Imperial College/Courses/2009/Synthetic Biology/Computer Modelling Practicals/Practical 3

From OpenWetWare

Jump to: navigation, search

Practical 3


Objectives:

  • To explore computationally some simple genetic motifs:
    • Constitutive Gene Expression.
    • Activated and Repressed Gene Expression.
    • A Mystery Circuit.



Part I: Constitutive Gene Expression


In this section, we explore a computational model to describe a constitutively expressed gene. The model is based on a simple interpretation of the central dogma: Gene -(transcription)-> mRNA -(translation)-> Protein, with both the mRNA molecules and the protein molecules being naturally degraded. If you need a quick refresh on the Central Dogma, have a look at the 'additional resources' section below.

The modelling parameters, used throughout this practical, are characteristic of the E.Coli bacteria.


Model CellDesigner Instructions
 Gene \rightarrow mRNA \rightarrow Protein

Define the topology of the reaction network:

  • Download this File, and Open it with CellDesigner.
  • This file contains the network topology describing a simple constitutively expressed gene model. No kinetics information is yet described.

Constitutive Gene Expression

Following the law of mass action, we can write:


\begin{alignat}{1}
\frac{d[mRNA]}{dt} & = k_{1} - d_{1}[mRNA] \\
\frac{d[Protein]}{dt} & =  k_{2}[mRNA] - d_{2}[Protein] \\
\end{alignat}
  • k1 is the transcription rate. It is considered to be constant, and it represents the number of mRNA molecules produced per gene, and per unit of time.
  • d1 is the mRNA degradation rate of the mRNA molecule. The typical half-life for the mRNAs, in E.Coli, has been measured to be between 2min and 8min (average 5min).
  • k2 is the translation rate. It is considered to be constant, and it represents the number of protein molecules produced per mRNA molecule, and per unit of time.
  • d2 is the protein degradation rate. In this practical, we will only consider very stable proteins, i.e. not engaged in any active degradation pathways. In that case, we can approximate the degradation of the protein to be only due to the dilution effect caused by the cell division.

The following questions constitue Part C of your coursework

  • Question 1:

From the ODE system given, write down the steady-state expression of the [mRNA] concentration and the [Protein] concentration, with regards to k1,k2,d1,d2. Remember that steady-state means that we consider d[mRNA]/dt=0 and d[protein]/dt=0.

Now let us define the simulation constants...

  • Question 2: The dilution rate of a protein corresponds to the growth rate of the volume of a bacterium. It can be safely assumed that a bacterium divides once its volume has doubled. Knowing that the typical cell division time for e-coli is 30-40min (use 35 minutes), what is the dilution rate?
  • Question 3: Knowing that average number of mRNA molecules per gene is 2.5 in E.Coli, what is the average transcription rate ? Keep in mind that the problem only gives the mRNA half life, not the actual degradation rate (see the exponential decay model for more info)
  • Question 4: Knowing that the average number of proteins per gene is 1000 in E.Coli, what is the average translation rate ?

You are now ready to run the simulations with CellDesigner. Define all the necessary kinetics laws for the model, and create all the appropriate parameters.

  • Question 5: Run a simulation, and comment on the simulation outputs (mRNA and Protein). Pay special attention to the transient phase and the steady-state.
  • Question 6:From a Synthetic Biology point of view, this motif can be seen as a 'Protein Generator'. One might be interested in controlling the steady-state protein output level of this device. Using the 'parameter scan' function, run a simple sensitivity analysis on each of the 4 parameters, within a 10% range of their default value. Illustrate, and describe briefly, how each parameter impacts the protein steady state.



Part II: Interlude :A Simplified Model for Constitutive Gene Expression


Before moving to more exciting models and some other types of genes, it is interesting to explore the quasi-steady-state assumption on the mRNA molecules expression. Please note: this section is for you to increase your knowledge only. It will be useful for your mini-iGEM project but it is not part of your coursework!!!


  • From the previous simulations, you might have noticed that the concentration of mRNA reaches steady-state very quickly, compared to the protein concentration.
  • This suggests that we could consider that the [mRNA] concentration is always at steady state , i.e. \frac{d[mRNA]}{dt}=0 all the time. This model means that we want to apply a quasi-steady-state assumption on the [mRNA] molecules.
Model CellDesigner Instructions
 Gene \rightarrow  Protein
  • Within the same CellDesigner file, build a new network.
  • Create a network that will allow you to directly synthesise a Protein from a Gene. Also, you wan to describe the fact that the protein is being naturally degraded.

Quasi-Steady-State Constitutive Gene Expression

Following the law of mass action, we can write:


\begin{alignat}{2}
\frac{d[Protein]}{dt} =  s - d[Protein] \\
\end{alignat}
  • Create the appropriate parameters and reaction kinetics laws.


