BIO-103

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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


Contents

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 Ym

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 β1, 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 Ym

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 − αt

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 αX 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(tT)

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 Xs.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,

Xs.s = K but now triggered by Z.

For a fair comparison let us pick βu,au 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.

Summary

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