1 CAP5510 – Bioinformatics Sequence Comparison Tamer Kahveci CISE Department University of...

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CAP5510 – BioinformaticsSequence Comparison

Tamer Kahveci

CISE Department

University of Florida

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Goals

• Understand major sequence comparison algorithms.

• Gain hands on experience

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Why Compare Sequences ?

• Prediction of function

• Construction of phylogeny

• Shotgun assembly

• Finding motifs

• Understanding of biological processes

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Question

• Q = AATTCGA

• X = ACATCGG• Y = CATTCGCC• Z = ATTCCGC

• Form groups of 2-3. Sort X, Y, and Z in decreasing similarity to Q. (5 min)

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Dot Plot

A A T T C G A

A

C

A

T

C

G

G

How can we compute similarity?

O(m+n) time

Is it a good scheme ?

Use longer subsequences (k-gram)

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Dot Plot

A A T T C G A

A

C

A

T

C

G

G

Use longer subsequences (k-gram)

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Sequence Comparison

• How to align– Global alignment: align entire sequences– Local alignment: align subsequences

• How to evaluate– Distance– Score

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Global Alignment

• Q = AATTCGA• |rr|||r• X = ACATCGG

• 4 match• 3 mismatch

• Q = A-ATTCGA• |i|d|||r• X = ACA-TCGG• 5 match• 1 insert• 1 delete• 1 mismatch

Similarity is defined in terms of Distance / Score of alignment

Many combinations of Insert / delete / (mis)match

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Each Alignment Maps to a Path

A A T T C G

A

C

A

T

C

G

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Edit Distance

• Minimum number of insert / delete / replace operators to transform one sequence into the other.

• Q = AATTCGA• | ||| => 3• X = ACATCGG

How do we find the minimum edit distance ?

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Global sequence alignment(Needleman-Wunsch)

• Compute distance recursively : dynamic programming.

Case 1 : match (0) or mismatch (1)

Case 2 : delete (1)

Case 3 : insert (1)

Case 0 : one string is empty (n)

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• Optimal string alignment• D(i,j) = edit distance between A(1:i) and B(1:j)• d(a,b) = 0 if a = b, 1 otherwise.• Recurrence relation

– D(i,0) = Σ d(A(k),-), 0 <= k <= i– D(0,j) = Σ d(-,B(k)), 0 <= k <= j– D(i,j) = Min {

• D(i-1,j) + d(A(i),-), • D(i,j-1) + d(-,B(j)),• D(i-1,j-1) + d(A(i),B(j))}

Global sequence alignment(Needleman-Wunsch)

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DP Example

A A T T C G

A

C

A

T

C

G

D(i,0) = D(i,0) = Σ d(A(k),-), 0 <= k <= iD(0,j) = D(0,j) = Σ d(-,B(k)), 0 <= k <= j

D(i,j) = D(i,j) = Min {•D(i-1,j) + d(A(i),-), •D(i,j-1) + d(-,B(j)),•D(i-1,j-1) + d(A(i),B(j))}

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DP Example: Backtracking

A A T T C G

0 1 2 3 4 5 6

1 0 1 2 3 4 5

2 1 1 2 3 3 4

3 2 1 2 3 4 4

4 3 2 1 2 3 4

5 4 3 2 2 2 3

6 5 4 3 3 3 2

A

C

A

T

C

G

•O(mn) time and space

•Reconstruct alignment

•O(max{m,n}) space if alignment not needed. How ?

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Number of Alignments

• N(n, m) = number of alignments of sequences of n and m letters (not necessarily optimal alignment).

• N(0, i) = N(i, 0) = 1• N(n, m) = N(n-1, m) + N(n, m-1) + N(n-1,m-1)• N(n, n) ~ (1 + 21/2)2n+1n-1/2.• N(1000, 1000) > 10767

• 1080 atoms in the universe !

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Edit Distance: a Good Measure?

• Compare these two alignments. Which one is better ?

• Q = AATTCGA• | |||• X = ACATCGG

• Q = A-ATTCGA• | | |||• X = ACA-TCGG

Scoring scheme: • +1 for each match• -1 for each mismatch/indel

Can be computed the same as edit distance by including +1 for each match

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More Trouble: Scoring Matrices

• Different mutations may occur at different rates in nature. Why ?

