Michaelis and Menten derived their rate law by assuming a rapid pre-equilibrium of E, S, and...

Michaelis and Menten derived their rate law by assuming a rapid pre-equilibrium
of E, S, and ES.

a) Derive the rate law in this manner: pre-equilibrium

b) Identify the conditions under which it becomes the same as that based on the steady·
state approximation.

Expert Solution

The rapid equilibrium that Michaelis and Menten considered during their Enzyme Kinetic model,the following equilibrium was considered-

Let's now come to the Steady-State approximation. Here we consider,the concentration of the intermediate in a multistep reaction doesn't chnage with time i.e. .Uning this concept,we now derive the rate law of the above equilibrium-

Now compare the two equations i.e equation 5 and eqn 7. The difference is in the expression of rate constant which is oversimplified in equation 5.So, the equation 5 can also be alike equation 7 if we did not impose the condition that " species 'E' 'S' and 'ES' are considered to be in rapid equilibrium during the entire course of reaction" i.e the Pre-equilibrium concept has to be lifted for the intermediate. The only option was then that the intermediate concentration is constant which is the Steady-Stare approximation itself.

queen_honey_blossom answered 3 years ago

For the reaction E+S > ES> P, the michaelis menten constant, Km, is actually a summary...

For the reaction E+S > ES> P, the michaelis menten constant, Km, is actually a summary of three terms. What are they? How is Km determined graphically?

Initial rate data for an enzyme that obeys Michaelis-Menten kinetics are shown in the following table....

Initial rate data for an enzyme that obeys Michaelis-Menten kinetics are shown in the following table. When the enzyme concentration is 3 nmol ml−1, a Lineweaver-Burk plot of this data gives a line with a y-intercept of 0.00426 (μmol−1ml s). [S] μM v0 (μmol ml−1 s−1) 320 169 160 132 80.0 92.0 40.0 57.2 20.0 32.6 10.0 17.5 a. Calculate Kcat for the reaction b. Calculate Km for the enzyme cWhen the reactions in part (B) are repeated in the...

IV) Many enzymes obey simple Michaelis-Menten kinetics, which are summarized by the equation: rate = (V...

IV) Many enzymes obey simple Michaelis-Menten kinetics, which are summarized by the equation: rate = (V max [S])/([S]+K m ); where V max = maximum velocity, [S] = concentration of substrate; and K m = the Michaelis constant. It is instructive to plug in a few values of [S] into the equation to see how rate is affected. What are the rates for [S] equal to zero, equal to K m , and equal to infinite concentration? Show all work.

1. Starting from the enzyme-catalyzed reaction: S -> P Derive the (a) Michaelis-Menten Equation (b) starting...

1. Starting from the enzyme-catalyzed reaction: S -> P Derive the (a) Michaelis-Menten Equation (b) starting from the Michaelis-Menten equation, derive the Lineweaver-Burker plot. Provide brief explanation in each step. 2. Predict the optimum pH and temperature for human saliat amylase. Why did you arrive on the prediction?

For a Michaelis-Menten reaction, k1=5107 M-1S-1, k-1=2104 S-1 and k2=4*102 S-1. Calculate the Km and Ks...

For a Michaelis-Menten reaction, k1=5*107 M-1S-1, k-1=2*104 S-1 and k2=4*102 S-1. Calculate the Km and Ks for this reaction. Does the substrate achieve equilibrium? How would inspection of the rate constant immediately have given you the same answer

Develop enzymatic rate expressions for competitive, noncompetitive, uncompetitive inhibitions using rapid equilibrium assumption.

S = 1/(?0) E X B How was the poynting vector derived? Please give a detailed...

S = 1/(?0) E X B How was the poynting vector derived? Please give a detailed derivation of equations that leads to the Poynting Vector.

Calculate the standard potential, E°, for this reaction from its equilibrium constant at 298 K. X(s)...

Calculate the standard potential, E°, for this reaction from its equilibrium constant at 298 K. X(s) + Y^2+(aq) ⇌ X^2+(aq) + Y(s) K= 1.43 x 10^-6 E^o = ? V

Part A Calculate the equilibrium constant at 25 ∘C for the reaction Ni(s)+2Ag+(aq)→Ni2+(aq)+2Ag(s), if Ni2+(aq)+2e−→Ni(s), E∘...

Part A Calculate the equilibrium constant at 25 ∘C for the reaction Ni(s)+2Ag+(aq)→Ni2+(aq)+2Ag(s), if Ni2+(aq)+2e−→Ni(s), E∘ = -0.26 V, Al+(aq)+e−→Al(s), E∘ = 0.80 V Express your answer using one significant figure. Part B Calculate the equilibrium constant at 25 ∘C for the reaction Hg2+2(aq)→Hg(l)+Hg2+(aq) See Appendix D for standard reduction potentials. Express your answer using one significant figure.

Part A Consider the second-order reaction: 2HI(g)→H2(g)+I2(g) Rate law: k[H]^2 k= 6.410^-9 (mols) at 500 K...

Part A Consider the second-order reaction: 2HI(g)→H2(g)+I2(g) Rate law: k[H]^2 k= 6.4*10^-9 (mol*s) at 500 K Initial rate = 1.6 * 10^-7 mol (l*s) What will be the concentration of HI after t = 3.65×1010 s ([HI]t) for a reaction starting under this condition? Part B In a study of the decomposition of the compound X via the reaction X(g)⇌Y(g)+Z(g) the following concentration-time data were collected: Time (min) [X](M) 0 0.467 1 0.267 2 0.187 3 0.144 4 0.117 5...

Question