· Feedback regulation Positive feedback Negative regulation A gene regulatory network Other regulatory elements Examples Claudio Altaﬁni, February 15, 2007 – p. 12/54 Elementary

Claudio Altafini, February 15, 2007 – p. 1/54

ODEs models in Systems Biology

Lecture 5 & 6

Claudio AltafiniSISSA

http://people.sissa.it/∼altafini

Mathematical modeling

● System Biology

Elementary modules

Examples


System Biology

■ Systems Biology = investigates function of genetic,molecular and cellular processes in a systematic way:→ making (biologically relevant) sense out of data

■ why “systems”?◆ biological processes→ mathematical models◆ many processes simultaneously→ systematic methods

■ Approach of today: detailed description of elementarydynamical moduli for biological processes by means ofODEs (Ordinary Differential Equations)

■ Inspiration: reaction kinetics■ basic building blocks in constructing networks of interactions

for known biochemical pathways

http://people.sissa.it/~altafini


● System Biology

Elementary modules

Examples


References for today’s material

■ Leah Edelstein-Keshet "Mathematical Models in Biology" (reprint SIAM 2005)

■ Eduardo Sontag “Lecture notes in Mathematical Biology” (lecture Notes available at

http://www.math.rutgers.edu/∼sontag)

■ Olaf Wolkenhauer “System Biology” (lecture notes available at

http://www.sbi.uni-rostock.de/teaching.html)

■ J.J. Tyson, K.C. Chen and B. Novak, “Sniffers, buzzers, toggles and blinkers:

dynamics of regulatory and signaling pathways in the cell”, Curr. Opin. Cell Biol.

15:221-231 (2003).


Elementary modules

● Biological moduli as ODEs

● Maltus Law

● Solving ODEs: Matlab

● Qualitative behavior: steady state

● Reaction kinetics

● Elementary reaction kinetics

● Enzyme catalyzed reactions

● Michaelis Menten kinetics

● Competitive inhibition

● Cooperativity

● Sigmoidal responses

● Multistability

● Enzymes and multistability

● Feedback regulation

● Positive feedback

● Negative regulation

● A gene regulatory network

● Other regulatory elements

Examples


Biological moduli as ODEs

■ what is a ODE?Consider x = concentration of a substrate X (e.g. mRNA,protein, small molecule, metabolite, any reagent)

■ at time t + ∆t, (∆t small) one can expand in Taylor series

x(t + ∆t) ' x(t) + kx(t)∆t

■ in the limit for small times lim∆t→0x(t+∆t)−x(t)

∆t=: dx

dt

=⇒dx

dt= kx

◆ 1st order, linear, autonomous ordinary differential equation◆ it involves two different quantities:

■ x = state variable■ k = parameter (does not describe an elementary

component of the process)◆ it describes a reaction rate, i.e., how the concentration of

X varies with time


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples


Maltus Law

■ solving this ODE: Maltus law (1798)

◆ separate the variables dxx

= kdt◆ integrate both sides

∫

dx

x=

∫

kdt

lnx(t)

x(0)= k t

ln x(t) = ln x(0) + k t

where x(0) = xo = constant◆ Exponentiating

x(t) = xoekt


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples


Maltus Law

x(t) = xoekt

The characteristic evolution of the Maltus law is exponentialand depends on the sign of k1. k > 0 exponential growth

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

5

10

15

20

25

30

35

t

x

2. k < 0 exponential decay

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

t


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples


Solving ODEs: Matlab

■ Apart from simple cases (like Maltus law) there is littlechance to find explicit solutions of an ODE or of a system ofODEs. How to proceed then?

■ Use a simulator to numerically integrate the ODEs.■ example: Matlab (standard software in all engineering fields)

◆ define a function for the ODEfunction dzdt = Maltus(t,z,k);dzdt=k*z;

◆ define parameter, initial condition, time intervalk = 0.7; % growth constantx0 = 1; % initial conditiontspan = [0, 5]; % time interval

◆ integration routine[t,x] = ode45(@Maltus,tspan,x0,[],k);


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples


Qualitative behavior: steady state

■ what is the long time behavior? x(∞) =???■ if one is lucky it coincides with the steady state solution of

the ODE: steady state(s) is the value(s) xss of x for which

dx

dt= 0

since dx/dt = 0, the rate does not change =⇒ ODE “staysthere” forever

■ a steady state xss

is said stable whenchanging the initialconditions the steadystate remains thesame 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