• E.g., each amino acid = three nucleotides. Transformation of one amino acid to other due to single nucleotide modification may be biased– E = GAA, GAG– D = GAU, GAC– F = UUU, UUC– E similar to D, not similar to F

• Mutation probability of different pairs of nucleotides may differ.

• PAM, BLOSUM matrices

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A R N D C Q E G H I L K M F P S T W Y VA 5 -2 -1 -2 -1 -1 -1 0 -2 -1 -1 -1 -1 -2 -1 1 0 -2 -2 0 R -2 7 0 -1 -3 1 0 -2 0 -3 -2 3 -1 -2 -2 -1 -1 -2 -1 -2 N -1 0 6 2 -2 0 0 0 1 -2 -3 0 -2 -2 -2 1 0 -4 -2 -3 D -2 -1 2 7 -3 0 2 -1 0 -4 -3 0 -3 -4 -1 0 -1 -4 -2 -3 C -1 -3 -2 -3 12 -3 -3 -3 -3 -3 -2 -3 -2 -2 -4 -1 -1 -5 -3 -1 Q -1 1 0 0 -3 6 2 -2 1 -2 -2 1 0 -4 -1 0 -1 -2 -1 -3 E -1 0 0 2 -3 2 6 -2 0 -3 -2 1 -2 -3 0 0 -1 -3 -2 -3 G 0 -2 0 -1 -3 -2 -2 7 -2 -4 -3 -2 -2 -3 -2 0 -2 -2 -3 -3H -2 0 1 0 -3 1 0 -2 10 -3 -2 -1 0 -2 -2 -1 -2 -3 2 -3 I -1 -3 -2 -4 -3 -2 -3 -4 -3 5 2 -3 2 0 -2 -2 -1 -2 0 3 L -1 -2 -3 -3 -2 -2 -2 -3 -2 2 5 -3 2 1 -3 -3 -1 -2 0 1 K -1 3 0 0 -3 1 1 -2 -1 -3 -3 5 -1 -3 -1 -1 -1 -2 -1 -2 M -1 -1 -2 -3 -2 0 -2 -2 0 2 2 -1 6 0 -2 -2 -1 -2 0 1 F -2 -2 -2 -4 -2 -4 -3 -3 -2 0 1 -3 0 8 -3 -2 -1 1 3 0 P -1 -2 -2 -1 -4 -1 0 -2 -2 -2 -3 -1 -2 -3 9 -1 -1 -3 -3 -3 S 1 -1 1 0 -1 0 0 0 -1 -2 -3 -1 -2 -2 -1 4 2 -4 -2 -1 T 0 -1 0 -1 -1 -1 -1 -2 -2 -1 -1 -1 -1 -1 -1 2 5 -3 -1 0 W -2 -2 -4 -4 -5 -2 -3 -2 -3 -2 -2 -2 -2 1 -3 -4 -3 15 3 -3 Y -2 -1 -2 -2 -3 -1 -2 -3 2 0 0 -1 0 3 -3 -2 -1 3 8 -1V 0 -2 -3 -3 -1 -3 -3 -3 -3 3 1 -2 1 0 -3 -1 0 -3 -1 5

The BLOSUM45 Matrix

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score(H,P) = -2, gap Penalty = –8

H E A G A W G H E E

0 -8 -16 -24 -32 -40 -48 -56 -64 -72 -80

P -8 -2

A -16

W -24

H -32

E -40

A -48

E -56

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Score(E,P) = 0, score(E,A) = -1, score(H,A) = -2

H E A G A W G H E E

0 -8 -16 -24 -32 -40 -48 -56 -64 -72 -80

P -8 -2 -8

A -16 -10 -3

W -24

H -32

E -40

A -48

E -56

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H E A G A W G H E E

0 -8 -16 -24 -32 -40 -48 -56 -64 -72 -80

P -8 -2 -8 -16 -24 -33 -42 -49 -57 -65 -73

A -16 -10 -3 -4 -12 -19 -28 -36 -44 -52 -60

W -24 -18 -11 -6 -7 -15 -4 -12 -21 -29 -37

H -32 -14 -18 -13 -8 -9 -12 -6 -2 -11 -19

E -40 -22 -8 -16 -16 -9 -12 -14 -6 4 -5

A -48 -30 -16 -3 -11 -11 -12 -12 -14 -4 2

E -56 -38 -24 -11 -6 -12 -14 -15 -12 -8 2

H E A G A W G H E - E- P - - A W - H E A E

Optimal alignment:

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Distance v.s. Similarity

• Similarity model: s(a,b), g’(k)• Distance model: d(a,b), g(k)If there is a constant c, such that

– S(a,b) = c – d(a,b)– G’(k) = g(k) – kc/2

ThenSimilarity optimal alignment = distance optimal

alignment

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Global Alignment ?

