Chapter 12 Part 1

For the reaction, the steady state assumption Entry field with correct answer implies that k1=k−1 implies that k−1 and k2 are such that the [ES] = k1[ES] [P]>>[E] [S] = [P] ES breakdown occurs at the same rate as ES formation

ES breakdown occurs at the same rate as ES formation

Which of the following describes the effect of an enzyme on the initial velocity of a given reaction? Entry field with correct answer No change Decrease Increase

Increase

Which of the following is true for the maximal velocity of an enzyme-catalyzed reaction? Entry field with correct answer Maximal velocity can be calculated from the initial rate of the reaction at a concentration of substrate that is equal to the KM. Maximal velocity increases when pH increases. Maximal velocity is reduced in the presence of a transition state analog competitive inhibitor. Maximal velocity is not affected by the presence of a noncompetitive inhibitor.

Maximal velocity can be calculated from the initial rate of the reaction at a concentration of substrate that is equal to the KM. Vmax is equal to twice the initial rate at the concentration of substrate equal to the KM.

Which of the following is true for the maximal velocity of an enzyme catalyzed reaction? Entry field with correct answer Maximal velocity is reduced in the presence of a transition state analog inhibitor. Maximal velocity may be used to determine KM. Maximal velocity increases when pH increases. Maximal velocity increases when pH increases and is reduced in the presence of a transition state analog inhibitor. Maximal velocity may be used to determine KM. Maximal velocity increases when pH increases and is reduced in the presence of a transition state analog inhibitor

Maximal velocity may be used to determine KM

For an enzyme that displays Michaelis-Menten kinetics, what is the effect on KM of doubling the concentration of substrate? Entry field with correct answer No effect Increase Decrease

No effect The KM of an enzyme is a concentration of substrate, the concentration at which the initial velocity (Vo) is half maximal velocity (Vmax). As such, it cannot be affected by changes in substrate concentration.

Pseudo-first-order reaction kinetics would be observed for the reaction A + B C Entry field with correct answer if [A] or [B] > [C]. if [C]>[A] and [C]>[B]. if [A] or [B] = 0. if [C] = 0. None of the above.

None of the above.

If A B is a zero-order reaction, the rate is dependent on ______.

The rate constant

Which of the following statements is false for an enzyme that follows Michaelis-Menten kinetics? Entry field with correct answer Maximal velocity occurs when the enzyme is entirely in the ES form. The relationship between substrate concentration and reaction rate is sigmoidal. The Michaelis-Menten equation assumes that ES maintains a steady state. The initial velocity of the reaction is dependent on substrate concentration

The relationship between substrate concentration and reaction rate is sigmoidal Enzymes which follow Michaelis-Menten kinetics have a rectangular hyperbolic relationship between substrate concentration and reaction rate.

Which of the following must be known to calculate the kcat of an enzyme? Entry field with incorrect answer Vmax for the substrate. Enzyme concentration. The KM for the substrate. Vmax for the substrate and enzyme concentration. Vmax for the substrate, enzyme concentration and KM for the substrate

Vmax for the substrate and enzyme concentration The kcat of an enzyme is also known as its turnover number. It is the number of molecules of substrate converted to product in a given time per molecule of enzyme, when the enzyme is saturated with substrate. KM is not needed to calculate this.

In order for an enzymatic reaction obeying the Michaelis-Menten equation to reach 3/4 of its maximum velocity, Entry field with correct answer [S] would need to be equal to KM [S] would need to be ½ KM [S] would need to be 3KM [S] would need to be ¾ KM not enough information is given to make this calculation

[S] would need to be 3KM

The E+S E+P reaction is ______.

bimolecular

Which expression containing the free energy of activation (∆G‡) is proportional to the rate of a reaction? Entry field with correct answer +∆G‡ /RT e(-∆G‡ /RT) ln(∆G‡ /RT) -∆G‡ /RT

e(-∆G‡ /RT)

Reaction that is first order with respect to A and B Entry field with correct answer is dependent on the concentration of A and B. is dependent on the concentration of A. has smaller rate constants than first-order reactions regardless of reactant concentration. is independent of reactant concentration. is always faster than first-order reactions due to loss of concentration dependence.

is dependent on the concentration of A and B.

