One of the enzymes involved in glycolysis, aldolase, requires Zn2+ for catalysis. Under conditions ofzinc deficiency, when the enzyme may lack zinc, it would be referred to as the:
Which one of the following is not among the six internationally accepted classes of enzymes?
Enzymes are potent catalysts because they:
The role of an enzyme in an enzyme-catalyzed reaction is to:
Which one of the following statements is true of enzyme catalysts?
Which one of the following statements is true of enzyme catalysts?
Which of the following statements is false?
Enzymes differ from other catalysts in that only enzymes:
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
Which of the following is true of the binding energy derived from enzyme-substrate interactions?
The concept of “induced fit” refers to the fact that:
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 _________.
The benefit of measuring the initial rate of a reaction V0 is that at the beginning of a reaction:
Which of the following statements about a plot of V0 vs. [S] for an enzyme that follows Michaelis-Menten kinetics is false?
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:
The steady state assumption, as applied to enzyme kinetics, implies:
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?
Which of these statements about enzyme-catalyzed reactions is false?
For enzymes in which the slowest (rate-limiting) step is the reaction
k2 ES → P
Km becomes equivalent to:
The Lineweaver-Burk plot is used to:
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:
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:
In a plot of l/V against 1/[S] for an enzyme-catalyzed reaction, the presence of a competitive inhibitorwill alter the:
Vmax for an enzyme-catalyzed reaction:
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:
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:
A transition-state analog:
Which of the following statements about allosteric control of enzymatic activity is false?
A small molecule that decreases the activity of an enzyme by binding to a site other than the catalyticsite is termed a(n):
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?
Which of the following has not been shown to play a role in determining the specificity of proteinkinases?
Define the terms “cofactor” and “coenzyme.”
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?
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?
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.
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?
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:
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.
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?
Why does pH affect the activity of an enzyme?
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?
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.
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.
Explain how a biochemist might discover that a certain enzyme is allosterically regulated.
What is a zymogen (proenzyme)? Explain briefly with an example.