Reversible states of the functioning of regulatory circuits discrete models of gene networks Текст научной статьи по специальности «Математика»

2011

ВЕСТНИК ТОМСКОГО ГОСУДАРСТВЕННОГО УНИВЕРСИТЕТА Управление, вычислительная техника и информатика

№ 1(14)

ПРОЕКТИРОВАНИЕ И ДИАГНОСТИКА ВЫЧИСЛИТЕЛЬНЫХ СИСТЕМ

УДК 519.7

E.O. Kutumova, A.A. Evdokimov

REVERSIBLE STATES OF THE FUNCTIONING OF REGULATORY CIRCUITS DISCRETE MODELS OF GENE NETWORKS1

The paper describes reversible states of the functional graphs identified by the discrete models of the gene network regulatory circuits with special functions at its vertices. The recurrent formula of the reversible states number as well as its asymptotic behavior is found.

Keywords: gene network, discrete model, regulatory circuit, reversible state.

We consider one of the methods for description and modeling of gene networks in terms of discrete models of the regulatory circuits functioning, which are an extension of random models of genetic regulatory networks [1,2]. The regulatory circuit is represented as a connected digraph G(V,D), where V e 1,n is the set of vertices, identified with the products of genetic elements (RNA, proteins, etc.), and D is the set of edges associated with the regulatory relations. Circulant digraphs Gn,k, where (k - 1) is the number of incoming (and outgoing) edges, k < n , are considered. Variables taking integer values with the threshold p are corresponded to all vertices. Each value indicates the weight of a given vertex in a given moment of time and represents a concentration of the product, identified with the vertex. The regulatory circuit functioning is characterized by the stepwise changing of states - n-vectors in the alphabet <0,...,(p - 1)>. The paper describes reversible states that means vectors with incoming edges in the functional graphs of the regulatory circuits, and reports their number estimation depending on values of the parameters n, k and p. It is not our goal to provide an overview of various approaches to the modeling of gene networks. We only note that [3] include the sections devoted to the discrete approach to modeling of biochemical networks and the extensive bibliography concerning results of analysis of the networks functioning.

1. The problem of characterization of reversible states of the functional graphs

Gene networks have an important role in the living systems functioning. They are the basis for modeling the processes occurring in cells. A characteristic feature of its organization is their ability to regulate itself through the regulatory circuits with positive and negative feedbacks. These two types of circuits make it possible to maintain a certain functional state or switch to another state of a gene network including the switch under the influence of environmental factors [4].

1 The work was supported by the Russian Foundation for Basic Research (08-01-00671, 09-01-00070) and the interdisciplinary grant 119 of SB RAS.

The regulatory circuit is determined by specifying of a circulant digraph G„k (V, D) with the vertex set V and the edge set D, where V = {vi \i e1,n},

D ={(vi ,vi+j (mod„))\ i 6 1,„,j 6 1,k _ k 6 2 „ .

Expect v0 = v„ by the condition of cyclicity and denote Bp ={0,...,p-1} and

B„ = {(X„ )\ Xi 6 Bp , i 61, „}.

Define the functions fv : Bkp_ ^ Bp in every vertex vi 6 V, i 6 1, „. Let values of

these functions are calculated for variables Xj, j 6 i - k + 1(mod „), i - 1(mod „), assigned to the vertices with edges incoming into vi.

The gene network functioning is characterized by the changing of substance concentrations, i.e. changing of „-vectors (states) in Bp , corresponding to the values of fv

in „ graph vertices at every moment of time [5]. Thus, dynamics of state changes for each initial state is determined by the following mapping A(fv ,..., fv , p): B„ ^ B„.

1 „ F F

Let the same function f: Bkp^ Bp is defined for all graph vertices. We will use

the notation A(X) = Y for the vector X of variables and the vector Y of values assuming that the mapping A is defined for the function f and the threshold p.

Definition 1.1. The sequence of states X',..., Xr 6 Bp is called a cycle of the le„gth

r of the mapping A : B„p ^ B„p , if

|A(X’) = Xi+1, i 6~r, j Xr+1 = X'.

