Lecture 7: Adaptation & Transient Responses - ethz.ch Adaptation Network Architectures for Adaptation Protein degradation and sequestration in Adaptation Transient Responses We now

Perfect AdaptationNetwork Architectures for Adaptation

Protein degradation and sequestration in Adaptation

Lecture 7: Adaptation & Transient Responses

Prof Dagmar Iber, PhD DPhil

Computational Biology (CoBi), D-BSSE, ETHZ

1



Outline

1 Perfect Adaptation

2 Network Architectures for Adaptation1-component network2-component network3-component network

3 Protein degradation and sequestration in Adaptation

2



Transient Responses

We now consider a situationwhere the steady state isstable and ideally its valuedoes not change in responseto the signal.

Yet, a transient response isobserved as the signal triggersa brief excursion before thesystem returns to its stablesteady state.

3



Transient Responses

To discuss transient responses we need to relate the signalinput I to the signal output or response O. Figure 1 shows atypical transient response.

Figure : Transient Responses. O1 is the output with pre-stimulationinput I1, Opeak the maximal output, and O2 the long-term output in thepresence of input I2.

4



Transient Responses

The response to a persisting stimulus I2− I1 can becharacterized by the sensitivity of the response,

S =

∣∣∣∣(Opeak −O1)/O1

(I2− I1)/I1

∣∣∣∣ , (1)

and the precision of long-term adaptation to pre-stimulusresponse levels,

P =

∣∣∣∣(O2−O1)/O1

(I2− I1)/I1

∣∣∣∣ . (2)

5



Types of Adaptation

Perfect adaptation: if a system responds only transiently to apersisting perturbation and returns to the pre-stimulussteady-state value (O2 = O1).

Near-perfect adaptation: if it returns to a steady-state nearby(O2−O1� 1).

Sustained response: If O2 ∼Opeak

6



Perfect Adaptation

7



Perfect Adaptation

To obtain perfect adaptation three conditions need to be met:the steady state needs to be stable,the value/position of the steady-state must be insensitive tothe perturbation,the kinetic parameters must be such that there is atransient response before the system returns to the steadystate.

8



Conditions for Perfect Adaptation

Consider the general ODE model

dxdt

= f (x , t ,σ). (3)

At the steady state xs, dxdt = f (xs, t ,σ) = 0.

We require that the steady state xs is stable (i.e. the real partsof the eigenvalues of the Jacobian are all negative) and that thevalue of xs does not change in response to a change in thesignal strength σ , i.e.

dxs

dt= f (xs, t ,σ) = f (xs, t ,σ + ∆σ) = 0. (4)

9




We expand f (xs, t ,σ + ∆σ) in a Taylor series and obtain

f (xs, t ,σ + ∆σ) = f (xs, t ,σ) +dfdσ

∣∣∣∣s

∆σ + h.o.t .. (5)

Since f (xs, t ,σ) = f (xs, t ,σ + ∆σ) = 0 and ∆σ 6= 0 we require

dfdσ

∣∣∣∣s

=∂ f∂σ

∣∣∣∣s

+∂ f∂x

∣∣∣∣s

dxs

dσ= 0 (6)

which can be rewritten as

dxs

dσ=− ∂ f

∂x

∣∣∣∣−1

s

∂ f∂σ

∣∣∣∣s

= 0. (7)

10




dxs

dσ=− ∂ f

∂x

∣∣∣∣−1

s

∂ f∂σ

∣∣∣∣s

= 0. (8)

The steady state thus does not change in response to aperturbation in σ if either

∂ f∂x

∣∣∣−1

s= 0: the inverse of the Jacobian cannot be entirely 0

∂ f∂σ

∣∣∣s

= 0: the perturbation would then not affect the kinetics ofthe system. This can be the case when gain and loss rates arenot at all affected, or are affected in the same way by σ .However, in this case the system will also not respondtransiently.

