CS262 Lecture 15, Win07, Batzoglou Multiple Sequence Alignments

CS262 Lecture 15, Win07, Batzoglou

Multiple Sequence Multiple Sequence AlignmentsAlignments


Saving cells in DP

1. Find local alignments

2. Chain -O(NlogN) L.I.S.

3. Restricted DP


Methods to CHAIN Local Alignments

Sparse Dynamic ProgrammingO(N log N)


The Problem: Find a Chain of Local Alignments

(x,y) (x’,y’)

requires

x < x’y < y’

Each local alignment has a weight

FIND the chain with highest total weight


Sparse Dynamic Programming

Back to the LCS problem:

• Given two sequences x = x1, …, xm

y = y1, …, yn

• Find the longest common subsequence Quadratic solution with DP

• How about when “hits” xi = yj are sparse?


Sparse Dynamic Programming

15 3 24 16 20 4 24 3 11 18

4

20

24

3

11

15

11

4

18

20

• Imagine a situation where the number of hits is much smaller than O(nm) – maybe O(n) instead


Sparse Dynamic Programming – L.I.S.

• Longest Increasing Subsequence

• Given a sequence over an ordered alphabet

x = x1, …, xm

• Find a subsequence

s = s1, …, sk

s1 < s2 < … < sk


Sparse Dynamic Programming – L.I.S.

Let input be w: w1,…, wn

INITIALIZATION:L: last LIS elt. array L[0] = -inf

L[1] = w1 L[2…n] = +inf

B: array holding LIS elts; B[0] = 0P: array of backpointers// L[j]: smallest jth element wi of j-long LIS seen so far

ALGORITHMfor i = 2 to n { Find j such that L[j – 1] < w[i] ≤ L[j] L[j] w[i]

B[j] iP[i] B[j – 1]

}

That’s it!!!• Running time?


Sparse LCS expressed as LIS

Create a sequence w

• Every matching point (i, j), is inserted into w as follows:

• For each column j = 1…m, insert in w the points (i, j), in decreasing row i order

• The 11 example points are inserted in the order given

• a = (y, x), b = (y’, x’) can be chained iff

a is before b in w, and y < y’

15 3 24 16 20 4 24 3 11 18

6

4

2 7

1 8

10

9

5

11

3

4

20

24

3

11

15

11

4

18

20

x

y


Sparse LCS expressed as LIS

Create a sequence w

w = (4,2) (3,3) (10,5) (2,5) (8,6) (1,6) (3,7) (4,8) (7,9) (5,9) (9,10)

Consider now w’s elements as ordered lexicographically, where

• (y, x) < (y’, x’) if y < y’

Claim: An increasing subsequence of w is a common subsequence of x and y

15 3 24 16 20 4 24 3 11 18

6

4

2 7

1 8

10

9

5

11

3

4

20

24

3

11

15

11

4

18

20

x

y

Why don’t we insert elements (i, j) in w in

increasing row i order?


Sparse Dynamic Programming for LIS

Example:w = (4,2) (3,3) (10,5) (2,5) (8,6)

(1,6) (3,7) (4,8) (7,9) (5,9) (9,10)

L = [L1] [L2] [L3] [L4] [L5] …

1. (4,2)2. (3,3)3. (3,3) (10,5)4. (2,5) (10,5)5. (2,5) (8,6)6. (1,6) (8,6)7. (1,6) (3,7)8. (1,6) (3,7) (4,8)9. (1,6) (3,7) (4,8) (7,9)10. (1,6) (3,7) (4,8) (5,9)11. (1,6) (3,7) (4,8) (5,9) (9,10)

Longest common subsequence:s = 4, 24, 3, 11, 18

15 3 24 16 20 4 24 3 11 18

6

4

2 7

1 8

10

9

5

11

3

4

20

24

3

11

15

11

4

18

20

x

y


Sparse DP for rectangle chaining

• 1,…, N: rectangles

• (hj, lj): y-coordinates of rectangle j

• w(j): weight of rectangle j

• V(j): optimal score of chain ending in j

• L: list of triplets (lj, V(j), j)

L is sorted by lj: smallest (North) to largest (South) value

L is implemented as a balanced binary tree

y

h

l



Main idea:

• Sweep through x-coordinates

• To the right of b, anything chainable to a is chainable to b

• Therefore, if V(b) > V(a), rectangle a is “useless” for subsequent chaining

• In L, keep rectangles j sorted with increasing lj-coordinates sorted with increasing V(j) score

V(b)V(a)



Go through rectangle x-coordinates, from lowest to highest:

1. When on the leftmost end of rectangle i:

a. j: rectangle in L, with largest lj < hi

b. V(i) = w(i) + V(j)

2. When on the rightmost end of i:

a. k: rectangle in L, with largest lk lib. If V(i) > V(k):

i. INSERT (li, V(i), i) in L

ii. REMOVE all (lj, V(j), j) with V(j) V(i) & lj li

i

j

k

Is k ever removed?


