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Calculations ofEnzyme Activity

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Enzyme Activity

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

Specific activity:Used to follow the increasing purity of anenzyme 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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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

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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

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Example 1

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

Work the problem.

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Example 1 Answer

The rate of an enzyme catalyzed reaction is 35

mol/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

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Example 2

An enzyme (1.84 gm, MW = 36800) catalyzes areaction in presence of excess substrate at a rate of4.2 mol substrate/min. What is the TON in min-1 ?

What is the TON in sec

-1

?

Work the problem.

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Example 2 Answer

An enzyme (1.84 gm, MW = 36800) catalyzes areaction in presence of excess substrate at a rate of4.2 mol substrate/min. What is the TON ?

1.84 gm mol E = ------------------------- = 5 x 10-5mol E36800gm/mol

4.2

mol/minTON = ------------------ = 84000 min-15 x 10-5mol

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Example 2 Answer

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

84000 min-1 1 sec-1TON E = ------------------ x ---------- = 1400sec-1

60 min-1

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Example 3

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

At [S] = 0.012 M the rate is 3.41 x 10

3

mol/min.

a. What is Vm ?b. What is KM ?c. What is k

2(kcat) ?

Work these.

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Example 3

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

Vmax = 6.86 x 103mol/min.

b. At [S] = 0.012 M the rate is 3.41 x 103mol/minwhich is Vmax so:

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

Michaelis-Menton equation.

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

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End of Enzyme Activity