gene perturbation and intervention in probabilistic boolean networks

Gene perturbation and intervention in probabilistic Boolean networks

Ilya Shmulevich, Edward R. Dougherty and Wei Zhang

2004. 4. 23Seoul National Univ.

BIBS Lab.Kim Ha Seong

Contents

Introduction

Definitions and notation

Perturbations

Intervention

Sensitivity of stationary distributions to gene perturbations

Conclusion


CONTENTS

1. Introduction

2. Probabilistic boolean networks : definitions and notation

3. Random gene perturbations

4. Intervention

5. Sensitivity of stationary distributions to gene perturbations

6. Conclustion

Contents

Introduction


Perturbations

Intervention


Conclusion


INTRODUCTION

Human Genone Project reveals 30~40,000 genes in human genome

Many genes each is linked to other genes Level of transcription regulation Level of protein interaction

Massive amounts of genetic information How to use it in an intellegent and comprehensive manner

Require a new models and powerful tools Find genes effectively Understanding and managing complex genetic networks

Contents

Introduction


Perturbations

Intervention


Conclusion


INTRODUCTION

Boolean network model (Kauffman, 1969, 1993; Glass and Kauffman, 1973) Gene expression is quantized (1,0) Each gene is related to the expression states Insights into the overall behavior of large genetic networks Structurally simple yet dynamically complex

Probabilistic Boolean networks (Shmulevich et al. 2002) Probabilistic nature cause genes to cope with uncertainty Probabilistic context of Markov chain

Gene regulatory networks Spontaneous emergence Find attractors in regulatory networks (Huang and Ingber, 2000) BN – fixed point and limit cycle attractor PBNs – absorbing states and irreducible sets of states

Contents

Introduction


Perturbations

Intervention


Conclusion


INTRODUCTION

Gene perturbation 특정 유전자를 mutation, over expression, inhibition of translation 등의 방법을 사용하여 유전자의 활동성 (activity) 을 측정하는 기술 .

Attractors are quite stable under most gene perturbations (Kauffman, 1993).A change of certain genes at certain states of the network may drastically affect the values of many other genes in the long-run and lead to different attractors.

Find good potential candidates for desirable stateStudy the effects of gene perturbation on long-run network

behaviorDevelop tools for discovering intervention targets

Distinguish between random gene perturbation and intentional gene intervention

PBN model provide a unified viewpointMake a distinction transient and permanent perturbation or intervention

transient – reversed by the network itselfpermanent – unchangeable or fixed

Contents

Introduction


Perturbations

Intervention


Conclusion


PBNs DEFINITIONS AND NOTATION

1 2,

( ) ( )1 1 ( )

Boolean network is a kind of graph ( , )

{ , ..., }

( ,..., ), { ,... }

1: gene i is expressed

0 : gene i is not expressed

( ) is number of passible functions for gene

N

n

i in i l i

i

i

i

i

G V F

V x x x

F F F F f f

x

x

l i x

( ) ( )

1

( ) ( )

( )

( ) ( ) ( )

:

( )

f ( ,..., ) f can take on all possible realizations of the PBN

The probability that predictor is used to predict gene i

{ } {f f }i i

jki

n

i n

ij

i i ij j k

k f f

l i

f f

f

c P f f P

Contents

Introduction


Perturbations

Intervention


Conclusion



cj(i) is the probability that fj(i) is used to predict gene I

Obtain cj(i) from gene expression data, using the coefficient of determination (Dougherty et al., 2000; Kim et al., 2000a,b)

( )

1

Pr{ }ij

nj

i Kj

P Network i is selected c

1 1 1 ... 1 1

1 1 1 ... 1 2

.. .. .. ... .. ( )

1 1 1 ... 2 1K =

1 1 1 ... 2

l n

2

.. .. .. ... 2 ( )

.. .. .. ... .. ..

