power points on computational biology

8/13/2019 Power Points on Computational Biology

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AMS/BIO 332

Spring 2014

Lectures 03-04

Mass-action kinetic models of generegulatory networks.


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Info and exhibits:

Melville Library Lobby

9AM3PM

Feb. 14

http://life.bio.sunysb.edu/darwinsbu/

Keynote speaker:

Dr. David Jablonski

University of Chicago

Evolution and Extinction:

Lessons from the Fossil Record

7:30 PM Earth & Space Sci 001Films and Discussion:Javits Conference Room 2nd floor Melville Library

"Great Transformations 9:30AM - 11:00AM

"Why Sex?" 11:00AM - 12:30PM

"Evolutionary Arms Race" 12:30PM - 2:00PM"Extinction" 2:00PM - 3:30PM


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Assignment #1

Due date shifted due to class cancellations:

New due date Wednesday, Feb. 19th.


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Quiz!

Answer on a blank sheet of paper

***DO NOT LOOK AT YOUR NOTES***

Explain the difference between lysis

and lysogeny in the life cycle of abacteriophage.


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Self-assessment!

Give yourself a score out of 10

based on your confidence in your

answer.

We will go over the answers next


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Quiz answers!

Lysis and lysogeny are the two possible outcomes from theinfection of a bacterial cell by a bacteriophage (virus).

In lysis, the bacteriophage reproduces quickly inside the bacterialcell, creating new phage; when many new phage have beensynthesized and assembled, the bacteria bursts (lyses), and the new

phage are free to infect new hosts. In lysogeny, the phage genome is inserted into the bacterial

genome, but no new phage are synthesized. However, as theinfected bacterium grows and divides, all daughter cells also containthe phage genome in a quiescent state.

Infected bacteria that are in a lysogenic state will remain stable (andnot undergo lysis) indefinitely (for many generations). However,when exposed to stress (such as UV radiation), the cell will switchinto a lytic state: new phage will be rapidly synthesized andreleased into the environment.


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Self-assessment!

Give yourself a second score (againout of 10) based on the answers

weve just discussed.

Turn in your graded papers for review

and recording of grades.


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Mechanism of lysis-lysogeny decision

making circuit.

cro and other genes for lysiscI and other genes for lysogeny

OR1 OR2 OR3

Phage genome is structured in two parts, with control elements between them.

The genes for lysogeny include the transcription factor cI.

The genes for lysis include the transcription factor cro.


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making circuit.

OR1 OR2 OR3

cro protein (dimer)

(plus other lytic proteins)

mRNA of cro (and other lytic genes)

cI protein (dimer)

(plus other lysogenic proteins)

mRNA of cI (and other lysogenic genes)

Transcription

Translation


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making circuit.

OR1 OR2 OR3

lysogeny

cI (dimer)

preferentially

binds to OR3

cro (dimer)

preferentially

binds to OR1

lysis

X X

cro binding to OR1 blocks

transcription of lysogenic genes

cI binding to OR3 blocks

trancsription of lytic genes(cro)2 (cI)2


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A double-negative feedback loop.

cI gene cI mRNA cI

cro gene cro mRNA cro

cI cro

Mutual inhibition:


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Examples of other types of gene-

regulatory structures.

gene (DNA) mRNA Protein

A

Autoregulatory (auto-induced):

Triple-negative feedback: Negative feedforward:

B

A

C

BA C


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Building a mathematical model of a

gene-regulatory circuit.

What are the fundamental processes

involved?

Transcription (synthesis of mRNA)

Translation (synthesis of protein)

Protein oligomerization (e.g. dimerization)

Protein binding to DNA

Degradation of mRNA and protein!

Without degradation, quantities can only increase.


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What type of mathematical expression might

we use?


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we use?

Chemical (mass-action) kinetics describe the rates

of reactions as the rate of change of concentration

over time.

][][ 1 XkdtXd ]][[][ 2 BAk

dtABd

YX ABBA


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we use?

A system of Ordinary Differential Equations

(ODEs) describes the rates of change ofconcentrations.

To be contrasted with Partial Differential

Equations (PDEs) and Stochastic Master Equationsthat we will discuss later.


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Building our equations.

At the most fundamental level we have two

species for each gene that we care about:

mRNA for gene A, protein for gene A, etc.

For each one, we will have a general rate

equation:

There is no single right answer to how we construct the equations; we

will consider one of the simplest possible models!

nRateDegradatioateSynthesisR][

dt

Xd


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Degradation:

The simplest model of degradation is a

unimolecular process, much like radioactive

decay:

The degradation rate constant may be different for

each species.

