PROBLEMS

Question 7.1

1.  Different orders. Differentiate between a first-order rate constant and a second-order rate constant. ✓ 3

Question 7.2

2.  A fraudulent reaction? What is a pseudo-first-order reaction? ✓ 3

Question 7.3

3.  Changing in concert. On the graph at the right, show how substrate and product concentrations of the simple enzyme-catalyzed reaction



change with time. ✓ 3

Unnumbered Figure

(a) For a reaction in which the equilibrium is in favor of product formation.

(b) For a reaction in which the equilibrium constant is 1.

Question 7.4

4.  Active yet responsive. What is the biochemical advantage of having a KM approximately equal to the substrate concentration normally available to an enzyme? ✓ 4

Question 7.5

5.  Defining attributes. What is the defining characteristic for an enzyme catalyzing a sequential reaction? A double-displacement reaction? ✓ 4

Question 7.6

6.  Affinity or not affinity? That is the question. The affinity between a protein and a molecule that binds to the protein is frequently measured as a dissociation constant KD.



Does KM measure the affinity of the enzyme complex? Under what circumstances might KM approximately equal KD? ✓ 4

Question 7.7

7.  Destroying the Trojan horse. Penicillin is hydrolyzed and thereby rendered inactive by penicillinase (also known as β-lactamase), an enzyme present in some 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. ✓ 4

[Penicillin] μM

Amount hydrolyzed (nanomoles)

 1

0.11

 3

0.25

 5

0.34

10

0.45

30

0.58

50

0.61

(a) 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?

(b) What is the value of Vmax?

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

Destroying the Trojan horse. Penicillin is hydrolyzed and thereby rendered inactive by penicillinase (also known as β -lactamase), an enzyme present in some 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. ✓ 4 (a) Plot V 0 versus [S] and 1/ V 0 versus 1/[S] for these data. Does penicillinase appear to obey Michaelis–Menten kinetics? If so, what is the value of K M? (b) What is the value of V max? (c) What is the turnover number of penicillinase under these experimental conditions? Assume one active site per enzyme molecule.

Question 7.8

8.  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 Escherichia 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. ✓ 4

(a) How many moles of substrate are hydrolyzed per second per milligram of enzyme when the substrate concentration is much greater than KM?

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

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

Question 7.9

9.  A fresh view. The plot of 1/V0 versus 1/[S] is sometimes called a double-reciprocal or 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. ✓ 4

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

(b) What is the significance of the slope, the vertical intercept, and the horizontal intercept in a plot of V0 versus V0/[S]?

Question 7.10

10.  More Michaelis–Menten. For an enzyme that follows simple Michaelis–Menten kinetics, what is the value of Vmax if V0 is equal to 1 mmol minute−1 when [S] = 1/10 KM? ✓ 4

Question 7.11

11.  Fractional occupation. Starting with the Michaelis–Menten equation, derive an equation that shows the fraction of active sites filled (fES) at any V0. ✓ 4

Question 7.12

12.  They go together like spaghetti and meatballs. Match the term with the proper description.

Enzyme
Kinetics
Michaelis–Menten equation
Michaelis constant (KM)
Lineweaver–Burk equation
Turnover number
kcat/KM
Sequential reaction
Double-displacement (ping-pong) reaction
Allosteric enzyme
Double-reciprocal plot
The study of reaction rates
Protein catalyst
A ternary complex is formed
Describes the kinetics of simple one-substrate reactions
k2 or kcat
Substrate concentrationthat yields 1/2Vmax
Responds to environmental signals
A measure of enzyme efficiency
Includes a substituted enzyme intermediate

Question 7.13

13.  A new view of cooperativity. Draw a double-reciprocal plot for a typical Michaelis–Menten enzyme and an allosteric enzyme that have the same Vmax. ✓ 5

Question 7.14

14.  Angry biochemists. Many biochemists go bananas, and justifiably, when they see a Michaelis–Menten plot like the one shown below. To see why they go bananas, determine the V0 as a fraction of Vmax when the substrate concentration is equal to 10 KM and 20 KM. Please control your outrage. ✓ 4

Unnumbered Figure

Question 7.15

15.  Knowing when to say when. What is feedback inhibition? Why is it a useful property? ✓ 5

Question 7.16

16.  Turned upside down. An allosteric enzyme that follows the concerted mechanism has a T/R ratio of 300 in the absence of substrate. Suppose that a mutation reversed the ratio. How would this mutation affect the relation between the velocity of the reaction and the substrate concentration? ✓ 5

Page 128

Question 7.17

17.  RT equilibrium. Differentiate between homotropic and heterotropic effectors. ✓ 5

Question 7.18

18.  Restoration project. Aspartate transcarbamoylase (ATCase) is an allosteric enzyme that regulates the synthesis of uridine triphosphate (UTP) and cytidine triphosphate (CTP). It can be separated into regulatory subunits and catalytic subunits. If isolated regulatory subunits and catalytic subunits of ATCase are mixed, the native enzyme is reconstituted. What is the biological significance of the observation? ✓ 5

Question 7.19

19.  Negative cooperativity. You have isolated a dimeric enzyme that contains two identical active sites. The binding of substrate to one active site decreases the substrate affinity of the other active site. Can the concerted model account for this negative cooperativity? ✓ 5

Data Interpretation Problems

Question 7.20

20.  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. ✓ 4

 

Maximum velocity

KM

Wild type

100 μmol/min

10 mM

Mutant

1 μmol/min

0.1 mM

(a) With the assumption that a two-step reaction in which k−1 is much larger than k2, which enzyme has the higher affinity for substrate?