When r = 1, we have A(X1) = X1 and X1 is called a fixed poi„t of the mapping

A : B„p ^ Bnp .

Definition 1.2. The directed graph is called a fu„ctio„al graph of the regulatory circuit if its vertices correspond to the elements of B„p and the edge from a state X 6 B„p

goes to a state Y 6 B„p if and only if A(X) = Y.

Definition 1.3. The state Y 6 B„p is reversible for the mapping A : B„ ^ B„ if there exists the state X 6 Bp for which A(X) = Y. This corresponds to the edge (X, Y) from X to Y in the functional graph.

Let X = (Xj,...,x„) 6 Bn„ . The weight of X is a value ||X|| = ^^xi . Note that in the

i=1

case p = 2,

\\X © Y\\ = \\X\\ + \\Y\\ - 2 <X, Y> for all X, Y 6 B„ , where <X, Y> is the scalar product of vectors X and Y.

Suppose 8: Bnp ^ B„ is the operator of a seque„ce cyclic shift. Let 8+ is the right shift operator, that means 8+ (X) = Y for states X, Y 6 B„p if and only if yi+1(mod „) = Xt for all i 6 1, „, and 8_ is the left shift operator, that means 8_ (X) = Y if and only if

yi = Xi+1(mod „). Define the function f compared to the vertices of G„,k for arbitrary values of the parameters „, k and p by the following way

Xi + 1, if ^ Xj = 0 and Xi < p - 1, (1.1)

j6Di

Xi - 1, if ^ Xj > 0 and Xt > 0, (1.2)

j6Dt

xi, otherwise, (1.3)

f (xi-k+1(mod „),..., Xi-1(mod „))

where Di ={i - j(mod„) \ j 6 1, k -1} represents the number of vertices with edges incoming into vi. Therefore, we specified the discrete model of the gene network regulatory circuits with the functions.

2. Necessary and sufficient conditions for invertibility of functional graph reversible states

This section describes the reversible states of the mapping A : B„ ^ Bp . At the beginning we consider the case p = 2 and then the case of an arbitrary p. We establish that a state is reversible if and only if it does not contain occurrences of the certain kind subvectors. We call such subvectors by prohibitio„s.

2.1. Reversible states for p = 2

Statement 2.1. Whe„ k =2, all states of the functional graph of the mapping A : B„ ^ B„ are reversible.

Proof. Let Y = (yj,...,y„) is an arbitrary state. According to the definition of the mapping A : B„ ^ B„ with k = 2, we have A(X) = Yfor some state X if and only

Ixi +1, if xi-1 = 0 a„d xi = 0, xi -1, if xi-1 = 1 a„d xi = 1, xi, else.

Since p = 2, then xi +1 = xi -1 = xi = xi-1. So, it is necessary that yi = xi-1. Thus, A(X) = Yfor the state X = (y2,...,y„,y1) whose coordinates are obtained by the cyclic shift and negation of Y coordinates. That means the state Y is reversible for the mapping

A : B2„ ^ B2„ .

Theorem 2.1. When k > 3 the state Y = (yj,...,y„) is reversible for the mapping A : B„ ^ B„ if and only if Y does not contain the prohibitions Yr =(1,0r ,1) for all

r 61, k - 2 .

Proof. Necessity. Let the state Y contains a prohibition Yr for some r 61, k - 2. Without loss of generality we may assume that the occurrence of this prohibition begins with the first coordinate of Y, i.e. Y = (Yr,*,...,*), where *6{0,1} and X = xr+2 -1 = 1. Because Y is the reversible state of the mapping A, there exists the state X such that A(X) = Y. Since yi = 1, then according to (1.1) - (1.3) we obtain

X = (*,...,*,0k-1). (2.1)

Since yr+2 = 1, then similar

X = (0r+1,*,...,*,0k-r - 2). (2.2)

From (2.1) and (2.2) we derive

X = (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.

