1. CSCI2950-C Lecture 2 DNA Sequencing and Fragment Assembly

2. DNA sequencing How we obtain the sequence of nucleotides of a species5’ 3’
…ACGTGACTGAGGACCGTG
CGACTGAGACTGACTGGGT
CTAGCTAGACTACGTTTTA
TATATATATACGTCGTCGT
ACTGATGACTAGATTACAG
ACTGATTTAGATACCTGAC
TGATTTTAAAAAAATATT… 3’ 5’

3. Outline DNA Sequencing Technology Fragment Assembly Problem
Overlap-Layout-Consensus
Sequencing by Hybridization
Eulerian and Hamiltonian Graphs
Next-generation DNA Sequencing

4. DNA Replication

5. DNA Sequencing – gel electrophoresis
Start at fixed location (primer)
Grow DNA chain
Include dideoxynucleoside (modified a, c, g, t)
Stops reaction at all possible points
Separate products with length, using gel electrophoresis

6. Technical Limitations
Need a lot of DNA
Reaction only works for 500bp
Solutions
Biology
Computer Science

7. DNA Sequencing – Cloning
Shake
DNA fragments
Known
location
(restriction
site)
Vector
Circular genome
(bacterium, plasmid) + =
Many host cells  DNA amplification

8. Different types of vectors
Size of insert
Plasmid
2,000-10,000
Can control the size
Cosmid
40,000
BAC (Bacterial Artificial Chromosome)
70, ,000
YAC (Yeast Artificial Chromosome)
> 300,000
Not used much recently

9. 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 ~500 letters at a time

10. Electrophoresis diagrams

11. Challenging to read answer

12. Challenging to read answer

13. Challenging to read answer

14. Reading an electropherogram
Filtering
Smoothening
Correction for length compressions
A method for calling the letters – PHRED
PHRED – PHil’s Read EDitor (by Phil Green)
Several better methods exist, but labs are reluctant to change

15. Output of PHRED: a read A read: 500-700 nucleotides
A C G A A T C A G …A
…21
Quality scores: -10log10Prob(Error)
Reads can be obtained from leftmost, rightmost ends of the insert
Double-barreled sequencing: (1990)
Both leftmost & rightmost ends are sequenced, reads are paired

16. Method to sequence longer regions
genomic segment
cut many times at random (Shotgun)
Get one or two reads from each segment
~500 bp
~500 bp

17. Whole Genome Shotgun Sequencing
cut many times at random
plasmids (2 – 10 Kbp)
forward-reverse paired reads
(mate pair)
known dist
cosmids (40 Kbp)
~500 bp
~500 bp

18. Sequencing and Fragment Assembly
AGTAGCACAGACTACGACGAGACGATCGTGCGAGCGACGGCGTAGTGTGCTGTACTGTCGTGTGTGTGTACTCTCCT
3x109 nucleotides

19. Fragment Assembly
Computational Challenge: assemble individual short fragments (reads) into a single genomic sequence (“superstring”)

20. Shortest Common Superstring Problem (SCS)
Problem: Given a set of strings, find a shortest string that contains all of them
Input: Strings s1, s2,…., sn
Output: A string s that contains all strings
s1, s2,…., sn as substrings, such that the length of s is minimized
Note: this formulation does not take into account sequencing errors

21. Shortest Common Superstring Problem: Example

22. Greedy Algorithm for SCS
Find pair of strings si, sj with longest overlap.
Merge(si, sj)
Recurse
How “good” is the greedy algorithm?