(1) (2) .. ... .. ( 1)

l n

l l l n

’

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Introduction


Perturbations

Intervention


Conclusion



(1) ( 2) ( )' ' '1 1 1 2 11 2

1 1

' '1 1

: ( , , ) , ( , , ) , , ( , , )

( )1

A( , ') is the probability of transitioning from

( ,..., ) to ' ( ' ,..., ' )

Pr{( , , ) ( , , )}

(1 | ( , ,

nn n n nK K Ki i in

ij

n n

n n

ii f x x x f x x x f x x x

ji K n

x x

x x x x x x

x x x x

P

P f x x

'

1 1

{0,1}

) |)nN

ji j

x

' '1 1

1

Pr{( , , ) ( , , ) | _ _ }N

n n ii

x x x x Network i selected P

Contents

Introduction


Perturbations

Intervention


Conclusion


PBNs DEFINITIONS AND NOTATIONEXAMPLE

PBN G = (V, F), given function table:

(1) (1)1 1 2

(2)2 1

(3) (3)3 1 2

{ , }

{ }

{ , }

F f f

F f

F f f

1 2 3

1 2 3

( , , )

( , , )

V x x x

F F F F

Since there are 2 functions for x1, 1 for x2, 2 for x3, so there are N = 4 possible networks for matrix K

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Conclusion



Since there are

– 2 functions for node x1



N = 2 * 1 * 2 = 4

K matrix: lexicographically ordered row

1 1 1

1 1 2

2 1 1

2 1 2

K

(1) (1)

1 1 2

(2)2 1

(3) (3)3 1 2

{ , }

{ }

{ , }

F f f

F f

F f f

N = 4

# of genes

Second Row of K (1,1,2) means that the predictors (f1

(1), f2(1) , f3

(2) ) will be used.

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Introduction


Perturbations

Intervention


Conclusion



4 3 2 1

2 4 1 3

2 4 1 3

1 0 0 0 0 0 0 0

0 0 0 0 0 0 1 0

0 0 0 0 0 0 1 0

0 0 0 0

0 0 1 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0 0 1

P P P PA

P P P P

P P P P

Consider Pr{(1,1,0) -> (1,0,0)} with corresponds to entry A7,5 .

In the functions table, we can see there are two combinations give (1,0,0).

– f1(1)=1,f1(2)=0,f2(3)=0 (1,1,2) or – f2(1)=1,f1(2)=0,f2(3)=0 (2,1,2)

And we can know the

path (1,1,2) or (2,1,2) in the K matrix.

1 1 1

1 1 2

2 1 1

2 1 2

K

The row indices of path( 1,1,2) and (2,1,2) of the K matrix are 2 and 4 , respectively.

So we can know that A(7,5) transition probability is

Pr{(1,1,0)->(1,0,0) } = P2 + P4

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Introduction


Perturbations

Intervention


Conclusion



4 3 2 1

2 4 1 3

2 4 1 3

1 0 0 0 0 0 0 0

0 0 0 0 0 0 1 0

0 0 0 0 0 0 1 0

0 0 0 0

0 0 1 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0 0 1

P P P PA

P P P P

P P P P

000

001

010

011

100

101

110

111

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Conclusion



1 1 1

1 1 2

2 1 1

2 1 2

K

1 0 0 0 0 0 0 0

0 0 0 0 0 0 1 0

0 0 0 0 0 0 1 0

0.2 0.2 0 0 0.3 0.3 0 0

0 0 1 0 0 0 0 0

0 0 0 0 0 0 0.5 0.5

0 0 0 0 0.5 0.5 0 0

0 0 0 0 0 0 0 1

A

(1) (2) (3)2 1 1 2 0.6 1 0.5 0.3P c c c

Results:

1 2 3 40.3, 0.3, 0.2, 0.2P P P P

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Intervention


Conclusion



• If D0=[ 1/8,1/8,…,1/8 ] of uniform distribution, the limiting probabilities are = [ 0.15, 0,0,0,0,0,0, 0.85 ]

• This means that in the long run, all three genes will be either OFF or ON , which are called absorbing.

• This concept corresponds to the attractor in Boolean networks

Contents

Introduction


Perturbations

Intervention


Conclusion


PBNs DEFINITIONS AND NOTATIONInfluence of genes in PBN

Definition 1 : partial derivative of a Boolean function

Definition 2 : Influence of the variable xj on the function f:

( ) ( )1 1 1

( ,0) ( ,1) ( , )1 1 1

( ) | ( ) ( ) | where ( , , , , , , )

( )( ) ( ) where ( , , , , , , ) 0,1

j jj j j n

j

j j j kj j n

j

f xf x f x x x x x x x

x

f xf x f x x x x k x x k

x

( )( ) ( )( ) Pr 1 Pr{ ( ) ( )}j

j Dj j

f x f xI f E f x f x

x x

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Introduction


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Intervention


Conclusion


PBNs DEFINITIONS AND NOTATIONInfluence of genes in PBN

Definition 3 : The whole influence of gene xk on gene xi

Definition 4 : influence matrix

i

( ) ( )1 2 ( )

( ) ( )

Let F be the set of predictors for gene with corresponding probabilities

, ,...,

Let ( ) be the influence of variable on the predictor

i

(i) i il i

i ik j k j

x

c c c

I f x f

( )( ) ( )

1

( ) ( )l i

i ik i k j j

j

I x I f c

( )ij i jI x

Contents

Introduction


Perturbations

Intervention


Conclusion


PBNs DEFINITIONS AND NOTATIONInfluence of genes in PBN EXAMPLE 2.