Sufficie„cy. Let the state Y does not contain the prohibitions Yr for all r 61, k - 2 . Noticing that Y = (1,...,1) is the state of the cycle (0,...,0) ^ (1,...,1) and, therefore, a reversible state, consider the case when Y contains zero coordinates. In this case it can

be represented as Y = (1s1, 0r1, 1s2, 0r2,..., 1sm ,0m) for some m > 1, where

X m=1(r- + si) = „ and ri > k -1, sf > 1. Form the state

X = (0s1 -1,1r1 -k+2,0s2 +k-2 ,..., 1rm-1 k+2,0sm +k-2,1rm ~k+2,0k-1).

It is implied from (1.1) - (1.3) that A(X) = Y and Y is a reversible state. Sufficiency is proved.

Corollary 2.1. The state Y 6 B„ is reversible for the mapping A : B„ ^ B„ if and

only if every block of units in Y is preceded by no less than (k - 1) zeros.

2.2. Reversible states for p > 3 Theorem 2.2. The state Y = (y1,..., y„) is reversible for the mapping A : B„p ^ B„p

if and only if Y does not contain the prohibitions (2.4) - (2.8).

1. (a, 0r, p - 1), where a 61, p -1, r 61, k - 2, (2.4)

2. (b, p - 1), where b 6 2, p -1, (2.5)

3. (a, 0r, 1s, p - 1), where a 61, p -1, r 6 1, k - 2, s 61, „ - k , (2.6)

4. (b, 1s, p - 1), where b 6 2, p -1, s 6 1,„ - k, (2.7)

5. (V-k+\ p - 1). (2.8)

Proof. Necessity. Let the state Y = (y1,...,y„) is reversible for the mapping

A : B„ ^ Bp , i.e. there exist the state X for which A(X) = Y. We will carry out the proof

by contradiction, considering all the prohibitions in order. It can be assumed that an occurrence of a prohibition in Y begins with the first coordinate.

1. Y = (a,0r, p -1,*,...,*), where a 61, p -1 and r 61, k - 2. Here and below

* 6 0, p -1. According to (1.1), (1.3) and the condition yr+2 = p -1 we have X = (0r+1,xr+ 2,*,...,*,0k-r-2), where xr+ 2 = p-2 or xr+2 = p-1. At the same time x1 = 0 and y1 = a > 0. So, it is necessary from (1.1) that X = (0r+1,xr+2,*,...,*,0k-) 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. Y = (b, p -1,*,...,*), where b 6 2, p -1. Since y1 = b > 2 and the mapping A applied to any states does not change coordinates of this state by more than one, then x1 > 1. From (1.2) we have the contradiction with the condition y2 = p -1. So, Y does not contain prohibition (2.5).

3. Y = (a,0r ,1s, p -1,*,...,*), where a 6 1, p -1, r 61, k - 2 and s 6 1, „ - k . Arguing as in the first item we derive X = (0s+r+1, xs+r+2,*,...,*), where xr+s+2 6 p -1, p - 2 and s + r < „ - k -1. So, we obtain the contradiction because A(X) ^ Y . Thus, Y does not contain prohibition (2.6).

4. Y = (b,1s,p-1,*,...,*), where b 6 2,p-1 and s 6 1,„-k . The condition y1 = 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 < p -1. We have the contradiction. That means Y does not contain prohibition (2.7).

5. Y = (1„-k+\p -1,*,...,*). Since y1 = 1 and X„=n_k+^j > y„-k+2 = p -^ then X1

= 2. Similar x2 =...= x„-k+1 = 2 . We conclude the contradiction as in the item 4. Therefore, Y does not contain prohibition (2.8).

Necessity is proved.

Sufficie„cy. Let the state Y = (y1,...,y„) does not contain prohibitions (2.4) - (2.8). Show that A(X) = Y for some state X. Let m is the number of Y coordinates which are

equal to p - 1. We have m < L„J , hence, the following cases are considered below: m =

k

0, m = 1 and 2 < m < \_„J .

k

1. m = 0. The condition yi < p - 2 implies that X < p -1. So, it is correctly to form the state X by the following way X = (x1,..., x„) = (y1 +1,..., y„ +1). Further A(X) = Y from (1.2) because xi > 1 for all i 6 1, „.

2. m = 1. Without loss of generality it can be assumed that i = „, i.e. y„ = p -1. So far as Y does not contain prohibition (2.5), we conclude that y„-1 6 {0,1} .

