A biotechnology company has cloned four different forms of the enzyme money synthetase which catalyzes the reaction:Garbage + ATP = Money + ADP + Phosphate + H+ The Km and Vmax values of these enzymes for the substrate garbage are as follows:Enzyme 1; Km = 0.1mmol Vmax = 5.0 mmol/minEnzyme 2; Km = 0.3 mmol Vmax = 2.0 mmol/minEnzyme 3; Km = 1.0 mmol Vmax = 5.0 mmol/minEnzyme 4; Km = 3.0 mmol Vmax = 20.0 mmol/minWhich of the four enzymes is fastest at a saturating ATP concentration and a garbage concentration of 0.01 mmol/L?A. Enzyme 1B. Enzyme 2C. Enzyme 3D. Enzyme 4E. None of the above

A. Enzyme 1v = (Vmax x [S])/Km + [S]Enzyme 1v = 5 x 0.01/0.1 + 0.01v = 0.05/0.11 = 0.45Enzyme 2v = 2 x 0.01/0.3 + 0.01v = 0.02/0.31 = 0.06Enzyme 3v = 5 x 0.01/1.0 + 0.01v = 0.05/1.01 = 0.05Enzyme 4v = 20 x 0.01/3.0 + 0.01v = 0.20/3.01 = 0.07

BCCB2000 Lecture 9 Questions Flashcards by Ana Lis

Enzymes are not as efficient as most catalysts used in organic chemistry, since they must function at body temperature. True or False?

False

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A biotechnology company has cloned four different forms of the enzyme money synthetase which catalyzes the reaction: Garbage + ATP = Money + ADP + Phosphate + H+ The Km and Vmax values of these enzymes for the substrate garbage are as follows: Enzyme 1; Km = 0.1mmol Vmax = 5.0 mmol/min Enzyme 2; Km = 0.3 mmol Vmax = 2.0 mmol/min Enzyme 3; Km = 1.0 mmol Vmax = 5.0 mmol/min Enzyme 4; Km = 3.0 mmol Vmax = 20.0 mmol/min Which of the four enzymes is fastest at a saturating ATP concentration and a garbage concentration of 0.01 mmol/L? A. Enzyme 1 B. Enzyme 2 C. Enzyme 3 D. Enzyme 4 E. None of the above

A. Enzyme 1 v = (Vmax x [S])/Km + [S] Enzyme 1 v = 5 x 0.01/0.1 + 0.01 v = 0.05/0.11 = 0.45 Enzyme 2 v = 2 x 0.01/0.3 + 0.01 v = 0.02/0.31 = 0.06 Enzyme 3 v = 5 x 0.01/1.0 + 0.01 v = 0.05/1.01 = 0.05 Enzyme 4 v = 20 x 0.01/3.0 + 0.01 v = 0.20/3.01 = 0.07

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What distinguishes reversible inhibitors from irreversible inhibitors? A. Reversible inhibitors are not covalently bound to enzymes but irreversible inhibitors are. B. There is an equilibrium between bound and unbound reversible inhibitor. C. There usually is very little reverse reaction for the binding of an irreversible inhibitor. D. Reversible inhibitors are easier to purify from solutions of enzymes than irreversible inhibitors. E. A, B, & C above F. A and B only

E. A, B, & C above

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Enzymes differ from other catalysts in that enzymes: A. lower the activation energy of the reaction catalyzed. B. fail to influence the equilibrium point of the reaction. C. form an activated complex with the reactants. D. usually display specificity toward a single reactant. E. are not consumed in the reaction.

D. usually display specificity toward a single reactant.

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What is the estimate of Vmax if the total enzyme concentration in each reaction mixture is doubled? Vo = 0.5micromol/min

Vmax is proportional to the enzyme concentration. Doubling the enzyme concentration will produce a new Vmax of 2 x 0.50 micromol per minute = 1.00 micromole per minute.

