Imperial College/Courses/Spring2008/Synthetic Biology/Computer Modelling Practicals/Practical 2: Difference between revisions

Latest revision as of 03:26, 3 September 2008

Synthetic Biology (Spring2008): Computer Modelling Practicals

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Practical 2

Objectives:

To explore computationally some simple genetic motifs:
- Synthesis Degradation Model
- Constitutive Gene Expression.
- Activated and Repressed Gene Expression.
- Positive and Negative Feedback Gene Expression.

Deliverables

A report is expected.

Instructors: Vincent Rouilly, Geoff Baldwin.

Part I: Synthesis-Degradation Model

In this section, we investigate a very common motif in biochemistry. It models the continuous and constant synthesis of a compound, and its natural degradation. From a Mathematical point of view, the model is described as a first-order linear ordinary differential equation.

Model

CellDesigner Instructions

[math]\displaystyle{ 0 \xrightarrow{k_{1}} A \xrightarrow{k_{2}} 0 }[/math]

Open a NEW document. Name it 'Synthesis_Degradation_Model'.

Build the topology of the reaction network

Create a 'Source' compound, thanks to the 'simple molecule' icon.
The same way, create a 'A' compound.
Create a reaction link between 'Source' and 'A', Reaction_1, using the 'state transition' icon.
Create a 'degradation reaction' linked to 'A', Reaction_2, using the 'degradation reaction' icon.
Save your file.

From the law of mass action, we can write:

[math]\displaystyle{ \frac{d[A]}{dt} = k_{1} - k_{2}*[A] }[/math]

Define the kinetics driving the reaction network

Edit Reaction_1, and define a new parameter k_1 = 1.0, and create the kinetic law according to the ODE system.
Edit Reaction_2, and define a new parameter k_2 = .01, and create the kinetic law according to the ODE system.
Save your model.

Simulate the dynamical behaviour

Open Simulation Panel
Set time for the simulation to be 1000 seconds, with 1000 points.
Press Execute, and check results.

Questions:(see report structure)
- Run a simulation over t=1000s, nb points=1000. Comment on the time evolution of 'A'. (illustration needed).
- Using the dynamical system definition, what is the steady state level of 'A' with regards to the parameters k1 and k2 ? (Steady state means that [math]\displaystyle{ \frac{d[A]}{dt}=0 }[/math]
- Using the 'Parameter Scan' feature, illustrate the influence of both parameters (k_1 and k_2), on the steady state level of 'A' (illustration needed).
- Bonus: Give the analytical solution of the ODE system.
- Now, consider that k_1=0, and [math]\displaystyle{ [A]_{t=0}=A_{0} \gt 0 }[/math]. Keep k_2=0.01. Illustrate the concept of half-life for the compound 'A'.
- Check Derivation of the Half-life expression: exponential decay model

Additional Resources:
- Exponential decay formula (from Wikipedia)
- Linear differential equation

Part II: 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

[math]\displaystyle{ Gene \rightarrow mRNA \rightarrow Protein }[/math]

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.

Following the law of mass action, we can write:

[math]\displaystyle{ \begin{alignat}{1} \frac{d[mRNA]}{dt} & = k_{1} - d_{1}[mRNA] \\ \frac{d[Protein]}{dt} & = k_{2}[mRNA] - d_{2}[Protein] \\ \end{alignat} }[/math]

[math]\displaystyle{ k_1 }[/math] 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.
[math]\displaystyle{ d_1 }[/math] 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).
[math]\displaystyle{ k_2 }[/math] 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.
[math]\displaystyle{ d_2 }[/math] 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. Cell division will be 40min.

Questions:

From the ODE system given, write down the steady-state expression of the [mRNA] concentration and the [Protein] concentration, with regards to [math]\displaystyle{ k_1, k_2, d_1, d_2 }[/math]. Remember that steady-state means that we consider d[mRNA]/dt=0 and d[protein]/dt=0.
Using the previously found equations, and 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)
In the same way, knowing that the average number of proteins per gene is 1000 in E.Coli, what is the average translation rate ? Once again, remember that the problem gives the protein half-life, and not the actual protein degradation rate.
In CellDesigner, define all the necessary kinetics laws for the model, and create all the appropriate parameters.
Run a simulation, and comment on the simulation outputs (mRNA and Protein), with regards to the transient phase and the steady-state.
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.

