From CYPHYNETS

(Difference between revisions)

Jump to: navigation, search

Current revision

These notes were prepared for the mathematical biology module for Bio-103, Spring 2009.

Instructor in-charge: Shahid Khan

Guest lecturers: Muhammad Sabieh Anwar, Abubakr Muhammad

Note taker: Mohammad Adil, BS-2012

Autoregulation in Transcription Networks

Gene transcription and translation

Talk about basic biology. I will fill this later.

Activation Dynamics

Please start here. Basic activation differential equations. Column 1 of Pages 1, 2

Suppose that gene X regulates gene Y via activation. This is represented symbolically as $X \rightarrow Y$

Let the concentration of mRNA be given by $Y m$

Let dynamics of the mRNA concentration is governed by the following simple differential equation.

$\frac{d Y_m}{dt} = - \gamma Y_m + \beta_0 + f(X^*)$

Substituting the hill function form for activation, we get

$\frac{d Y_m}{dt} = - \gamma Y_m + \beta_0 + \frac{\beta_1 {X^*}^n}{K^n + {X^*}^n}$

When the concentration of $X *$ is high to begin with, then the hill function saturates to $β 1$ , thus giving us a simpler looking differential equation,

$\frac{d Y_m(t)}{dt} = - \gamma Y_m(t) + \beta_0 + \beta_1$

One can write a solution for this simple linear differential equation,

$Y_m(t) = \frac{\beta_1 + \beta_0}{\gamma}(1 - e^{-\gamma t})$

The steady state of this reaction is $Y_{ss} = \frac{\beta_1 + \beta_0}{\gamma}$

This is the fast part of the reaction.

For translation of mRNA into protein Y, we study the slow reaction,

$\frac{dY(t)}{dt} = - \alpha Y(t) + \beta_2 Y_m(t)$

Since the fast reaction, reaches its steady state value very early, we can safely substitute the steady state value compute above.

$\frac{dY(t)}{dt} = - \alpha Y(t) + \beta_2 \frac{\beta_1 + \beta_0}{\gamma}$

Let us combine various constants into a new one, $β$ ,

$\frac{dY(t)}{dt} = - \alpha Y(t) + \beta$

Again, the solution is given by

$Y(t) = \frac{\beta}{\alpha}(1 - e^{-a t})$

and the steady state concentration of protein is given by $Y_{ss} = \frac{\beta}{\alpha}$

Note again that this is a slow reaction, because the time constant $\frac{1}{\alpha}$ for the translation reaction is much larger than the time constant $\frac{1}{\gamma}$ for the slow transcription reaction.

Repression Dynamics

Suppose that gene X regulates gene Y via repression. This is represented symbolically as $X \dashv Y$

Let the concentration of mRNA be given by $Y m$

The dynamics of the mRNA concentration is governed by the following simple differential equation

$\frac{d Y_m}{dt} = - \gamma Y_m + \beta_0 + f(X^*)$

Substituting the Hill function for repression, we get

$\frac{d Y_m}{dt} = - \gamma Y_m + \beta_0 + \frac{\beta_1 K^n}{K^n + (X^*)^n}$

When the concentration of $X *$ is high to begin with, the Hill function goes to 0, thus giving us a simpler looking differential equation

$\frac{d Y_m}{dt} = - \gamma Y_m + \beta_0$

One can write a solution for this simple linear differential equation

$Y_m(t) = \frac{ \beta_0}{\gamma}(1 - e^{-\gamma t})$

The steady state of this reaction is $Y_{ss} = \frac{\beta_0}{\gamma}$

This is the fast part of the reaction.

For translation of mRNA into protein, we study the slow reaction,

$\frac{dY}{dt} = - \alpha Y + \beta_2 Y_m(t)$

Since the fast reaction reaches its steady state very early, we can substitute the steady state value,

$\frac{dY}{dt} = - \alpha Y + zero(negligible)$

The solution is given by,

$Y (t) = Y (0) e - α t$

and the steady state concentration of protein is given by $Y(t) \rightarrow 0$

Note again that this is a slow reaction, because the time constant $\frac{1}{\alpha}$ is much larger for the transcription reaction than the time constant $\frac{1}{\gamma}$ for the translation reaction.

Autoregulation as a Network Motif

Suppose that gene X regulates itself via activation or repression. This is represented symbolically as $X \rightarrow X$ for activation and $X \dashv X$ for repression.

