Team:Groningen/Modelling/Arsenic

From 2009.igem.org

A schematic representation of the processes involved in arsenic filtering (keep in mind that ArsR represses the expression of the genes behind ars). Note that MBPArsR and fMT are not shown for clarity.

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Based on the quasi-steady-state derivation below we have made the simplified version of our model shown below. The simplification is based on two key assumptions (which are also illustrated below, next to the table "Breakdown of core substances"):

• Binding and unbinding of arsenic to/from the transporters occurs on a much smaller time scale than changes in the concentration of arsenic inside and outside the cell. And similarly, we assume that (un)binding of ArsR to/from the ars promoter is much faster than the production of ArsR (for example).
• The concentration of transporters is insignificant compared to the concentration of arsenic inside and outside the cell.

This leads to the Michaelis-Menten equation for import, but also some more general equations for export using ArsB and accumulation with ArsR (for example, the Hill equation can be recognized in the activity of the ars promoter).

A schematic representation of the processes involved in arsenic filtering (keep in mind that ArsR represses the expression of the genes behind ars). Note that MBPArsR and fMT are not shown for clarity.

Breakdown of core substances Core substance Component Relative abundance ArsBT ArsB K7 ArsBAs As(III)in As(III)inT As(III)in 1 ArsRAs ArsR / KRd MBPArsRAs MBPArsRT / (KMd + As(III)in) fMTAs nf fMTT As(III)innf-1 / (KFdnf + As(III)innf) arsT ars KAd² ArsRars ArsR² ars ars1 ars1T ars2 ars2T ArsRT ArsR 1 ArsRAs As(III)in / KRd ArsRars 2 ArsR ars / KAd² MBPArsRT MBPArsR KMd MBPArsRAs As(III)in fMTT fMT KFdnf fMTAs As(III)innf

Circles correspond to core substances. We consider the reactions between the overlapping substances so fast that we model them by determining the ratios between the substances when the reactions between them are in equilibrium. Also, the complexes formed with ars, GlpF and ArsB (the small circles) are considered to have such a low concentration that they are of no importance to the concentrations of As(III)in/-ex and ArsR (the large circles).TODO: Make ars a big circle and possibly add fMT and ArsRMBP.

The raw model

The variables above can be related to each other through the following "reactions" (color coding is continued below to show which parts of the differential equations refer to which groups of reactions):

• Transport (based on Rosen1996, Meng2004 and Rosen2009)

• Accumulation (mostly based on Chen1997)

Using the following constants/definitions:

See Chen1997 for the interplay between ArsR and ArsD (the latter has a role similar to ArsR, but we do not treat it, as it is not present in our system).

Quasi steady state

First of all, we assume the concentration of transporters is quite low compared to the concentration of the transported substances. After all, if this were not the case the transporters would act more like "storage" proteins than transporters (note that this can be even more rigorously justified if, for example, GlpFT<<K5). This leads to:

As(III)exT ≈ As(III)ex
As(III)inT ≈ As(III)in + ArsRAs + MBPArsRAs + nf fMTAs

Also, we assume the binding and unbinding of molecules to the transporters occurs on a much finer time-scale than any actual changes to the concentrations inside and outside the cell. Similarly, within the cell we assume diffusion processes are very fast and binding/unbinding of substances is quite fast compared to the production of proteins. This leads us to assume that the following ratios between substances are constantly in equilibrium:

arsT = ars + ArsRars
ars1 / ars1T = ars2 / ars2T

 ars = ars1 + ars2
 ars = ars1 (1 + ars2T / ars1T)
ars1 = ars / (1 + ars2T / ars1T)
ars1 = ars ars1T / arsT

ars2 = ars ars2T / arsT

As(III)ex : GlpFAs ≈ As(III)ex : 0
GlpF : GlpFAs
ArsB : ArsBAs
As(III)in : ArsRAs : MBPArsRAs : nf fMTAs : ArsBAs ≈ As(III)in : ArsRAs : MBPArsRAs : nf fMTAs : 0
ArsR : ArsRAs : 2 ArsRars
ars : ArsRars

To determine what the unknown ratios are we can set the following derivatives to zero (these are the derivatives of the complexes corresponding to the four overlapping regions in the diagram):

0 = (d/dt) GlpFAs = k5on As(III)ex GlpF - (k5off+k6) GlpFAs
0 = (d/dt) ArsBAs = k7on As(III)in ArsB - (k7off+k8) ArsBAs
0 = (d/dt) ArsRars = kAon ArsR² ars - kAoff ArsRars
0 = (d/dt) ArsRAs = kRon ArsR As(III)in - kRoff ArsRAs
0 = (d/dt) MBPArsRAs = kMon MBPArsR As(III)in - kMoff MBPArsRAs
0 = (d/dt) fMTAs = kFon fMT As(III)in^nf - kFoff fMTAs

The first two derivates let us determine the ratios between bound and unbound transporters:

0 = (d/dt) GlpFAs = k5on As(III)ex GlpF - (k5off+k6) GlpFAs

k5on As(III)ex GlpF = (k5off+k6) GlpFAs
               GlpF = (k5off+k6)/k5on GlpFAs / As(III)ex
               GlpF = K5 GlpFAs / As(III)ex

                 GlpF : GlpFAs
K5 GlpFAs / As(III)ex : GlpFAs
                   K5 : As(III)ex

0 = (d/dt) ArsBAs = k7on As(III)in ArsB - (k7off+k8) ArsBAs

k7on As(III)in ArsB = (k7off+k8) ArsBAs
               ArsB = (k7off+k8)/k7on ArsBAs / As(III)in
               ArsB = K7 ArsBAs / As(III)in

                 ArsB : ArsBAs
K7 ArsBAs / As(III)in : ArsBAs
                   K7 : As(III)in

The next two differential equations can be used to determine the relative abundances of ArsR and ArsRAs, and ars and ArsRars:

0 = (d/dt) ArsRAs = kRon ArsR As(III)in - kRoff ArsRAs

kRon ArsR As(III)in = kRoff ArsRAs
           ArsRAs = kRon/kRoff ArsR As(III)in
           ArsRAs = ArsR As(III)in / KRd

ArsR : ArsRAs
ArsR : ArsR As(III)in / KRd
KRd  : As(III)in

0 = (d/dt) ArsRars = kAon ArsR² ars - kAoff ArsRars

kAon ArsR² ars = kAoff ArsRars
       ArsRars = kAon/kAoff ArsR² ars
       ArsRars = ArsR² ars / KAd²

ArsR : 2 ArsRars
ArsR : 2 ArsR² ars / KAd²
KAd² : 2 ArsR ars

ars : ArsRars
ars : ArsR² ars / KAd²
KAd² : ArsR²

For MBPArsR and fMT we find:

0 = (d/dt) MBPArsRAs = kMon MBPArsR As(III)in - kMoff MBPArsRAs

MBPArsR : MBPArsRAs = KMd : As(III)in

0 = (d/dt) fMTAs = kFon fMT As(III)in^nf - kFoff fMTAs

fMT : fMTAs = KFd^nf : As(III)in^nf

And finally the relative abundances of arsenic:

ArsRAs = ArsR As(III)in / KRd

As(III)in : ArsRAs               : MBPArsRAs                            : n fMTAs
As(III)in : ArsR As(III)in / KRd : MBPArsRT As(III)in / (KMd+As(III)in) : n fMTT As(III)in^nf / (KFd^nf+As(III)in^nf)
1         : ArsR / KRd           : MBPArsRT / (KMd+As(III)in)           : n fMTT As(III)in^(nf-1) / (KFd^nf+As(III)in^nf)

Summarizing:

GlpF : GlpFAs = K5 : As(III)ex
ArsB : ArsBAs = K7 : As(III)in
As(III)in : ArsRAs : MBPArsRAs : n fMTAs ≈ 1 : ArsR / KRd : MBPArsRT / (KMd+As(III)in) : n fMTT As(III)in^(nf-1) / (KFd^nf+As(III)in^nf)
ars : ArsRars = KAd² : ArsR²
ArsR : ArsRAs : 2 ArsRars ≈ 1 : As(III)in / KRd : 2 ArsR ars / KAd²
MBPArsR : MBPArsRAs = KMd : As(III)in
fMT : fMTAs = KFd^nf : As(III)in^nf

Now we can look at the differential equations for the totals of ArsB (so ArsBT=ArsB+ArsBAs), ArsR, As(III)in and As(III)ex (GlpFT and arsT are assumed to be constant):

(d/dt) As(III)exT = (d/dt) As(III)ex + (d/dt) GlpFAs
                  = - (d/dt) GlpFAs - k6 GlpFAs + (Vc/Vs) k8 ArsBAs + (d/dt) GlpFAs
                  = (Vc/Vs) k8 ArsBAs - k6 GlpFAs
                  = (Vc/Vs) k8 ArsBAs - (Vc/Vs) v5 GlpFAs / GlpFT
                  = (Vc/Vs) k8 ArsBAs - (Vc/Vs) v5 As(III)ex / (K5+As(III)ex)
                  = (Vc/Vs) k8 ArsBAs - (Vc/Vs) v5 As(III)exT / (K5+As(III)exT)
(d/dt) ArsBT = (d/dt) ArsB + (d/dt) ArsBAs
             = - (d/dt) ArsBAs + βB ars1 - ln(2)/τB ArsB + (d/dt) ArsBAs
             = βB ars1 - ln(2)/τB ArsB
(d/dt) As(III)inT = -(Vs/Vc) (d/dt) As(III)exT
                  = v5 As(III)exT / (K5+As(III)exT) - k8 ArsBT As(III)in / (K7+As(III)in)
(d/dt) ArsRT = (d/dt) ArsR + (d/dt) ArsRAs + 2 (d/dt) ArsRars
             = βRN ars1 + βR proR - (ln(2)/τR) ArsR - (d/dt) ArsRAs - 2 (d/dt) ArsRars + (d/dt) ArsRAs + 2 (d/dt) ArsRars
             = βRN ars1 + βR proR - (ln(2)/τR) ArsR
(d/dt) MBPArsRT = (d/dt) MBPArsR + (d/dt) MBPArsRAs
                = βM proM - (ln(2)/τM) MBPArsR
(d/dt) fMTT = (d/dt) fMT + (d/dt) fMTAs
            = βF proF - (ln(2)/τF) fMT

