Next-generation DNA sequencing 10-100’s million short reads Short read: 100 nucleotides

… GGTAGTTAG … … TATAATTAG … … AGCCATTAG … … CGTACCTAG … … CATTCAGTAG … … GGTAAACTAG …

Question: How to account for discrepancy between lengths of reference and short read? Human reference genome is 3,300,000,000 nucleotides, while a short read is 100 nucleotides. Global sequence alignment will not work!

Fitting Alignment

22

For short read alignment, we want to align complete short read 𝐰 ∈ Σ` to substring of reference genome 𝐱 ∈ Σa. Note that 𝑛 ≪ 𝑜. Fitting Alignment problem: Given strings 𝐰 ∈ Σ` and 𝐱 ∈ Σa and scoring function 𝜀, find a alignment of 𝐰 and a substring of 𝐱 with maximum global alignment score 𝑡∗ among all global alignments of 𝐰 and all substrings of 𝐱

𝐰 ∈ Σ` 𝐱 ∈ Σa

Fitting Alignment – Naive Approach

• Consider all contiguous non-empty substrings of 𝐱, defined by 1 ≤ 𝑗 ≤ 𝑘 ≤ 𝑜
• How many?

23

Fitting Alignment problem: Given strings 𝐰 ∈ Σ` and 𝐱 ∈ Σa and scoring function 𝜀, find an alignment of 𝐰 and a substring of 𝐱 with maximum global alignment score 𝑡∗ among all global alignments of 𝐰 and all substrings of 𝐱

𝐰 ∈ Σ` 𝐱 ∈ Σa

Fitting Alignment – Naive Approach

• Consider all contiguous non-empty substrings of 𝐱, defined by 1 ≤ 𝑗 ≤ 𝑘 ≤ 𝑜
• How many? Answer: 𝑜 +

a Z

• What are their total lengths?
• What is the running time?

24

Fitting Alignment problem: Given strings 𝐰 ∈ Σ` and 𝐱 ∈ Σa and scoring function 𝜀, find an alignment of 𝐰 and a substring of 𝐱 with maximum global alignment score 𝑡∗ among all global alignments of 𝐰 and all substrings of 𝐱

𝐰 ∈ Σ` 𝐱 ∈ Σa

Fitting Alignment – Dynamic Programming

25

Fitting Alignment problem: Given strings 𝐰 ∈ Σ` and 𝐱 ∈ Σa and scoring function 𝜀, find an alignment of 𝐰 and a substring of 𝐱 with maximum global alignment score 𝑡∗ among all global alignments of 𝐰 and all substrings of 𝐱

s[i, j] = max          0, if i = 0, s[i − 1, j] + δ(vi, −), if i > 0, s[i, j − 1] + δ(−, wj), if i > 0 and j > 0, s[i − 1, j − 1] + δ(vi, wj), if i > 0 and j > 0. s∗ = max{s[m, 0], . . . , s[m, n]}

A G G T A C G G C

𝐰\𝐱

Fitting Alignment – Dynamic Programming

26

Fitting Alignment problem: Given strings 𝐰 ∈ Σ` and 𝐱 ∈ Σa and scoring function 𝜀, find an alignment of 𝐰 and a substring of 𝐱 with maximum global alignment score 𝑡∗ among all global alignments of 𝐰 and all substrings of 𝐱

s[i, j] = max          0, if i = 0, s[i − 1, j] + δ(vi, −), if i > 0, s[i, j − 1] + δ(−, wj), if i > 0 and j > 0, s[i − 1, j − 1] + δ(vi, wj), if i > 0 and j > 0. s∗ = max{s[m, 0], . . . , s[m, n]}

A G G T A C G G C

𝐰\𝐱

Start anywhere on first row End anywhere on last row

Fitting Alignment – Dynamic Programming

27

Fitting Alignment problem: Given strings 𝐰 ∈ Σ` and 𝐱 ∈ Σa and scoring function 𝜀, find an alignment of 𝐰 and a substring of 𝐱 with maximum global alignment score 𝑡∗ among all global alignments of 𝐰 and all substrings of 𝐱

s[i, j] = max          0, if i = 0, s[i − 1, j] + δ(vi, −), if i > 0, s[i, j − 1] + δ(−, wj), if i > 0 and j > 0, s[i − 1, j − 1] + δ(vi, wj), if i > 0 and j > 0. s∗ = max{s[m, 0], . . . , s[m, n]}

Question: Let match score be 1, mismatch/indel score be -1. What is 𝑡∗? Question: Same scores. What is optimal global alignment and score? A G G T A C G G C

