Chapter 8. Enzymes: Basic Concepts and Kinetics

Question 1

Question 8.1

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

Answer 1

1. Rate enhancement and substrate specificity.

Question 2

Question 8.2

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

Answer 2

2. A cofactor.

Question 3

Question 8.3

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

Answer 3

3. Coenzymes and metals.

Question 4

Question 8.4

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

Answer 4

4. Vitamins are converted into coenzymes.

Question 5

Question 8.5

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

Answer 5

5. Enzymes facilitate the formation of the transition state.

Question 6

Question 8.6

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

Answer 6

6. The intricate three-dimensional structure of proteins allows the construction of active sites that will recognize only specific substrates.

Question 7

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?

Answer 7

9. Protein hydrolysis has a large activation energy. Protein synthesis must require energy to proceed.

Question 8

Question 8.10

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

Answer 8

10. The enzymes help protect the fluid that surrounds eyes from bacterial infection.

Question 9

Question 8.11

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

Answer 9

11. Binding energy is the free energy released when two molecules bind together, such as when an enzyme and a substrate interact.

Question 10

Question 8.12

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

Answer 10

12. Binding energy is maximized when an enzyme interacts with the transition state, thereby facilitating the formation of the transition state and enhancing the rate of the reaction.

Question 11

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?

Answer 11

13. There would be no catalytic activity. If the enzyme–substrate complex is more stable than the enzyme–transition-state complex, the transition state would not form and catalysis would not take place.

Question 12

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.

Answer 12

14. Transition states are very unstable. Consequently, molecules that resemble transition states are themselves likely to be unstable and, hence, difficult to synthesize.

Question 13

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.

Answer 13

18. The three-dimensional structure of an enzyme is stabilized by interactions with the substrate, reaction intermediates, and products. This stabilization minimizes thermal denaturation.

Question 14

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?

Answer 14

19. At substrate concentrations near the KM, the enzyme displays significant catalysis yet is sensitive to changes in substrate concentration.

Question 15

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.

Answer 15

21. When [S] = 10 KM, V0 = 0.91 Vmax. When [S] = 20 KM, V0 = 95 Vmax.

So any Michaelis–Menten curves showing that the enzyme actually attains Vmax are pernicious lies.

Question 16

Question 8.24

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

Answer 16

24. Penicillinase, like glycopeptide transpeptidase, forms an acyl-enzyme intermediate with its substrate but transfers the intermediate to water rather than to the terminal glycine residue of the pentaglycine bridge.

Question 17

Question 8.27

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

Answer 17

27. Sequential reactions are characterized by the formation of a ternary complex consisting of the enzyme and both substrates. Double-displacement reactions always require the formation of a temporarily substituted enzyme intermediate.

Question 18

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?

Answer 18

29. The mutation slows the reaction by a factor of 100 because the activation free energy is increased by +11.42 kJ mol−1 (+2.73 kcal mol−1). Strong binding of the substrate relative to the transition state slows catalysis.

Question 19

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?

Answer 19

31. (a) This piece of information is necessary for determining the correct dosage of succinylcholine to administer.
    (b) The duration of the paralysis depends on the ability of the serum cholinesterase to clear the drug. If there were one-eighth the amount of enzyme activity, paralysis could last eight times as long.
    (c) KM is the concentration needed by the enzyme to reach ½ Vmax. Consequently, for a given concentration of substrate, the reaction catalyzed by the enzyme with the lower KMwill have the higher rate. The mutant patient with the higher KM will clear the drug at a much lower rate.

Question 20

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?

Answer 20

33. Enzyme 2. Despite the fact that enzyme 1 has a higher Vmax than enzyme 2, enzyme 2 shows greater activity at the concentration of the substrate in the environment because enzyme 2 has a lower KM for the substrate.

Question 21

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.

Answer 21

35. If the total amount of enzyme (ET) is increased, Vmax will increase, because Vmax = k2[E]T. But KM = (k−1 + k2)/k1; that is, it is independent of substrate concentration. The middle graph describes this situation.

Question 22

Answer 22

Question 23

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.

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

Suggest a plausible explanation for these kinetic results.

Answer 23

Question 24

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?

Answer 24

38. The first step will be the rate-limiting step. Enzymes EB and EC are operating at ½ Vmax, whereas the KM for enzyme EA is greater than the substrate concentration. EA would be operating at approximately 10−2 Vmax.

Question 25

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?

Answer 25

39. The fluorescence spectroscopy reveals the existence of an enzyme–serine complex and of an enzyme–serine–indole complex.

Question 26

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.

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

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

Answer 26

41. (a) Incubating the enzyme at 37°C leads to a denaturation of enzyme structure and a loss of activity. For this reason, most enzymes must be kept cool if they are not actively catalyzing their reactions.
    (b) The coenzyme apparently helps to stabilize enzyme structure, because enzyme from PLP-deficient cells denatures faster. Cofactors often help stabilize enzyme structure.

Question 27

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.

Answer 27

42. The slow second order step occurs when complementary sequences on two different molecules bind with one another. Once the initial sequence alignment occurs, the remainder of the molecule can quickly locate complementary sequences from this nucleation site and reanneal, a first order reaction.