Next, we want to explore the quasi-steady-state assumption on the mRNA molecules expression. From the previous simulations, you might have noticed that the concentration of mRNA reaches steady-state very quickly, compared to the protein concentration. In that case, we want to explore a model where we would consider that the [mRNA] concentration is always considered at steady state , i.e. [math]\displaystyle{ \frac{d[mRNA]}{dt}=0 }[/math] all the time. This model means that we want to apply a quasi-steady-state assumption on the [mRNA] molecules.

Model

CellDesigner Instructions

[math]\displaystyle{ Gene \rightarrow Protein }[/math]

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.

Following the law of mass action, we can write:

[math]\displaystyle{ \begin{alignat}{2} \frac{d[Protein]}{dt} = s - d[Protein] \\ \end{alignat} }[/math]

Create the appropriate parameters and reaction kinetics laws.

Questions:

Taking into account the quasi-steady-state assumption on the [mRNA], work-out the value of 's', and 'd' with regards to [math]\displaystyle{ k_1, k_2, d_1, d_2 }[/math], 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 ?

Additional Resources:

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: [math]\displaystyle{ f(R)=\frac{\beta.{K_m}^n}{{K_m}^n+R^n} }[/math]. 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: [math]\displaystyle{ \beta, n, K_m }[/math]:

[math]\displaystyle{ \beta }[/math] is the maximal expression rate when there is no repressor, i.e. [math]\displaystyle{ f(R=0)=\beta }[/math].
[math]\displaystyle{ K_m }[/math] 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 [math]\displaystyle{ f(K_m)=\frac{\beta}{2} }[/math]
[math]\displaystyle{ n }[/math] is the Hill Coefficient. It controls the steepness of the switch between no-repression to full-repression.

Model

CellDesigner Instructions

[math]\displaystyle{ \begin{align} & Repressor \\ & \bot \\ Gene &\rightarrow mRNA \rightarrow Protein \end{align} }[/math]

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:

[math]\displaystyle{ transcriptionRate=\frac{k_1.{K_m}^n}{{K_m}^n+R^n} }[/math]

[math]\displaystyle{ k_1 }[/math]: maximal transcription rate
[math]\displaystyle{ K_m }[/math]: repression coefficient
[math]\displaystyle{ n }[/math]: Hill coefficient

Following the law of mass action, we can write:

[math]\displaystyle{ \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} }[/math]

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.

Questions:

We want to establish the transfer function between the [repressor] concentration and the protein steady-state level. To do so plot [math]\displaystyle{ [Protein]_{steady-state} = F([Repressor]) }[/math] using the results of the simulations where the repressor concentration varies from [0 to 1000] by steps of 10.
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: [math]\displaystyle{ f(R)=\frac{\beta.{A}^n}{{K_m}^n+A^n} }[/math]. The Hill function can be derived from considering the transcription factor binding/unbinding on the promoter region to be at equilibrium (similar to the enzyme-substrate assumption in the Michaelis-Menten formula). This function has 3 parameters: [math]\displaystyle{ \beta, n, K_m }[/math]:

[math]\displaystyle{ \beta }[/math] is the maximal expression rate when there is a lot of activators, i.e. [math]\displaystyle{ f(A \gg K_m)=\beta }[/math].
K_m is the activation coefficient (units of concentration), it is equal to the concentration of activator needed to activate by 50% the overall expression, i.e [math]\displaystyle{ f(K_m)=\frac{\beta}{2} }[/math]
[math]\displaystyle{ n }[/math] is the Hill Coefficient. It controls the steepness of the switch between no-repression to full-repression.

Model

CellDesigner Instructions

[math]\displaystyle{ \begin{align} & Activator \\ & \downarrow \\ Gene & \rightarrow mRNA \rightarrow Protein \end{align} }[/math]

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:

[math]\displaystyle{ transcriptionRate=\frac{k_1.{A}^n}{{K_m}^n+A^n} }[/math]

[math]\displaystyle{ k_1 }[/math]: maximal transcription rate
[math]\displaystyle{ K_m }[/math]: activation coefficient
[math]\displaystyle{ n }[/math]: Hill coefficient

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:

[math]\displaystyle{ \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} }[/math]

We want to establish the transfer function between the [Activator] concentration and the protein steady-state level. To do so plot [math]\displaystyle{ [Protein]_{steady-state} = F([Activator]) }[/math] using the results of the simulations where the Activator concentration varies from [0 to 1000] by steps of 10.
Suggest an application where this genetic circuit might be useful.

