Question

When a peeled apple Is exposed to air, it turns brown. This is because apple contains the enzyme o-diphenol oxidase which catalyzes the oxidation of phenols in the apple to quinines, which are darker in color. The rate law for this reaction follows Michaelis-Menten kinetics, with the following constants: V_max = 0.10 minutes^-1 and K_M = 0.00125 M. a) If the concentration of catechol (a phenol substrate) in a slice of apple is initially 0.10 M, how long will it take for the concentration to fall to 0.025 M? b) Para-hydroxybenzoic acid (PHBA) is a competitive inhibitor of o-diphenol oxidase, which means it also reversibly binds to the active site of this enzyme. It is found that when this inhibitor is introduced, K_M increases, but V_max is unchanged. Why does V_max not change? Why does K_m increase? c) When a high concentration of PHBA is introduced, K_M is measured to be 1.0 M. How long will it take for the concentration of catechol (initially at 0.1 M) to fall to 0.025M?

Answer 1

From Michaelis- Menten kinetics

-dS/dt= VmaxS/(KM+S)

(KM+S)*dS/S= -Vmax*dt

(KM/S) dS + dS= -Vmax*dt

when integrated noting that t=0 S =So and t=t S= S

KM* ln (So/S)+ (S-SO) = Vmax*t

KM= 0.00125M, Vmax = 0.1/min

SO= 0.1M, SO= 0.025

hence 0.00125* ln (0.1/0.025)+ (0.1-0.025)= 0.1*t

t=0.89 min =0.89*60 sec=53.4 sec

2. The presence of competitive inhibitor increases Km, because more substrate is needed to reach Vmax due because of the inhibitor occupies the active sires and slows the reaction.

3. from Km* ln (So/S)+ (SO-S)= Vmax*t

1* ln (0.1/0.025)+ (0.1-0.025)= 0.1*t

t=14.6 minutes