Going back to the equation for regulation,

$\frac{dY}{dt} = - \alpha Y + \beta (X)$

Here we replace Y by X. Also we replace $β(X)$ with $f (X)$ because $X_m \rightarrow X$ is a fast reaction. We get the following differential equation for autoregulation,

$\frac{dX}{dt} = - \alpha X + f(X)$

where the type of hill function $f (X)$ depends on whether the reaction is $X \rightarrow X$ or $X \dashv X$ . We call this an autoregulated or self-regulated circuit/network.

Q. Why is this arrangement important in transcription networks?

For this we introduce the idea of "network motif".

Network Motif

Take a transcription network .Try to spot a "motif".

By evolutionary processes, different edges are being generated or killed at random.

Q. Which patterns are "significant" or "accidental"?

We can generate a random network ( Edros-renyi) to see what a randomly generated transcription network looks like

Recipe/Algorithm for creating random network

Given N nodes ( Here proteins/genes) and E edges ( E will be the # of edges in the network),

1) Pick a node randomly.

2) Pick another node randomly.

3) Put an edge from the 1st to 2nd node.

4) Repeat the process E times.

Consider the probability of having a self-edge,

$P_{self} = \frac{1}{N}$ (A node has N choices to chose between)

Average number of self-nodes of E-edges $= E P_{self} = \frac{E}{N}$

Standard Deviation $= \sqrt{\frac{E}{N}}$ (Assume Poisson Distribution)

Example: For $N = 424$ , $μ = 1.2$

For $E = 519$ , $σ = 1.1$

In this case, expected self-edges = 0 , 1 , 2

But let us consider a real E.Coli Network with the same number of nodes and edges,

Here, # of self-edges = 40!

So, there must be a reason nature keeps auto-regulation in transcription networks. Patterns such as auto-regulation that are extremely hard to explain as "evolutionary accidents" are called "Network Motifs". Examples of other network motifs are:

1) Feed Forward Loops.

2) Two node feedback loops.

There are many more network motifs (see ref). Motifs can also be discovered in other types of networks.

What is feedback control?

I will add this.

Robustness via Autoregulation

Pages 6,7,8

We know that autoregulation is important but what is it's purpose? Refer back to equation for $X \dashv X$

$\frac{dX}{dt} = f(X) - \alpha X$ where $f(X) = \frac{\beta K^n}{K^n + X^n}$

Suppose that initially $X$ is zero or $\ll K$ . Hence $f(X) \rightarrow \beta$ and the we get the following equation,

$\frac{dX}{dt} = \beta - \alpha X$ while $X < K$

In fact when X is too small $α X$ also goes to zero. So we have,

$\frac{dX}{dt} = \beta$

This differential equation has a simple solution,

$X(t) \cong \beta t$ while $X < K , X \ll \beta$

When X crosses K, $f(X) \rightarrow 0$ , so $β$ is turned off. Hence X decays by,

$\frac{dX}{dt} = - \alpha X$

Solving the differential equation we get,

$X (t) = X (T) e - α(t - T)$

If X overshoots K, then X decays back to K.

If X undershoots K, then X again triggers to $f(X) \rightarrow \beta$ .

Moral of this analysis

$X(t) \rightarrow X_{s.s} = K (Steady State)$

Q. What is the response time?

Q. What is the time taken to reaches $X s . s / 2$ ?

Using linear approximation, $\frac{K}{2} = \beta {T_{1/2}}^{a}$ , Where a $\rightarrow$ autoregulation.

${T_{1/2}}^{a} = \frac{k}{2 \beta}$ $\rightarrow$ Response time for Negative autoregulation.

Stronger the value of $β$ , shorter the time.

Using autoregulation : Use a stronger promoter $β$ , to give an initial fast production.

Nature can tune the following parameters,

1) K : By mutations in the binding site of the X in the promoter.

2) $β$ : By mutations in the binding site of the RNA in the promoter.

Thus tuning the steady state and response time independently.

Comparing with an unregulated/simply regulated Transcription Network

$Z \rightarrow X \$ and $\frac{dX}{dt} = \beta_{u} - \alpha_{u} X$ where $u \rightarrow unregulated$

Again objective is to produce,

$X s . s = K$ but now triggered by Z.

For a fair comparison let us pick $β u, a u$ such that

$X_{s.s} = \frac{\beta_{u}}{\alpha_{u}} = K$

What is the response time?

From the last lecture,

${T_{1/2}}^{u} = \frac{log 2}{\alpha_{u}}$

${T_{1/2}}^{a} = \frac{K}{\beta 2}$

Comparing the two,

$\frac{{T_{1/2}}^{u}}{{T_{1/2}}^{a}} = \frac{K \alpha_{u}}{2 log2 \beta}$

By tuning (maths) (i.e making it big), response of autoregulator can be made much faster than simple control by Z.

