1. DNA Sequencing and Assembly

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

3. 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 – 1/10,000
Other organisms have much higher polymorphism rates

4. DNA sequencing – vectors
Shake
DNA fragments
Known
location
(restriction
site)
Vector
Circular genome
(bacterium, plasmid) + =

5. 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

6. DNA sequencing – gel electrophoresis
Start at primer
(restriction site)
Grow DNA chain
Include dideoxynucleoside
(modified a, c, g, t)
Stops reaction at all
possible points
Separate products with
length, using
gel electrophoresis

7. Electrophoresis diagrams

8. Output of gel electrophoresis: a read
A read: nucleotides
A C G A A T C A G …. A
Quality scores: -10log10Prob(Error)
Reads can be obtained from leftmost, rightmost ends of the insert
Double-barreled sequencing:
Both leftmost & rightmost ends are sequenced

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

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

11. Definition of Coverage
Length of genomic segment: L
Number of reads: n
Length of each read: l
Definition: Coverage C = nl/L
How much coverage is enough?
(Lander-Waterman model):
Assuming uniform distribution of reads, C=10 results in 1 gapped region /1,000,000 nucleotides

12. Challenges with Fragment Assembly
Sequencing errors
~1-2% of bases are wrong
Repeats
Computation: ~ O( N2 ) where N = # reads
false overlap due to repeat

13. 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)
Common Repeat Families
SINE (Short Interspersed Nuclear Elements)
(e.g. ALU: ~300-long, 106 copies)
LINE (Long Interspersed Nuclear Elements)
~500-5,000-long, 200,000 copies
MIR
LTR/Retroviral
Other
-Genes that are duplicated & then diverge (paralogs)
-Recent duplications, ~100,000-long, very similar copies

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

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

16. Strategies for sequencing a whole genome
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, Fugu

17. Hierarchical Sequencing

18. Hierarchical Sequencing Strategy
a BAC clone
map
genome
Obtain a large collection of BAC clones
Map them onto the genome (Physical Mapping)
Select a minimum tiling path
Sequence each clone in the path with shotgun
Assemble
Put everything together

19. Methods of physical mapping
Goal:
Make a map of the locations of each clone relative to one another
Use the map to select a minimal set of clones to sequence
Methods:
Hybridization
Digestion

20. 1. Hybridization p1 pn
Short words, the probes, attach to complementary words
Construct many probes
Treat each BAC with all probes
Record which ones attach to it
Same words attaching to BACS X, Y  overlap

21. Hybridization – Computational Challenge
p1 p2 …………………….pm
Matrix:
m probes  n clones
(i, j): 1, if pi hybridizes to Cj
0, otherwise
Definition: Consecutive ones matrix
A matrix 1s are consecutive
Computational problem:
Reorder the probes so that matrix is in consecutive-ones form
Can be solved in O(m3) time (m >> n)
Unfortunately, data is not perfect
0 0 1 …………………..1
C1 C2 ……………….Cn
1 1 0 …………………..0
1 0 1…………………...0
pi1pi2…………………….pim
……………..0
……………..0
……………..0
Cj1Cj2 ……………….Cjn
………
………

22. 2. Digestion Restriction enzymes cut DNA where specific words appear
Cut each clone separately with an enzyme
Run fragments on a gel and measure length
Clones Ca, Cb have fragments of length { li, lj, lk }  overlap
Double digestion:
Cut with enzyme A, enzyme B, then enzymes A + B

23. Whole-Genome Shotgun Sequencing

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

25. The Overlap-Layout-Consensus approach
1. Find overlapping reads
2. Merge good pairs of reads into longer contigs
3. Link contigs to form supercontigs
4. Derive consensus sequence
..ACGATTACAATAGGTT..
+ many heuristics

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

27. 1. Find Overlapping Reads
One caveat: repeats
A k-mer that appears N times, initiates N2 comparisons
ALU: 1,000,000 times
Solution:
Discard all k-mers that appear more than c  Coverage, (c ~ 10)

28. 1. Find Overlapping Reads
Create local multiple alignments from the overlapping reads
TAGATTACACAGATTACTGA
TAGATTACACAGATTACTGA
TAG TTACACAGATTATTGA
TAGATTACACAGATTACTGA
TAGATTACACAGATTACTGA
TAGATTACACAGATTACTGA
TAG TTACACAGATTATTGA
TAGATTACACAGATTACTGA

29. 1. Find Overlapping Reads (cont’d)
Correct errors using multiple alignment
C: 20
C: 20
C: 35
C: 35
T: 30
C: 0
C: 35
C: 35
TAGATTACACAGATTACTGA
C: 40
C: 40
TAGATTACACAGATTACTGA
TAG TTACACAGATTATTGA
TAGATTACACAGATTACTGA
TAGATTACACAGATTACTGA
A: 15
A: 15
A: 25
A: 25 -
A: 0
A: 40
A: 40
A: 25
A: 25
Score alignments
Accept alignments with good scores

30. Basic principle of assembly
Repeats confuse us
Ability to merge two reads  ability to detect repeats
We can dismiss as repeat any overlap of < t% similarity
Role of error correction:
Discards ~90% of single-letter sequencing errors
 Threshold t% increases

31. 2. Merge Reads into Contigs (cont’d)
repeat region
Merge reads up to potential repeat boundaries
(Myers, 1995)

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

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

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

35. 3. Link Contigs into Supercontigs
Normal density
Too dense: Overcollapsed?
(Myers et al. 2000)
Inconsistent links: Overcollapsed?

36. 3. Link Contigs into Supercontigs (cont’d)
Find all links between unique contigs
Connect contigs incrementally, if  2 links

37. 3. Link Contigs into Supercontigs
Fill gaps in supercontigs with paths of overcollapsed contigs

38. 3. Link Contigs into Supercontigs
d ( A, B )
Contig A
Contig B
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

39. 3. Link Contigs into Supercontigs
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

40. 4. 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

41. Mouse Genome Several heuristics of iteratively:
Breaking supercontigs that are suspicious
Rejoining supercontigs
Size of problem: 32,000,000 reads
Time: 15 days, 1 processor
Memory: 28 Gb
N50 Contig size: Kb  24.8 Kb
N50 Supercontig size: Mb  16.9 Mb

42. Mouse Assembly

43. Sequencing in the (near) future…
CMOS Chip
Photodiodes
10mm
4mm
Microfluidic Chip
Inlet
Outlet
6mm