CptS 572 Fall 2013 Lecture 3

1. CptS 572 Fall 2013Lecture 3 Setting Up Kinetic Simulations of Cellular Signaling Networks Example: Boris model

2. Derive kinetic equations for reaction scheme 1 2 3 L + R ↔ LR ↔ LR* + A ↔ LR*A, where L denotes ligand R denotes receptor LR is the bound receptor LR*is the bound and phosphorylated receptor A denotes adaptor protein LR*A is the bound, phosphorylated, and adapted receptor

3. In reaction scheme 1 2 3 L + R ↔ LR ↔ LR* + A ↔ LR*A 1 is a reversible 2nd order reaction 2 is a combination of enzyme reactions forward catalyzed by a kinase treat in linear approximation back catalyzed by a phosphatase treat as Michaslis-Menten 3 is a reversible 2nd order reaction

4. In reaction scheme 1 2 3 L + R ↔ LR ↔ LR* + A ↔ LR*A Define: V1 = kf1[L][R] – kb1[LR] V2 = kf2 [LR] – Vmax2 [LR*] Km2 + [LR*] V3 = kf3 [LR*][A] – kb3[LR*A]

5. In reaction scheme 1 2 3 L + R ↔ LR ↔ LR* + A ↔ LR*A Net rate of change of species X equals sum of rate all reactions that create X minus sum of rate all reactions that destroy X d[L] = d[R] = -V1 d[A] = -V3 d[LR*A] = V3 dt dt dt dt d[LR] = V1 – V2 d[LR*] = V2 – V3 dt dt

6. L V1 V3 V2 LR LR* LR”A R A V5 V4 V7 L L V6 R* R*A A Reaction scheme 7 reactions 8 reactants How do we find the kinetic equations?

7. L V1 V3 V2 LR LR* LR”A R A V5 V4 V7 L L V6 R* R*A A Directed graph with places and edges Reactions have unique edge Reactants do not have a unique place

8. - - - L + - + + - + - V3 V1 V2 LR* LR”A LR R - - A - - V5 V4 V7 R* V6 + - - R*A + - Each reactant has a unique place. Negative term in ODE for each arrow leaving a place Positive term in ODE for each arrow approaching a place

9. - - - L + + + + - + - V3 V1 V2 LR* LR”A LR R - - A - - V5 V4 V7 R* V6 + - - R*A + - Let X’ = dX dt L’ = -V1 - V5 - V7 R’ = -V1 - V4 A’ = -V3 - V6 R*’ = V4 - V5 - V6 LR’ = V1 - V2 LR*’ = V2 - V3 + V5 R*A’ = V6 - V7 LR*A = V3 + V7 All that remains is to define the Vs.

10. L V1 V3 V2 LR LR* LR*A R A V5 V4 V7 L L V6 R* R*A A V1 = kf1[L] [R] – kb1[LR] V2 = kf2 [LR] – Vmax2 [LR*]/(Km2 + [LR*]) V3 = kf3 [LR*] [A] – kb3 [LR*A] V4 = kf4 [R] –Vmax4 [R*]/(Km4 + [R*]) V5 = kf5 [L] [R*] – kb5 [LR*] V6 = kf6 [R*] [A] – kb6 [R*A] V7 = kf7 [L] [R*A] –kb7 [LR*A]

11. receptor ligand extracellular medium plasma membrane kinase phosphatase cytoplasm Concentration of receptors depends on compartment Number of receptors per cell is well defined Define a volumn for each compartment Vex = total volume of medium / number of cells ~ 1ml / 106 Vcy = total volume of cell x faction cytoplasm = 70% volume sphere radius 10mm ~ 2 x 10 -12 liters Vex and Vcy are constants of the model.

12. receptor ligand extracellular medium plasma membrane kinase phosphatase cytoplasm Receptors react with ligands in Vex. [R]ex = NR / Vex Receptors react with kinase and phosphatase in Vcy. [R]cy = NR / Vcy Use NR as dynamic variable in place of concentrations.

13. Rate laws in terms of number of molecules Given Nx molecules of species X in volume V, then [X] = Nx / V 1st order X→Y d[X] = -k1 [X] dNx = -k1 Nx dt dt 2nd order X+Y→Z d[X] = -k2 [X][Y] dNx = -k2 Nx Ny dt dt V enzymatic S→P d[P] = vmax [S]dNp = V vmax Ns dt Km + [S] dt V Km + Ns All terms have units of 1/time {vmav} are concentration/time

14. L V1 V3 V2 LR LR* LR*A R A V5 V4 V7 L L V6 R* R*A A - - In the plasma membrane dNR = -V1ex - V4cy dt dNR = -kf1 NR NL + kb1 NLR – kf4 NR + Vcy vmax4 NR* dt Vex Vcy Km4 + NR*

15. Short Term Signaling from EGFR Using Cell Designer Boris Kolodenko et al. J. Biol. Chem. 1999 Example that includes receptor signaling

