Subject Categories: Biological Science --- Synthetic biology, Signal Transduction Supplemental Materials Reverse engineering of bacterial chemotaxis pathway via frequency domain analysis Junjie Luo1, Jun Wang2, Ting Martin Ma1, and Zhirong Sun1* 1 Department of Biological Sciences and Biotechnology, Tsinghua University, Ministry of Education Key Laboratory of Bioinformatics, Beijing, P.R.China 100084 2 Department of Computer Science, Tsinghua University, Beijing, P.R.China 100084 *Correspondent author: Zhirong Sun, Department of Biological Sciences and Biotechnology, Tsinghua University, Ministry of Education Key Laboratory of Bioinformatics, Room 216, New Biological Building, Tsinghua University, Beijing, P.R. China 100084 E-mail: sunzhr@mail.tsinghua.edu.cn Tel: +86-10-62772237 Fax: +86-10-62772237 Transfer Function of Pathway near Equilibrium The ordinary differential equation model for a signaling pathway can be written in following standard form: dx F ( x , u ); y G ( x , u ); (0.0.1) dt u is the input of this pathway, which can be the concentration of ligand or activated receptor. x is the molecular concentration in this pathway. y is pathway output, represents concentration of CheY-P in our research. Set u0 , x0 , y0 are the equilibrium values of u , x , y . So the three values obey equation group(1.1.2). (0.0.2) 0 F ( x0 , u0 ); y0 G( x0 , u0 ); One order Taylor expansion of equation group (0.0.1) is: F F d (x x0 ) F ( x0 , u0 ) x j u j o(x ) o(u ); dt j x j x x0 , j u j ux ux0 , u u0 0 G G y y0 G ( x0 , u0 ) x j u j o(x ) o(u ); j x j x x0 , j u j ux ux0 , u u0 0 (0.0.3) Where u u u0 , x x x0 , y y y0 . Use equation group (0.0.2) to simplify above equation group and ignore high order term of it, we can get following linear equation group: d x Ax Bu ; y C x Du (0.0.4) dt F F Gi Gi In which, Aij i ; Bij i ; Cij ; Dij x j u j x j u j x x ,u u x x ,u u x x ,u u x x ,u u 0 0 0 0 0 0 0 0 We will get the following formula from the Laplace transformation of (0.0.4): sx Ax Bu ; y C x Du (0.0.5) Transfer function describing the dynamic relationship between u and y can be calculated from these equations by following formula: (0.0.6) H (s) C ( sI A)1 B D In our research, we use MATLAB function “fsolve” to solve the equilibrium equation group (0.0.2). This function also return the Jacobi matrix at equilibrium, which just correspond A, B, C, D in (0.0.4). This method enables us to get the transfer function directly from the standardized differential equation model shown in (0.0.1). The ordinary differential of equation group (0.0.2) is: F 0 i j x j F dx j 0 i x x0 , j u j u u0 G dyi 0 i j x j du j 0 ; x x0 , u u0 G dx j 0 i x x0 , j u j u u0 du j 0 ; x x0 , u u0 (0.1.1) That is: 0 Adx0 Bdu0 dy0 Cdx0 Ddu0 Eliminate dx0 in above equation group: (0.1.2) dy0 (CA1 B D)du0 (0.1.3) If we solve equation group (0.0.2), we can get y0 as a function of u0. The partial differential of this function is: (0.1.4) yi 0 (u0 ) CA1 B D H ij (0) ij u j 0 Therefore, yi0 is not relative to uj0 is equivalent to Hij(0)=0. This provides a litmus stone to test whether a pathway is adaptive. Systematic Design a Pathway Basing on Defined Transfer Function The transfer function