0

5

10

15

20

25

30

35

40

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.2

0.4

0.6

0.8

1

1.2

1.4


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples


Stability of steady states

■ when is a steady state stable?■ for the Maltus law dx/dt = kx

dx

dt= 0 ⇐⇒ x = 0

1. k > 0 :

{

x(0) = ε < 0 =⇒ dxdt

< 0

x(0) = ε > 0 =⇒ dxdt

> 0

◆ =⇒ xss = 0 is unstable◆ x grows and dx

dtgrows =⇒ unbounded solution

2. k < 0 :

{

x(0) = ε < 0 =⇒ dxdt

> 0

x(0) = ε > 0 =⇒ dxdt

< 0

◆ =⇒ xss = 0 is stable


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples


Reaction kinetics

■ the decay dx/dt = kx, k < 0 can be seen as a kineticreaction (in which I am not interested in the product of thedegradation):

Xk−−→

■ to analyze more complex reactions: law of mass action

when 2 or more reactants are involved in a reactionstep, the reaction rates are proportional to the productof their concentrations

◆ justification: macroscopic version of collision theory◆ validity:

■ constant temperature■ medium must be well-mixed■ # of molecules must be high


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples


Elementary reaction kinetics

■ bimolecular reaction

X + Yk−−→ Z ODEs:

dxdt

= −k x ydydt

= −k x ydzdt

= k x y

■ dissociation

Zk−−→ X + Y ODEs:

dxdt

= k zdydt

= k zdzdt

= −k z

■ reversible dissociation

X + Y

k+

−−−→←−−−

k−

Z ODEs:

dxdt

= −k+xy + k−zdydt

= −k+xy + k−zdzdt

= k+xy − k−z


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples


Elementary reaction kinetics

■ conservation laws (e.g. mass conservation) can be used toreduce the number of equations involved:

■ example:

dxdt

= −k+xy + k−zdydt

= −k+xy + k−zdzdt

= k+xy − k−z

⇒

d(x+z)dt

= 0d(y+z)

dt= 0

⇒x(t) + z(t) = xo + zo = ao

y(t) + z(t) = yo + zo = bo

hence the system of 3 ODEs reduces to the scalar ODE

dz

dt= k+(ao − z)(bo − z)− k−z

once I solve this ODE (homework: write your Matlab routine)for z(t) I can recover

x(t) = ao − z(t)

y(t) = bo − z(t)


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples


Enzyme catalyzed reactions

■ most reactions need to be catalyzed to take place atinteresting rates

■ enzymes = proteins that convert specific reactants (calledsubstrates) into products while remaining basicallyunchanged

substrate

product

enzyme


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples



■ rate of production depends nonlinearly on the concentrationof the substrate

S + Ek1−−−→←−−−−

k−1

Ck2−−−→ P + E

◆ S = substrate◆ E = enzyme◆ C = complex (“ = [ES]”)◆ P = product

S

C

E

P

k

k −1

k1k 2

kk 2

1−1 +

■ ODEs

dsdt

= −k1se + k−1cdedt

= −k1se + (k−1 + k2)cdcdt

= k1se− (k−1 + k2)cdpdt

= k2c


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples



■ simplifications:◆ last equation does not feedback =⇒ I can ignore it and get

p(t) by integration once I have c(t)◆ conservation of mass for the enzyme:

de

dt+

dc

dt= 0 =⇒ e(t) + c(t) = const = eo

often at t = 0 one has c(0) = 0 and e(0) = eo

◆ =⇒ another equation can be eliminated

ds

dt= −k1s(eo − c) + k−1c

dc

dt= k1s(eo − c)− (k−1 + k2)c

■ how do you solve these?


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples


Michaelis Menten kinetics

■ quasi steady state approximation: after a transient period inwhich the enzyme “fills up”, the amount of complex C stays(almost) the same:

dc

dt= 0 =⇒ c =

eos

θ + swhere θ =

k−1 + k2

k1

=⇒ system reduces to a scalar ODE

ds

dt= −

Vms

θ + swhere Vm = k2eo

■ meaning of θ and Vm:◆ Vm = upper bound for ds

dt◆ θ = value of s for which

dsdt

= 12Vm

dsdt

s

mV

V2

θ

m


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples


Michaelis Menten kinetics

■ meaning of a Michaelis Menten reaction kinetics

dx

dt=

Vmx

θ + x

◆ it introduces a saturating behavior in the dynamics◆ often more realistic than Maltus law even when the

substrate is abundantMaltus law

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

x

dxdt

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

5

10

15

20

25

30

35

t

x

MM law

0 5 10 150

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

dxdt

x0 5 10 15

0

2

4

6

8

10

12

x

t


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples


“True” vs MM reaction

■ looking again at the ODE, there are 2 time scales◆ “slow time scale” in which c can be thought of as constant