(1)(1) 1

2 12

( )( ) 0.5D

f xI f E

x

(1) (1) (2,0) (1) (2,1)2 1 1 1( ) ( ) ( )DI f E f x f x

X(2,0)

=000 001 100 101

X(2,1)

=010 011 110 111

x1

0 1

x30

0 1 01 1 1

10 1 01 1 1

x2XOR

1010

So, (1+1) /4 = 0.5

The influence of gene x2 on f1(1)

x1

We want to compute the influence of variable x2 on variable x1.

Contents

Introduction


Perturbations

Intervention


Conclusion


PBNs DEFINITIONS AND NOTATIONInfluence of genes in PBN EXAMPLE 2.

(1)(1) 2

2 22

( )( ) 0.75D

f xI f E

x


2 1( ) 0.5 0.6 0.75 0.4 0.6I x

The total predictors of gene x1 consist of f1(1) and f2

(1), so we can obtain

( )( ) ( )

1

( ) ( )l i

i ik i k j j

j

I x I f c

Remind:

We can calculate other Influences by the same way.

0.1 0.75 0.375

0.6 0.75 0.375

0.6 0.75 0.375

Fromx1

x2

x3To x1 x2 x3

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Perturbations

Intervention


Conclusion


RANDOM GENE PERTURBTIONS

For Boolean networks, such random gene perturbations can be implemented with the popular DDLab software.Many states converge on one attractor

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Introduction


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Conclusion



Genome is not closed systemGenome system has inputs from the outsideGenes may become either activated or inhibited due to

external stimuli. (mutagens, heat stress, etc)A network model should be able to capture this phenomenon.Gene can sometimes change value with a small probability p

n

k 1

1

: probability of perturbation

{0,1} : perturbatoin vector

Pr{ 1} [ ] for all i=1,...,n

Pr{ =(0,...,0)} = (1- )

f ( ,..., ), k = 1, 2, ..., N

Possible transition function in entire PBN

( ,

i i

n

n

p

E p

p

x x

x x

1

..., ) state of network at some given time

, with probability 1 (1 )'

( ,... ), with probability (1 )

n

n

nk n

x

x px

f x x p

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Conclusion



PROPOSITION 1.

For p > 0, the Markov chain corresponding the the PBN is ergodic

PROOF.

Markov chain is ergodic implies that it possesses a steady-state distribution equal to stationary distribution.The convergence rate will depend on the parameter p.

A simulation-based analysis of the network involving gene perturbation may require the transition probability A(x,x’) = Pr{(x1, …, x n) (x’1, …, x’n) } between any two aritrary states of the network.

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THEOREM 2. Given a PBN G(V, F) with genes and a list of Boolean predictors, as well as a gene perturbation probability p>0,

Where is the Hamming distance between vectors x and x’, Pi is given in (2), and is an indicator function that is equal to 1 only when

PROOF.

1V={x ,..., } nx(i) ( )

1 n i 1 ( )F=(F ,..., F ) of sets F ={f ,..., } il if

( ) '1

1 1

( , ') ( , ')[ ']

( , ') (1 | ( ,..., ) |)

(1 ) (1 ) 1 ,

ij

nNj

i K n ji j

n x x n x xx x

A x x P f x x x

p p p

'

1

( , ') ( )n

i ii

x x x x

[ ']1 x x

'x x

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Intervention


Conclusion



If perturbation vector is not identically distributed, transition probabilities become more complicated.Require products of individual probabilities Pr{ = 1 }.Theorem 2 shows that two different states cannot be 0 so long as p>0.

Need stationary distribution of Markov chain to quantify ‘long-term’ influence.

Markov chain may consist of a number of irreducible subchains.These probabilities depend on the initial starting state.Obtaining long-run behavior directly from the state-transition matrix A also be impractical.

Require simulation-based analysis.Random gene perturbation.

