Calculations of Enzyme 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 = k cat )

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] = M, (K M = 2 x ). Calculate the velocity at [S] = 2 x M. Work the problem.

Example 1 Answer The rate of an enzyme catalyzed reaction is 35 μmol/min at [S] = M, (K M = 2 x ). Calculate the velocity at [S] = 2 x M. First calculate V M using the Michaelis-Menton eqn: V M [S] V M (10 -4 ) V M (10 -4 ) v = , so: 35 = = K M + [S] 2 x x V M = 1.2(35) = 42  mol/min;then calculate v: 42 (2 x ) 84 x v = = = 3.8  mol/min 2 x x 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 μmol E μgm/ μmol 4.2 μmol/min TON = = min -1 5 x μmol

Example 2 Answer What is the value of this TON (84000 min -1 ) in units of sec -1 ? min -1 1 sec -1 TON E = x = 1400 sec 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 10 3 μmol/min. At [S] = M the rate is 3.41 x 10 3 μmol/min. a. What is Vm ? b. What is K M ? c. What is k 2 (kcat) ? Work these.

Example 3 a.The rate in presence of excess substrate is Vmax so: Vmax = 6.86 x 10 3 μmol/min. b. At [S] = M the rate is 3.41 x 10 3 μmol/min which is ½ Vmax so: K M = M. This may also be determined using the Michaelis-Menton equation. c. Divide Vmax by μmol of E T to find kcat. kcat = 2.05 x 10 7 min -1

End of Enzyme Activity