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Information Theory of DNA Sequencing

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## Information Theory of DNA Sequencing

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**Information Theory of DNA Sequencing**David Tse Dept. of EECS U.C. Berkeley LIDS Student Conference MIT Feb. 2, 2012 Research supported by NSF Center for Science of Information. AbolfazlMotahari Guy Bresler TexPoint fonts used in EMF: AAAAAAAAAAAAAAAA**DNA sequencing**DNA: the blueprint of life Problem: to obtain the sequence of nucleotides. …ACGTGACTGAGGACCGTG CGACTGAGACTGACTGGGT CTAGCTAGACTACGTTTTA TATATATATACGTCGTCGT ACTGATGACTAGATTACAG ACTGATTTAGATACCTGAC TGATTTTAAAAAAATATT… courtesy: Batzoglou**Impetus: Human Genome Project**1990: Start 2001: Draft 3 billion basepairs 2003: Finished courtesy: Batzoglou**Sequencing Gets Cheaper and Faster**Cost of one human genome • HGP: $ 3 billion • 2004: $30,000,000 • 2008: $100,000 • 2010: $10,000 • 2011: $4,000 • 2012-13: $1,000 • ???: $300 courtesy: Batzoglou**But many genomes to sequence**100 million species (e.g. phylogeny) 7 billion individuals (SNP, personal genomics) 1013 cells in a human (e.g. somatic mutations such as HIV, cancer) courtesy: Batzoglou**Whole Genome Shotgun Sequencing**Reads are assembled to reconstruct the original DNA sequence.**Sequencing Technologies**• HGP era: single technology (Sanger) • Current: multiple “next generation” technologies (eg. Illumina, SoLiD, Pac Bio, Ion Torrent, etc.) • All provide massively parallel sequencing. • Each technology has different read lengths, noise profiles, etc**Assembly Algorithms**• Many proposed algorithms. • Different algorithms tailored to different technologies. • Each algorithm deals with the full complexity of the problem while trying to scale well with the massive amount of data. • Lots of heuristics used in the design.**A Basic Question**• What is the minimum number of reads needed to reconstruct with a given reliability? • A benchmark for comparing different algorithms. • An algorithm-independent basis for comparing different technologies and designing new ones.**Coverage Analysis**• Pioneered by Lander-Waterman • What is the minimum number of reads to ensure there is no gap between the reads with a desired prob.? • Only provides a lower bound. • Can one get a tight lower bound?**Communication and Sequencing: An Analogy**Communication: Sequencing:**Communication: Fundamental Limits**Given statistical models for source and channel: Shannon 48 Asymptotically reliable communication at rate R source symbols per channel output symbol if and only if:**DNA Sequencing: Fundamental Limits?**• Define: sequencing rate R = G/N basepairs per read • Question: can one define a sequencing capacity C such that: asymptotically reliable reconstruction is possible if and only if R < C?**A Simple Model**• DNA sequence: i.i.d. with distribution p. • Starting positions of reads are i.i.d. uniform on the DNA sequence. • Read process is noiseless. Will extend to more complex source model and noisy read process later.**The read channel**• Capacity depends on • read length: L • DNA length: G • Normalized read length: • Eg. L = 100, G = 3 £109 : AGGTCC AGCTTATAGGTCCGCATTACC read channel**Result: Sequencing Capacity**Renyi entropy of order 2**Coverage Constraint**G T L N reads Starting positions of reads ~ Poisson(1/R)**No-Duplication Constraint**L L L L The two possibilities have the same set of length L subsequences.**Achievability**no-duplication constraint coverage constraint achievable?**Greedy Algorithm**Input: the set of N reads of length L • Set the initial set of contigs as the reads. • Find two contigs with largest overlap and merge them into a new contig. • Repeat step 2 until only one contig remains or no more merging can be done. Algorithm progresses in stages: at stage merge reads at overlap**Greedy algorithm: the beginning**gap Most reads have large overlap with neighbors probability two disjoint reads are equal Expected # of errors in stage L-1: Very small since no-duplication constraint is satisfied.**Greedy algorithm: stage**Expected # of errors at stage probability two disjoint reads appear to overlap This may get larger, but no larger than when Very small since coverage constraint is satisfied.**Summary: Two Regimes**coverage-limited regime duplication-limited regime**Relation to Earlier Works**• Coverage constraint: Lander-Waterman 88 • No-duplication constraint: Arratia et al 96 • Arratia et al focused on a model where all length L subsequences are given (seq. by hybridization) • Our result: the two constraints together are necessary and sufficient for shotgun sequencing.**Rest of Talk**• Impact of read noise. • Impact of repeats in DNA sequence**Read Noise**A A Model: discrete memoryless channel defined by transition probabilities C T T T A T ACGTCCTATGCGTATGCGTAATGCCACATATTGCTATGCGTAATGCGT**Modified Greedy algorithm**X Y Y Do we merge the two reads at overlap ? We observe two strings: X and Y. (merge) Are they noisy versions of the same DNA subsequence? (do not merge) Or from two different locations? This is a hypothesis testing problem!**Impact on Sequencing Rate**H0: noisy versions of the same DNA subsequence (merge) H1: from disjoint DNA subsequences (do not merge) X Y • Hypothesis test: Y MAP rule: declare H0 if Two types of error: • false positive (same as before) • missed detection (new type of error)**Impact on Sequencing Rate**no-duplication constraint coverage constraint obtained by optimizing MAP threshold**More Complex DNA Statistics**• i.i.d. is not a very good model for the DNA sequence. • More generally, we may want to model it as a correlated random process. • For short-scale correlation, H2(p) can be replaced by the Renyi entropy rate of the process. • But for higher mammals, DNA contains long repeats, repeat length comparable or longer than reads. • This is handled by paired-end reads in practice.**A Simple Model for Repeats**K Model: M repeats of length K placed uniformly into DNA sequence If repeat length K>> read length L, how to reconstruct sequence? Use paired-end reads: J These reads can bridge the repeats reads come in pairs with known separation**Impact on Sequencing Rate**K= repeat length J = paired-end separation no-duplication constraint constant indepof K coverage of repeats constraint coverage constraint If J > 2d + K then capacity is the same as without repeats**Conclusion**• DNA sequencing is an important problem. • Many new technologies and new applications. • An analogy between sequencing and communication is drawn. • A notion of sequencing capacity is formulated. • A principled design framework?