=⇒ quasi steady state approximation holds◆ “fast time scale” (in which the approximation does not

hold)■ in short: c(t) adjusts very fast to the steady state solution

slow time scale:

0 5 10 15 20 250

1

2

3

4

5

6

7

8

9

10

t

scs

MM

■ more rigorous analysis:singular pertubation theory


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples


Matlab code used

■ function for a system of ODEs: same as scalar ODEs% function for enzyme reaction% with Michaelis-Menten kinetics% system of 2 ODEs

function dzdt = f enz mich ment1(t,z);% pass the parameters to the functionglobal k1 k1m k2 e0% extract the states from the vector zs=z(1);c=z(2);% ODEsdsdt=k1m*c-k1*s*(e0-c);dcdt=k1*s*(e0-c)-(k1m+k2)*c;% output vectordzdt=[dsdt; dcdt];

■ integration routine[t,z] = ode45(@f enz mich ment1,tspan,z0,[]);


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples


Competitive inhibition

■ A second substrate (called inhibitor) can bind to the enzyme,thus preventing the primary substrate from doing it. Theresult is that no product can be created.

enzyme

substrate

competitive inhibitor

■ this is for example a mechanicsdescribing the action of a drug ona receptor (protein on the cellsurface that interact with theexternal world)

■ chemical kinetic model

S + Ek1−−−→←−−−−

k−1

C1k2−−−→ P + E I + E

k3−−−→←−−−−

k−3

C2


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples



■ ODEs

dsdt

= −k1se + k−1c1dedt

= −k1se− k3ie + (k−1 + k2)c1 + k−3c2dc1

dt= k1se− (k−1 + k2)c1

dc2

dt= k3ie− k−3c2

didt

= −k−3ie + k−3c2dpdt

= k2c1

■ reduce #: conservation laws

c1 + c2 + e = eo

i + c2 = io=⇒

dsdt

= −k1s(eo − c1 − c2) + k−1c1dc1

dt= k1s(eo − c1 − c2)− (k−1 + k2)c1

dc2

dt= k3(io − c2)(eo − c1 − c2)− k−3c2


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples



■ quasi steady state approximation: dc1

dt= dc2

dt= 0

c1 =θieos

θmi + θis + θmθi

, c2 =θmeoi

θmi + θis + θmθi

where θm = k−1+k2

k1, θi = k

−3

k3

■ product formation rate

dp

dt=

Vms

θm(1 + i/θi) + s

is smaller than in the no-inhibition case (i = 0)


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples


Cooperativity

■ when n molecules of substrate must fit together with theenzyme in order for the reaction to take place

substrates

enzyme

product

■ ex:◆ ligand binding to cell surface receptors◆ binding of transcriptor factors to DNA to control gene

expression■ kinetics

nS + Ek1−−−→←−−−−

k−1

Ck2−−−→ P + E


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples


Cooperativity

■ ODEsdsdt

= −k1sne + nk−1c

dedt

= −k1sne + (k−1 + k2)c

dcdt

= k1sne− (k−1 + k2)c

dpdt

= k2c

■ after the quasi steady state approximation dpdt

= Vmsn

θ+sn

■ modifying θ → θn: Hill function h+(s, θ, n)

dp

dt= Vm

sn

θn + sn=: Vm h+(s, θ, n)


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples


Sigmoidal responses

■ common modules for saturated growth rate and saturateddecay rate

■ positive growth

h+(s, θ, n) =sn

θn + sn12

sθ

+h

1

■ negative growth

h−(s, θ, n) = 1−sn

θn + sn=

θn

θn + sn12

sθ

−h

1

■ θ = threshold= value of s at which h± reaches 1/2 of the saturation value

n = Hill parameter


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples


Hyperbolic vs Sigmoidal responses

■ consider dpdt

= Vmsn

θn+sn

◆ for n = 1 the graph of theformation rate is hyperbolic

◆ for n > 1 it is sigmoidal

dpdt

s

mV

V2

θ

mhyperbolic

sigmoidal

■ difference:◆ n = 1 graph is concave◆ for n > 1 the graph starts with “concavity up” and finishes

with “concavity down”=⇒ different stability properties

■ when n grows the sharpnessof the transition increases

dpdt

s

mV

V2

θ

m

=⇒ tends to a boolean switch=⇒ ultrasensitive response (“all or nothing” behavior)


