the effect of randomness in chemical reactions …...the effect of randomness in chemical reactions...

The Effect of Randomness in ChemicalReactions on Decision-Making:

Two Short StoriesJ. P. Keener

Department of Mathematics

University of Utah

[email protected]

Length Regulation of Flagellar Hooks – p.1/46

The Fundamental Problem

All living organisms make decisions (when to divide, when todifferentiate, when to destroy, when to repair, when to grow,when to die)

• Question 1: What is the chemistry of decision making? Howis information about size, age, resource availability, etc.transduced into chemistry?

• Question 2: When the organism or organelle is small, thenumber of molecules involved in the decision makingprocess is necessarily small. How does the resultantrandomness affect the reliability of the decision?

To be or not or not to be, that is the question. – p.2/46

Story 1: Length Regulation

Observation: Every structure which could in principle be of anysize but instead is found in a narrow distribution of sizes must beendowed with a size control mechanism.

• Question 1: How are the specifications for the size of astructure set?

• Question 2: How are the decisions regarding manufacturemade? (When, what and how much to build?)

The first of many slides – p.3/46

Example: Cilia

Hair cell Steriocilia bronchial cilia

The first of many slides – p.4/46

Example:Tails

phage lambda chlamydomonas

The third of many slides – p.5/46

Injectosomes

sType III Secretion System - Injectosome

The fourth of many slides – p.6/46

Specific Example: Salmonella


Control of Flagellar Growth

The motor is built in a precise step-by-step fashion.

• Step 1: Basal Body

• Step 2: Hook (FlgE secretion)

• Step 3:Filament (FliC secretion)Basal Body






• Step 3: Filament (FliC secretion)

FlgE

Hook

Basal Body






• Step 3: Filament (FliC secretion)FlgE

Hook−filamentjunction

Filament

Basal Body

FliC

Hook

Click to see movies






• Step 3: Filament (FliC secretion)FlgE

Hook−filamentjunction

Filament

Basal Body

FliC

Hook

Click to see movies

• How are the switches between steps coordinated?

• How is the hook length regulated (55 ±6 nm)?


Proteins of Flagellar Assembly


Hook Length Regulation

• Hook is built by FlgE secretion.




• FliK is the "hook length regulatory protein".




• FliK is the "hook length regulatory protein".• FliK is secreted only during hook production.




• FliK is the "hook length regulatory protein".• FliK is secreted only during hook production.• Mutants of FliK produce long hooks; overproduction of

FliK gives shorter hooks.





FliK gives shorter hooks.• Lengthening FliK gives longer hooks.





FliK gives shorter hooks.• Lengthening FliK gives longer hooks.• 5-10 molecules of FliK are secreted per hook (115-120

molecules of FlgE).


Hook Length Data

Mutants of FliK produce long hooks; overproduction of FliK givesshorter hooks:

0 20 40 60 80 1000

50

100

150

200

250

Num

ber

of h

ooks

(a) L(nm)0 20 40 60 80 100

0

50

100

150

200

250

Num

ber

of h

ooks

(b) L(nm)0 50 100 150 200 250

0

50

100

150

200

250

Num

ber

of h

ooks

(c) L(nm)

Wild type Overexpressed Underexpressed(M = 55nm) (M = 47nm) (M = 76nm)


The Secretion Machinery

• Secreted molecules arechaperoned to preventfolding.

Step 1

FliI

FliJ

Cring

MS ring

CM FlhA FlhB

C

N

FliH

componentsmembrane




• Insertion is facilitated byhydrolysis of ATP.

FliJ

Step 2

N

C

Cring

MS ring

CM FlhA FlhB

membranecomponents

FliH

FliI





• FlhB is the gatekeeperrecognizing the Nterminus of secretants.

N

C

Step 3

Cring

MS ring

CM FlhA FlhB

FliJ

ATP ADP+Pi

FliH

membranecomponents

FliI





• FlhB is the gatekeeperrecognizing the Nterminus of secretants.

• Once inside, molecularmovement is by diffu-sion.

C

Step 4

FliJ

N

Cring

MS ring

CM FlhA FlhB

membranecomponents

FliI

FliH


Secretion Control

Secretion is regulated by FlhB

• During hook formation, only FlgE and FliK can be secreted.

• After hook is complete, FlgE and FliK are no longersecreted, but other molecules can be secreted (thoseneeded for filament growth.)

• The switch occurs when the C-terminus of FlhB is cleavedby FlK.

Question: Why is the switch in FlhB length dependent?


Hypothesis: How Hook Length isdetermined

• The Infrequent Molecular Ruler Mechanism. FliK issecreted once in a while to test the length of the hook.