• A
• G

G

• T

A C G G C

𝐰 𝐱

𝐰\𝐱

Start anywhere on first row End anywhere on last row

Fitting Alignment – Dynamic Programming

• Online:

https://valiec.github.io/AlignmentVisualizer/index.html

28

Question: Let match score be 1, mismatch/indel score be -1. What is 𝑡∗? Question: Same scores. What is optimal global alignment and score? A G G T A C G G C

• A
• G

G

• T

A C G G C

𝐰 𝐱

𝐰\𝐱

s[i, j] = max          0, if i = 0, s[i − 1, j] + δ(vi, −), if i > 0, s[i, j − 1] + δ(−, wj), if i > 0 and j > 0, s[i − 1, j − 1] + δ(vi, wj), if i > 0 and j > 0. s∗ = max{s[m, 0], . . . , s[m, n]}

Outline

• 1. Edit distance
• 2. Global alignment
• 3. Fitting alignment
• 4. Local alignment
• 5. Gapped alignment

Reading:

• Jones and Pevzner. Chapters 6.6, 6.8 and 6.9
• Lecture notes

29

Local Alignment – Biological Motivation

30

ABL1 SHKA

From Pfam database (http://pfam.sanger.ac.uk/)

Proteins are composed of functional units called domains. Such domains may occur in different proteins even across species.

Local Alignment problem: Given strings 𝐰 ∈ Σ` and 𝐱 ∈ Σa and scoring function 𝜀, find a substring of 𝐰 and a substring of 𝐱 whose alignment has maximum global alignment score 𝑡∗ among all global alignments of all substrings of 𝐰 and 𝐱

Global, Fitting and Local Alignment

31

Local Alignment problem: Given strings 𝐰 ∈ Σ` and 𝐱 ∈ Σa and scoring function 𝜀, find a substring of 𝐰 and a substring of 𝐱 whose alignment has maximum global alignment score 𝑡∗ among all global alignments of all substrings of 𝐰 and 𝐱 Fitting Alignment problem: Given strings 𝐰 ∈ Σ` and 𝐱 ∈ Σa and scoring function 𝜀, find an alignment of 𝐰 and a substring of 𝐱 with maximum global alignment score 𝑡∗ among all global alignments of 𝐰 and all substrings of 𝐱 Global Alignment problem: Given strings 𝐰 ∈ Σ` and 𝐱 ∈ Σa and scoring function 𝜀, find alignment of 𝐰 and 𝐱 with maximum score.

Local Alignment – Naive Algorithm

Brute force:

• 1. Generate all pairs (𝐰V, 𝐱V) of substrings of 𝐰 and 𝐱
• 2. For each pair (𝐰V, 𝐱V), solve global alignment problem.

32

Question: There are `

Z a Z pairs of substrings.

But they have different lengths. What is the running time?

Local Alignment problem: Given strings 𝐰 ∈ Σ` and 𝐱 ∈ Σa and scoring function 𝜀, find a substring of 𝐰 and a substring of 𝐱 whose alignment has maximum global alignment score 𝑡∗ among all global alignments of all substrings of 𝐰 and 𝐱

Key Idea

33

Local alignment:

• Start and end anywhere

Global alignment:

• Start at (0,0) and end at (𝑛, 𝑜)

Local Alignment Recurrence

34

s[i, j] = max          0, if i = 0 and j = 0, s[i − 1, j] + δ(vi, −), if i > 0, s[i, j − 1] + δ(−, wj), if j > 0, s[i − 1, j − 1] + δ(vi, wj), if i > 0 and j > 0. s∗ = max

i,j s[i, j]

Local Alignment problem: Given strings 𝐰 ∈ Σ` and 𝐱 ∈ Σa and scoring function 𝜀, find a substring

• f 𝐰 and a substring of 𝐱 whose alignment has

maximum global alignment score 𝑡∗ among all global alignments of all substrings of 𝐰 and 𝐱

Local Alignment Recurrence

35

s[i, j] = max          0, if i = 0 and j = 0, s[i − 1, j] + δ(vi, −), if i > 0, s[i, j − 1] + δ(−, wj), if j > 0, s[i − 1, j − 1] + δ(vi, wj), if i > 0 and j > 0. s∗ = max

i,j s[i, j]

Local Alignment problem: Given strings 𝐰 ∈ Σ` and 𝐱 ∈ Σa and scoring function 𝜀, find a substring

• f 𝐰 and a substring of 𝐱 whose alignment has

maximum global alignment score 𝑡∗ among all global alignments of all substrings of 𝐰 and 𝐱

Start anywhere End anywhere

Running time: 𝑃(𝑛𝑜)

Local Alignment – Dynamic Programming

• Online:

https://valiec.github.io/AlignmentVisualizer/index.html

36

Question: Let match score be 2, mismatch score be -2 and indel be -4. What is 𝑡∗? A G G T A C G G C G G G G

