Reversible states of the functioning of regulatory circuits discrete models of gene networks.pdf We consider one of the methods for description and modeling of gene networks interms of discrete models of the regulatory circuits functioning, which are an extensionof random models of genetic regulatory networks [1,2]. The regulatory circuit is representedas a connected digraph G(V,D), where e V1,n is the set of vertices, identifiedwith the products of genetic elements (RNA, proteins, etc.), and D is the set of edges associatedwith the regulatory relations. Circulant digraphs Gn,k, where (k - 1) is the numberof incoming (and outgoing) edges, k≤n, are considered. Variables taking integervalues with the threshold p are corresponded to all vertices. Each value indicates theweight of a given vertex in a given moment of time and represents a concentration ofthe product, identified with the vertex. The regulatory circuit functioning is characterizedby the stepwise changing of states - n-vectors in the alphabet . Thepaper describes reversible states that means vectors with incoming edges in the functionalgraphs of the regulatory circuits, and reports their number estimation dependingon values of the parameters n, k and p. It is not our goal to provide an overview of variousapproaches to the modeling of gene networks. We only note that [3] include thesections devoted to the discrete approach to modeling of biochemical networks and theextensive bibliography concerning results of analysis of the networks functioning.1. The problem of characterization of reversible states of the functional graphsGene networks have an important role in the living systems functioning. They arethe basis for modeling the processes occurring in cells. A characteristic feature of its organizationis their ability to regulate itself through the regulatory circuits with positiveand negative feedbacks. These two types of circuits make it possible to maintain a certainfunctional state or switch to another state of a gene network including the switchunder the influence of environmental factors [4].The regulatory circuit is determined by specifying of a circulant digraphGn,k (V, D) with the vertex set V and the edge set D, where = { | V vi i1,n},D= {(vi,vi+j(modn)) |i1,n,j1,k−1}, k 2,n.Expect v0 = vn by the condition of cyclicity and denote Bp = {0,...,p−1}andn = {( 1,..., ) | , 1, }Bp x xn xiBpi n.Define the functions : k 1vi p pf B− B in every vertex vi V, i1,n. Let values ofthese functions are calculated for variables xj, ji−k+1(mod n),i−1(mod n) , assignedto the vertices with edges incoming into vi.The gene network functioning is characterized by the changing of substance concentrations,i.e. changing of n-vectors (states) in npB , corresponding to the values of vifin n graph vertices at every moment of time [5]. Thus, dynamics of state changes foreach initial state is determined by the following mapping1( ,..., , ) : n n.v vn p pAf f p B B Let the same function : k 1f Bp− Bp is defined for all graph vertices. We will usethe notation A(X) = Y for the vector X of variables and the vector Y of values assumingthat the mapping A is defined for the function f and the threshold p.Definition 1.1. The sequence of states 1,..., r nX X Bp is called a cycle of the lengthr of the mapping : n nABpBp, if11 1( )= , 1, ,= .i irAX X i rX X++⎧ ⎨⎩When r = 1, we have A(X1) = X1 and X1 is called a fixed point of the mapping: n nABpBp.Definition 1.2. The directed graph is called a functional graph of the regulatory circuitif its vertices correspond to the elements of npB and the edge from a state npX  Bgoes to a state npYB if and only if A(X) = Y.Definition 1.3. The state npYB is reversible for the mapping : n nABpBp if thereexists the state npX  B for which A(X) = Y. This corresponds to the edge (X, Y) from Xto Y in the functional graph.Let = ( 1,..., ) nX x xnBp. The weight of X is a value1niiX x== ƒ . Note that in thecase p = 2,||X ⊕ Y|| = ||X|| + ||Y|| - 2 for all , npX Y B  , where is the scalar product of vectors X and Y.Suppose : n nƒBpBp is the operator of a sequence cyclic shift. Let ƒ+ is the rightshift operator, that means ( )= X Y + ƒ for states , npX YB if and only if yi+1(mod n) = xifor all i1,n, and ƒ− is the left shift operator, that means ƒ−(X)=Y if and only ifyi = xi+1(mod n). Define the function f compared to the vertices of Gn,k for arbitrary valuesof the parameters n, k and p by the following way1(mod ) 1(mod f(xi−k+ n ,...,xi− n) )⎧⎪⎪=⎨⎪⎪⎩xi + 1, if 0ijj Dxƒ = and xi < p - 1, (1.1)xi - 1, if 0ijj Dxƒ > and xi > 0, (1.2)xi, otherwise, (1.3)where Di = {i−j(modn) | j1, k −1} represents the number of vertices with edges incominginto vi. Therefore, we specified the discrete model of the gene network regulatorycircuits with the functions.2. Necessary and sufficient conditions for invertibilityof functional graph reversible statesThis section describes the reversible states of the mapping : n nABpBp. At the beginningwe consider the case p = 2 and then the case of an arbitrary p. We establish thata state is reversible if and only if it does not contain occurrences of the certain kind subvectors.We call such subvectors by prohibitions.2 . 