][][

Xdt

XdX0X


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Protein synthesis (translation):

The simplest model of translation assumes that

the rate of protein synthesis is directly

proportional to the amount of mRNA (of the givenprotein) that is present:

This assumes that the other materials involved in protein

synthesis (ribosomes, tRNA, etc) are present in abundance.

][][

mRNAX

proteinX

dt

XdproteinmRNAmRNA XXX


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mRNA synthesis (transcription):

Our model of transcription needs to consider the

control elementsthis is more complicated!

There are two extremes:

If fully activated, transcription will occur at a constant

(maximum) rate.

If completely unactivated, the rate will be at a

minimum (typically zero).

XmRNA

dt

Xd

][max 0

][min

dt

Xd mRNA


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mRNA synthesis (transcription): We might assume that the degree of activation is

proportional to the fraction of the time that a transcriptionfactor is bound to the regulatory site.

Enhancer: If the fraction bound is 0, we have the minimalrate (0), and if the fraction bound is 1, we have themaximal rate:

Repressor: If the fraction bound is 0, we have the maximalrate, and if the fraction bound is 1, we have the minimalrate (0):

XX

mRNA

dt

Xd )1(

][

XXmRNA

dt

Xd

][


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In most cases, the transcription factor is a protein

encoded by some other gene:

22

2

proteinXY

protein

XYK

Y

XXmRNA

dt

Xd )1(

][

Repression


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In some cases, the transcription factor is the

protein encoded by the very same gene:

22

2

proteinXX

protein

XXK

X

XXmRNA

dt

Xd

][

Auto-regulation


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In networks, we need to keep track of multiple

genes, and their effects on each other:

22

2

proteinXY

protein

XYK

Y

XXmRNA

dt

Xd )1(

][

222

proteinYX

proteinY

XKX

YYmRNA

dtYd )1(][

Cross-regulation (repression) of two genes


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Putting it all togetheran auto-regulatory

gene:

][][

22

2

mRNAXmRNAX

proteinXY

proteinmRNA X

XK

X

dt

Xd

][][][ proteinXproteinmRNAXprotein XXdt

Xd

Synthesis Degradation


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Putting it all togethermutual repression:

][1][

22

2

mRNAXmRNAX

proteinXY

proteinmRNA XYK

Y

dt

Xd

][1][

22

2

mRNAYmRNAY

proteinYX

proteinmRNA YXK

X

dt

Yd

][][][

proteinXproteinmRNAX

proteinXX

dt

Xd

][][][ proteinYproteinmRNAYprotein YYdt

Yd


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Modeling the behavior of a gene-

regulatory circuit.

So now what?

We have been able to write down a system of

ordinary differential equations: the rate of

change in the concentrations of both mRNA andprotein of each species of interest.

How do we use this?

Are the rates of change what we really careabout?


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regulatory circuit.

What we really care about is the AMOUNT

(concentration) of each species at any given

point of time!

Our equations tell us how these amounts

change.

If we know how much we begin with, we can

compute the changes over time!


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regulatory circuit.

Boundary value problem:

We know the values of the variables at a certain

point in time (typically t=0).

We have a general expression for the rate ofchange of each variable.

We want to know the values of the variable at an

arbitrary time.),( yxf

dt

dx

),( yxgdt

dy

0)0( xx

0)0( yy

?)( tx

?)( ty


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regulatory circuit.

Solving the problem for a small time step:

The challenge is that the derivatives change over

time, because the variables themselves change.

If we only consider a small period of time, thechange will be small; we can approximate this as a

constant.

),( 00 yxft

x

tyxfxtx ),()( 000

),( 00 yxgt

y

tyxfx ),( 00

tyxgy ),( 00 tyxgyty ),()( 000


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regulatory circuit.

An iterative approach (Forward Euler):

We can find an approximate value of x(t) as

described; the smaller t is, the better.

Taking x(t) as our starting point, we can find anapproximate value of x(2t) the same way.

Repeat to find x(3t), x(4t), x(5t) etc!

ttytxftxttx )(),()()( )(),( tytxftx

)(),( tytxgt

y

ttytxgtytty )(),()()(


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regulatory circuit.

Forward Eulerin Matlab:

xedt

dx Integrate the equation:

for 100 s, with a time step of 0.1 s,

beginning at x(0) = 2.7.

Plot the results as x vs. t.


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regulatory circuit. Forward Eulerin Matlab:

numsteps = 1000;deltat = 0.1;x = zeros(numsteps,1);t = zeros(numsteps,1);x(1) = 2.7;t(1) = 0;for (i=1:numsteps-1)

dx_dt = exp(-x(i));x(i+1) = x(i) + dx_dt*deltat;t(i+1) = t(i) + deltat;

endplot(t,x,k-);xlabel(Time);ylabel(Value of x);title(exp(-x) vs Time, using Forward Euler)

1. Initialization of variables

2. Integration loop

3. Visualization of results


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regulatory circuit.