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In a first order chemical reaction, the velocity of the reaction is proportional to the ________, while in a zero order reaction, the velocity of the reaction is proportional to the ________.

concentration of substrate; amount of enzyme

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Which of the following statements does not apply to the effect of temperature on an enzyme-catalysed reaction? A. Enzymes are inactivated both by being held for long periods of time at moderate temperatures and by being held for short periods of time at high temperatures B. Denatured enzymes always regain their lost activity on cooling C. The temperature at which the maximum occurs on the activity versus temperature graph is called the optimum temperature D. At high temperatures the enzyme may be denatured E. At low temperatures the rate doubles (approximately) for each 10°C rise

B. Denatured enzymes always regain their lost activity on cooling

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Phosphorylation that changes an enzymeʹs activity is an example of ________. A. covalent modification B. allosteric regulation C. sequential modification D. site-directed mutagenesis

A. covalent modification

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The maximum velocity of an enzyme-catalysed reaction is: A. the maximum velocity that can be achieved at a given temperature and pH B. twice the Km C. the limiting velocity for a given temperature, pH, and concentration of enzyme as the substrate concentration is increased D. the initial velocity that can be achieved at the Km for a given temperature and pH for a specified amount of enzyme E. the turnover number

C. the limiting velocity for a given temperature, pH, and concentration of enzyme as the substrate concentration is increased

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The Lineweaver-Burk plot and other linear transformation of the Michaelis-Menten curve of kinetics are valuable for: A. determination of Km. B. determination of Vmax. C. determination of kcat. D. determination of types of enzyme inhibition. E. All of the above

E. All of the above

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The release of free energy as a result of the interaction between enzyme and substrate is sometimes called the binding energy. Which of the following is true of the binding energy derived from enzyme-substrate interactions? A. Most of it is used for enzyme stability B. Most of it is derived from covalent bonds between enzyme and substrate. C. It is sometimes used to hold two substrates in the optimal orientation for reaction. D. It is the difference in initial and final free energy between reactants and products. E. It is used to provide more kinetic energy to the reactants

C. It is sometimes used to hold two substrates in the optimal orientation for reaction.

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The Michaelis-Menton model of enzyme action: A. explains the mechanism of stereospecificity of enzyme reactions B. explains the mechanism for the saturation effect observed at high substrate concentrations C. explains the mechanism for the catalytic efficiency of enzymes D. assumes that the mechanism for the formation of an enzyme and substrate complex is a covalent bond E. explains the mechanism of allosteric behaviour of certain regulatory enzymes

B. explains the mechanism for the saturation effect observed at high substrate concentrations

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The reason to rewrite the Michaelis-Menten equation (such as the Lineweaver-Burk plot) is to: A. visualize reactions better. B. form enzyme kinetic data as a hyperbolic curve. C. calculate catalytic proficiency. D. calculate Vmax and Km.

D. calculate Vmax and Km.

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The role of an enzyme in an enzyme-catalyzed reaction is to: A. ensure that the product is more stable than the substrate. B. make the free-energy change for the reaction more favorable. C. increase the rate at which substrate is converted into product. D. ensure that all the substrate is converted to product. E. do none of the above.

C. increase the rate at which substrate is converted into product.

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To calculate the turnover number of an enzyme you need to know the: A. initial velocity of the catalyzed reaction at low [S]. B. initial velocity of the catalyzed reaction at [S]&raquo_space; Km. C. Km for the substrate. D. enzyme concentration. E. both B and D.

E. both B and D.

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The Km values for enzyme reactions such as A + B → C + D A. cannot be determined using the Lineweaver-Burk plot analysis. B. can be determined by holding one (A or B) at high concentration, while varying the concentration of the other substrate. C. can be determined for one substrate and not the other. D. do not indicate the efficiency of the enzyme.

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B. can be determined by holding one (A or B) at high concentration, while varying the concentration of the other substrate.

The concept of “induced fit” refers to the fact that: A. when a substrate binds to an enzyme, the enzyme induces a loss of water (desolvation) from the substrate. B. substrate binding may induce a conformational change in the enzyme, which then brings catalytic groups into proper orientation. C. enzyme-substrate binding induces an increase in the reaction entropy, thereby catalyzing the reaction. D. enzyme specificity is induced by enzyme-substrate binding. E. enzyme-substrate binding induces movement along the reaction coordinate to the transition state.

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B. substrate binding may induce a conformational change in the enzyme, which then brings catalytic groups into proper orientation.

The following data were obtained in a study of an enzyme known to follow Michaelis-Menten kinetics: V0 Substrate added (μmol/min) (mmol/L) ——————————————————————- 217 0.8 325 2 433 4 488 6 647 1,000 The Km for this enzyme is approximately: A. 488 μmol/min B. 325 μmol/min C. 1 mM. D. 6 mM. E. 2 mM.

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E. 2 mM. Km is defined as the substrate concentration that is half maximum velocity. This data shows very high substrate concentration and a rate of 647 μmol/min this is probably the close to the limiting value of Vmax. Hence the approximate Km would be the substrate concentration at half of this value = 2mM

When varying the substrate concentration at a fixed concentration of enzyme it is observed that at low concentrations of substrate the reaction is ________,while at high concentrations of substrate the reaction is ________.

