modelling with odes a complex pathway - unibo.it · modelling with odes a complex pathway how do...

Modelling with ODEs a complex pathwayHow do the response of an edge change as the input nodes changes?

Metabolism:

A Enzyme B Afdt

dB

0

1

0

[A]

d[B

]/dt

Michaelis-Menten kinetics

A Enzyme B EBAEEAk

k

k

2

1

1

AK

Av

dt

dB

MM

Max

KMM KMM

vMax

vMax/2

Michaelis-Menten kinetics

A Enzyme B EBAEEAk

k

k

2

1

1

AK

Av

dt

dB

MM

Max

vMax

1° order kinetics0° order kinetics

Maxvdt

dBA

K

v

dt

dB

MM

Max

Cooperative kinetics

The enzyme (the DNA) has different binding sites. The affinity to further ligands may change after binding one or more ligands:

Positive the affinity increases after bindingNegative the affinity decreases after binding

Homotropic cooperativity between ligands of the same typeHeterotropic cooperativity between ligands of different types

Hill kinetics

Hill kinetics allow one to include the

effects of cooperative binding events.

nn

B

n

Max

AK

Av

dt

dB

~

Hill functions

nn

B

n

AK

AY

n=2 n=5 n=20

Hill functions

nn

B

n

AK

AY

KB=0.2 KB=0.7 KB=1

n=2 Y=1/2 for A=KB-1/2

C.Piazza, Università di Udine http://iclp08.dimi.uniud.it/PRESENTAZIONI/Piazza.pdf

Edges of the network: response function

How do the response of an edge change as the input nodes changes?

Regulation

A X A X

Afdt

dX XAf

dt

dX,

Ordinary Differential Equations

),,( tpdt

dxf

x

x: variable vector (x1,x2,..,xn)p: parameterst: timef: function vector (f1,f2,..fn)

If f does not depend on time, the ODE is called autonomous

),( pdt

dxf

x

We search for a general solution

or for a particular solution (given some initial condition xo)

),()( ptt Fx

),,()( 0xFx ptt

Ordinary Differential Equations: Steady state

Consider an autonomous ODE

),( pdt

dxf

x

The steady state, when existing, is the value of x that gives:

0),(0 pdt

d SSxf

x

If x reaches the value xSS at time t0, it will not vary after t0

Ordinary Differential Equations, 1D

adt

dx The variation rate is constant

The sign of a determines the family of the solution

For all constants C, the solution satisfies the ODE

NO STEADY STATE

Cattx )(

General solution

-20

-15

-10

-5

0

5

10

15

-4 -3 -2 -1 0 1 2 3 4

t

x(t)

a=1, C= 0 a=1, C= -10a=1, C=5 a=-1, C=0a=-1, C=-10 a=-1, C=5


adt

dx

CCaxx 0)0(0

For determining a particular solution we need some condition on the value of x(t) a some t

Usually the value x(0) = x0

0)( xattx

So


axdt

dx The variation rate is proportional to x

The sign of a determines the family of the solution

For all constants C, the solution satisfies the ODE

STEADY STATE: xSS=0

atCetx )(General solution

-600

-500

-400

-300

-200

-100

0

100

200

300

400

-4 -3 -2 -1 0 1 2 3 4

t

x(t)

a=1, C= 0

a=1, C= -10

a=1, C=5

a=-1, C=0

a=-1, C=-10

a=-1, C=5


axdt

dx

CCexx a 0

0 )0(



atextx 0)(

So


2axdt

dx The variation rate is proportional to x2

General solution

Cat

tx

Catx

dtax

dx

1

12

-1.5

-1

-0.5

0

0.5

1

1.5

-44

-38

-32

-26

-20

-14 -8 -2 4 10 16 22 28 34 40

t

x(t)