∂ f∂x

∣∣∣−1

s∂ f∂σ

∣∣∣s

= 0: Adaptation is possible because ∂ f∂σ

∣∣∣s

willgenerally be a sparse vector with few non-zero entries since inmost cases only one (or few) elements serve as a receiver.

11




dxs

dσ=− ∂ f

∂x

∣∣∣∣−1

s

∂ f∂σ

∣∣∣∣s

= 0. (9)

Suppose that for simplicity the first entry is the only non-zeroentry in ∂ f

∂σ

∣∣∣s. Then it would suffice that the first column of the

inverse of the Jacobian was zero to obtain dxsdσ

= 0.

In general, there are only few response units that are critical inthe network. In that case only the element in the 1st column inthe row(s) of the response unit has/have to be zero in theinverse of the Jacobian.

In the following we will study the implications of this idea byexample of 1-, 2-, and 3-component networks.

12



1-component network2-component network3-component network

Network Architectures for Adaptation

13




Network Architectures for Adaptation

We will now analyse what network architectures can give rise totransient responses to a sustained stimulus if componentscannot be degraded and if there is no explicit delay (i.e. nodelay equations).

For more details, please see Ma, W., et al. (2009). "Definingnetwork topologies that can achieve biochemical adaptation."Cell 138(4): 760-773.

14




1-component network

15




1-component network

Perfect adaptation is impossible in a 1-component network -mainly because it would be impossible to generate the delaythat is necessary to obtain a transient response.

It is instructive, however, to ask under what condition theposition of the steady state would be insensitive to fluctuationsin the signal strength σ .

16




1-component network

For an insensitive steady state of a 1-component network

dxdt

= f (x , t ,σ). (10)

we requiredxs

dσ=− df

dx

∣∣∣∣−1

s

∂ f∂σ

∣∣∣∣s

= 0. (11)

dfdx

∣∣∣−1

s< 0 is required for the steady state to be stable.

We therefore require ∂ f∂σ

∣∣∣s

= 0 for dxsdσ

= 0.

However, this implies that the perturbation does not affect thekinetics of the system.

17




Example for steady state insensitivity

Forf (x , t ,σ) = σ −σx (12)

we have∂ f∂σ

∣∣∣∣s

= 0. (13)

Here, σ affects gain and loss rates the same.

This is the only way (and rather a trivial one) of achieving aninsensitivity of the steady state to perturbations (input-outputrobustness) in a 1-component system.

18




2-component network

19




2-component network

In a 2-component system J = ∂ f∂x

∣∣∣s

refers to the Jacobian withelements

J =

(J11 J12J21 J22

)(14)

We need to distinguish the cases when receiver and responseelement are the same or distinct elements.

20




Case 1: distinct components

1st component: sensor2nd component: response unit.

Input Signal // x1

J21

##

J11

��

x2

J22

GG

J12

cc

// Output

21




Case 1: distinct components

Condition for perfect adaptation of the response unit:

dx2s

dσ=−J−1

21∂ f∂σ

∣∣∣∣s

= 0. (15)

where J−121 denotes the entry on the second row and first

column of the inverse of the Jacobian.

22




Inverse

Recall that for a 2×2 matrix

J =

(J11 J12J21 J22

)(16)

the inverse is given by

J−1 =1

detJ

(J22 −J12−J21 J11

). (17)

23




2-component network

dx2s

dσ=−J−1

21∂ f∂σ

∣∣∣∣s

= 0. (18)

We thus require J21 = 0, i.e. the sensor element must notimpact on the response element.

It is intuitively clear that isolation of the response unit from thesensor will permit the response unit to remain unaffected.

However, it is also clear that there will also not be any transientresponses and thus no perfect adaptation.

24




Case 2: sensor and response unit same element

1st component: sensor & response unit

Input Signal // x1

J21

%%

J11

��// Output

x2

J22

GG

J12

ff

25




Case 2: sensor and response unit same element

Assuming dual function element is the first entry x1:

dx1s

dσ=−J−1

11∂ f∂σ

∣∣∣∣s

= 0. (19)

Since J−111 = J22

det(J) we require J22 = 0.

