What is the use of assessing the Km value of an enzyme? What is the application?

What is the use of assessing the Km value of an enzyme? What is the application?
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Determination of Km value is also useful to understand the natural substrate of an enzyme. Study of Km value will also differentiate the competitive and non-competitive inhibitions.
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Km value of enzyme is substrate concentration at (A) ½ Vmax (B) 2 Vmax (C) ½ Vmax (D) 4 Vmax

Description : Km value of enzyme is substrate concentration at (A) ½ Vmax (B) 2 Vmax (C) ½ Vmax (D) 4 Vmax

Last Answer : Answer : D

Velocity maximum for an enzyme at half the substrate concentration gives (A) The molecular weight of the enzyme (B) Km value (C) Isoelectric pH (D) Pk value

Description : Velocity maximum for an enzyme at half the substrate concentration gives (A) The molecular weight of the enzyme (B) Km value (C) Isoelectric pH (D) Pk value

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The polysaccharide used in assessing the glomerular fittration rate (GFR) is (A) Glycogen (B) Agar (C) Inulin (D) Hyaluronic acid CARBOHYDRATES AND CARBOHYDRATE METABOLISM 7

Description : The polysaccharide used in assessing the glomerular fittration rate (GFR) is (A) Glycogen (B) Agar (C) Inulin (D) Hyaluronic acid CARBOHYDRATES AND CARBOHYDRATE METABOLISM 7

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The rate of an enzyme catalyzed reaction was measured using several substrate concentrations that were much lower than Km, the dependence of reaction velocity on substrate concentration can best be described as (A) Independent of enzyme concentration (B) A constant fraction of Vmax (C) Equal to Km (D) Proportional to the substrate concentration

Description : The rate of an enzyme catalyzed reaction was measured using several substrate concentrations that were much lower than Km, the dependence of reaction velocity on substrate concentration can best be ... constant fraction of Vmax (C) Equal to Km (D) Proportional to the substrate concentration

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The Michaelis constant, Km is (A) Numerically equal to ½ Vmax (B) Dependent on the enzyme concentration (C) Independent of pH (D) Numerically equal to the substrate concentration that gives half maximal velocity

Description : The Michaelis constant, Km is (A) Numerically equal to ½ Vmax (B) Dependent on the enzyme concentration (C) Independent of pH (D) Numerically equal to the substrate concentration that gives half maximal velocity

Last Answer : Answer : D

A competitive inhibitor of an enzyme (A) Increases Km without affecting Vmax (B) Decreases Km without affecting Vmax (C) Increases Vmax without affecting Km (D) Decreases both Vmax and Km

Description : A competitive inhibitor of an enzyme (A) Increases Km without affecting Vmax (B) Decreases Km without affecting Vmax (C) Increases Vmax without affecting Km (D) Decreases both Vmax and Km

Last Answer : Answer : A

In competitive inhibition of enzyme action (A) The apparent Km is decreased (B) The apparent Km is increased (C) Vmax is decreased (D) Apparent concentration of enzyme molecules decreased

Description : In competitive inhibition of enzyme action (A) The apparent Km is decreased (B) The apparent Km is increased (C) Vmax is decreased (D) Apparent concentration of enzyme molecules decreased

Last Answer : Answer : B

Which of the following statements about an enzyme exhibiting allosteric kinetics with cooperative interaction is false? (A) A plot of V-Vk [s] has a sigmaidal shape (B) An inhibitor may increase the apparent Km (C) Line weaver Bnrk plot is useful for determining Km and Vmax (D) Removal of allosteric inhibitor may result in hyperbolic V-S [s] plot

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When [s] is equal to Km, which of the following conditions exist? (A) Half the enzyme molecules are bound to substrate (B) The velocity of the reaction is equal to Vmax (C) The velocity of the reaction is independent of substrate concentration (D) Enzyme is completely saturated with substrate

Description : When [s] is equal to Km, which of the following conditions exist? (A) Half the enzyme molecules are bound to substrate (B) The velocity of the reaction is equal to Vmax (C) The velocity of the reaction is independent of substrate concentration (D) Enzyme is completely saturated with substrate

