Question

Chymotrypsin is a serine protease that cleaves on the carboxyl side of aromatic side chains. Esters, including p-nitrophenyl acetate (KM of 0.002 mM for chymotrypsin), can serve as substrates for chymotrypsin. When supplied with 2.5 mM p-nitrophenyl acetate, a certain amount of chymotrypsin can hydrolyze it at a rate of 0.1 mM/sec. With this in mind what concentration of p-nitrophenyl acetate would be required to give a rate that is 80% of Vmax? Explain. (Hint: Chymotrypsin follows Michaelis-Menten kinetics)

Answer #1

For chymotrypsin, K_{M} = 0.002 mM

Subtrate, p-nitrophenyl acetate concentration, [S] = 2.5 mM

Rate, *v* = 0.1 mM.sec-1

Applying Michaelis-Menten equation:

*v* = 0.1 mM.sec-1 = Vmax[S] / (K_{M} + [S])

=> Vmax = 0.1 mM.sec-1 x (K_{M} + [S]) / [S] m

=> Vmax = 0.1 mM.sec-1 x (0.002 mM + 2.5 mM) / (2.5 mM)

=> Vmax = 0.10008 mM.sec-1

80% of Vmax = 0.10008 mM.sec-1 x (80/100) = 0.080064 mM.sec-1

Hence new rate, *v* = 0.080064 mM.sec-1 =
Vmax[S] / (K_{M} + [S])

=> 0.080064 mM.sec-1 = 0.10008 x [S] / (0.002 mM + [S])

=> 1.60128x10^{-4} + 0.080064x[S] = 0.10008x[S]

=> [S] = 1.60128x10^{-4} / 0.020016 = 0.008 mM/sec
(answer)

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