a=1, C= 0 a=1, C= -10

a=1, C=5 a=-1, C=0

a=-1, C=-10 a=-1, C=5

STEADY STATE: xSS=0


2axdt

dx



So

0

0

11

0

10

xC

CCaxx

tax

x

xat

tx0

0

0

11

1


x

a

dt

dx The variation rate is inversely proportional to x

General solution

a

CtCattx

Catx

dtadxx

,22

2

2

0

2

4

6

8

10

12

-44

-37

-30

-23

-16 -9 -2 5 12 19 26 33 40

t

x(t)

a=1, C= 0 a=1, C= -10a=1, C=5 a=-1, C=0a=-1, C=-10 a=-1, C=5

NO STEADY STATE




So

x

a

dt

dx

CCaxx 220200

2

00 2)( xatxsigntx

Linear Ordinary Differential Equations

baxdt

dx

If b≠0 : non-homogeneous case. Find the steady state

a

bx

dt

dx SS 0

Then consider the transformation: SSxxx ˆIt comes that:

xaxxaa

bxabax

dt

dx

dt

xd SS ˆ)(ˆ

So

Homogeneous

a

bCexCex atat ˆ

Linear Ordinary Differential Equations

baxdt

dx

a

bxCC

a

bCe

a

bxx a

0

0

0 )0(



atat ea

bx

a

bCe

a

btx

0)(

So

-4

-2

0

2

4

6

8

10

12

14

-1 0 1 2 3 4 5 6 7 8

t

x(t)

a=-1, b=1, x0= 0

a=-1, b=1, x0=1

a=-1, b=1, x0=5

0,)(

0,)(

0

0

aexxxtx

aea

bx

a

btx

atSSSS

at

0

2

4

6

8

10

12

0 5 10 15 20 25 30 35 40 45t

x(t)

a=-1, b=1, x0= 0

a=-1, b=10, x0=0

a=-0.1, b=1, x0=0

0,)( 0

ae

a

bx

a

btx at

The solution converges to the value -b/a; the value a determines the velocity of the response

0,)1()( 0 aexextx atatSS

0

2

4

6

8

10

12

0 5 10 15 20 25 30 35 40 45t

x(t)

a=-1, b=10, x0=0

a=-0.1, b=1, x0=0

It is the time for reaching xSS/2 starting from 0

ate

xextx ta

SStaSS 2ln

5.02

)1()(

Response time

-6

-4

-2

0

2

4

6

8

10

12

0 5 10 15 20 25 30 35 40 45

t

x(t)

a=-0.1, b=1, x0=0

a=-0.1, b=1, x0=-4

In general, it is the time for reaching half of the distance between x0 and xSS

a

txx

xxe

xxxexexxtx

SSSSta

SS

tataSS

2ln

21

2)1()(

00

0000

Response time

Linear systems of autonomous ODEs

nnnnnn

nn

nn

xaxaxadt

dx

xaxaxadt

dx

xaxaxadt

dx

.....

....

.....

.....

2211

22221212

12121111

nx

x

x

...

2

1

x Axx

dt

d


Axx

dt

d

The general solution can be written as

n

t

b

b

b

et..

where)(2

1

bbx

0 bIAAbbx

tt eedt

d

So:

That is satisfied for all i that are eigenvalues of A and b(i) the corresponding eigenvectors

tn

i

iiect

1

)( ibx

The value of cis are fixed to satisfy initial conditions

Example in 2 dimensions

2

1

2

1

23

21

x

x

x

x

dt

d

Eigenvalues-Eigenvectors

That is satisfied for all i eigenvalues of A and b(i) the corresponding eigenvectors

