Computational Molecular Biology

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Computational Molecular Biology. Pooling Designs – Inhibitor Models. An Inhibitor Model. In sample spaces, exists some inhibitors Inhibitor = anti-positive (Positives + Inhibitor) = Negative. _. _. _. _. _. Inhibitor. +. _. x. +. Negative. An Example of Inhibitors. - PowerPoint PPT Presentation

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Computational Molecular Biology

Pooling Designs – Inhibitor Models

My T. Thaimythai@cise.ufl.edu

2

An Inhibitor Model

In sample spaces, exists some inhibitors Inhibitor = anti-positive (Positives + Inhibitor) = Negative

_+__

___x

Inhibitor

Negative

+

My T. Thaimythai@cise.ufl.edu

3

An Example of Inhibitors

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Inhibitor Model

Definition: Given a sample with d positive clones, subject to at

most r inhibitors Find a pooling design with a minimum number of

tests to identify all the positive clones (also design a decoding algorithm with your pooling design)

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Inhibitors with Fault Tolerance Model

Definition: Given n clones with at most d positive clones and at

most r inhibitors, subject to at most e testing errors Identify all positive items with less number of tests

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Preliminaries

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2-stages Algorithm

What is AI? The set AI should contains all the inhibitors and

no positives.

Hence the set PN contains all positives (and some negatives) but no inhibitors

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2-stages Algorithm

At this stage, the problem become the e-error-correcting problem.

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Non-adaptive Solution (1 stage)

1. P contains all positives

2. N contains all negatives

3. O contains all inhibitors and no positives

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Non-adaptive Solution

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Generalization

The positive outcomes due to the combination effect of several items

Items are molecules Depends on a complex: subset of molecules Example: complexes of Eukaryotic DNA transcription

and RNA translation

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A Complex Model

Definition Given n items and a collection of at most d positive

subsets Identify all positive subsets with the minimum

number of tests

Pool: set of subsets of items Positive pool: Contains a positive subset

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What is Hypergraph H?

H = (V,E ) where: V is a set of n vertices (items) E a set of m hyperedges Ej where Ej is a subsets of V

Rank: r = max {| Ej| s.t Ej inE }

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Group Testing in Hypergraph H

Definition: Given H with at most d positive hyperedges Identify all positive hyperedges with the minimum number

of tests

Hyperedges = suspect subsets Positive hyperedges = positive subsets Positive pool: contains a positive hyperedge Assume that Ei Ej

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d(H)-disjunct Matrix

Definition: M is a binary matrix with t rows and n columns For any d + 1 edges E0, E1, …, Ed of H, there exists

a row containing E0 but not E1, …, Ed

Decoding Algorithm: Remove all negatives edges from the negative pools Remaining edges are positive

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Construction Algorithms

Consider a finite field GF(q). Choose k, s, and q:

Step 1: for each v in V

associate v with pv of degree k -1 over GF(q)

kqnqskrd and1)1(

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Step 2: Construct matrix Asxm as follows:

for x from 0 to s -1 (rkd <=s < q)

for each edge Ej inE

A[x,Ej] = PE(x) = {pv(x) | v in Ej}

E1 E2 Ej Em

0

1

A =

x PE2(x) PEj(x)

s-1

A Proposed Algorithm

)}(),(),({)(

then },,{

3212

3212

xpxpxpxP

vvvE

vvvE

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Step 3: Construct matrix Btxn from Asxm as follows:

for x from 0 to s -1

for each PEj(x)

for each vertex v in V

if pv(x) in PEj(x), then B[(x, PEj(x)),v] = 1

else B[(x, PEj(x)),v] = 0

E1 E2 Ej Em

0

1

A =

x PEj(x)

s-1

A Proposed Algorithm

v1 v2 vj vn

(0, PE0(0))

(0, PE1(0))

B =

(x, PEj(x))

(s-1, PEm(s-1))

(x)P (x)pjE2v

0 1

(x)P (x)p Ejv j

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Analysis

Theorem: If rd (k -1) + 1≤ s ≤ q, then B is d(H)-disjunct

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Proof of d(H)-disjunct Matrix Construction

Matrix A has this property: For any d + 1 columns C0, …, Cd, there exists a row

at which the entry of C0 does not contain the entry of Cj for j = 1…d

Proof: Using contradiction method. Assume that that row does not exist, then there exists a j (in 1…d) such that entries of C0 contain corresponding entries of Cj at least r(k-1)+1 rows. Then PEj(x) is in PE0(x) for at least r(k-1)+1 distinct values of x. This means that Ej is in E0

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Proof of d(H)-disjunct Matrix Construction (cont)

Prove B is d(H)-disjunct Proof: A has a row x such that the entry F in

cell (x, E0) does not contain the entry at cell (x, Ej) for all j = 1…d. Then the row <x,F> in B will contain E0 but not Ej for all j = 1…d

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