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CSE182-L16

CSE182-L16. LW statistics/Assembly. Silly Quiz. Who are these people, and what is the occasion?. Genome Sequencing and Assembly. Sequencing. A break at T is shown here. Measuring the lengths using electrophoresis allows us to get the position of each T

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CSE182-L16

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  1. CSE182-L16 LW statistics/Assembly

  2. Silly Quiz • Who are these people, and what is the occasion?

  3. Genome Sequencing and Assembly

  4. Sequencing • A break at T is shown here. • Measuring the lengths using electrophoresis allows us to get the position of each T • The same can be done with every nucleotide. Fluorescent labeling can help separate different nucleotides Bafna

  5. Automated detectors ‘read’ the terminating bases. • The signal decays after 1000 bases. Bafna

  6. Sequencing Genomes: Clone by Clone • Clones are constructed to span the entire length of the genome. • These clones are ordered and oriented correctly (Mapping) • Each clone is sequenced individually Bafna

  7. Shotgun sequencing of clones was considered viable However, researchers in 1999 proposed shotgunning the entire genome. Shotgun Sequencing Bafna

  8. Create vectors of the sequence and introduce them into bacteria. As bacteria multiply you will have many copies of the same clone. Library Bafna

  9. Whole Genome Shotgun • Break up the entire genome into pieces • Sequence ends, and assemble using a computer • LW statistics & Repeats argue against the success of such an approach Alternative: build a roadmap of the genome, with physical clones mapped for each region. Sequence each of the clones, and put them together Bafna

  10. Whole Genome Shotgun • Break up the entire genome into pieces • Sequence ends, and assemble using a computer • LW statistics & Repeats argue against the success of such an approach Alternative: build a roadmap of the genome, with physical clones mapped for each region. Sequence each of the clones, and put them together

  11. Shotgun sequencing of clones was considered viable for small genomes However, researchers in 1999 proposed shotgunning the entire genome. Shotgun Sequencing

  12. Massively parallel sequencing • Sanger sequencing allows us to sequence <=1000 bp in one lane, up to 96 lanes, in one run. • Today, we can sequence many Mbp in a single run

  13. Algorithmic: How do you put the genome back together from the pieces? Statistical? How many pieces do you need to sequence, etc.? The answer to the statistical questions had already been given in the context of mapping, by Lander and Waterman. Questions

  14. Lander Waterman Statistics • The fragments are falling randomly on the genome • Overlapping fragments form islands of contiguous sequence. • Ideally, we want one island for each chromosome. How many fragments should we sequence? L G

  15. Lander Waterman Statistics L G

  16. LW statistics: questions • As the coverage c increases, more and more areas of the genome are likely to be covered. Ideally, you want to see 1 island of overlapping contigs. • Q1: What is the expected number of islands? • The number should increase at first, and gradually decrease.

  17. Analysis: Expected Number Islands • Computing Expected # islands. • Let Xi=1 if an island ends at position i, Xi=0 otherwise. • Number of islands = ∑i Xi • Expected # islands = E(∑i Xi) = ∑i E(Xi)

  18. Prob. of an island ending at i L i T • E(Xi) = Prob (Island ends at pos. i) • =Prob(clone began at position i-L+1 AND no clone began in the next L-T positions)

  19. Computing # islands • As the coverage c increases, more and more areas of the genome are likely to be covered. Ideally, you want to see 1 island. • Q1: What is the expected number of islands? • Ans: N exp(-c) • The number increases at first, and gradually decreases.

  20. Expected # of clones in an island • Expected # of clones in an island = Q: How? Why do we care? Often, at the beginning of a genome project, we do not know the length of the genome. This equation helps us determine the length.

  21. Problem 1: size of contigs • Islands might simply be too small in length •  = (1-T/L) = (1-50/500) = 0.9, c = 8. • #Islands = N e-c = 36K • Size of an island = 54K • Not enough to make it an acceptable assembly! • PLUS, there is the problem of Repeats, Chimerism etc.

  22. Assembly Basics • Three main components: • Overlap • Layout • Consensus

  23. Yes, if a prefix of s2 matches a suffix of s1 Overlap • Given a pair of fragments s1 and s2, do they belong together? • How would you compute such a match?

  24. Overlap • S[i,j] = optimum score of an alignment of s1[1..i] against a substring of that starts anywhere, but ends in j. s2[*..j] j i • The best prefix-suffix alignment is given by: • Maxi {S[i,n] }

  25. Overlap Detection • Compute the best prefix-suffix alignments between each pair of fragments. • Keep the “high-scoring” ones as evidence of true overlap. • What is the problem?

