Question

The Michaelis-Menten equation models the hyperbolic relationship between [S] and the initial reaction rate (V0) for an enzyme catalyzed, single substrate reaction: E S ES E P. The model can be more readily understood when comparing three conditions: [S]Km. Match each statement with the condition that it describes. Note: \"Rate\" refers to initial velocity (V0) where steady state conditions are assumed; [Etotal] refers to the total enzyme concentration, and [Efree] refers to the concentration of free enzyme.

categories: [S]<<Km, [S]=Km, [S]>>Km, not true for any of the conditions

Which category doe these fit into?:

-[ES] is much lower than [Efree]

-rate is directly proportional to [S]

-almost all active sites will be filled

-[Efree] is equal to [ES]

-adding more S will not increase the rate

-increasing [Etotal] will lower Km

Answer 1

Ans. Km is the [S] at which half-V_max is achieved. So, if [S] is lower than Km, the rate will be very low because there is very little amount of substrate than required to reach V_max. Because of low [S], many enzymes remain unbound ( E _total >> [ES]) to the substrate and thus, the active site also remains unoccupied.

At [S] >> Km, there is excess of substrate, thus all active sites are occupied by the substrate molecules.   

1. [S] << K_m : [ES] is much lower than [E_total], rate is directly proportional to [S]- MM curve gives linear curve at very low [S]; at higher [S] it becomes hyperbolic.

2. [S] = K_m : E_free is equal to [ES]

3. [S] >> K_m : Adding more S will not increase the rate. Almost all active sites will be filled.

4. Not true: Increasing [E_total] will lower K_m. Km is a constant raminaing unaffected of [E] or [S], however, V₀ depends on them.