Transcript Enzyme Activity - Kigali Institute of Science and Technology

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 = kcat)
Enzyme Activity
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
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 Answer
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
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 Vmax
so:
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