Question

Problem 6: (Goldbeter Koshland switch) Goldbeter & Koshland (PNAS 78:6840, 1981) considered the phosphorylation and dephosphorylation of a substrate S by a kinase E and phosphatase H, respectively: Η S→Sp Sp →S Assuming that both reactions obey Michaelis-Menten kinetics, we can write d[S]_ VE [S] VH [Sp] dt Kme+ [S] K + [S] Define the dimensionless variable ** [9]+8,] where [S]+[SP] = [ST] = constant, and show that dr_v. (1-x) Vix dt Joh 1-x Jx + x Tell me the definitions of ve, Vh, Je, Jh in terms of Vmaxe, VmaxH, Kme, Kmh and (ST). Show that the steady-state solution of Eq. (1.2) satisfies a quadratic equation of the form ax? + bx + c = 0, and solve this equation (using the quadratic formula) for Xss as a function of ve, Vh, Je, Jh. Show that your solution can also be written as Xss = G(ve, Vn, J,JW)=- 20.Ja ** v. -v. +v.Jn +vyJo+V/v-v. +v.Jr +v,J.)? - 4v, – v.)V.Ja Show that G(V, VR, JO, JD)=G(* 1,J., Jn) Plot Xss as a function of ve/vh for fixed values of {Je, Jh} = {.01, 01}, {.1,1}, {1,1}, {.01,1).

Answer 1

the Goldbeter- koshland kinetics is a steady-state solution for a two state biological system in which the inconversion between two states takes place by the two enzymes having the opposing effect.

Therefore, The Goldbeter - Koshland stated that kinetics Where S- Substrate ex-protein S= protein Sp- phosphorylated protein J E = kinase IL H= phosphatase 4 These both are used to interconvert the forms
  
When these hoth reactions obey Michaelis - Meuten kinetics we can say d[sp] - Vmax E[S] - Vmaxi [sp] KmE+ [s] KmH + [sp] cut & Where Sp- phosphorylated substrate Sto unphosphorylated The Goldbeter-koshland kinetics is described by G-k junction. which is (n in this case, a dimensich less vanable. и и & definitions of ve, Un, Je, In in termsg Vmaxt Vmax , kmf and KmH
  
As we know, Ve Una Je, Ih constants which Written as: are the can be Te = Umax[E] where Umax is a rate constant for the cataly red reaction : Un = Umax [H] where Umax is a rate constant for the catalyzed reaction. Je = Km E where km stands for the Michaelis - Menten censtant which describes the binding
  
of enzymes and hand catalynes the conversion In = km H) where km is the Michaelis - Menten Const. Proving sclution that the forms a stedy-state quadratic Guation U from Michaelis-Menten kinetics > the rate at which Sp is dephospherylatented us (1) rate at which s 7 the is
  
- Where r = Vmax H[ Sp] KmH + [sp] 82= Vmax € [s KME+ [s] de assuming total concentration g s is constant [5] = [sp] + [s] id[s] ani-82 - at a ng - Vmax M ( [SJO-[s]) km H+ [[s] - [s] 82 = Vanax E[s Kmet [S]
  
=) Vmaxh ([s] - [s] - U maxEls] Km H + (( 5 Jo - [s]) KME+ [s] seVmax (1 – 1973.)Vmore [152 [S M Vh u Tht n as we know > n=15] [So > Ve=VmaxlH Un=Vmax € Je =kmH = KmH/iso In = km E 2) KME/ (slo

of me selve this egn © for In , we get ve (1-2) = whn. Je +(1-2) In tn Thue tuve - Juven-n²ve is equal to : = ln vh Jet n Vh -n²vh & & n²(un-ve)-n (un-vet Je vht In ve) J & ve Jh = 0 is n² + bn te= 0 hence proved.
  

Ve/Vh Xes Ve/Vh

the graph showing Xss vs Ve/Vh

the whole answer is written in handwritten notes and all the images are uploaded serial wise.

hope it will help.