1 . R e v e r s i b l e s t a t e s f o r p = 2Statement 2.1. When k = 2, all states of the functional graph of the mappingA:B2nB2n are reversible.Proof. Let Y= (y1,...,yn ) is an arbitrary state. According to the definition of themapping A:B2nB2n with k = 2, we have A(X) = Y for some state X if and onlyif111, = 0 = 0,= 1, =1 =1,, .i i ii i i iix ifx and xy x if x and xx else−−+ ⎧⎪− ⎨⎪⎩Since p = 2, then 1 1= 1= = i i i i x x x x− + − . So, it is necessary that yi=xi−1 . Thus,A(X) = Y for the state X = (y2,...,yn ,y1) whose coordinates are obtained by the cyclicshift and negation of Y coordinates. That means the state Y is reversible for the mappingA:B2nB2n.Theorem 2.1. When k ≥ 3 the state Y= (y1,...,yn ) is reversible for the mappingA:B2nB2n if and only if Y does not contain the prohibitions = (1,0r ,1)Yr for allr1,k−2 .Proof. Necessity. Let the state Y contains a prohibition Yr for some r1,k−2 .Without loss of generality we may assume that the occurrence of this prohibition beginswith the first coordinate of Y, i.e. Y= (Yr ,*,...,*) , where *{0,1} andx1=xr+2−1=1. Because Y is the reversible state of the mapping A, there exists thestate X such that A(X) = Y. Since y1 = 1, then according to (1.1) - (1.3) we obtainX = (*,...,*,0k −1). (2.1)Since yr+2 = 1, then similarX = (0r+1,*,...,*,0k−r −2). (2.2)From (2.1) and (2.2) we deriveX = (0r+1,*,...,*,0k−1). (2.3)So, we have a contradiction with (2.3) because 1r 2+  Yr for r > 0,A(X) = (1r +2 ,*,...,*) Y. Necessity is proved.Sufficiency. Let the state Y does not contain the prohibitions Yr for all r1,k−2 .Noticing that Y = (1,…,1) is the state of the cycle (0,…,0) ↔ (1,…,1) and, therefore, areversible state, consider the case when Y contains zero coordinates. In this case it canbe represented as =(11,01,12,02,...,1 ,0 ) Y s r s r sm rm for some m ≥ 1, where=1( )= mi i i ƒ r +s n and ri ≥k−1 , si ≥ 1. Form the state= (011,11 2,02 2,...,1 1 2,0 2,1 2 ,0 1). X s r k s k rm k smk rm k k − − + + − − − + + − − + −It is implied from (1.1) - (1.3) that A(X) = Y and Y is a reversible state. Sufficiencyis proved.Corollary 2.1. The state 2YBn is reversible for the mapping A:B2nB2n if andonly if every block of units in Y is preceded by no less than (k - 1) zeros.2 . 2 . R e v e r s i b l e s t a t e s f o r p ≥ 3Theorem 2.2. The state Y= (y1,...,yn ) is reversible for the mapping : n nABpBpif and only if Y does not contain the prohibitions (2.4) - (2.8).1. (a, 0r, p - 1), where a1,p−1, r1,k−2 , (2.4)2. (b, p - 1), where b2,p−1 , (2.5)3. (a, 0r, 1s, p - 1), where a1,p−1, r1,k−2 , s1,n−k, (2.6)4. (b, 1s, p - 1), where b2,p−1 , s1,n−k, (2.7)5. (1n−k+1, p - 1). (2.8)Proof. Necessity. Let the state Y= (y1,...,yn ) is reversible for the mapping: n nABpBp, i.e. there exist the state X for which A(X) = Y. We will carry out the proofby contradiction, considering all the prohibitions in order. It can be assumed that an occurrenceof a prohibition in Y begins with the first coordinate.1. Y= (a,0r ,p−1,*,...,*) , where a1,p−1 and r1,k−2 . Here and below*0,p−1. According to (1.1), (1.3) and the condition yr +2=p−1 we have1 2= (0r, 2 ,*,...,*,0kr ),X +xr −−+ where xr+2=p−2 or xr+2=p −1. At the same timex1 = 0 and y1=a>0. So, it is necessary from (1.1) that 1 1= (0r, 2 ,*,...,*,0k)X +xr −+and a = 1. Therefore, A(X) = (1r +1,p−1,*,...,*)Y, and we obtain the contradiction.That means Y does not contain prohibition (2.4).2. = ( , 1,*,...,*Y bp− ) , where b2,p−1 . Since y1=b≥2 and the mapping A appliedto any states does not change coordinates of this state by more than one, thenx1 ≥ 1. From (1.2) we have the contradiction with the condition y2=p−1. So, Y doesnot contain prohibition (2.5).3. Y= (a,0r,1s,p−1,*,...,*) , where a1,p−1, r1,k−2 and s1,n−k. Arguingas in the first item we derive 1= (0s r , 2,*,...,*),X + + xs r+ + wherexr+s+2 p−1,p−2 and s +r≤n−k−1. So, we obtain the contradiction becauseA(X)Y. Thus, Y does not contain prohibition (2.6).4. Y= (b,1s ,p−1,*,...,*) , where b2,p−1 and s1,n−k. The conditiony1=b≥2 according to (1.2) implies the expressions x1 ≥ 1 and x2=...=xs+1= 2 . Then(1.2) leads to the inequality ys+2

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