16. Network: EGFR activation and 3 downstream cycles. Shc-dependent Ras activation Shc-independent Ras activation Phospholipase C activation Phosphorylated receptor is forming complexes with species in the cytoplasm Species appreciations are not consistent. ShP denotes phosphorylated Shc. Sometimes P is dropped. Always true for phosphorylated receptor

17. Total phosphorylated EGFR is sum of concentrations of all species that contain RP Your primary computational task is to reproduce this figure. This result must be included in your report. Model must be run at different EGF concentrations to generate these results Eqs( 13 ) in Boris paper defines the quantity displayed in Figure 4A

18. Boris’ method of dealing with receptor concentrations L + R → LR only 2nd order reaction in the extracellular medium kf1 [L] [R] only term in system affected by ambiguity in [R] Define all concentrations relative to volume of cytoplasm and correct the kf1 [L] [R] term Substitution of NL = [L] Vcy and NR = [R] Vcy into dNR = -kf1NRNL = -kf1 Vcy [R][L] dt Vex Vex In JBC paper, Vcy / Vex = 1 / 33.3

19. My calculations differ slightly from JBC paper 107 cells are suspended in 1ml of growth medium Vex = growth medium volume per cell = 10-7ml = 10-13 m3 Cell diameter = 20 mm 70% of cell volume is cytoplasm Vcy = 2.93x10-15 m3 Vex / Vcy = 34.1 EFG = 20 nM is concentration in growth medium NEGF = [EGF]cy Vcy = [EGF]ex Vex Therefore: [EGF]cy = [EGF]ex Vex / Vcy = 20nM x 34.1 = 682 nM

20. System of ODEs Refer to Table II for definition of V’s ATP, ADP, And Pi are not reactants ODE for all other species (but one) in the network

21. Table II: Database of rate laws and reaction rates d[EGF]/dt = -k1 (Vcy/Vex) [R][EGF] + k-1[Ra] footnote indicates that k1 has already been multiplied by Vcy /Vex = 1/ 33.3

22. Read the caption of Table II carefully. [RPL] should be [PLC]. [EGF] is relative to Vcy. As shown in slide #19, 20 nM EGF in growth medium becomes 682 nM when referred to the volume of the cytoplasm [EGF]cy = [EGF]ex Vex / Vcy = 20nM x 34.1 = 682 nM

23. Basic assumptions in Boris • Binding of cytoplasmic proteins to RP is competitive. • As long as one protein is bound to a receptor it cannot bind another protein • Avoids a combinatorial explosion of chemical species in the model • We have populations of particular types of complexes between RP and cytoplasmic proteins

24. Basic assumptions in Boris 2 • Phosphatase activity is blocked on complexes between RP an cytoplasmic proteins • RP-A can decay to RP + A but not to R-A • Phosphatase activity is restricted to RP, PLCgP, and ShP (V4, V8, and V16 in table II) • Reasonable values of rate constants are known (3rd column of table II) • Qualitative agreement between experiment (Figs 2B and 3A-3C) and theory (Figs 4Aand 5A-5D) supports this hypothesis

25. Question on Lecture 3 You decide to model the system L+R ↔ LR ↔ LR* with number per cell as dynamic variables. You treat the forward reaction of LR ↔ LR* by linear enzyme kinetics and the reverse reaction by full Michaelis–Menten (M-M) kinetics. The cell has 104 receptors and your experiments has 20 nM of L in 1 ml of growth medium containing 106 cells. The dissociation rate of L and R is 0.06/s and the equilibrium constants is 20 nM. The rate parameter for activation of LR is 1/s and the M-M parameters for inactivation of LR are vmax=450 nM/s and Km=50 nM. Write the ODEs for L, R, LR, and LR* with number per cell as the dynamic variables. Calculate the rate constants and initial conditions that are appropriate for solution of the kinetics in these dynamic variables.

26. Solve by answering the following: • Where does the reaction occur? • What rate laws apply? • What are the reaction volumes? • What are the rate constants ? • What are the initial conditions? 1 2 L+R ↔ LR ↔ LR*

27. dNL = dNR = -kb1 NR NL + kb1 NLR dt dt KdVex dNLR = kb1 NR NL - kb1 NLR – kf2 NLR + Vcy vmax NLR* dt KdVex Vcy Km + NLR* dNLR* = kf2 NLR - Vcy vmax NLR* dt Vcy Km + NLR* Rate laws

28. Look for dimensionless combinations x=kb1t

29. Reaction volumes 106 cells are suspended in 1ml of growth medium Vex = growth medium volume per cell = 10-6 ml = 10-12 m3 Cell diameter = 20 mm 70% of cell volume is cytoplasm Vcy = 2.93x10-15 m3

30. Dimensionless combinations KdVex = 1.2x107 Vcyvmax /kf2 = 8.1x105 VcyKm = 9x104 kf2/kb1 = 16.67

31. Initial conditions in number per cell NL = # L/ml = 20x10-9moles x 6x1023 x liter # cells/ml 106 liter mole 1000ml NL =1.2x107 NR = 104 NLR = 0 NLR* = 0