• The probability of FlhB cleavage is length dependent.


Binding Probability

Suppose the probability of FlhB cleavage by FliK is a function oflength Pc(L). Then, the probability of cleavage on or before timet, P (t), is determined by

dP

dt= αr(L)Pc(L)(1− P )

where r(L) is the secretion rate, α is the fraction of secretedmolecules that are FliK, and

dL

dt= βr(L)∆

where β = 1− α fraction of secreted FlgE molecules, ∆ lengthincrement per FlgE molecule.


Binding Probability

It follows thatdP

dL=

α

β∆Pc(L)(1− P )

or

− ln(1− P (L)) = κ

∫ L

0Pc(L)dL

Remark: The only difference between mutant strains should bein the parameter κ = α

β∆ .


Check the Data

0 20 40 60 80 1000

50

100

150

200

250

Num

ber

of h

ooks

(a) L(nm)0 20 40 60 80 100

0

50

100

150

200

250

Num

ber

of h

ooks

(b) L(nm)0 50 100 150 200 250

0

50

100

150

200

250

Num

ber

of h

ooks

(c) L(nm)

Wild type Overexpressed Underexpressed

0 10 20 30 40 50 60 70 80 90 1000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Hook Length (nm)

P(L

)

OverexpressedWTUnderexpressed


Check the Data

0 10 20 30 40 50 60 70 80 90 1000

2

4

6

8

10

Hook Length (nm)

−ln

(1−

P)

0 10 20 30 40 50 60 70 80 90 1000

2

4

6

8

10

Hook Length (nm)

−κ

ln(1

−P

)

OverexpressedWTUnderexpressed

κ = 0.9κ = 1κ = 4

− ln(1− P (L)) = κ

∫ L

0Pc(L)dL?


Use the Data

to estimate Pc(L):

0 10 20 30 40 50 60 70 80 90 1000

1

2

3

4

5

6

Hook Length (nm)0 10 20 30 40 50 60 70 80 90 100

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Hook Length (nm)

Pc(L

)


Test #2: 3 Cultures

25 30 35 40 45 50 55 60 65 700

5

10

15

hook length (nm)

num

ber

of h

ooks

0 200 400 600 800 1000 12000

2

4

6

8

10

12

14

hook length (nm)

num

ber

of h

ooks

1: WT 2: No FliK

0 50 100 150 200 250 300 350 400 4500

10

20

30

40

50

60

70

80

hook length (nm)

num

ber

of h

ooks

3: FliK induced at 45 min

Question: Can 3 be predicted from 1 and 2?


Test #2

Suppose that FliK becomes available only after the hook islength L0. Then,

dP

dL= βPc(L)(1− P ).

with P (L|L0) = 0 so that

P (L|L0) = 1− exp(−κ

∫ L

L0

Pc(L)dL).

Let’s see how this formula can be used:


Step 1: Analysis of WT Data

1) Determine Pc(L) from WT data

25 30 35 40 45 50 55 60 65 700

5

10

15

hook length (nm)

num

ber

of h

ooks

25 30 35 40 45 50 55 60 65 700

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Hook Length (nm)

P(L

)

25 30 35 40 45 50 55 60 65 700

1

2

3

4

5

6

7

Hook Length (nm)

−ln

(1−

P)

20 25 30 35 40 45 50 55 60 65 700

0.05

0.1

0.15

0.2

0.25

Hook Length (nm)

β P

c(L)

(nm

−1 )


Step 2: Analysis of Polyhook Data

2) Determine distribution of polyhooks Pp(L) (hooks grown withno hook length control)

0 200 400 600 800 1000 12000

2

4

6

8

10

12

14

hook length (nm)

num

ber

of h

ooks

0 200 400 600 800 1000 12000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

hook lengthP

p(L)


Step 3: Predict Lengths after FliKinduction

Pi(L) = P (L|0)Pp(L∗) +

∫ L

0P (L|L0)pp(L0 + L∗)dL0.

where P (L|L0) = 1− exp(−κ∫ L

L0Pc(L)dL).

0 50 100 150 200 250 300 350 400 4500

10

20

30

40

50

60

70

80

hook length (nm)

num

ber

of h

ooks

0 50 100 150 200 250 300 350 400 4500

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

hook length (nm)

Pi(L

)


Hypothesis: How Hook Length isdetermined

• The Infrequent Molecular Ruler Mechanism.

• The probability of FlhB cleavage is length dependent. Whatis the mechanism that determines Pc(L)?


Secretion Model

Hypothesis: FliK binds to FlhB during translocation to causeswitching of secretion target by cleaving a recognition sequence.