When y„-1=0, prohibition (2.4) implies y„-k+1 = y„-k+2=...= y„-1=0, ie. X = (y1,..., y„-k ,0k-1, p-1), where y1,..., y„-k 6 0, p - 2. Since y1 < p - 2 for all i 61, „ - k, then it is correctly to form the state X = (y1 +1,..., y„-k + 1,0k-1, p - 1).Thus, A(X) = Y from conditions (1.2) - (1.3).

Consider the case y„-1 =1. Let l is the index of the rightmost but not the last coordinate of Y, which is not equal to 1. From prohibition (2.8) we have l > k -1 and further (2.7) implies yt =0. So, according to (2.6) yt-k+2 =...= yt =0 and, therefore,

Y = (y1,...,yt-k+1,0k-1A„-1 -1, p -1), where yj 6 0, p - 2 for j 6 1,l -k +1. If l = k - 1, then (1.1) - (1.3) induce A(X) = Y for state X = (0„-1,p -1). If l > k, we form the state X = (y1 +1,..., yl - k+1 +1,0„+k-I - 2, p -1) and according to (1.2) - (1.3) obtain A(X) = Y.

3. 2 < m < L—J . As in the item 2 the state Y contains coordinates equal to (p - 1) if

k

and only if the subvector (0k 1,1S), 5 e 0, n - k, precedes them. Form Y by follows

Y = (* * Y2 * *Y)

where * e 0, p - 2 and Yj is one of the subvectors (0k 1,15, p -1) such that

Xm_l(ij +1 Yj |) = n for all j e 1,m . Now form the state

X = (•,...,• ,0|Y1 \p-1,•,...,• ,0|Y2' \p -1,...,•,...,• ,0|Ym 1 \p-1),

i1 i2 im

with the symbol {•} denoting coordinates of the state Y increased by one. From (1.1) -

(1.3) we obtainA(X) = Y. Sufficiency is proved.

Corollary 2.2. The state Y e Bnp is reversible for the mapping A : Bnp ^ B— if and

only if the subvector (0k-1,15), 5 e 1, n - k, precedes all Y coordinates equal to (p - 1).

Theorem 2.3. All connected components of the functional graph of the mapping A : B— ^ Bnp are cycles if and only ifp = k = 2.

Proof. Necessity follows from theorems 2.1 and 2.2, and sufficiency follows from statement 2.1.

3. The number of reversible states for p = 2

In the case p = 2 the mapping A : B— ^ B— is defined on the set B— of all binary states. According to formulas (1.1) - (1.3) the mapping A applied to any state X e B— gives the state X' e B— whose coordinates are calculated by formulas

1, if X = 0, (3.1)

jeDi

0, if X Xj > 0, (3.2)

jeDt

As before Di is the numbers of (k - 1) vertices which have the output edges leading to the vertex with the number i e 1, — in the graph Gn k, k < — .

If k = 2 then by theorem 2.3 the set of reversible states coincides with the set B— . Therefore, their number in this case is equal to 2—.

3.1. The recurrent formula for the reversible states number in the case k > 3

Assume further k > 3 . The state X = (Xj,...,x—) is called a (k, —)-state if it does not

contain the linear occurrences of subvectors (1,0r ,1) for all r e1, k - 2. Denote the subset of such states in B— by Akn and note that Ak+1 n cAk— for all values of — and k.

Define the function Q : K^-K calculating the (k, n)-states as follows. For any fixed value k e 3,n

Q(0) = 1, Q(1) = 2, Q(2) = 4 and Q(n) = | AM |.

Lemma 3.1. For all n > 3 the function Q(n) satisfies the recurrent relation

Q(n) = 2Q(n - 1) - Q(n - 2) + Q(n - k). (3.3)

Proof. The proof is by the induction on n.

The induction base. n = k = 3. In this case the value Q(3) is equal to the number of states of the length 3 that do not contain the linear occurrences of the subvector (1,0,1). There are exactly 23 - 1 = 7 of such states. Whereas relation (3.3) gives Q(3)=2Q(2) -Q(1) + Q(0) = 7.