23. Algorithm Analysis
S = {s1, s2,…., sn } 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

24. SCS and Overlap Graph Build directed graph G = (V,E)
V = {s1, s2,…., sn }
e = (si, sj) if prefix of sj matches suffix of si
w(si, sj) = -length of overlap b/w si, sj
Goal: Find a minimum weight path visiting every VERTEX exactly once in the OVERLAP graph:
Travelling Salesman (Hamiltonian path) problem

25. SCS to TSP: An Example S = { ATC, CCA, CAG, TCC, AGT } SCS AGT CCA ATC
ATCCAGT
TCC
CAG
TSP
ATC 2 1 1
AGT 1
CCA 1 2 2 2 1
CAG
TCC
ATCCAGT

26. We need to use a linear-time algorithm
Fragment Assembly
Given N reads…
Where N ~ 30 million…
We need to use a linear-time algorithm

27. Strategies for whole-genome sequencing
Hierarchical – Clone-by-clone
Break genome into many long pieces
Map each long piece onto the genome
Sequence each piece with shotgun
Example: Yeast, Worm, Human, Rat
Online version of (1) – Walking
Start sequencing each piece with shotgun
Construct map as you go
Example: Rice genome
Whole genome shotgun
One large shotgun pass on the whole genome
Example: Drosophila, Human (Celera),
Neurospora, Mouse, Rat, Dog
Until late 1990s the shotgun fragment assembly of human genome was viewed as intractable problem

28. Reconstructing the Sequence (Fragment Assembly)
reads
Cover region with ~7-fold redundancy (7X)
Overlap reads and extend to reconstruct the original genomic region

29. 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

30. 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 c
P(covered) 1
0.632 2
0.864 4
0.982 8
0.999

31. 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
..ACGATTACAATAGGTT..

32. Challenges in Fragment Assembly
Repeats: A major problem for fragment assembly
> 50% of human genome are repeats:
- over 1 million Alu repeats (about 300 bp)
- about 200,000 LINE repeats (1000 bp and longer)
Repeat
Green and yellow fragments are interchangeable when
assembling repetitive DNA

33. Repeat Types 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
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

34. Triazzle: A Fun Example
The puzzle looks simple
BUT there are repeats!!!
The repeats make it very difficult.
Try it – only $7.99 at

35. Sequencing and Fragment Assembly
AGTAGCACAGACTACGACGAGACGATCGTGCGAGCGACGGCGTAGTGTGCTGTACTGTCGTGTGTGTGTACTCTCCT
3x109 nucleotides
50% of human DNA is composed of repeats
Error!
Glued together two distant regions

36. What can we do about repeats?
Two main approaches:
Cluster the reads
Link the reads

37. Sequencing and Fragment Assembly
AGTAGCACAGACTACGACGAGACGATCGTGCGAGCGACGGCGTAGTGTGCTGTACTGTCGTGTGTGTGTACTCTCCT
3x109 nucleotides A R B
ARB, CRD or
ARD, CRB ? C R D

38. 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
..ACGATTACAATAGGTT..

39. Overlap
Find the best match between the suffix of one read and the prefix of another
Due to sequencing errors, need to use dynamic programming to find the optimal overlap alignment O(N2 L2) for N reads of length L
Apply a filtration method to filter out pairs of fragments that do not share a significantly long common substring

40. Overlapping Reads Sort all k-mers in reads (k ~ 24)
Find pairs of reads sharing a k-mer via hashing
Extend to full alignment – throw away if not >95% similar
TACA
TAGT ||
TAGATTACACAGATTAC
T GA
TAGA
| ||
|||||||||||||||||
TAGATTACACAGATTAC

41. Overlapping Reads and Repeats
A k-mer that appears M times, initiates M2 comparisons
For an Alu that appears 106 times  1012 comparisons – too much
Solution:
Discard all k-mers that appear more than
t  Coverage, (t ~ 10)

42. Finding Overlapping Reads
Create local multiple alignments from the overlapping reads
TAGATTACACAGATTACTGA
TAGATTACACAGATTACTGA
TAG TTACACAGATTATTGA
TAGATTACACAGATTACTGA
TAGATTACACAGATTACTGA
TAGATTACACAGATTACTGA
TAG TTACACAGATTATTGA
TAGATTACACAGATTACTGA

43. Find Overlapping Reads
Correct errors using multiple alignment
TAGATTACACAGATTACTGA
TAGATTACACAGATTACTGA
TAGATTACACAGATTACTGA
TAGATTACACAGATTACTGA
TAGATTACACAGATTATTGA
TAG-TTACACAGATTATTGA
TAGATTACACAGATTACTGA
TAGATTACACAGATTACTGA
TAG-TTACACAGATTACTGA
TAG-TTACACAGATTATTGA
insert A
correlated errors—
probably caused by repeats
 disentangle overlaps
replace T with C
TAGATTACACAGATTACTGA
TAGATTACACAGATTACTGA
TAGATTACACAGATTACTGA
In practice, error correction removes
up to 98% of the errors
TAG-TTACACAGATTATTGA
TAG-TTACACAGATTATTGA