Additional Resources:

Part IV: Positive and Negative Feedback

We have just seen that some genes can be controlled by DNA-binding proteins. This type of interactions will enable the construction of genetic networks, called transcription networks. One gene produces a protein, which then binds to one or more other genes, and control them (positively or negatively), and so on. During the practical 3, we will explore some of those circuits.

At the same time, it has been observed that some genes can also regulate themselves (positively or negatively). When a gene is regulated by the very same protein it produces, the motif is called a feedback. It can be a positive feedback, or a negative, depending on the type of interaction there is between the protein and the promoter region. In this section, we will explore those 2 options.

Model

CellDesigner Instructions

do not consider this graph ...

[math]\displaystyle{ \begin{align} Gene & \rightarrow mRNA \rightarrow Protein \\ & \bot \\ Gene & \rightarrow mRNA \rightarrow Protein \\ & \downarrow \\ Gene & \rightarrow mRNA \rightarrow Protein \end{align} }[/math]

Define the topology of the reaction network:

Create a new CellDesigner Project where you will build 3 independent genetic motif:
- Constitutive expression
- Negative feedback
- Positive feedback
Save your file.

Questions:

Provide the 3 ODE systems you need to describe the 3 different motifs:constitutive, negative feedback, and positive feedback.
Create all the necessary kinetic laws considering the parameters as follow:
- Constitutive Expression: K_1=.346; d_1=.138 ; K_2=6.931 ; d_2=0.017 .
- Negative Feedback Expression: K_1=33.847 ; n=1; Km=10 ; d_1=.138 ; K_2=6.931 ; d_2=0.017
- Positive Feedback Expression: K_1=.346 ; n=1; Km=10 ; d_1=.138 ; K_2=6.931 ; d_2= 0.017
Compare the different models with regards to their protein steady-state level, and also with regards to how fast each model is reaching its steady-state.
Comment on the benefits and drawbacks of using a negative, or positive, feedback.

Imperial College/Courses/Spring2008/Synthetic Biology/Computer Modelling Practicals/Practical 2: Difference between revisions

