Ch04 dna mapping

DNA Mapping and Brute
Force Algorithms

1. Restriction Enzymes
2. Gel Electrophoresis
3. Partial Digest Problem
4. Brute Force Algorithm for Partial Digest Problem
5. Branch and Bound Algorithm for Partial Digest Problem
6. Double Digest Problem
Outline

Section 1:
Restriction Enzymes

Discovery of Restriction Enzymes
• HindII: First restriction enzyme.
• Was discovered accidentally in 1970 while scientists were
studying how the bacterium Haemophilus influenzae takes
up DNA from the virus.
• Recognizes and cuts DNA at sequences:
• GTGCAC
• GTTAAC

Discovering Restriction Enzymes
Werner Arber
My father has discovered a
servant who serves as a pair of
scissors. If a foreign king invades
a bacterium, this servant can cut
him in small fragments, but he
does not do any harm to his own
king. Clever people use the
servant with the scissors to find
out the secrets of the kings. For
this reason my father received
the Nobel Prize for the discovery
of the servant with the scissors.”
Daniel Nathans’ daughter
(from Nobel lecture)
Hamilton SmithDaniel Nathans
Werner Arber – discovered restriction
enzymes
Daniel Nathans - pioneered the application
of restriction for the
construction of genetic
maps
Hamilton Smith - showed that restriction
enzyme cuts DNA in the
middle of a specific sequence

Molecular Cell Biology, 4thEdition
Molecular Scissors

Restriction Enzymes: Common Recognition Sites
Molecular Cell Biology, 4thEdition

• Recombinant DNA technology
• Cloning
• cDNA/genomic library construction
• DNA mapping
Uses of Restriction Enzymes

• A restriction map is a map showing positions of restriction sites
in a DNA sequence.
• If DNA sequence is known then construction of restriction map
is trivial exercise.
• In early days of molecular
biology DNA sequences
were often unknown.
• Biologists had to solve
the problem of constructing
restriction maps without
knowing DNA sequences.
Restriction Maps

Full Restriction Digest
• Cutting DNA at each restriction site creates multiple
restriction fragments:
• Full Restriction Digest: Is it possible to reconstruct the order
of the fragments from the sizes of the fragments?
• Example: Say the fragments have lengths {3,5,5,9} as in
the above sequence.

Full Restriction Digest: Multiple Solutions
• For the set of fragment lengths {3, 5, 5, 9} we have the
original segment as a possible solution:
• However, we could also have the following segment:

Section 2:
Gel Electrophoresis

• Restriction enzymes break DNA into restriction fragments.
• Gel electrophoresis:A process for separating DNA by size
and measuring sizes of restriction fragments.
• Modern electrophoresis machines can separate DNA
fragments that differ in length by 1 nucleotide for fragments
up to 500 nucleotides long.
Gel Electrophoresis: Measure Segment Lengths

• DNA fragments are injected into a gel positioned in an
electric field.
• DNA are negatively charged near neutral pH.
• The ribose phosphate backbone of each nucleotide is
acidic; DNA has an overall negative charge.
• Thus DNA molecules move towards the positive electrode.
Gel Electrophoresis: How It Works

Gel Electrophoresis
• DNA fragments of different lengths are separated according to
size.
• Smaller molecules move through the gel matrix more readily
than larger molecules.
• The gel matrix restricts random diffusion so molecules of
different lengths separate into different bands.

Direction of DNA
movement
Detecting DNA: Autoradiography
• Separated DNA bands on a
gel can be viewed via
autoradiography:
1. DNA is radioactively
labeled.
2. The gel is laid against a
sheet of photographic
film in the dark,
exposing the film at the
positions where the
DNA is present.
Molecular Cell Biology, 4th edition

• Another way to visualize DNA bands in gel is through
fluorescence:
• The gel is incubated with a solution containing the
fluorescent dye ethidium.
• Ethidium binds to the DNA.
• The DNA lights up when the gel is exposed to ultraviolet
light.
Detecting DNA: Fluorescence

Section 3:
Partial Digest Problem

• The sample of DNA is exposed to the restriction enzyme for
only a limited amount of time to prevent it from being cut at
all restriction sites; this procedure is called partial
(restriction) digest.
• This experiment generates the set of all possible restriction
fragments between every two (not necessarily consecutive)
cuts.
• This set of fragment sizes is used to determine the positions of
the restriction sites in the DNA sequence.
Partial Restriction Digest

• Partial Digest results in the following 10 restriction fragments:
Partial Digest: Example

• We assume that multiplicity of a fragment can be detected, i.e.,
the number of restriction fragments of the same length can be
determined.
• Here we would detect two
fragments of length 5 and
two of length 14.

