Ste11 synthesis/degradation

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Ste11 Degradation

  • Ste11-Myc expressed from the Gal promoter (so presumably overexpressed) has an apparent half-life of ~200 min in the absence of pheromone. Esch and Errede. 2002 PMID 12077316
  • Ste11-Myc expressed from the Gal promoter (so presumably overexpressed) has an apparent half-life of ~50 min in the presence of pheromone. Esch and Errede. 2002 PMID 12077316
  • The decrease in Ste11-Myc half-life in the presence of pheromone appears to be due to ubiquitinylation mediated by active Fus3 and/or Kss1. Esch and Errede. 2002 PMID 12077316
    • Disruption of ubiquitin degradation processes results in an apparent half-life of >200 min for Ste11-Myc in the presence of pheromone.
    • Ste11-Myc half-life is >200 min in fus3Δkss1Δ cells.
    • Thus the 50 min half-life in the presence of pheromone is likely a composite rate between the rate of Fus3/Kss1 phosphorylation (I'm assuming that degradation is due to MAPK phosphorylation) of Ste11, the ubiquitinylation rate of phosphorylated Ste11, and the degradation of ubiquitinylated Ste11.
  • Ste11-Myc half-life is not affected by response to high osmolarity. Esch and Errede. 2002 PMID 12077316
  • A constitutively active Ste11 mutant, Ste11-4, cannot be seem in by western blot, even when overexpressed from the Gal promoter. Ste11-4 can be detected when Fus3 and Kss1 are deleted(?). Source unknown - maybe a Sprague paper?
    • This suggests that active Ste11 might have a much shorter half-life than the 50 minutes measured above, and this number may be artificially low because only a small portion of Ste11 is activated in these experiments.
    • If these facts are verified, this might necessitate changing the reactions below such that only active Ste11 is targeted for degradation, and that the half-time of degradation is shorter than 50 minutes.

Reaction Definition

In the absence of pheromone, Ste11 has a half-life of 200 minutes, so kdeg_Ste11 = ln(2) / 200min = 5.8 * 10-5 s-1.

See Assumptions to re-test with full model for a discussion on retesting the assumptions about Ste11 degradation using simulations.

Assumptions:

  • Fus3 and Kss1 phosphorylate Ste11 with equal efficiency, tagging Ste11 for ubiquitination and rapid degradation.
  • The rate of ubiquitination is lumped into the phosphorylation and degradation rates.
  • The MAPKs and Ste11 cannot interact when any of them are bound to Ste5.
  • Phosphorylation of Ste11 by Ste20 does not affect its degradation rate.
  • Phosphorylation is relatively fast compared to the degradation half-life of 50 min, so we can estimate the degradation rate of Ste11 that has been phosphorylated by Fus3/Kss1 as kdeg_Ste11_PO4 = ln(2) / 50min = 2.3 * 10-4 s-1. 
Fus3(docking_site, T180~none, Y182~PO4) + Ste11(Ste5_site, MAPK_site, Feedback_PO4~none) <->
   Fus3(docking_site!1, T180~none, Y182~PO4).Ste11(Ste5_site, MAPK_site!1, Feedback_PO4~none)




Kss1(docking_site, T183~none, Y185~PO4) + Ste11(Ste5_site, MAPK_site, Feedback_PO4~none) <->
    Kss1(docking_site!1, T183~none, Y185~PO4).Ste11(Ste5_site, MAPK_site!1, Feedback_PO4~none)



Fus3(docking_site, T180~PO4, Y182~none) + Ste11(Ste5_site, MAPK_site, Feedback_PO4~none) <->
   Fus3(docking_site!1, T180~PO4, Y182~none).Ste11(Ste5_site, MAPK_site!1, Feedback_PO4~none)




Kss1(docking_site, T183~PO4, Y185~none) + Ste11(Ste5_site, MAPK_site, Feedback_PO4~none) <->
    Kss1(docking_site!1, T183~PO4, Y185~none).Ste11(Ste5_site, MAPK_site!1, Feedback_PO4~none)



Fus3(docking_site, T180~PO4, Y182~PO4) + Ste11(Ste5_site, MAPK_site, Feedback_PO4~none) <->
   Fus3(docking_site!1, T180~PO4, Y182~PO4).Ste11(Ste5_site, MAPK_site!1, Feedback_PO4~none)




Kss1(docking_site, T183~PO4, Y185~PO4) + Ste11(Ste5_site, MAPK_site, Feedback_PO4~none) <->
    Kss1(docking_site!1, T183~PO4, Y185~PO4).Ste11(Ste5_site, MAPK_site!1, Feedback_PO4~none)



Fus3(docking_site!1, T180~none, Y182~PO4).Ste11(Ste5_site, MAPK_site!1, Feedback_PO4~none) ->
   Fus3(docking_site, T180~none, Y182~PO4) + Ste11(Ste5_site, MAPK_site, Feedback_PO4~PO4)




Kss1(docking_site!1, T183~none, Y185~PO4).Ste11(Ste5_site, MAPK_site!1, Feedback_PO4~none) ->
    Kss1(docking_site, T183~none, Y185~PO4) + Ste11(Ste5_site, MAPK_site, Feedback_PO4~PO4)



Fus3(docking_site!1, T180~PO4, Y182~none).Ste11(Ste5_site, MAPK_site!1, Feedback_PO4~none) ->
   Fus3(docking_site, T180~PO4, Y182~none) + Ste11(Ste5_site, MAPK_site, Feedback_PO4~PO4)




Kss1(docking_site!1, T183~PO4, Y185~none).Ste11(Ste5_site, MAPK_site!1, Feedback_PO4~none) ->
    Kss1(docking_site, T183~PO4, Y185~none) + Ste11(Ste5_site, MAPK_site, Feedback_PO4~PO4)



Fus3(docking_site!1, T180~PO4, Y182~PO4).Ste11(Ste5_site, MAPK_site!1, Feedback_PO4~none) ->
   Fus3(docking_site, T180~PO4, Y182~PO4) + Ste11(Ste5_site, MAPK_site, Feedback_PO4~PO4)




Kss1(docking_site!1, T183~PO4, Y185~PO4).Ste11(Ste5_site, MAPK_site!1, Feedback_PO4~none) ->
    Kss1(docking_site, T183~PO4, Y185~PO4) + Ste11(Ste5_site, MAPK_site, Feedback_PO4~PO4)



Ste11 + Cell -> Cell

  • In order for the degradation reaction to selectively degrade Ste11 out of a complex, we need to include the parameter DeleteMolecules

  • Dummy species Cell used because BioNetGen requires at least one species on each side of reaction 
Ste11(Feedback_PO4~PO4) + Cell -> Cell

  • In order for the degradation reaction to selectively degrade Ste11 out of a complex, we need to include the parameter DeleteMolecules

  • Dummy species Cell used because BioNetGen requires at least one species on each side of reaction

Ste11 Synthesis

Reaction Definition

Ste11 synthesis must balance degradation (in the absence of pheromone). Therefore, ksynth_Ste11 = Ste11_tot_conc * kdeg_Ste11. 

Cell -> Cell + Ste11(Ste5_site, MAPK_site, S302_S306_T307~none, Feedback_PO4~none)

  • Dummy species Cell used because BioNetGen requires at least one species on each side of reaction
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