Steady state

By looking at the steady state of the system we can say something about its long-term behaviour. This also makes it easier to analyze relations between variables. To derive the steady state solution we take the quasi steady state solution and simplify it further by setting additional derivatives to zero:

0 = (d/dt) ArsBT = βB ars1 - ln(2)/τB ArsB
0 = (d/dt) As(III)inT = v5 As(III)exT / (K5+As(III)exT) - k8 ArsBAs
0 = (d/dt) ArsRT = βRN ars1 + βR pro - (ln(2)/τR) ArsR
0 = (d/dt) MBPArsRT = βM proM - (ln(2)/τM) MBPArsR
0 = (d/dt) fMTT = βF proF - (ln(2)/τF) fMT
0 = (d/dt) GV = βG ars2 - ln(2)/τG GV

This directly leads to:

   0 = βB ars1 - ln(2)/τB ArsB
ArsB = βB (τB/ln(2)) ars1
ArsB = βB (τB/ln(2)) ars1T KAd²/(KAd²+ArsR²)

      0 = βM proM - (ln(2)/τM) MBPArsR
MBPArsR = βM (τM/ln(2)) proM

  0 = βF proF - (ln(2)/τF) fMT
fMT = βF (τF/ln(2)) proF

 0 = βG ars2 - ln(2)/τG GV
GV = βG (τB/ln(2)) ars2
GV = βG (τB/ln(2)) ars2T KAd²/(KAd²+ArsR²)

         0 = v5 As(III)exT / (K5+As(III)exT) - k8 ArsBAs
         0 = v5 As(III)exT / (K5+As(III)exT) - k8 ArsB As(III)in / K7
         0 = v5 As(III)exT - k8 ArsB As(III)in / K7 (K5+As(III)exT)
         0 = v5 As(III)exT - k8 ArsB As(III)in As(III)exT / K7 - k8 ArsB As(III)in K5 / K7
         0 = As(III)exT (v5 - k8 ArsB As(III)in / K7) - k8 ArsB As(III)in K5 / K7
As(III)exT = k8 ArsB As(III)in K5 / (v5 K7 - k8 ArsB As(III)in)
As(III)exT = K5 As(III)in / (K7 v5/(k8 ArsB) - As(III)in)

As we can safely assume arsenic neither disappears into nothingness nor appears from nothingness, we can use this to derive (As(III)T is the total amount of arsenic):

As(III)inT = As(III)in (1 + ArsR/KRd + MBPArsR/KMd + fMT As(III)in^(nf-1)/KFd^nf)

As(III)T = Vs As(III)exT + Vc As(III)inT
0 = Vs As(III)exT + Vc As(III)inT - As(III)T
0 = Vs K5 As(III)in / (K7 v5/(k8 ArsB) - As(III)in) + Vc As(III)in (1 + ArsR/KRd + MBPArsR/KMd + fMT As(III)in^(nf-1)/KFd^nf) - As(III)T

As the function on the right-hand side is non-decreasing for As(III)in∈[0,K7 v5/(k8 ArsB)⟩ it at most has one zero on this interval (and it has one, as it starts at a negative value and gets arbitrarily large as As(III)in approaches the end of its range). So this zero can safely be found using any number of numerical methods.

Finally, for ArsR we can find the following third-order equation:

0 = βRN ars1 + βR pro - (ln(2)/τR) ArsR
0 = βRN ars1T KAd²/(KAd²+ArsR²) + βR pro - (ln(2)/τR) ArsR
0 = βRN ars1T KAd² + βR pro (KAd²+ArsR²) - (ln(2)/τR) ArsR (KAd²+ArsR²)
0 = βRN ars1T KAd² + βR pro KAd² + βR pro ArsR² - (ln(2)/τR) ArsR KAd² - (ln(2)/τR) ArsR³
0 = (βRN ars1T + βR pro) KAd² - (ln(2)/τR) KAd² ArsR + βR pro ArsR² - (ln(2)/τR) ArsR³
0 = (βRN ars1T + βR pro) (τR/ln(2)) KAd² - KAd² ArsR + βR (τR/ln(2)) pro ArsR² - ArsR³

According to Mathematica's solution of Reduce[eq && KAd > 0 && arsT >= 0 && pro >= 0 && β1 > 0 && β3 > 0 && τR > 0, ArsR, Reals] (where eq is the equation shown above) there is only one real solution (examining the discriminant of eq confirms this), so we can solve the equation safely using Newton's (or Halley's) method.