Team:Groningen/Modelling/Arsenic

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The raw model

A schematic representation of the processes involved in arsenic filtering (keep in mind that ArsR represses the expression of the genes behind OpG). Note that ArsD is not shown here, as it is not present in our E. coli and has a role analogous to ArsR.

The following variables play an important role in our system (these can be concentrations of substances, the density of the cell, etc.):

• Extracellular:
  • As(III)_ex
  • As(V)_ex
• Membrane:
  • GlpF (concentration w.r.t. the exterior of the cell)
  • GlpF_As (concentration w.r.t. the exterior of the cell)
  • ArsB (concentration w.r.t. the interior of the cell)
  • ArsB_As (concentration w.r.t. the interior of the cell)
• Intracellular:
  • As(III)
  • OpG (concentration of unbound operators)
  • OpH (concentration of ArsR producing operators that are always on)
  • As(V)
  • ArsC
  • ArsR
    ArsR binds to OpG to repress production of the genes they regulate, and binds to As(III) to make it less of a problem for the cell.

     i 
  • ArsR_As (bound to As(III))
    • At equilibrium: ArsR As(III) = (k1_off/k1_on) ArsR_As
  • ArsR_opg (bound to opg)

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)
  • As(III)_ex + GlpF ↔ GlpF_As
  • GlpF_As → GlpF + As(III)
  • As(III) + ArsB ↔ ArsB_As
  • ArsB_As → ArsB + As(III)_ex
  • ArsB → null (degradation)
• Accumulation (based on Chen1997)
  • As(III) + ArsR ↔ ArsR_As
  • OpG + 2 ArsR ↔ ArsR_opg
  • OpG → OpG + ArsR + ArsB (transcription + translation)
  • OpH → OpH + ArsR (transcription + translation)
  • ArsR → null (degradation)

Resulting in the following differential equations (please note that some can be formed by linear combinations of the others), using color coding to show the correspondence to the reactions above:

• (d/dt) As(III)_ex = -k5_on As(III)_ex GlpF + k5_off GlpF_As + (Vc/Vs) k8 ArsB_As
• (d/dt) GlpF = -k5_on As(III)_ex GlpF + (k5_off+k6) GlpF_As
• (d/dt) GlpF_As = k5_on As(III)_ex GlpF - (k5_off+k6) GlpF_As
• (d/dt) ArsB = -k7_on As(III) ArsB + (k7_off+k8) ArsB_As + β4 OpG - ln(2)/τB ArsB
• (d/dt) ArsB_As = k7_on As(III) ArsB - (k7_off+k8) ArsB_As
• (d/dt) As(III) = - (d/dt) ArsR_As - k7_on As(III) ArsB + k7_off ArsB_As + (Vs/Vc) k6 GlpF_As
• (d/dt) OpG = - (d/dt) ArsR_opg
• (d/dt) ArsR = β1 OpG + β3 OpH - (ln(2)/τR) ArsR - (d/dt) ArsR_As - 2 (d/dt) ArsR_opg
• (d/dt) ArsR_As = k1_on ArsR As(III) - k1_off ArsR_As
• (d/dt) ArsR_opg = k3_on ArsR² OpG - k3_off ArsR_opg

Using the following constants/definitions:

• k1on, k2on, etc. = reaction rate constants for reactions to a complex
• k1off, k2off, etc. = reaction rate constants for reactions from a complex
• K1d - K4d = dissociation constants
  • K1_d = k1_off/k1_on
  • K2_d = k2_off/k2_on = 60µM (Chen1997)
  • K3_d = k3_off/k3_on = 0.33µM (Chen1997, suspect as the relevant reference doesn't actually seem to give any value for this)
  • K4_d = k4_off/k4_on = 65µM (Chen1997)
• degradation rate = ln(2)/τ
  If you take just the degradation into account you will have the equation dC/dt = -k*C, which leads to C(t) = C(0) e^{-k t}. So if k = ln(2)/τ we get C(t) = C(0) e^{-ln(2)/τ t} = C(0) 2^-t/τ. In other words τ is the time it takes for the concentration to half.

   i 
  • ArsR half-life time = τR
• β1, β2, etc. = production rates
  • β1 = the production rate for ArsR behind the ars promoter
  • β3 = the production rate for ArsR behind our constitutive promoter
  • β4 = the production rate for ArsB behind the ars promoter
• Vs = volume of solution (excluding cells), Vc = total volume of cells (in solution) (so Vs+Vc is the total volume)
• v5, v7 = maximum reaction rates (see Michaelis-Menten equation)
  • v5 = k6 GlpFT
  • v7 = k8 ArsBT
• v7'·OpG = reaction rate when As(III) = K7
  • v7' = k8 (β4 τB/ln(2))
• K5, K7 = concentration at which the reaction reaches half its maximum reaction rate (see Michaelis-Menten equation)
  • K5 = (k5off+k6) / k5on
  • K7 = (k7off+k8) / k7on (in our model, with regulation of ArsB, this is the concentration As(III) at which the reaction rate equals v7'·OpG)

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).