Investigating the Simplified Model:

  • Taking into account the quasi-steady-state assumption on the [mRNA], work-out the value of 's', and 'd' with regards to k1,k2,d1,d2, so that the two models are equivalent.
  • Simulate the full model, alongside with the quasi-steady state approximation. When using the parameters from the previous section, comment on how good this approximation seems to be.
  • Why do you think such a reduced model would be useful ?


The is the end of the interlude. The following two sections actually count for your coursework.

Part III: Activated and Repressed Gene Expression


Very few genes are known to have a purely constitutive expression, most genes have their expression controlled by some outside signals (DNA-binding proteins, Temperature, metabolites, RNA molecules ...). In this section, we will particularly focus on the study of DNA-binding proteins, called transcription factors. These proteins, when binding to a promoter region, can either have an activation effect on the gene (positive control), or a repression effect (negative control). In prokaryotes, control of transcriptional initiation is considered to be the major point of regulation. In this part of the tutorial, we investigate one of the most common model used to describe this type of interactions.


Let's first consider the case of a transcription factor acting as a repressor. A repressor will bind to the DNA so that it prevents the initiation of transcription. Typically, we expect the transcription rate to decrease as the concentration of repressor increases. A very useful family of functions to describe this effect is the Hill function: f(R)=\frac{\beta.{K_m}^n}{{K_m}^n+R^n}.

  • The Hill function can be derived from considering the transcription factor binding/unbinding to the promoter region to be at equilibrium (similar to the enzyme-substrate assumption in the Michaelis-Menten formula).
  • This function has 3 parameters: β,n,Km:
    • β is the maximal expression rate when there is no repressor, i.e. f(R = 0) = β.
    • Km is the repression coefficient (units of concentration), it is equal to the concentration of repressor needed to repressed by 50% the overall expression, i.e f(K_m)=\frac{\beta}{2}
    • n is the Hill Coefficient.
Model CellDesigner Instructions

\begin{align}
& Repressor \\
& \bot \\
Gene &\rightarrow mRNA \rightarrow Protein 
\end{align}

Define the topology of the reaction network:

  • From the previous constitutive expression model, create a new 'simple molecule' compound for the repressor.
  • Create an inhibitory link between the repressor, and the transcription reaction.
  • Save your file under a new name.

Hill function for transcriptional repression:

 
transcriptionRate=\frac{k_1.{K_m}^n}{{K_m}^n+R^n}
  • k1: maximal transcription rate
  • Km: repression coefficient
  • n: Hill coefficient
  • R = [repressor]

Hill Function (Repressor)

Following the law of mass action, we can write:


\begin{alignat}{1}
\frac{d[mRNA]}{dt} & = \frac{k_{1}.{K_m}^n}{{K_m}^n+R^n} - d_{1}[mRNA] \\
\frac{d[Protein]}{dt} & =  k_{2}[mRNA] - d_{2}[Protein] \\
\end{alignat}

Repressed Gene Expression

  • Create all the necessary kinetics laws and parameters.
  • Consider K_m=100, n=2. k_1, k_2, d_1, d_2 as found previously.


The following questions are part of section D of your coursework:

  • Question 1: Analysis of the Hill function

the parameter n of the Hill function determines the profile of the Hill function. Plot the Hill function for well chosen values of n (we recommend n=1/2,n=1, n=2 and n=3). All the other parameters of the function can be set to 1. Describe the evolution of the Hill function. Why can we regard it as a switch function?

  • Question 2: Analysis of the Repression

We assume that the concentration of repressor remains constant. Plot [Protein] = f(time) for a few well-chosen repressor concentrations between 0 and 1000. Describe your results

  • Question 3: The Transfer Function

We still assume that the concentration of repressor remains constant. We want to establish the transfer function between the [repressor] concentration and the protein steady-state level. To do so plot [Protein]steadystate = F([Repressor]) where the repressor concentration varies from [0 to 1000] (steps of 50 are enough).

  • Question 4: Suggest an application where this genetic circuit might be useful.


Now, let's consider the case of a transcription factor acting as an activator. An activator will bind to the DNA so that it promotes the initiation of transcription. Typically, we expect the transcription rate to increase as the concentration of activator increases. Once again, the Hill type function will be useful to describe the interaction effect. It is slightly different from the previous one: f(R)=\frac{\beta.{A}^n}{{K_m}^n+A^n}.


Model CellDesigner Instructions
 
\begin{align}
& Activator \\
& \downarrow \\
Gene & \rightarrow mRNA \rightarrow Protein 
\end{align}

Define the topology of the reaction network:

  • From the previous constitutive expression model, create a new 'simple molecule' for the activator.
  • Create an catalytic reaction link between the activator, and the transcription reaction.
  • Save your file under a new name.

