Question

Acetylcholinesterase catalyzes the hydrolysis of the neurotransmitter acetylcholine: Acetylcholine + H2O  acetate + choline The Km of acetylcholinesterase for its substrate acetylcholine is 9.5x10-5M. In a reaction mixture containing 5 nanomoles/mL of acetylcholinesterase and 150μM acetylcholine, a velocity vo=40μmol/mLsec was observed for the acetylcholinesterase reaction.

a. Calculate Vmax for this amount of enzyme

b. Calculate kcat for acetylcholinesterase

c. Calculate the catalytic efficiency (kcat/Km) for acetylcholinesterase

d. Does acetylcholinesterase approach catalytic perfection?

e. What determines the ultimate speed limit of an enzyme-catalyzed reaction? That is, what is it that imposes the physical limit on catalytic perfection?

Answer 1

a. We know from Michaelis-Menten equation that

V = Vmax * S / ( S + Km)

So, Vmax = V * ( S + Km) /S

= 40 μmol/mLsec * (9.5*10-5 M + 150 μM) / 150 μM

We know, 1 μmol/mL = 0.001 M

So, 40 μmol/mL = 40*0.001 M = 0.04 M

We know, 1 μM = 10-6 M

So, 150 μM = 150 * 10-6 M = 15*10-5 M

So, Vmax = 0.04 M/sec * (9.5*10-5 M + 15*10-5 M ) / 15*10-5 M = 0.0653 M/sec = 65.3 μmol/ml sec.

b. We know, Vmax = Kcat * Enzyme concentration

So, Kcat = Vmax / enzyme concentration

= 65.3 μmol/ml sec / 5 nanomoles/ml

= 65.3 μmol/ml sec / 5*10-3 μmol/ml

= 1.306*10^4 sec-1

c. catalytic efficiency (kcat/Km) for acetylcholinesterase = 1.306*10^4 sec-1 / 9.5x10-5M

= 1.37*10^8 sec-1 M-1

d. Acetylcholinesterase does approach catalytic perfection. Enzymes possessing catalytic efficiency in the range of 10^8 to 10^9 M-1 sec-1 are said to have achieved catalytic perfection.

e. Km determines the ultimate speed limit of an enzyme-catalyzed reaction. An enzyme having low Km will reach its catalytic perfection very fast and vice versa.