• FliK molecules move through the growingtube by diffusion.

L

x

0


Secretion Model



• They remain unfolded before and duringsecretion, but begin to fold as they exit thetube.

L

x

0


Secretion Model




• Folding on exit prevents back diffusion,giving a brownian ratchet effect.

L

x

0


Secretion Model





• For short hooks, folding prevents FlhBcleavage.

0

L

x


Secretion Model





• For short hooks, folding prevents FlhBcleavage.

• For long hooks, movement solely by diffu-sion allows more time for cleavage.

0

L

x


Stochastic Model

Follow the position x(t) of the C-terminus us-ing the langevin equation

νdx = F (x)dt+√

2kbTνdW,

where F (x) represents the folding force act-ing on the unfolded FliK molecule, W (t) isbrownian white noise.

L

x

0


Fokker-Planck Equation

Let P (x, t) be the probability density of being at position x attime t with FlhB uncleaved, and Q(t) be the probability of beingcleaved by time t. Then

∂P

∂t= −

∂

∂x(F (x)P ) +D

∂2P

∂x2− g(x)P,

anddQ

dt=

∫ b

a

g(x)P (x, t)dx.

where g(x) is the rate of FlhBcleavage at position x.

g(x)

0 L x


Probability of Cleavage

To determine the probability of cleavage πc(x) starting fromposition x, solve

Dd2πc

dx2+ F (x)

dπc

dx− g(x)πc = −g(x)

subject to π′c(a) = 0 and πc(b) = 0.Then Pc(L) = πc(a).

Pc(L)

L


Results

0 20 40 60 80 1000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

(a) L(nm)

CD

F

0 20 40 60 80 1000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

(b) L(nm)

CD

F

0 50 100 150 200 2500

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

(c) L(nm)

CD

F

0 20 40 60 80 1000

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

(a) L(nm)

PD

F

0 20 40 60 80 1000

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

(b) L(nm)

PD

F

0 50 100 150 200 2500

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

(c) L(nm)P

DF

Wild type Overexpressed Underexpressed

The data are matched extremely well by the model predictions.Length Regulation of Flagellar Hooks – p.30/46

Story 2: Action Potential initiation

• Deterministic models have been used with great success todescribe many important decision making biologicalprocesses. However, the discrepancies between the modelpredictions and what actually happens are becoming moreapparent.

• The cause of this discrepancy is often that the number ofmolecules involved is small and so stochastic behavior mustbe taken into account.

• Question 1: Why do deterministic and stochasticdescriptions sometimes give different answers?

• Question 2: Specifically, what is the mechanism for thespontaneous (noise driven) firing of an action potential?

Spontaneous Action Potentials – p.31/46

Example: Keizer’s Paradox

Consider the autocatalytic reaction

E +Xk1−→←−k2

2X, Xk3→ E

Law of mass action: Let e = [E], x = [X], then

dx

dt= k1ex− k2x

2 − k3x, e+ x = E0.

This is the logistic equation with x = k1E0−k3k1+k2

> 0 stable steady

solution, x = 0 unstable steady solution.


Keizer’s Paradox

The stochastic version of this reaction: Let Sj be the state inwhich there are j molecules of X and N − j molecules of E,

Sj

αj

−→←−βj+1

Sj+1, αj = j(N − j)k1

v, βj = jk3 + j

k2

v.

Notice that α0 = 0 so that S0 is an absorbing state. Furthermore,the master equation for this reaction has a unique steadysolution with p0 = 1 and pj = 0 for all j 6= 0. So, for all initial data,

p(t) → e0 as t → ∞

i.e., j → 0 with probability 1 (i.e., extinction is certain), even though0 is an unstable rest point of the deterministic dynamics.


Keizer’s Paradox Visualized

0 10 20 30 40 50 60 70 800

1

2

3

4

5

6

7

8

9

10

time

# of

x m

olec

ules

1 2 3 4 5 6 7 8 9 100

50

100

150

200

250

300

350

400

There appears to be (is) a slow manifold, extinction is a rare

event.


QSS Analysis

Let ǫ = k3vk1N

≪ 1. Then, we can write the master equation as

dp0

dt= k3p1,

dp

dt= k1N(A0 + ǫA1)p

where A0 represents the reactions E +Xk1−→←−

k22X and A1

represents the reactions Xk3→ E.

If p0 satisfies A0p0 = 0, 1Tp0 = 1, set p = αp0 + ǫp1 and projectto find

dα

dt= k3v1

TA1p0α,

an exponential process.


