CSCI2950-C DNA Sequencing and Fragment Assembly

Transcription

1 CSCI2950-C DNA Sequencing and Fragment Assembly Lecture 2: Sept. 7, DNA sequencing How we obtain the sequence of nucleotides of a species 5 3 ACGTGACTGAGGACCGTG CGACTGAGACTGACTGGGT CTAGCTAGACTACGTTTTA TATATATATACGTCGTCGT ACTGATGACTAGATTACAG ACTGATTTAGATACCTGAC TGATTTTAAAAAAATATT 3 5 1

2 Outline DNA Sequencing Technology: Old and New Fragment Assembly Problem Overlap-Layout-Consensus Sequencing by Hybridization Eulerian and Hamiltonian Graphs Goal: DNA Sequencing Find the complete sequence of A, C, G, T s in DNA molecule. Challenge: There is no machine that takes long DNA as an input, and gives the complete sequence as output Can only sequence: letters at a time (Sanger sequencing) letters (Next-gen sequencing) 2

3 How to Sequence DNA? Co-opt machinery of the cell. Every time a cell divides it copies its DNA sequence. DNA Replication One strand used as template to make a copy of other strand. $!! $! $ " # $! # " # "! $ " # $! " # $!! # " # " # " # "!! $! $ " " # " #! $ # " " # " # # $ $! " # " $! " # # " $! 3

4 DNA Replication DNA Replication Suppose we could: 1) Stop this reaction and measure the last nucleotide added. 2) Measure the length of DNA molecules. 4

5 DNA Sequencing 1. Start at fixed location (primer) 2. Grow DNA chain 3. Include mixture of normal nucleotides (A,C,T,G) and modified (fluorescent/ radioactive) nucleotides. 4. Modified nucleotides stop reaction at all possible points 5. Separate products with length, using gel electrophoresis Sanger sequencing (F. Sanger Nobel prize 1980) Technical Limitations 1. Need a lot of DNA 2. Reaction only works for bp 1. Biology 2. Computer Science Solutions 5

6 DNA Sequencing Cloning Many host cells DNA amplification Different types of vectors VECTOR Plasmid (bacteria) Size of insert 2,000-10,000 Can control the size Fosmid (bacteria) 40,000 BAC (Bacterial Artificial Chromosome) YAC (Yeast Artificial Chromosome) 70, ,000 > 300,000 Not used much recently 6

7 Disadvantages of Traditional (Sanger) Sequencing 1. Cloning process is laborious. Grow (bacterial) cells, each containing a single fragment. 2. Only one fragment sequenced at a time. Next-generation DNA sequencing 1. Massively parallel sequencing reactions. 2. In situ amplification of DNA. No cloning required. 7

8 1. PREPARE GENOMIC DNA SAMPLE 2. ATTACH DNA TO SURFACE 3. BRIDGE AMPLIFICATION Illumina Sequencing DNA Adapters Adapter Adapter DNA fragment Dense lawn of primers Randomly fragment genomic DNA and ligate adapters to both ends of the fragments. Bind single-stranded fragments randomly to the inside surface of the flow cell channels. Add unlabeled nucleotides and enzyme to initiate solid-phase bridge amplification. 4. FRAGMENTS BECOME DOUBLE-STRANDED 5. DENATURE THE DOUBLE- STRANDED MOLECULES 6. COMPLETE AMPLIFICATION Attached Free terminus terminus Attached terminus Attached Attached Clusters The enzyme incorporates nucleotides to build double-stranded bridges on the solid-phase substrate. Denaturation leaves single-stranded templates anchored to the substrate. Several million dense clusters of double-stranded DNA are generated in each channel of the flow cell. Illumina Sequencing 7. DETERMINE FIRST BASE 8. IMAGE FIRST BASE 9. DETERMINE SECOND BASE Laser Laser The first sequencing cycle begins by adding four labeled reversible terminators, primers, and DNA polymerase. After laser excitation, the emitted fluorescence from each cluster is captured and the first base is identified. The next cycle repeats the incorporation of four labeled reversible terminators, primers, and DNA polymerase. 10. I MAGE SECOND CHEMISTRY CYCLE 11. SEQUENCING OVER MUL- TIPLE CHEMISTRY CYCLES 12. A LIGN DATA GCTGA... After laser excitation, the image is captured as before, and the identity of the second base is recorded. The sequencing cycles are repeated to determine the sequence of bases in a fragment, one base at a time. The data are aligned and compared to a reference, and sequencing differences are identified. 8