• Q = A-ATTCGA• | | |||• X = ACA-TCGG

• Q = AATTCGA-• ||||| • Y = CATTCGCC

Which one is more similar to Q ?

Local alignment: highest scoring subsequence alignment. How can we find it ?

Brute force: O(n3m3)Gotoh (Smith-Waterman): O(nm)

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Local Suffix Alignment

• V(i, 0) = v(0, j) = 0• V(i,j) = max{0, v(i-1, j-1) + s(x(i), y(j)), v(i-1, j) + s(x(i), -) v(i, j-1) + s(-, y(j))}

X[1: i]

Y[1: j]

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Local Alignment

• The prefixes with highest local suffix alignment

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-- G C T G G A A G G C A T

-- 0 0 0 0 0 0 0 0 0 0 0 0

G 0 5 1 0 5 5 1 0 5 5 1 0 0

C 0 1 10 6 2 1 1 0 1 1 10 6 2

A 0 0 6 6 2 0 6 6 2 0 6 15 11

G 0 5 2 2 11 7 3 2 11 7 3 11 11

A 0 1 1 0 7 7 11 8 7 7 3 8 7

G 0 5 1 0 5 11 7 7 13 12 8 4 4

C 0 0 10 6 2 7 7 3 9 8 17 13 9

A 0 0 6 6 2 3 11 12 8 5 13 22 18

C 0 0 5 2 2 0 7 8 8 4 18 18 18

G 0 5 1 1 7 7 5 4 13 13 14 14 14

P’s subsequence: G C A G A G C AQ’s subsequence: G A A G – G C A

P

Q

Match = +5Mismatch = -4

Local Alignment Example

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Goals

• Other important sequence comparison problems– banded alignment– end free search– pattern search– non-overlapping alignments– gaps– linear-space algorithms– bitwise operations– neighborhood searching– NFAs– Approximate alignment

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Banded Global Alignment

• Two sequences differ by at most w edit operations (w<<n).

• How can we align ?

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Banded Alignment Example

• O(wn) time and space.

• Example: – w=3.– Match = +1– Mismatch = -1– Indel = -2

A C C A C A C A0 -2 -4 -6

A -2 1 -1 -3 -5

C -4 -1 2 0 -2 -4

A -6 -3 0 1 1 -1 -3

C -5 -2 1 0 2 0 -2

C -4 -1 0 1 1 1 -1

A -3 0 -1 2 0 2

T -2 -1 0 1 0

A -1 0 -1 2

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End space free alignment

• --CCA-TGAC• TTCCAGTG--• How can we find it ?

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End space free alignment

• --CCA-TGAC• TTCCAGTG--

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Pattern search

• AAGCAGCCATGACGGAAAT• CCAGTG• How can we find it ?

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Pattern search

• AAGCAGCCATGACGGAAAT• CCAGTG

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Non-overlapping Local Alignments

• GCTCTGCGAATA• GCTCTGCGAATA

• CGTTGAGATACT• CGTTGAGATACT

• Find all non-overlapping local alignments with score > threshold.

• Two alignments overlap if they share same letter pair.

• How do we find ?

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Non-overlapping Local Alignments

1. Compute DP matrix

2. Find the largest scoring alignment > threshold

3. Report the alignment

4. Remove the effects of the alignment from the matrix

5. Go to step 2

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Next: Closer look into gaps

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Gaps

• Q = AATTCGAG• ||||| • Y = -ATTCGC-

• Q = AATTCGAG• ||||| • Z = AATTCC--

Which one is more similar to Q ?

Starting an indel is less likely than continuing an indel.

Affine gap model: Large gap open and smaller gap extend penalty.

How can we compute it ?