023

21det

01)4(0430621 2

1

11

3

24 2

21

)((1)bb

tt ecectx

x

1

1

3

2)( 2

4

1

2

1

Example in 2 dimensions

2

1

2

1

23

21

x

x

x

x

dt

dWith

1

1

3

2)0(

0

521

2

1cc

x

x

0

50

2

1

x

x

3

1

52

3

2

1

21

12

c

c

cc

cc

tt

tt

eetx

eetx

33

32

4

2

4

1

0

1000

2000

3000

4000

5000

6000

7000

0 0,20,40,60,8 1 1,21,41,61,8

X1 X2


zAxx

dt

d

The Steady state solution is zAxzAx10 ssss

xAzAxAzAxxx 1 ˆ)(

ˆ

dt

d

dt

d

So

zAbxbx

1ii

t

n

i

i

tn

i

iii ectect

11

)()(ˆ

Define the transformationss

xxx ˆ

Higher order ODEs

Can be reduced to linear systems, by introducing extra variables

1

22

21

21

2

2

2

,

xkdt

dx

xdt

dx

dt

dxxxx

xkdt

xd

Axx

dt

d

0

102k

A

Solutions

kk

kk

kk

1,

,1

,

0)(1

detdet

2

1

22

2

(2)

(1)

b

b

IA

ktkt

ktkt

tt

tt

ekcekc

ecec

ebcebc

ebcebc

tx

tx

21

21

2)2(

22)1(

1

1)2(

21)1(

1

2

1

21

21

)(

)(

Solutions with initial conditions

5.00

1

)0(

)0(21

21

21

2

1

cc

kckc

cc

x

x

)sinh(

)cosh(5.0

)(

)(

2

1

kt

kt

ekek

ee

tx

txktkt

ktkt

Higher order ODEs

Can be reduced to linear systems, by introducing extra variables

1

22

21

21

2

2

2

,

xkdt

dx

xdt

dx

dt

dxxxx

xkdt

xd

Axx

dt

d

0

102k

A

Harmonic force

Imaginary solutions

ikik

ikik

kk

1,

,1

,

0)(1

detdet

2

1

22

2

(2)

(1)

b

b

IA

iktikt

iktikt

tt

tt

eikceikc

ecec

ebcebc

ebcebc

tx

tx

21

21

2)2(

22)1(

1

1)2(

21)1(

1

2

1

21

21

)(

)(

1i

Euler’s formula

)sin()cos( ktikteikt

Proof: the exponential eax function is DEFINED as the function satisfying:

1)0( with xaxdt

dx

Indeed:

1)0( with xikxdt

dx

)cos()sin()cos()sin()sin()cos( ktktiikktikktkktiktdt

d

So we have to proof that the given expression satisfies:

1)0sin()0cos( kik

Example

2

1

0

121

21

21

cc

ikcikc

cc

With initial condition

0

1

)0(

)0(

dt

dx

x

iktikt

iktikt

tt

tt

eikceikc

ecec

ebcebc

ebcebc

tx

tx

21

21

2)2(

22)1(

1

1)2(

21)1(

1

2

1

21

21

)(

)(

kt

kt

ktii

kt

eikeik

ee

tx

txiktikt

iktikt

sin

cos

sin2

cos2

2

1

2

1

)(

)(

2

1

xkdt

xd 2

2

2

Time evolution plot

kt

kt

tx

tx

sin

cos

)(

)(

2

1

-1,5

-1

-0,5

0

0,5

1

1,5

00,

81,

62,

43,

2 44,

85,

66,

47,

2 88,

89,

6

t

x1(t

),x2(t

)

x1(t)

x2(t)

X1 and x2 are plotted with respect to time

Phase space plot

kt

kt

tx

tx

sin

cos

)(

)(

2

1

Relation between x1 and x2 are plotted without reference to time

Stability of steady states in ODEs

0),( pSSxf

Given an autonomous ODE ),( pdt

dxf

x

The steady state is given by the condition

If x=xss the solution does not varyWe want to analyse the behaviour of the solution when

with small eThe steady state is:STABLE if the system returns to this state upon perturbationUNSTABLE if the system leaves this state upon perturbationMETASTABLE if the system behaviour is indifferent

