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

CAP5510 – BioinformaticsSequence Comparison

Tamer Kahveci

CISE Department

University of Florida

• Understand major sequence comparison algorithms.

• Gain hands on experience

Why Compare Sequences ?

• Prediction of function

• Construction of phylogeny

• Shotgun assembly

• Finding motifs

• Understanding of biological processes

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)

Dot Plot

A A T T C G A

How can we compute similarity?

O(m+n) time

Is it a good scheme ?

Use longer subsequences (k-gram)

Dot Plot

A A T T C G A

Use longer subsequences (k-gram)

Sequence Comparison

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

• How to evaluate– Distance– Score

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

Each Alignment Maps to a Path

A A T T C G

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 ?

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)

• 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)

DP Example

A A T 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))}

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

•O(mn) time and space

•Reconstruct alignment

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

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 !

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

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

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

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

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

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:

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

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)

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]

Local Alignment

• The prefixes with highest local suffix alignment

-- 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

Match = +5Mismatch = -4

Local Alignment Example

• 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

Banded Global Alignment

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

• How can we align ?

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

End space free alignment

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

End space free alignment

• --CCA-TGAC• TTCCAGTG--

Pattern search

• AAGCAGCCATGACGGAAAT• CCAGTG• How can we find it ?

Pattern search

• AAGCAGCCATGACGGAAAT• CCAGTG

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 ?

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

Next: Closer look into 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 ?

Computing affine gaps

• 3 cases

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}

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}

Recursions

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

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)}

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

DP in Linear Space ?

Linear Space DP

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

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

A A T T C G

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

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

Next: inversions

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 ?

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

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)

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

Pattern Searching with Bitwise Operations

UM-92 (A3)

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

• 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

• AATAACAATACAT• AATAC

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

A C G T

A 1 0 0 0

T 0 0 0 1

A 1 0 0 0

C 0 1 0 0

• AATAACAATACAT• AATAC

A C G T

A 1 0 0 0

T 0 0 0 1

A 1 0 0 0

C 0 1 0 0

• 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.

• 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

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

• 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).

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)

Dirichlet (pigeonhole) Principle

• NK balls

• K+1 boxes

• Put balls in boxes

• At least one box contains < N balls

• For searching k edit distance threshold

• Partition y into k+1 pieces

• At least one of them is an exact match.

• Why ? (Dirichlet principle)

• 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

Neighborhood Searching

Myers-94 (A4)

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

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

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 -> #

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 ?

• 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

Using NFA for Sequence Matching

Baeza-Yates, Navarro – 99 (A6)

Using NFA for Sequence Matching (1)

mismatch

deletepattern

insertpattern

1: active state

0: inactive state

NFA for pattern “patt” Search inside “waitt”

• 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)

• NFA gets large for long y and large error threshold

• How can we manage long y ?– Dirichlet principle

• Extension: Searching multiple patterns in parallel

Approximate Global Alignment of Sequences

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

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.

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

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).

Distance Prediction using Frequency Vectors

A C T - - T A G

A A T G A T A G

A C T T A G C

* * * *

A A T G A T A

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]

Mean :

Variance :

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)

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

Unbounded Alignment: upper bounding the alignment

•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

How to Compute the Upper Bound?

Dij = distance for a sample alignment (suffix)

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

Di,j = 4

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

A A T C T G

Optimization 3: Path Prune

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

• Remove entry (i,j)

Optimization 3: Path Prune

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

• Remove entry (i,j)

Unbounded Alignment: Time

Unbounded Alignment: Space

Bounded Alignment: Time

Bounded Alignment: Space

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

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

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.

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.

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.

Paralog, Ortholog, Xenolog

ParalogOrthologXenolog

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

Documents

CAP 5510: Introduction to Bioinformaticsgiri/teach/Bioinf/S08/Lec6.pdf · Research Tool. 2/7/07 CAP5510 5 Multiple Alignment. 2/7/07 CAP5510 6 How to Score Multiple Alignments?

CISE Annual Report

1 CAP5510 – Bioinformatics Fall 2015 Tamer Kahveci CISE Department University of Florida

Kuru Kahveci Mehmet Efendi

CAP5510 – Bioinformatics Database Searches for Biological Sequences

CISE REU Evaluation Toolkit

CISE 2013 Newsletter.indd

1 CAP5510 – Bioinformatics Protein Structures Tamer Kahveci CISE Department University of Florida

1 CAP5510 – Bioinformatics Fall 2009 Tamer Kahveci CISE Department University of Florida

Introduction to CISE

Boletín n° 4 Investiga CISE

CISE 2012-2013 Annual Report

CISE and ToNC

Department of CISE

Júramelo.es reconocida por el CISE (PDF)

CISE IAB Meeting March 25, 2010

Moroccan CISE

CISE 412 Mechatronics Lab Manual

1 CAP5510 – Bioinformatics Substitution Patterns Tamer Kahveci CISE Department University of Florida

Mustafa Kahveci - ugr.com.tr