Question

3) (10 marks) 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 contain 5 nanomoles/mL of acetylcholinesterase and 150uM acetylcholine, a velocity Ve=40umol/mLsec was observed for the acetylcholinesterase reaction. a. Calculate Vmax for this amount of enzyme b. Calculate kcat for acetylcholinesterase 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? C.

Answer 1

Answer:

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.

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