Chapter 5: DNA Sequence Assembly

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Chapter 5 DNA Sequence Assembly
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• Natura enim simplex est, et rerum causis
  superfluis non luxuriat.
• 
  Principia
• I.
  Newton

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DNA Sequencing

• Two ways of copying DNA
• 1. Polymerase Chain Reaction (1986) make many
  copies of a fragment (500). Needs primers (end
  segments). Cleave into two. Anneal primers (5-
  3, and 3- 5). Make two double chains. Repeat
  1,2,4,8,16,

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5
5
3
5
3

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3
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• 2. Cloning. Insert a large piece of DNA into a
  cloning vector (virus, bacteria, yeast -YAC).
  Then the vector is inserted into a host cell to
  duplicate naturally.
• DNA (shotgun) Sequencing
• Make many copies (single strand)
• Cut them into fragments of lengths 500.
• For each fragment of length L, use some process
  like PCR, generating all lengths 1 L with some
  fluorescence dye. In old scheme, you generate all
  fragments end with A, then with C, then G, then
  T, run them in 4 lanes of electrophoresis gel. In
  the new scheme, you have 4 colors (of the dye)
  all fragments in 1 lane.
• Then assemble all fragments into the shortest
  common superstring by GREEDY repeatedly merge
  the pair with max overlap until finish.

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Mouse Whole Genome Assembly

• Multiple copies of the genome are broken into
  pieces
• Both ends of every piece are read.
• Length (and orientation) of each piece form
  constraints.
• Reads 500-1000 bp
• Quality array for each position.
• Reconstruct genome from reads and constraints.
• Issues both ends of a read usually low quality,
  chimeric reads, repetitive regions.

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Mouse Genome Data Set

• Dec. 2001 release, 33 million reads. After
  removing low quality reads, 30.5 million reads.
• 12.9 million constraints.
• After removing repeats, if two reads overlap
  large enough, merge (with max score)
• A contig is an ordered and oriented list of
  overlapping reads.
• A scaffold is an ordered and oriented list of
  contigs.
• Covered about 2.5 GB.

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Shortest Common Superstring

• In FOCS 1990, we formulated and analyzed the
  following learning problem Infer original DNA
  sequence from fragments. Or given n strings,
  find the shortest common superstring.
• The problem has been proved to be NPC, 1980.
• Open for 10 years does GREEDY give linear
  approximation ratio --- greedy/opt lt constant.
• We solved this, proving approximation ratio 3,
  STOC91 (Blum, Jiang, Li, Tromp, Yannakakis).
• Improvements by many 2.89, 2.81, 2.79, 2.75,
  2.66,
• Formal Statement Given Ss1, sn, find a
  shortest s such that for all i, si is a substring
  of s.
• E.g. alf ate half lethal alpha alfalfa ?
  lethalphalfalfate

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Many Systems

• PHREP
• CAP
• Euler
• TIGR Assembler
• Arachne
• LSA
• We will touch on some fundamental facets of the
  sequence assembly problem.

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Why Shortest superstring?

• Theorem. Given a DNA sequence S. Random sample
  substring (of length L) from it according to
  distribution D for 6nlogn/Le times. Then given a
  query is s a substring of D, with probability gt
  1-e, we can answer approximately correctly (with
  probD gt 1-e, under distribution D).
• Remarks
• I.e., polynomial time pac learnable.
• L 500 700
• Weak theorem, but it does show nontrivially why
  we need the shortest superstring the shorter
  the more likely it is correct.

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Theorem. GREEDY achieves 4.

• Proof. Given Ss1, ,sm, construct G
• Nodes are s1, ,sm
• Edges if then
  add edge
• where pref is the pref length. I.e.
  siprefoverlap length with sj
• SCS(S) length shortest Hamilton cycle in G
• (Modified) Greedy restated find all cycles with
  minimum weights in G, then open cycle,
  concatenate to obtain the final superstring.

sj
pref
si
pref
si
sj
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This minimum cycle exists

• Assuming initial Hamilton cycle has w(C) n
• Then merging si with sj is equivalent to breaking
  into two cycles. We have
• w(C1) w(C2) n
• Proof We merged (si, sj) because they have max
  overlap. Picture shows
• d(si,sj)d(s,s)ltd(si,s)d(s,sj)
• Continue this process,end with self-cycles C1,
  C2, C3, C4,
• Sw(Ci) n.

C
si
sj
s
s

S
sj
C1
si
S
si
sj
s
s
C2
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• Then we open cycles concat
• Let Wiw(Ci)
• Li longest string in Ci
• open Ci Wi Li
• n SWi
• Lemma. S1 and S2 overlap w1w2
• S(Li-2Wi) n, by lemma, since Lis must be in
  final SCS.
• Greedy(S)ltS(LiWi)
• S(Li-2Wi)S3Wi
• n 3n
• 4n.
• QED

w1
w1
w1
w1
s1
w2
w2
w2
s2
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Dealing with Repeats

• When there are repeats, no system can handle
  them.
• Are they unsolvable?
• Observations repeated sequences are often not
  exact repeats. We can use such small differences
  to separate them.
• Formalizing this Given a collection of repeats
  (approximately similar strings). Separate them
  into k piles each with a center, minimizing max
  Hamming distance. This problem is NP-hard. An
  interesting research problem. For kO(1), similar
  techniques (as in Motif finding PTAS lecture) can
  be used to give PTAS. However, for larger k, this
  becomes exponential in k.

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Projects

• Comparative study CAP, PHREP, Arachne, Euler.

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Open Problems

• Prove the 2 approximation ratio for Greedy.
• Solve the repeat resolution problem.