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

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?

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

Which of the following statements is false?

For S → P, a catalyst shifts the reaction equilibrium to the right.

Enzymes differ from other catalysts in that only enzymes:

display specificity toward a single reactant.

Compare the two reaction coordinate diagrams below and select the answer that correctly describes
their relationship. In each case, the single intermediate is the ES complex.

Accompanied by image

(a) describes a strict “lock and key” model, whereas (b) describes a transition-state complementarity model.

Which of the following is true of the binding energy derived from enzyme-substrate interactions?

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

The concept of “induced fit” refers to the fact that:

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

In the following diagram of the first step in the reaction catalyzed by the protease chymotrypsin, theprocess of general base catalysis is illustrated by the number ________, and the process of covalent
catalysis is illustrated by the number _________.

1; 2

The benefit of measuring the initial rate of a reaction V0 is that at the beginning of a reaction:

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

Which of the following statements about a plot of V0 vs. [S] for an enzyme that follows Michaelis-Menten kinetics is false?

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

Michaelis and Menten assumed that the overall reaction for an enzyme-catalyzed reaction could bewritten as
k1 k2E + S ES → E + P k-1 Using this reaction, the rate of breakdown of the enzyme-substrate complex can be described by theexpression:

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

The steady state assumption, as applied to enzyme kinetics, implies:

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

An enzyme-catalyzed reaction was carried out with the substrate concentration initially a thousandtimes greater than the Km for that substrate. After 9 minutes, 1% of the substrate had been convertedto product, and the amount of product formed in the reaction mixture was 12 µmol. If, in a separateexperiment, one-third as much enzyme and twice as much substrate had been combined, how longwould it take for the same amount (12 µmol) of product to be formed?

27 min

Which of these statements about enzyme-catalyzed reactions is false?

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.

For enzymes in which the slowest (rate-limiting) step is the reaction
k2 ES → P
Km becomes equivalent to:

the dissociation constant, Kd, for the ES complex.

The Lineweaver-Burk plot is used to:

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

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:

multiply the reciprocal of the x-axis intercept by −1.

To calculate the turnover number of an enzyme, you need to know:

both A and B.

The number of substrate molecules converted to product in a given unit of time by a single enzymemolecule at saturation is referred to as the:

turnover number.

In a plot of l/V against 1/[S] for an enzyme-catalyzed reaction, the presence of a competitive inhibitorwill alter the:

intercept on the l/[S] axis.

In competitive inhibition, an inhibitor:

binds reversibly at the active site.

Vmax for an enzyme-catalyzed reaction:

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

Enzyme X exhibits maximum activity at pH = 6.9. X shows a fairly sharp decrease in its activitywhen the pH goes much lower than 6.4. One likely interpretation of this pH activity is that:

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

Both water and glucose share an —OH that can serve as a substrate for a reaction with the terminalphosphate of ATP catalyzed by hexokinase. Glucose, however, is about a million times more reactiveas a substrate than water. The best explanation is that:

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.

A good transition-state analog:

binds to the enzyme more tightly than the substrate.

A transition-state analog:

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

The role of the metal ion (Mg2+) in catalysis by enolase is to

facilitate general base catalysis

Which of the following statements about allosteric control of enzymatic activity is false?

Heterotropic allosteric effectors compete with substrate for binding sites.

A small molecule that decreases the activity of an enzyme by binding to a site other than the catalyticsite is termed a(n):

allosteric inhibitor.

Allosteric enzymes:

usually have more than one polypeptide chain.

A metabolic pathway proceeds according to the scheme, R → S → T → U → V → W. A regulatoryenzyme, X, catalyzes the first reaction in the pathway. Which of the following is most likely correctfor this pathway?

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

Which of the following has not been shown to play a role in determining the specificity of proteinkinases?

Disulfide bonds near the phosphorylation site

How is trypsinogen converted to trypsin?

Proteolysis of trypsinogen forms trypsin.

Define the terms “cofactor” and “coenzyme.”

A cofactor is any chemical component required for enzyme activity; it includes both organicmolecules, called “coenzymes,” and inorganic ions.

The difference in (standard) free energy content, ∆G'°, between substrate S and product P may varyconsiderably among different reactions. What is the significance of these differences?