Since in biology all components decay, the decay rate wouldeither need to be limited by another factor (i.e. component 2would need to be highly abundant), or there would need to besome auto-activation of component 2 to yield net J22 = 0.

26




2-component network

For stability of the steady state we require

tr(J) < 0, det(J) > 0. (20)

This implies that J11 < 0 and that the off-diagonals must haveopposite signs, i.e.

J =

(− + | −− |+ 0

). (21)

There is thus a negative feedback between the two nodes andthe second node functions as a buffering node.

27




2-component network

J =

(− + | −− |+ 0

). (22)

However, since the kinetics of the second element are affectedonly by the first element in steady state, the first componentmust be zero. If the first component is zero in steady state so isthe second component.

Non-zero steady states can be obtained if we allow constantfluxes that do not depend on either variable. However, in thiscase we consider an open, non-conserved system.

Adaptation is thus not possible in 1- or 2-node networks withconserved protein concentrations.

28




Example: 2-component TGF-β model

We return to our 2-component model for TGF-β signaling withthe variables R for the R-Smad concentration and I for theI-Smad concentration.

As before we linearize the system at the steady state and studythe dynamics of the perturbed steady state concentrations ∆Rand ∆I, i.e.(

d∆R \dtd∆I \dt

)=

(fR fIgR gI

)(∆R∆I

)+

(∂ f \∂σ

0

)∆σ = 0

(23)and solve for ∆R.

29





(d∆R \dtd∆I \dt

)=

(fR fIgR gI

)(∆R∆I

)+

(∂ f \∂σ

0

)∆σ = 0

(24)After some rearrangements we obtain

∆R \Rs

∆σ \σ=−∆σ

RsJ−1

(∂ f \∂σ

0

)=−∆σ

Rs

gI

detJ∂ f∂σ

. (25)

Insensitivity of Rs to perturbations in σ therefore requiresgI = 0.

30





As gI 6= 0 in our model, the steady state value is sensitive to thesignal:

31





∆R \Rs

∆σ \σ=−∆σ

RsJ−1

(∂ f \∂σ

0

)=−∆σ

Rs

gI

detJ∂ f∂σ

. (26)

Insensitivity of Rs to perturbations in σ therefore requiresgI = 0, i.e. I must have no net effect on itself.

However, since R has a positive impact on I (gR > 0) thisimplies that in steady state R = 0 and thus I = 0.

While this steady state is stable and insensitive to perturbationsin σ it is not a very interesting system state.

32




3-component network

33




3-component network

We now consider a 3-component network of the form

I // S

fMS

fRS

##

fSS��

MfMM 88

fSM

II

fRM33 R

fRR

EE

fSR

cc

fMRss // O

Here the sensor S senses the input I and transfers theinformation either directly or indirectly via the mediator M to theresponse element R that generates the output O.

34




3-component network

We can now analyse the dynamic network by writing

S = fS(S,M,R) (27)M = fM(S,M,R) (28)R = fR(S,M,R). (29)

The Jacobian J of the system is given by

J =

fSS fSM fSRfMS fMM fMRfRS fRM fRR

. (30)

The subscript denotes fXY = ∂ fX∂Y , where X = fX (X ,Y ,Z ).

35




3-component network

We can then linearize the system around the steady state andwrite for the time evolution of small perturbations ∆X aroundthe steady state X ∗

ddt

∆S∆M∆R

=


∆S∆M∆R

+

fSI00

∆I.

For the steady state to be stable we require that the real partsof all eigenvalues are negative.

36




3-component network

In a 3-component system we have for the characteristicpolynomial

P(λ ) = λ3 + α2λ

2 + α1λ + α0 = 0. (31)

The coefficients are determined by the Jacobian J of thedynamical system, i.e. α0 =−det(J), α2 =−tr(J).