Example

x

y

a: 5

c: 3

b: 6

d: 4e: 2

2

56

9101112141516

1. When on the leftmost end of rectangle i:

a. j: rectangle in L, with largest lj < hi

b. V(i) = w(i) + V(j)

2. When on the rightmost end of i:

a. k: rectangle in L, with largest lk lib. If V(i) > V(k):

i. INSERT (li, V(i), i) in L

ii. REMOVE all (lj, V(j), j) with V(j) V(i) & lj li

a b c d eV

5

L

li

V(i)

i

5

5

a

8

11

8

c

11 12

9

11

b

15

12

d

13

16

13

3


Time Analysis

1. Sorting the x-coords takes O(N log N)

2. Going through x-coords: N steps

3. Each of N steps requires O(log N) time:

• Searching L takes log N• Inserting to L takes log N• All deletions are consecutive, so logN per deletion• Each element is deleted at most once: < N logN for all deletions

• Recall that INSERT, DELETE, SUCCESSOR, take O(log N) time in a balanced binary search tree


Whole-genome Alignment Pipelines

Given N species, phylogenetic tree:

1. Local Alignment between all pairs – BLAST

2. In the order of the tree:1. Synteny mapping: find long regions with lots of collinear alignments

2. In each synteny region,1. Chaining

2. Global alignment

Alternatively, all species are mapped to one reference (e.g., human)

Then, in each unbroken synteny region between multiple species, perform chaining & progressive multiple alignment


Examples

Human Genome BrowserABC


Whole-genome alignment Rat—Mouse—Human


Next 2 years: 20+ mammals, & many other animals, will be sequenced & aligned


Gene Recognition

Credits for slides:Serafim BatzoglouMarina AlexanderssonLior PachterSerge Saxonov


The Central Dogma

Protein

RNA

DNA

transcription

translation

CCTGAGCCAACTATTGATGAA

PEPTIDE

CCUGAGCCAACUAUUGAUGAA


Gene structure

exon1 exon2 exon3intron1 intron2

transcription

translation

splicing

exon = protein-codingintron = non-coding

Codon:A triplet of nucleotides that is converted to one amino acid


Finding Genes in Yeast

Start codonATG

5’ 3’

Stop codonTAG/TGA/TAA

Intergenic Coding Intergenic

Mean coding length about 1500bp (500 codons)

Transcript


Finding Genes in Yeast

Yeast ORF distribution


Introns: The Bane of ORF Scanning

Start codonATG

5’ 3’

Stop codonTAG/TGA/TAA

Splice sites

Intergenic Exon Intron IntergenicExon ExonIntron

Transcript


Introns: The Bane of ORF Scanning

• Drosophila:

• 3.4 introns per gene on average

• mean intron length 475, mean exon length 397

• Human:

• 8.8 introns per gene on average

• mean intron length 4400, mean exon length 165

• ORF scanning is defeated


Where are the genes?Where are the genes?



Needles in a Haystack


Signals for Gene Finding

• We need to use more information to help recognize genes

1. Regular gene structure

2. Exon/intron lengths

3. Nucleotide composition

4. Motifs at the boundaries of exons, introns, etc.Start codon, stop codon, splice sites

5. Patterns of conservation


Regular Gene Structure

• Start, Stop of translation region: Protein-coding starts with ATG ends with TAA / TAG / TGA

• Exon – Intron – Exon – Intron … – Exon

• g[ GT/GC ]gag – Intron – cAGt

• Exon reading frame: NNN – NNN – NNN – NNN – NN… NN – NNN – NNN – NNN – NN… N – NNN – NNN – NNN – NNN…


Next Exon:Frame 0

Next Exon:Frame 1


Exon/Intron Lengths


Nucleotide Composition

• Base composition in exons is characteristic due to the genetic code

Amino Acid SLC DNA CodonsIsoleucine I ATT, ATC, ATALeucine L CTT, CTC, CTA, CTG, TTA, TTGValine V GTT, GTC, GTA, GTGPhenylalanine F TTT, TTCMethionine M ATGCysteine C TGT, TGCAlanine A GCT, GCC, GCA, GCG Glycine G GGT, GGC, GGA, GGG Proline P CCT, CCC, CCA, CCGThreonine T ACT, ACC, ACA, ACGSerine S TCT, TCC, TCA, TCG, AGT, AGCTyrosine Y TAT, TACTryptophan W TGGGlutamine Q CAA, CAGAsparagine N AAT, AACHistidine H CAT, CACGlutamic acid E GAA, GAGAspartic acid D GAT, GACLysine K AAA, AAGArginine R CGT, CGC, CGA, CGG, AGA, AGG


Biological Signals

• How does the cell recognize start/stop codons and splice sites? In part, from characteristic base composition

• Donor site (start of intron) is recognized by a section of U1 snRNA

U1 snRNA: GUCCAUUCADonor site consensus: MAGGTRAGT

M means “A or C”, R means “A or G”


atg

tga

ggtgag

ggtgag

ggtgag

caggtg

cagatg

cagttg

caggccggtgag


5’ 3’Donor site

Position

-8 … -2 -1 0 1 2 … 17

A 26 … 60 9 0 0 54 … 21C 26 … 15 5 0 1 2 … 27G 25 … 12 78 100 0 41 … 27T 23 … 13 8 0 99 3 … 25

Splice Sites


Splice Sites

(http://www-lmmb.ncifcrf.gov/~toms/sequencelogo.html)


• WMM: weight matrix model = PSSM (Staden 1984)• WAM: weight array model = 1st order Markov (Zhang & Marr 1993)

• MDD: maximal dependence decomposition (Burge & Karlin 1997) Decision-tree algorithm to take pairwise dependencies into account

Starting with a training set of known splice sites:

• For each position I, calculate Si = ji2(Ci, Xj)

• Choose i* such that Si* is maximal and partition into two subsets, until• No significant dependencies left, or• Not enough sequences in subset

Train separate WMM models for each subset

All donor splice sites

G5

not G5

G5G-1

G5

not G-1

G5G-1

A2

G5G-1

not A2

G5G-1

A2U6

G5G-1A2

not U6

Splice Sites

Documents

CS262 Lecture 15, Win07, Batzoglou Multiple Sequence Alignments