Team:SCU-WestChina/Model
The model mainly contain two sections, one is the model of proper promoters choice due to the problem of rate-determining enzymes, another is the model of blood flow in the dialysis device. In section”II, first we will introduce the chemical equations and the relative molecular mass of crucial substrates and products and their proportion in the metobolic pathway of the urate. Then the differential equations system can be listed with the advanced Michaelis-Menten equation which describe the relatioship between the reaction rate and the concentration of substrate and enzyme. The equations system can be solved numerically assuming the rule of distribution of energy and equilibrium, from which the proper proportion of the concentration of the enzyme can be obtained. Besides, the relative capacity of the constitutive promoter family Anderson Promoter in the iGEM database can be obtained according to the experiments of eGFP which called the scale of promoters. Finally, with the scale of promoters and proper distribution of enzymes, the proper promoters upstream of different genes can be determined which optimize the expression with least energy consumed.In section”III,we will talk about the modeling of the hemodialysis device. The viscosity of the fluid in our experiments cannot be precisely the same to blood in real condition. To deal with this, we have proposed an equation enabling us to determine the flow density of a fluid given its dynamic viscosity as long as the power of our pump is controlled. Besides, the cleaning-out rate of urate acid is depicted as a function of flow density by solving the diffusion equation or namely Fick’s Law.
A. Urate metabolic pathway
In the metabolic pathway of uric acid of microorganisms, urate is catalyzed into hydroxyisourate by uricase, then hydroxyisourate can react spontaneously and form allantoin. Allantoin is highly solubility and can be easily excreted from the body. The above mentioned new uricase drug is based on this mechanism. There are anther two enzymes that can speed up this process
The three chemical equations in Fig.1 which have been balanced are:
from which we can see the proportions of the substrate and the product are all 1 : 1 In E. coli, the Ygfu gene encodes a uric acid transporter. We plan to overexpress the protein to achieve greater efficiency of uric acid utilization. We found detailed information about these three enzymes and the transporter from the BRENDA database.Maybe the data are not very accurate.
B. Protein producing relative to promoter activity
The model is a made mathematical model of ordinary defferential equations describing the eGFP mRNA concentretion (Eq.1), immature eGFP protein concentration (Eq.2) and mature eGFP protein concentration (Eq.3).
The equations system can be solved with Mathematica
C. Scale of different promoters
In subsectionII D, the concentration of enzyme was required. So these serise of experiments in which we can get the relationship between the fluorescence and the concentration show the scale of promoters.
First the time invariance can be proved by the datas. The promoters producted in different time. And the relationship is independent with time after calculating the variance ratio.
After fitting the data, the function of concentration and fluorescence can be obtained.Now the similarity of standard line in differernt time is analysed. Intuitively, the ratio of overlapping area to the mean area above the x-axis will determine the similarity. s is used to represent the similarity and it’s formular could be written as
in which Soverlap represent the overlapping area,S represent the mean area above the x-axis. Now these two areas are calculating.
The sum of areas between two lines is
in which 1/2 comes from the repeat of counting. And the mean area can be written as
from above we can get
The definition of similarity is relative which scale of is [0,1] . The closer to 1 the value is, the more similar two lines are. This means the lines in our experience are stable.
The MreGFP is 29371.03Da, molarity of eGFP can be writen as
in which ε is a constant that we regard it as 108. So the promoter activity A can be writen as
There are amount of experiments results which could lead more accuracy scale. For exmpale, we make N colony and measure the fluorescence and OD for M times for each colony and LB for Q times.
The scale is showed as follow
According the scale and formula(Eq.4), the protein producing can be draw as fig.2.
D. Kinetic equations of metabolic pathway
First we will obtain the function of the reaction rate and the concentration of substrate and enzyme from Michaelis-Menten equation.
The chemical equation can be writed as:
Assuming the equilibrium of each part of the reaction has been reached which means the producing rate of [ES] and degradating rate of [ES] are equal.
Simply this equation, we can get
Assuming that
Noticing that the reaction of [P] and [E] was ignored. However the concentration of [P] will influence the equilibrium which will affect the producing rate. So as time going, the Eq(14) may not so proper. This will be discussed later.