𝐰 𝐱

𝐰\𝐱

s[i, j] = max          0, if i = 0 and j = 0, s[i − 1, j] + δ(vi, −), if i > 0, s[i, j − 1] + δ(−, wj), if j > 0, s[i − 1, j − 1] + δ(vi, wj), if i > 0 and j > 0. s∗ = max

i,j s[i, j]

Global, Fitting and Local Alignment

37

Local Alignment problem: Given strings 𝐰 ∈ Σ` and 𝐱 ∈ Σa and scoring function 𝜀, find a substring of 𝐰 and a substring of 𝐱 whose alignment has maximum global alignment score 𝑡∗ among all global alignments of all substrings of 𝐰 and 𝐱 Fitting Alignment problem: Given strings 𝐰 ∈ Σ` and 𝐱 ∈ Σa and scoring function 𝜀, find an alignment of 𝐰 and a substring of 𝐱 with maximum global alignment score 𝑡∗ among all global alignments of 𝐰 and all substrings of 𝐱 Global Alignment problem: Given strings 𝐰 ∈ Σ` and 𝐱 ∈ Σa and scoring function 𝜀, find alignment of 𝐰 and 𝐱 with maximum score.

Outline

• 1. Edit distance
• 2. Global alignment
• 3. Fitting alignment
• 4. Local alignment
• 5. Gapped alignment

Reading:

• Jones and Pevzner. Chapters 6.6, 6.8 and 6.9
• Lecture notes

38

Scoring Gaps

39

Let 𝐰 = AAC and 𝐱 = ACAGGC Match 𝜀 𝑑, 𝑑 = 1; Mismatch 𝜀 𝑑, 𝑒 = −1 (where 𝑑 ≠ 𝑒); Indel 𝜀 𝑑, − = 𝜀 −, 𝑑 = −2 Both alignments have 3 matches and 2 indels. Score: 3 ∗ 1 + 2 ∗ −2 = −1

A

• A

C A C A A C

𝐰 𝐱

A

• A
• C

A C A A C

𝐰 𝐱

Scoring Gaps

40

Let 𝐰 = AAC and 𝐱 = ACAGGC Match 𝜀 𝑑, 𝑑 = 1; Mismatch 𝜀 𝑑, 𝑒 = −1 (where 𝑑 ≠ 𝑒); Indel 𝜀 𝑑, − = 𝜀 −, 𝑑 = −2 Question: Which alignment is better?

A

• A

C A C A A C

𝐰 𝐱

A

• A
• C

A C A A C

𝐰 𝐱 Both alignments have 3 matches and 2 indels. Score: 3 ∗ 1 + 2 ∗ −2 = −1

Scoring Gaps – Affine Gap Penalties

41

Desired: Lower penalty for consecutive gaps than interspersed gaps. Why: Consecutive gaps are more likely due to slippage errors in DNA replication (2-3 nucleotides), codons for protein sequences, etc.

A

• A

C A C A A C

𝐰 𝐱

A

• A
• C

A C A A C

𝐰 𝐱

Scoring Gaps – Affine Gap Penalties

42

Desired: Lower penalty for consecutive gaps than interspersed gaps. Why: Consecutive gaps are more likely due to slippage errors in DNA replication (2-3 nucleotides), codons for protein sequences, etc. Affine gap penalty: Two penalties: (i) gap open penalty 𝜍 ≥ 0 and (ii) gap extension penalty 𝜏 ≥ 0. Stretch of 𝑙 consecutive gaps has score −(𝜍 + 𝜏𝑙).

A

• A

C A C A A C

𝐰 𝐱

A

• A
• C

A C A A C

𝐰 𝐱

Scoring Gaps – Affine Gap Penalties

43

Desired: Lower penalty for consecutive gaps than interspersed gaps. Why: Consecutive gaps are more likely due to slippage errors in DNA replication (2-3 nucleotides), codons for protein sequences, etc. Affine gap penalty: Two penalties: (i) gap open penalty 𝜍 ≥ 0 and (ii) gap extension penalty 𝜏 ≥ 0. Stretch of 𝑙 consecutive gaps has score −(𝜍 + 𝜏𝑙). Let 𝜍 = 10 and 𝜏 = 1. Left: 3 ∗ 1 − 10 + 1 ∗ 2 = −9. Right: 3 ∗ 1 − (10 + 1 ∗ 1) − 10 + 1 ∗ 1 = −19.

