Phy Pharm Assignment
4
Deriving the Michaelis-Menten Equation (proving for page 20.) A simple model of enzyme action: For this model, let V 0 be the initial velocity of the reaction. Then V 0 = k cat [ES]. (2) The maximum velocity V max occurs when the enzyme is saturated -- that is, when all enzyme molecules are tied up with S, or [ES] = [E] total . So V max = k cat [E] total . (3) Rate of formation of ES = k 1 [E][S]. Rate of consumption of ES = k -1 [ES] + k cat [ES]. So in the steady state, k -1 [ES] + k cat [ES] = k 1 [E][S]. (4) Remember that we are trying to solve for [ES] in terms of measurables, so that we can replace it in (2). First, collect the kinetic constants in (4): (k -1 + k cat ) [ES] = k 1 [E][S], and (k -1 + k cat )/k 1 = [E][S]/[ES]. (5) To simplify (5), first group the kinetic constants by defining them as K m : 1
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Deriving the Michaelis-Menten Equation
Transcript of Phy Pharm Assignment
Deriving the Michaelis-Menten Equation (proving for page 20.)A simple model of enzyme action:
For this model, let V0be the initial velocity of the reaction. ThenV0= kcat[ES]. (2)The maximum velocity Vmaxoccurs when the enzyme is saturated -- that is, whenall enzyme molecules are tied up with S, or[ES] = [E]total.So Vmax= kcat[E]total. (3)Rate of formation of ES = k1[E][S].Rate of consumption of ES = k-1[ES] + kcat[ES].So in the steady state, k-1[ES] + kcat[ES] = k1[E][S]. (4)Remember that we are trying to solve for [ES] in terms of measurables, so that we can replace it in (2). First, collect the kinetic constants in (4):(k-1+ kcat) [ES] = k1[E][S],and (k-1+ kcat)/k1= [E][S]/[ES]. (5)To simplify (5), first group the kinetic constants by defining them as Km:Km= (k-1+ kcat)/k1(6)and then express [E] in terms of [ES] and [E]total:[E] = [E]total- [ES] (7)Substitute (6) and (7) into (5):Km = ([E]total- [ES]) [S]/[ES] (8)Solve (8) for [ES]: First multiply both sides by [ES]:[ES] Km= [E]total[S] - [ES][S]Then collect terms containing [ES] on the left:[ES] Km+ [ES][S] = [E]total[S]Factor [ES] from the left-hand terms:[ES](Km+ [S]) = [E]total[S]and finally, divide both sides by (Km+ [S]):[ES] = [E]total[S]/(Km+ [S]) (9)dP/dt = Kcat [ES]
dP/dt=Kcat [E]total[S]/(Km+ [S])
Deriving The Lindemann-Hinshelwood mechanism(Proving for pg 21)A simple model of enzyme action:
k4k3k1
A B + C D P
k2
(1st &2nd) (3rd) (4th) steps steps steps
rate of formation of B + C = K1[A]
rate of consumption of B + C = K2[B ][ C] + K3[B ][ C]
steady state => rate of formation = rate of consumption
K1[A] = K2[B ][ C] + K3[B ][ C]K1[A] = (K2+K3) [B ][ C][B]SS = K1[A]/ K2+K3 C
rate of formation of D = K3[B][C]
rate of consumption of D = K4[D]
steady state => rate of formation = rate of consumption
K3[B][C] = K4[D]*[B] SS = K1[A]/ K2+K3 C
K3 (K1[A]/ K2+K3 C)[C] = K4[D]K1 K3 A C/ K2+K3 C = K4[D]
[D]SS = K1 K3 A C / K4 (K2+K3 C)
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