  26. Overlap detection problem • Consider the number of fragments. The LW statistics say that we need good coverage (c=8, 10) to get most of the base-pairs. • G = 3000Mb, L=500 • Coverage LN/G = 10 • N = 10*3*109/500 = 6*107 • Number of comparisons needed = 3.6 * 1015 • Number of alignments per minute=6 • Number of compute nodes = 100 • Time needed (Number of years) = • Not good! (Only a small fraction are true overlaps)

  27. k-mer based overlap (Piegeonhole principle again) • Consider a 25bp sequence. • Expected number of occurrences in the genome • 3*109*4-25 = 2*10-6 • A 25-bp sequence appears is unique to the genome! • Two overlapping sequences should share a 25-mer • Two non-overlapping sequences should not! 25bp

  28. Sorting k-mers • Build a list of k-mers that appear in the sequences and their reverse complements • Create a record with 4 entries: • K-mer • Sequence number • Position in the sequence • Reverse complementation flag • Sort a vector of these according to k-mer • How many records per k-mer are expected? • If number of records exceeds threshold, discard (why?) K-mer S.id Pos.

  29. Alignment module • Coalesce k-mer hits into longer, gap-free partial alignments. • These extended k-mer hits are saved. • For each pair of sequences, form a directed graph. • For each maximal path in the graph, construct an alignment. • Refine alignment via banded DP

  30. Problem2: Size • Islands might simply be too small in length •  = (1-T/L) = (1-50/500) = 0.9, c = 8. • #Islands = N e-c = 36K • Size of an island = 54K • Not enough to make it an acceptable assembly! • PLUS, there is the problem of Repeats, Chimerism etc.

  31. Recall that we sequence about 1000bp of the end of a clone If we sequenced both ends, we get extra information, particularly if we know the length of the original clone. Solution 2: Clones can have mate-pairs

  32. Mate-pairs allow you to merge islands (contigs) into super-contigs Mate Pairs

  33. Make clones of truly predictable length. EX: 3 sets can be used: 2Kb, 10Kb and 50Kb. The variance in these lengths should be small. Use the mate-pairs to order and orient the contigs, and make super-contigs. Super-contigs are quite large

  34. Problem 3: Repeats

  35. 40-50% of the human genome is made up of repetitive elements. Repeats can cause great problems in the assembly! Chimerism causes a clone to be from two different parts of the genome. Can again give a completely wrong assembly Repeats & Chimerisms

  36. Repeat detection • Lander Waterman strikes again! • The expected number of clones in a Repeat containing island is MUCH larger than in a non-repeat containing island (contig). • Thus, every contig can be marked as Unique, or non-unique. In the first step, throw away the non-unique islands. Repeat

  37. Detecting Repeat Contigs 1: Read Density • Compute the log-odds ratio of two hypotheses: • H1: The contig is from a unique region of the genome. • The contig is from a region that is repeated at least twice

  38. Detecting Chimeric reads • Chimeric reads: Reads that contain sequence from two genomic locations. • Good overlaps: G(a,b) if a,b overlap with a high score • Transitive overlap: T(a,c) if G(a,b), and G(b,c) • Find a point x across which only transitive overlaps occur. X is a point of chimerism

  39. Whole genome shotgun • Input: • Shotgun sequence fragments (reads) • Mate pairs • Output: • A single sequence created by consensus of overlapping reads • First generation of assemblers did not include mate-pairs (Phrap, CAP..) • Second generation: CA, Arachne, Euler

  40. Assembly • Use k-mers to detect potential overlaps • Use alignments to build contig graphs • Decide the unique contigs based on LW statistics • Discard repeat contigs • Break chimeric contigs • Use mate-pairs to build scaffolds • Fill gaps

  41. Assembly • Use k-mers to detect potential overlaps • Use alignments to build contig graphs • Decide the unique contigs based on LW statistics • Discard repeat contigs • Break chimeric contigs • Use mate-pairs to build scaffolds

  42. Consensus Derivation • Consensus sequence is created by converting pairwise read alignments into multiple-read alignments

  43. Summary • Whole genome shotgun is now routine: • Human, Mouse, Rat, Dog, Chimpanzee.. • Many Prokaryotes (One can be sequenced in a day) • Plant genomes: Arabidopsis, Rice • Model organisms: Worm, Fly, Yeast • A lot is not known about genome structure, organization and function. • Comparative genomics offers low hanging fruit

  44. Final exam syllabus • Take home • The entire course, but emphasis will be given to post-midterm lectures • HMMs, • Gene-finding, • mass spectrometry, • Micro-array analysis, • genome sequencing and assembly

  45. What we did not cover

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