Physical Mapping of DNA

1
Physical Mapping of DNA

• Shanna Terry
• March 2, 2004

2
Overview

• Background
• Types of Mapping
• Mathematical Models
• Enhanced Double Digest Problem

3
Background

• Given a sequence of DNA, how do we figure out
  where on some larger chromosome the sequence
  lies?
• ?

4
Background II

• Look for markers that match in both the
  chromosome and the shorter sequence.
• Markers Usually short, precisely defined
  sequences
• match!

5
Background III

• How do we create the original map?
• Generate fingerprints (markers) with
• Restriction site mapping
• Hybridization

6
Background IV

• But why? Cant we just expand the sequence
  assembly techniques weve already learned?
• NO! (with one exception)
• Why not?
• A chromosome isnt just 150k bps long.
• Human chromosomes range in length from 51
  million to 245 million base pairs.

7
Overview

• Background
• Types of Mapping
• Mathematical Models
• Enhanced Double Digest Problem

8
Restriction Site Mapping

• In this situation, the fingerprint is the length
  between restriction sites of given enzymes
  (recall from previous lectures).
• Make three copies of target DNA strings A, B, C.
• Apply one enzyme (a) to string A, another (ß) to
  string B, and both (a and ß) to string C.
• Line up the fragments in A and B so they match C
  this is the double digest problem.

6 14
7
5
8 3
15
4
6 2 3
9 7 1
4
9
Restriction Site Mapping II

• A variant is the partial digest approach
• Use only one enzyme, but allow it to act for
  different time periods. Different restriction
  sites will be recognized.
• Fragment sites 6, 20, 27, 32 14, 21, 26 7,
  12 and 5

6 14
7
5
10
Restriction Site Mapping III

• 6
• 20
• 14
• 21
• etc
• Fragment sites 6, 20, 27, 32 14, 21, 26 7, 12
  and 5

6
6 14

14
14
7
6
14 7
5
11
Hybridization Mapping

• Check whether specific small sequences (called
  probes) bind (hybridize) to fragments (clones)
• The fingerprint is the subset of probes that
  successfully hybridize to the clone.
• If some portion of one clones fingerprint
  matches another, they are likely to be from
  overlapping regions of the target.

12
Hybridization Mapping II

• Probes x, y, z, bound to clone A x, w and z
  bound to clone B overlap in x and z.
• y x z w
• Except we dont know that much. We only know
  which probes bind to which clones. Not ordering
  or even relative lengths!

13

• Background
• Types of Mapping
• Mathematical Models
• Enhanced Double Digest Problem

14
Restriction Site Models

• Back to the double digest problem weve split
  the strings A, B, C into fragments with two
  enzymes.
• We have the multisets made up of the fragment
  lengths
• From previous example
• A 5, 6, 7, 14
• B 3, 4, 8, 15
• C 1, 2, 3, 4, 6, 7, 9
• Find permutations of A and B such that there is a
  one-to-one correspondence between all the
  subintervals and C. Not too bad, right?

15
Restriction Site Models II

• BAD NEWS
• The double digest problem is NP-complete. It is
  a generalization of the set-partition problem,
  already known to be NP-complete.
• To give you an ideathe number of solutions is
  (k-1)! for k number of restriction sites.
• BUT we will see a heuristic later

16
Interval Graph Models

• Model hybridization mapping in terms of interval
  graphs
• Interval graph A graph G which is mapped from a
  series of intervals. For each interval there is
  a vertex in G. For each intersection of
  intervals there is an edge in G.

17
Interval Graph Models II

• Ex a
• b
  
• 
  c
• d
  e
• b
• a c
  e
• d

18
Interval Graph Models III

• To apply this to the hybridization mapping
  problem
• We create graphs with vertices representing
  clones (fragments), and edges representing
  overlaps between clones.
• Two graphs one for known overlaps and one with
  known and unknown overlap information (neither
  are necessarily interval graphs).

19
Interval Graph Models IV

• Now find the true interval graph (a subgraph of
  the known and unknown graph) given the two
  graphs.
• Known overlaps Known/Unknown
  Actual Interval
• Overlaps Graph
• Hmm Not too easy.

20
Interval Graph Models V

• MORE BAD NEWS!
• This is NP-hard.
• Maybe another model?
• There are two other possible models for
  hybridization mapping (described in the book).
  But.
• Those are NP-hard too!

21
Consecutive Ones Property

• Were sick of NP hard problems. Give us
  something a little easier.
• The Consecutive Ones Property Model (C1P) can be
  solved in linear time!
• Assumptions
• The probes are unique.
• There are no errors. (!!)
• All of the correspondences of clones and probes
  have been found. (!!!)

