Biochem: Chp 6 - Allosteric enzymes

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

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

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.

Which of the following statements about allosteric control of enzymatic activity is false? E) Heterotropic allosteric effectors compete with substrate for binding sites.

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.

Which of the following statements is false? C) For S-->P, a catalyst shifts the reaction equilibrium to the right.

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.

Which of these statements about enzyme-catalyzed reactions is false? 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. *ENZYMES DO NOT CHANGE EQUILIBRIUM, ONLY REACTION RATES!*

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.

Which one of the following statements is true of enzyme catalysts? 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.

Which one of the following statements is true of enzyme catalysts? Enzymes differ from other catalysts in that only enzymes: D) They lower the activation energy for the conversion of substrate to product.

Define the terms "cofactor" and "coenzyme."

cofactor-any chemical component required for enzyme activity; includes both organic molecules (coenzymes) and inorganic ions coenzyme-an organic cofactor required for enzyme activity

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 are not among one of the 6 internationally accepted classes of enzymes.

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.

A good transition-state analog: B) binds to the enzyme more tightly than the substrate.

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 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. (Heterotropic allosteric inhibitors have different specific binding sites in enzymes. Homotropic allosteric inhibitors share a regulatory and active site.)

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.

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

What is a zymogen (proenzyme)? Explain briefly with an example.

A zymogen is an inactive form of an enzyme that is activated by one or more proteolytic cleavages in its sequence. Chymotrypsinogen, trypsinogen, and proelastase are all zymogens, becoming chymotrypsin, trypsin, and elastase, respectively, after proper cleavage.

In the following diagram of the first step in the reaction catalyzed by the protease chymotrypsin, the process of general base catalysis is illustrated by the number ________, and the process of covalent catalysis is illustrated by the number _________. A) 1; 2 B) 1; 3 C) 2; 3 D) 2; 3 E) 3; 2

A) 1; 2

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

Michaelis-Menten kinetics is sometimes referred to as "saturation" kinetics. Why?

According to the Michaelis-Menten model of enzyme-substrate interaction, when [S] becomes very high, an enzyme molecule's active site will become occupied with a new substrate molecule as soon as it releases a product. Therefore, at very high [S], V0 does not increase with additional substrate, and the enzyme is said to be "saturated" with substrate.

K-system effectors

Affect K0.5 Activators decrease K0.5 Inhibitors increase K0.5

V-system effectors

Affect Vmax Activators increase Vmax Inhibitors decrease Vmax

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.

Allosteric enzymes: C) usually have more than one polypeptide chain.

Homotropic effectors

Almost always positive (activators) A substrate binds to the active site and increases affinity of substrate to other active sites

Allosteric inhibitor

An effector that binds the allosteric site and decreases the enzyme's activity

Allosteric activator

An effector that binds the allosteric site and enhances the enzyme's activity

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.

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: 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.

Heterotropic effectors

Can be either negative (inhibitors) or positive (activators) Bind to allosteric site A regulatory molecule (NOT SUBSTRATE) that binds to an allosteric site and causes a conformational change of the enzyme Also changes the kinetic properties of the enzyme

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].

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.

Allosteric effectors

Effectors may be positive (activators) or negative (inhibitors) 2 types of allosteric effectors

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.

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: B) a His residue on the enzyme is involved in the reaction. (His residues are sensitive to pH due to the fact that the residue carries a pH of 6.0. Relatively small changes in pH changes the overall charge of His.)

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.

Enzymes are potent catalysts because they: E) lower the activation energy for the reactions they catalyze.

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.

Enzymes differ from other catalysts in that only enzymes: B) display specificity toward a single reactant.

Allosteric enzyme

Enzymes that undergo a conformational change upon binding to an effector. This results in a change in binding affinity at a different ligand binding site More than 1 binding site More than one subunit

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.

For enzymes in which the slowest (rate-limiting) step is the reaction: k2 ES --> P Km becomes equivalent to: C) the dissociation constant, Kd, for the ES complex.

