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Heuristic Local Aligners

Heuristic Local Aligners. The basic indexing & extension technique Indexing: techniques to improve sensitivity Pairs of Words, Patterns Systems for local alignment. Indexing-based local alignment. ……. Dictionary: All words of length k (~10)

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Heuristic Local Aligners

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  1. Heuristic Local Aligners • The basic indexing & extension technique • Indexing: techniques to improve sensitivity • Pairs of Words, Patterns • Systems for local alignment

  2. Indexing-based local alignment …… Dictionary: All words of length k (~10) Alignment initiated between words of alignment score  T (typically T = k) Alignment: Ungapped extensions until score below statistical threshold Output: All local alignments with score > statistical threshold query …… scan DB query

  3. Indexing-based local alignment—Extensions A C G A A G T A A G G T C C A G T Gapped extensions until threshold • Extensions with gaps until score < C below best score so far Output: GTAAGGTCCAGT GTTAGGTC-AGT C T G A T C C T G G A T T G C G A

  4. Sensitivity-Speed Tradeoff X% Sens. Speed Kent WJ, Genome Research 2002

  5. Sensitivity-Speed Tradeoff Methods to improve sensitivity/speed • Using pairs of words • Using inexact words • Patterns—non consecutive positions ……ATAACGGACGACTGATTACACTGATTCTTAC…… ……GGCACGGACCAGTGACTACTCTGATTCCCAG…… ……ATAACGGACGACTGATTACACTGATTCTTAC…… ……GGCGCCGACGAGTGATTACACAGATTGCCAG…… TTTGATTACACAGAT T G TT CAC G

  6. Measured improvement Kent WJ, Genome Research 2002

  7. Non-consecutive words—Patterns Patterns increase the likelihood of at least one match within a long conserved region Consecutive Positions Non-Consecutive Positions 6 common 5 common 7 common 3 common On a 100-long 70% conserved region: ConsecutiveNon-consecutive Expected # hits: 1.07 0.97 Prob[at least one hit]: 0.30 0.47

  8. Advantage of Patterns 11 positions 11 positions 10 positions

  9. Multiple patterns TTTGATTACACAGAT T G TT CAC G T G T C CAG TTGATT A G • K patterns • Takes K times longer to scan • Patterns can complement one another • Computational problem: • Given: a model (prob distribution) for homology between two regions • Find: best set of K patterns that maximizes Prob(at least one match) How long does it take to search the query? Buhler et al. RECOMB 2003 Sun & Buhler RECOMB 2004

  10. Variants of BLAST • NCBI BLAST: search the universe http://www.ncbi.nlm.nih.gov/BLAST/ • MEGABLAST: http://genopole.toulouse.inra.fr/blast/megablast.html • Optimized to align very similar sequences • Works best when k = 4i  16 • Linear gap penalty • WU-BLAST: (Wash U BLAST) http://blast.wustl.edu/ • Very good optimizations • Good set of features & command line arguments • BLAT http://genome.ucsc.edu/cgi-bin/hgBlat • Faster, less sensitive than BLAST • Good for aligning huge numbers of queries • CHAOS http://www.cs.berkeley.edu/~brudno/chaos • Uses inexact k-mers, sensitive • PatternHunter http://www.bioinformaticssolutions.com/products/ph/index.php • Uses patterns instead of k-mers • BlastZ http://www.psc.edu/general/software/packages/blastz/ • Uses patterns, good for finding genes • Typhon http://typhon.stanford.edu • Uses multiple alignments to improve sensitivity/speed tradeoff

  11. DNA sequencing How we obtain the sequence of nucleotides of a species …ACGTGACTGAGGACCGTG CGACTGAGACTGACTGGGT CTAGCTAGACTACGTTTTA TATATATATACGTCGTCGT ACTGATGACTAGATTACAG ACTGATTTAGATACCTGAC TGATTTTAAAAAAATATT…

  12. Which representative of the species? Which human? Answer one: Answer two: it doesn’t matter Polymorphism rate: number of letter changes between two different members of a species Humans: ~1/1,000 Other organisms have much higher polymorphism rates • Population size!