Latest revision as of 03:26, 3 September 2008

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@@ Line 1: / Line 1: @@
 {{Template:ImperialCollege Spring2008 SyntheticBiologyCourse ComputerModellingPractical‎}}
-<font color=red> UNDER DEVELOPMENT </font>
 <div style="padding: 10px; width: 720px; border: 5px solid #B3CD4E;">
 <div style="padding: 10px; width: 680px; border: 5px solid #B3CD4E; color: #2171B7; background-color: #FFFFFF">
@@ Line 15: / Line 12: @@
 </div>
 * To explore computationally some simple genetic motifs:
-** Constitutive gene expression.
+** Synthesis Degradation Model
-** Activated and Repressed gene expression.
+** Constitutive Gene Expression.
-** Positive and Negative feedback gene expression.
+** Activated and Repressed Gene Expression.
+** Positive and Negative Feedback Gene Expression.
 <div style="padding: 0px; width: 680px; color: #2171B7; background-color: #B3CD4E">
@@ Line 24: / Line 22: @@
 </div>
-* A report is expected.
+* A [[Imperial College/Courses/Spring2008/Synthetic Biology/Computer Modelling Practicals/Practical 2/Report |report]] is expected.
-* When you find in the text '''(illustration needed)''', it means that you will have to provide an image export of your simulation results in your report.
 * '''Instructors:'''[[User:Vincent Rouilly| Vincent Rouilly]], Geoff Baldwin.
@@ Line 33: / Line 30: @@
 <div style="padding: 0px; width: 710px; color: #2171B7; background-color: #B3CD4E">
 <font face="trebuchet ms" style="color:#2171B7" size="3">
-'''Part I: Constitutive Gene Expression '''
+'''Part I: Synthesis-Degradation Model
 </font><br>
 </div>
-In this section, we explore a simple model to represent a gene expressed constitutively. The model is based on an interpretation of the central dogma: '''Gene -(transcription)-> mRNA -(translation)-> Protein''', with both the mRNA and the Protein being also naturally degraded.
+In this section, we investigate a very common motif in biochemistry. It models the continuous and constant synthesis of a compound, and its natural degradation.
-The parameters used throughout this practical will be characteristic of the E.Coli bacteria.
+From a Mathematical point of view, the model is described as a first-order linear ordinary differential equation.
+{| border="1" cellpadding="10" cellspacing="2" style="width:700px"
+!Model
+!CellDesigner Instructions
+|-
+| <center><math>
+\xrightarrow{k_{1}} A \xrightarrow{k_{2}} 0
+</math></center>
+[[Image:CellDesigner_SynthesisDegradation_Model_Network.png|200px|CellDesigner Synthesis-Degradation Model]]
+|
+* Open a NEW document. Name it 'Synthesis_Degradation_Model'.
+'''Build the topology of the reaction network'''
+* Create a 'Source' compound, thanks to the 'simple molecule' icon.
+* The same way, create a 'A' compound.
+* Create a reaction link between 'Source' and 'A', Reaction_1, using the 'state transition' icon.
+* Create a 'degradation reaction' linked to 'A', Reaction_2, using the 'degradation reaction' icon.
+* Save your file.
+|-
+| From the law of mass action, we can write:
+* <math>\frac{d[A]}{dt} = k_{1} - k_{2}*[A]</math>
+|
+'''Define the kinetics driving the reaction network'''
+* Edit Reaction_1, and define a new parameter k_1 = 1.0, and create the kinetic law according to the ODE system.
+* Edit Reaction_2, and define a new parameter k_2 = .01, and create the kinetic law according to the ODE system.
+* Save your model.
+|-
+| Simulate the dynamical behaviour
+|
+* Open Simulation Panel
+* Set time for the simulation to be 1000 seconds, with 1000 points.
+* Press Execute, and check results.
+|}
+* '''Questions''':([[Imperial College/Courses/Spring2008/Synthetic Biology/Computer Modelling Practicals/Practical 1/Report|see report structure]])
+** Run a simulation over t=1000s, nb points=1000. Comment on the time evolution of 'A'. (illustration needed).
+** Using the dynamical system definition, what is the steady state level of 'A' with regards to the parameters k1 and k2 ? (Steady state means that <math> \frac{d[A]}{dt}=0</math>
+** Using the 'Parameter Scan' feature, illustrate the influence of both parameters (k_1 and k_2), on the steady state level of 'A' (illustration needed).
+** '''Bonus''': Give the analytical solution of the ODE system.
+** Now, consider that k_1=0, and <math>[A]_{t=0}=A_{0} > 0 </math>. Keep k_2=0.01. Illustrate the concept of half-life for the compound 'A'.
+**  '''Check Derivation of the Half-life expression''': [[Imperial College/Courses/Spring2008/Synthetic Biology/Computer Modelling Practicals/Exponential_Decay_Model |exponential decay model]]
+* '''Additional Resources:'''
+** [http://en.wikipedia.org/wiki/Exponential_decay Exponential decay formula (from Wikipedia)]
+** [http://en.wikipedia.org/wiki/Linear_differential_equation Linear differential equation]
+<div style="padding: 0px; width: 710px; color: #2171B7; background-color: #B3CD4E">
+<font face="trebuchet ms" style="color:#2171B7" size="3">
+'''Part II: Constitutive Gene Expression '''
+</font><br>