of a two-molecule pathway Suppose a pathway is only made up of two molecules: receptor and CheY, whose concentrations are u and y. The dynamic of it can be described by following differential equation: dy (C11 C1u )( yt y ) ( D11 D1u ) y (1.1.1) dt Let Eij=Cij+Dij. The above equation can be simplified into: dy (C11 C1u ) yt ( E11 E1u ) y (1.1.2) dt The equilibrium of this equation is: C C1u0 (1.1.3) y0 11 yt E11 E1u0 Calculation Taylor Series and Laplace Transformation near the equilibrium, the differential equation (1.1.2) is approximated into following linear form: sy C1uyt E1uy0 ( E11 E1u0 )y (1.1.4) Where Δu=u-u0; Δy=y-y0. Therefore, the transfer function of the pathway is: C1 yt E1 y0 y (1.1.5) H ( s) u s ( E11 E1u0 ) This transfer function is one rank, thus the pathway it represents can only be a low-pass filter with cut-off frequency E11+E1u0. And this pathway is not adaptive. The transfer function of three-molecule pathway Suppose a pathway is made up of molecule u, v and CheY. The dynamic of it can be described by following differential equation group: dv dt (C11 C12 y C1u )(vt v) ( D11 D12 y D1u )v (1.2.1) dy (C v C C u )( y y ) ( D v D D u ) y 21 22 2 t 21 22 2 dt Let Eij=Cij+Dij. The above equation group can be simplified into: dv dt (C11 C12 y C1u )vt ( E11 E12 y E1u )v (1.2.2) dy (C v C C u ) y ( E v E E u ) y 21 22 2 t 21 22 2 dt Calculation Taylor Series and Laplace Transformation near the equilibrium (u0, v0, y0), the differential equation group (1.2.2) is approximated into following linear form: sv (C12 y C1u )vt ( E12 y E1u )v0 ( E11 E12 y0 E1u0 )v (1.2.3) sy (C21v C2 u ) yt ( E21v E2 u ) y0 ( E21v0 E22 E2u0 )y Where Δu=u-u0; Δv=v-v0; Δy=y-y0. Substitute H(s)Δu for Δy, and eliminate Δu and Δv in (1.2.2): s ( E11 E12 y0 E1u0 ) (C12 H (s) C1 )vt ( E12 H (s) E1 )v0 (1.2.4) C21 yt E21 y0 sH (s) C2 yt E2 y0 ( E21v0 E22 E2u0 )H (s ) Therefore, the transfer function of the pathway is: s ( E11 E12 y0 E1u0 ) C2 yt E2 y0 C21 yt E21 y0 C1vt E1v0 H ( s) s ( E11 E12 y0 E1u0 ) s ( E21v0 E22 E2u0 ) C21 yt E21 y0 C12 vt E12 v0 (1.2.5) This transfer function is two-rank, thus the pathway it represents can be a band-pass filter or a low-pass filter. Let C1vt=k1, E11=dv, C21=E21=k3, E2=k2, and other Cij and Eij equal to 0. The equilibrium and transfer function can be simplified into: d v k2 y0 yt 1 k1 k3 (1.2.6) k y 1 d k k k s y 2 t v 2 1 3 H (s) u s d v s u0 k2 k3 k1 d v This is the positive chemotaxis pathway we design. Similarly, we can design a negative pathway by letting C1vt=k1, E11=dv, C2=E2=k2, E21=k3, and other Cij and Eij equal to 0: k1 k3 y0 yt 1 d v k2 (1.2.7) k2 yt 1 dv k2 k1k3 s H ( s ) y u s d v s u0 k2 k3 k1 d v And a model of pseudochemotaxis pathway, by letting C1vt=k1, E11=dv, C22=E22=k2, E21=k3, and other Cij and Eij equal to 0: k1 k3 u0 y0 yt 1 d k v 2 (1.2.8) y 1 k k u d k y t 3 1 0 v 2 H (s) u s dv s k2 u0 k3 k1 dv In our simulation, k1=dv=0.02s-1, k2=k3=20μM-1s-1(in (1.2.8), k2=10.9s-1), so the chemotaxis effects of the three pathway are comparable. And in (1.2.8), dv and (k2+u0k3k1/dv), the two polar points of the transfer function correspond to the speed of CheZ mRNA degradation and CheY-P dephosphorylation in