Hill function for transcriptional activation:

 
transcriptionRate=\frac{k_1.{A}^n}{{K_m}^n+A^n}
  • k1: maximal transcription rate
  • Km: activation coefficient
  • n: Hill coefficient
  • A=[activator]

Hill Function (Activator)

  • Create all the necessary kinetics laws and parameters.
  • Consider K_m=100, n=2. k_1, k_2, d_1, d_2 as found previously.

Following the law of mass action, we can write:


\begin{alignat}{1}
\frac{d[mRNA]}{dt} & = \frac{k_{1}.A^n}{{K_m}^n+A^n} - d_{1}[mRNA] \\
\frac{d[Protein]}{dt} & =  k_{2}[mRNA] - d_{2}[Protein] \\
\end{alignat}

Activated Gene Expression

The following questions are also part of section D of your coursework:

  • Question 5: Analysis of the Activation

We assume that the concentration of activator remains constant. Plot [Protein] = f(time) for a few well-chosen activator concentrations between 0 and 1000. Describe your results

  • Question 6:The new transfer function

We still assume that the concentration of activator remains constant. We want to establish the transfer function between the [Activator] concentration and the protein steady-state level. To do so plot [Protein]steadystate = F([Activator]) where the Activator concentration varies from [0 to 1000] (steps of 50 are enough).

  • Question 7: Suggest an application where this genetic circuit might be useful.


Part IV: A Mystery Circuit


So far the circuits you have seen have always behaved the same. But you should not think that gene circuits always do. New useful behaviours emerge when genes interact with each other in a non-linear fashion. To finish, let us consider a simple example of a circuit where new, exciting things happen.

Download this File, and Open it with CellDesigner.


Model CellDesigner Instructions

It can be shown that after some normalisation the ODE system can be written as:


\begin{alignat}{1}
\frac{d[mRNA]_{i}}{dt} & = \frac{a}{1+{[Protein]_{j}}^n} - [mRNA]_{i} \\
\frac{d[Protein]_{i}}{dt} & =  b[mRNA]_{i} - b[Protein]_{i} \\
\ i=1,2,3; \\
\ j=3,1,2; \\
\end{alignat}


Repressilator Genetic Circuit

The following questions constiture the last part of your coursework (Section E):

  • Question 1 : Before studying the general properties of the mystery circuit, let us study a simplified version of it. The ODE system indeed contains a lot of symmetries that can be exploited and yield surprising properties. Let us consider a particular choice of initial conditions: the initial conditions of the mRNA terms are all equal and the initial conditions of the protein terms are also all equal.
    • For Bioengineers only: It can be shown that for such a choice of initial conditions leads, the mRNA concentrations all remain equal (but vary with time) and so do the protein concentrations.Can you explain briefly why this is the case? Note: you do not have to submit a full mathematical proof.
    • We now call X the mRNA concentration and Y the protein concentration. What system of ODE does X and Y satisfy? Show that this system corresponds to an auto-regulated sytem.
    • What is the natural choice for the initial conditions of the system (justify)?
    • Simulate the new simplified system for a=10, b=1000 and n=2 for these intial conditions and comment.
    • Now simulate the new simplified system for a=200, b=5 and n=2 for these intial conditions and comment.
    • For either of these choice of parameters, let the initial conditions of the simplified system vary and comment on your results.
  • Question 2 : Now we return to the case a=10, b=1000 and n=2. The simplified system studied in the previous question is of course not representative of the overall behaviour of the mystery circuit.
    • Let us assume that we can purify one of the proteins so that its initial condition is 1. Run the simulation again. What happens?
    • Describe the properties of the system (the simulation you have run is representative)
    • Can you give a qualitative explanation for the behaviour of the system?
  • Question 3 : Now we return to the case a=200, b=5 and n=2 and we seek to investigate
    • Again we assume that we can purify one of the proteins so that its initial condition is 1. Run the simulation again. What happens?
    • Describe the properties of the system
    • Can you give a qualitative explanation for the behaviour of the system?
  • Question 4: The interesting property takes some time to emerge. From an experimental point of view, this is a problem ( the synthetic plasmid is liable to be rejected, the growth medium may run out of nutrients etc...). In iGEM 2007 the Imperial College team investigated playing on the initial conditions of their simple synthetic system so that their system had better proporties. We can do the same here.
    • Let us assume that we can purify all the proteins so that you can set up their initial conditions. You want to determine the initial conditions that will make the property emerge fastest. How would you do this?
    • Please note: You are not asked to run all the relevant simulations, just explain how you can solve such a problem. A few simulations or drawings would be welcome though...
  • Question 5: The circuit is called the repressilator.Can you explain why?
  • Hint: This circuit appeared was in an article published in 2000 by M.Elowitz. Track it if you need help!

Part V: Additional Resources


Personal tools