+</div>
+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.
@@ Line 52: / Line 106: @@
 '''Define the topology of the reaction network''':
 * Download this [[Media:Constitutive_Gene_Expression_No_Kinetics.xml| File]], and Open it with CellDesigner.
-* This file contains the network topology to describe a simple constitutive gene expression model. No Kinetic information is yet described.
+* This file contains the network topology describing a simple constitutively expressed gene model. No kinetics information is yet described.
 [[Image:CellDesigner Constitutive Gene Expression Network.png|300px|Constitutive Gene Expression]]
 |-
@@ Line 64: / Line 118: @@
 </math></center>
 |
-* k_1 is the transcription rate. It is considered constant, and depends on ...
+* <math>k_1</math> 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.
-* d_1 is the mRNA degradation rate of the mRNA. The typical half-life of mRNAs in E.Coli has been measured to be between 2min and 8min (average 5min).
+* <math>d_1</math> 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).
-* k_2 is the translation rate. It is considered constant, and depends on ...
+* <math>k_2</math> 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.
-* d_2 is the protein degradation rate. Here, we will only consider very stable proteins, 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 happening during cell division. Cell division will be 30min.
+* <math>d_2</math> 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. Cell division will be 40min.
 |}
 '''Questions:'''
-* From the ODE system given, write down the steady-state expression of [mRNA] and [Protein], with regards to <math> K_1, K_2, d_1, d_2</math>.
+* From the ODE system given, write down the steady-state expression of the [mRNA] concentration and the [Protein] concentration, with regards to <math> k_1, k_2, d_1, d_2</math>. ''Remember that steady-state means that we consider d[mRNA]/dt=0 and d[protein]/dt=0''.
-* Knowing that average mRNA per gene is 2.5 in E.Coli, what is the average transcription rate ?
+* Using the previously found equations, and 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 [[Imperial College/Courses/Spring2008/Synthetic Biology/Computer Modelling Practicals/Exponential_Decay_Model |exponential decay model]] for more info)''
-* Knowing that the average number of proteins per gene is 1000 in E.Coli, what is the average translation rate ?
+* In the same way, knowing that the average number of proteins per gene is 1000 in E.Coli, what is the average translation rate ? ''Once again, remember that the problem gives the protein half-life, and not the actual protein degradation rate.''
-* Define all the necessary kinetics laws for the model, and create all the appropriate parameters.
+* In CellDesigner, define all the necessary kinetics laws for the model, and create all the appropriate parameters.
-* Run a simulation, and comment on the simulation outputs.
+* Run a simulation, and comment on the simulation outputs (mRNA and Protein), with regards to the transient phase and the steady-state.
-* From a Synthetic Biology point of view, this motif can be seen as a 'Protein Generator'. One might be interested in controlling the Protein output of this device. Using the 'parameter scan' function, proceed to a simple sensitivity analysis on each of the 4 parameters, within a 10% range. Describe how each parameter impacts the Protein steady state.
+* 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.
-Next, we want to explore the quasi-steady-state assumption on the mRNA expression. From the previous simulations, you might have noticed that the concentration of mRNA is reaching steady-state very quickly, compared to the Protein concentration. In that case, we can approximate the previous model by considering that <math>\frac{d[mRNA]}{dt}=0</math>.
+Next, we want to explore the quasi-steady-state assumption on the mRNA molecules expression. From the previous simulations, you might have noticed that the concentration of mRNA reaches steady-state very quickly, compared to the protein concentration. In that case, we want to explore a model where we would consider that the [mRNA] concentration is always considered at steady state , i.e. <math>\frac{d[mRNA]}{dt}=0</math> all the time. This model means that we want to apply a quasi-steady-state assumption on the [mRNA] molecules.
 {| border="1"  cellpadding="10" cellspacing="2" style="width:700px; border:3px green"
@@ Line 88: / Line 142: @@
 <center><math> Gene \rightarrow  Protein </math></center>
 |
 * Within the same CellDesigner file, build a new network.
-* Create a network that would allow you to directly synthesis a Protein from a Gene.
+* 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.
 [[Image:CellDesigner QSS Constitutive Gene Expression Network.png|300px|Quasi-Steady-State Constitutive Gene Expression]]
 |-
@@ Line 105: / Line 159: @@
 '''Questions:'''
