dynamic programming (cont’d)

Dynamic Programming (cont’d)

CS 466Saurabh Sinha

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:

Normal scoring would give the same score for both alignments

This is more likely.

This is less likely.

ATA__GGCATGATCGC

ATA_G_GCATGATCGC

Accounting for Gaps• 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

• 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

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

Continue Gap in w (deletion)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

Optional Reading Section 6.10 (J & P)Multiple Alignment

Gene Prediction

• Gene: A sequence of nucleotides coding for protein

• Gene Prediction Problem: Determine the beginning and end positions of genes in a genome

Gene Prediction: Computational Challenge

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• In 1961 Sydney Brenner and Francis Crick discovered frameshift mutations

• Systematically deleted nucleotides from DNA– Single and double deletions dramatically

altered protein product– Effects of triple deletions were minor– Conclusion: every triplet of nucleotides,

each codon, codes for exactly one amino acid in a protein

Codons

• In 1964, Charles Yanofsky and Sydney Brenner proved colinearity in the order of codons with respect to amino acids in proteins

• As a result, it was incorrectly assumed that the triplets encoding for amino acid sequences form contiguous strips of information.

Great Discovery Provoking Wrong Assumption

Exons and Introns• In eukaryotes, the gene is a combination of

coding segments (exons) that are interrupted by non-coding segments (introns)

• This makes computational gene prediction in eukaryotes even more difficult

• Prokaryotes don’t have introns - Genes in prokaryotes are continuous

Splicingexon1 exon2 exon3

intron1 intron2

transcription

translation

splicing

exon = codingintron = non-coding

Batzoglou

Gene prediction

• More difficult in eukaryotes than in prokaryotes (due to introns).

• In human genome, ~3% of DNA sequence is genes

• Lot of “junk” DNA between genes, and even inside genes (between exons).

• Gene prediction must deal with this.

Gene prediction: broadly speaking

• Statistical approaches:look for features than appear frequently in genes and infrequently elsewhere

• Similarity based approaches: a newly sequenced gene may be similar to a known gene.– even this is not so simple. The exon structures

may be different between otherwise similar genes

Statistical approaches

• Let us consider gene prediction in prokaryotes (no introns)

• Detect potential coding regions by looking at ORFs– A region of length n is comprised of (n/3) codons– Stop codons break genome into segments between

consecutive Stop codons– The subsegments of these that start from the Start codon

(ATG) are ORFs

Genomic Sequence

Open reading frame

ATG TGA

Open Reading Frames (ORFs)

ORFs

• 6 reading frames in any given sequence– 6 ways to map the DNA sequence to codon

sequence (+1,+2,+3,-1,-2,-3)– 3 on either strand

• Look at all 6 reading frames for ORFs

• Long open reading frames may be a gene– At random, we should expect one stop codon

every (64/3) ~= 21 codons– However, genes are usually much longer than

this• A basic approach is to scan for ORFs whose length

exceeds certain threshold– This is naïve because some genes (e.g. some

neural and immune system genes) are relatively short

Long vs.Short ORFs

Codon usage• In a given sequence (e.g., an ORF), compute

frequency distribution of codons (64 element array): codon usage array

• Codon usage array for coding sequences is different from that for non-coding sequences

• If the codon usage array for an ORF is much more similar to that of coding sequences than to that of non-coding sequences, the ORF could be a gene

Codon usage

• Codons coding for “Arg” in human:– CGU: 37%, CGC: 38%, CGA: 7%, CGG:

10%, AGA: 5%, AGG: 3%– In a coding sequence, codon CGC is 12

times more likely than codon AGG– An ORF preferring CGC over AGG is likely

to be a gene

Codon Usage in Human Genome

Codon usage• One way to test if an ORF is a gene is to

compute– Pr(ORF sequence under a coding sequence

model)– Pr(ORF sequence under a non-coding model)– Ratio of the two.

• These methods work best in prokaryotes• The exon-intron trouble is not handled yet

Promoter Structure in Prokaryotes (E.Coli)

Transcription starts at offset 0.

