Dynamics of filament Growth
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Transcript of Dynamics of filament Growth
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Dual Degree Project Presentation
Dynamics of filament growthUnder the supervision of
Prof.Mandar Inamdar
Neeraj KookadaDepartment of Civil Engineering
IIT Bombay
May 2, 2013
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Flow of Presentation
Introduction Objectives
Stochastic Simulations of Chemical Reactions
Gillespie Algorithm
Degradation Reaction
Degradation and Production Reaction
Diffusion using Compartment-based approach Force generation by filament polymerization
Introduction
2-State Model
Theoretical Stall force for a single filament
Stall force for a two filament model
Pseudo code Simulation Results and Discussions
Conclusion and Future work
References
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Introduction Optimization problems in dynamic, stochastic environments are an increasingly
important part of biological systems.
Simulation methods can provide solutions for complex partial differential
equations and problems which involve uncertainty about models to draw
conclusions about the related variables.
The growth dynamics of fibres like microtubules and actin filaments play an
important role in cellular biology. These fibres generate forces that play a role in cellular motility processes such as
the motion of chromosomes during mitosis.
Recent experimental advances have allowed to determine the dynamics of growth
of cytoskeleton proteins with a single-molecule precision at different conditions.
These experiments suggest that the complex biochemical transitions and
intermediate states in the cytoskeleton proteins influence their dynamic properties
and functions and thus the control the force that a cell can produce.
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Objectives To develop stochastic models for processes like degradation, production and
diffusion which mimic the interactions between different components in molecularbiology
To apply the Gillespie algorithm as it turns out to be a more efficient model
reducing the computational time when compared with Monte Carlo simulations to
predict the time evaluation of processes.
To develop a two-state model for a single actin filament which assumes thepolymerization and depolymerisation, as well as switching of states, where the
filament is restrained to grow against a wall (movable) and the resulting force
generated is studied
The maximum force generated by the filaments stall force and its relations with
the velocity of the wall and length fluctuations of the filament is plotted using
Monte Carlo Simulations. The one filament model is extended to a two filament model and numerically it is
derived that the stall force for N filaments is N times the stall force for a single
filament.
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Gillespie Algorithm
Introduction
Gillespie Algorithm generates a statistically correct trajectory (possible solution) of
a stochastic equation
Gillespie algorithm allows a discrete and stochastic simulation of a system with fewreactants because every reaction is explicitly simulated.
A Gillespie realization represents a random walk that exactly represents the
distribution of the Master equation.
The physical basis of the algorithm is the collision of molecules within a reaction
vessel (well mixed). All reactions within the Gillespie framework must involve at most two molecules.
Reactions involving three molecules are assumed to be extremely rare and are
modeled as a sequence of binary reactions.
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Flowchart for Gillespie Algorithm
Initialize Start Initialize System time to zero
Event list
Create event list of all possible transitions
Calculate the time to-event outcomes
Time step
If events available, perform next event, otherwise stop
Advance time
Terminate
Re-compute the event list
Time reached
Stop
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Algorithm
1. Choose a small time step t. We compute the number of molecules A(t) at times t = i t, i =
1, 2, 3, . . . , as follows. Starting with t = 0 and A(0) = n0, we perform two steps at time t,
2. Generate a random number r uniformly distributed in the interval (0, 1).
3. If r < A(t)kt, then put A(tt) = A(t)1; otherwise, A(tt) = A(t).
4. Continue with step (1) for time t +t.
5. Generate a random number r uniformly distributed in the interval (0,1)6. is computed from eq (1.3)
7. Compute the number of molecules at time t by A (t) = A (t)-1.
8. Continue with step (4).
since r is a random number uniformly distributed in the interval (0, 1), the probability that
r < A(t)kt is equal to A(t)kt.
We use k = 0.1sec-1 and t = 0.005 sec. A(0) = 20, Time = 30s.
For the given input values, A(t)kt (0, 0.01).