-* Taking into account the quasi-steady-state assumption on the [mRNA], work-out the value of 's', and 'd' with regards to <math> K_1, K_2, d_1, d_2 </math>.
+* Taking into account the quasi-steady-state assumption on the [mRNA], work-out the value of 's', and 'd' with regards to <math> k_1, k_2, d_1, d_2 </math>, so that the two models are equivalent.
-* Simulate the full model, along with the quasi-steady state approximation. Comment on how good this approximation seems to be.
+* 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.
-* '''Bonus''': Derive Hill Equation in the case of a single binding Protein to the promoter site. Consider the equilibrium ...
+* Why do you think such a reduced model would be useful ?
 '''Additional Resources:'''
 * [http://en.wikipedia.org/wiki/Central_dogma_of_molecular_biology Central Dogma in Molecular Biology (from Wikipedia)]
 * [http://en.wikipedia.org/wiki/Gene_expression Gene Expression (from Wikipedia)]
+* [http://en.wikipedia.org/wiki/Steady_state_(chemistry) Steady-State in Chemistry]
 * [http://redpoll.pharmacy.ualberta.ca/CCDB/cgi-bin/STAT_NEW.cgi E.Coli Statistics]
@@ Line 118: / Line 173: @@
 <div style="padding: 0px; width: 710px; color: #2171B7; background-color: #B3CD4E">
 <font face="trebuchet ms" style="color:#2171B7" size="3">
-'''Part II: Activated and Repressed Gene Expression
+'''Part III: Activated and Repressed Gene Expression
 </font><br>
 </div>
@@ Line 124: / Line 179: @@
-Very few genes are known to have a purely constitutive expression, most of genes have their expression controlled by some outside signals. 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 this part of the tutorial, we investigate one of the most common model used to describe this type of interactions.
+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:
-<math>
+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:
-...
+<math>f(R)=\frac{\beta.{K_m}^n}{{K_m}^n+R^n}</math>.
-</math>
+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: Vmax, n, Km
+This function has 3 parameters: <math> \beta, n, K_m </math>:
+* <math>\beta</math> is the maximal expression rate when there is no repressor, i.e. <math>f(R=0)=\beta</math>.
+* <math>K_m</math> 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 <math>f(K_m)=\frac{\beta}{2}</math>
+* <math>n</math> is the Hill Coefficient. It controls the steepness of the switch between no-repression to full-repression.
@@ Line 138: / Line 197: @@
 |-
 |
-<center><math> Gene \rightarrow mRNA \rightarrow Protein </math></center>
+<center><math>
+\begin{align}
+& Repressor \\
+& \bot \\
+Gene &\rightarrow mRNA \rightarrow Protein
+\end{align}
+</math></center>
 |
 '''Define the topology of the reaction network''':
-* Download and open this [[Media:SimpleRepressedGeneExpression.xml | file]]
+* 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 repressors:
+Hill function for transcriptional repression:
-<math>
+<center><math>
-\frac{Vmax.{K_m}^n}{{K_m}^n+R^n}
+transcriptionRate=\frac{k_1.{K_m}^n}{{K_m}^n+R^n}
-</math>
+</math></center>
+* <math>k_1</math>: maximal transcription rate
+* <math>K_m</math>: repression coefficient
+* <math>n</math>: Hill coefficient
 |
 [[Image:Repressed_Hill_Function.png|300px|Hill Function (Repressor)]]
@@ Line 161: / Line 233: @@
 |
 [[Image:CellDesigner_Repressed_Gene_Expression_Network.png|300px|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.
 |}
+'''Questions:'''
+* We want to establish the transfer function between the [repressor] concentration and the protein steady-state level. To do so plot <math> [Protein]_{steady-state} = F([Repressor])</math> using the results of the simulations where the repressor concentration varies from [0 to 1000] by steps of 10.
+* 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: <math>f(R)=\frac{\beta.{A}^n}{{K_m}^n+A^n}</math>.
+The Hill function can be derived from considering the transcription factor binding/unbinding on the promoter region to be at equilibrium (similar to the enzyme-substrate assumption in the Michaelis-Menten formula).
+This function has 3 parameters: <math> \beta, n, K_m </math>:
+* <math>\beta</math> is the maximal expression rate when there is a lot of activators, i.e. <math>f(A \gg K_m)=\beta</math>.
+* K_m is the activation coefficient (units of concentration), it is equal to the concentration of activator needed to activate by 50% the overall expression, i.e <math>f(K_m)=\frac{\beta}{2}</math>
+* <math>n</math> is the Hill Coefficient. It controls the steepness of the switch between no-repression to full-repression.
-Then, let's consider the case of a transcription factor acting as an activator. A 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 repressor increases. Once again, q very useful family of functions to describe this effect is the Hill function. It is slightly different from the previous one:
-<math>
-...
-</math>
-This function has 3 parameters: Vmax, n, Km
 {| border="1"  cellpadding="10" cellspacing="2" style="width:700px; border:3px green"
@@ Line 175: / Line 255: @@
 |-
 |