εxxSS

Stability of steady states

FORCE

Unstable Stable Metastable

On phase space

In 1D, linear case 0 , aaxdt

dx

)( axxf

point steady , 0dt

dxx

The phase space is 1D and the f(x) locally defines a vectorial field that represents the rate of change of x in each point

The steady point defines two subspaces with opposite behaviour

The steady state is unstable 0 if

0 if

0

0

0

xx

xxexx(t)

t

tat

More on phase space

In 1D, linear case 0 , aaxdt

dx

)( axxf

point steady , 0dt

dxx

The steady point divides two subspaces with opposite behaviour

The steady state is stable 0 0, 0

t

at xaexx(t)

Stability of steady states in 1D linear ODEs

Given an autonomous linear ODE, in 1 dimension

baxdt

dx

The steady state is

The particular solution starting with x0=xss+e is

a

bxss

atatssat ea

be

a

bx

a

be

a

bx

a

btx

ee0)(

The steady state isSTABLE if a<0UNSTABLE if a>0METASTABLE if a=0


0

0,5

1

1,5

2

2,5

3

3,5

0 1 2 3t

x(t)

a=-1 a=0 a=1

xss

Non linear equations

In 1D, general case xfdt

dx

)(xf

pointssteady , 0dt

dx

x

The steady point defines five subspaces

The steady states A and C are stable, B and D are not.

A B DC

Basin of attraction

In 1D, general case xfdt

dx

)(xf

x

Three basins of attraction can be defined:Evolution of points in Basin1 tends to the stable point AEvolution of points in Basin2 tends to the stable point CEvolution of points in Basin3 tends to infinite

A B DC

Basin3Basin2Basin1

Stability of steady states in 1D non linear ODEs


xfdt

dx

The steady states are the solutions

In proximity of each steady state, the function can be linearized

0ssxf

SSSS x

SS

x

SSSS

dx

df

dt

xd

dx

dfxfxf e

eee

The steady state isSTABLE if df/dx <0UNSTABLE if df/dx >0METASTABLE if df/dx =0

Logistic model

Population increase with competition on resources

)(xf

x0

0,0 , 1

Kr

K

PrP

dt

dPP:populationr: growth rateK: carrying capacity

0,0 ,with , 1 KrK

Pxxrx

dt

dx

1

Logistic model: steady states

Steady state stability:x=0: x=1

Unstable Stable

)(xf

x0

0,0 ,with , 1 KrK

Pxxrx

dt

dx

The whole domain x>0 forms a unique basin of attraction converging to the point x=1

1

020

r rrxdx

dfx

021

r rrxdx

dfx

Logistic model: explicit solution

0,0 ,with , 1 KrK

Pxxrx

dt

dx

tx

x

rdtxx

dxrdt

xx

dx

001

1

11)(

1ln

0

0

0

rt

rtx

xex

extxrt

x

x

x(0)=5

x(0)=1/5



2222121

1212111

2

1

2

1

2221

1211

2

1

zxaxa

zxaxa

z

z

x

x

aa

aa

x

x

dt

d

1212111 zxaxa

1x

2x

2222121 zxaxa

For each point in the phase space a vector is defined

Phase space in 2D

2

1

2

1

01

10

x

x

x

x

dt

d

1212

2211

,

,

xxxv

xxxv

Components of the vectors

Phase space in 2D

2

1

2

1

20

11

x

x

x

x

dt

d

2212

21211

2,

,

xxxv

xxxxv

Components of the vectors



2222121

1212111

2

1

2

1

2221

1211

2

1

zxaxa

zxaxa

z

z

x

x

aa

aa

x

x

dt

d

Steady state

01212111 zxaxa

02222121 zxaxaNull clines

Phase space in 2D

2

1

2

1

01

10

x

x

x

x

dt

d

0

0

2

1

x

x

Null clines

Steady state(0,0)

Phase space in 2D

2

1

2

1

20

11

x

x

x

x

dt

d

0

0

2

21

x

xx

Null clines

Steady state(0,0)