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Result
Results of ten realizations of SSA (a2)(c2)(solid lines; different colours show different
realizations) and stochastic mean plotted by the dashed line
Since the function A(t)
has only integer values
{0, 1, 2, . . . , 20}, someof the computed
curves A(t) partially
overlap.
It would be very
unlikely that there
would be two
realizations giving
exactly the same
results.
Details of realization
A(t) depend on thesequence of random
numbers)
Average values give a
reproducible
characteristic of the
system
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Degradation & Production Reactions
Consider the chemical reactions
The first reaction describes the degradation of chemical A with the rate constant k1. It was
already studied previously as reaction (1.1). We couple it with the second reaction which
represents the production of chemical A with the rate constant k2. This reaction is similarto a random walk carried by a particle in the following case.
The particle in the given figure has three mechanisms r1,r2 and r3 possible at different
possibilities of p1,p2 and p3 respectively, where r1,r2 and r3 are set of mutually exclusive
events.
(2.4)
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Algorithm
(starting with A(0) = n0 at time t = 0):
1. Generate two random numbers r1, r2 uniformly distributed in (0, 1).
2. Compute 0 = A(t)k1 + k2.
3. Compute the time when the next chemical reaction takes place as t where from
(1.3)
4. = 1/0 ln(1/r1)
5. Compute the number of molecules at time t by
A(t ) =A(t) 1 if r2 < k2/ 0;
= A(t) 1 if r2 k2/ 0.
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Graphs
frequency over 10000 simulationsDistance v/s Time
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Diffusion using Compartment-based
approachDiffusion is the random migration of molecules (or small particles) arising from motion due tothermal energy. As shown by Einstein, the kinetic energy of a molecule (e.g. protein) isproportional to the absolute temperature. In particular, the protein molecule has a non-zeroinstantaneous speed at, for example, room temperature or at the temperature of the humanbody. A typical protein molecule is immersed in the aqueous medium of a living cell.Consequently, it cannot travel too far before it bumps into other molecules (e.g. water molecules)in the solution. As a result, the trajectory of the molecule is not straight but it executes a randomwalk.
Consider diffusion of 1000 molecules
.
We simulate directly the time evolution of 40 compartments. Considering a computational
domain [0,L] and dividing it into K = 40 compartments of length h = L/K = 25 m. Denote thenumber of molecules of chemical species A in the i-th compartment [(i 1)h, ih) by Ai, i = 1, . . .
,K. The rate of diffusion is assumed as d. Applying the Gillespie SSA provides a correct model of
diffusion provided that the rate constant d is chosen as d = D/h2 where D is the diffusion constant
and h is the compartment length.
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AlgorithmGenerate two random numbers r1, r2 uniformly distributed in (0, 1).
1. Compute propensity functions of reactions by i= A
i(t)d for i = 1, 2, . . . ,K. Compute
+
2. Compute the time at which the next chemical reaction takes place as
3. If < = , then for j = (1,2,,k-1),
<
=
=
4. The number of molecules at time t is given by:
Aj(t ) = Aj (t)-1
Aj+1(t ) = Aj+1 (t) + 1
Ai(t ) = A
i(t) for I j, I j 1
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5. If
= , then for j = (2,3,,k-1),
+
= <
+
=
=
=
6. The number of molecules at time t is given by:
Aj(t ) = Aj (t)-1
Aj-1(t ) = Aj-1 (t) + 1
Ai(t ) = Ai(t) for I j, I j 1
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We simulate 1000 molecules starting from position 0.4 mm in the interval [0,L]for L = 1 mm. We use K = 40.
Since 0.4 mm is exactly a boundary between the 16th and 17th compartment,the initial condition is given by A16(0) = 500, A17(0) = 500 and Ai(0) = 0 fori 16, i 17.
As D = 104 mm2 sec1, we have d = D/h2 = 0.16 sec1. The numbers Ai(t),
i = 1, . . . ,K, at time t = 4 min, are plotted. For example, only one chemical
reaction occurs per time step. Consequently, only two propensity functions change and need to be updated
here N = 1000 is the total number of molecules in the simulation (this numberis conserved because there is no creation or degradation of the molecules inthe system).