-<center><math> Gene \rightarrow mRNA \rightarrow Protein </math></center>
+<center><math>
+\begin{align}
+& Activator \\
+& \downarrow \\
+Gene & \rightarrow mRNA \rightarrow Protein
+\end{align}
+</math></center>
 |
 '''Define the topology of the reaction network''':
-* Download and open this [[Media:SimpleActivatedGeneExpression.xml | file]]
+* 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 activators:
+Hill function for transcriptional activation:
-<math>
+<center><math>
-\frac{Vmax.A^n}{{K_m}^n+A^n}
+transcriptionRate=\frac{k_1.{A}^n}{{K_m}^n+A^n}
-</math>
+</math></center>
+* <math>k_1</math>: maximal transcription rate
+* <math>K_m</math>: activation coefficient
+* <math>n</math>: Hill coefficient
 |
 [[Image:Activated_Hill_Function.png|300px|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.
 |-
 |
@@ Line 201: / Line 295: @@
 |}
-'''Questions:'''
+* We want to establish the transfer function between the [Activator] concentration and the protein steady-state level. To do so plot <math> [Protein]_{steady-state} = F([Activator])</math> using the results of the simulations where the Activator concentration varies from [0 to 1000] by steps of 10.
-* Draw the transfer function between the Activator concentration and the steady-state protein concentration. And illustrate the effect of the parameters 'Km' and 'n'.
+* Suggest an application where this genetic circuit might be useful.
-* Imagine an application where this behaviour might be useful.
-* Draw the transfer function between the Repressor concentration and the steady-state protein concentration. And illustrate the effect of the parameters 'Km' and 'n'.
-* Imagine an application where this behaviour might be useful.
@@ Line 212: / Line 303: @@
 * [http://math.jccc.net:8180/webMathematica/MSP/mmartin/hill Hill function simulator]
 * [http://en.wikipedia.org/wiki/Hill_function Hill Function (from Wikipedia)]
+* [http://books.google.co.uk/books?id=pAUdPQlCZ54C Book:"Introduction to Systems Biology", Uri Alon]
 <div style="padding: 0px; width: 710px; color: #2171B7; background-color: #B3CD4E">
 <font face="trebuchet ms" style="color:#2171B7" size="3">
-'''Part III: Positive and Negative Feedbacks
+'''Part IV: Positive and Negative Feedback
 </font><br>
 </div>
-we have just seen that some genes can be controlled by DNA-binding proteins. This type of interactions will enable the construction of genetic networks, with one gene producing a protein, which will then control one, or more, genes.
+We have just seen that some genes can be controlled by DNA-binding proteins. This type of interactions will enable the construction of genetic networks, called transcription networks. One gene produces a protein, which then binds to one or more other genes, and control them (positively or negatively), and so on. During the practical 3, we will explore some of those circuits.
-However, it has been observed that some gene are also self-regulating themselves (positvely or negatively). In this section, we will explore ....
+At the same time, it has been observed that some genes can also regulate themselves (positively or negatively). When a gene is regulated by the very same protein it produces, the motif is called a feedback. It can be a positive feedback, or a negative, depending on the type of interaction there is between the protein and the promoter region. In this section, we will explore those 2 options.
 {| border="1"  cellpadding="10" cellspacing="2" style="width:700px; border:3px green"
@@ Line 229: / Line 322: @@
 |-
 |
-<center><math> Gene \rightarrow mRNA \rightarrow Protein </math></center>
+* do not consider this graph ...
+<center><math>
+\begin{align}
+Gene & \rightarrow mRNA \rightarrow Protein \\
+& \bot \\
+Gene & \rightarrow mRNA \rightarrow Protein \\
+& \downarrow \\
+Gene & \rightarrow mRNA \rightarrow Protein
+\end{align}
+</math></center>
 |
 * '''Define the topology of the reaction network''':
-**
+[[Image:CellDesigner_FeedBack_Gene_Expression_Network.png|300px|Activated Gene Expression]]
-|-
+* Create a new CellDesigner Project where you will build 3 independent genetic motif:
-|
+** Constitutive expression
-Following the law of mass action, we can write:
+** Negative feedback
-|
+** Positive feedback
-* ...
+* Save your file.
 |}
 '''Questions:'''
-* ...
+* Provide the 3 ODE systems you need to describe the 3 different motifs:constitutive, negative feedback, and positive feedback.
+* Create all the necessary kinetic laws considering the parameters as follow:
+** '''Constitutive Expression''': K_1=.346; d_1=.138  ; K_2=6.931 ; d_2=0.017  .
+** '''Negative Feedback Expression''': K_1=33.847 ; n=1; Km=10 ; d_1=.138  ; K_2=6.931 ; d_2=0.017
+** '''Positive Feedback Expression''': K_1=.346 ; n=1; Km=10 ; d_1=.138  ; K_2=6.931 ; d_2= 0.017
+* Compare the different models with regards to their protein steady-state level, and also with regards to how fast each model is reaching its steady-state.
+* Comment on the benefits and drawbacks of using a negative, or positive, feedback.