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples


Multistability

■ consider the enzymatic reaction in the quasi steady stateapproximation

dp

dt= Vm

sn

θn + sn, n > 1

■ assume this is an autocatalytic process: p enhances its ownconcentration =⇒ s = αp

■ add a degradation rate = rate at which the product is beingused (or naturally degradates, or diffuses, etc.)

dp

dt= Vm

(αp)n

θn + (αp)n− γp, γ > 0

■ for simplicity: α = γ = Vm = 1

dp

dt=

pn

θn + pn− p


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples


Multistability

■ what are the possible steady states? (for which p dpdt

= 0?)1. hyperbolic case

◆ A = unstable equilibrium point◆ B = stable equilibrium point=⇒ p(t) converges uniquelyto an intermediate value

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

rates

degradation rate

p

BA

production ratehyperbolic

2. sigmoidal case◆ A = stable equilibrium point◆ B = unstable equilibrium point◆ C = stable equilibrium point=⇒ p(t) converges eitherat A or at C depending onthe initial condition=⇒ bistability (and a bifurcation)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

sigmoidal production rate

rates

degradation rate

=0.4θθ =0.7

p

A B C


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples


Enzyme-catalyzed reactions and multistabilityG. Craciun, Y. Tang, and Martin Feinberg, Understanding bistability in complex

enzyme-driven reaction networks, PNAS 103(23): 8697-8702, 2006■ which enzyme-catalyzed

reactions may admita multiple steady state?

■ according to mass-action laws (no quasi steady-stateapproximantion involved) this “has the capacity of a multiplesteady state”


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples


Enzyme-catalyzed reactions and multistability


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples


Feedback regulation

■ autocatalytic reactions: concentration of p modifies dpdt

◆ p regulates itself◆ p enhances itself =⇒ positive regulation

■ another example of positive feedback: gene regulation◆ state variables

■ x1 = [ mRNA]■ x2 = [protein]■ k1, k2 production rates■ γ1, γ2 degradation rates

mRNA

protein

gene

k 1

k 2

◆ ODEs:dx1

dt= k1 h+(x2, θ, n)− γ1 x1

dx2

dt= k2 x1 − γ2 x2

◆ assuming this is a stand alone system: is the geneexpressed or not at “steady state”?


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples


Positive feedback and multistability

■ simulation # 1:◆ x1(0) = 0.1 (low)◆ x2(0) = 0.2 (low)

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

t

x1x2

■ simulation # 2:◆ x1(0) = 0.1 (low)◆ x2(0) = 0.8 (high)

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

t

x1x2

■ important in differentiation mechanisms


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples


Negative regulation

■ if instead the protein inhibits the transcription then we havean example of negative feedback

mRNA

protein

gene

k 1

k 2

■ ODEs:

dx1

dt= k1 h−(x2, θ, n)− γ1 x1

dx2

dt= k2 x1 − γ2 x2

■ how about “steady state”?

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

t

x1x2

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

t

x1x2

■ negative feedback is synonymous of homeostasis


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples


A gene regulatory network

■ the simplest possible gene regulatory network is

protein

gene

mRNAmRNA

protein

gene

k gB

k pB

A

A

A

B

B

B

kgA

pAk

■ ODEs:dxgA

dt= kgA h+(xpB , θpB, npB)− γgA xgA

dxgB

dt= kgB h−(xpA, θpA, npA)− γgB xgB


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples


Other regulatory elements

■ delayed regulation

◆ ODEs:

dx1

dt= k1 h+(xτ

2 , θ, n)− γ1 x1dx2

dt= k2 xτ

1 − γ2 x2

where■ xτ

1 = x1(t− τ1)■ xτ

2 = x2(t− τ2)

τ 1

τ 2

mRNA

protein

gene

■ multiple regulation

◆ ODE

dxg

dt= kg h+(xpA, θpA, npA)+kg h−(xpB , θpB, npB)−γg xg

proteinA

gene

promoter

protein B


Elementary modules


● Maltus Law








● Cooperativity


● Multistability







Examples


Indirect regulation

mRNA

gene

k 1

k 2

metabolite M

inactiverepressor

repressoractive

enzyme E

substrate

■ ODEdxg

dt= kg h+(xM , θM , nM )− γg xg

dxE

dt= kE xg − γE xE

dxM

dt= kM xE − γM xM


Elementary modules

Examples

● Example: negative feedback

● Example: repressilator

● Example: bistability

● Example: Genetic oscillator

● Example: yeast

● Software packages


Example: autoregulatory negative feedback

A. Becskei and L. Serrano Engineering stability in gene

networks by autoregulation, Nature 405, 590 - 593, 2000J. Hasty, D. McMillen, J. Collins Engineered gene circuits,