Based on Descartes’ Rule of signs we know that the polynomialhas at most as many roots which are real and positive and thecoefficients have sign changes. To have no roots which are realand positive we require α0,α1,α2 > 0, and thus as before in the2-component system tr(J) < 0, but unlike in the 2-componentsystem det(J) < 0.

37




3-component network

To obtain adaptation, the stability of the steady state is notsufficient. We also require that the value of the steady statedoes not change as the input changes. The change in thesteady state value in response to a change in the input can becalculated by setting

ddt

∆S∆M∆R

=


∆S∆M∆R

+

fSI00

∆I = 0

and solving for ∆R. We obtain

∆R \Rs

∆I \ I=

IRs

(J−1)31∂ f∂ I

. (32)

where ∂ f∂ I = fSI .

38




3-component network

∆R \Rs

∆I \ I=

IRs

(J−1)31∂ f∂ I

. (33)

Note that ∆R = 0 would be trivially obtained if there is nochange in input, i.e. ∆I = 0 as we would expect for a stablesteady state.

If ∂ f∂ I ∆I 6= 0 then it depends on the inverse of Jacobian (J−1)31

and thus on the particular reaction kinetics whether or not∆R = 0.

39




3-component network

The inverse of matrix J is

J−1 =adj(J)

det(J)

with adj(J) denoting the adjugate matrix of J. It is defined by

adj(J)ij = (−1)i+jMji

where Mji is the (i , j) minor of J, i.e., the determinat of the(n−1)× (n−1) matrix that results from deleting row j andcolumn i of J.

40




3-component network

Taking all together, (J−1

)31

=N

det(J)

where N refers to the subdeterminant of the Jacobian J that isobtained by removing the row that describes the dynamics ofthe sensor and by removing the column of the response unit,i.e.

N = det(J−131 ) = det

(J21 J22J31 J32

)= fMSfRM − fMM fRS

det(J) = fSSfMM fRR + fSM fMRfRS + fSRfMSfRM

−fRSfMM fSR− fRM fMRfSS− fRRfMSfSM

41




3-component network

Condition for Perfect Adaptation

∆R \Rs

∆I \ I=

IRs

NdetJ

∂ f∂ I

= 0. (34)

For the steady state to be stable, we require det(J) < 0.Moreover, Rs > 0 and ∂ f

∂ I > 0.

To achieve perfect adaption, we thus require

N = det(J−131 ) = det

(J21 J22J31 J32

)= fMSfRM − fMM fRS = 0.

Condition for Perfect Adaptation1 J21J32 = J22J31 = 02 J21J32 = J22J31 6= 0.

42




Case I: Negative Feedback with a buffer node

I // S

fMS

fRS

##

fSS��

MfMM 88

fSM

II

fRM33 R

fRR

EE

fSR

cc

fMRss // O

J21J32 = J22J31 = 0fMSfRM = fMM fRS = 0

Condition I rules out afeed-forward loop since either

1 J21 = fMS = 0 (no impact ofsensor on buffer element)

2 J32 = fRM = 0 (no impact ofbuffer element onresponse element).

Since J31 = fRS 6= 0 must thenhold (path from S to R),J22 = fMM = 0 (no impact ofbuffer node on itself).

43





There are thus two possible compatible network designs:a. fMM = 0 and fMS = 0 b. fMM = 0 and fRM = 0

// S

MfRM

33 R��

fRS

//

// S

fMS

M R

��

fRS

//

We immediately notice that there is in both cases only oneconnection between S and R. This means that there is nopossibility of a feed-forward, i.e. there is no possibility to relaytwo separate signals from S on R at different speeds.

44





Let us now look at the second constrain that arises from theneed to obtain a stable steady state:

det(J) < 0.

Network I: fMS = 0⇒ det(J) = fSM fMRfRS− fRM fMR fSS < 0.

// S

M

fSM

II

fRM33 RfMRss��

fRS

//

To obtain a negative determinant, eitherfSM fMRfRS < 0 and/or fRM fMR fSS > 0. Thefirst implies a net negative feedback.For stability we further require negativeself-regulation, i.e. fSS, fRR < 0. Thisimplies for the second term thatfRM fMR < 0.

