IGEM:Imperial/2010/Overview

Output Amplification Model

Why do we want to model the output amplification?

• To determine whether simple production is better than 1- or 2-step amplification.

Model pre-A: Simple production of Dioxygenase

This model involves the simple transcription and translation of Dioxygenase. It can be modelled by using Michaelis-Menten kinetics. Assumptions of Michaelis-Menten kinetics:

• No intermediate or product inhibition
• No allostericity or cooperativity
• Quasi-steady-state assumption
• Total enzyme concentration does not change over time ([E] = [E0] + [ES])

We then have one differential equation:

• [p'] = s - d*[p]
• [p]...concentration of dioxygenase
• s...production rate of dioxygenase
• d...degradation rate of dioxygenase

Using the built-in Matlab function ode45, which is a solver for differential equations, we obtain the following graph:

• Here is the code for Model pre-A.

Model A: 1-step amplification

The 1-step amplification model involves the following enzymatic reaction:

• TEV + sD <--> TEV-sD --> TEV + D
• TEV...Tobacco Etch Virus
• sD...split Dioxygenase
• TEV-sD...intermediate complex
• D...Dioxygenase

Note that this enzymatic reaction cannot be simplified using Michaelis-Menten kinetics because of the assumptions of Michaelis-Menten. Michaelis-Menten assumptions that our system does not meet:

• Vmax is proportional to the overall concentration of the enzyme.

But we are producing enzyme, so Vmax will change! Therefore, the conservation E0 = E + ES does not hold for our system.

• Substrate >> Enzyme

Since we are producing both substrate and enzyme, we have roughly the same amount of substrate and enzyme.

Therefore, we have to solve from 1st principle by using the law of mass action. Assumptions of mass action:

• Chemical reaction rate is directly proportional to the concentration of the reacting substances

The reaction can then be rewritten in these four sub-equations:

1. [T'] = -k1[T][sD] + (k2+k3)[TsD] + sT - dT[T]
2. [sD']= -k1[T][sD] + k2[TsD] + ssD - dsD[sD]
3. [TsD'] = k1[T][sD] - (k2+k3)[TsD] - dTsD[TsD]
4. [D'] = k3[TsD] - dD[D]

Implementing these differenital equations in the ode45-solver does not give reasonable results (e.g. concentrations fall below zero). We found out that this is due to our set of differential equations being stiff, hence, we use a different Matlab solver (ode15s). The results from the Matlab simulation are highly dependent on the initial concentration of split Dioxygenase, as well as other constants that we have approximated. Here is the Matlab result for an initial concentration of 0.01 mol/dm^3: It can be seen that the 1-step amplification is more efficient if we wait for longer than 1000s. However, the most important information for this graph is the threshold concentration, for which Dioxygenase becomes visible. If this threshold concentration corresponds to a time greater than 1000s, then the 1-step amplification will be more efficient.

• Here is the code for Model A.

Varying constants

We want to determine how our system reacts if different parameters are changed. This is to find out which parameters our system is very sensitive to.

Model B: 2-step amplification

The only difference between this model and the 1-step amplification is the added step for amplification. Similar to 1-step amplification, this system can be modelled using mass action.

Implementing the 2-step amplification model in Matlab (using ode15s) gives the following graph and comparing all three different models: It can be seen that 2-step amplification is only marginally more efficient than 1-step amplification. Therefore, it would be more sensible to implement the 1-step amplification.

• Here is the code for Model B.

Constants used for Modelling

Type of constant	Derivation of value
TEV Enzyme dynamics	Enzymatic Reaction: E + S <-> ES -> E + P Let k1 = rate constant for E + S -> ES k2 = rate constant for E + S <- ES kcat = rate constant for ES -> E + P We know that Km = (kcat + k2)/k1 Assuming that kcat << k2 << k1, we can rewrite Km ~ k2/k1 From this paper constants for TEV can be found: e.g. wildtype TEV Km = 0.062 +/- 0.010 mM kcat = 0.16 +/- 0.01 s^-1 These values correspond with our assumption that kcat ~ 0.1 s^-1 and Km ~ 0.01 mM. Hence, we can estimate the following orders of magnitude for the rate constants: k1 ~ 10^5 M^-1 s^-1 k2 ~ 10^3 s^-1 Using these values should be a good approximation for our model.
Degradation rate (common for all)	Assumption: To be approximated by cell division (dilution of media) as none of the proteins are involved in any active degradation pathways Growth rate (divisions/h): 0.53<G.r.<2.18 Hence on average, g.r.=1.5 divisions per hour => 1 division every 1/1.5=0.6667 of an hour (40mins) To deduce degradation rate use the following formula: τ_(1⁄2)=ln2/k Where τ_(1⁄2)=0.667 hour, k – is the degradation rate k=ln2/τ_(1⁄2) = 0.000289 s^(-1)
Production rate (TEV and dioxygenase)	We had real trouble finding the production rate values in the literature and we hope to be able to perform experiments to obtain those values for (TEV protease and catechol 2,3-dioxygenase). The experiments will not be possible to be carried out soon, so for the time being we suggest very simplistic approach for estimating production rates. We have found production rates for two arbitrary proteins in E.Coli. We want to get estimates of production rates by comparing the lengths of the proteins (number of amino-acids). As this approach is very vague, it is important to realise its limitations and inconsistencies: Found values are taken from E.Coli not B.sub. The two found rates are of the same value for quite different amino-acid number which indicates that protein folding is limiting the production rates (we use the chosen approach as the only way of getting the estimate of order of reaction) LacY production = 100 molecules/min (417 Amino Acids) LacZ production = 100 molecules/min (1024 AA) Average production ≈ 100molecules/min 720 AA That gives us: TEV production ≈ 24 molecules/min = 0.40 molecules/s (3054 AA) As production rate needs to be expressed in concentration units per unit volume, the above number is converted to mols/s and divided by the cell volume → 2.3808*10^(-10) mol*dm^(-3)*s^(-1) C23D production ≈ 252 molecules/min = 4.2 molecules/s (285 AA) → 2.4998*10^(-9) mol*dm^(-3)*s^(-1) We will treat these numbers as guiding us in terms of range of orders of magnitudes. We will try to run our models for variety of values and determine system’s limitations.
Kinetic parameters of dioxygenase	Initial velocity of the enzymatic reaction was investigated at pH 7.5 and 30 °C. Wild type (we use that one) Km = 10μM kcat = 52 s^(−1) Mutated type Km = 40μM kcat = 192 s^(−1) Consequently, the kcat/Km 4.8 of the mutant was slightly lower than kcat/Km 5.2 of the wild type, indicating that the mutation has little effect on catalytic efficiency. reference
Dimensions of Bacillus subtillis cell	Dimensions of Basillus subtilis (cylinder/rod shape) in reach media (ref. bionumbers): diameter: d= 0.87um length: l=4.7 um in rich media This gives: Volume= π∙(d/2)^2∙l=2.793999 μm^3≈ 2.79∙10^(-15) dm^3
Split TEV production rates	*Assume the both parts of split TEV are half of size of the whole TEV (3054/2=1527 AA) The length of the coil is 90 AA The whole construct is then: 1617 AA → split TEV production rate ≈ 1.2606*10^(-10) mol*dm^(-3)*s^(-1)