Hence, we need to re-compute only when there is a change in or ,i.e. whenever the boundary compartments were involved in the previousreaction
0 1 k
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Results
Compartment-based approach
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Force generation by filament
polymerizationIntroduction
Actin filaments and microtubules are key components of the cytoskeleton of
eukaryotic cells.
In order to understand the dynamical behaviour of single filaments such as actin or
microtubules and the force they can generate, discrete stochastic models havebeen developed that incorporate at the molecular level the coupling of hydrolysis
and polymerization.
Single polymerizing filaments can generate forces in the pico-newton range, as has
been demonstrated experimentally for microtubules (MTs). Such force generation
mainly relies on the gain in chemical bonding energy upon monomer attachment. An opposing force slows down filament growth, which finally stops at the stall
force representing the maximal polymerization force a filament can generate.
Therefore, the stall force is the essential quantity to characterize polymerization
forces.
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2-state Model
Assumptions
Consider two rigid surfaces: one fixed where filaments are nucleated (nucleating
wall) and one movable (barrier) whose position is defined to be the position of the
filament furthest away from the nucleating wall .
We do not model the internal structure of the filaments, and in particular we do
not account for hydrolysis.
After nucleation, the filaments grow or shrink by exchanging monomers with the
surrounding pool of monomers, which acts as a reservoir.
The filaments are coupled only through mechanical contact with the barrier.
The filament consists of two species of monomers (type blue and type orange, as
depicted in fig.(4.1). The filament is modelled as a two-state model depending onwhich type of monomer is in contact with the barrier. Only the filaments in contact
feel the force exerted by the barrier on them, and as a result, this changes their
polymerization rates as compared with free filaments.
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2-state Model
Assumptions
We assume that a type blue and type orange monomers can be added to any free
filament with a rate u0and removed with a rate w and w respectively. Similarly,
the monomers can be added to the filament in contact with the barrier with a rate
u(f) and removed with a rate w(f) and w(f) respectively. Assume that the barrierexerts a constant forcefon the filaments in contact
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2-State Model
Dynamics of a single filament
For a filament that has exchanged work with the barrier through addition or loss of
monomers, we use the following relation:
00
Where: u0 and w0 represent the polymerization and depolymerisation rates for monomers
attaching to free filaments (not bounded by the barrier),
u, w represent the polymerization and depolymerisation rates for monomers
attaching to bounded ends of the filaments.
Consider the switching of states (blue/orange = 1/2) with rates k12 and k21
respectively.
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For a single filament, the polymerization and depolymerisation rates are as
follows:
0 0 1
where, the parameter [0, 1] depends on the location of the energy barrierwhich the wall has to overcome in moving one monomeric step.
We have defined two states 1 and 2 for the filament. The state actually depends
on the last molecule in the filament. We define the rate of conversion from state 1
to 2 as k12 and vice versa. Let the probability of a filament of length l being in
state 1 or 2 at time t be P1(l,t) and P2 (l,t) respectively. The probabilities obey the
following :
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1 ,
1 1, + 11 + 1 , + 212 , + 1 + 12 1 ,
2 ,
2 1, + 22( + 1 , ) + 121(, ) + 2 + 21 2(,)
(1())/212 121
2 121 212()
Summing over all l, we get
For steady state, probabilities are independent of time.
1 212 121 0212 121
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The average position is given by
, + ,
=
The velocity is given by
(1) 1 ,
+2 ,
0
1 1 , + ( 2)2 , 1 1 + ( 2)2
In steady state
(1) 1 21 + 2 1212 + 21
Putting V(1) = 0 and u = ue-Fstall, we find the stall force for one filament system
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Stall force expression for a single
filament The stall force is defined as the value of the force applied on the barrier for which
the velocity given by equation (4.7) vanishes.