44. Layout Repeats are a major challenge
Do two aligned fragments really overlap, or are they from two copies of a repeat?
Solution: repeat masking – hide the repeats!!!
Masking results in high rate of misassembly (up to 20%)
Misassembly means alot more work at the finishing step

45. Merge Reads into Contigs
repeat region
Unique Contig
Overcollapsed Contig
We want to merge reads up to potential repeat boundaries

46. Repeats, errors, and contig lengths
Repeats shorter than read length are OK
Read that spans across a repeat disambiguates order of flanking regions
Repeats with more base pair diffs than sequencing error rate are OK
We throw overlaps between two reads in different copies of the repeat
To make the genome appear less repetitive, try to:
Increase read length
Decrease sequencing error rate
Role of error correction:
Discards up to 98% of single-letter sequencing errors
decreases error rate
 decreases effective repeat content
 increases contig length

47. Merge Reads into Contigs (cont’d)
repeat region
Ignore non-maximal reads
Merge only maximal reads into contigs

48. Merge Reads into Contigs (cont’d)
repeat boundary???
sequencing error b a
Ignore “hanging” reads, when detecting repeat boundaries

49. Merge Reads into Contigs (cont’d)
?????
Unambiguous
Insert non-maximal reads whenever unambiguous

50. Removing Repeats A-statistic (Myers) used in Celera assembler
Use Lander-Waterman statistics to estimate likelihood ratio of unique region vs. over-collapsed repeat
Normal density
Too dense
 Overcollapsed

51. Link Contigs into Supercontigs
Normal density
Too dense
 Overcollapsed
Inconsistent links
 Overcollapsed?
Scaffolding Problem: various heuristic approaches

52. Link Contigs into Supercontigs (cont’d)
Find all links between unique contigs
Connect contigs incrementally, if  2 links
supercontig
(aka scaffold)

53. Link Contigs into Supercontigs
Fill gaps in supercontigs with paths of repeat contigs
Use length/distance constraints

54. Link Contigs into Supercontigs (cont’d)
d ( A, B )
Contig A
Define G = ( V, E )
V := contigs
E := ( A, B ) such that d( A, B ) < C
Reason to do so: Efficiency; full shortest paths cannot be computed
Contig B

55. Link Contigs into Supercontigs (cont’d)
Contig A
Contig B
Define T: contigs linked to either A or B
Fill gap between A and B if there is a path in G passing only from contigs in T

56. Consensus
A consensus sequence is derived from a profile of the assembled fragments
A sufficient number of reads is required to ensure a statistically significant consensus
Reading errors are corrected

57. Derive Consensus Sequence
TAGATTACACAGATTACTGA TTGATGGCGTAA CTA
TAGATTACACAGATTACTGACTTGATGGCGTAAACTA
TAG TTACACAGATTATTGACTTCATGGCGTAA CTA
TAGATTACACAGATTACTGACTTGATGGCGTAA CTA
TAGATTACACAGATTACTGACTTGATGGGGTAA CTA
TAGATTACACAGATTACTGACTTGATGGCGTAA CTA
Derive multiple alignment from pairwise read alignments
Derive each consensus base by weighted voting
(Alternative: take maximum-quality letter)

58. Some Assemblers PHRAP Celera Arachne Phusion Euler
Early assembler, widely used, good model of read errors
Overlap O(n2)  layout (no mate pairs)  consensus
Celera
First assembler to handle large genomes (fly, human, mouse)
Overlap  layout  consensus
Arachne
Public assembler (mouse, several fungi)
Phusion
Overlap  clustering  PHRAP  assemblage  consensus
Euler
Indexing  Euler graph  layout by picking paths  consensus

