biochem exam 3 homework chapter 12

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 2H, 14C, 32P, 35S 3H, 14C, 31P, 35S

3H, 14C, 32P, 35S

A molecule decreases the activity of an enzyme by binding to a site other than the active site. What is the name of this molecule? A covalent catalyst. A competitive inhibitor. A transition state analog. An allosteric inhibitor. A cofactor.

An allosteric inhibitor.

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 Increase Decrease

Increase

For an enzyme that displays Michaelis-Menten kinetics what is the effect of an uncompetitive inhibitor on Vmax? Entry field with correct answer It may increase or decrease It increases It decreases There is no effect

It decreases

For an enzyme that displays Michaelis-Menten kinetics, what is the effect of a competitive inhibitor on KM? Entry field with correct answer It increases It decreases There is no effect It may increase or decrease

It increases

he 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

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

Assume a first order reaction, the rate of the reaction 2A B is dependent on ______. isozymes [A] the rate constant Ping Pong bimolecular ES complex random ordered unimolecular [A]2 competitive inhibition phosphorylation small KS large KS uncompetitive inhibition [B]

[A]

[S] = KM for a simple enzymatic reaction. When [S] is doubled the initial velocity is Entry field with correct answer 2 Vmax equal to Vmax (1/3) Vmax 0.5 Vmax 2 KM/[S]

(1/3) Vmax

What is the DIFFERENCE between an allosteric inhibitor and a competitive inhibitor? Entry field with correct answer Allosteric inhibitors act only on multisubunit enzymes; competitive inhibitors act only on monomeric enzymes. An allosteric inhibitor bound to one subunit alters substrate binding to other subunits; a competitive inhibitor bound at one active site alters binding at only that active site. A competitive inhibitor reduces Vmax; an allosteric inhibitor does not affect Vmax. Competitive inhibitors are substrate analogs; allosteric inhibitors are transition state analogs.

An allosteric inhibitor bound to one subunit alters substrate binding to other subunits; a competitive inhibitor bound at one active site alters binding at only that active site.

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] c

ES breakdown occurs at the same rate as ES formation

This diagram refers to a (an) Entry field with incorrect answer Ping Pong reaction. ordered bisubstrate reaction. random bisubstrate reaction. double order ping pong reaction. X, Y, and Z must be provided in order to answer correctly.

Random bisubstrate reaction

Which of the following features in a Lineweaver-Burk plot is altered by a competitive inhibitor in an enzyme-catalyzed reaction? Entry field with correct answer Shape of the plot Intercept on the y-axis which is the 1/Vo axis Intercept on the 1/Vo axis Intercept on the x-axis Intercept on the y-axis

Intercept on the x-axis

A two-substrate enzymatic reaction in which one product is produced before the second substrate binds to the enzyme has a ______ mechanism. Entry field with correct answer isozymes [A] the rate constant Ping Pong bimolecular ES complex random ordered competitive inhibition unimolecular [A]2 phosphorylation small KS large KS uncompetitive inhibition [B]

Ping pong

Irreversible enzyme inhibitors Entry field with correct answer inactivate the enzyme. inhibit competitively. maximize product by minimizing ES E+S. behave allosterically. function via Ping Pong mechanism.

inactivate the enzyme.

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.

Which of the following does NOT alter enzyme activity in a cell by changing the exact geometry of the active site? A. Allostery. B. Competitive inhibition. C. Phosphorylation. D. Increased gene expression. B and D.

B and D.

The E+S E+P reaction is ______. Entry field with correct answer isozymes [A] the rate constant Ping Pong bimolecular ES complex random ordered competitive inhibition unimolecular [A]2 phosphorylation small KS large KS uncompetitive inhibition [B]

Bimolecular

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 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 increases when pH increases and is reduced in the presence of a transition state analog inhibitor. Maximal velocity increases when pH increases.

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 Increase No effect Decrease

No effect

Fourth-order reactions. Entry field with correct answer have three or more sequential rate determining steps. require a 'Ping Pong' mechanism. are best analyzed using Lineweaver-Burk plots. exist only when enzymatically catalyzed. none of the above.

None of the above

Which of the following methods of altering enzyme activity is irreversible? Binding of transition state analogs to enzymes. Binding of competitive inhibitors to enzymes. Binding of allosteric effectors to enzymes. Phosphorylation of enzymes. None of the above.