the pseudochemotaxis pathway. For this reason, this model can simulate the pathway build by Topp S and Gallivan JP (Topp & Gallivan, 2007). Transfer function of a pathway with discretionary number of molecules Similarly, the differential equation describing the dynamic of the pathway can be written in following form (Goldstein & Soyer, 2008): dyi (1.3.1) (Cii Cij y j Ci u )( yit yi ) ( Dii Dij y j Di u ) yi dt j i j i Where u is the receptor occupation, yti is the total concentration of other proteins in the pathway except the receptor, and yi is the concentration of activated protein i. Cij (Dij) is the rate at which activated protein j activates (deactivates) protein i, Cii (Dii) is protein i’s rate of self-activation (deactivation), Ci (Di) represents the rate at which the receptor activates (deactivates) the protein i. Let Eij=Cij+Dij. The above equation group can be simplified into: dyi (1.3.2) (Cii Cij y j Ci u ) yit ( Eii Eij y j Ei u ) yi dt j i j i Calculation Taylor Series and Laplace Transformation near the equilibrium (u0, v0, y0), the differential equation group (1.3.2) is approximated into following linear form: syi (Cij yit Eij yi 0 )y j ( Eii Eij y j 0 Ei u0 )yi (Ci yit Ei yi 0 )u j i (1.3.3) j i Where Δu=u-u0; Δyi=yi-yi0. Let Aij=Cijyit-Eijyi0, (i≠j); Aii=-(Eii+∑k≠iEikyk0+Eiu0); Bi1=Ciyit-Eiyi0; ΔY=(y1,y2,…,yn)T. (1.3.4) sY AY Bu Therefore, the transfer function of the pathway is: H ( s ) (0, , 0,1) Y / u (0, , 0,1)( sI A) 1 B A1, n 1 A1n s A11 A12 A21 det An 2,n 1 An 1,n 2 s An 1,n 1 An 1,n An 1,1 (1.3.5) B Bn1 11 A1n s A11 A12 A21 det An 1,n An ,n 1 s Ann An ,1 The number of the molecules in the pathway is (n+1) (yi (i=1 to n), and u). Equation (1.3.5) shows that the rank of the transfer function is n, and the highest power of s in numerator is (n-1). Consequently, if we want to design an adaptive pathway and make the transfer function of it contain a differential link, n should be 2 at least. That is the smallest pathway fulfilling above properties has 3 molecules. The Model for Molecular Motor in E. coli We assume the change of molecular motor in E. coli is a Markov process (Block et al, 1983). That is the probability of motor rotation state at a time point is determined by the probability at the time point before. In following derivation, we use 0 to represent clockwise (CW) rotation and 1 to represent count-clockwise (CCW) rotation. The state transfer matrix of this Markov chain is: p10 1 p01 (2.1.1) P 1 p10 p01 The p01 stands for the probability of the motor changes from clockwise rotation to count-clockwise rotation. And p10 stands for the probability of the motor changes from count-clockwise rotation to clockwise rotation. In our simulation, the Markov process is assumed to be discrete, which means that the state of the motor does not change in a short time scale (Δt). The research done by Cluzel P. et al measured the rotation bias (pCW) and switching frequency (fswitch) of the molecular motor under different CheY-P concentrations in single cell. We suppose [CheY-P] influence the behavior of the motor by changing the parameter in the state transfer matrix. In this experiment, the time they monitoring the motor state