Computational Molecular Biology Pooling Designs – Inhibitor Models

My T. Thai 2 An Inhibitor Model  In sample spaces, exists some inhibitors  Inhibitor = anti-positive  (Positives + Inhibitor) = Negative _ + _ _ _ _ _ x Inhibitor Negative +

My T. Thai 3 An Example of Inhibitors

My T. Thai 4 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)

My T. Thai 5 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

My T. Thai 6 Preliminaries

My T. Thai 7 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

My T. Thai 8 2-stages Algorithm At this stage, the problem become the e-error- correcting problem.

My T. Thai 9 Non-adaptive Solution (1 stage) 1.P contains all positives 2.N contains all negatives 3.O contains all inhibitors and no positives

My T. Thai 10 Non-adaptive Solution

My T. Thai 11 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

My T. Thai 12 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

My T. Thai 13 What is Hypergraph H?  H = (V, E ) where:  V is a set of n vertices (items)  E a set of m hyperedges E j where E j is a subsets of V  Rank: r = max {| E j | s.t E j in E }

My T. Thai 14 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 E i E j

My T. Thai 15 d(H)-disjunct Matrix  Definition:  M is a binary matrix with t rows and n columns  For any d + 1 edges E 0, E 1, …, E d of H, there exists a row containing E 0 but not E 1, …, E d  Decoding Algorithm:  Remove all negatives edges from the negative pools  Remaining edges are positive

My T. Thai 16 Construction Algorithms Consider a finite field GF(q). Choose k, s, and q: Step 1: for each v in V associate v with p v of degree k -1 over GF(q)

My T. Thai 17 Step 2: Construct matrix A sxm as follows: for x from 0 to s -1 (rkd <=s < q) for each edge E j in E A[x,E j ] = P E (x) = {p v (x) | v in E j } E 1 E 2 E j E m 0 1 A = x P E2 (x) P Ej (x) s-1 A Proposed Algorithm

My T. Thai 18 Step 3: Construct matrix B txn from A sxm as follows: for x from 0 to s -1 for each P Ej (x) for each vertex v in V if p v (x) in P Ej (x), then B[(x, P Ej (x)),v] = 1 else B[(x, P Ej (x)),v] = 0 E 1 E 2 E j E m 0 1 A = x P Ej (x) s-1 A Proposed Algorithm v 1 v 2 v j v n (0, P E0 (0)) (0, P E1 (0)) B = (x, P Ej (x)) (s-1, P Em (s-1)) 01

My T. Thai 19 Analysis  Theorem: If rd (k -1) + 1≤ s ≤ q, then B is d(H)-disjunct

My T. Thai 20 Proof of d(H)-disjunct Matrix Construction  Matrix A has this property:  For any d + 1 columns C 0, …, C d, there exists a row at which the entry of C 0 does not contain the entry of C j 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 C 0 contain corresponding entries of C j at least r(k-1)+1 rows. Then P Ej (x) is in P E0 (x) for at least r(k- 1)+1 distinct values of x. This means that E j is in E 0

My T. Thai 21 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, E 0 ) does not contain the entry at cell (x, E j ) for all j = 1…d. Then the row in B will contain E 0 but not E j for all j = 1…d