Nature 420, 224, 2002

■ a: natural negative feedback from bacteriophage λ on E-coli■ b: synthetic negative

feedback


Elementary modules

Examples





● Example: yeast



Example: repressilator

■ M. B. Elowitz, S. Leibler A synthetic oscillatory network of transcriptional regulators

Nature 403, 335 - 338, 2000

■ model◆ states

■ mi = mRNA of cI, lacI, tetR■ pi # of proteins cI, lacI, tetR

◆ ODEs

dmi

dt= kmi

θ2i

θ2i + p2

i+1

− γmimi + kli

dpi

dt= kpimi − µipi


Elementary modules

Examples





● Example: yeast



Example: positive feedback

■ genetic switch of the bacteriophage λF. J. Isaacs, J. Hasty, C. R. Cantor and J. J. Collins Prediction and

measurement of an autoregulatory genetic module PNAS,100:7714-7719,

2003

■ gene for the λ repressor (cI857)and its 3 upstream promoters wereintroduced into E.coli

■ the λ repressor is temperature sensitive=⇒changing temperature the steadystate reached is different◆ if at most a single repressor binds

=⇒basal rate (low steady state)◆ if a repressor protein binds also at OR2

=⇒amplification of the transcription◆ if there is binding to all 3 OR

=⇒negative feedback again


Elementary modules

Examples





● Example: yeast



Example: positive feedback■ variables:

◆ X = repressor◆ X2 = repressor dimer◆ D0, D1, D2, D3 = DNA promoter sites with i dimers

bound

■ fast reactions: dimerization & binding at the promoter region

X + X

k1/V−−−−−→

←−−−−

k−1

X2, x2 =k1

k−1V

x2

D0 + X2

k2/V−−−−−→

←−−−−

k−2

D1, d1 =k2

k−2V

d0x2

D1 + X2

k3/V−−−−−→

←−−−−

k−3

D2, s2 =k3

k−3V

d1x2

D2 + X2

k4/V−−−−−→

←−−−−

k−4

D3, d3 =k4

k−4V

d2x2

■ fast variables: quasi steady state assumption→ replace withalgebraic relations


Elementary modules

Examples





● Example: yeast




■ slow, irreversible reactions: transcription, protein degradation

D0kt−−−→ D0 + X

D1kt−−−→ D1 + X

D2αkt−−−−→ D2 + X

Xkd−−−→

■ α = enhancement factor for D2 transcription (i.e. with OR1

and OR2 occupied)■ slowly changing variable of interest: total number of cI

molecules (in any form)

z = x + 2x2 + 2d1 + 4d2 + 6d3■ ODE

dz

dt= β(d0 + d1) + αβd2 − γxx, β = τ0kt, γx = τ0kd


Elementary modules

Examples





● Example: yeast




■ conservation law: copy number m is constant =⇒eliminated0

m = d0 + d1 + d2 + d3

■ up to some parameter reshuffling....

=⇒ d0 =m

1 + cx2/v2 + σ1c2x4/v4 + σ1σ2c3x6/v6' const

and (for the full derivation: see PNAS-Supplement)

dz

dt=

mβ(1 + cx2/v2 + ασ1c2x4/v4)

1 + cx2/v2 + σ1c2x4/v4 + σ1σ2c3x6/v6− γxx

from which the ODE for x is derived


Elementary modules

Examples





● Example: yeast




ODEs■ x = [λ]

■ g = [GFP ](measured protein)

dx

dt=

1

h(x, v)(β f(x, v)− γxx)

dg

dt= ηβ f(x, v)− γgg

f(x, v) =m(1 + cx2/v2 + ασ1c

2x4/v4)

