Chapter 6

Which of the following has not been shown to play a role in determining the specificity of protein kinases? A) Disulfide bonds near the phosphorylation site B) Primary sequence at phosphorylation site C) Protein quaternary structure D) Protein tertiary structure E) Residues near the phosphorylation site

A) Disulfide bonds near the phosphorylation site

Phenyl-methane-sulfonyl-fluoride (PMSF) inactivates serine proteases by binding covalently to the catalytic serine residue at the active site; this enzyme-inhibitor bond is not cleaved by the enzyme. This is an example of what kind of inhibition? A) Irreversible B) Competitive C) Non-competitive D) Mixed E) pH inhibition

A) Irreversible

A small molecule that decreases the activity of an enzyme by binding to a site other than the catalytic site is termed a(n): A) allosteric inhibitor. B) alternative inhibitor. C) competitive inhibitor. D) stereospecific agent. E) transition-state analog.

A) allosteric inhibitor.

One of the enzymes involved in glycolysis, aldolase, requires Zn2+ for catalysis. Under conditions of zinc deficiency, when the enzyme may lack zinc, it would be referred to as the: A) apoenzyme. B) coenzyme. C) holoenzyme. D) prosthetic group. E) substrate.

A) apoenzyme.

Phenyl-methane-sulfonyl-fluoride (PMSF) inactivates serine proteases by binding covalently to the catalytic serine residue at the active site; this enzyme-inhibitor bond is not cleaved by the enzyme. This is an example of what kind of inhibition? A) irreversible B) competitive C) non-competitive D) mixed E) pH inhibition

A) irreversible

The double-reciprocal transformation of the Michaelis-Menten equation, also called the Lineweaver-Burk plot, is given by 1/V0 = Km /(Vmax[S]) + 1/Vmax. To determine Km from a double-reciprocal plot, you would: A) multiply the reciprocal of the x-axis intercept by −1. B) multiply the reciprocal of the y-axis intercept by −1. C) take the reciprocal of the x-axis intercept. D) take the reciprocal of the y-axis intercept. E) take the x-axis intercept where V0 = 1/2 Vmax.

A) multiply the reciprocal of the x-axis intercept by -1

The double-reciprocal transformation of the Michaelis-Menten equation, also called the Lineweaver-Burk plot, is given by 1/V0 = Km /(Vmax[S]) + 1/Vmax. To determine Km from a double-reciprocal plot, you would: A) multiply the reciprocal of the x-axis intercept by -1. B) multiply the reciprocal of the y-axis intercept by -1. C) take the reciprocal of the x-axis intercept. D) take the reciprocal of the y-axis intercept. E) take the x-axis intercept where V0 = 1/2 Vmax.

A) multiply the reciprocal of the x-axis intercept by -1.

Which of the following statements about a plot of V0 vs. [S] for an enzyme that follows Michaelis-Menten kinetics is false? A) As [S] increases, the initial velocity of reaction V0 also increases. B) At very high [S], the velocity curve becomes a horizontal line that intersects the y-axis at Km. C) Km is the [S] at which V0 = 1/2 Vmax. D) The shape of the curve is a hyperbola. E) The y-axis is a rate term with units of μm/min.

B) At very high [S], the velocity curve becomes a horizontal line that intersects the y-axis at Km.

Enzyme X exhibits maximum activity at pH = 6.9. X shows a fairly sharp decrease in its activity when the pH goes much lower than 6.4. One likely interpretation of this pH activity is that: A) a Glu residue on the enzyme is involved in the reaction. B) a His residue on the enzyme is involved in the reaction. C) the enzyme has a metallic cofactor. D) the enzyme is found in gastric secretions. E) the reaction relies on specific acid-base catalysis.

B) a His residue on the enzyme is involved in the reaction.

A good transition-state analog: A) binds covalently to the enzyme. B) binds to the enzyme more tightly than the substrate. C) binds very weakly to the enzyme. D) is too unstable to isolate. E) must be almost identical to the substrate.

B) binds to the enzyme more tightly than the substrate.

The benefit of measuring the initial rate of a reaction V0 is that at the beginning of a reaction: A) [ES] can be measured accurately. B) changes in [S] are negligible, so [S] can be treated as a constant. C) changes in Km are negligible, so Km can be treated as a constant. D) V0 = Vmax. E) varying [S] has no effect on V0.

