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Calculations of Enzyme Activity

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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)

Classical units:

Unit of enzyme activity:

mmol substrate transformed/min = unit

Specific activity:

mmol substrate/min-mg E = unit/mg E

Molecular activity:

mmol substrate/min- mmol E = units/mmol E

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

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.

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.

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 mmol/min;then calculate v:

42 (2 x 10-6) 84 x 10-6 v = ------------------------ = ------------ = 3.8 mmol/min 2 x 10-5 + 2 x 10-6 22 x 10-6

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.

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 10-5μmol E

36800μgm/μmol

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

5 x 10-5μmol

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

84000 min-1 1 sec-1

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

60 min-1

Ten micrograms of carbonic anhydrase (MW = 30000) in the presence of excess substrate exhibits a reaction rate of 6.82 x 103μmol/min.

At [S] = 0.012 M the rate is 3.41 x 103μmol/min.

a. What is Vm ?

b. What is KM ?

c. What is k2 (kcat) ?

Work these.

- The rate in presence of excess substrate is Vmax
- so:
- Vmax = 6.86 x 103μmol/min.
- b. At [S] = 0.012 M the rate is 3.41 x 103μmol/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