Multiset: {3, 5, 5, 8, 9, 14, 14, 17, 19, 22}
• We therefore have a multiset of fragment lengths.

• We now provide a basic mathematical framework for the
partial digest process.
The multiset of integers representing lengths of each of the
DNA fragments produced from a partial digest; formed
from X by taking all pairwise differences.
The set of n integers representing the location of all cuts in
the restriction map, including the start and end.
X:
DX:
Partial Digest: Mathematical Framework

Return to Partial Digest Example

0 5 14 19 22

• n = 5
0 5 14 19 22

• n = 5
• X = {0, 5, 14, 19, 22}
0 5 14 19 22

0 5 14 19 22
• n = 5
• X = {0, 5, 14, 19, 22}
• DX = {3,5,5,8,9,14,14,17,19,22}

• n = 5
• X = {0, 5, 14, 19, 22}
• DX = {3,5,5,8,9,14,14,17,19,22}
• Represent DX as a table, with
elements of X along both the top and
left sides.

• n = 5
• X = {0, 5, 14, 19, 22}
• DX = {3,5,5,8,9,14,14,17,19,22}
elements of X along both the top
and left sides.
X 0 5 14 19 22
0
5
14
19
22

• n = 5
• X = {0, 5, 14, 19, 22}
• DX = {3,5,5,8,9,14,14,17,19,22}
elements of X along both the top
and left sides.
• We place xj – xi into entry (i,j) for all
1 ≤ i<j ≤ n
X 0 5 14 19 22
0 5 14 19 22
5 9 14 17
14 5 8
19 3
22

• Goal: Given all pairwise distances between points on a line,
reconstruct the positions of those points.
• Input: The multiset of pairwise distances L, containing n(n-
1)/2 integers.
• Output: A set X, of n integers, such that ∆X = L.
Partial Digest Problem (PDP): Formulation

• It is not always possible to uniquely reconstruct a set X based
only on ∆X.
• Example:The sets
X = {0, 2, 5} (X + 10) = {10, 12, 15}
both produce ∆X = ∆(X + 10) ={2, 3, 5} as their partial
digest.
• Two sets X and Y are homometric if ∆X = ∆Y.
• The sets {0,1,2,5,7,9,12} and {0,1,5,7,8,10,12} present a less
trivial example of homometric sets. They both digest into:
{1, 1, 2, 2, 2, 3, 3, 4, 4, 5, 5, 5, 6, 7, 7, 7, 8, 9, 10, 11, 12}
Multiple Solutions to the PDP

X = {0,1,2,5,7,9,12}
Homometric Sets: Example

0 1 2 5 7 9 12
0 1 2 5 7 9 12
1 1 4 6 8 11
2 3 5 7 10
5 2 4 7
7 2 5
9 3
12
X = {0,1,2,5,7,9,12}

0 1 2 5 7 9 12
0 1 2 5 7 9 12
1 1 4 6 8 11
2 3 5 7 10
5 2 4 7
7 2 5
9 3
12
X = {0,1,2,5,7,9,12} Y = {0,1,5,7,8,10,12}

0 1 2 5 7 9 12
0 1 2 5 7 9 12
1 1 4 6 8 11
2 3 5 7 10
5 2 4 7
7 2 5
9 3
12
0 1 5 7 8 10 12
0 1 5 7 8 10 12
1 4 6 7 9 11
5 2 3 5 7
7 1 3 5
8 2 4
10 2
12
X = {0,1,2,5,7,9,12} Y = {0,1,5,7,8,10,12}

Section 4:
Brute Force Algorithm for
Partial Digest Problem

• Brute force algorithms, also known as exhaustive search
algorithms, examine every possible variant to find a solution.
• Efficient only in rare cases; usually impractical.
Brute Force Algorithms

1. Find the restriction fragment of maximum length M. Note: M
is the length of the DNA sequence.
2. For every possible set X={0, x2, … ,xn-1, M}
compute the corresponding DX.
3. If DX is equal to the experimental partial digest L, then X is
a possible restriction map.
Partial Digest: Brute Force

Partial Digest: Brute Force
1. AnotherBruteForcePDP(L, n)
2. M  maximum element in L
3. for every set of n – 2 integers 0 < x2 < … xn-1 < M
4. X { 0,x2,…,xn-1,M }
5. Form DX from X
6. if DX = L
7. return X
8. output “no solution”

• BruteForcePDP takes O(M n-2) time since it must examine all
possible sets of positions. Note: the number of such sets is
• One way to improve the algorithm is to limit the values of xi
to only those values which occur in L, because we are
assuming for the sake of simplicity that 0 is contained in X.
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Efficiency of BruteForcePDP