Eq.14 shows that the reaction rate is relative to [S], [P] and properties of enzyme. No mater first order reaction or zero order reaction, the rate only determined by the concentrationof enzyme when the type of enzyme and the initial concentration of substrate are given. Apparently, the higher concentration of enzyme is, the faster reaction will be. However, there is a globally best solution for muti-steps enzyme catalysis to optimize reactions. If there is a big difference between Vmax = Kcat [E] of two reactions, the product of one of them cannot be reacted which make the metabolic limited by it(fig.3)
On the contrary, if Vmax are proper, the product will not accumulate(fig.4).So proper promoters need to be chosed to make Vmax equal.
E. The experiment out of the bacteria
This subsection is about the model of out of bacteria. For convenience, we call the reaction that transports urate reaction 1, the reaction which substrate of is urate (subsectionII A) reaction 2 and the two reactions after reaction 2 (subsectionII A) reaction 3 and reaction 4. Because the enzymes was put in urate derictly, there is no need to consider the reaction 1.
According to the subsectionII D, the reaction rate of each reaction are
Without considering the spontaneous degradation, rate of one thing equals its producing rate minus reacting rate. And the proportions of each reaction are 1 : 1.
These differential equations are not ordinary, we will numerically solve them with some initial conditions and assumptions later.
F. The experiment in the bacteria
Considering the urate acid transporter as a reaction speeded by enzyme of which the substrate is urate acid out of bacteria and the product is urate acid in the bacteria. The Eq.18 becomes
Besides, there will be two more equaions:
The other equaions are same as Eq.15 Eq.17 and Eq.19 Eq.21.
G. Solution of equation
The equations can be sovled by fourth Runge-Kutta method. Here are two condition we must consider in the algorithm.
1. The total concentration of enzymes is constant because the protein producing ability is limited which means the energy and re-source are distributing determined by different promoters;
2. After some time of reaction, the rates are affect by the equilibrium. So in the algorithm before the equilibrium, the rates suit to the equations in subsectionII E and II F. However, when the proportion of concentration of urate and C5H4N4O4 satisfy equilibrium conditions, the rate becomes to 0.
First the equations system of pathway out of bacteria is solved. According to discussion about Vmax in subsection IID and value given in table II A, we choose J23100 to express PucL, J23113 to express 4.1.1.97 and varied promoters to express PucM(fig.6. The data in table IIA are not very accuracy so may be the values of progress’ result have deviation. However, the trends they predict can be trusted.
1. The fluid is incompressible
2. The fluid is isotropic or the viscosity is a constant at a given temperature
3. The temperature is a constant
4. The flow is a steady laminar flow and its velocity decreases linearly outwards
Flow steady, the viscous dissipation is responsible for the consumption of the input power Newton’s law of viscosity
where we’ve defined
Consider the area element. It‘s a slender strip connecting two ends of the tube which corresponds to infinitesimal angle dθ
The velocity of the mass element at radius r
Now we can obtain the power of the mass element at radius r
or explicitly
Integrate equation (30), we have
In equation (31), vm is still not an observable. The following is dedicated to figure out its connection with flow density which can be easily measured via simple devices.
Consider the flow density element between radius r and r + dr
Thus, the total flow density can be attained by an integration
Solving equation (31) and (33) simultaneously, we have
Equation (34) says that the power consumed(which in our case equals the the input one) is proportional to the viscosity η flow density Q and the total length L.
This is highly reasonable because the more viscous the fluid is, the intenser the fluid are forced to flow and the longer the tube is which leads to grater dissipation, the more power is demanded.
A closer look at equation (34) indicates that flow velocity is linear to the rotational speed of the pump. This theoretical prediction is already confirmed by the former experiments.
We can predict the behaviour of the practical medium beyond purified water in simulation.
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[3] French J B, Ealick S E. Structural and kinetic insights into the mechanism of 5-hydroxyisourate hydrolase from Klebsiella pneumoniae.[J]. Acta Crystallographica, 2011, 67(8):671–677.
[4] French J B, Ealick S E. Structural and mechanistic studies on Klebsiella pneumoniae 2-Oxo-4-hydroxy-4-carboxy-5-ureidoimidazoline decarboxylase.[J]. Journal of Biological Chemistry, 2010, 285(46):35446-35454.
[5] https://2015.igem.org/Team:TU_Delft/Modeling#gene