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Computing affine gaps

• 3 cases

E

F

G

i

j

i

j

i

j

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Recursions

• E(i, 0) = gap_open + i x gap_extend

• E(i,j) = max{E(i, j-1) + gap_extend, V(i, j-1) + gap_open +

gap_extend}

Ei

j

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Recursions

• F(0, j) = gap_open + j x gap_extend

• F(i,j) = max{F(i-1, j) + gap_extend, V(i-1, j) + gap_open +

gap_extend}

Fi

j

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Recursions

• G(i,j) = G(i-1, j-1) + s(x(i), y(j))

Gi

j

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Recursions

• V(i, 0) = gap_open + i x gap_extend

• V(0, j) = gap_open + j x gap_extend

• V(i, j) = max{E(i, j), F(i, j), G(i, j)}

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Other Gap Models

• Constant: fixed gap penalty per gap regardless of length

• Non-linear: Gap cost increase is non-linear.– E.g., g(n) = -(1 + ½ + 1/3 + … + 1/n)

• Arbitrary

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DP in Linear Space ?

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Linear Space DP

• Keep two vectors at a time: – Two columns or two

rows

• O(min{m,n}) space• O(mn) time• No backtracking

A A T T C G

A

C

A

T

C

G

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Linear Space DP with Backtracking

• Find midpoint of the alignment– Align the first half– Align the second half– Choose the point with

best sum of score/distance

• Search the upper left and lower right of mid point

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Linear Space DP with Backtracking: Time Complexity

• 2(n/2 x m) = nm

• 2(n/4 x k) + 2(n/4 x (m-k)) = nm/2

• …

• nm/2i

• Adds up to 2nm

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Next: inversions

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Alignment with Inversions

• A’ = T and G’ = C

• ACTCTCTCGCTGTACTG• AATCT-ACTACTGCTTG

• Each letter is inverted only once.• An inversion cost (inv) for each inverted block.• How to find the alignment ?

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Alignment with Inversions

1. For i=1:m1. For j=1:n

1. For g=1:I1. For h=1:j

1. Compute Z(g,h; I,j)2. V(I,j) = max{

» Max{v(i-1,j-1) + z(g,h; I,j)} + inv» V(i-1,j-1) + s(xi, yj)» V(i-1, j) + ins» V(I, j-1) + del}

• O(n6) time

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Alignment with Inversions: Faster Method

1. Find all local alignments of x and y’ (Z)2. V(I,j) = max{

1. max{V(g-1, h-1) + Z(g, h; I, j)} + inv,

2. V(i-1, j-1) + s(xi, yj),3. V(i-1, j) + ins4. V(I, j-1) + del }

• O(nmL) time, where L is the average number of inverse alignments ending at (i,j)

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Recap & Goals

• Other important sequence comparison problems– banded alignment– end free search– pattern search– non-overlapping alignments– gaps– linear-space algorithms– inversions– bitwise operations– neighborhood searching– NFAs– Approximate alignment– Homology

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Pattern Searching with Bitwise Operations

UM-92 (A3)

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Pattern Searching with Bitwise Operations (1)

• Simple case : Find all exact matches to y in x• Rj[i] = 1 if first i letters of y matches last i letters of x.

• R0[i] = 1 (if i = 0) 0 (if 0 < i <= m)• Rj+1[i] =

– 1 (if Rj[i-1] = 1 and y[i] = x[j]) 0 (else)

• Match if Rj[m] = 1

x

y mn

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• Si[k] = – 1 if y[i] = kth letter in the alphabet– 0 else– (for i = 1, 2, …, m)

• Rj+1 = (right shift of Rj) AND (Si)– where x[j+1] = ith letter in the alphabet

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• AATAACAATACAT• AATAC

A A T A A C A A T A C A T

A 1

A 0

T 0

A 0

C 0

A C G T

A 1 0 0 0

A 1 0 0 0

T 0 0 0 1

A 1 0 0 0

C 0 1 0 0

11000

11010

11000

AND

R S

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• AATAACAATACAT• AATAC


A 1 1

A 0 1

T 0 0

A 0 0

C 0 0

A C G T

A 1 0 0 0

A 1 0 0 0

T 0 0 0 1

A 1 0 0 0

C 0 1 0 0

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• Harder case: one edit distance allowed

• Use R and R1

• R for exact match

• R1j[i] = 1 if first i letters of y matches last i letters of x with at most one edit operation.

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• Insertion1. y[1:i] matches x[:j] exactly

• insert x[j+1]

2. y[1:i-1] matches x[:j] with one insertion• match y[i] with x[j+1] if they are equal

• R1j+1 = (Rj) OR ((right shift of R1j) AND (Si))– where x[j+1] = ith letter in the alphabet

• Similar reasoning for delete and replace

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A 1 1 0 1

A 0 1 0 0

T 0 0 1 0

A 0 0 0 1

C 0 0 0 0


A 1 1 1 1

A 0 1 1 1

T 0 0 1 1

A 0 0 0 1

C 0 0 0 0

R R1

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• General problem: k edit operations are allowed

• Use R1, R2, …, Rk

• Update Rz+1 using Rz and Rz+1

• Improve running time by partitioning y into k+1 pieces (next slide).