Local stability of a steady point

Stable Unstable Center

Particular directions should be analysed

Saddle

zAxx

dt

d zAbx

1i

t

n

i

iiect

1

)(

with λ, eigenvalues and b, eigenvectors

Phase space in 2D

2

1

2

1

01

10

x

x

x

x

dt

d

1

1

2

1

Eigenvalues

Eigenvectors

1

1;

1

1

λ=1

λ=-1

Phase space in 2D

2

1

2

1

20

11

x

x

x

x

dt

d

2

1

2

1

Eigenvalues

Eigenvectors

1

1;

0

1

λ=2

λ=1

Phase space in 2D

2

1

2

1

01

10

x

x

x

x

dt

d

i

i

2

1

Eigenvalues

ii

1;

1

Stability of steady states in nD linear ODEs

Given an autonomous linear ODE, in n dimension

zAx

xdt

d

The steady state is

The general solution is

The constant ci for a particular solution starting with x0=x

ss+e are given by:

zAxss 1

zAbx

1i

t

n

i

iiect

1

)(

εbεzAεxzAbx

i1ss1i

n

i

i

n

i

i cc11

)0(


So:

Consider the case in which all the eigenvalues are real.

Considering different e,parallel to the different eigenvectors b(i) we conclude that the system returns to the steady state if i<0 and leaves the steady state if i>0

So the steady state is STABLE if i<0 i (stable node)UNSTABLE if i>0 i (unstable node)UNSTABLE if i>0 for some i (unstable saddle point)

zAbx

1i

t

n

i

iiect

1

)( εb

i

n

i

ic1


(stable node) (unstable node) (unstable saddle point)


So:

Consider the case in which the eigenvalues are complex.

There are oscillatory parts, but the stability is given by the real part of eigenvalues

So the steady state is STABLE if Re(i)<0 i (stable focus)UNSTABLE if Re(i)>0 i (unstable focus)UNSTABLE if Re(i)=0 i (stable center)

zAbx

1i

t

n

i

iiect

1

)( εb

i

n

i

ic1


(stable center) (stable focus) (unstable focus)

General rules of stability in 2D

Given a matrix A (2x2)

We must compute

dc

ba

AAbcaddadc

badetTrdet 22

So:

2

)det(4)(Tr)(Tr 2 AAA

)det(

)(Tr

21

21

A

A

General rules of stability in 2D

2

)det(4)(Tr)(Tr 2 AAA

If we obtain 2 complex eigenvalues

if Tr(A) > 0: positive real part: UNSTABLE FOCUSif Tr(A) < 0: negative real part: STABLE FOCUSif Tr(A) = 0: null real part: CENTER

If we obtain 2 real eigenvalues

if det(A) < 0: two opposite eigenvevtors: SADDLEif det(A) > 0 and Tr(A) < 0:

two negative eigenvectors: STABLE NODEif det(A) > 0 and Tr(A) > 0:

two positive eigenvectors: UNSTABLE NODE

2)(Tr4

1)det( AA

2)(Tr4

1)det( AA

=Det (A)

=Tr (A)

Linear approximation around the steady state

0),( pSSxf

Given an autonomous ODE ),( pdt

dxf

x

The steady state is given by the condition

The stability around the steady state can be analysed considering the linear approximation of the system around it

dt

d

dt

d

dt

dt

dt

d

dt

d εεxεx

xss

ss

εxSS

)(

For each component xi the following condition holds:

n

j

j

j

ik

n

j

n

k

j

kj

in

j

j

j

ii

ii

x

f

xx

f

x

ff

fdt

dx

11 1

2

1

..2

1)(

)(

eeeeSS

SS

εx

x

εxSS

Linear approximation around the steady state

Jεε

dt

dThen:

with the Jacobian

The eigenvalues of J determine the stability of the system around the steady state

n

nnn

n

n

x

f

x

f

x

f

x

f

x

f

x

f

x

f

x

f

x

f

..

