The Michaelis-Menten equation models the hyperbolic relationship between [S] and
ID: 1049976 • Letter: T
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
Explanation / Answer
Ans. Km is the [S] at which half-Vmax 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 Vmax. 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] << Km : [ES] is much lower than [Etotal], rate is directly proportional to [S]- MM curve gives linear curve at very low [S]; at higher [S] it becomes hyperbolic.
2. [S] = Km : Efree is equal to [ES]
3. [S] >> Km : Adding more S will not increase the rate. Almost all active sites will be filled.
4. Not true: Increasing [Etotal] will lower Km. Km is a constant raminaing unaffected of [E] or [S], however, V0 depends on them.
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