IGEM:Imperial/2010/Variables1: Difference between revisions
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| Split TEV | | Split TEV | ||
production rates | production rates | ||
|*Assume the both parts of split TEV are half of size of the whole TEV (3054/2=1527 AA) | | | ||
*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 length of the coil is 90 AA | ||
Revision as of 07:04, 3 September 2010
Constants for Modelling
Type of constant | Derivation of value |
---|---|
TEV Enzyme dynamics | Enzymatic Reaction:
[math]\displaystyle{ E+S\rightleftarrows ES \rightarrow E+P }[/math] Let
We know that [math]\displaystyle{ K_m = \dfrac{k_{cat} + k_2}{k_1} }[/math] Assuming that [math]\displaystyle{ k_{cat} \lt \lt k_2 \lt \lt k_1 }[/math], we can rewrite [math]\displaystyle{ K_m }[/math]≈[math]\displaystyle{ k_2/k_1 }[/math] From this paper constants for TEV can be found:
These values correspond with our assumption that [math]\displaystyle{ k_{cat} = 0.1 s^{-1} }[/math] and [math]\displaystyle{ K_m = 0.01 mM }[/math]. Hence, we can estimate the following orders of magnitude for the rate constants:
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, gr (divisions/h): 0.53<gr<2.18 Hence on average, gr = 1.5 divisions per hour => 1 division every 40mins To deduce degradation rate use the following formula: [math]\displaystyle{ \tau_{\tfrac{1}{2}}=\frac{ln2}{k} }[/math] Where [math]\displaystyle{ \tau_{\tfrac{1}{2}}=0.667 hour }[/math] k...the degradation rate [math]\displaystyle{ k=\frac{ln2}{\tau_\tfrac{1}{2}}=0.000289s^{-1} }[/math] |
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:
LacZ production = 100 molecules/min (1024 AA) Average production ≈ 100molecules/min 720 AA That gives us:
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 → [math]\displaystyle{ 2.3808*10^{-10} mol/dm^3/s }[/math]
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.
[math]\displaystyle{ K_m = 10\mu M }[/math] [math]\displaystyle{ k_{cat} = 52s^{-1} }[/math]
[math]\displaystyle{ K_m = 40\mu M }[/math] [math]\displaystyle{ k_{cat} = 192s^{−1} }[/math] Consequently, the [math]\displaystyle{ \tfrac{k_{cat}}{K_m} = 4.8 }[/math] of the mutant was slightly lower than [math]\displaystyle{ \tfrac{k_{cat}}{K_m} = 5.2 }[/math] of the wild type, indicating that the mutation has little effect on catalytic efficiency. |
Dimensions of
Bacillus subtillis cell |
Dimensions of Basillus subtilis (cylinder/rod shape) in reach media (ref. bionumbers):
This gives: [math]\displaystyle{ {Volume}= \pi\tfrac{d^2}{4}l=2.793999\mu m^3\approx 2.79∙10^{-15} dm^3 }[/math] |
Split TEV
production rates |
The whole construct is then: 1617 AA → split TEV production rate ≈ [math]\displaystyle{ 1.2606*10^{-10} mol/dm^3/s }[/math] |
Relevant concentrations of catechol | We have catechol in the lab in powder form so we're limited only by catechol's solubility.
For concentration of 0.1 M with built up levels of dioxygenase the colour change happens within seconds! We will run our models for 0.1M ± several orders of magnitude to determine the smallest catechol concentration that will give appreciable difference between simple production response and the amplified response. |