9 Goal: DNA Sequencing Find the complete sequence of A, C, G, T s in a DNA molecule. Challenge: There is no machine that takes long DNA as an input, and gives the complete sequence as output Can only sequence: letters at a time (Sanger sequencing) letters (Next-gen sequencing) Issues 1. How to process data from sequencing machine? Base-calling ACTCGT 2. How to obtain sequence for longer DNA molecules? De novo assembly. 3. How to find differences from a reference genome: resequencing. 9

10 Sequence Trace Challenging to read answer 10

11 Challenging to read answer Challenging to read answer 11

12 Issues 1. How to process data from sequencing machine? Base-calling ACTCGT 2. How to obtain sequence for longer DNA molecules? De novo assembly. 3. How to find differences from a reference genome: resequencing. Sequencing Longer Regions 12

13 Fragment Assembly Fragment Assembly Computational Challenge: assemble individual short sequences (reads) into a single genomic sequence ( superstring ). Objective function? 13

14 Shortest Common Superstring Problem (SCS) Problem: Given a set of strings, find a shortest string that contains all of them Input: Strings s 1, s 2,., s n Output: A string s that Contains all strings s 1, s 2,., s n as substrings Has minimum length. Note: this formulation does not take into account errors in strings s i Shortest Common Superstring Problem: Example 14

15 Greedy Algorithm for SCS 1. Find pair of strings s i, s j with longest overlap. 2. Merge(s i, s j ) 3. Recurse How good is the greedy algorithm? Algorithm Analysis S = {s 1, s 2,., s n } Opt(S ) = length of SCS Claim: Greedy solution 4 Opt(S ). Conjecture: Greedy solution 2 Opt(S ). [Best known bound 2.5] Theorem: SCS is NP complete 15

16 SCS and Overlap Graph Build directed graph G = (V,E) V = {s 1, s 2,., s n } e = (s i, s j ) if prefix of s j matches suffix of s i w(s i, s j ) = length of overlap b/w s i, s j Goal: Find a maximum weight path visiting every VERTEX exactly once in the OVERLAP graph: Travelling Salesman (Hamiltonian path) problem Convert max to min by w -w SCS to TSP: An Example S = { ATC, CCA, CAG, TCC, AGT } SCS AGT CCA ATC ATCCAGT TCC CAG TSP 0 AGT 1 2 CAG ATC TCC CCA ATCCAGT 16

17 Problems 1. What if there is not a solution to SCS for real data? No superstring due to missing overlaps. 2. Is shortest string the correct objective function? Repeated sequences common in most genomes. Questions 1. How many reads to sequence? 2. How to assemble in presence of missing data and/or repeated sequences? 17

18 Definition of Coverage 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: Assuming uniform distribution of reads, C=10 results in 1 gapped region /1,000,000 nucleotides Lander-Waterman Statistics Given: N reads of length L from a genome of size G. Assume left end of reads uniformly distributed on genome. P(read R starts at ζ) = 1/G P(ζ covered by read R) = L/G ζ 18

19 Lander-Waterman Statistics Given: N reads of length L from a genome of size G P(ζ covered by read) = 1 (1 L/G) N 1 e -c, where c = N L / G is coverage P(ζ covered by read) 1 e -c Poisson approximation Rescale g = G/L, l = L/L. Left ends of reads are Poisson process with rate c = N L / G. Pr[ k reads in interval of length t] = e -ct (ct) k / k! c P(covered) Questions 1. How many reads to sequence? 2. How to assemble in presence of missing data and/or repeated sequences? 19

