Chapter Review in Chapter 6 Enzymes

Chapter Review

KEY TERMS

Terms in bold are defined in the glossary.

enzyme
cofactor
coenzyme
prosthetic group
holoenzyme
apoenzyme
apoprotein
active site
substrate
ground state
standard free-energy change, $mathml alt text636$ $bold upper Delta bold-italic upper G Superscript bold degree$
biochemical standard free-energy change, $mathml alt text637$ $bold upper Delta bold-italic upper G bold prime degree$
transition state
activation energy $mathml alt text638$ $left-parenthesis bold upper Delta bold-italic upper G Superscript bold double-dagger Baseline right-parenthesis$
reaction intermediate
rate-limiting step
equilibrium constant $mathml alt text639$ $left-parenthesis bold-italic upper K Subscript bold eq Baseline right-parenthesis$
rate constant
binding energy $mathml alt text640$ $left-parenthesis bold upper Delta bold-italic upper G Subscript bold upper B Baseline right-parenthesis$
specificity
desolvation
induced fit
specific acid-base catalysis
general acid-base catalysis
covalent catalysis
enzyme kinetics
pre–steady state
steady state
steady-state kinetics
$V_{0}$ $bold-italic upper V bold 0$
$V_{max}$ $bold-italic upper V Subscript bold max$
Michaelis-Menten equation
steady-state assumption
Michaelis constant $mathml alt text643$ $left-parenthesis bold-italic upper K Subscript bold m Baseline right-parenthesis$
Lineweaver-Burk equation
Michaelis-Menten kinetics
dissociation constant $mathml alt text644$ $left-parenthesis bold-italic upper K Subscript bold d Baseline right-parenthesis$
$k_{cat}$ $k Subscript cat$
turnover number
Cleland nomenclature
reversible inhibition
competitive inhibition
uncompetitive inhibition
mixed inhibition
noncompetitive inhibition
irreversible inhibitors
suicide inactivator
transition-state analog
serine proteases
retrovirus
regulatory enzyme
allosteric enzyme
allosteric modulator (allosteric effector)
regulatory protein
homotropic
heterotropic
protein kinases
protein phosphatases
zymogen
proproteins (proenzymes)
regulatory cascade
platelets
fibrin
fibrinogen
thromboxane
thrombin
intrinsic pathway
extrinsic pathway

Problems

1. Keeping the Sweet Taste of Corn The sweet taste of freshly picked corn (maize) is due to the high level of sugar in the kernels. Store-bought corn (several days after picking) is not as sweet, because about 50% of the free sugar is converted to starch within one day of picking. To preserve the sweetness of fresh corn, the husked ears can be immersed in boiling water for a few minutes (“blanched”), then cooled in cold water. Corn processed in this way and stored in a freezer maintains its sweetness. What is the biochemical basis for this procedure?
2. Intracellular Concentration of Enzymes To approximate the concentration of enzymes in a bacterial cell, assume that the cell contains equal concentrations of 1,000 different enzymes in solution in the cytosol and that each protein has a molecular weight of 100,000. Assume also that the bacterial cell is a cylinder (diameter 1.0 μm, height 2.0 μm), that the cytosol (specific gravity 1.20) is 20% soluble protein by weight, and that the soluble protein consists entirely of enzymes. Calculate the average molar concentration of each enzyme in this hypothetical cell.
3. Rate Enhancement by Urease The enzyme urease enhances the rate of urea hydrolysis at pH 8.0 and $mathml alt text646$ $20 Superscript degree Baseline upper C$ by a factor of $mathml alt text647$ $10 Superscript 14$ . Suppose that a given quantity of urease can completely hydrolyze a given quantity of urea in 5.0 min at $mathml alt text648$ $20 Superscript degree Baseline upper C$ and pH 8.0. How long would it take for this amount of urea to be hydrolyzed under the same conditions in the absence of urease? Assume that both reactions take place in sterile systems so that bacteria cannot attack the urea.
4. Protection of an Enzyme against Denaturation by Heat When enzyme solutions are heated, there is a progressive loss of catalytic activity over time due to denaturation of the enzyme. A solution of the enzyme hexokinase incubated at $mathml alt text649$ $45 Superscript degree Baseline upper C$ lost 50% of its activity in 12 min, but when incubated at $mathml alt text650$ $45 Superscript degree Baseline upper C$ in the presence of a very large concentration of one of its substrates, it lost only 3% of its activity in 12 min. Suggest why thermal denaturation of hexokinase was retarded in the presence of one of its substrates.
5. Quantitative Assay for Lactate Dehydrogenase The muscle enzyme lactate dehydrogenase catalyzes the reaction

Pyruvate has a central carbon double bonded to O above, single bonded to C O O minus on the right, and bonded to C H 3 on the left. Pyruvate plus N A D H plus H plus yield lactate plus N A D plus. Lactate has a central carbon bonded to O H above, C O O minus on the right, H below, and C H 3 on the left.

