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iGEM Grenoble 2012

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Preliminary

We will use the quasi steady state approximation (QSSA) then. The idea is that there are quick reactions, such as enzymatic ones, complexations, etc… And there are slow reactions such as protein production. We assume that the evolution speed of an element that is created only by quick reaction is null.

Indeed, we have these types of evolution for the biological elements. The ones involved only in quick reactions are most of the time in a steady state, and there jump from one steady state to an other has an infinite speed, which doesn’t interest us.

Goal

In this part, we want to answer to three questions:

The system

Here is the schema of the real system, in orange are the reactions which didn’t appear in the simplified system of the overview:

cAMP is the quorum sensing molecule. When we put some cAMP out of the system, it enters into the system. Then, it complexes with CRP to create (CRP-cAMP), which is the transcription factor of the gene arac. When some Arac is created, it will complex with arabinose to create Arac*. Arac* is the active form of Arac. Arac* with (CRP-cAMP) are the transcription factors of the gene cya. Then when some protein of adenylate cyclase is produced, it will catalyze the production of cAMP.

We first have the complexation of cAMP with CRP to get (CRP-CAMP). It is modeled by this biological reaction:

Thus, we get the evolution speed of (CRP-CAMP), r_(CRP-cAMP):

Then, we have the conservation equations:

Thus, we get:

Eventually, we get:

where

We have

Eventually, we have to choose between the two solutions

These two solutions are positive, because:

To chose which solution is the good one, we know that:

Eventually, we get:

(CRP-cAMP) is the transcription factor of the gene arac. When it appears in the network, it activates the production of the protein Arac. This is modeled by a Hill function. In addition, there is some outflow linked to the promoter pAraBAD, which is the promoter regulating arac, thus there is a basal production of Arac. We take into account this basal production, because we need to know if because of them our system will always be turned on, thus useless. Arac is also naturally degraded by the bacterium. Thus, we get as the equation of evolution of Arac concentration:

Then, the protein Arac complexes with arabinose to create Arac active, written Arac*. It is modeled by the following chemical equation:

We get the evolution of Arac* r_Arac*:

With the qssa, we get:

In addition we have the conservation equation of Arac:

We assume that we have

Then, Arac* with (CRP-cAMP) are the transcription factors of of the gene ca. When they appear in the network the protein Ca is produced. The product of two hill functions models this. For the same reasons as for Arac we take into account the basal production of the adenylate cyclase. In addition it is degraded by the bacterium. Thus, we get the equation:

Eventually, Ca catalyzes the production of cAMP with ATP. We have the following enzymatic reaction:

We use the Michaelis Menten model for an enzymatic reaction, thus we get the evolution of cAMP r_cAMP:

In addition, we assume that [ATP]>>K_M. Eventually we get:

In addition, the bacterium naturally degrades cAMP. Finally there is a quorum sensing term. However, we are not modeling the quorum sensing here, thus we assume that [cAMP_out] is constant. Thus, we get:

The QSSA enables us to have r_cAMP=0. Then, we have:

Remark:

We don’t take into account in the equations the growth of the bacterium. Indeed, the bacterium grows as long as it has some glucose. However, as long as there is glucose the gene cya is not expressed. Cya begins to be expressed only when the bacterium doesn’t grow anymore. Indeed, the biologists in order to check the “AND gate” behavior, the biologists built, see protocol protocol "AND gate test" . Here we give the biological graphs of the absorbance and the graph of the RFU in function of the time for arabinose and cAMP maximum:

We can observe that until 100 minutes the bacterium is growing (the absorbance increases), and we don’t get any RFU signal. Then, the bacterium almost stops growing, and thus we begin to get a signal.

Sensitivity

The first question was: what is the sensitivity of our system?

Even if we don’t know the exact value of all the parameters, we have enough information on them to be able to have a good evaluation of the sensitivity of our system.

To answer this question, we plot the evolution of the steady state of adenylate cyclase in function of [cAMP_out], the initial concentration of cAMP, which is assumed to be constant. Here are the scripts that enable us to plot the graphs. We solve the differential equations to get the steady state, because if we wanted to solve a set of equations we would have had to give an initial point. If we had given 0, matlab would have stayed at this point, and we couldn’t give another initial point without solving the equations.

Conclusion:

The sensitivity of our system is 10^-6 mol/L of initial cyclic AMP. When we introduced this quantity in the system, the bacteria will turn on. Then, the next question is to know when we are under this value in how much time we will be able to observe that one bacteria turned on.

Temporal evolution

To evaluate the time it will take to be able to detect a signal, we need to plot the evolution of the adenylate cyclase in the time for an initial concentration of cAMP_out≥10^-6 mol/L. We first give the graph with cAMP_out=10^-3 mol/L:

We notice that after 200 minutes, we will be able to begin to detect a signal.

Then, we want to see the behavior of the system around the threshold. We give the evolution of the adenylate cyclase in the time in function with cAMP_out=10^-6 mol/L :

Here, to be able to begin to detect a signal, we should wait around 1300 minutes. So even if our system can detect 〖cAMP〗_out=〖10〗^(-6) mol.L^(-1), we may not be able to say if it’s a real detection or a false positive. We will be able to answer this question with the stochastic part.