  13. Why humans are so similar Out of Africa N A small population that interbred reduced the genetic variation Out of Africa ~ 40,000 years ago Heterozygosity: H H = 4Nu/(1 + 4Nu) u ~ 10-8, N ~ 104  H ~ 410-4

  14. DNA Sequencing Goal: Find the complete sequence of A, C, G, T’s in DNA Challenge: There is no machine that takes long DNA as an input, and gives the complete sequence as output Can only sequence ~150 letters at a time

  15. Method to sequence longer regions genomic segment cut many times at random (Shotgun) Get one or two reads from each segment ~100 bp ~100 bp

  16. Definition of Coverage C Length of genomic segment: G Number of reads: N Length of each read: L Definition:Coverage C = N L / G How much coverage is enough? Lander-Waterman model: Prob[ not covered bp ] = e-C Assuming uniform distribution of reads, C=10 results in 1 gapped region /1,000,000 nucleotides

  17. Repeats Bacterial genomes: 5% Mammals: 50% Repeat types: • Low-Complexity DNA (e.g. ATATATATACATA…) • Microsatellite repeats (a1…ak)N where k ~ 3-6 (e.g. CAGCAGTAGCAGCACCAG) • Transposons • SINE(Short Interspersed Nuclear Elements) e.g., ALU: ~300-long, 106 copies • LINE(Long Interspersed Nuclear Elements) ~4000-long, 200,000 copies • LTRretroposons(Long Terminal Repeats (~700 bp) at each end) cousins of HIV • Gene Families genes duplicate & then diverge (paralogs) • Recent duplications ~100,000-long, very similar copies

  18. Two main assembly problems • De Novo Assembly • Resequencing

  19. Human Genome Variation TGCTGAGA TGCCGAGA TGCTCGGAGA TGC - - - GAGA SNP Novel Sequence Mobile Element or Pseudogene Insertion Inversion Translocation Tandem Duplication TGC - - AGA TGCCGAGA Microdeletion Transposition TGC Novel Sequence at Breakpoint Large Deletion

  20. Read Mapping • Want ultra fast, highly similar alignment • Detection of genomic variation CATCGACCGAGCGCGATGCTAGCTAGGTGATCGT...... TGCCGCATCGACCGAGCGCGATGCTAGCTAGGTGATCGT... GCATGCCGCATCGACCGAGCGCGATGCTAGCTAGGTGATCGT GTGCATGCCGCATCGACCGAGCGCGATGCTAGCTAGGTGATC ......AGGTGCATGCCGCATCGATCGAGCGCGATGCTAGCTAGCTGATCGT......

  21. Read Mapping • Modern fast read aligners: BWA, Bowtie, SOAP • Based on Burrows-Wheeler transform CATCGACCGAGCGCGATGCTAGCTAGGTGATCGT...... TGCCGCATCGACCGAGCGCGATGCTAGCTAGGTGATCGT... GCATGCCGCATCGACCGAGCGCGATGCTAGCTAGGTGATCGT GTGCATGCCGCATCGACCGAGCGCGATGCTAGCTAGGTGATC ......AGGTGCATGCCGCATCGATCGAGCGCGATGCTAGCTAGCTGATCGT......

  22. Burrows-Wheeler Transform ANA BANANA ANANA NANA ANA NA A BANANA ANANA NANA ANA NA A X = BANANA suffixes of BANANA

  23. Burrows-Wheeler Transform ANA BANANA$ ANANA$ NANA$ ANA$ NA$ A$ $ BANANA$ ANANA$ NANA$ ANA$ NA$ A$ $ X = BANANA$

  24. Burrows-Wheeler Transform ANA BANANA$ ANANA$B NANA$BA ANA$BAN NA$BANA A$BANAN $BANANA BANANA$ ANANA$B NANA$BA ANA$BAN NA$BANA A$BANAN $BANANA BANANA$ ANANA$B NANA$BA ANA$BAN NA$BANA A$BANAN $BANANA X = BANANA$