Positive cooperativity

Greater change in activity over a narrower range of [S] Enzyme without S bound is TENSE Enzyme bound to S is RELAXED

2 types of allosteric effectors

Homotropic effectors Heterotropic effectors

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.

In a plot of l/V against 1/[S] for an enzyme-catalyzed reaction, the presence of a competitive inhibitor will alter the: B) intercept on the l/[S] axis. (Because competitive inhibitors compete with [S]).

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.

In competitive inhibition, an inhibitor: D) binds reversibly at the active site.

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

It would take 27 minutes for the same amount of product to be formed. (Think about it: The first time it takes 9 minutes for only 1% to be formed. Therefore, if the amount of substrate is twice the amount as the previous but there was 1/3 less of the amount of enzyme, then it would take about 3 times as long for the reduced amount of enzyme to bind to the substrate. Even though there is more substrate, there is 3 times less enzyme and the rate of an enzyme-catalyzed reaction is proportional to the amount of [E].)

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

Penicillin and related drugs inhibit the enzyme _______; this enzyme is produced by _______. C) transpeptidase; bacteria

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.

The Lineweaver-Burk plot is used to: D) solve, graphically, for the rate of an enzymatic reaction at infinite substrate concentration. (Remember, Vmax is how much substrate is necessary for an enzymatic reaction to occur at its maximum rate, and the Lineweaver-Burk plot measures 1/Km and 1/[S] to get enzymatic rates.)

Give the Michaelis-Menten equation and define each term in it. Does this equation apply to all enzymes? If not, to which kind does it not apply?

The Michaelis-Menten equation is: V0 = Vmax [S]/( Km + [S]) -V0 is the initial velocity at any given concentration of S. -Vmax is the velocity when all enzyme molecules are saturated with S. -[S] is the concentration of S. -Km is a constant characteristic for the enzyme. This equation does not apply to enzymes that display sigmoidal V0 vs. [S] curves, but only to those giving hyperbolic kinetic plots.

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.

The benefit of measuring the initial rate of a reaction V0 is that at the beginning of a reaction: (B) changes in [S] are negligible, so [S] can be treated as a constant. ([S] is normally far greater than [Et] (the sum of free and substrate-bound enzyme), so the amount of substrate bound at any given time is negligible compared with the total [S].)

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.

Cooperativity

The binding of one substrate molecule facilitates the binding of subsequent molecules to the enzyme MOST ALLOSTERIC ENZYMES DISPLAY POSITIVE COOPERATIVITY

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.

The concept of "induced fit" refers to the fact that: D) substrate binding may induce a conformational change in the enzyme, which then brings catalytic groups into proper orientation.

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.

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.

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.

The most likely correct for the pathway R-->S-->T-->U-->V-->W is: C) The last product, W, is likely to be a negative modulator of X, leading to feedback inhibition.

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.

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: E) turnover number.

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.

The role of an enzyme in an enzyme-catalyzed reaction is to: D) increase the rate at which substrate is converted into product.

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

The role of the metal ion (Mg2+) in catalysis by enolase is to: D) facilitate general base catalysis

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.

The statement that is false concerning a plot of V0 vs. [S] for an enzyme that follows Michaelis-Menten kinetics is: (B) At very high [S], the velocity curve becomes a horizontal line that intersects the y-axis at Km. -At very high [S], the velocity curve becomes a plateau-like and is close to to Vmax.

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.

The steady state assumption, as applied to enzyme kinetics, implies: C) the ES complex is formed and broken down at equivalent rates.

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.

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

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.

Trypsinogen is converted to trypsin by: C) Proteolysis of trypsinogen forms trypsin.

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.

Under conditions of zinc deficiency, when the enzyme may lack zinc, it would be referred to as the: A) apoenzyme. (holoenzyme=enzyme + coenzyme)

2 types of heterotropic effectors

V-system effectors K-system effectors

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.

Vmax for an enzyme-catalyzed reaction: D) is twice the rate observed when the concentration of substrate is equal to the Km. ([S] and therefore Km are increased, but Vmax stays the same.)