Calculations ofEnzyme Activity

Enzyme Activity

Unit of enzyme activity: Used to measure total units of activity in a given volume of solution.

Specific activity: Used to follow the increasing purity of an enzyme through several procedural steps.

Molecular activity: Used to compare activities of different enzymes. Also called the turn-over number (TON = kcat)

Enzyme Activity

Classical units:

Unit of enzyme activity:mol substrate transformed/min = unit

Specific activity:mol substrate/min-mg E = unit/mg E

Molecular activity:mol substrate/min- mol E = units/mol E

Enzyme Activity

New international units:

Unit of enzyme activity:mol substrate/sec = katal

Specific activity:mol substrate/sec-kg E = katal/kg E

Molecular activity:mol substrate/sec-mol E = katal/mol E

Example 1

The rate of an enzyme catalyzed reaction is 35 μmol/min at [S] = 10-4 M, (KM = 2 x 10-5).Calculate the velocity at [S] = 2 x 10-6 M.

Work the problem.

Example 1 AnswerThe rate of an enzyme catalyzed reaction is 35 μmol/min at [S] = 10-4 M, (KM = 2 x 10-5).Calculate the velocity at [S] = 2 x 10-6 M.

First calculate VM using the Michaelis-Menton eqn: VM [S] VM (10-4) VM (10-4)v = -----------, so: 35 = ------------------ = -------------- KM + [S] 2 x 10-5 + 10-4 1.2 x 10-4

VM = 1.2(35) = 42 mol/min; then calculate v:

42 (2 x 10-6) 84 x 10-6 v = ------------------------ = ------------ = 3.8 mol/min

2 x 10-5 + 2 x 10-6 22 x 10-6

Example 2

An enzyme (1.84 μgm, MW = 36800) catalyzes a reaction in presence of excess substrate at a rate of 4.2 μmol substrate/min. What is the TON in min-1 ? What is the TON in sec-1 ?

Work the problem.

Example 2 Answer

An enzyme (1.84 μgm, MW = 36800) catalyzes a reaction in presence of excess substrate at a rate of 4.2 μmol substrate/min. What is the TON ?

1.84 μgm μ mol E = ------------------------- = 5 x 10-5 μmol E

36800 μgm/ μmol

4.2 μmol/min TON = ------------------ = 84000 min-1

5 x 10-5 μmol

Example 2 Answer

What is the value of this TON (84000 min-1) in units of sec-1 ?

84000 min-1 1 sec-1

TON E = ------------------ x ---------- = 1400 sec-1

60 min-1

Example 3

Ten micrograms of carbonic anhydrase (MW = 30000) in the presence of excess substrate exhibits a reaction rate of 6.82 x 103 μmol/min.At [S] = 0.012 M the rate is 3.41 x 103 μmol/min.

a. What is Vm ? b. What is KM ? c. What is k2 (kcat) ?

Work these.

Example 3

a. The rate in presence of excess substrate is Vmaxso:

Vmax = 6.86 x 103 μmol/min.

b. At [S] = 0.012 M the rate is 3.41 x 103 μmol/min which is ½ Vmax so:

KM = 0.012 M. This may also be determined using the

Michaelis-Menton equation.

c. Divide Vmax by μmol of ET to find kcat. kcat = 2.05 x 107 min-1

End of Enzyme Activity