The difference in free energy content between substrate (or reactant) and product for eachreaction reflects the relative amounts of each compound present at equilibrium. The greater thedifference in free energy, the greater the difference in amounts of each compound at equilibrium.

For a reaction that can take place with or without catalysis by an enzyme, what would be the effect ofthe enzyme on the:

(a) standard free energy change of the reaction?
(b) activation energy of the reaction?
(c) initial velocity of the reaction?
(d) equilibrium constant of the reaction?

(a) no change;
(b) decrease;
(c) increase;
(d) no change

Sometimes the difference in (standard) free-energy content, ∆G'°, between a substrate S and a productP is very large, yet the rate of chemical conversion, S → P, is quite slow. Why?

The rate of conversion from substrate to product (or the reverse reaction, from product tosubstrate) does not depend on the free-energy difference between them. The rate of the reactiondepends upon the activation energy of the reaction ∆G'‡, which is the difference between the free-energy content of S (or P) and the reaction transition state.

Write an equilibrium expression for the reaction S → P and briefly explain the relationship betweenthe value of the equilibrium constant and free energy.

Keq' = [P]/[S].
The value of Keq' reflects the difference between the free energy content of S andP. Free energy and equilibrium constant are related by the expression: ∆G'° = -RT ln Keq' For each change in Keq' by one order of magnitude, ∆G'° changes by 5.7 Kjoule/mole

What is the difference between general acid-base catalysis and specific acid-base catalysis? (Assumethat the solvent is water.)

Specific acid-base catalysis refers to catalysis by the constituents of water, i.e., the donation of aproton by the hydronium ion, H3O+ or the acceptance of a proton by the hydroxyl ion OH-. General acid-base catalysis refers to the donation or acceptance of a proton by weak acids and bases other than water.

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

According to the Michaelis-Menten model of enzyme-substrate interaction, when [S] becomesvery 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 additionalsubstrate, and the enzyme is said to be “saturated” with substrate.

Two different enzymes are able to catalyze the same reaction, A → B. They both have the same Vmax,but differ their Km the substrate A. For enzyme 1, the Km is 1.0 mM; for enzyme 2, the Km is 10 mM.When enzyme 1 was incubated with 0.1 mM A, it was observed that B was produced at a rate of0.0020 mmoles/minute.

a) What is the value of the Vmax of the enzymes?
b) What will be the rate ofproduction of B when enzyme 2 is incubated with 0.1 mM A?
c) What will be the rate of productionof B when enzyme 1 is incubated with 1 M (i.e., 1000 mM) A?

a) 0.022 mmol/min;
b) 0.0022 mmol/min;
c) 0.022 mmol/min

An enzyme can catalyze a reaction with either of two substrates, S1 or S2. The Km for S1 was found tobe 2.0 mM, and the Km, for S2 was found to be 20 mM. A student determined that the Vmax was thesame for the two substrates. Unfortunately, he lost the page of his notebook and needed to know thevalue of Vmax. He carried out two reactions: one with 0.1 mM S1, the other with 0.1 mM S2. Unfortunately, he forgot to label which reaction tube contained which substrate. Determine the valueof Vmax from the results he obtained:

Vmax = 101

For the reaction E + S → ES → P the Michaelis-Menten constant, Km, is actually a summary of threeterms. What are they? How is Km determined graphically?

Km = (k2 + k-1)/ k1, where k-1 and k1 are the rate constants for the breakdown and association,

respectively, of the ES complex and k2 is the rate constant for the breakdown of ES to form E + P. Kmcan be determined graphically on a plot of V0 vs. [S] by finding the [S] at which V0 = 1/2 Vmax. Moreconveniently, on a double-reciprocal plot, the x-axis intercept = –1/ Km.

An enzyme catalyzes a reaction at a velocity of 20 µmol/min when the concentration of substrate (S)is 0.01 M. The Km for this substrate is 1 × 10-5 M. Assuming that Michaelis-Menten kinetics arefollowed, what will the reaction velocity be when the concentration of S is (a) 1 × 10-5 M and (b) 1 ×10-6 M?

The velocity of 20 µmol/min is essentially Vmax because it is measured at [S] >> Km.

(a) When[S] = 10-5 M = Km, V = 1/2 Vmax, or 10 µmol/min.

(b) When [S] is 10-6 M, velocity can be calculatedfrom the Michaelis-Menten equation

V0 = Vmax [S]/( Km + [S]) = (20 µmol/min)(10-6 M)/(10-5 + 10-6) = 1.8 µmol/min.