At stall force, the velocity can be assumed to be zero, which gives:
( ++ )
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Stall force for a two filament model
Dynamics of a two-filament system
Here, two filaments have been modeled as a two state model with the rates of
polymerization and depolymerization same as had been considered in the previouscase.
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Pseudo Code : Stall force v/s length of filament
function [ final_length ] = biofil( F )
%UNTITLED Summary of this function goes here
% Detailed explanation goes here
iter_max=10000;
k1=0.3; k2=0.15;
u0=1;
w1=0.2; w2=0.1; a1=50e3;
b1=50.1e3;
t_max=1; t=2;
iter=1;
stateA = 1; stateB=1; A=zeros(iter_max,2);
while(t>t_max)
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Pseudo Code
if(a1>b1)
s1=0;s2=1;
elseif(b1>a1)
s1=1;s2=0;
else
s1=u0*exp(-F)/(u0+u0*exp(-F));
s2=u0*exp(-F)/(u0+u0*exp(-F));
end
%molecules entering
r=rand*2;
if(s2*u0*exp(-F)+s1*u0>r)
a1=a1+1;
end
r=rand*2;
if(s1*u0*exp(-F)+s2*u0>r)
b1=b1+1;
end
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Pseudo Code
%state change
r=rand*2;
if(stateA==1)
if(k1>r)
stateA=2;
end
else
if(k2>r)
stateA=1;
end
end
r=rand*2;
if(stateB==1)
if(k1>r)
stateB=2;
end
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Pseudo Code
else
if(k2>r)
stateB=1;
end
end
%molecules leaving
r=rand*2;
if(stateA==1)
if(w1>r && a1>0)
a1=a1-1;
end
else
if(w2>r && a1>0)
a1=a1-1;
end
end
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Pseudo Code
r=rand*2;
if(stateB==1)
if(w1>r && b1>0)
b1=b1-1;
end
else
if(w2>r && b1>0)
b1=b1-1;
end
end
%loop exit conditions
if(iter>iter_max-1)
t=0;
end
A(iter,1) = a1; A(iter,2) = b1;
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Pseudo Code Stall Force v/s Velocity
function [ output_args ] = velocity( )
%UNTITLED Summary of this function goes here
% Detailed explanation goes here
v=zeros(23,1);
i=1;
f=linspace(0,4.4,23);
for fs=0:0.2:4.4;
v(i)=(biofil(fs)-50e3)/1000;
i=i+1;
end
plot(f,v);
end
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Simulation Results and Discussions
From the graph, it is evident that the position of the barrier isnt changing much with the
iterations. The numerical value of the force used in the code is 4.23 units which is nearly
equal to twice the stall force for a single element. Thus in absence of hydrolysis, the stall
force of a system of N filaments is equal to N times the stall force for each filament
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For forces below the stall force, the length of the filament goes on increasing with
time.
On the other hand, for forces above the stall force, the length of the filament goes on
decreasing with time.
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The mean velocity of the barrier becomes zero at stall force which from the monte carlo
simulations comes out to be around 4.23
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Conclusion and Future Work
We presented SSAs for systems of chemical reactions and molecular diffusion. Thealgorithms for simulating systems of chemical reactions were based on the Gillespie
algorithm.
We presented a model of diffusion in this paper based on the chain of chemical
reactions computing the time evolution of the numbers of molecules in compartments.
Based on the models, we obtained the exact result for the stall force of polymerizing
ensembles of rigid filaments with lateral interactions. The stall force is a linear functionof the number N of filaments in the ensemble and increases linearly with the lateral
interaction.
These results have been confirmed by simulations using the Gillespie algorithm.
Our results are relevant for the interpretation of experimental data on the force-
velocity relation in microtubule polymerization and in the polymerization of bundles of
interacting actin filaments or microtubules.
Future work would include extending the parameters like stall force, velocity for N-
filament structures. The study and effects of hydrolysis on the polymerization and
depolymerisation should be understood and a new exhaustive model should be
prepared
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Thank You