Comparison II

${X_{s.s}}^{u} = K$ , ${X_{s.s}}^{a} = \frac{\beta_{u}}{\alpha_{u}}$

1) $β,α$ (Fluctuations in the metabolic capacity) vary from cell to cell.

2) $K$ (Strength of chemical bonds between X and RNA) vary little.

Thus we can conclude that autoregulation is much more robust.

@@ Line 1: / Line 1: @@
+These notes were prepared for the mathematical biology module for Bio-103, Spring 2009.
+'''Instructor in-charge:''' Shahid Khan
+'''Guest lecturers''': Muhammad Sabieh Anwar, Abubakr Muhammad
+'''Note taker:''' Mohammad Adil, BS-2012
 = Autoregulation in Transcription Networks  =
@@ Line 223: / Line 232: @@
 ===Moral of this analysis===
+<math> X(t) \rightarrow X_{s.s} = K (Steady State) </math>
 Q. What is the response time?
-Q. What is the time taken to reach (Maths)?
+Q. What is the time taken to reaches <math> X_{s.s/2}</math>?
 Using linear approximation,
+<math> \frac{K}{2} = \beta {T_{1/2}}^{a} </math>,
+Where a <math>\rightarrow </math> autoregulation.
-Stronger the value of (Maths), shorter the time.
+<math>{T_{1/2}}^{a} = \frac{k}{2 \beta} </math>
+<math> \rightarrow </math>
+Response time for Negative autoregulation.
+Stronger the value of <math> \beta </math>, shorter the time.
-Using autoregulation : Use a stronger promoter (maths), to give an initial fast production.
+Using autoregulation : Use a stronger promoter <math> \beta </math>, to give an initial fast production.
 Nature can tune the following parameters,
@@ Line 238: / Line 255: @@
 ) K : By mutations in the binding site of the X in the promoter.
-) (maths) : By mutations in the binding site of the RNA in the promoter.
+) <math> \beta </math> : By mutations in the binding site of the RNA in the promoter.
 Thus tuning the steady state and response time independently.
 ====Comparing with an unregulated/simply regulated Transcription Network====
+<math> Z \rightarrow X \ </math> and
+<math> \frac{dX}{dt} = \beta_{u} - \alpha_{u} X  </math>
+where <math> u \rightarrow unregulated</math>
 Again objective is to produce,
-For a fair comparison let us pick (maths),
+<math> X_{s.s} = K </math>
+but now triggered by Z.
-Q. What is the response time?
+For a fair comparison let us pick <math> \beta_{u}, a_{u} </math> such that
+<math> X_{s.s} = \frac{\beta_{u}}{\alpha_{u}} = K </math>
+What is the response time?
 From the last lecture,
+<math> {T_{1/2}}^{u} = \frac{log 2}{\alpha_{u}} </math>
+<math> {T_{1/2}}^{a} = \frac{K}{\beta 2} </math>
 Comparing the two,
+<math> \frac{{T_{1/2}}^{u}}{{T_{1/2}}^{a}} = \frac{K \alpha_{u}}{2 log2 \beta} </math>
 By tuning (maths) (i.e making it big), response of autoregulator can be made much faster than simple control by Z.
@@ Line 258: / Line 290: @@
 ==== Comparison II====
-) (maths) (Fluctuations in the  metabolic capacity) vary from cell to cell.
+<math> {X_{s.s}}^{u} = K </math> , <math> {X_{s.s}}^{a} = \frac{\beta_{u}}{\alpha_{u}} </math>
-) (math) (Strength of chemical bonds between X and RNA) vary little.
+) <math> \beta , \alpha </math> (Fluctuations in the  metabolic capacity) vary from cell to cell.
+) <math> K </math> (Strength of chemical bonds between X and RNA) vary little.
 Thus we can conclude that autoregulation is much more robust.
 == Summary ==
-===== Testing Please Ignore It =======
-<math>\int dx</math>

BIO-103

From CYPHYNETS

Current revision

Contents

Autoregulation in Transcription Networks

Gene transcription and translation

Activation Dynamics

Repression Dynamics

Autoregulation as a Network Motif

Network Motif

Recipe/Algorithm for creating random network

What is feedback control?

Robustness via Autoregulation

Moral of this analysis

Comparing with an unregulated/simply regulated Transcription Network

Comparison II

Summary

Views

Personal tools

Links

Search

Toolbox