  25. Burrows-Wheeler Transform ANA BANANA$ ANANA$B NANA$BA ANA$BAN NA$BANA A$BANAN $BANANA BANANA$ ANANA$B NANA$BA ANA$BAN NA$BANA A$BANAN $BANANA $BANANA A$BANAN ANA$BAN ANANA$B BANANA$ NA$BANA NANA$BA X = BANANA$

  26. Burrows-Wheeler Transform ANA BANANA$ ANANA$B NANA$BA ANA$BAN NA$BANA A$BANAN $BANANA BANANA$ ANANA$B NANA$BA ANA$BAN NA$BANA A$BANAN $BANANA $BANANA A$BANAN ANA$BAN ANANA$B BANANA$ NA$BANA NANA$BA X = BANANA$

  27. Burrows-Wheeler Transform ANA BANANA$ ANANA$B NANA$BA ANA$BAN NA$BANA A$BANAN BANANA$ ANANA$B NANA$BA ANA$BAN NA$BANA A$BANAN $BANANA A$BANAN ANA$BAN ANANA$B BANANA$ NA$BANA NANA$BA X = BANANA$ BWT matrix of string ‘BANANA’ BWT(BANANA) = ANNB$AA

  28. Suffix Arrays $BANANA A$BANAN ANA$BAN ANANA$B BANANA$ NA$BANA NANA$BA 1 $BANANA 2 A$BANAN 3 ANA$BAN 4 ANANA$B 5 BANANA$ 6 NA$BANA 7 NANA$BA Suffixes are sorted in the BWT matrix S(i) = j, where Xj…Xn is the i-th suffix lexicographically X BWT(X) constructed from S: At each position, take the letter to the left of the one pointed by S S BWT(X)

  29. Reconstructing BANANA A$ NA NA BA $B AN AN $B A$ AN AN BA NA NA A$B NA$ NAN BAN $BA ANA ANA $BANANA A$BANAN ANA$BAN ANANA$B BANANA$ NA$BANA NANA$BA A N N B $ A A $ A A A B N N $BA A$B ANA ANA BAN NA$ NAN BWT matrix of string ‘BANANA’ append BWT append BWT sort sort sort

  30. Reconstructing BANANA - faster Lemma. The i-th occurrence of character c in last column is the same text character as the i-th occurrence of c in the first column $BANANA A$BANAN ANA$BAN ANANA$B BANANA$ NA$BANA NANA$BA $BANANA A$BANAN ANA$BAN ANANA$B BANANA$ NA$BANA NANA$BA BWT matrix of string ‘BANANA’

  31. Reconstructing BANANA - faster Lemma. The i-th occurrence of character c in last column is the same text character as the i-th occurrence of c in the first column $BANANA A$BANAN ANA$BAN ANANA$B BANANA$ NA$BANA NANA$BA A N N B $ A A $BANAN A$BANA ANA$BA ANANA$ BANANA NA$BAN NANA$B BWT matrix of string ‘BANANA’

  32. Reconstructing BANANA - faster Lemma. The i-th occurrence of character c in last column is the same text character as the i-th occurrence of c in the first column $BANANA A$BANAN ANA$BAN ANANA$B BANANA$ NA$BANA NANA$BA A N N B $ A A $BANAN A$BANA ANA$BA ANANA$ BANANA NA$BAN NANA$B A$BANAN ANA$BAN ANANA$B Same words, same sorted order BWT matrix of string ‘BANANA’

  33. Reconstructing BANANA - faster Lemma. The i-th occurrence of character ‘a’ in last column is the same text character as the i-th occurrence of ‘a’ in the first column LF(): Map the i-th occurrence of character ‘a’ in last column to the first column LF(r): Let row r contain the i-th occurrence of ‘a’ in last column Then, LF(r) = r’; r’: i-th row starting with ‘a’ $BANANA A$BANAN ANA$BAN ANANA$B BANANA$ NA$BANA NANA$BA BWT matrix of string ‘BANANA’

