Team:Groningen/Modelling/Characterization

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Δv5 = (v5(1) - v5(0))/S
Δv5 = (v5(1) - v5(0))/S
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[[Team:Groningen/Literature#Meng2004|Meng 2004]]was able to knock out all efflux of arsenic. If there is no efflux of arsenic the dervative of the accumulation graph is the speed at wich arsenic is pumped inside the cell. The maximum speed would be v5. In such a senario two measurements would be enough to determine the relative promotor strength. One could even determine the reaction rate k6 and GlpF with a simple calculation, since v5 = k6 GlpFT (Vs/Vc). However we do have efflux, not only do we have efflux, but we have efflux that is dependent on the total amount of arsenic inside the cell. Also a portion arsenic gets bound to ArsR so we have a problem.
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For both MBPArsR and fMT we assume the amount of bound As(III) for a given relative promoter strength S obeys (for MBPArsR n=1):
For both MBPArsR and fMT we assume the amount of bound As(III) for a given relative promoter strength S obeys (for MBPArsR n=1):

Revision as of 09:32, 21 October 2009

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We have four kinds of parts we would like to characterize (RPS stands for Relative Promoter Strength):

Importers Accumulators Sensors GVP cluster
RPS → Δvmax RPS → Asbound(As(III)in) metal(t) → RPS(t) RPS → GV

We can measure how much v5 (vmax for As(III) import via GlpF) is in wild-type E.coli and when we over express GlpF at a certain promoter strength S (measured in RPUs). As v5 is a constant times the amount of (active) GlpF this leads to a simple equation for Δv5, if we assume the amount of (active) GlpF produced by our construct is linearly dependent on the promoter strength (v5(0) and v5(1) would be measured):

v5(RPS) = v5wt + Δv5*RPS

v5(0) = v5wt + Δv5*0
v5(S) = v5wt + Δv5*S

Δv5 = (v5(1) - v5(0))/S

Meng 2004was able to knock out all efflux of arsenic. If there is no efflux of arsenic the dervative of the accumulation graph is the speed at wich arsenic is pumped inside the cell. The maximum speed would be v5. In such a senario two measurements would be enough to determine the relative promotor strength. One could even determine the reaction rate k6 and GlpF with a simple calculation, since v5 = k6 GlpFT (Vs/Vc). However we do have efflux, not only do we have efflux, but we have efflux that is dependent on the total amount of arsenic inside the cell. Also a portion arsenic gets bound to ArsR so we have a problem.

For both MBPArsR and fMT we assume the amount of bound As(III) for a given relative promoter strength S obeys (for MBPArsR n=1):

Asbound(As(III)in)
 = S Bmax As(III)in^n
    / (K^n + As(III)in^n)
The constants Bmax, K and n can be determined from uptake experiments comparing E. coli with and without fMT expression. Of course this can be done in general by fitting our model to experimental data, if enough data is provided the fit will be tight enough to allow this. However, even without fitting the full model it should be possible to make a fair estimation from equilibrium measurements.
If the total cell volume is much smaller than the volume of the solution it is reasonable to assume a constant import rate. Also, regardless of whether they feature fMT or not, in equilibrium the amount of ArsR is the same, as is the amount of ArsB, leading to the same amount of unbound arsenic being present. This means that any difference in uptake of arsenic is completely due to arsenic being bound to fMT or MBPArsR. By measuring the amount of arsenic in equilibrium in wild-type cells as well as in cells expressing fMT/MBPArsR for several different (inital) concentrations of As(III), at one or more (known) levels of expression, it is possible to determine the constants Bmax, K and n.
 i 

The ars promoter is part of a feedback loop, so it is not a simple matter of defining the (instantaneous) promoter strength. Instead we suggest using the relevant equations from our model. The necessary parameters can be determined by fitting uptake measurement data to our model. Specifically, if the RPS is measured without arsenic present and with enough arsenic present to keep the promoter fully active during the experiment we can determine βRN τR as follows (under the assumption that the RPS is linearly dependent on arsT/ars and using the fact that without any arsenic present the cells will be in equilibrium):

S(max) / S(0) = ars(max) / ars(0)
S(max) / S(0) = arsT / ars(0)
S(max) / S(0) = 1 + ArsR(0)²/KAd²
      ArsR(0) = KAd √(S(max)/S(0) - 1)

0 = βRN ars1(0) - (ln(2)/τR) ArsR(0)
0 = βRN ars1T S(0)/S(max)
     - (ln(2)/τR) KAd √(S(max)/S(0) - 1)
βRN ars1T S(0)/S(max)
 = (ln(2)/τR) KAd √(S(max)/S(0) - 1)
βRN τR = (ln(2)/ars1T) KAd
   (S(max)/S(0)) √(S(max)/S(0) - 1)
RPS → GV

The amount of gas vesicles can be expressed in terms of buoyant density, as volume fraction, using the total mass of the vesicles, etc. No matter how it is expressed, we assume a simple linear dependency between the RPS and the amount of gas vesicles. By taking (T)EM pictures of slices the amount of gas vesicles formed under influence of different RPSes can be determined and a straightforward fit made.

Uptake measurements

To fit our model to experimental data from different uptake experiments and/or papers we have implemented an optimization procedure that allows for experiments with different genotypes and circumstances by letting constants be overridden per experiment. It aims to optimize the sum of the RMS errors for each experiment using Simulated Annealing. By clicking the button "Fit" the optimization is started and its progress can be followed by looking at the table of constants and the graphs shown below the table (which are updated in real-time as the best solution is improved).

Sampling scheme
Time (min)
0 10 20 40 60
As(III)exT(0)
(µM)
0 x
10 x x x x x
20 x
50 x
100 x

To efficiently look at both time and concentration dependent processes we took samples as in the table on the right. Below we list all results, which have been used for fitting all necessary parameters.

best cur gradient solved
v5/K5
v5
K5
k8/K7
k8
K7
tauBbetaB
tauB
betaB
tauR
betaRN
tauFbetaF
tauF
betaF
tauKbetaK
tauK
betaK
ars2T
E

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