22
Consecutive Ones Property II

• Build a matrix (n x m), n of clones, m of
  probes. Entry i,j is a binary code for whether
  probe j hybridized to clone i.
• above probe 1 hybridized to clone 1, probe 2
  hybridized to clone 1, probe 1 hybridized to
  clone 3, probe 4 hybridized to clone 3.

23
Consecutive Ones Property III

• Find a permutation of the columns (probes) such
  that all the 1s in each row (clone) are
  consecutive.

24
Consecutive Ones Property IV

• This algorithm can be run in linear time!
• Unfortunately, the assumption that there are no
  errors isnt useful because biology isnt a
  mathematical model. Probes may not bind, DNA may
  be replicated incorrectly.
• And generalizations make the problem NP-hard
  again!
• We need a good heuristic

25

• Background
• Types of Mapping
• Mathematical Models
• Enhanced Double Digest Problem

26
Enhanced Double Digest Problem

• The Enhanced Double Digest (EDD) problem is
  NP-hard in the general case, but if the lengths
  of fragments in C (the string acted upon by both
  types of enzymes) are distinct, it can be solved
  in linear time!
• Why do the fragments have to be distinct?
• What if all the fragments are the same length?

27
Problem Formulation

• We have the multisets A and B.
• A 6, 14, 7, 5
• B 8, 3, 15, 4
• Take the actual fragments corresponding to each
  member of the either set (since the sets are only
  lengths). Apply the other enzyme to the fragment
  (i.e. apply enzyme ß to fragments from A, and
  vice versa) to create subfragments.
• ABi is the multiset of subfragments created by
  applying enzyme ß to fragments from A BAj from
  applying enzyme a to fragments of B.

28
Problem Formulation II

• Example
• 8 2,6 5 1,4
• A5, 6, 7, 14
• B3, 4, 8, 15
• AB11,4, AB26, AB37, AB42,3,9
• BA13, BA24, BA32,6, BA41,7,9

6 14
7
5
8 3
15
4
6 2 3
9 7 1
4
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Algorithm

• Given A, B, ABi and BAj for all i, j, construct
  an undirected graph that connects each element of
  A and B to its corresponding AB/BA. Note that
  all elements in C will be covered
• A 5 6 7 14
• C 1 2 3 4 6 7 9
• B 3 4 8 15

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Algorithm II

• Create a spanning tree
• Start at random node, follow all paths from the
  node, dont repeat edges.
• 6A 6C 8B 2C 14A 9C
  15B 1C 5A 4C 4B
• 3C 7C
• 3B 7A

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Properties

• The graph (G) will always be connected, and every
  node in A and B will only be adjacent to nodes
  from C. Each node from C connects to only one
  node each from A and B.
• If the problem can be solved G will be a
  spanning tree, and any subtree that hangs on
  the longest path will be a 2-node length path
  (dangler).

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Properties II

• Danglers
• 6A 6C 8B 2C
  14A 9C 15B 1C 5A
  4C 4B
• 3C
  7C
• 3B
  7A

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Algorithm III

• If the graph G is not a spanning tree, and not
  all subtrees hanged off the longest path are
  danglers, then there is no valid permutation.
• Perform Dangler-first search on the graph G

34
Dangler-First Search

• Traverse G starting at one end of a path S with
  the largest number of edges, reading only the
  nodes from C. Whenever reaching a node with
  degree greater than 2 (must have a dangler), read
  the nodes in C from the hanging danglers first,
  then continue to traverse S. This sequence is pc.

35
Dangler-First Search II

• pc 6, 2, 3, 9, 7, 1, 4.
• 6A 6C 8B 2C
  14A 9C 15B 1C 5A
  4C 4B
• 3C
  7C
• 3B
  7A

36
Algorithm IV

• The elements in each ABi form a consecutive
  subsequence in pc. Likewise, the elements in
  each BAj also form a consecutive subsequence in
  pc.
• This permutation is a valid permutation meaning
  you have the answer!

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Solution

• A 6, 14, 7, 5
• pc 6, 2, 3, 9, 7, 1, 4
• B 8, 3, 15, 4
• 6A 6C 8B 2C
  14A 9C 15B 1C 5A
  4C 4B
• 3C
  7C
• 3B
  7A

6 14
7 5
8 3
15 4
6 2 3 9
7 1 4
38
Enhanced Double Digest Problem II

• This can be solved in O(n) time!
• A generalization (assuming a constant number of
  duplicates) can also be solved in O(n) time.
• The general enhanced double digest problem is
  still NP-Hard. This can be shown by reduction
  from the Hamiltonian Path problem.