is long enough, so we consider the probability of the motor on each state has become stable: pCW pCW (2.1.2) P 1 pCW 1 pCW This Equation can be simplified into following equation: p120 pCW ( p021 p120 ) (2.1.3) Suppose a motor has changed its state n times in a ti (s=0 or 1). We use P(n,k,s) to represent the probability of this event. This probability follows equation group(3.1.4). P(n, k , 0) (1 p01 ) P(n, k 1, 0) p10 P(n 1, k 1,1) (2.1.4) P(n, k ,1) (1 p10 ) P(n, k 1,1) p01 P(n 1, k 1, 0) In above equation group, if n>k or n < 0 or k<0, P(n,k,s)=0. Sum the two equations in(2.1.4), we can get: ( P(n, k , 0) P(n, k ,1)) ( P(n, k 1, 0) P(n, k 1,1)) (2.1.5) p10 ( P(n 1, k 1,1) P(n, k 1,1)) p01 ( P(n 1, k 1, 0) P(n, k 1, 0)) k k n( P(n, k , 0) P(n, k ,1)) n( P(n, k 1, 0) P(n, k 1,1)) n 0 n 0 k k n 0 n 0 p10 n( P (n 1, k 1,1) P (n, k 1,1)) p01 n( P (n 1, k 1, 0) P (n, k 1, 0)) k k n( P(n, k , 0) P(n, k ,1)) n( P(n, k 1, 0) P(n, k 1,1)) n 0 n 0 p10 P (n 1, k 1,1) ((n 1) P(n 1, k 1,1) nP(n, k 1,1)) n0 n 0 k k k k p01 P (n 1, k 1, 0) ((n 1) P(n 1, k 1, 0) nP(n, k 1, 0)) n 0 n 0 k k n( P(n, k , 0) P(n, k ,1)) n( P(n, k 1, 0) P(n, k 1,1)) n 0 n 0 k k n 0 n 0 p10 P(n 1, k 1,1) p01 P(n 1, k 1, 0) (2.1.6) Use the following 3 equations to eliminate P(n,k,s) in equation(2.1.6). k n( P(n, k , 0) P(n, k ,1)) f n 0 switch k t; k P(n, k ,1) 1 p CW n 0 ; (2.1.7) k P(n, k , 0) p n 0 CW ; Then we can get the equation describing the correlation of fswitch, Δt, p10, p01, pCW: f switch t p10 (1 pCW ) p01 pCW (2.1.8) The p10, p01 in the state transfer matrix can be derived from equation (2.1.3) and (2.1.8): f switch t p01 2 p CW (2.1.9) f p switch t 10 2(1 pCW ) The CW bias pCW and switching frequency fswitch as functions of CheY-P are cited from the research of Cluzel P. et al (Cluzel et al, 2000): ([CheY P] / K M ) H p ; CW ([CheY P] / K M ) H 1 (2.1.10) pCW 4([CheY P] / K M ) H 1 f fM ; switch (([CheY P] / K M ) H 1) 2 [CheY P ] Where Hill coefficient H=10.3, KM=3.1 μM, fM=1.7 Hz. Figure S6. Characteristic response of motors against CheY-P concentrations. Each point describes a simulation of the motor under a certain CheY-P concentration. The states of motor in 2500s are recorded and CW bias and switching frequency are calculated from them. The solid lines are the curve of functions (2.1.10). The simulation results fit the curve from literature very well. The probability density of bacteria with parameter x, y, D, s: f ( x , y, D, s, t ) x : position of bacteria; y : concentration of molecules related to chemotaxis in bacteria; D : direction vector of bacteria; s : state of motor on bacteria; t: time. Suppose the change of the molecules in chemotaxis pathway can be represented by following standard differential equations: y F ( y , L( x )); t This equation group contains the information about concentration field and the interaction of molecules in cell. Let the average speed of the bacteria is v, we can get the derivation of x: x vD; t In our simulation, we suppose during a small time scale Δt, the state of a bacterium changes from ( x ', y ', D ', s ', t ) to ( x , y, D, s ', t ) , and at the end of this time scale, the bacterium’s state switches into ( x , y, D, s, t ) If the motor state s'=0: x ' x; y ' y y y dx ' dy ' dD ' y t y F t ; D ' D D t I 0 D( x ', y ', D ') F dxdydD t x D( x , y, D) 0 I F t y 0 0 0 dxdydD I F t dxdydD y I If the motor state s'=1: x ' x x x Dvt ; y ' y y y dx ' dy ' dD ' y t y F t ; D ' D t I 0 Ivt D( x ', y ', D ') F dxdydD t x D( x , y, D) 0 I F t y 0 0 dxdydD I F t dxdydD y I Then we can get a equation group describing the change of the probability density of bacteria parameters in Δt: F t dxdydD dP(D) f ( x , y y, D D, 0, t ) I y p10 ( y ) f ( x , y, D, 0, t t )dxdydD 1 p01 ( y ) F f ( x , y, D,1, t t )dxdydD p01 ( y ) 1 p10 ( y ) t dxdydD f ( x x , y y, D,1, t ) I y p10 ( y ) dP(D) f ( x , y y, D D, 0, t ) f ( x , y, D, 0, t t ) 1 p01 ( y ) F t I p ( y ) 1 p ( y ) y 10 f ( x x , y y, D,1, t ) f ( x , y, D,1, t t ) 01 f 2 f0 f 1 Di D j 0 yi dP(D) f ( x , y, D, 0, t ) 0 Di k10 ( y )t 2 i , j Di D j f ( x , y, D, 0, t t ) 1 k01 ( y )t i Di i yi 1 Fi t f ( x , y, D,1, t t ) k01 ( y )t 1 k10 ( y )t i yi f1 f1 f ( x , y, D,1, t ) xi yi i xi i yi f ( x , y, D, 0, t t ) 1 k01 ( y )t f ( x , y, D,1, t t ) 1 k01 ( y )t f 0 2 f0 1 E Di Cov Di , D j Fi 2 i , j Di D j k10 ( y )t f ( x , y, D, 0, t ) i Di i k10 ( y )t f ( x , y, D,1, t ) f1 D v t Fi i x i i i f 0 t yi 1 Fi t f1 i yi t yi f f 0 2 f0 1 E Di Cov Di , D j Fi 0 t yi 2 i , j Di D j k ( y ) k10 ( y ) f ( x , y, D, 0, t ) f ( x , y, D, 0, t t ) f ( x , y, D, 0, t ) i Di i f ( x , y, D, 0, t ) Fi t 01 t k01 ( y ) k10 ( y ) f ( x , y, D,1, t ) f1 f ( x , y, D,1, t ) i yi f1 f ( x , y, D,1, t t ) f ( x , y, D,1, t ) t Fi i x Di vt yi i i f E Di 1 2 f 0 Cov Di , D j ( Fi f 0 ) 0 yi 1 f ( x , y, D, 0, t ) i Di t 2 i , j Di D j t i k01 ( y ) k10 ( y ) f ( x , y, D, 0, t ) ( F f ) 1 t f ( x , y, D,1, t ) i 1 f1 f ( x , y, D,1, t ) D v i y i i xi i The CW bias and switching frequency of the motor can be measured in experiment, let they are the function of [CheY-P]: fswitch(yn), PCW(yn). As mentioned above: k01 f switch ( yn ) f switch ( yn ) ; k10 ; 2 pCW ( yn ) 2(1 pCW ( yn )) The differential coefficient of some parameters can be derived from above equation group: E Di L( x ) Di 1 s xi svDi ; yi Fi ( y, L( x )) ; ; L( x ) sv Di t t t t t xi i These equations consist to the dynamics relations of a single bacterium and our simulation results. If we simulate the movement of bacteria in 2D space, their moving direction can be represented by the angle θ, and Δθ can be considered as a random walk process: E 0; D k t ; cos x v ; t sin Use testing the chemotaxis effects of the band-pass filter as an example: 12 y1 0 x1 2 L y 0 ; L L0 exp ; r t 1 (1 2 ) y2 y0 A(1 2 ) L K L Consequently, the total equation describing the change of the distribution of bacteria population is: k 2 f ( x , y, , 0, t ) 2 f ( x, y, , 0, t ) 2 ( y y ) f ( x, y, , 0, t ) 0 y1 1 2 2 f ( x, y, ,1, t ) y2 t f ( x, y, ,1, t ) v sin v cos f ( x , y, ,1, t ) x1 x2 f switch ( y2 ) 1 1 2 1 pCW ( y2 ) f ( x, y, , 0, t ) A ) y1 (1 2 )( y2 y0 x1 2 f ( x, y, ,1, t ) KL 1 exp r L 0 f ( x, y, , 0, t ) 1 pCW ( y2 ) f ( x, y, ,1, t ) 1 Reference Block SM, Segall JE, Berg HC (1983) Adaptation kinetics in bacterial chemotaxis. 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