Add lines for additional variables:

% initialize variables (not shown)

% numsteps, x, y and t arrays, x(1) and y(1), etc

for (i=1:numsteps-1)% calculate derivatives using values at (i)

dx_dt = y(i)-x(i); % For example given

dy_dt = 3*x(i)^2; % For example given

x(i+1) = x(i) + dx_dt*deltat;

y(i+1) = y(i) + dy_dt*deltat;t(i+1) = t(i) + deltat;

end

23xdt

dy xydt

dx


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regulatory circuit.

The Forward Euler algorithm is simple, bothconceptually and to implement.

It gives reasonable answers for many systems, if thetime step is small enough.

If the time step is too big, the solution can be quiteinaccurate!

Many other algorithms exist that are based on similarideas, but that use a better approximation (e.g.Runge-

Kutta) Many of these are implemented as Matlab functions:

More discussion of these later


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regulatory circuit.

Our implementation of the Forward Euler algorithmwas built around a for loop:

for (i=1:numsteps-1)

x(i+1) = x(i) + dx_dt*deltat;

end

What is the right number to steps?

What is the right value of deltat? Knowing how to answer these questions is essential

to being able to model a system!


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regulatory circuit.

N = 10

T = 1


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regulatory circuit.

N = 50

T = 0.2


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regulatory circuit.

N = 100

T = 0.1


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regulatory circuit.

N = 500

T = 0.02

f


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regulatory circuit.

When the time step is particularly large, thetrajectory can be obviously coarse.

More subtle differences can occur at

moderate step sizes. Check the reasonableness of the time step by

repeating the simulation with a smaller value(divide by 2 or more):

Any differences should be very difficult to see.

But what about the length of the simulation?

d l h b h f


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regulatory circuit.

N = 500

T = 0.02

d li h b h i f


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regulatory circuit.

N = 1000

T = 0.02

M d li h b h i f


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regulatory circuit.

N = 2000

T = 0.05

M d li h b h i f


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regulatory circuit.

Does the simulation reach a steady-state, or appear asit may be getting close to one?

Confirm your guess by doubling the length of thesimulation:

There should be virtually no change in any value of thesecond half of the extended simulation.

Note: A system doesnt have to reach a steady-state;it could, for example, display persistent oscillations. How to assess a long enough simulation will be different

in these cases; we may want to make sure that oursimulation lasts multiple periods of oscillation to that anyvariations in the period and/or amplitude would benoticed.

S d i i d


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Steady-states, stationary points and

equilibrium.

What does is mean if we observe that the

system reaches a steady-state?

St d t t t ti i t d


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equilibrium.

What does is mean if we observe that thesystem reaches a steady-state?

It does NOT mean that no reactions are

occurring; but simply that the overallconcentrations of all species are not changingover time:

Synthesis and degradation rates are balance from

all species.

The system is in equilibrium.

St d t t t ti i t d


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equilibrium.

What does is mean if we observe that the systemreaches a steady-state?

It means that the rate of change of all species is

zero, e.g.:

We are at a root of the system of ODEs:

A stationary point (or critical point).

0][ dt

Xd mRNA0][ dt

Xd protein

0][

dt

Yd protein 0][

dt

Yd mRNA

St d t t t ti i t d


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equilibrium.

In some cases, we can solve directly for the existence

of stationary points.

][][

22

2

mRNAXmRNAX

proteinXY

proteinmRNA XXK

X

dt

Xd

][][][

proteinXproteinmRNAX

proteinXX

dt

Xd

][][ proteinXproteinmRNAX XX 0][ dt

Xd protein

0][

dt

Xd mRNA

][

22

2

mRNAXmRNAX

proteinXY

proteinX

XK

X

St d t t t ti i t d


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equilibrium.

The root of a single equation is called a null-cline;these are generally a line (or curve) through thestate space.

Stationary points occur at the intersection of thenull clines (one null cline per variable).

][][ proteinX

XproteinmRNA XX

0][

dt

Xdprotein

0][

dt

Xd mRNA

XmRNAX

proteinXY

protein

mRNAXK

XX

22

2

][

St d t t t ti i t d


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equilibrium.

][][ proteinX

Xprotein

mRNA XX

XmRNA

X

proteinXY

protein

mRNAXK

XX

22

2

][

St d t t t ti i t d


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equilibrium.

Null clines can indicate switching lines for the sign

of the rate of change for one variable.

The general direction of motion in a region can be

qualitatively sketched. Stationary points can be of different types:

Stable, unstable, saddle.

This analysis doesnt tell us the details of HOW a

system reaches the stationary state, just a

qualitative picture, simulation is still essential.

It CAN help a lot with interpretation!

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