Keizer’s Paradox Resolved

The nullspace of A0 is the binomial distribution

p0j =

(

N

j

)

(

k1k2

)j

(1 + k1k2)N − 1

, j = 1, 2, · · ·N,

Consequently, the exit time is approximately

τe =1

1TA1p0

=k2

k1k3

(1 + k1k2)N − 1

N

which is exponentially large in N . Thus, even though extinctionis certain, the expected extinction time is exponentially large.


Example: Action Potential Dynamics

eCm Iion

v

v

v = v − v

e

i

Extracellular Space

Intracellular Space

INa

Il

IK

i

The Morris-Lecar equations

dv

dt= gNam∞(v)(vNa − v) + gKw(vK − v) + gl(vl − v) + I

where w satisfies

dw

dt= αw(v)(1− w)− βw(v)w.


Action Potential Dynamics

• This is a fast-slow system

• with excitable dynamics and threshold behavior.

Question: How does noise affect the initiation of action poten-

tials?


Upstroke Dynamics

To study upstroke dynamics, we hold w fixed (assuming w is a"slow" variable), and equation reduces to

dv

dt= gNam∞(v)(vNa − v) + gKw0(vK − v) + gl(vl − v) + I

= m∞(v)f(v)− g(v) ≡ G′(v)

which is the bistable equation.

G(v)

−100 −50 0 50 100

v [mV]

I < I∗

I = I∗

v0

v∗

One might conclude that spontaneous action potentials occur

when noisy Na+ currents push v out of one potential well into

another.Spontaneous Action Potentials – p.39/46

Discrete Ion Channels

Since ion channels are noisy and discrete, a more appropriatemodel is

dv

dt= gNa

n

N(vNa − v) + gK

m

M(vK − v) + gl(vl − v) + I

where n and m are the number of open Na+ and K+ channels,respectively, and transition rates are voltage dependent. Inparticular, assume channels open and close via stochasticprocess

Cα(v)−→←−β(v)

O.

If there are N such channels, the ion channel state diagram is

S0

Nα−→←−β

S1

(N−1)α−→←−2β

S2 · · ·SN−1

α−→←−Nβ

SN Spontaneous Action Potentials – p.40/46

Stochastic Simulations

0 2 4 6 8 10 12 14 16−60

−40

−20

0

20

V

0 2 4 6 8 10 12 14 160

0.5

1

Na+

Cha

nnel

s

0 2 4 6 8 10 12 14 160

0.2

0.4

0.6

0.8K

+ C

hann

els

−50 −40 −30 −20 −10 0 10 20 300

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

V

w

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

0.8

Na+

K+ C

hann

els

Observation: Action potentials are initiated by randomfluctuations in K+ channels, not Na+ channels.


The Chapman-Kolmogorov Equation

Master equation for jump-velocity process p = {p(v, n,w, t)}

∂p

∂t= −

∂

∂v

(

(dvnm

dt). ∗ p

)

+1

ǫANa+p+AK+p,

where

dvnw

dt= gNa

n

N(vNa − v) + gK

w

M(vK − v) + gl(vl − v) + I

and ANa+ , AK+ are transition operators for Na+ and K+

channels. ǫ ≪ 1 since Na+ channels are "fast".

Goal: Exploit the fact that ǫ ≪ 1 and that ANa+ represents a

birth-death process.


Analysis

WKB: Seek equilibrium solution of the form

p(v, w, n) = r(n|v,w) exp(φ(v,w)

ǫ),

substitute and find to leading order in ǫ an equation of the form

H(v, w, p, q, ) = 0, p =∂φ

∂v, q =

∂φ

∂w

a Hamilton-Jacobi equation. (H is too complicated to display.)This can be solved by method of characteristics:

dφ

dt= p

dv

dt+ q

dw

dtalong

d

dt

(

p

q

)

= −

(

∂H∂v∂H∂w

)

d

dt

(

v

w

)

=

(

∂H∂p∂H∂q

)


The Likelihood Surface

The likelihood surface: (a "stochastic phase plane")

• Trajectories are most"most likely path"

• There is a caustic/separatrix

• there is an approxiamte"saddle-node" (SN)


Conclusion

• WKB analysis produces a maximum likelihood surface thathas saddle-point behavior.

• Trajectories leading to action potentials may be Na+

induced, if N is small, or K+-induced if M is small.

• The fast-slow analysis of the deterministic model ismisleading with respect to stochastic, channelnoise-induced action potentials.


Thanks

Thanks to• Kelly Hughes (hooks)

• Berton Earnshaw(Channel Kinetics)

• Parker Childs (Keizer)

• Jay Newby (SAC’s)

• Paul Bressloff (SAC’s)

• NSF (funding!!)


the effect of randomness in chemical reactions …...the effect of randomness in chemical reactions...

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