None of the above.

f ATP is a substrate for an enzyme, and ADP was found to be a more potent competitive inhibitor than adenosine for the reaction, which of the following would be a plausible explanation? Entry field with correct answer The ribose ring must be important for binding. Adenosine is too large to bind at the active site. Adenosine must be binding outside of the active site. The phosphate groups of ADP must be important for binding.

The phosphate groups of ADP must be important for binding.

Which of the following is (are) true? Entry field with incorrect answer The [ES] will remain constant if k2 > k1 and k−1 < k2. The reaction is zero order with respect to [S] if [S] >> [E]. It describes a double displacement reaction. All of the above are true. None of the above is true.

The reaction is zero order with respect to [S] if [S] >> [E].

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 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.

The relationship between substrate concentration and reaction rate is sigmoidal.

Which of the following is correct in regards to the diagram below? Entry field with incorrect answer X=A, Y=B, Z=P X=B, Y=A, Z=Q X=E, Y=A, Z=E X=E, Y=B, Z=Q X=E, Y=B, Z=P

X=B, Y=A, Z=Q

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

e(-∆G‡ /RT)

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.

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.

From the graph below plotting data that was collected under steady state conditions, velocity on the y-axis in units of μM/s and substrate concentration of the x-axis in units of μM, what is the Vmax? Entry field with incorrect answer 0.24 μM/s 18 μM 0.2 μM 0.24 μM 0.12 μM/s

.24 um/s

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

A new drug has been discovered which inhibits the reaction catalyzed by enzyme A. Based on the information shown below, what is this drug? Entry field with correct answer competitive inhibitor uncompetitive inhibitor mixed inhibitor allosteric activator More information is required to answer the question.

Competitive inhibitor

From the graph below plotting data that was collected under steady state conditions, velocity on the y-axis in units of μM/s and substrate concentration of the x-axis in units of μM, what is the KM? Entry field with correct answer 0.24 μM/s 18 μM 0.2 μM 0.24 μM 0.12 μM/s

18 μM

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 calculatio

4.8 × 102 M·sec-1

Which of the following is TRUE about competitive inhibitors? Entry field with incorrect answer A. Competitive inhibitors lower the KM and the Vmax of the enzyme. B. Competitive inhibitors structurally resemble the substrate and so they bind to the active site and become covalently attached to the enzyme. C. Transition state analogs often make better competitive inhibitors than do substrate analogs. A and C are both true. A, B and C are all true.

C. Transition state analogs often make better competitive inhibitors than do substrate analogs.

The type of enzyme inhibition in which Vmax is unaffected is ______. isozymes [A] the rate constant Ping Pong bimolecular ES complex random ordered unimolecular [A]2 competitive inhibition phosphorylation small KS large KS uncompetitive inhibition [B]

Competitive Inhibition

In ______, the inhibitor binds to a site involved in both substrate binding and catalysis. Entry field with correct answer isozymes [A] the rate constant Ping Pong bimolecular ES complex random ordered competitive inhibition unimolecular [A]2 phosphorylation small KS large KS uncompetitive inhibition [B]

Competitive inhibition

Which of the following is TRUE in competitive inhibition? Entry field with correct answer The inhibitor binds reversibly at the active site. The inhibitor lowers the characteristic Vmax of the enzyme. The inhibitor binds covalently to the enzyme. The inhibitor binds at several different sites on an enzyme.

The inhibitor binds reversibly at the active site.

For an enzyme that displays Michaelis-Menten kinetics, what is the effect of a competitive inhibitor on Vmax? It may increase or decrease There is no effect It increases It decreases

There is no effect

Following several experiments, the data presented on the graph below was obtained. What can you determine from this graph? This data may have been collected both in the absence (solid line) and presence (dashed line) of a competitive inhibitor. This data may have been collected both in the absence (solid line) and presence (dashed line) of a mixed (noncompetitive) inhibitor. This data may have been collected both in the absence (solid line) and presence (dashed line) of mechanism based inhibitor. This data may have been collected both in the absence (solid line) and presence (dashed line) of an inhibitor which binds the active site. More than one of the above are correct.

This data may have been collected both in the absence (solid line) and presence (dashed line) of a mixed (noncompetitive) inhibitor.