B) changes in [S] are negligible, so [S] can be treated as a constant.

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

B) display specificity toward a single reactant.

In a plot of l/V against 1/[S] for an enzyme-catalyzed reaction, the presence of a competitive inhibitor will alter the: A) curvature of the plot. B) intercept on the l/[S] axis. C) intercept on the l/V axis. D) pK of the plot. E) Vmax.

B) intercept on the l/[S] axis.

The allosteric enzyme ATCase is regulated by CTP, which binds to the T-state of ATCase. CTP is a: A) positive regulator. B) negative regulator. C) co-factor. D) competitive inhibitor. E) coenzyme.

B) negative regulator.

Both water and glucose share an —OH that can serve as a substrate for a reaction with the terminal phosphate of ATP catalyzed by hexokinase. Glucose, however, is about a million times more reactive as a substrate than water. The best explanation is that: A) glucose has more —OH groups per molecule than does water. B) the larger glucose binds better to the enzyme; it induces a conformational change in hexokinase that brings active-site amino acids into position for catalysis. C) the —OH group of water is attached to an inhibitory H atom, while the glucose —OH group is attached to C. D) water and the second substrate, ATP, compete for the active site resulting in a competitive inhibition of the enzyme. E) water normally will not reach the active site because it is hydrophobic.

B) the larger glucose binds better to the enzyme; it induces a conformational change in hexokinase that brings active-site amino acids into position for catalysis.

Both water and glucose share an —OH that can serve as a substrate for a reaction with the terminal phosphate of ATP catalyzed by hexokinase. Glucose, however, is about a million times more reactive as a substrate than water. The best explanation is that: F) glucose has more —OH groups per molecule than does water. G) the larger glucose binds better to the enzyme; it induces a conformational change in hexokinase that brings active-site amino acids into position for catalysis. H) the —OH group of water is attached to an inhibitory H atom, while the glucose —OH group is attached to C. I) water and the second substrate, ATP, compete for the active site resulting in a competitive inhibition of the enzyme. J) water normally will not reach the active site because it is hydrophobic.

B) the larger glucose binds better to the enzyme; it induces a conformational change in hexokinase that brings active-site amino acids into position for catalysis.

The following data were obtained in a study of an enzyme known to follow Michaelis-Menten kinetics: V0 Substrate added (micromol/min) (mmol/L) ————————————— 217 0.8 325 2 433 4 488 6 647 1,000 ————————————— The Km for this enzyme is approximately: A) 1 mM. B) 1000 mM. C) 2 mM. D) 4 mM. E) 6 mM.

C) 2 mM.

An enzyme-catalyzed reaction was carried out with the substrate concentration initially a thousand times greater than the Km for that substrate. After 9 minutes, 1% of the substrate had been converted to product, and the amount of product formed in the reaction mixture was 12 μmol. If, in a separate experiment, one-third as much enzyme and twice as much substrate had been combined, how long would it take for the same amount (12 μmol) of product to be formed? A) 1.5 min B) 13.5 min C) 27 min D) 3 min E) 6 min

C) 27 min

Which of the following statements is false? A) A reaction may not occur at a detectable rate even though it has a favorable equilibrium. B) After a reaction, the enzyme involved becomes available to catalyze the reaction again. C) For S-->P, a catalyst shifts the reaction equilibrium to the right. D) Lowering the temperature of a reaction will lower the reaction rate. E) Substrate binds to an enzyme's active site.

C) For S-->P, a catalyst shifts the reaction equilibrium to the right.

How is trypsinogen converted to trypsin? A) A protein kinase-catalyzed phosphorylation converts trypsinogen to trypsin. B) An increase in Ca2+ concentration promotes the conversion. C) Proteolysis of trypsinogen forms trypsin. D) Trypsinogen dimers bind an allosteric modulator, cAMP, causing dissociation into active trypsin monomers. E) Two inactive trypsinogen dimers pair to form an active trypsin tetramer.

C) Proteolysis of trypsinogen forms trypsin.

