Biological networks can also be modeled as a manufacturing system. In this aproach the biological network is analogous to a manufacturing line. The biological network considered consists of a series of enzymatic reactions which convert raw substrates into products.
Two stochastic independent processes play a role in enzyme kinetics: search and reconfiguration. During the search process substrates and enzymes search for each other until they collide, characterized by search time t_{s}. The reconfiguration process is the actual reconfiguration of substrate into product, characterized by reconfiguration time t_{r}. In contrary to ODE models described above, a distinction is made between these two stochastic independent processes.
In this chapter first a deterministic discrete-event model (DEM) model is described and validated with the ODEs. subsequently the stochastic behavior is implemented.
The simple substrate-enzyme reaction is considered again, the reactions are:
In the first reaction, substrate S and enzyme E form the substrate-enzyme complex C. This is presented in Figure 4. The blocks S, E and C represent buffers. The buffers contain the concentrations of respectively the substrate, enzyme and complex. The reaction process is described by M.
^{Figure 4: Reaction 1 presented as an manufacturing line }
Maximal velocity of this reaction Vmax_{1} is denoted by the product of reaction rate constant k_{1}, total enzyme concentration [E]_{t} and total substrate concentration [S]_{t}:
With the maximal velocity known, reconfiguration time t_{r1} can be calculated from:
The specific reaction rate V_{1} is denoted by:
The total reaction time for the first reaction t_{t1} is a product of the specific rate V and the total concentrations of the reactants (E]_{t} and [S]_{t} for the first reaction).
The search time t_{s} can be calculated from:
The Discrete Event Model representation of the complete substrate-enzyme reaction is presented in Figure 5. The results of the simulation are presented in Figure 6. The results correspond with the results of the deterministic ODEs in Figure 1. The chi-code can be found here:chi.
^{Figure 5:The Discrete Event Model representation.}
^{Figure 6: Simulation results}
by introducing Michaelis-Menten kinetics this system can be reduced to the system presented in Figure 7. Herein denotes S the substrate buffer, M the reaction process and P the product buffer. The results of the simulation are shown in Figure 8 and corresponds with the results in Figure 2.
Similar to the ODE way of modeling we now present a stochastic model of the substrate-enzyme reaction. The model is the same as the model in the deterministic case, see Figure 5. Random number distributions are used for the search time t_{s} and the reconfiguration time t_{r}. In this experiment we have had exponential dustributions. This implies that mean m and standard deviation are equal, so the coefficient of variation c = 1. Figure 9 presents the result of a stochastic simulation with exponential distributions. The parameters used in the model are given in Table 1.
^{Figure 9: Exp - Exp distribution.}
t_{s} | t_{r} | |
---|---|---|
M_{1} | 1/ V_{1} - t_{r1} | 1e^{-7} |
M_{2} | 1/ V_{2} - t_{r2} | 2e^{-11} |
M_{3} | 1/ V_{3} - t_{r3} | 2e^{-8} |
^{Table 1: Parameters used in simulation with exponential distributions.}
In the previous examples initial concentrations of substrate and enzyme molecules were used without arrival of new molecules. To show the effects of molecular timing we now look at a single enzyme with arriving substrates in a fixed order of time. A substrate-enzyme can also be represented with the enzyme as a machine, see Figure 10. Substrate S connects with enzyme E after search time t_{s}, then the enzyme reconfigures the substrate to product P. The duration of the reconfiguration step is t_{r}.
_{Figure 10: representation of a enzyme reaction with the enzyme as machine.}
If the enzyme is not reconfiguring (or processing) a substrate it can start processing the arriving substrate. If the enzyme is already processing a substrate and a new substrate arrives, the substrate moves on and the enzyme finishes his work. This is a pure loss system. The substrate molecules arrive at rate λ (1/t_{s}) and are processed by the enzyme with rate μ (1/t_{r}) [Kuhl2006].
A symbolic representation of a manufacturing system is A/B/C/D, A represents the interarrival time distribution, B is manufacturing time distribution, C is the number of machines and D is the buffer length. In this case the symbolic representation is G/G/1/0, G stands for General (arbitrary distribution). We want to present the utilisation of the machine versus the mean arrival rate over the mean process rates, ρ = λ/μ. Utilisation u denotes the fraction a machine is not idle. A machine is considered idle if it could start processing a new product. Utilisation has no dimension and can never exceed 1.0.
With the ODE theory only exponential distributions of the arrival and process rates are possible (M/M/1/0, M is for markovian/exponential). Results of simulations with only Markovian distributions are presented in Figure 11. The chi-code can be found here:chi. The curve of the M/M/1/0 system can be described by λ/(λ+1).
_{Figure 11: Molecular timing with exponential distributions.}
The utilisation of this system with deterministic arrival and process times, D/D/1/0, is equal to ρ if there is no loss of products ( ρ ≤ 1 ). If ρ exceeds 1 with a infinitesimal value the utilisation drops to 0.5. In this case, the timing is half-optimal since every second product is rejected by the machine, see Figure 12. Since not all reactions have exponential distributions some other distributions are applied. Results of simulations with gamma-gamma and gamma-deterministic distributions are shown in Figure 13.
_{Figure 12: Molecular timing with deterministic distributions.}
A receptor may be defined as a molecular sensor that receives chemical or physical signals and reacts to them in a way determined by its structural and kinetic properties, the environment into which it is embedded, and the kind, intensity, and temporal pattern of the signals received. When a signal or a stream of signals hits a receptor, the response at the receptor level may be different, depending on the state or phase in which the receptor happens to be at the instant of the signal’s arrival. These responses are [Kuhl2006]:
In the previous examples the substrate-enzyme reactions only considered response 1, if the enzyme is not processing, and response 5, if the enzyme is processing. In other words, we ignore the different possibilities by which the processing of an enzyme can be shortened, prolonged, or restructured by molecules arriving during that period.
In Figure 14 the results of implementing the responses to a deterministic system are shown. If deactivation occurs, response 2, the enzyme processes the substrate 8 times faster than normal and continuous normal production. If the enzyme is inactivated, response 3, it pauses 20 seconds before continuing its work. Response 4, superactivation, triggers the enzyme to process 4 times faster during 10 seconds. The occurrence of responses 1, 2, 3 and 4 while the enzyme is processing are shown in the lower figure and the number of substrate molecules are presented in the upper figure. The chi-code for this model can be found here:chi.
_{Figure 14: Simulation results of a deterministic system with the different responses while the enzyme is processing.}
The effect of introducing a single response to the system with only activation and letting the receptor unaffected is shown next. In Figure 15 the effect of response 2 on the throughput of the system is shown for different ρ's. On the horizontal axis is the probability of the response plotted. If the enzyme receives response 2, deactivation, it finishes the product faster than normal and therefore has a positive effect on the throughput.
_{Figure 15: Effect of response 2 on the throughput.}
The effect of introducing a response 3 to the system is presented for different ρ's in Figure 16. If the enzyme is processing and receives response 3, inactivation, it pauses for a while before continuing processing the substrate. This pause has got a negative effect on the throughput.
_{Figure 16: Effect of response 3 on the throughput.}
Response 4, superactivation, has got a positive effect on the throughput, see Figure 17. If the enzyme is processing and receives response 4, superactivation, it increases the process rate for a period of time and can therefore process more substrates.
_{Figure 17: Effect of response 4 on the throughput.}
A combination of all responses, compared to the previous system is presented in Figure 18. The arrival and process rates are exponentially distributed.