20 Triazzle: A Fun Example Repeats make it very difficult! Try it (even on your IPhone): Repeats and Fragment Assembly Repeats a major problem for fragment assembly >50% of human (mammalian) genomes are repeats. ~5% for most bacterial genomes. 20

21 Repeat Types in Human Low-Complexity DNA (e.g. ATATATATACATA ) Microsatellite repeats (a 1 a k ) N where k ~ 3-6 (e.g. CAGCAGTAGCAGCACCAG) Transposons SINE (Short Interspersed Nuclear Elements) e.g., ALU: ~300-long, 10 6 copies LINE (Long Interspersed Nuclear Elements) ~4000-long, 200,000 copies LTR retroposons (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 Repeats and Fragment Assembly Repeats: A major problem for fragment assembly Repeat Repeat Repeat Green and yellow fragments are interchangeable when assembling repetitive DNA 21

22 Questions 1. How many reads to sequence? 2. How to assemble in presence of missing data and/or repeated sequences? Dominant paradigms 1. Overlap-Layout-Consensus 2. debruijn graphs Overlap-Layout-Consensus Assemblers: ARACHNE, PHRAP, CAP, TIGR, CELERA Overlap: find potentially overlapping reads Layout: Merge reads into contigs and contigs into supercontigs Consensus: derive the DNA sequence and correct read errors 22

23 Overlap Find best match between suffix of one read and the prefix of another r suffix(r) prefix(r ) r' Due to sequencing errors, need to use dynamic programming to find the optimal overlap alignment Given 2 DNA sequences v and w: v : w : ATGTTAT ATCGTAC m = 7 n = 7 Alignment : 2 * k matrix ( k > m, n ) letters of v letters of w A T -- G T T A T A T C G T -- A C 5 matches 1 insertions 1 deletions 1 mismatch 23

24 Edit Distance Levenshtein (1966) introduced edit distance between two strings as the minimum number of elementary operations (insertions, deletions, and substitutions) to transform one string into the other d(v,w) = MIN number of elementary operations to transform v w Computing Edit Distance Let v i = prefix of v of length i: v 1 v i w j = prefix of w of length j: w 1 w j Let a i,j = edit distance between v i and w j a i, j = min a i-1, j + 1 a i, j a i-1, j if v i w j Insert Delete Mismatch a i-1, j-1 if v i = w j Match Can replace +1 with different penalties/scores for different types of changes 24

25 Computing Alignment i-1,j -1 i-1,j 0 or 1 1 i,j -1 1 i,j a i, j = min a i-1, j + 1 a i, j Insert Delete a i-1, j if v i w j Mismatch a i-1, j-1 if v i = w j Match Every Path in the Grid Corresponds to an Alignment W A T C G V A T G V = A T - G T W= A T C G T 4 25

26 Edit Graph Weighted graph G with two distinct vertices, one labeled source one labeled sink Every alignment is path from source to sink. Alignment as a Path in the Edit Graph - Corresponding path - 26

27 Alignments in Edit Graph (cont d) and represent indels in v and w. represent matches or mismatches. Every path in the edit graph corresponds to an alignment: 27

28 Alignment Runtime O(nm) time to fill in the mxn dynamic programming matrix. Overlap Goal: find best match between the suffix of one read and the prefix of another Thus far: global alignment. How to fix? 28

29 Overlap Goal: find best match between the suffix of one read and the prefix of another Finding optimal overlap alignment for all pairs of N reads of length L: O(N 2 L 2 ) Improve efficiency: Apply a filtration method to filter out pairs of fragments that do not share a significantly long common substring Overlapping Reads Identify all k-mers in reads (k ~ 24) TACA TAGATTACACAGATTAC T GA TAGT TAGATTACACAGATTAC TAGA 29

30 Sources Genomes sequenced: Serafim Batzoglou CS262_2006/ (Sequencing slides) 30