A metabolic pathway proceeds according to the scheme, R --> S --> T --> U --> V --> W. A regulatory enzyme, X, catalyzes the first reaction in the pathway. Which of the following is most likely correct for this pathway? A) Either metabolite U or V is likely to be a positive modulator, increasing the activity of X. B) The first product S, is probably the primary negative modulator of X, leading to feedback inhibition. C) The last product, W, is likely to be a negative modulator of X, leading to feedback inhibition. D) The last product, W, is likely to be a positive modulator, increasing the activity of X. E) The last reaction will be catalyzed by a second regulatory enzyme.

C) The last product, W, is likely to be a negative modulator of X, leading to feedback inhibition.

A transition-state analog: A) is less stable when binding to an enzyme than the normal substrate. B) resembles the active site of general acid-base enzymes. C) resembles the transition-state structure of the normal enzyme-substrate complex. D) stabilizes the transition state for the normal enzyme-substrate complex. E) typically reacts more rapidly with an enzyme than the normal substrate.

C) resembles the transition-state structure of the normal enzyme-substrate complex.

The steady state assumption, as applied to enzyme kinetics, implies: A) Km = Ks. B) the enzyme is regulated. C) the ES complex is formed and broken down at equivalent rates. D) the Km is equivalent to the cellular substrate concentration. E) the maximum velocity occurs when the enzyme is saturated.

C) the ES complex is formed and broken down at equivalent rates.

For enzymes in which the slowest (rate-limiting) step is the reaction: ES (k2) --> P Km becomes equivalent to: A) kcat. B) the [S] where V0 = Vmax. C) the dissociation constant, Kd, for the ES complex. D) the maximal velocity. E) the turnover number.

C) the dissociation constant, Kd, for the ES complex.

Penicillin and related drugs inhibit the enzyme ________; this enzyme is produced by _________. A) β-lacamase; bacteria B) transpeptidase; human cells C) transpeptidase; bacteria D) lysozyme; human cells E) aldolase; bacteria

C) transpeptidase; bacteria

Allosteric enzymes: A) are regulated primarily by covalent modification. B) usually catalyze several different reactions within a metabolic pathway. C) usually have more than one polypeptide chain. D) usually have only one active site. E) usually show strict Michaelis-Menten kinetics.

C) usually have more than one polypeptide chain.

The Lineweaver-Burk plot is used to: A) determine the equilibrium constant for an enzymatic reaction. B) extrapolate for the value of reaction rate at infinite enzyme concentration. C) illustrate the effect of temperature on an enzymatic reaction. D) solve, graphically, for the rate of an enzymatic reaction at infinite substrate concentration. E) solve, graphically, for the ratio of products to reactants for any starting substrate concentration.

D solve, graphically, for the rate of an enzymatic reaction at infinite substrate concentration. Page: 197

Which one of the following is not among the six internationally accepted classes of enzymes? A) Hydrolases B) Ligases C) Oxidoreductases D) Polymerases E) Transferases

D) Polymerases

Which of these statements about enzyme-catalyzed reactions is false? A) At saturating levels of substrate, the rate of an enzyme-catalyzed reaction is proportional to the enzyme concentration. B) If enough substrate is added, the normal Vmax of a reaction can be attained even in the presence of a competitive inhibitor. C) The rate of a reaction decreases steadily with time as substrate is depleted. D) The activation energy for the catalyzed reaction is the same as for the uncatalyzed reaction, but the equilibrium constant is more favorable in the enzyme-catalyzed reaction. E) The Michaelis-Menten constant Km equals the [S] at which V = 1/2 Vmax.

D) The activation energy for the catalyzed reaction is the same as for the uncatalyzed reaction, but the equilibrium constant is more favorable in the enzyme-catalyzed reaction.

Which one of the following statements is true of enzyme catalysts? A) Their catalytic activity is independent of pH. B) They are generally equally active on D and L isomers of a given substrate. C) They can increase the equilibrium constant for a given reaction by a thousand fold or more. D) They can increase the reaction rate for a given reaction by a thousand fold or more. E) To be effective, they must be present at the same concentration as their substrate.

D) They can increase the reaction rate for a given reaction by a thousand fold or more.