Which of the following is true regarding the effect of inhibitors on Michaelis-Menten reactions? Entry field with correct answer Noncompetitive inhibitors have no effect on apparent KM. Competitive inhibitors make it impossible to calculate the Vmax of an enzyme for its substrate. Competitive inhibitors decrease the apparent KM. Uncompetitive inhibitors decreases the apparent KM and decrease the apparent Vmax. Competitive inhibitors have no effect on Vo.

Uncompetitive inhibitors decreases the apparent KM and decrease the apparent Vmax.

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

Vmax for the substrate and enzyme concentration.

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.

Based on the figures below, which of the following expressions would correctly define KM? Entry field with correct answer A= KM KM = A/2 B = KM C = - KM D= 1/ KM

B=KM

Which of the following methods of altering enzyme activity depends on the activity of protein kinases and protein phosphatases? Entry field with incorrect answer Allostery. Product inhibition. Feedback inhibition. Covalent modification

Covalent modification

Based on the figures below, which of the following expressions would be correct? Entry field with correct answer Vmax = 1/B C = 1/ Vmax D= Vmax D = 1/ Vmax A = 1/ Vmax

D = 1/ Vmax

A compound that distorts the active site, rendering the enzyme catalytically inactive is called Entry field with incorrect answer a uncompetitive inhibitor an allosteric effector an inactivator a competitive inhibitor none of the above

uncompetitive inhibitor

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). 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

I propose to design a new drug which will act as an inhibitor for an enzyme. If I have used all current information about the mechanism of this enzyme to design this inhibitor and I carefully engineer it with similar chemical properties of the transition state, what type of inhibitor am I attempting to engineer and how will I know if I have succeeded? Entry field with correct answer A competitive inhibitor, collect kinetic data both in the presence and absence of inhibitor and watch for a change in Vmax. A competitive inhibitor, collect kinetic data both in the presence and absence of inhibitor and watch for a change in KM. A uncompetitive inhibitor, collect kinetic data both in the presence and absence of inhibitor and watch for a change in KM. A uncompetitive inhibitor, collect kinetic data both in the presence and absence of inhibitor and watch for a change in Vmax. None of the above.

A competitive inhibitor, collect kinetic data both in the presence and absence of inhibitor and watch for a change in KM.

Determine the KM and Vmax from the following graph. (Note: On the x-axis the minor tick mark spacing is 0.005; on the y-axis the minor tick mark spacing is 0.002) Entry field with correct answer KM = [0.006]; Vmax = 0.0075/s KM = [0.196]; Vmax = 0.0075/s KM = [165]; Vmax = 33/s KM = [33]; Vmax = 167/s KM = [270]; Vmax = 68/s

KM = [33]; Vmax = 167/s

Which of the following is true for the maximal velocity of an enzyme-catalyzed reaction? 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. Maximal velocity increases when pH increases.

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

The presence of an uncompetitive inhibitor in an enzyme-catalyzed reaction will alter which of the following in a Lineweaver-Burk plot? Entry field with correct answer Intercept on the x-axis Slope of the plot The intercept on both axes Shape of the plot Intercept on the y-axis

The intercept on both axes

If A B is a zero-order reaction, the rate is dependent on ______. Entry field with correct answer isozymes [A] the rate constant Ping Pong bimolecular ES complex random ordered competitive inhibition unimolecular [A]2 phosphorylation small KS large KS uncompetitive inhibition [B]

The rate constant

An extremely efficient enzyme called "efficase" catalyzes the conversion of "A" to "B." A researcher decides to mutate the enzyme in order to try to improve its performance. Following active site mutations, a significant reduction in the value of KM and Vmax was observed. Which of the following may have occurred? Entry field with correct answer The affinity of the enzyme for the substrate was increased to a point which did not favor propagation (continuation) of the reaction. The decrease in Vmax was not related to the decrease in KM. If the reaction was first-order, the change in KM cannot have affected Vmax. The stability of E+S (E+A as written above) was increased, thereby increasing the KM. The reverse reaction (breakdown of EA to E+A) was favored, slowing the Vmax.

The reverse reaction (breakdown of EA to E+A) was favored, slowing the Vmax.

In the plot below, can the KM be determined? If so, what is its value? Entry field with incorrect answer Yes, it is 30 mM. Yes, it is 30 mM/sec. Yes, it is 60 mM/sec Yes, it is 60 mM No this data does not follow Michaelis-Menten kinetics

Yes, it is 30 mM.

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

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