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Improving Running Time of Approximate Pattern Search

• For searching k edit distance threshold

• Partition y into k+1 pieces

• At least one of them is an exact match.

• Why ? (Dirichlet principle)

k = 3

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Dirichlet (pigeonhole) Principle

• NK balls

• K+1 boxes

• Put balls in boxes

• At least one box contains < N balls

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• For searching k edit distance threshold

• Partition y into k+1 pieces

• At least one of them is an exact match.

• Why ? (Dirichlet principle)

k = 3

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• Search each partition for an exact match.• Align around the exact matches only

• Is it a good idea ? • (k+1)n/(Am/(k+1)) random matches, where A is the

alphabet size

k = 3

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Neighborhood Searching

Myers-94 (A4)

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Neighborhood Searching (1)

• Find all subsequences of x within D edit distance to y.• Assumption: m = logAn

• D-neighborhood of y = D-N(y) = set of all sequences within D edit distance to y

1. Find D-N(y)2. Find exact matches to all the sequences in D-N(y) in x

x

y mn

68


• Condensed D-neighborhood of y = D-N’(y) = Sequences in D-N’(y) which do not contain a prefix in D-N’(y)

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Searching Neighbors: Hash Table

• Fix the length of the sequence to search (say m)

• Create a hash table for all subsequences of x of length m.

• Lookup for query sequence of length m

• 01234567• CACACATGGTA

AAAA -> #…ACAC -> 1…CACA -> 0, 2…TTTT -> #

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Hash Table

• {A, C, G, T}• {0, 1, 2, 3}• {00, 01, 10, 11}

• GTCAT– 101101 = 29– (((2 x 4) + 3) x 4) + 1

• GTCAT– (((3 x 4) + 1) x 4) + 0– 52

• O(n) space and construction time

• What happens when query is– shorter ?– longer ?

• What happens when– alphabet size is large ?– m is large ?

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• O(Dn) worst case

• O(Dnf(D/m)log n) expected time, where f(D/m) is an increasing concave function

• What if m is large (i.e., m > logAn) ?

– Dirichlet principle

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Using NFA for Sequence Matching

Baeza-Yates, Navarro – 99 (A6)

x

y mn

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Using NFA for Sequence Matching (1)

match

mismatch

deletepattern

insertpattern

1: active state

0: inactive state

NFA for pattern “patt” Search inside “waitt”

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• If A(i,j) = 1 then A(i+d,j+d) = 1 for d>0

• Keep each diagonal’s first active node.

• Di = k if the first active node in diagonal I is k.

• Computation of D (next slide)

A’(i,j) = (A(i,j-1) AND x(k) == y(j)) OR A(i-1,j) OR A(i-1,j-1) OR A’(i-1,j-1)

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• NFA gets large for long y and large error threshold

• How can we manage long y ?– Dirichlet principle

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• Extension: Searching multiple patterns in parallel

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Approximate Global Alignment of Sequences

T. Kahveci, V. Ramaswamy, H. Tao, T. Li - 2005

79

The problem

• Given sequences X and Y– Bounded: Find global alignment of X and Y

with at most k edit ops.– Unbounded: Find global alignment of X and Y

with p% approximation

• p = 100 % = optimal alignment.

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Frequency Vectors [KS’01]

• Frequency vector is the count of each letter.

– f(s = AATGATAG) = [4, 0, 2, 2].

• Edit operations & frequency vectors:– (del. G), s = AAT.ATAG => f(s) = [4, 0, 1, 2]– (ins. C), s = AACTATAG => f(s) = [4, 1, 1, 2]– (AC), s = ACCTATAG => f(s) = [3, 2, 1, 2]

• Use frequency vectors to measure distance!

nA nGnC nT

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An Approximation to ED:Frequency Distance (FD)

• s = AATGATAG => f(s)=[4, 0, 2, 2]• q = ACTTAGC => f(q)=[2, 2, 1, 2]

– dec = (4-2) + (2-1) = 3– inc = (2-0) = 2– FD(f(s),f(q)) = 3– ED(q,s) = 4

• FD(f(s1),f(s2))=max{inc,dec}.• FD(f(s1),f(s2)) ED(s1,s2).