Which one of the following statements is true of enzyme catalysts? A) They bind to substrates, but are never covalently attached to substrate or product. B) They increase the equilibrium constant for a reaction, thus favoring product formation. C) They increase the stability of the product of a desired reaction by allowing ionizations, resonance, and isomerizations not normally available to substrates. D) They lower the activation energy for the conversion of substrate to product. E) To be effective they must be present at the same concentration as their substrates.

D) They lower the activation energy for the conversion of substrate to product.

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

D) binds reversibly at the active site

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

D) binds reversibly at the active site.

The role of the metal ion (Mg2+) in catalysis by enolase is to: A) act as a general acid catalyst B) act as a general base catalyst C) facilitate general acid catalysis D) facilitate general base catalysis E) stabilize protein conformation

D) facilitate general base catalysis

The role of an enzyme in an enzyme-catalyzed reaction is to: A) bind a transition state intermediate, such that it cannot be converted back to substrate. B) ensure that all of the substrate is converted to product. C) ensure that the product is more stable than the substrate. D) increase the rate at which substrate is converted into product. E) make the free-energy change for the reaction more favorable.

D) increase the rate at which substrate is converted into product.

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

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

Michaelis and Menten assumed that the overall reaction for an enzyme-catalyzed reaction could be written as E + S (k1,k-1)> ES(k2) > E + P Using this reaction, the rate of breakdown of the enzyme-substrate complex can be described by the expression: A) k1 ([Et] - [ES]). B) k1 ([Et] - [ES])[S]. C) k2 [ES]. D) k-1 [ES] + k2 [ES]. E) k-1 [ES].

D) k-1 [ES] + k2 [ES].

For the simplified representation of an enzyme-catalyzed reaction shown below, the statement "ES is in steady-state" means that: E + S (k1,k-1) > ES (k2, k-2) > E + P A) k2 is very slow. B) k1= k2. C) k1= k-1. D) k1[E][S] = k-1[ES] + k2[ES]. E) k1[E][S] = k-1[ES].

D) k1[E][S] = k-1[ES] + k2[ES].

The Lineweaver-Burk plot is used to: A) determine the equilibrium constant for an enzymatic reaction. B) extrapolate for the value of reaction rate at infinite enzyme concentration. C) illustrate the effect of temperature on an enzymatic reaction. D) solve, graphically, for the rate of an enzymatic reaction at infinite substrate concentration. E) solve, graphically, for the ratio of products to reactants for any starting substrate concentration.

D) solve, graphically, for the rate of an enzymatic reaction at infinite substrate concentration.

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

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

Blood coagulation involves: A) a kinase cascade. B) zymogen activation. C) serine proteases. D) A and B. E) B and C.

E) B and C.

Which of the following statements about allosteric control of enzymatic activity is false? A) Allosteric effectors give rise to sigmoidal V0 vs. [S] kinetic plots. B) Allosteric proteins are generally composed of several subunits. C) An effector may either inhibit or activate an enzyme. D) Binding of the effector changes the conformation of the enzyme molecule. E) Heterotropic allosteric effectors compete with substrate for binding sites.

E) Heterotropic allosteric effectors compete with substrate for binding sites.

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

E) both A and B, which are:

Enzymes are potent catalysts because they: A) are consumed in the reactions they catalyze. B) are very specific and can prevent the conversion of products back to substrates. C) drive reactions to completion while other catalysts drive reactions to equilibrium. D) increase the equilibrium constants for the reactions they catalyze. E) lower the activation energy for the reactions they catalyze.

E) lower the activation energy for the reactions they catalyze.

The number of substrate molecules converted to product in a given unit of time by a single enzyme molecule at saturation is referred to as the: A) dissociation constant. B) half-saturation constant. C) maximum velocity. D) Michaelis-Menten number. E) turnover number.

E) turnover number.

Which of the following is true of the binding energy derived from enzyme-substrate interactions? A) It cannot provide enough energy to explain the large rate accelerations brought about by enzymes. B) It is sometimes used to hold two substrates in the optimal orientation for reaction. C) It is the result of covalent bonds formed between enzyme and substrate. D) Most of it is derived from covalent bonds between enzyme and substrate. E) Most of it is used up simply binding the substrate to the enzyme.

The binding energy derived from enzyme-substrate interactions is B) sometimes used to hold two substrates in the optimal orientation for reaction.