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Unformatted text preview: exp k
d degradation of
“each protein X”
at a constant rate Degradation regulation " feedback mechanism used to regulate the concentration
of a protein by destroying protein molecules “in excess” Example #1: Degradation Regulation
x(t) = concentration of protein X at time t x k dx x exp k
d degradation of
“each protein X”
at a constant rate production of X at
constant rate Degradation regulation " feedback mechanism used to regulate the concentration
of a protein by destroying protein molecules “in excess”
Suppose:! Gene G produces an enzyme that
tags proteins for destruction
(e.g., ubiquitination for subsequent
degradation by the proteasome) angiebiotech.com x
x k
k dx G oﬀ
G on protein is only
degraded when the
Gene is on Example #1: Degradation Regulation
Negative feedback " when the protein X is a transcription factor that activates the gene
X binds to G and activates it
(Xdependent activation rate) λon x dt
g 0 x k g
x 1
k dx λoﬀ dt
X unbinds to the gene
(Xindependent deactivation rate) Is this enough to keep the variance bounded? Small?
For which gene activation rates λon pxq ?
What about higher order moments? Example #1: Degradation Regulation
Negative feedback " when the protein X is a transcription factor that activates the gene
X binds to G and activates it
(Xdependent activation rate) λon x dt
g 0 x g k x 1
k dx λoﬀ dt
X unbinds to the gene
(Xindependent deactivation rate) What if the degradation is constrained by the enzyme concentration?
λon x dt
g
x k g 0
x k λoﬀ dt 1
d xh
α xh bounded
decay rate ODE " Lie Derivative Given scalarvalued function V : Rn → R
dV x(t)
∂ V x ( t)
=
f x ( t)
dt
∂x derivative
along solution
to ODE Lf V
Lie derivative of V Basis of “Lyapunov” formal arguments to establish boundedness and stability… E.g., picking V (x) := x2
dV x t
dt V
fx
x 0 V xt xt 2 x0 2 x2 remains bounded along trajectories ! Generator of a Stochastic Hybrid System
λ q, x dt x (q, x) → φ (q, x) fq x Given scalarvalued function V : Q × Rn → R
d
E V q (t), x(t) = E (LV ) q (t), x(t)
dt x & q are discontinuous,
but the expected value is
differentiable
Dynkin’s formula
(in differential form) where
LV
(extended)
generator of
the SHS q, x V
q, x fq x
x Lie derivative
instantaneous variation m λ q, x V φ q, x
1 intensity V q, x Reset term Example #1: Degradation Regulation λon x dt
g 0 x g k x 1
k dx λoﬀ dt
LV g, x V g, x
k
x gd λon x 1 g λ o ﬀ g V 1, x V 0, x L yapunov Analysis " SHSs
λ(x)dt x expectedvalue
notions samplepath
notions gxw
d
E V x(t) = E (LV ) x(t)
dt fx
x → φ(x) probability of x(t) exceeding any given bound M,
can be made arbitrarily small by making x0 small classK functions:
(zero at zero & mon. increasing) α1 (x) ≤ V (x) ≤ α2 (x)
LV (x) ≤ −α3 (x)
V ( x)...
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 Fall '13
 Smith

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