PROBLEMS

Question 8.1

Raisons d’être. What are the two properties of enzymes that make them especially useful catalysts?

Question 8.2

Partners. What does an apoenzyme require to become a holoenzyme?

Question 8.3

Different partners. What are the two main types of cofactors?

Question 8.4

One a day. Why are vitamins necessary for good health?

Question 8.5

A function of state. What is the fundamental mechanism by which enzymes enhance the rate of chemical reactions?

Question 8.6

Nooks and crannies. What is the structural basis for enzyme specificity?

Question 8.7

Give with one hand, take with the other. Why does the activation energy of a reaction not appear in the final ΔG of the reaction?

Question 8.8

The more things change, the more they stay the same. Suppose that, in the absence of enzyme, the forward rate constant (kF) for the conversion of S into P is 10−4 s−1 and the reverse rate constant (kR) for the conversion of P into S is 10−6 s−1.

  1. What is the equilibrium for the reaction? What is the ΔG°′?

  2. Suppose an enzyme enhances the rate of the reaction 100 fold. What are the rate constants for the enzyme-catalyzed reaction? The equilibrium constant? The ΔG°′?

Question 8.9

Mountain climbing. Proteins are thermodynamically unstable. The ΔG of the hydrolysis of proteins is quite negative, yet proteins can be quite stable. Explain this apparent paradox. What does it tell you about protein synthesis?

Question 8.10

Protection. Suggest why the enzyme lysozyme, which degrades cell walls of some bacteria, is present in tears.

Question 8.11

Mutual attraction. What is meant by the term binding energy?

Question 8.12

Catalytically binding. What is the role of binding energy in enzyme catalysis?

Question 8.13

Sticky situation. What would be the result of an enzyme having a greater binding energy for the substrate than for the transition state?

Question 8.14

Stability matters. Transition-state analogs, which can be used as enzyme inhibitors and to generate catalytic antibodies, are often difficult to synthesize. Suggest a reason.

Question 8.15

Match’em. Match the values with the appropriate ΔG°′ values.

 

ΔG°′(kJ mol−1)

(a)

1

   28.53

(b)

10−5

−11.42

(c)

104

    5.69

(d)

102

         0

(e)

10−1

−22.84

Question 8.16

Free energy! Assume that you have a solution of 0.1 M glucose 6-phosphate. To this solution, you add the enzyme phosphoglucomutase, which catalyzes the following reaction:

The ΔG°′ for the reaction is +7.5 kJ mol−1 (+1.8 kcal mol−1).

  1. Does the reaction proceed as written? If so, what are the final concentrations of glucose 6-phosphate and glucose 1-phosphate?

  2. Under what cellular conditions could you produce glucose 1-phosphate at a high rate?

Question 8.17

Free energy, too! Consider the following reaction:

After reactant and product were mixed and allowed to reach equilibrium at 25°C, the concentration of each compound was measured:

Calculate Keq and ΔG°′.

Question 8.18

Keeping busy. Many isolated enzymes, if incubated at 37°C, will be denatured. However, if the enzymes are incubated at 37°C in the presence of substrate, the enzymes are catalytically active. Explain this apparent paradox.

Question 8.19

Active yet responsive. What is the biochemical advantage of having a KM approximately equal to the substrate concentration normally available to an enzyme?

Question 8.20

Affinity or not affinity? That is the question. The affinity between a protein and a molecule that binds to the protein is frequently expressed in terms of a dissociation constant Kd.

Does KM measure the affinity of the enzyme complex? Under what circumstances might KM approximately equal Kd?

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Question 8.21

Angry biochemists. Many biochemists go bananas, and justifiably, when they see a Michaelis–Menten plot like the one shown below. To see why, determine the V0 as a fraction of Vmax when the substrate concentration is equal to 10 KM and 20 KM. Please control your outrage.

Question 8.22

Hydrolytic driving force. The hydrolysis of pyrophosphate to orthophosphate is important in driving forward biosynthetic reactions such as the synthesis of DNA. This hydrolytic reaction is catalyzed in E. coli by a pyrophosphatase that has a mass of 120 kDa and consists of six identical subunits. For this enzyme, a unit of activity is defined as the amount of enzyme that hydrolyzes 10 μmol of pyrophosphate in 15 minutes at 37°C under standard assay conditions. The purified enzyme has a Vmax of 2800 units per milligram of enzyme.

  1. How many moles of substrate is hydrolyzed per second per milligram of enzyme when the substrate concentration is much greater than KM?

  2. How many moles of active sites are there in 1 mg of enzyme? Assume that each subunit has one active site.

  3. What is the turnover number of the enzyme? Compare this value with others mentioned in this chapter.

Question 8.23

Destroying the Trojan horse. Penicillin is hydrolyzed and thereby rendered inactive by penicillinase (also known as β-lactamase), an enzyme present in some penicillin-resistant bacteria. The mass of this enzyme in Staphylococcus aureus is 29.6 kDa. The amount of penicillin hydrolyzed in 1 minute in a 10-ml solution containing 10−9 g of purified penicillinase was measured as a function of the concentration of penicillin. Assume that the concentration of penicillin does not change appreciably during the assay.