Another BruteForcePDP
• Limiting the members of X to those contained in L is almost
identical to BruteForcePDP, except for line 3:
3. for every set of n – 2 integers 0 < x2 < … xn-1 < M
4. X  { 0,x2,…,xn-1,M }
5. Form DX from X
6. if DX = L
7. return X

Another BruteForcePDP
• Limiting the members of X to those contained in L is almost
identical to BruteForcePDP, except for line 3:
3. for every set of n – 2 integers 0 < x2 < … xn-1 < M from L
4. X  { 0,x2,…,xn-1,M }
5. Form DX from X
6. if DX = L
7. return X

• More efficient than BruteForce PDP, but still slow.
• If L = {2, 998, 1000} (n = 3, M= 1000), BruteForcePDP will
be extremely slow, but AnotherBruteForcePDP will be quite
fast.
• Fewer sets are examined, but runtime is still exponential:
O(n2n-4).
Another BruteForcePDP: Efficiency

Section 5:
Branch and Bound
Algorithm for Partial
Digest Problem

1. Begin with X = {0}.
Branch and Bound Algorithm for PDP

2. Remove the largest element in L and place it in X.

3. See if the element fits on the right or left side of the restriction
map.

map.
4. When if fits, find the other lengths it creates and remove those
from L.

map.
from L.
5. Go back to step 1 until L is empty.

map.
from L.
5. Go back to step 1 until L is empty.
WRONG ALGORITHM

• Before describing PartialDigest, first define D(y, X) as the
multiset of all distances between point y and all other points in
the set X.
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Defining D(y, X)

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PartialDigest: Pseudocode

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { }
PartialDigest: Example

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0 }
• Remove 10 from L and insert it (along with 0) into X. We know
this must be the length of the DNA sequence because it is the
largest fragment.

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0, 10 }

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0, 10 }
• Take 8 from L and make y = 2 or 8. But since the two cases are
symmetric, we can assume y = 2.

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0, 10 }
• We find that the distances from y=2 to other elements in X are
D(y, X) = {8, 2}, so we remove {8, 2} from L and add 2 to X.

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0, 2, 10 }

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0, 2, 10 }
• Take 7 from L and make y = 7 or y = 10 – 7 = 3. We will
explore y = 7 first, so D(y, X ) = {7, 5, 3}.

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0, 2, 10 }
• For y = 7 first, D(y, X ) = {7, 5, 3}. Therefore we remove {7, 5
,3} from L and add 7 to X.
D(y, X) = {7, 5, 3} = {½7 – 0½, ½7 – 2½, ½7 – 10½}

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0, 2, 7, 10 }

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0, 2, 7, 10 }
• Take 6 from L and make y = 6.
• Unfortunately D(y, X) = {6, 4, 1 ,4}, which is not a subset of L.
Therefore we won’t explore this branch.
6

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0, 2, 7, 10 }
• This time make y = 4. D(y, X) = {4, 2, 3 ,6}, which is a subset
of L so we will explore this branch. We remove {4, 2, 3 ,6}
from L and add 4 to X.

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0, 2, 4, 7, 10 }

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0, 2, 4, 7, 10 }
• L is now empty, so we have a solution, which is X.

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0, 2, 7, 10 }
• To find other solutions, we backtrack.

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0, 2, 10 }
• More backtrack.

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0, 2, 10 }
• This time we will explore y = 3. D(y, X) = {3, 1, 7}, which is
not a subset of L, so we won’t explore this branch.

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 }
X = { 0, 10 }
• We backtracked back to the root. Therefore we have found all
the solutions.

• Still exponential in worst case, but is very fast on average.
• Informally, let T(n) be time PartialDigest takes to place n cuts.
• No branching case: T(n) < T(n-1) + O(n)
• Quadratic
• Branching case: T(n) < 2T(n-1) + O(n)
• Exponential
Analyzing the PartialDigest Algorithm

Section 6:
Double Digest Problem

• Double Digest is yet another experimentally method to
construct restriction maps
• Use two restriction enzymes; three full digests:
1. One with only first enzyme
2. One with only second enzyme
3. One with both enzymes
• Computationally, Double Digest problem is more complex
than Partial Digest problem
Double Digest Mapping

• Without the information about X (i.e. A+B), it is impossible to
solve the double digest problem as this diagram illustrates
Double Digest: Example

• Input:
• dA – fragment lengths from the digest with enzyme A.
• dB – fragment lengths from the digest with enzyme B.
• dX – fragment lengths from the digest with both A and B.
• Output:
• A – location of the cuts in the restriction map for the
enzyme A.
• B – location of the cuts in the restriction map for the
enzyme B.
Double Digest Problem

Double Digest: Multiple Solutions

Ch04 dna mapping

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