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Calculations of Enzyme Activity - PowerPoint Slideshow


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).

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Calculations of Enzyme Activity

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Calculations of enzyme activity

Calculations ofEnzyme Activity


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 activity1

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 activity2

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

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

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

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

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 10-5μmol E

36800μgm/μmol

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

5 x 10-5μmol


Example 2 answer1

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 ---------- = 1400sec-1

60 min-1


Example 3

Example 3

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.


Example 31

Example 3

  • 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


End of enzyme activity

End of Enzyme Activity