[Penicillin] μM

Amount hydrolyzed (nmol)

  1

0.11

  3

0.25

  5

0.34

10

0.45

30

0.58

50

0.61

  1. Plot V0 versus [S] and 1/V0 versus 1/[S] for these data. Does penicillinase appear to obey Michaelis–Menten kinetics? If so, what is the value of KM?

  2. What is the value of Vmax?

  3. What is the turnover number of penicillinase under these experimental conditions? Assume one active site per enzyme molecule.

Question 8.24

Counterpoint. Penicillinase (β-lactamase) hydrolyzes penicillin. Compare penicillinase with glycopeptide transpeptidase.

Question 8.25

A different mode. The kinetics of an enzyme is measured as a function of substrate concentration in the presence and absence of 100 μM inhibitor.

  1. What are the values of Vmax and KM in the presence of this inhibitor?

  2. What type of inhibition is it?

  3. What is the dissociation constant of this inhibitor?

     

    Velocity (μmol minute−1)

    [S] (μM)

    No inhibitor

    Inhibitor

      3

    10.4

    2.1

      5

    14.5

    2.9

    10

    22.5

    4.5

    30

    33.8

    6.8

    90

    40.5

    8.1

  4. If [S] = 30 μM, what fraction of the enzyme molecules have a bound substrate in the presence and in the absence of 100 μM inhibitor?

Question 8.26

A fresh view. The plot of 1/V0 versus 1/[S] is sometimes called a Lineweaver–Burk plot. Another way of expressing the kinetic data is to plot V0 versus V0/[S], which is known as an Eadie–Hofstee plot.

  1. Rearrange the Michaelis–Menten equation to give V0 as a function of V0/[S].

  2. What is the significance of the slope, the y-intercept, and the x-intercept in a plot of V0 versus V0/[S]?

  3. Sketch a plot of V0 versus V0/[S] in the absence of an inhibitor, in the presence of a competitive inhibitor, and in the presence of a noncompetitive inhibitor.

Question 8.27

Defining attributes. What is the defining characteristic for an enzyme catalyzing a sequential reaction? A double-displacement reaction?

Question 8.28

Competing substrates. Suppose that two substrates, A and B, compete for an enzyme. Derive an expression relating the ratio of the rates of utilization of A and B, VA/VB, to the concentrations of these substrates and their values of kcat and KM. (Hint: Express VA as a function of kcat/KM for substrate A, and do the same for VB.) Is specificity determined by KM alone?

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Question 8.29

A tenacious mutant. Suppose that a mutant enzyme binds a substrate 100 times as tightly as does the native enzyme. What is the effect of this mutation on catalytic rate if the binding of the transition state is unaffected?

Question 8.30

More Michaelis–Menten. For an enzyme that follows simple Michaelis–Menten kinetics, what is the value of Vmax if V0 is equal to 1 μmol minute−1 at 10 KM?

Question 8.31

Controlled paralysis. Succinylcholine is a fast-acting, short-duration muscle relaxant that is used when a tube is inserted into a patient’s trachea or when a bronchoscope is used to examine the trachea and bronchi for signs of cancer. Within seconds of the administration of succinylcholine, the patient experiences muscle paralysis and is placed on a respirator while the examination proceeds. Succinylcholine is a competitive inhibitor of acetylcholinesterase, a nervous system enzyme, and this inhibition causes paralysis. However, succinylcholine is hydrolyzed by blood-serum cholinesterase, which shows a broader substrate specificity than does the nervous system enzyme. Paralysis lasts until the succinylcholine is hydrolyzed by the serum cholinesterase, usually several minutes later.

  1. As a safety measure, serum cholinesterase is measured before the examination takes place. Explain why this measurement is good idea.

  2. What would happen to the patient if the serum cholinesterase activity were only 10 units of activity per liter rather than the normal activity of about 80 units?

  3. Some patients have a mutant form of the serum cholinesterase that displays a KM of 10 mM, rather than the normal 1.4 mM. What will be the effect of this mutation on the patient?

Data Interpretation Problems

Question 8.32

A natural attraction, but more complicated. You have isolated two versions of the same enzyme, a wild type and a mutant differing from the wild type at a single amino acid. Working carefully but expeditiously, you then establish the following kinetic characteristics of the enzymes.

 

Maximum velocity

KM

Wild type

100 μmol/min

10 mM

Mutant

    1 μmol/min

0.1 mM

  1. With the assumption that the reaction occurs in two steps in which k−1 is much larger than k2, which enzyme has the higher affinity for substrate?