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first order; zero order

The Michaelis constant, Km, is equal to the ________.

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substrate concentration when the rate is equal to half its maximal value

It is difficult to determine either Km or Vmax from a graph of velocity verses substrate concentration because: A. too much substrate is required to determine them. B. the graph is sigmoid. C. an asymptotic value must be determined from the graph. D. the points on the graph are often not spread out on the hyperbola.

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C. an asymptotic value must be determined from the graph.

Vmax for an enzyme-catalyzed reaction: A. generally increases when pH increases. B. increases in the presence of a competitive inhibitor. C. is unchanged in the presence of a uncompetitive inhibitor. D. is about twice the rate observed when the concentration of substrate is equal to the Km. E. is limited only by the amount of substrate supplied.

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D. is about twice the rate observed when the concentration of substrate is equal to the Km.

In competitive inhibition, an inhibitor: A. binds at several different sites on an enzyme. B. binds reversibly at the active site. C. binds only to the ES complex. D. binds covalently to the enzyme. E. lowers the characteristic Vmax of the enzyme.

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B. binds reversibly at the active site.

An inhibitor binds to a site other than the active site of the enzyme. Which statement below correlates with this observation? A. It must be a competitive inhibitor. B. The inhibition must be irreversible. C. It could be noncompetitive or uncompetitive inhibition. D. It could be irreversible, competitive, noncompetitive or uncompetitive.

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C. It could be noncompetitive or uncompetitive inhibition.

Which of the following parameters remains the same for S → P, regardless of whether the reaction is enzyme-catalyzed or uncatalyzed?

A. rate constant k
B. velocity V
C. Initial velocity V0
D. Equilibrium constant Keq
E. Michaelis constant Km

D. Equilibrium constant Keq

The following irreversible enzymatic reaction is inhibited by high concentrations of its own product citrate. This product inhibition can be overcome and a normal Vmax can be restored when the oxaloacetate concentration is raised but not when the acetylCoA concentration is raised. This observation suggests that citrate is:

oxaloacetate + acetyl-CoA = citrate + CoA-SH

A. An irreversible inhibitor reacting with the oxaloacetate binding site of the enzyme
B. A competitive inhibitor binding to the acetyl-CoA binding site of the enzyme
C. A competitive inhibitor binding to the oxaloacetate binding site of the enzyme
D. A noncompetitive inhibitor binding to acetyl-CoA binding site of the enzyme
E. A noncompetitive inhibitor binding to the oxaloacetate binding site of the enzyme

C. A competitive inhibitor binding to the oxaloacetate binding site of the enzyme

Chemical reactions in the absence of a catalyst may be distinguished from enzyme-catalysed reactions because:

A. chemical reactions proceed at lower temperatures
B. chemical reactions are saturable with substrate
C. enzyme-catalysed reactions produce a variety of products from a single substrate
D. the free energy of activation is lower for an enzyme-catalysed reaction
E. enzyme-catalysed reactions are not first-order with respect to substrate concentration below the Km value

D. the free energy of activation is lower for an enzyme-catalysed reaction

The initial velocity of an enzyme reaction (vo) describes:

A. the concentration of the enzyme at maximal velocity.
B. the concentration of substrate at maximal velocity.
C. the concentration of enzyme and substrate at the start of the reaction.
D. the rate of the reaction at when the substrate and enzyme are first mixed

D. the rate of the reaction at when the substrate and enzyme are first mixed

Which of the following is NOT true of enzyme-catalysed reactions?

A. The substrate concentration which gives a half-maximum rate is sometimes a good measure of the binding affinity of the enzyme for the substrate
B. The maximum rate of reaction is proportional to the concentration of the substrate
C. At low substrate concentrations, relative to the Km, the initial rate is proportional to the substrate concentration
D. At high substrate concentrations, relative to the Km, the initial rate is independent of the substrate concentration
E. At high substrate concentration, relative to the Km, the initial rate is proportional to the concentration of the enzyme

B. The maximum rate of reaction is proportional to the concentration of the substrate

What are two essential things to consider when you want to determine the total amount of enzyme in a sample (given that the temperature, ionicstrength, and pH optimum have been determined)?

A. Max velocity is measured and [S]>>Km
B. Maximum velocity is measured and [S]<
C. Initial velocity is measured and [S] << Km
D. Initial velocity is measured and [S] >>Km
E. None of the above

D. Initial velocity is measured and [S] >>Km

BCCB2000 Lecture 9 Questions Flashcards

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