Transport

By looking at the system in equilibrium we can more easily assess the impact of parameters and derive formulas for obtaining them. To this end we consider the system when all derivatives are zero (just taking the equations relevant for transport):

0 = -k5on As(III)ex GlpF + k5off GlpFAs + (Vc/Vs) k8 ArsBAs
0 = -k5on As(III)ex GlpF + (k5off+k6) GlpFAs
0 =  k5on As(III)ex GlpF - (k5off+k6) GlpFAs
0 = -k7on As(III) ArsB + (k7off+k8) ArsBAs + β4 OpG - ln(2)/τB ArsB
0 =  k7on As(III) ArsB - (k7off+k8) ArsBAs
0 = -k7on As(III) ArsB + k7off ArsBAs + (Vs/Vc) k6 GlpFAs

The third equation is equivalent to the second and the fifth can be eliminated from the fourth, leading to:

0 = -k5on As(III)ex GlpF + k5off GlpFAs + (Vc/Vs) k8 ArsBAs
0 = -k5on As(III)ex GlpF + (k5off+k6) GlpFAs
0 = β4 OpG - ln(2)/τB ArsB
0 =  k7on As(III) ArsB - (k7off+k8) ArsBAs
0 = -k7on As(III) ArsB + k7off ArsBAs + (Vs/Vc) k6 GlpFAs

By combining the last two equations a rather obvious equation can be derived that essentially expresses that the import rate equals the export rate:

    0 - 0 =    (-k7on As(III) ArsB + k7off ArsBAs + (Vs/Vc) k6 GlpFAs)
             + ( k7on As(III) ArsB - (k7off+k8) ArsBAs)
        0 = (Vs/Vc) k6 GlpFAs - k8 ArsBAs
k8 ArsBAs = (Vs/Vc) k6 GlpFAs

Using the equations above (and the fact that the total amount of importers doesn't change in our model) GlpFAs can be expressed as follows (working towards a Michaelis-Menten equation):

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

 GlpFT = GlpF + GlpFAs
 GlpFT = GlpFAs (1 + (k5off+k6) / (k5on As(III)ex))
GlpFAs = GlpFT / (1 + (k5off+k6) / (k5on As(III)ex))
GlpFAs = GlpFT As(III)ex / (As(III)ex + K5)

Without ArsB regulation

If we disregard ArsB regulation ArsBAs can be determined similarly to GlpFAs:

ArsBAs = ArsBT As(III) / (As(III) + K7)

By substituting the above equations for GlpFAs and ArsBAs we can now derive a relation between the extracellular and intracellular concentrations of arsenic (where we recognize the constants of the well-known Michaelis-Menten equation):

                                              k8 ArsBAs = (Vs/Vc) k6 GlpFAs
                      k8 ArsBT As(III) / (As(III) + K7) = (Vs/Vc) k6 GlpFT As(III)ex / (As(III)ex + K5)
                           v7 As(III) / (As(III) + K7) = (Vs/Vc) v5 As(III)ex / (As(III)ex + K5)
                           v7 As(III) (As(III)ex + K5) = (Vs/Vc) v5 As(III)ex (As(III) + K7)
As(III) (v7 As(III)ex + v7 K5 - (Vs/Vc) v5 As(III)ex) = (Vs/Vc) v5 K7 As(III)ex
                                                As(III) = (Vs/Vc) v5 K7 As(III)ex / (v7 As(III)ex + v7 K5 - (Vs/Vc) v5 As(III)ex)
                                                As(III) = (Vs/Vc) v5 K7 As(III)ex / ((v7 - (Vs/Vc) v5) As(III)ex + v7 K5)

An important flaw in this model is that the production of ArsB is dependent on the concentration of arsenic in the cell (via regulation by ArsR, see our transport page). This could be one of the reasons that this model is unable to fit the curve shown in figure 3A in Kostal2004 (if we try a least squares fit with the equation above, filling in v5 and K5 from what we computed for figure 1B in Meng2004, we get negative values for v7 and K7). (And something similar is true of figure 1 from Singh2008.) This led us to consider the regulation of ArsB, as is described in the following section.