There is thus at least one negativefeedback.

45





Let us now look at the second constrain that arises from theneed to obtain a stable steady state:

det(J) < 0.

Network II: fRM = 0⇒ det(J) = fSM fMRfRS− fRRfMSfSM < 0.

// S

fMS

M

fSM

II

RfMRss��

fRS

//

To obtain a negative determinant, eitherfSM fMRfRS < 0 and/or fRRfMSfSM > 0. Thefirst implies a net negative feedback.

For stability we further require negativeself-regulation, i.e. fSS, fRR < 0. Thisimplies for the second term thatfMSfSM < 0.

There is thus at least one negativefeedback.

46





// S

M

fSM

II

fRM33 RfMRss��

fRS

//

// S

fMS

M

fSM

II

RfMRss��

fRS

//

fSM fMRfRS < 0, fRM fMR < 0 and fMSfSM < 0 all imply a negativefeedback node since as the net action from R on itself via M isnegative. We therefore speak of a negative feedback designwith a buffer node (M).

47





To achieve perfect adaptationwe require that the nullcline forthe buffer node B does notchange with regard to theresponse node C.

The adaptation precision isdirectly related to the flatnessof the B nullcline near theintersection of the twonullclines.

48





Saturation (zerothorderultrasensitivity)leads to flatness inB nullcline: i.e.small change in Bdoes not affect C,but other wayround large effect.

49




RECALL: Ultrasensitive Switch

Consider a protein X that can exist in a phosphorylated (Xp)and an un-phosphorylated (X ) state. If E1 and E2 are thekinase and phosphatase of X , respectively we have

X + E1k1

GGGGGGBFGGGGGG

k−1

C1k2−→ E1 + Xp

Xp + E2p1

GGGGGGBFGGGGGG

p−1C2

p2−→ E2 + X

and thus for the kinetics of the phosphorylated andunphosphorylated forms, Xp and X respectively,

d [Xp]

dt=−d [X ]

dt= kphosS

XT − [Xp]

KM1 + XT − [Xp]−kdephos

[Xp]

KM2 + [Xp]

Here kphos and kdephos are the vmax of the enzymatic reactions.S refers to an external signal that is assumed to only affect thekinase and thus the phosphorylation reaction.

50




RECALL: Ultrasensitive Switch

d [Xp]

dt=−d [X ]

dt= kphosS

XT − [Xp]

KM1 + XT − [Xp]−kdephos

[Xp]

KM2 + [Xp]

If both enzymes are saturated with substrate then a smallchange in vmax via a signal S can shift the system from beingmainly phosphorylated to being mainly dephosphorylated.

Figure : Ultrasensitivity in the Goldbeter-Koshland Kinetics. (a)Reversable reactions can give rise to ultrasensitive responses. (b) Ifthe enzymes are saturated with substrate (XT � ET ) then a smallchange in the vmax of one of the enzymes can greatly alter theresponse. (c) Without enzyme saturation the response is much lesssensitive to changes in the vmax of one of the enzymes. Note that theratio of the the vmax of the kinase and the phosphatase are given onthe horizontal axis. The vmax includes the turn-over rate constant aswell as the total enzyme concentration. Regulatory interactions cantherefore change the vmax .

51





Consider saturating conditions for enzymes acting on node B(KCB, K ′FBB � 1): dB

dt = CkCB−FBk ′FBB

52





dBdt

= CkCB−FBk ′FBB (35)

In steady-state we then have

C∗ =FBk ′FBB

kCB(36)

allowing us to write

dBdt

= kCB(C−C∗) (37)

B = B∗(I0) + kCB

∫ t

0(C−C∗)dτ (38)

53




Case I: Integral Control

B = B∗(I0) + kCB

∫ t

0(C−C∗)dτ (39)

Integral ControlThis network design leads to adaptation by integral control.