82

Distance Prediction using Frequency Vectors

A C T - - T A G

R I I

A A T G A T A G

A C T T A G C

* * * *

A A T G A T A

ED

GED

Given frequency vectors of two strings x and y, GED(x,y) is normally distributed.

Q = [12, 10, 3, 5]

U = [11, 11, 4, 4]

V = [6, 5, 9, 10]

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Mean :

Variance :

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Bounded Alignment: lower bounding the alignment

• Mi,j = Edit distance between prefixes of X and Y

• d = lower bound to ED between suffixes of X and Y with at least p% probability.

• If (Mi,j + d > cutoff) then – No solution exists from (i,j)

with p% probability.– Remove entry (i,j)

i

j

X

Y

d

Mi,j

p %d

85

Cost of computing lower bound?

• Frequency vectors can be computed in O(1) time incrementally.

A A T T C

[2 1 0 2][A C G T]

[1 1 0 2]

A A T T C

86

Unbounded Alignment: upper bounding the alignment

i

jX

Y

Dij

Mi,j

•Dij = upper bound to the distance between suffixes.

•Use mini,j {Mi,j + Dij} as cutoff.

•Prune if (Mi,j + d > cutoff)

•Dij: desirable if it is•Computed quickly•Tight

87

How to Compute the Upper Bound?

i

jX

Y

Dij

Mi,j

X

YDi,j

Dij = distance for a sample alignment (suffix)

- A A C C T CG C A T C T A

e.g.

Di,j = 4

88

Cost of Computing Upper Bound?

• Upper bound can be computed in O(1) time incrementally.

A A T C T G- C T C A G

A A T C T G- - T C A G

A T C T GC T C A G

D = 3 D = 3

D = 2

A A T C T G

TCAG

C

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Optimization 3: Path Prune

X

X

• No solution exists from entry (i, j) if its path to entry (0, 0) is blocked.

• Remove entry (i,j)

90

Optimization 3: Path Prune

X X

X X

X X X

X X X

• No solution exists from entry (i, j) if its path to entry (0, 0) is blocked.

• Remove entry (i,j)

91

Unbounded Alignment: Time

92

Unbounded Alignment: Space

93

Bounded Alignment: Time

94

Bounded Alignment: Space

95

Recap & Goals

• Other important sequence comparison problems– banded alignment– end free search– pattern search– non-overlapping alignments– gaps– linear-space algorithms– inversions– bitwise operations– neighborhood searching– NFAs– Homology

96

What is Similarity Anyway ?

• Similar: have similar letters

• Homolog: have common ancestor

• Not exactly the same !• Three types of homology

– Paralog– Ortholog– Xenolog

Organism A Organism B

Parent Organism

97

Paralog & Ortholog (1)• "Two genes are said to be

paralogous if they are derived from a duplication event, but orthologous if they are derived from a speciation event.“

W-H Li

1. A gene called A in species w 2. is duplicated producing initially two

copies of A.3. With time the two copies diverge by

evolution forming related genes A1 and A2. These two genes are said to be paralogous to one another. Paralogy typically involves comparisons within a species.

98

Paralog & Ortholog (2)

• Two species, x and y evolve from species w, their common ancestor. The descendants of the A1 and A2 genes are now called A1x, A1y, and A2x, A2y to reflect which species they now occupy. A1x is orthologous to A1y and A2x is orthologous to A2y. The comparison is between two species.

99

Xenolog

• Xenology is defined as that condition (horizontal transfer) where the history of the gene involves an interspecies transfer of genetic material. It does not include transfer between organelles and the nucleus. It is the only form of homology in which the history has an episode where the descent is not from parent to offspring but, rather, from one organism to another.

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Paralog, Ortholog, Xenolog

ParalogOrthologXenolog

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Recommended Reading

• Fitch, WM, “Homology a personal view on some of the problems”, Trends. Genet., 2000, 16: 227-231

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Overview

• Dot plots• Dynamic programming solutions

– Local, global alignments and their extensions

• Distance and similarity models• Gap models• Improvements and different on computation of

sequence similarity• Similarity versus homology

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Next: Substitution Patterns

•Predict substitutions

•What are scoring matrices and how are they derived?

1 CAP5510 – Bioinformatics Sequence Comparison Tamer Kahveci CISE Department University of...

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