With ArsB regulation

As explained above ArsBAs needs a slightly different approach, as ArsB can be produced in response to arsenic ArsBT is not a (known) constant and we have a formula dependent on OpG instead:

β4 OpG = ln(2)/τB ArsB
  ArsB = (β4 τB/ln(2)) OpG

(k7off+k8) ArsBAs = k7on As(III) ArsB
(k7off+k8) ArsBAs = (β4 τB/ln(2)) OpG k7on As(III)
           ArsBAs = (β4 τB/ln(2)) OpG (k7on/(k7off+k8)) As(III)
           ArsBAs = (β4 τB/ln(2)) OpG As(III) / K7

This now leads to the following relation between intra- and extracellular As(III), note that v7' (in contrast to v7) is not the maximum export rate (it has units of 1/second):

                        k8 ArsBAs = (Vs/Vc) k6 GlpFAs
k8 (β4 τB/ln(2)) OpG As(III) / K7 = (Vs/Vc) k6 GlpFT As(III)ex / (As(III)ex + K5)
               v7'/K7 OpG As(III) = (Vs/Vc) v5 As(III)ex / (As(III)ex + K5)
                          As(III) = (Vs/Vc) v5/(v7' OpG) K7 As(III)ex / (As(III)ex + K5)

To get rid of the unknown OpG in this equation we can use two equations that are derived below for accumulation:

              0 = β1 OpG + β3 OpH - (ln(2)/τR) ArsR
(ln(2)/τR) ArsR = β1 OpG + β3 OpH
           ArsR = (β1 OpG + β3 OpH) (τR/ln(2))

 OpG = OpGT/(1 + ArsR²/K3d)
OpGT = (1 + ArsR²/K3d) OpG
OpGT = OpG + ArsR² OpG / K3d
   0 = K3d OpG + ArsR² OpG - K3d OpGT
   0 = K3d OpG + (β1 OpG + β3 OpH)² (τR/ln(2))² OpG - K3d OpGT
   0 = K3d OpG + (β1 OpG + β3 OpH) (β1 OpG² + β3 OpH OpG) (τR/ln(2))² - K3d OpGT
   0 = K3d OpG + β1 OpG β1 OpG² (τR/ln(2))² + β1 OpG β3 OpH OpG (τR/ln(2))² + β3 OpH β1 OpG² (τR/ln(2))² + β3 OpH β3 OpH OpG (τR/ln(2))² - K3d OpGT
   0 = β1² (τR/ln(2))² OpG³ + 2 β1 β3 (τR/ln(2))² OpH OpG² + (β3² (τR/ln(2))² OpH² + K3d) OpG - K3d OpGT

According to Mathematica's solution of Reduce[eq && K3d > 0 && OpGT >= 0 && OpH >= 0 && β1 > 0 && β3 > 0 && τR > 0, OpG, Reals] (where eq is the equation above) there is only one real root, so we solve the equation safely using Newton's (or Halley's) method.

Accumulation

For many purposes, like determining the total amount of accumulated arsenic, it can be quite useful to consider the system in equilibrium. That is, when the derivatives of all variables to time are zero (just taking the equations relevant for accumulation here, assuming the As(III) concentration to be constant):

0 = (d/dt) ArsR = β1 OpG + β3 OpH - (ln(2)/τR) ArsR - (d/dt) ArsRAs - 2 (d/dt) ArsRopg
0 = (d/dt) ArsRAs = k1on ArsR As(III) - k1off ArsRAs
0 = (d/dt) ArsRopg = k3on ArsR² OpG - k3off ArsRopg

By eliminating the last two equations from the first and dividing the last two by k?on we are left with:

0 = β1 OpG + β3 OpH - (ln(2)/τR) ArsR
0 = ArsR As(III) - K1d ArsRAs
0 = ArsR² OpG - K3d ArsRopg

Using the fact that the "concentration" of operators remains constant the last equation can be used to derive an equation for OpG:

      0 = ArsR² OpG - K3d ArsRopg
ArsRopg = OpG ArsR² / K3d

OpGT = OpG + ArsRopg
OpGT = OpG + OpG ArsR² / K3d
OpGT = OpG (1 + ArsR²/K3d)
 OpG = OpGT/(1 + ArsR²/K3d)

This leads to the following:

0 = β1 OpG + β3 OpH - (ln(2)/τR) ArsR
0 = β1 OpGT + (β3 OpH - (ln(2)/τR) ArsR) (ArsR²/K3d + 1)
0 = β1 OpGT + β3 OpH ArsR²/K3d + β3 OpH - (ln(2)/τR) ArsR³/K3d - (ln(2)/τR) ArsR
0 = K3d (τR/ln(2)) β1 OpGT + (τR/ln(2)) β3 OpH ArsR² + K3d (τR/ln(2)) β3 OpH - ArsR³ - K3d ArsR
0 = ArsR³ - (τR/ln(2)) β3 OpH ArsR² + K3d ArsR - K3d (τR/ln(2)) (β1 OpGT + β3 OpH)

According to Mathematica's solution of Reduce[eq && K3d > 0 && OpGT >= 0 && OpH >= 0 && β1 > 0 && β3 > 0 && τR > 0, ArsR, Reals] (where eq is the equation shown above) there is only one real solution, so we can find it safely using Newton's (or Halley's) method.

Analogously to the derivation of the equation for OpG an equation can be derived for the fraction of unbound arsenic in the cell:

As(III)T = As(III) + ArsRAs
As(III)/As(III)T = 1/(ArsR/K1d + 1)