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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

Q. What is the response time?

Q. What is the time taken to reach (Maths)?

Using linear approximation,

Stronger the value of (Maths), shorter the time.

Using autoregulation : Use a stronger promoter (maths), 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) (maths) : 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

Again objective is to produce,

For a fair comparison let us pick (maths),

Q. What is the response time?

From the last lecture,

Comparing the two,

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

Comparison II

1) (maths) (Fluctuations in the metabolic capacity) vary from cell to cell.

2) (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 ==

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$\int dy$ $\frac{1}{2}$

@@ Line 267: / Line 267: @@
 ===== Testing Please Ignore It =======
-<math>\frac{1}{2}=0.5</math>
+<math>\frac{1}[2}</math>
 <math>\int dy</math>
 <math>\frac{1}{2}</math>

BIO-103