The induction step. Suppose that Q(n) = 2Q(n - 1) - Q(n - 2) + Q(n - k) for any fixed value k e 3,n . Fix k e 3,n +1 and find the value Q(n + 1). In the case k e 3,n we have

Ak, n+1 c {X = (X1,..., xn ,*)|* e 0,1,(X1,..., xn) e Ak, n }.

Denote by Q,(s), i e{0,1}, s e 3,n +1, the number of states in Aks with the last coordinate equal to i. Then

Q0(s) = | Ak,s-11 = Q(s - 1) and Q^s) = Q(s) - Q(s - 1). (3.4)

Because only when x— = 1, x„-r+i = .. .= xn = 0 and x„+i = 1 for any r e 1, k - 2, then X = (x1,..., xn, xn+1) g Ak, n+1 for the subvector (x1,..., xn) eAk ,n, therefore,

Q(n + 1) = | Ak,n+11 = 2| Ak — | - Q:(n - 1) -• • •- Q:(n - k + 2). (3.5)

From (3.4) and (3.5) we obtain

Q(n + 1) = 2Q(n) - Q(n - 1) + Q(n + 1 - k). (3.6)

Now let k = n + 1. So far as An+1 n+1 c A nn+1, then deleting the state (1,0n-1,1) from the set An n+1 and taking into account (3.6) we conclude

| An+1,n+11 = | An,n+11 - 1 = 2Q(n) - Q(n - 1) + Q(1) - Q(0) = 2Q(n) - Q(n - 1) + Q(0).

Lemma 3.2. The number of (k, n)-states with the first and last coordinates equal to 1 is given by the value F(n) = Q(n - k).

Proof. The number of (k, n)-states of the type (0,*,...,*,0), where *e{0,1}, is equal to Q(n - 2). The number of (k, n)-states of the type (0,*,.,*,1) is equal to the number of (k, n)-states of the type (1,*,...,*,0) and is given by the expression Q(n - 1) - Q(n - 2). Therefore, the number of (k, n)-states of the type (1,*,.,*,1) is equal to

F(n) = Q(n) - Q(n - 2) - 2(Q(n - 1) - Q(n - 2)) = Q(n) - 2Q(n - 1) + Q(n - 2). (3.7)

According to lemma 3.1 Q(n) = 2Q(n - 1) - Q(n - 2) + Q(n - k). Thus, from (3.7) we obtain F(n) = Q(n - k).

Lemmas 3.1 and 3.2 provide the number F(n) to satisfy the recurrent relation

F(n) = 2F(n - 1) - F(n - 2) + F(n - k). (3.8)

Denote the number of reversible states of the mapping A : Bnp ^ Bnp by H(n, k, p).

Lemma 3.3. The number H(n, k, 2) of reversible states of the mapping A : B?, ^ Bn is equal to H(n, k,2) = k • F(n +1) - (k - 2) • F(n), where F(n) satisfies the recurrent relation (3.8).

Proof. By theorem 2.1 the value H(n, k, 2) is equal to the number of n-states which do not contain subvectors (1,0r,1) for all r e 1,k -2 . Meanwhile lemma 3.1 gives the recurrent relation for the number Q(n) of (k, n)-vectors which do not contain the linear occurrences of the subvectors (1,0r,1) for any r e 1, k - 2 . Therefore, in order to find H(n, k, 2), all states of the type (O'5,1,*,...,*,1,0r-5), where * e {0,1}, 5 e 0,r,

r e 1,k -2, should be deleted from the set Ak n . By lemma 3.2 the number of such

states for any r e 1, k - 2 is equal to (r +1) • F(n - r), where the coefficient (r + 1) appears due to the parameter 5 e 0,r . Hence, we obtain the following sequence of equalities