1 + cx2/v2 + σ1c2x4/v4 + σ1σ2c3x6/v6

h(x, v) = 1 +4c1x

v+ m

4cx/v2 + 16σ1c2x3/v4 + 36σ1σ2c

3x5/v6

1 + cx2/v2 + σ1c2x4/v4 + σ1σ2c3x6/v6


Elementary modules

Examples





● Example: yeast



Example: genetic oscillator

J. M. G. Vilar, H. Y. Kueh, N. Barkai, and S. Leibler, Mechanisms of

noise-resistance in genetic oscillators PNAS,99:5988-5992, 2002

■ genetic oscillator = example of circadian clock■ variables

◆ C = protein complex (inactive)◆ A = activator protein◆ R = repressor protein◆ MA = activator mRNA◆ MR = repressor mRNA◆ DA (D′

A) = activator genewithout (with) A bound to thepromoter (DA + D′

A = 1)◆ DR (D′

R) = repressor genewithout (with) A bound to thepromoter (DR + D′

R = 1)


Elementary modules

Examples





● Example: yeast




■ A binds to A and R promoters increasing their transcriptionrates→ activator

■ R “sequestrates” A in the complex C → repressor


Elementary modules

Examples





● Example: yeast




■ 2 feedback loops:1. A on A: positive feedback2. A on R:

◆ postive effect on the transcription of R◆ negative effect on A (through the complex C formation)

AD DA’ DR D ’R

M A

A R

RM

■ oscillatory behavior: dynamics of R is slower than thedynamics of A


Elementary modules

Examples





● Example: yeast



Example: genetic oscillator■ to study oscillations: quasi steady state assumptions

◆ fast variables: A, MA, MR, DA, D′A, DR, D′

R

◆ slow variables: R and C=⇒reduced dynamical system

such that

■ time evolution of R and Cin the “true” system andin the reduced systemare quite similar→ good!


Elementary modules

Examples





● Example: yeast




■ limit cycle and stability analysis:◆ for 2-variable systems: ∃ of oscillations can be inferred

from The Poincaré-Bendixson theorem◆ Theorem: Given a 2-variable system, assume that

1. ∃ a region D in the plane s.t. the trajectories that enterD never leave it

2. @ stable steady states in D for the systemThen ∃ a periodic orbit in D.

◆ D : obtained by construction◆ equilibrium: use linear stability

analysis, Ro, Co: is unstable(real part of the eigenvalues is > 0)


Elementary modules

Examples





● Example: yeast



Example: modeling yeast cell cycle

■ Budding yeast(Saccharomyces cerevisiae)cell cycle regulation hasbeen studied in detail.

Chen, K.C., Calzone, L., Csikasz-Nagy, A., Cross, F.R., Novak, B., andTyson, J.J. (2004). Integrative analysis of cell cycle control in budding yeast.Mol. Biol. Cell 15:3841-3862

http://mpf.biol.vt.edu/research/budding yeast model/pp/

■ building a model1. wiring diagram (pathways)2. mathematical equations3. assign values to the parameters


Elementary modules

Examples





● Example: yeast




■ wiring diagram: network of proteins entering into the process


Elementary modules

Examples





● Example: yeast




■ ODEsTable 1. Equations

d

dt�mass� � kg � �mass�

d�Cln2�

dt� �k�s,n2 � k�s,n2 � �SBF�� mass� � kd,n2 � �Cln2�

d�Clb5�

dt� �k�s,b5 � k�s,b5 � �MBF�� mass� � �kd3,c1 � �C5P� � kdi,b5 � �C5�� kd3,f6 � �F5P� � kdi,f5 � �F5�� Vd,b5 � kas,b5 � �Sic1� � kas,f5 � �Cdc6�� Clb5�

d�Clb2�

dt� �k�s,b2 � k�s,b2 � �Mcm1�� mass� � �kd3,c1 � �C2P� � kdi,b2 � �C2�� kd3,f6 � �F2P� � kdi,f2 � �F2�� Vd,b2 � kas,b2 � �Sic1� � kas,f2 � �Cdc6�� Clb2�

d�Sic1�

dt� �k�s,c1 � k�s,c1 � �Swi5�� Vd,b2 � kdi,b2� � �C2� � �Vd,b5 � kdi,b5� � �C5� � kpp,c1 � �Cdc14� � �Sic1P� � �kas,b2 � �Clb2� � kas,b5 � �Clb5�

� Vkp,c1) � �Sic1�

d�Sic1P�

dt� Vkp,c1 � �Sic1� � �kpp,c1 � �Cdc14� � kd3,c1� � �Sic1P� � Vd,b2 � �C2P� � Vd,b5 � �C5P�

d�C2�

dt� kas,b2 � �Clb2� � �Sic1� � kpp,c1 � �Cdc14� � �C2P� � �kdi,b2 � Vd,b2 � Vkp,c1� � �C2�

d�C5�

dt� kas,b5 � �Clb5� � �Sic1� � kpp,c1 � �Cdc14� � �C5P� � �kdi,b5 � Vd,b5 � Vkp,c1� � �C5�

d�C2P�

dt� Vkp,c1 � �C2� � �kpp,c1 � �Cdc14� � kd3,c1 � Vd,b2� � �C2P�

d�C5P�

dt� Vkp,c1 � �C5� � �kpp,c1 � �Cdc14� � kd3,c1 � Vd,b5� � �C5P�

d�Cdc6�

dt� �k�s,f6 � k�s,f6 � �Swi5� � k�s,f6 � �SBF�� Vd,b2 � kdi,f2� � �F2� � �Vd,b5 � kdi,f5� � �F5� � kpp,f6 � �Cdc14� � �Cdc6P� � �kas,f2 � �Clb2�

� kas,f5 � �Clb5� � Vkp,f6) � �Cdc6�

d�Cdc6P�

dt� Vkp,f6 � �Cdc6� � �kpp,f6 � �Cdc14� � kd3,f6� � �Cdc6P� � Vd,b2 � �F2P� � Vd,b5 � �F5P�

d�F2�

dt� kas,f2 � �Clb2� � �Cdc6� � kpp,f6 � �Cdc14� � �F2P� � �kdi,f2 � Vd,b2 � Vkp,f6� � �F2�

d�F5�

dt� kas,f5 � �Clb5� � �Cdc6� � kpp,f6 � �Cdc14� � �F5P� � �kdi,f5 � Vd,b5 � Vkp,f6� � �F5�

d�F2P�

dt� Vkp,f6 � �F2� � �kpp,f6 � �Cdc14� � kd3,f6 � Vd,b2� � �F2P�

d�F5P�

dt� Vkp,f6 � �F5� � �kpp,f6 � �Cdc14� � kd3,f6 � Vd,b5� � �F5P�

d�Swi5�T

dt� k�s,swi � k�s,swi � �Mcm1� � kd,swi � �Swi5�T

d�Swi5�

dt� k�s,swi � k�s,swi � �Mcm1� � ka,swi � �Cdc14� � ��Swi5�T � �Swi5�� kd,swi � ki,swi � �Clb2�� Swi5�

d�APC-P�

dt�

ka,apc � �Clb2� � �1 � �APC-P��

Ja,apc � 1 � �APC-P��

ki,apc � �APC-P�

Ji,apc � �APC-P�

d�Cdc20�T

dt� k�s,20 � k�s,20 � �Mcm1� � kd,20 � �Cdc20�T

d�Cdc20�A

dt� �k�a,20 � k�a,20 � �APC-P�� Cdc20�T � �Cdc20�A� � �kmad2 � kd,20� � �Cdc20�A

(continued)

Integrative Analysis of Cell Cycle Control

Vol. 15, August 2004 3843

Table 1. (Continued).

d�Cdh1�T

dt� ks,cdh � kd,cdh � �Cdh1�T

d�Cdh1�

dt� ks,cdh � kd,cdh � �Cdh1� �

Va,cdh � ��Cdh1�T � �Cdh1��

Ja,cdh � �Cdh1�T � �Cdh1��

Vi,cdh � �Cdh1�

Ji,cdh � �Cdh1�

d�Tem1�

dt�

klte1 � ��Tem1�T � �Tem1��

Ja,tem � �Tem1�T � �Tem1��

kbub2 � �Tem1�

Ji,tem � �Tem1�

d�Cdc15�

dt� �k�a,15 � ��Tem1�T � �Tem1�� k�a,15 � �Tem1� � k�a,15 � �Cdc14�� Cdc15�T � �Cdc15�� ki,15 � �Cdc15�

d�Cdc14�T

dt� ks,14 � kd,14 � �Cdc14�T

d�Cdc14�

dt� ks,14 � kd,14 � �Cdc14� � kd,net � ��RENT� � �RENTP�� kdi,rent � �RENT� � kdi,rentp � �RENTP� � �kas,rent � �Net1�

� kas,rentp � �Net1P�) � �Cdc14�

d�Net1�T

dt� ks,net � kd,net � �Net1�T

d�Net1�

dt� ks,net � kd,net � �Net1� � kd,14 � �RENT� � kdi,rent � �RENT� � kas,rent � �Cdc14� � �Net1� � Vpp,net � �Net1P� � Vkp,net � �Net1�

d�RENT�

dt� � �kd,14 � kd,net� � �RENT� � kdi,rent � �RENT� � kas,rent � �Cdc14� � �Net1� � Vpp,net � �RENTP� � Vkp,net � �RENT�

d�PPX�

dt� ks,ppx � Vd,ppx � �PPX�

d�Pds1�

dt� k�s,pds � k�s1,pds � �SBF� � k�s2,pds � �Mcm1� � kdi,esp � �PE� � �Vd,pds � kas,esp � �Esp1�� Pds1�

d�Esp1�

dt� � kas,esp � �Pds1� � �Esp1� � �kdi,esp � Vd,pds� � �PE�

d�ORI�

dt� ks,ori � ��ori,b5 � �Clb5� � �ori,b2 � �Clb2�� kd,ori � �ORI�

d�BUD�

dt� ks,bud � ��bud,n2 � �Cln2� � �bud,n3 � �Cln3� � �bud,b5 � �Clb5�� kd,bud � �BUD�

d�SPN�

dt� ks,spn �

�Clb2�

Jspn � �Clb2�� kd,spn � �SPN�

G�Va,Vi,Ja,Ji� �

2JiVa

Vi � Va � JaVi � JiVa � ��Vi � Va � JaVi � JiVa�2

� 4�Vi � Va�JiVa

�SBF� � �MBF� � G�Va,sbf, Vi,sbf, Ja,sbf, Ji,sbf�

�Mcm1� � G�ka,mcm � �Clb2�, ki,mcm, Ja,mcm, Ji,mcm�

�Cln3� � C0 � Dn3 � �mass�/�Jn3 � Dn3 � �mass��

�Bck2� � B0 � �mass�

�Clb5�T � �Clb5� � �C5� � �C5P� � �F5� � �F5P�

�Clb2�T � �Clb2� � �C2� � �C2P� � �F2� � �F2P�

�Sic1�T � �Sic1� � �Sic1P� � �C2� � �C2P� � �C5� � �C5P�

�Cdc6�T � �Cdc6� � �Cdc6P� � �F2� � �F2P� � �F5� � �F5P�

�CKI�T � �Sic1�T � �Cdc6�T

(continued)

K.C. Chen et al.

Molecular Biology of the Cell3844


Elementary modules

Examples





● Example: yeast




■ the model has◆ 36 ODE (mostly nonlinear: quadratic, MM,

Goldeberg-Koshland functional)◆ 24 algebraic equations (linear and nonlinear)◆ ∼ 130 parameters

■ simulated behavior:◆ fits the cell cycle of the wild type◆ predicts correctly the phenotype behavior of 120 (out of

131 studied) mutants (obtained by gene inhibition oroverexpression)

◆ is not crucially sensitive to the values chosen for theparameters


Elementary modules

Examples





● Example: yeast



A few (!) software packages

BALSA BASIS BIOCHAM BioCharon biocyc2SBML

BioGrid BioModels BioNetGen BioPathway Explorer PathArt

Bio Sketch Pad BioSens BioSPICE Dashboard BioSpreadsheet BioTapestry

BioUML BSTLab CADLIVE CellDesigner Cellerator

CellML2SBML Cellware CL-SBML COPASI Cytoscape

DBsolve Dizzy E-CELL ecellJ ESS

FluxAnalyzer Fluxor Gepasi INSILICO discovery JACOBIAN

JDesigner JWS Online Karyote& KEGG2SBML Kinsolver&

libSBML MathSBML MesoRD MetaboLogica MetaFluxNet

MMT2 Modesto Moleculizer Narrator NetBuilder

PANTHER Pathway PathScout PathwayLab Pathway Tools PathwayBuilder

PaVESy Reactome ProcessDB PROTON pysbml

PySCeS runSBML SBML ODE Solver SBMLeditor SBMLR

SBMLSim SBMLToolbox SBToolbox SCIpath Sigmoid&

SigPath SIMBA SimBiology Simpathica SimWiz

SmartCell SRS Pathway Editor StochSim STOCKS TERANODE Suite

Trelis Virtual Cell WinSCAMP XPPAUT


Elementary modules

Examples





● Example: yeast



One for all: SBML

■ SBML= System Biology Markup Language■ made by:

◆ Caltech◆ University of Hertfordshire◆ Systems Biology Institute of Japan

Documents

· Feedback regulation Positive feedback Negative regulation A gene regulatory network Other regulatory elements Examples Claudio Altaﬁni, February 15, 2007 – p. 12/54 Elementary