KM Entry field with correct answer is the concentration of substrate where the enzyme achieves ½Vmax. is equal to Ks. measures the stability of the product is high if the enzyme has high affinity for the substrate. All of the above are correct.

is the concentration of substrate where the enzyme achieves ½Vmax.

The catalytic efficiency of an enzyme can never exceed Entry field with correct answer k2. k1. k-1. k-1 + k2. (k-1 + k2)/k1.

k1

The KM can be considered to be the same as the dissociation constant KS for E + S binding if Entry field with correct answer the concentration of [ES] is unchanged. ES E + P is fast compared to ES E + S. k1 >> k2. k2 << k-1. this statement cannot be completed because KM can never approximate KS.

k2 << k-1.

An enzyme is near maximum efficiency when Entry field with correct answer its turnover number is near Vmax. kcat/KM is near 108 M-1s-1. k1 << k-1. kcat/KM is equal to kcat. KM is large when k2 exceeds k1.

kcat/KM is near 108 M-1s-1

For a reaction A + B --> C, if the concentration of B is much larger than A so that [B] remains constant during the reaction while [A] is varied, the kinetics will be Entry field with correct answer sigmoidal. pseudo-first-order. unimolecular. zero-order. hyperbolic.

pseudo-first-order.

KM is Entry field with correct answer a measure of the catalytic efficiency of the enzyme. equal to half of Vmax. the rate constant for the reaction ES E + P. the [S] that half-saturates the enzyme. a ratio of substrate concentration relative to catalytic power.

the [S] that half-saturates the enzyme.

At substrate concentrations much lower than the enzyme concentration, Entry field with correct answer the rate of reaction is expected to be inversely proportional to substrate concentration. the rate of reaction is expected to be directly proportional to substrate concentration. first order enzyme kinetics are not observed. the KM is lower. the rate of reaction is independent of substrate concentration.

the rate of reaction is expected to be directly proportional to substrate concentration.

The breakdown of dopamine is catalyzed by the enzyme monoamine oxidase (MAO). What is the final concentration of product if the starting dopamine concentration is 0.050 M and the reaction runs for 5 seconds. (Assume the rate constant for the reaction is 0.249 s−1.) Entry field with correct answer 0.050 M 0.014 M 0.018 M 1.2 M 0.025 M

0.014 M

When [S] = KM, ν0 = (_____)× (Vmax). Entry field with correct answer [S] 0.75 0.5 KM kcat

0.5

Find the initial velocity for an enzymatic reaction when Vmax = 6.5 × 10-5 mol·sec-1, [S] = 3.0 × 10-3 M, KM = 4.5 × 10-3 M and the enzyme concentration at time zero is 1.5 × 10-2 μM. Entry field with correct answer 3.9 × 10-5 mol·sec-1 2.6 × 10-5 mol·sec-1 1.4 × 10-2 mol·sec-1 8.7 × 10-3 mol·sec-1 Not enough information is given to make this calculation.

2.6 × 10-5 mol·sec-1

Which of the following groups of isotopes includes only radioactive isotopes that are commonly used in biochemical experiments? Entry field with correct answer 3H, 14C, 32P, 35S 3H, 12C, 32P, 35S 3H, 14C, 31P, 35S 2H, 14C, 32P, 35S

3H, 14C, 32P, 35S

What is the velocity of a first-order reaction at 37oC when the reactant concentration is 6 × 10-2 M and the rate constant is 8 × 103 sec-1? Entry field with correct answer 1.33 × 105 M-1·sec-1 1.33 × 105 M·sec 7.5 × 10-2 M·sec 4.8 × 102 M·sec-1 Not enough data are given to make this calculation

4.8 × 102 M·sec-1

Find kcat for a reaction in which Vmax is 4 × 10-4 mol·min-1 and the reaction mixture contains one microgram of enzyme (the molecular weight of the enzyme is 200,000 D). Entry field with correct answer 2 × 10-11 min-1 8 × 107 min-1 8 × 109 min-1 2 × 10-14 min-1 4 × 108 min-1

8 × 107 min-1