(b) What is the initial velocity of the reaction catalyzed by the wild-type enzyme when the substrate concentration is 10 mM?

(c) Which enzyme alters the equilibrium more in the direction of product?

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. ✓ 4 (a) With the assumption that a two-step reaction in which k −1 is much larger than k 2, which enzyme has the higher affinity for substrate? (b) What is the initial velocity of the reaction catalyzed by the wild-type enzyme when the substrate concentration is 10 mM? (c) Which enzyme alters the equilibrium more in the direction of product?

Question 7.21

21.  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? ✓ 4

Unnumbered Figure

Question 7.22

22.  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.

Unnumbered Figure

Question 7.23

23.  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? ✓ 4

Question 7.24

24.  Enzyme specificity. Catalysis of the cleavage of peptide bonds in small peptides by a proteolytic enzyme is described in the following table. ✓ 4

Substrate*

KM (mM)

kcat (s−1)

EMTA↓G

4.0

24

EMTA↓A

1.5

30

EMTA↓F

0.5

18

*See Chapter 2 for amino acid abbreviations.

The arrow indicates the peptide bond cleaved in each case.

(a) 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 effectively? Least effectively? Briefly explain your reasoning, if any.

(b) The experiment is performed again on another peptide with the below results:

EMTI↓ F

9

18

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

Enzyme specificity. Catalysis of the cleavage of peptide bonds in small peptides by a proteolytic enzyme is described in the following table. ✓ 4 The arrow indicates the peptide bond cleaved in each case. (a) 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 effectively? Least effectively? Briefly explain your reasoning, if any. (b) The experiment is performed again on another peptide with the below results: On the basis of these data, suggest the features of the amino acid sequence that dictate the specificity of enzyme.

Challenge Problems

Question 7.25

25.  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? ✓ 4

Question 7.26

26.  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? ✓ 5

Unnumbered Figure

Question 7.27

27.  Too much of a good thing. A simple Michaelis–Menten enzyme, in the absence of any inhibitor, displayed the following kinetic behavior. The expected value of Vmax is shown on the y axis. ✓ 4 ✓ 5

Unnumbered Figure

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

(b) Provide an explanation for the kinetic results.

Question 7.28

28.  Paradoxical at first glance. Phosphonacetyl-L-aspartate (PALA) is a potent inhibitor of the allosteric enzyme aspartate transcarbamoylase (ATCase), which has six active sites, because PALA mimics the two physiological substrates. ATCase controls the synthesis of pyrimidine nucleotides. However, low concentrations of this unreactive bisubstrate analog increase the reaction velocity. On the addition of PALA, the reaction velocity increases until an average of three molecules of PALA are bound per molecule of enzyme. This maximal velocity is 17-fold greater than it is in the absence of PALA. The reaction velocity then decreases to nearly zero on the addition of three more molecules of PALA per molecule of enzyme. Why do low concentrations of PALA activate ATCase? ✓ 5

Unnumbered Figure

Question 7.29

29.  Distinguishing between models. The following graph shows the fraction of an allosteric enzyme in the R state (fR) and the fraction of active sites bound to substrate (Y) as a function of substrate concentration. Which model, the concerted or sequential, best explains these results? ✓ 5

Unnumbered Figure

Question 7.30

30.  Reporting live from ATCase 1. The allosteric enzyme aspartate transcarbamoylase (ATCase) has six active sites, arranged as two catalytic trimers. ATCase was modified with tetranitromethane to form a colored nitrotyrosine group (λmax = 430 nm) in each of its catalytic chains. The absorption by this reporter group depends on its immediate environment. An essential lysine residue at each catalytic site also was modified to block the binding of substrate. Catalytic trimers from this doubly modified enzyme were then combined with native trimers to form a hybrid enzyme. The absorption by the nitrotyrosine group was measured on addition of the substrate analog succinate. What is the significance of the alteration in the absorbance at 430 nm? ✓ 5

Unnumbered Figure

Page 130

Question 7.31

31.  Reporting live from ATCase 2. A different ATCase hybrid was constructed to test the effects of allosteric activators and inhibitors. Normal regulatory subunits were combined with nitrotyrosine-containing catalytic subunits. The addition of ATP, an allosteric activator of ATCase, in the absence of substrate increased the absorbance at 430 nm, the same change elicited by the addition of succinate (see the graph in problem 30). Conversely, CTP, an allosteric inhibitor, in the absence of substrate decreased the absorbance at 430 nm. What is the significance of the changes in absorption of the reporter groups? ✓ 5

Unnumbered Figure

Selected Readings for this chapter can be found online at www.whfreeman.com/tymoczko3e.

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