A

• A

C A C A A C

𝐰 𝐱

A

• A
• C

A C A A C

𝐰 𝐱

Affine Gap Penalty Alignment – Naive Approach

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Idea: Insert horizontal (deletion) and vertical (insertion) edges spanning 𝑙 > 1 gaps with score − (𝜍 + 𝜏𝑙).

new edges

• ld edges

Affine gap penalty: Two penalties: (i) gap open penalty 𝜍 ≥ 0 and (ii) gap extension penalty 𝜏 ≥ 0. Stretch of 𝑙 consecutive gaps has score −(𝜍 + 𝜏𝑙).

... ... ... ... ... ... ... ... ...

Affine Gap Penalty Alignment – Naive Approach

45

Idea: Insert horizontal (deletion) and vertical (insertion) edges spanning 𝑙 > 1 gaps with score − (𝜍 + 𝜏𝑙).

new edges

• ld edges

Question: What’s the running time? Question: What’s the recurrence? Affine gap penalty: Two penalties: (i) gap open penalty 𝜍 ≥ 0 and (ii) gap extension penalty 𝜏 ≥ 0. Stretch of 𝑙 consecutive gaps has score −(𝜍 + 𝜏𝑙).

... ... ... ... ... ... ... ... ...

46

Affine Gap Penalty Alignment

Idea: Three separate recurrences: (i) Gap in first sequence 𝑡→ 𝑗, 𝑘 (ii) Match/mismatch 𝑡↘[𝑗, 𝑘] (iii) Gap in second sequence 𝑡↓[𝑗, 𝑘]

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Affine Gap Penalty Alignment

Idea: Three separate recurrences: (i) Gap in first sequence 𝑡→ 𝑗, 𝑘 (ii) Match/mismatch 𝑡↘[𝑗, 𝑘] (iii) Gap in second sequence 𝑡↓[𝑗, 𝑘]

s![i, j] = max ( s![i, j − 1] − σ, if j > 1, s&[i, j − 1] − (σ + ρ), if j > 0, s&[i, j] = max 8 > > > < > > > : 0, if i = 0 and j = 0, s![i, j], if j > 0, s#[i, j], if i > 0, s&[i − 1, j − 1] + δ(vi, wj), if i > 0 and j > 0, s#[i, j] = max ( s#[i − 1, j] − σ, if i > 1, s&[i − 1, j] − (σ + ρ), if i > 0.

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Affine Gap Penalty Alignment

Idea: Three separate recurrences: (i) Gap in first sequence 𝑡→ 𝑗, 𝑘 (ii) Match/mismatch 𝑡↘[𝑗, 𝑘] (iii) Gap in second sequence 𝑡↓[𝑗, 𝑘] Running time: 𝑃(𝑛𝑜)

s![i, j] = max ( s![i, j − 1] − σ, if j > 1, s&[i, j − 1] − (σ + ρ), if j > 0, s&[i, j] = max 8 > > > < > > > : 0, if i = 0 and j = 0, s![i, j], if j > 0, s#[i, j], if i > 0, s&[i − 1, j − 1] + δ(vi, wj), if i > 0 and j > 0, s#[i, j] = max ( s#[i − 1, j] − σ, if i > 1, s&[i − 1, j] − (σ + ρ), if i > 0.

Affine Gap Penalty Alignment – Example

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𝐰 = AAC 𝐱 = ACAAC

Let 𝜍 = 10 and 𝜏 = 1. Match = 1. Mismatch = -1

s![i, j] = max ( s![i, j − 1] − σ, if j > 1, s&[i, j − 1] − (σ + ρ), if j > 0, s&[i, j] = max 8 > > > < > > > : 0, if i = 0 and j = 0, s![i, j], if j > 0, s#[i, j], if i > 0, s&[i − 1, j − 1] + δ(vi, wj), if i > 0 and j > 0, s#[i, j] = max ( s#[i − 1, j] − σ, if i > 1, s&[i − 1, j] − (σ + ρ), if i > 0.

Gapped Alignment – Additional Insights

• Naive approach supports arbitrary gap penalties given two

sequences 𝐰 ∈ Σ` and 𝐱 ∈ Σa. This results in an 𝑃(𝑛𝑜 𝑛 + 𝑜 ) algorithm.

• Alignment with convex gap

penalties given two sequences 𝐰 ∈ Σ` and 𝐱 ∈ Σa can be computed in 𝑃(𝑛𝑜 log 𝑛) time.

See: Dan Gusfield. 1997. Algorithms on Strings, Trees, and Sequences: Computer Science and Computational Biology. Cambridge University Press, New York, NY, USA.

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Take Home Messages

• 1. Edit distance
• 2. Global alignment
• 3. Fitting alignment
• 4. Local alignment
• 5. Gapped alignment

Reading:

• Jones and Pevzner. Chapters 6.6, 6.8 and 6.9
• Lecture notes

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Global alignment is longest path in DAG Small tweaks enable different extensions Edit distance is shortest path in DAG