Since all state communicate, the steady-state distribution is the same as the stationary distribution.Compute the distribution D(x) by keeping track of the proportion of time each combination of values of the genes in the domain of the predictor occurs.

i

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Intervention


Conclusion


RANDOM GENE PERTURBTIONS EXAMPLE 1.

PBN G = (V, F)

(1) (1)1 1 2

(2)2 1

(3) (3)3 1 2

{ , }

{ }

{ , }

F f f

F f

F f f

1 2 3

1 2 3

( , , )

( , , )

V x x x

F F F F

X1

X’1

X3 X2

X’3 X’2

V

V’

K(1) = 2 K(3) = 2K(2) = 1

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Conclusion



Compute the influence matrix

3( ) 1/ 8 for all {0,1}D x x

( )( ) ( )( ) Pr 1 Pr{ ( ) ( )}j

j Dj j

f x f xI f E f x f x

x x

( )( ) ( )

1

( ) ( )l i

i ik i k j j

j

I x I f c

( )ij i jI x

( ,0) ( ,1)( )( ) ( ) ( ) j j

j D Dj

f xI f E E f x f x

x

( ) ( )1 1 1

( ,0) ( ,1) ( , )1 1 1

Definition 1 : partial derivative of a Boolean function

( ) | ( ) ( ) | where ( , , , , , , )

( )( ) ( ) where ( , , , , , , )

j jj j j n

j

j j j kj j n

j

f xf x f x x x x x x x

x

f xf x f x x x x k x x k

x

0,1

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Intervention


Conclusion



(1)(1) 1

2 12

( )( ) 0.5D

f xI f E

x

(1) (1) (2,0) (1) (2,1)2 1 1 1( ) ( ) ( )DI f E f x f x

X(2,0)

=000 001 100 101

X(2,1)

=010 011 110 111

x1

0 1

x30

0 1 01 1 1

10 1 01 1 1

x2XOR

1010

So, (1+1) /4 = 0.5


x1

We want to compute the influence of variable x2 on variable x1.

Contents

Introduction


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Intervention


Conclusion



(1)(1) 2

2 22

( )( ) 0.75D

f xI f E

x


2 1( ) 0.5 0.6 0.75 0.4 0.6I x

The total predictors of gene x1 consist of f1(1) and f2

(1), so we can obtain

( )( ) ( )

1

( ) ( )l i

i ik i k j j

j

I x I f c

Remind:

We can calculate other Influences by the same way.

0.1 0.75 0.375

0.6 0.75 0.375

0.6 0.75 0.375

Fromx1

x2

x3To x1 x2 x3

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Conclusion



At the next step the distribution of all the states is no longer uniform.

The long-term influences are guaranteed to be independent of the initial starting state or distribution because a non-zero gene perturbation probability was used.

Fig.1. The trajectories of the influence I2(xi) for i=1,2,3 plotted as a function of the time-steps taken by the PBN given in Example1. p=0.01

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.20 10 20 30 40 50 60 70 80 90 100

I2(x1)I2(x2)

I2(x3)

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INTERVENTION

One of the key goals of PBN modeling is the determination of possible intervention targets(genes) such that the network can be “persuaded” to transition into a desired state or set of state.

When p=0, the cellular state cannot be alterd.P>0, than there is a chance that the current cellular state may switch to another state.

Clearly, perturbation of certain genes is more likely to achieve the desired result than that of some other genes.

The goal is to discover which genes are the best potential “lever points” in the sense of having the greatest possible impact on desired network behavior.

The problem of intervention is posed as : reaching a desired state as early as possible

Use first passage times

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INTERVENTIONEXAMPLE

P=0p1=0.3, p2=0.3, p3=0.2, p4=0.2(111) (000)Make x1=0 with probability P4=0.2, than it will transition into (000)it will be impossible for us to end up in (000) if we make x2=0 or x3=0eventually come back to (111)only intervening with gene x1

P>0entire Markov chain is ergodic and thus every state will eventually be visited.Reaching a desired state as early as possiblex1=0 much more likely to get us (000) faster.