59. EULER - A New Approach to Fragment Assembly
Traditional “overlap-layout-consensus” technique has a high rate of mis-assembly
EULER uses the Eulerian Path approach borrowed from “sequencing by hybridization” (SBH)
Fragment assembly without repeat masking can be done in linear time with greater accuracy

60. DNA Basepairing

61. DNA Microarrays

62. Sequencing by Hybridization (SBH)
Build a microarray with all 4l DNA sequences of length l (l ~ 20)
For DNA sequence s, measure l-mer composition

63. l-mer composition
Def: Given string s, the Spectrum ( s, l ) is unordered multiset of all possible (n – l + 1) l-mers in a string s of length n
The order of individual elements in Spectrum ( s, l ) does not matter
For s = TATGGTGC all of the following are equivalent representations of
Spectrum ( s, 3 ):
{TAT, ATG, TGG, GGT, GTG, TGC}
{ATG, GGT, GTG, TAT, TGC, TGG}
{TGG, TGC, TAT, GTG, GGT, ATG}

64. The SBH Problem Goal: Reconstruct a string from its l-mer composition
Input: A multiset S, representing all l-mers from an (unknown) string s
Output: String s such that Spectrum ( s,l ) = S

65. SBH: An Example S = { ATC, CCA, CAG, TCC, AGT } DNA Sequencing AGT CCA
ATCCAGT
TCC
CAG
S = { ATC, CCA, CAG, TCC, AGT }

66. SBH: Hamiltonian Path Approach
S = { ATG AGG TGC TCC GTC GGT GCA CAG } H
ATG
AGG
TGC
TCC
GTC
GGT
GCA
CAG
ATG C A G G T C C
Path visited every VERTEX once: Hamiltonian Path

67. SBH: Hamiltonian Path Approach
A more complicated graph:
S = { ATG TGG TGC GTG GGC GCA GCG CGT }

68. SBH: Hamiltonian Path Approach
S = { ATG TGG TGC GTG GGC GCA GCG CGT }
Path 1:
ATGCGTGGCA
Path 2:
ATGGCGTGCA

69. SBH: Eulerian Path Approach
S = { ATG, TGG, TGC, GTG, GGC, GCA, GCG, CGT }
Vertices correspond to ( l – 1 ) – mers : { AT, TG, GC, GG, GT, CA, CG }
Edges correspond to l – mers from S AT GT CG CA GC TG GG
Path visited every EDGE once

70. SBH: Eulerian Path Approach
S = {ATG, TGG, TGC, GTG, GGC, GCA, GCG, CGT }
Two different paths give different sequence reconstructions: GT CG GT CG AT TG GC AT TG GC CA CA GG GG
ATGGCGTGCA
ATGCGTGGCA

71. Euler Theorem
A graph is balanced if for every vertex the number of incoming edges equals to the number of outgoing edges:
in(v)=out(v)
Theorem: A connected graph is Eulerian if and only if each of its vertices is balanced.

72. Euler Theorem: Proof Eulerian → balanced
for every edge entering v (incoming edge) there exists an edge leaving v (outgoing edge). Therefore
in(v)=out(v)
Balanced → Eulerian
???

73. Algorithm for Constructing an Eulerian Cycle
a. Start with an arbitrary vertex v and form an arbitrary cycle with unused edges until a dead end is reached. Since the graph is Eulerian this dead end is necessarily the starting point, i.e., vertex v.

74. Algorithm for Constructing an Eulerian Cycle (cont’d)
b. If cycle from (a) above is not an Eulerian cycle, it must contain a vertex w, which has untraversed edges. Perform step (a) again, using vertex w as the starting point. Once again, we will end up in the starting vertex w.

75. Algorithm for Constructing an Eulerian Cycle (cont’d)
c. Combine the cycles from (a) and (b) into a single cycle and iterate step (b).

76. Euler Theorem: Extension
Theorem: A connected graph has an Eulerian path if and only if it contains at most two semi-balanced vertices and all other vertices are balanced.

77. Sources Genomes sequenced: http://www.genome.gov/10002154
Serafim Batzoglou (Sequencing slides)
(Euler slides)