........

..

..

21

2

2

2

1

2

1

2

1

1

1

J

Lotka-Volterra equations

Non linear system of ODEs that describes the approximation of the prey-predator dynamics

Prey population (x) tends to increase (infinite supply of food) and it decreases only because of the predators

Predator population (y) tends to decrease and it increases only because of the predators

The probability that preys and predators meet is equal to xy

DxyCydt

dy

BxyAxdt

dx

Lotka-Volterra equations: Steady state

with A, B, C, D > 0Two steady states are present

xSS1=(0;0) xSS2=(C/D;A/B)

The Jacobian matrix is

0

0

DxyCydt

dy

BxyAxdt

dx

0)(

0)(

DxCy

ByAx

DxCDy

BxByAJ

Lotka-Volterra equations: stability

xSS1=(0;0)

with A, C > 0: saddle point

xSS2=(C/D;A/B)

Tr(J) = 0 ; Det (J) = ACDet (J) > Tr(J)2/4 : Complex solution;Tr(J)=0 : Center

C

AJ

0

0

0

0

B

ADD

BC

J

Lotka-Volterra equations: stability

xSS1=(0;0)

with A, C > 0: saddle point

xSS2=(C/D;A/B)

The eigenvector are

The system has a center and cyclic orbits around it

C

AJ

0

0

0

0

B

ADD

BC

J

ACi

Lotka-Volterra equations: Time evolution

Lotka-Volterra equations: Phase space

Some other system

1

2

xydt

dy

yxdt

dx

xydt

dy

ydt

dx1)ln(

Dependence on parameters

Logistic model with harvesting

Population increase, with competition on resources and constant harvesting

0,0,0 , 1

HKr -H

K

PrP

dt

dPP:populationr: growth rateK: carrying capacityH= Harvesting rate

K

Hh

K

Pxhxrx

dt

dx ,with , 1

Stable points:

r

h

r

rhrrx

hxrx

4112

1

2

4

0 1

2

Logistic model with harvesting

)(xf

x

0 1

)(xf

x

0

1

r

h

r

rhrrx

hxrx

4112

1

2

4

0 1

2

Stable points:

0,

Stable point: 0

r

hx 411

2

1

4rh 4rh

Bifurcation

0

0,2

0,4

0,6

0,8

1

1,2

r/4 h

x

Stable node

Unstable node

Bifurcation in 2D: Saddle-node

ydt

dy

xdt

dx

2 Fixed points )0,();0,(

http://www.egwald.ca/nonlineardynamics/bifurcations.php#hopfbifurcation

y

x

0

y

x

0

Stable Saddle

Bifurcation in 2D: Saddle-node

ydt

dy

xdt

dx

2Fixed points )0,();0,(

Jacobian

10

02x


1,2:)0,(

1,2:)0,(

21

21

Bifurcation in 2D: Pitchfork

ydt

dy

xxdt

dx

3

Fixed points )0,();0,();0,0(


y

0

y

x

0

Stable StableStable

Bifurcation in 2D: Pitchfork

ydt

dy

xxdt

dx

3

Fixed points )0,();0,();0,0(


Jacobian

10

03 2x

1,:)0,0(

1,2:)0,(

21

21

Bifurcation in 2D: Hopf

22

22

yxyxdt

dy

yxxydt

dx

Analyze fixed point (0,0)

Jacobian

1

1


i

A limit cycle emerges

Prey-Predator Holling-Tanner Model

Alexander Panfilov: Introduction to Differential Equations


R: predator with logistic growth competing on preys

P: preys with logistic growth limiting for resources K and with

harvesting depending on predators. When preys increases,

the harvesting saturates to a.