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INTERVENTIONEXAMPLE

P=0.01 steady state distribution [0.0752 0.0028 0.0371 0.0076 0.0367 0.0424 0.0672 0.7310]Fk((011),(000)), Fk((101),(000)), Fk((110),(000))

For the states x of interest ad for a sufficiently large K0

1{0,1} { }

( , ) : first passage time from state x to state y at time k

( , ) ( , )

( , ) ( , ) ( , )n

k

kk

k kz y

F x y

M x y kF x y

F x y A x z F z y

0

01

( , ) ( , )k

k kk

H x y F x y

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INTERVENTIONEXAMPLE

0.25

0.2

0.15

0.1

0.05

00 2 4 6 8 10 12 14 16 18 20

(110)

(101)

(011)

There are several possibilities:Find the gene that

Minimizes the mean first passage time

Maximizes the probability of reaching a particular state before a certain fixed time

Minimizes the time needed to reach a certain state with a given fixed probability

0

( ) arg max i ( , )iopt ki H x y

( ) arg min i ( , )iopti M x y

0

( ) ( ) ( )0 0 0 arg min i min { : ( , ) }i i i

opt ki K K H x y h

for k0=1,...,20, for starting states(011),(101), and (110) corresponding to perturbations of first, second, and third genes, respectively.

0

( )( , )ikH x y

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InterventionSets of states, avoidance of states, and permanent intervention

Single start state and single destination state Sets of states

For example Proliferation, quiescence (attractors in BN)This role is played by irreducible subchains when p=0 in PBN.When p>0, the sets of states could be referred to as implicitly irreducible subchains.

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1 2

: a state

: set of states { , ,...}

First passage probability from x to y

( , ) ( , )k ky Y

x

Y y y

F x Y F x y

1 2

1 2

x



is believed to correspond to the current functional cellular state

: weight of x by their respective probablilities of occurrence.

( , )k

X x x

Y y y

X

F X Y

( , )k x

x X y Y

xx X

F x y

Avoiding a unwanted functional cellular state or set of states (proliferation)

Maximize the mean first passage time to the state

Mean first passage time from state x to state y at time k

( , ) ( , )kk

M x y kF x y

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Transient intervention or perturbation

(111)(101) (111),(110)(111) : absorbing state

This resistance to perturbations is a keyfactor for stability and robustness of PBNs

Permanent intervention or perturbationA gene changes value and remains at that value forever.Removing gene or ‘transplanting’ a gene, as done in gene therapy.(genetic)Reduce the state space by half (network)

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Sets of states, avoidance of states, and permanent interventionEXAMPLE

Simian Virus 40 (SV40)Discovered in the 1950s during the development of vaccine for poliovirus

SV40 could transform monkey kidney cells and develop tumors when injected into rodents.

SV40 may have a tumorigenic effect in humans too although with a long latent period. (kouhata et al., 2001)

SV40 encode large T-antigen which interacts with host cell molecules.And large T-antigen triggers a series of events. (bad for the host cells)

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Cell cycle checkpoints Several checkpoints function to ensure that complete genomes are transmitted to daughter cells. One major checkpoint arrests cells in G2 in response to damaged or unreplicated DNA. The presence of damaged DNA also leads to cell cycle arrest at a checkpoint in G1. Another checkpoint, in M phase, arrests mitosis if the daughter chromosomes are not properly aligned on the mitotic spindle. Role of p53 in G1 arrest induced by

DNA damage DNA damage, such as that resulting from irradiation, leads to rapid increases in p53 levels. The protein p53 then signals cell cycle arrest at the G1 checkpoint.

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Properties of S. cerevisiae cdc28 mutants The temperature-sensitive cdc28 mutant replicates normally at the permissive temperature. At the nonpermissive temperature, however, progression through the cell cycle is blocked at START.

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Accumulation and degradation of cyclins in sea urchin embryos The cyclins were identified as proteins that accumulate throughout interphase and are rapidly degraded toward the end of mitosis

Complexes of cyclins and cyclin-dependent kinases In yeast, passage through START is controlled by Cdc2 in association with G1 cyclins (Cln1, Cln2, and Cln3). Complexes of Cdc2 with distinct B-type cyclins (Clb's) then regulate progression through S phase and entry into mitosis. In animal cells, progression through the G1 restriction point is controlled by complexes of Cdk4 and Cdk6 with D-type cyclins. Cdk2/cyclin E complexes function later in G1 and are required for the G1 to S transition. Cdk2/cyclin A complexes are then required for progression through S phase, and Cdc2/cyclin B complexes drive the G2 to M transition.

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Structure of MPF MPF is a dimer consisting of cyclin B and the Cdc2 protein kinase.