• Pribnow Box (-10)

• Gilbert Box (-30)

• Ribosomal Binding Site (+10)

Ribosomal Binding Site

Splicing Signals: an additional statistical clue, for eukaryotes

Exons are interspersed with introns and typically flanked by GT and AG

Splice site detection5’ 3’

Donor site

Position% -8 … -2 -1 0 1 2 … 17A 26 … 60 9 0 1 54 … 21C 26 … 15 5 0 1 2 … 27G 25 … 12 78 99 0 41 … 27T 23 … 13 8 1 98 3 … 25

From lectures by Serafim Batzoglou (Stanford)

Consensus splice sites

Statistical approaches: summary

• Codon usage

• Promoter motifs

• Ribosome binding site

• Splicing sites

Similarity based approaches

Similarity based approaches

• Some genomes may be very well-studied, with many genes having been experimentally verified.

• Closely-related organisms may have similar genes

• Unknown genes in one species may be compared to genes in some closely-related species

The basic approach• Given a protein sequence, and a genomic sequence,

find a set of substrings of the genomic sequence whose concatenation best fits the protein sequence

• Deals with the exon-intron problem

• First cut: Find fragments in the genomic sequence that match portions of the protein sequence (local alignment)

• Then find the “optimal” subset of non-overlapping fragments

Exon chaining

• Each of the fragments of the genomic sequence that somewhat match the protein (locally) is a putative exon

• The “goodness” of the match is the “weight” assigned to this putative exon

• Thus, we have a set of weighted intervals (l,r,w): for a fragment from l to r, with weight w representing how well it matches (a portion of) the protein

Exon Chaining Problem

• Input: A set of weighted intervals (l,r,w)• Output: A maximum weight chain of

non-overlapping intervals from this set

Exon Chaining Problem: Graph Representation

• This problem can be solved with dynamic programming in O(n) time.

21

edge from every li to ri

edge between every two successive vertices

Assumptions

• No two intervals have a common boundary point. So the (li,ri) define 2n distinct points, if there are n intervals

Exon Chaining AlgorithmExonChaining (G, n) //Graph, number of intervalsfor i ← to 2n si ← 0for i ← 1 to 2n if vertex vi in G corresponds to right end of the interval

I j ← index of vertex for left end of the interval I w ← weight of the interval I si ← max {sj + w, si-1}else si ← si-1

return s2n

Not very helpful

• A chain is a set of non-overlapping exons in order (left to right)

• But the matching protein portions may not be in the same order !

Spliced Alignment• Begins by selecting either all putative exons

between potential acceptor and donor sites or by finding all substrings similar to the target protein (as in the Exon Chaining Problem).

• This set is further filtered in a such a way that attempt to retain all true exons, with some false ones.

• Then find the chain of exons such that the sequence similarity to the target protein sequence is maximized

Spliced Alignment Problem: Formulation

• Input: Genomic sequences G, target sequence T, and a set of candidate exons (blocks) B.

• Output: A chain of exons Γ such that the global alignment score between Γ* and T is maximized

Γ* - concatenation of all exons from chain Γ

Dynamic programming

• Genomic sequence G = g1g2…gn

• Target sequence T = t1t2…tm

• As usual, we want to find the optimal alignment score of the i-prefix of G and the j-prefix of T

• Problem is, there are many i-prefixes possible (since multiple blocks may include position i)

Idea

• Find the optimal alignment score of the i-prefix of G and the j-prefix of T assuming that this alignment uses a particular block B at position i

• S(i, j, B) • For every block B that includes i

Recurrence If i is not the starting vertex of block B:• S(i, j, B) =

max { S(i – 1, j, B) – indel penalty S(i, j – 1, B) – indel penalty

S(i – 1, j – 1, B) + δ(gi, tj) }

If i is the starting vertex of block B:• S(i, j, B) =

max { S(i, j – 1, B) – indel penalty maxall blocks B’ preceding block B S(end(B’), j, B’) – indel penalty maxall blocks B’ preceding block B S(end(B’), j – 1, B’) + δ(gi, tj)}