Question

(10 points) Michaelis-Menten kinetics. Consider ving enzymatic reaction, E + A H EA → E+ P where E, A, and P are the concentrations of enzyme, substrate and product reen Show that rate and product respectively. V = "max [A] Km + [A]

Answer 1

Answer:

Refer to the images below for the handwritten answer:

The Enzymatic reaction E + A to EAE + P - in Now, for ean. Xiy the rate of product formation is D = d[P] - K, LEA] - ily - dt here, a= rate of product formation or substrate consumption in moles/i-s. a k₂= rate constant. and Now, the rate of variation of the EA complex is d [£A] - \, [ɛ] [A] - K., [£A] - \, [£a] – jüly dt Since the enzyme is not complotsly consumed, therefore the conservation equation on the enzyme we gets [E] = [Eb]- [Ex] - ivy Now, Assuming a rapid equilibrium between the enzyme and the substrate to form an [EA] complex, we can use the equilibrium coefficient to express [EA] in terms of [A].

So, the equilibrium constant is Km = k- [] [A] - vy ki [EA ] Since [€]= [eo] – [EA] if the enzyme is conson then, by rearranging wy we got [EA] = [€] [A] ( ) + [a] or [en] - [£.] [4] - vily km +[A whoro which is the dissociation constant of the EA complex. Now, substituting Eq. (vily into Eq kiiy we get D = d[P] = k, [5.] [4] Kin + [A dt OY Vmax [A] kin'+ [A where, Vmax = K₂ [Eo

Note: This equation is formed by using the Rapid Equilibrium Assumption.

It can be done through another Assumption known as the Quasi-Steady State Assumption to obtain the same result.