Then we give one temporal evolution of the adenylate cyclase for cAMP_out=10^-8 mol/L. It is bellow the threshold, but because of the basal values, we want to see exactly what happens.

We notice that Ca is expressed, and that half of the steady state is quickly reached (around 100 minutes). However the value of the concentration of the adenylate cyclase at steady state is too low to enable us to get a signal.

Conclusion: The more cAMP_out is introduced in the system, the fastest the answer is. The quickest answer would be 200 minutes to reach half of the maximal expression of ca enabling us to get a signal. Because of the basal values, the adenylate cyclase is always expressed. Thus, we will make a steady state study of the system. This is useful seeing that with the sensitivity graph we couldn’t see the low expression of adenylate cyclase and it’s only in the temporal part that we could notice it, so we need a real study. Then, the stochastic part will be really important to be sure that there won’t be too many false positives because of these basal values.

Steady state study

Here, we want to find the steady states of the system, because we want to know where our system will always tend to stay in function of the initial cyclic AMP. At the equilibrium points, the derivatives of our system are null:

If we solve this system of equations, we will get the steady states of the system. Because of the non linearity they are not obvious. However, we need to give an initial point to the matlab solver, and it will return only the closest steady state. Because we want to know all the steady states, we will use a different method.

Equation (1) gives us Arac in function of Ca. If we plot this function, we plot all the steady states of Arac, while Ca is varying. Equation (2) gives us Ca in function of Arac. If we plot this function, we plot all the steady states of Ca while Ca is varying. These functions are called the isoclines of the system. The intersections of these two graphs represent the steady states of the system.

Here you can find the scripts we worked with in this part. First, I give the isoclines with cAMP_init=10^-5 mol/L.

Isoclines of Ca and Arac with cAMP_init=10^-5 mol/L.

In black we have Ca isocline, in purple, Arac one, it will always be the same color code. We notice that there is only one steady state. It is high enough to enable us to get a signal.

Then I give the isoclines with cAMP_init=10^-6 mol/L.

Ca and Arac isoclines with cAMP_init=10^-6 mol/L.

It seems that there are two steady state. However, with a zoom around 0, we have:

Ca and Arac isoclines with cAMP_init=10^-6 mol/L, zoom around 0.

We notice that there is no low steady state, but only a high one.

Eventually, I give the isoclines with cAMP_init=10^-7 mol/L:

Ca and Arac isoclines with cAMP_init=10^-7 mol/L.

We have the same graph as previously. We zoom around 0 again:

Ca and Arac isoclines with cAMP_init=10^-7 mol/L, zoom around 0.

We can still see a high steady state, but this time there also is a low one!

To actually answer the question to know where our system will always tend to stay we need to be sure that these steady states are stable. If they are when our system stops to one of these steady states, it will always stay there. It works like this:

To answer the question of the stability, we follow the classic dynamical system method. We first linearize the system around the steady states. We have:

The equilibrium points are

We compute the matrix

Thus, we have

Let us call λ₁ and λ₂ the eigenvalues of the matrix. Then, we get the solution of the system around the equilibrium points:

We give the computed values of the steady states and of the eigenvalues for:

- cAMP_init=10^-5 mol/L:

Arac steady state =10^-4 *0.167058129527727 mol/L

Ca seady states = 10^-6*0.1837444563636 mol/L

λ₁= -0.006000000912526 λ₂= -0.005763188664176

- cAMP_init=10^-6 mol/L:

Arac steady state =10^-4 *0.166879570344986 mol/L

Ca seady states = 10^-6*0.1832826298080 mol/L

λ₁= -0.006000000910603 λ₂= -0.005745344108236

- cAMP_init=10^-7 mol/L:

Arac steady state = 10^-6 *0.182361098919416 mol/L

Ca seady states = 10^-9*0.249177541683 mol/L

λ₁= -0.006000006994365 λ₂= -0.002117175391388

The steady states are globally stable if and only if the real part of the proper values is < 0 .

Their real part are always < 0 ! The steady states are stable.

Conclusion: Now, we can be sure that our system won’t always be turned on. In function of the quantity of initial cAMP, our system will stay at a low or a high steady state.

Conclusion

The sensitivity of our system is around 10^-7 mol/L. To be able to know if it’s satisfying, we need to link it with the signaling part. In addition, to know if our system is fast we need to link this part with the signaling too. That’s what we are going to do in the next part.

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 </br>
 <b>Conclusion:</b>
-</br>
-</br>
 The more cAMP<SUB>out</SUB> is introduced in the system, the fastest the answer is. The quickest answer would be 200 minutes to reach half of the maximal expression of ca enabling us to get a signal.
 Because of the basal values, the adenylate cyclase is always expressed. Thus, we will make a steady state study of the system. This is useful seeing that with the sensitivity graph we couldn’t see the low expression of adenylate cyclase and it’s only in the temporal part that we could notice it, so we need a real study.

Team:Grenoble/Modeling/Amplification/ODE