  2. What is the initial velocity of the reaction catalyzed by the wild-type enzyme when the substrate concentration is 10 mM?

  3. Which enzyme alters the equilibrium more in the direction of product?

Question 8.33

KM matters. The amino acid asparagine is required by cancer cells to proliferate. Treating patients with the enzyme asparaginase is sometimes used as a chemotherapy treatment. Asparaginase hydrolyzes asparagine to aspartate and ammonia. The adjoining illustration shows the Michaelis–Menten curves for two asparaginases from different sources, as well as the concentration of asparagine in the environment (indicated by the arrow). Which enzyme would make a better chemotherapeutic agent?

Question 8.34

Enzyme specificity. Catalysis of the cleavage of peptide bonds in small peptides by a proteolytic enzyme is described in the following table.

Substrate

KM (mM)

kcat(s−1)

EMTA↓G

4.0

24

EMTA↓A

1.5

30

EMTA↓F

0.5

18

The arrow indicates the peptide bond cleaved in each case.

  1. If a mixture of these peptides were presented to the enzyme with the concentration of each peptide being the same, which peptide would be digested most rapidly? Most slowly? Briefly explain your reasoning, if any.

  2. The experiment is performed again on another peptide with the following results.

    On the basis of these data, suggest the features of the amino acid sequence that dictate the specificity of the enzyme.

Question 8.35

Varying the enzyme. For a one-substrate, enzyme-catalyzed reaction, double-reciprocal plots were determined for three different enzyme concentrations. Which of the following three families of curve would you expect to be obtained? Explain.

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Question 8.36

Mental experiment. Picture in your mind the velocity vs. substrate concentration curve for a typical Michaelis-Menten enzyme. Now, imagine that the experimental conditions are altered as described below. For each of the conditions described, fill in the table indicating precisely (when possible) the effect on Vmax and KM of the imagined Michaelis-Menten enzyme.

Experimental condition

Vmax

KM

a. Twice as much enzyme is used.

 

 

b. Half as much enzyme is used

 

 

c. A competitive inhibitor is present.

 

 

d. An uncompetitive inhibitor is present.

 

 

e. A pure non-competitive inhibitor is present.

 

 

Question 8.37

Too much of a good thing. A simple Michaelis–Menten enzyme, in the absence of any inhibitor, displayed the following kinetic behavior.

  1. Draw a double-reciprocal plot that corresponds to the velocity-versus-substrate curve.

  2. Suggest a plausible explanation for these kinetic results.

Question 8.38

Rate-limiting step. In the conversion of A into D in the following biochemical pathway, enzymes EA, EB, and EC have the KM values indicated under each enzyme. If all of the substrates and products are present at a concentration of 10−4 M and the enzymes have approximately the same Vmax, which step will be rate limiting and why?

Question 8.39

Colored luminosity Tryptophan synthetase, a bacterial enzyme that contains a pyridoxal phosphate (PLP) prosthetic group, catalyzes the synthesis of l-tryptophan from l-serine and an indole derivative. The addition of l-serine to the enzyme produces a marked increase in the fluorescence of the PLP group, as the adjoining graph shows. The subsequent addition of indole, the second substrate, reduces this fluorescence to a level even lower than that produced by the enzyme alone. How do these changes in fluorescence support the notion that the enzyme interacts directly with its substrates?

Chapter Integration Problems

Question 8.40

Titration experiment. The effect of pH on the activity of an enzyme was examined. At its active site, the enzyme has an ionizable group that must be negatively charged for substrate binding and catalysis to take place. The ionizable group has a pKa of 6.0. The substrate is positively charged throughout the pH range of the experiment.

  1. Draw the V0-versus-pH curve when the substrate concentration is much greater than the enzyme KM.

  2. Draw the V0-versus-pH curve when the substrate concentration is much less than the enzyme KM.

  3. At which pH will the velocity equal one-half of the maximal velocity attainable under these conditions?

Question 8.41

A question of stability. Pyridoxal phosphate (PLP) is a coenzyme for the enzyme ornithine aminotransferase. The enzyme was purified from cells grown in PLP-deficient media as well as from cells grown in media that contained pyridoxal phosphate. The stability of the two different enzyme preparations was then measured by incubating the enzyme at 37°C for different lengths of time and then assaying for the amount of enzyme activity remaining. The following results were obtained.

250

  1. Why does the amount of active enzyme decrease with the time of incubation?

  2. Why does the amount of enzyme from the PLP-deficient cells decline more rapidly?

Question 8.42

Not just for enzymes. Kinetics is useful for studying reactions of all types, not just those catalyzed by enzymes. In Chapters 4 and 5, we learned that DNA could be reversibly melted. When melted double-stranded DNA is allowed to renature, the process can be described as consisting of two steps, a slow second order reaction followed by a rapid first order reaction. Explain what is occurring in each step.