1 / 12

Calculations of Enzyme Activity

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:

maya-saunders
Télécharger la présentation

Calculations of Enzyme Activity

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Calculations ofEnzyme Activity

  2. 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)

  3. 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

  4. 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

  5. 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.

  6. 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

  7. 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.

  8. 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

  9. 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

  10. 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.

  11. 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

  12. End of Enzyme Activity

More Related