  34. Reconstructing BANANA - faster LF(r): Let row r be the i-th occurrence of ‘a’ in last column Then, LF(r) = r’; r’: i-th row starting with ‘a’ $BANANA A$BANAN ANA$BAN ANANA$B BANANA$ NA$BANA NANA$BA $BANANA A$BANAN ANA$BAN ANANA$B BANANA$ NA$BANA NANA$BA $BANANA A$BANAN ANA$BAN ANANA$B BANANA$ NA$BANA NANA$BA BWT matrix of string ‘BANANA’ LF[] = [2, 6, 7, 5, 1, 3, 4] Row LF(r) is obtained by rotating row r one position to the right

  35. Reconstructing BANANA - faster $BANANA A$BANAN ANA$BAN ANANA$B BANANA$ NA$BANA NANA$BA $BANANA A$BANAN ANA$BAN ANANA$B BANANA$ NA$BANA NANA$BA $BANANA A$BANAN ANA$BAN ANANA$B BANANA$ NA$BANA NANA$BA LF[] = [2, 6, 7, 5, 1, 3, 4] • Computing LF() is easy: • Let C(a): # of characters smaller than ‘a’ • Example: C($) = 0; C(A) = 1; C(B) = 4; C(N) = 5 • Let row r end with the i-th occurrence of ‘a’ in last column • Then, LF(r) = C(a) + i(why?) BWT matrix of string ‘BANANA’

  36. Reconstructing BANANA - faster A N N B $ A A $BANANA A$BANAN ANA$BAN ANANA$B BANANA$ NA$BANA NANA$BA C() copied for convenience C() 1 554011 indicating this is i-th occurrence of ‘c’ index i 1 1 2 1 1 2 3 LF() 2 6 7 5 1 3 4 LF() = C() + i Reconstruct BANANA: S := “”; r := 1; c := BWT[r]; UNTIL c = ‘$’ { S := cS; r := LF(r); c := BWT(r); } BWT matrix of string ‘BANANA’ Credit: Ben Langmead thesis

  37. Searching for ANA L(W): lowest index in BWT matrix where W is prefix U(W): highest index in BWT matrix where W is prefix Example: L(“NA”) = 6 U(“NA”) = 7 Lemma (prove as exercise) L(aW) = C(a) + i +1, where i = # ‘a’s up to L(W) – 1 in BWT(X) U(aW) = C(a) + j, where j = # ‘a’s up to U(W) in BWT(X) Example: L(“ANA”) = C(‘A’) + # ‘A’s up to (L(“NA”) – 1) + 1 = 1 + (# ‘A’s up to 5) + 1 = 1 + 1 + 1 = 3 U(“ANA”) = 1 + # ‘A’s up to U(“NA”) = 1 + 3 = 4 $BANANA A$BANAN ANA$BAN ANANA$B BANANA$ NA$BANA NANA$BA BWT matrix of string ‘BANANA’

  38. Searching for ANA Let LFC(r, a) = C(a) + i, where i = #’a’s up to r in BWT ExactMatch(W[1…k]) { a := W[k]; low := C(a) +1; high := C(a+1); // a+1: lexicographically next char i := k – 1; while (low <= high && i >= 1) { a = W[i]; low = LFC(low – 1, a) + 1; high = LFC(high, a); i := i – 1; } return (low, high); } $BANANA A$BANAN ANA$BAN ANANA$B BANANA$ NA$BANA NANA$BA BWT matrix of string ‘BANANA’ Credit: Ben Langmead thesis

  39. Summary of BWT algorithm Suffix array of string X: S(i) = j, where Xj…Xn is the j-thsuffix lexicographically • BWT follows immediately from suffix array • Suffix array construction possible in O(n), many good O(n log n) algorithms • Reconstruct X from BWT(X) in time O(n) • Search for all exact occurrences of W in time O(|W|) • BWT(X) is easier to compress than X

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