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

V0 = Vmax [S]/( Km + [S]), in which V0 is the initial velocity atany given concentration of S, Vmax is the velocity when all enzyme molecules are saturated with S, [S] isthe concentration of S, and Km is a constant characteristic for the enzyme. This equation does not applyto enzymes that display sigmoidal V0 vs. [S] curves, but only to those giving hyperbolic kinetic plots.

Why is the Lineweaver-Burk (double reciprocal) plot (see Box 6, p. 206) more useful than thestandard V vs. [S] plot in determining kinetic constants for an enzyme? (Your answer shouldprobably show typical plots.)

The plot of V vs. [S] is hyperbolic; maximum velocity is never achieved experimentally,because it is impossible to do experiments at infinitely high [S]. The Lineweaver-Burk transformation of the Michaelis-Menten equation produces a linear plot that can be extrapolated toinfinite [S] (where 1/[S] becomes zero), allowing a determination of Vmax.

The turnover number for an enzyme is known to be 5,000 min-1. From the following set of data,calculate the Km and the total amount of enzyme present in these experiments.

Km = about 2 mM (the concentration of S needed to achieve one-half of Vmax, which is about500). The total enzyme present is producing about 500 µmol of product per minute. Because theturnover number is 5,000/min, the amount of enzyme present must be 0.1 µmol; 1 µmol of enzymewould produce 5,000 µmol product/min.

When 10 µg of an enzyme of Mr 50,000 is added to a solution containing its substrate at a concentration one hundred times the Km, it catalyzes the conversion of 75 µmol of substrate intoproduct in 3 min. What is the enzyme's turnover number?

Because the velocity measured occurs far above Km, it represents Vmax. Ten µg of the enzymerepresents 10 × 10-6 g/(5 × 104 g/mol), or 2 × 10-10 mol of enzyme. In 3 minutes, this amount ofenzyme produced 75 µmol of product, equivalent to 25 × 10-6 mol of product per minute. Theturnover number is therefore

(25 × 10-6 mol/min)/(2 × 10-10 mol) = 12.5 × 104 min-1

Fifteen µg of an enzyme of Mr 30,000 working at Vmax catalyzes the conversion of 60 µmol ofsubstrate into product in 3 min. What is the enzyme's turnover number?

The amount of enzyme present is 15 × 10-6 g, which is (15 × 10-6 g)/(3 × 104 g/mol) = 5 × 10-10mol of enzyme. The rate of product formation is 60 × 10-6 mol/3 min, or 20 × 10-6 mol of product perminute. The turnover number is therefore (20 × 10-6 mol/min)/(5 × 10-10 mol of enzyme), or 4 × 10-4min-1.

How does the total enzyme concentration affect turnover number and Vmax?

The turnover number, kcat, is the number of substrate molecules converted to product in a giventime by a single enzyme molecule, so turnover number is not affected by the total enzyme concentration, [Et]. For any given reaction, however, Vmax can change because Vmax is the product ofturnover number × the total enzyme concentration, or Vmax = kcat [Et].

Enzymes with a kcat / Km ratio of about 108 M-1s-1 are considered to show optimal catalytic efficiency.Fumarase, which catalyzes the reversible-dehydration reaction fumarate + H2
O malate has a ratio of turnover number to the Michaelis-Menten constant, (kcat / Km) of 1.6 × 108 for thesubstrate fumarate and 3.6 × 107 for the substrate malate. Because the turnover number for bothsubstrates is nearly identical, what factors might be involved that explain the different ratio for thetwo substrates?

If the turnover number is nearly identical for both substrates, then the Km for malate must bemuch larger than for fumarate. Similar turnover numbers suggest no significant differences in rate ofconversion of substrate to product, but the different Km values could possibly be explained by astronger binding affinity of the enzyme for fumarate than for malate or some other aspect of thereaction mechanism that affects Km.

Methanol (wood alcohol) is highly toxic because it is converted to formaldehyde in a reactioncatalyzed by the enzyme alcohol dehydrogenase:

NAD+ + methanol → NADH + H+ + formaldehyde

Part of the medical treatment for methanol poisoning is to administer ethanol (ethyl alcohol) inamounts large enough to cause intoxication under normal circumstances. Explain this in terms ofwhat you know about examples of enzymatic reactions.