k-2

H(n, k,2) = Q(n) - X (r +1) • F(n - r) =

r=1

k-2

= Q(n) -X (r +1) • (Q(n - r) - 2Q(n - r -1) + Q(n - r + 2)) =

r=1

k - 2 k - 2 k - 2

= Q(n)-X(r +1) • Q(n - r) + X2 • (r +1) • Q(n - r -1) -X(r +1) • Q(n - r - 2) =

r =1 r =1 r =1

= Q(n) - X(r + 1) • Q(n - r) + X2r • Q(n - r) - X(r -1) • Q(n - r) =

r=1 r=2 r=3

= Q(n) - 2Q (n -1) - 3Q(n - 2) - X(r +1) • Q(n - r) + 4Q(n - 2) + 2 • (k -1) • Q(n - k +1) +

r=3

k-2 k-2

+X2r • Q(n - r) - (k - 2) • Q(n - k +1) - (k -1) • Q(n - k) -X(r -1)-Q(n - r) =

r=3 r=3

= Q(n) - 2Q(n -1) + Q(n - 2) + k • Q(n - k +1) - (k -1) • Q(n - k).

Since by lemma 3.1 Q(n) = 2Q(n -1) - Q(n - 2) + Q(n -k), we conclude

H(n, k,2) = k • F(n +1) - (k - 2) • F(n).

3.2. The reversible states number asymptotic

for k>3

Theorem 3.1 The asymptotic behavior of the function H(n, k, 2) for the fixed parameter k > 3 is given by the relation H(n,k,2) ~ ckRn, where ck are the constants depending only on k and 1 < R < 2 is the largest by module real root of the characteristic equation

Xk -2Xk-1 +Xk-2 -1 = 0 (3.9)

of recurrent relation (3.8).

Proof. Since (3.8) is the linear recurrent relation of k order with constant coefficients, its solution is sought in the form fn = cXn with constants c and X. This leads to the characteristic equation Xk - 2Xk-1 + Xk-2 -1 = 0, k roots of which give solutions of the form X” for i e 1, k. Show that these roots are different. Denote

g (X) = Xk - 2Xk-1 + Xk-2 -1. Then

g (X) = k Xk-1 - 2(k - 1)Xk-2 + (k - 2)Xk-3.

- k - 2 So far as g (X) = 0 if and only if X = 0, X = 1 or X =------------- and these values are

k

not the roots of g(X), therefore, all roots of equation (3.9) are different. Thus, k roots of

equation (3.9) form the basis of the solution space of recurrent relation (3.8) whose gen-

k

eral solution can be written as F (n) = X XT. Analyze the asymptotic behavior of the

i=1

sequence F(1), F(2),... It is sufficient to consider the term ct X” with the largest absolute value of the root and the coefficient ct ^ 0 . Denote this root by R. Then the asymptotic behavior of the function F(n) is given by the relation F(n) ~ cRn, where c is constant. In view of lemma 3.3 this implies H(n,k,2) ~ ckRn, where ck is a constant depending only on k.

It is remain to show that 1 < R < 2 is the real root of (3.9). Let X = RX (cos^ + i sin^) is an arbitrary root of equation (3.9). Then Xk-2(X-1)2=1 and

| X-1|2 = —. On the other hand

| Xk-2 | rX-2

| X-1|2= RX2 -2RX cos ^ + 1 and (RX -1)2 <|X-1|2 < (RX +1)2.

So, we obtain

(RX-1)2 <-1T < (Rx+ 1)2. (3.10)

rX

Since g (1) < 0 and g(2) > 0 for k > 3 , there exist the real root of equation (3.9) which is greater than 1 by the function g (X) continuity on the real axis. Meanwhile (3.10) implies that there are not roots less than -1. Thus, R > 1. Then X reaches the

maximum absolute value R = RX at the minimum value of 1 _ which is equal to

RXk-2

(RX -1)2, i.e. when cos^ = 1 and sin^ = 0 . Hence, R is the real root of equation (3.9). The estimation R < 2 is obvious because the total number of n-states forp = 2 is 2n.

Conclusion

Investigation of the regulatory circuits functioning represented in this paper provides an opportunity to understand the regulatory mechanisms of the processes under the control of gene networks and possibility for a directional impact on them.

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Кутумова Елена Олеговна

Евдокимов Александр Андреевич

Институт математики СО РАН

E-mail: e.o.kutumova@gmail.com; evdok@math.nsc.ru Поступила в редакцию 26 октября 2010 г.