MPF regulation Cdc2 forms complexes with cyclin B during S and G2. Cdc2 is then phosphorylated on threonine-161, which is required for Cdc2 activity, as well as on tyrosine-15 (and threonine-14 in vertebrate cells), which inhibits Cdc2 activity. Dephosphorylation of Thr14 and Tyr15 activates MPF at the G2 to M transition. MPF activity is then terminated toward the end of mitosis by proteolytic degradation of cyclin B.

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Mechanisms of Cdk regulation The activities of Cdk's are regulated by four molecular mechanisms.

Induction of D-type cyclins Growth factors regulate cell cycle progression through the G1 restriction point by inducing synthesis of D-type cyclins via the Ras/Raf/ERK signaling pathway.

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Control of the G2 checkpoint A complex of checkpoint proteins recognizes unreplicated or damaged DNA and activates the protein kinase Chk1, which phosphorylates and inhibits the Cdc25 protein phosphatase. Inhibition of Cdc25 prevents dephosphorylation and activation of Cdc2.

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Interaction of Rb with oncogene proteins of DNA tumor viruses The oncogene proteins of several DNA tumor viruses (e.g., SV40 T antigen) induce transformation by binding to and inactivating Rb protein.

The SV40 genome The genome is divided into early and late regions. Large and small T antigens are produced by alternative splicing of early-region pre-mRNA

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cdk7

Cycin H

p21/WAF1Cyclin E

cdk2

Rb

DNA synthesis

CAK

Rb

cdk2cdk7

cyclin H

cyclin E

p21/WAF1

Process of Cells move from G1 phase to S phase:

•Cyclin E and cdk2 work together to phosphorylate the Rb protein and inactivate it

•Cdk2/Cyclin E is regulated by two switches:

•Positive switch complex called CAK;

•Negative switch P21/WAF1;

•The CAK complex can be composed of two gene products:

•Cyclin H;

•Cdk7

•When cyclin H and cdk7 are present, the complex can activate cdk2/cyclin E.

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Cell cycle regulation of Rb and E2F In its underphosphorylated form, Rb binds to members of the E2F family, repressing transcription of E2F-regulated genes. Phosphorylation of Rb by Cdk4, 6/cyclin D complexes results in its dissociation from E2F in late G1. E2F then stimulates expression of its target genes, which encode proteins required for cell cycle progression.

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Induction of p21 by DNA damage DNA damage results in the elevation of intracellular levels of p53, which activates transcription of the gene encoding the Cdk inhibitor p21. In addition to inhibiting cell cycle progression by binding to Cdk/cyclin complexes, p21 may directly inhibit DNA synthesis by interacting with PCNA (a subunit of DNA polymerase d).

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It characterizes the effect of perturbations on long-term network behavior

: original Markov matrix, : perturbed Markov matrix, : perturbation,

: original stationary distribution, : perturbed stationary distribution

, or ,

for some matrix norm , and

j jj

j

A A A E E

k E k E

, are called condition

numbers and are used as measures of sensitivity.

jk k

Contents

Introduction


Perturbations

Intervention


Conclusion



Proof.

P=0

P>0 (Theorem2)

Contents

Introduction


Perturbations

Intervention


Conclusion



Contents

Introduction


Perturbations

Intervention


Conclusion



Theorem 4 allows us to bound the sensitivity of the limiting probabilities of and state of the PBN, relative to the probability of random gene perturbation.

Implication of Theorem 4

If a particular state of a PBN can be ‘easily reached’ from other states, the mean first passage times are small.And the steady-state probability will be relatively unaffected by perturbations.Such sets of state are thus relatively insensitive to random gene perturbations.

Contents

Introduction


Perturbations

Intervention


Conclusion


Conclusion

Two questions.

1. Given the possibility of a random gene perturbations with a certain probability, to what extent do such perturbations affect the long-term behavior of the entire network?

In Theorem 4. steady-state probabilities of states of the network to which it is easy to transition from other states are more resilient to random gene perturbations.

2. Given a desire to elicit certain behavior from the network, what genes world make the best candidates for intervention so as to increase the likelihood of this behavior?

First passage time.Develop the tools for finding the best candidate genes for intervention.

Contents

Introduction


Perturbations

Intervention


Conclusion


Conclusion

Future works

Large-size network is difficult to capturing long-run network behavior because it is owing to the exponential increase of the state space.

Focus on obtaining steady-state behavior through simulation and efficient data structure.

Select small sub-networks out of a large network that function more or less independently of the rest of the network.

Algorithims for efficiently finding such sub-networks inferred from real gene-expression data.

gene perturbation and intervention in probabilistic boolean networks

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