Alexander Panfilov: Introduction to Differential Equations


Alternative paradigms for

dynamical modelling of systems

Prof. Yechiam Yemini (YY) Columbia University

Biochemical network that controls the cell cycle progression in fission

yeast S.pombe

Davidich, Bornholdt (2008) PLoS ONE 3(2): e1672


yeast S.pombe

Davidich, Bornholdt

(2008) PLoS ONE

3(2): e1672

Figure 2. Network state space. State

space of the 1024 possible network

states (green circles) and their

dynamical trajectories, all converging

towards fixed point attractors. Each

circle corresponds to one specific

network state with each of the ten

proteins being in one specific

activation state (active/inactive). The

largest attractor tree corresponds to

all network states flowing to the G1

fixed point (blue node). Arrows

between the network states indicate

the direction of the dynamical flow

from one network state to its

subsequent state. The fission yeast

cell-cycle sequence is shown with

blue arrows.

Davidich, Bornholdt (2008) PLoS ONE 3(2): e1672


yeast S.pombe

The study indicates that the regulatory robustness of biological chemical

networks may allow for ‘‘robust’’ modeling approaches

Regulatory network that controls formation and distruction of extra-

cellular matrix

Extracellular matrix (ECM) is the extracellular part of animal tissue that usually

provides structural support to the animal cells in addition to performing various

other important functions. The extracellular matrix is the defining feature of

connective tissue in animals. Extracellular matrix includes the interstitial matrix

and the basement membrane


cellular matrix

Wollbold et al (2009) BMC Systems Biology 3:77


cellular matrix

BMC Systems Biology 2009, 3:77


cellular matrix


TEMPORAL MODELLING

time steps as follows:

Transcription 1 (NFKB1: 2),

Translation: 1,

mRNA lifespan: 1,

protein lifespan: 2.

Since TGFβ1 and TNFα have to be released into the extracellular medium

after translation, they were assumed to take effect three time units after

induction.


cellular matrix



cellular matrix


Literature derived interactions are not in agreement with new experimental

data


cellular matrix



cellular matrix


Conclusion

The analyses in the present study were based on literature

data valid for healthy human synovial fibroblasts (SFB). These

findings were fine-tuned and adapted to gene expression time course

data triggered by TGFβ1 and TNFα in SFB from RA and OA patients.

Both the assembly of previous knowledge and the adaptation of the

Boolean functions gave detailed insight into disease-related

regulatory processes. To the best of our knowledge, this is the first

dynamical model of ECM formation and degradation by human SFB.

Wikipedia: Petri net

Firing example

2H2 + O2 2H2O

H2

O2

H2O

t

2

2

Modeling concurrency

t2

t3

t1 t4

concurrency

In computer science, concurrency is a property of systems in which several

computations are executing simultaneously, and potentially interacting with

each other.

Simple and intuitive representation of a metabolic network:

stoichiometry of each reaction encoded by the arc weights of its

transition node.

Execution of Petri nets is nondeterministic: when multiple transitions

are enabled at the same time, any one of them may fire. If a transition is

enabled, it may fire, but it doesn't have to.

Petri nets are well suited for modeling the concurrent behavior of

distributed system

http://en.wikipedia.org/wiki/Concurrency_(computer_science)

Heiner et al, LNCS 5016, pp. 215-264, 2008.

ODE description


PETRI NET DESCRIPTION


NOTE that transition states (association

between enzyme and substrate) are

explicitly modeled

Sub+Enz↔ SubEnz Prod+Enz

Regev A, Shapiro: The -calculus as an abstraction for biomolecular systems

Stochastic Pi-Calculus

Each channel is associated with a base rate. The channel's

base rate is identical to the mesoscopic rate constant of the

corresponding elementary reaction.

At each state in the pi-calculus system we determine the

actual rate of a channel based on that channel's base rate,

and the number of input and output offers on the channel at

that state (which represent the number of reactant molecules

in the corresponding reaction).

Stochastic selection of communication according to the

probability of a reaction

The resulting state evolution of the -calculus system

corresponds to the state evolution of a (statistically

representative) trajectory of the chemical system.

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