Ethanol is a structural analog of methanol, and competes with methanol for the binding site ofalcohol dehydrogenase, slowing the conversion of methanol to formaldehyde, and allowing itsclearance by the kidneys. The effect of ethanol is that of a competitive inhibitor.

The enzymatic activity of lysozyme is optimal at pH 5.2 and decreases above and below this pH value. Lysozyme contains two amino acid residues in the active site essential for catalysis: Glu35 andAsp52. The pK value for the carboxyl side chains of these two residues are 5.9 and 4.5, respectively.What is the ionization state of each residue at the pH optimum of lysozyme? How can the ionizationstates of these two amino acid residues explain the pH-activity profile of lysozyme?

For the enzyme to be active, it is likely that Asp52 is unprotonated and Glu35 is protonated.When the pH is below 4.5, Asp52 becomes protonated, and when it is above 5.9, Glu35 isdeprotonated, either of which decreases the activity of the enzyme. (See Fig. 6-20, p. 215.)

Why does pH affect the activity of an enzyme?

The state of ionization of several amino acid side chains is affected by pH, and the activity of manyenzymes requires that certain of the amino acid residue side chains be in a specific ionization state. (SeeFig 6-20, p. 215.)

Chymotrypsin belongs to a group of proteolytic enzymes called the “serine proteases,” many of whichhave an Asp, His, and Ser residue that are crucial to the catalytic mechanism. The serine hydroxylfunctions as a nucleophile. What do the other two amino acids do to support this nucleophilicreaction?

In chymotrypsin, histidine functions as a general base, accepting a proton from the serinehydroxyl, thereby increasing serine's reactivity as a nucleophile. The negatively charged Aspstabilizes the positive charge that develops on the His.

For serine to work effectively as a nucleophile in covalent catalysis in chymotrypsin a nearby aminoacid, histidine, must serve as general base catalyst. Briefly describe, in words, how these two aminoacids work together.

The serine is a polar hydroxyl, with the oxygen functioning as an electronegative nucleophile.A nearby histidine residue, with pKa ≈ 6.0, however, functions as a base to abstract the proton fromthe serine hydroxyl group. The result is to substantially increase the electronegativity of the serineoxygen, making it a much stronger nucleophile. This, in turn, lowers the activation energy of thecovalent catalysis between serine and the carbonyl carbon of the substrate peptide bond.

On the enzyme hexokinase, ATP reacts with glucose to produce glucose 6-phosphate and ADP fiveorders of magnitude faster than ATP reacts with H2O to form phosphate and ADP. The intrinsic chemical reactivity of the —OH group in water is about the same as that of the glucose molecule, andwater can certainly fit into the active site. Explain this rate differential in two sentences or less.

The binding of glucose to hexokinase induces a conformation change that brings the amino acidresidues that facilitate the phosphoryl transfer into position in the active site. Binding of water alone does not induce this conformational change.

Why is a transition-state analog not necessarily the same as a competitive inhibitor?

The structure of a competitive inhibitor may be similar to the structure of the free substrate.Similar structure will mean that the competitive inhibitor can associate with the enzyme at the activesite, effectively blocking the normal substrate from binding. A transition-state analog, however, issimilar in structure to the transition-state of the reaction catalyzed by the enzyme. Often a transition-state analog will bind tightly to an enzyme, and is not easily competed away by substrate.

The scheme S → T → U → V → W → X → Y represents a hypothetical pathway for the metabolicsynthesis of compound Y. The pathway is regulated by feedback inhibition. Indicate where theinhibition is most likely to occur and what the likely inhibitor is.

S → T → U → V → W → X → Y (most likely inhibitor) ↑ ↓ – – – – – – – ← – – – – – – – –(most likely inhibited step)

Explain how a biochemist might discover that a certain enzyme is allosterically regulated.

The enzyme would show kinetics that do not fit the Michaelis-Menten equation; the plot of Vvs. [S] would be sigmoidal, not hyperbolic. The enzyme kinetics would be affected by moleculesother than the substrate(s)

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 proteolyticcleavages in its sequence. Chymotrypsinogen, trypsinogen, and proelastase are all zymogens,becoming chymotrypsin, trypsin, and elastase, respectively, after proper cleavage.