All minimal negative feedback topologies with buffering nodefollow this integral control mechanism to achieve adaptation.

54




Case I: Sensitivity

The ability to mount an appropriatetransient response to the inputchange before achieving steady-stateadaptation depends on the vectorfields (dB/dt , dC/dt) in the phaseplane.

A large excursion (large sensitivity)requires large initial |dC/dt | and asmall initial |dB/dt | near thepre-stimulus steady-state.

55




Case I: Sensitivity

A large excursion (large sensitivity)requires that the response time ofnode C to the input change is fasterthan the adaptation time.

Slower adaptation time would requiresmaller rate constants in the C-Bloop.

56




Case I: Summary

PrecisionDetermined by Michaelis-MentenconstantsNeed to be tuned to achieveoperation in saturated regimes

SensitivityDetermined by timescales of thesystem

These two objectives can be tunedindependently.

57




Case II: Incoherent Feedforward

We now require J21,J22,J31,J32 6= 0 and J21J32 = J22J31.

In terms of the partial derivatives we thus havefMSfRM = fMM fRS 6= 0. This case implies a feedforward loopsince fMS, fRM , fMM , fRS 6= 0:

// S

fMS

M

fRM33 R��

fRS

//

58




Case II: Incoherent Feedforward

// S

fMS

M

fRM33 R��

fRS

//

In order to have a stable steady state we require J22 < 0 andthus J22 = J21J32

J31< 0 or in terms of the partial derivatives

fMSfRM

fRS= fMM < 0.

This means that the two branches that connect S and R haveopposite signs.

59




Case II: Incoherent Feedforward = proportional control

Mechanism: I activates C via A, but also activates B via A,which then counteracts the positive effect on C.

60





If the steady-state concentration of B is proportional to that ofthe positive regulatory A, then the equation determining thesteady-state value of C, dC/dt = 0, is independent of A andhence of the input I.

61





In order for the steady-state value of B to be proportional to thesteady-state value of A we require the first term to be in thesaturated region ((1−B)� KAB), and the second term tolinearly depend on B (K ′FBB � B).

62




Case II: Incoherent Feed-forward loop

Precision: Maintaining aconstant C requires theB nullcline to move thesame distance as the Cnullcline in response toan input change.

Sensitivity of the circuit:depends on the ratio ofthe speeds of the twosignal transductionbranches: A→ C, andA→ B→ C. These canbe independently tunedfrom the adaptationprecision.

63




Comparison of adaptation topologies

Negative FeedbackOutput node must notdirectly feedback to theinput nodeFeedback needs to feedthrough intermediate nodethat serves as buffer

Incoherent feed-forwardPerforms adaptation morerobustly than negativefeedback loops: capable ofhigher sensitivity higherprecisionIntermediate node needsto serve as proportioner

64




Summary: Adaptation in a 3-node network

To achieveadaptation in a3-node networkwith conservedproteinconcentrations,sensor andresponse elementneed to beseparate elementsand knownnetwork topologiesbelong to either oftwo networkarchitectures.

65





To enable the formationof a transient signal it isimportant that

the response istriggered rapidlythe second rectifyingprocess isresponding with adelay. This isachieved by theappropriate reactionkinetics.

66





only 0.01% ofall testedpossible3-nodenetworksshowadaptationno 2-link 3nodenetworksfound capableof adaptation

67




68




If protein concentrations are allowed to change then perfectadaptation can be achieved by delayed degradation orsequestration.

Please see Behar et al. (2007) for more details.

69



Thanks!!

Thanks for your attention!

Slides for this talk will be available at:http://www.bsse.ethz.ch/cobi/education

Lecture 7: Adaptation & Transient Responses

Prof Dagmar Iber, PhD DPhil

Computational Biology (CoBi), D-BSSE, ETHZ

70

http://www.bsse.ethz.ch/cobi/education

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Lecture 7: Adaptation & Transient Responses - ethz.ch Adaptation Network Architectures for Adaptation Protein degradation and sequestration in Adaptation Transient Responses We now