Last Answer : Answer : A

In non-competitive enzyme action (A) Vmax is increased (B) Apparent km is increased (C) Apparent km is decreased (D) Concentration of active enzyme molecule is reduced

Description : In non-competitive enzyme action (A) Vmax is increased (B) Apparent km is increased (C) Apparent km is decreased (D) Concentration of active enzyme molecule is reduced

Last Answer : Answer : C

If the substrate concentration is much below the km of the enzyme, the velocity of the reaction is (A) Directly proportional to substrate concentration (B) Not affected by enzyme concentration (C) Nearly equal to Vmax (D) Inversely proportional to substrate concentration

Description : If the substrate concentration is much below the km of the enzyme, the velocity of the reaction is (A) Directly proportional to substrate concentration (B) Not affected by enzyme concentration (C) Nearly equal to Vmax (D) Inversely proportional to substrate concentration

Last Answer : Answer : A

In non competitive enzyme activity inhibition, inhibitor (A) Increases Km (B) Decreases Km (C) Does not effect Km (D) Increases Km

Description : In non competitive enzyme activity inhibition, inhibitor (A) Increases Km (B) Decreases Km (C) Does not effect Km (D) Increases Km

Last Answer : Answer : C

In competitive enzyme activity inhibition (A) Apparent Km is decreased (B) Apparent Km is increased (C) Vmax is increased (D) Vmax is decreased

Description : In competitive enzyme activity inhibition (A) Apparent Km is decreased (B) Apparent Km is increased (C) Vmax is increased (D) Vmax is decreased

Last Answer : Answer : B

In enzyme kinetics Km implies (A) The substrate concentration that gives one half Vmax (B) The dissocation constant for the enzyme substrate comples (C) Concentration of enzyme (D) Half of the substrate concentration required to achieve Vmax

Description : In enzyme kinetics Km implies (A) The substrate concentration that gives one half Vmax (B) The dissocation constant for the enzyme substrate comples (C) Concentration of enzyme (D) Half of the substrate concentration required to achieve Vmax

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In competitive enzyme activity inhibition (A) The structure of inhibitor generally resembles that of the substrate (B) Inhibitor decreases apparent Km (C) Km remains unaffective (E) Inhibitor decreases Vmax without affecting Km

Description : In competitive enzyme activity inhibition (A) The structure of inhibitor generally resembles that of the substrate (B) Inhibitor decreases apparent Km (C) Km remains unaffective (E) Inhibitor decreases Vmax without affecting Km

Last Answer : Answer : A

In reversible non-competitive enzyme activity inhibition (A) Inhibitor bears structural resemblance to substrate (B) Inhibitor lowers the maximum velocity attainable with a given amount of enzyme (C) Km is increased (D) Km is decreased

Description : In reversible non-competitive enzyme activity inhibition (A) Inhibitor bears structural resemblance to substrate (B) Inhibitor lowers the maximum velocity attainable with a given amount of enzyme (C) Km is increased (D) Km is decreased

Last Answer : Answer : B

In reversible non-competitive enzyme activity inhibition (A) Vmax is increased (B) Km is increased (C) Km is decreased (D) Concentration of active enzyme is reduced

Description : In reversible non-competitive enzyme activity inhibition (A) Vmax is increased (B) Km is increased (C) Km is decreased (D) Concentration of active enzyme is reduced

Last Answer : Answer : D

The kinetic effect of purely competitive inhibitor of an enzyme (A) Increases Km without affecting Vmax (B) Decreases Km without affecting Vmax (C) Increases Vmax without affecting Km (D) Decreases Vmax without affecting Km

Description : The kinetic effect of purely competitive inhibitor of an enzyme (A) Increases Km without affecting Vmax (B) Decreases Km without affecting Vmax (C) Increases Vmax without affecting Km (D) Decreases Vmax without affecting Km

Last Answer : Answer : A

The Km of the enzyme giving the kinetic data as below is (A) –0.50 (B) –0.25 (C) +0.25 (D) +0.33

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