NADH and $mathml alt text651$ $NAD Superscript plus$ are the reduced and oxidized forms, respectively, of the coenzyme NAD. Solutions of NADH, but not $mathml alt text652$ $NAD Superscript plus$ , absorb light at 340 nm. This property is used to determine the concentration of NADH in solution by measuring spectrophotometrically the amount of light absorbed at 340 nm $mathml alt text653$ $left-parenthesis upper A Subscript 340 Baseline right-parenthesis$ by the solution. Explain how these properties of NADH can be used to design a quantitative assay for lactate dehydrogenase.
6. Effect of Enzymes on Reactions Consider this simple reaction:

$S ⇌_{k_{2}}^{k_{1}} P where K_{eq}^{'} = \frac{[P]}{[S]}$ $upper S right harpoon over left harpoon Underscript k Subscript 2 Baseline Overscript k Subscript 1 Baseline Endscripts upper P where upper K prime Subscript eq Baseline equals StartFraction left-bracket upper P right-bracket Over left-bracket upper S right-bracket EndFraction$

Which of the listed effects would be brought about by an enzyme catalyzing the simple reaction?
1. increased $mathml alt text655$ $k 1$
2. increased $mathml alt text656$ $upper K prime Subscript eq$
3. decreased $mathml alt text657$ $upper Delta upper G Superscript double-dagger$
4. more negative $mathml alt text658$ $upper Delta upper G prime degree$
5. increased $mathml alt text659$ $k 2$
7. Relation between Reaction Velocity and Substrate Concentration: Michaelis-Menten Equation The $mathml alt text660$ $upper K Subscript m$ of an enzyme is 5.0 mm.
1. Calculate the substrate concentration when this enzyme operates at one-quarter of its maximum rate.
2. Determine the fraction of $mathml alt text661$ $upper V Subscript max$ that would be obtained when the substrate concentration, [S], is $mathml alt text662$ $0.5 upper K Subscript m Baseline$ , $mathml alt text663$ $2 upper K Subscript m Baseline$ , and 10 $mathml alt text664$ $upper K Subscript m$ .
3. An enzyme that catalyzes the reaction $mathml alt text665$ $upper X right harpoon over left harpoon upper Y$ is isolated from two bacterial species. The enzymes have the same $mathml alt text666$ $upper V Subscript max$ but different $mathml alt text667$ $upper K Subscript m$ values for the substrate X. Enzyme A has a $mathml alt text668$ $upper K Subscript m$ of 2.0 μm, and enzyme B has a $mathml alt text669$ $upper K Subscript m$ of 0.5 μm. Kinetic experiments used the same concentration of each enzyme and 1 μm substrate X. The graph plots the concentration of product Y formed over time. Which curve corresponds to which enzyme?
  
  Both curves begin at the origin. The red curve increases almost diagonally upward and begins to slow at about two-thirds of the height of the vertical axis shortly before the midpoint on the vertical axis. It then levels off near the top of the vertical axis to become almost horizontal for most of the right side of the graph. The black curve begins almost diagonally beneath the first curve and increases much more slowly, ending at about three-quarters of the height of the vertical axis on the right of the graph. It does not show a dramatic slowdown and leveling off like the first graph. Instead, it begins to slow slightly on the right of the graph.
8. Applying the Michaelis-Menten Equation I An enzyme has a $mathml alt text670$ $upper V Subscript max$ of $mathml alt text671$ $1.2 mu upper M s Superscript negative 1$ . The $mathml alt text672$ $upper K Subscript m$ for its substrate is 10 μm. Calculate the initial velocity of the reaction, $mathml alt text673$ $upper V 0$ , when the substrate concentration is
1. 2 μm
2. 10 μm
3. 30 μm.
9. Applying the Michaelis-Menten Equation II An enzyme is present at a concentration of 1 nm and has a $mathml alt text674$ $upper V Subscript max$ of $mathml alt text675$ $2 mu upper M s Superscript negative 1$ . The $mathml alt text676$ $upper K Subscript m$ for its primary substrate is 4 μm.
1. Calculate $mathml alt text677$ $k Subscript cat$ .
2. Calculate the apparent (measured) $mathml alt text678$ $upper V Subscript max$ and apparent (measured) $mathml alt text679$ $upper K Subscript m$ of this enzyme in the presence of sufficient amounts of an uncompetitive inhibitor to generate an $mathml alt text680$ $alpha prime$ of 2. Assume that the enzyme concentration remains at 1 nm.
10. Applying the Michaelis-Menten Equation III A research group discovers a new version of happyase, which they call happyase*, that catalyzes the chemical reaction $mathml alt text681$ $HAPPY right harpoon over left harpoon SAD$ . The researchers begin to characterize the enzyme.
1. In the first experiment, with $mathml alt text682$ $left-bracket upper E Subscript t Baseline right-bracket$ at 4 nm, they find that the $mathml alt text683$ $upper V Subscript max$ is $mathml alt text684$ $1.6 mu upper M s Superscript negative 1$ . Based on this experiment, what is the $mathml alt text685$ $k Subscript cat$ for happyase*? (Include appropriate units.)
2. In the second experiment, with $mathml alt text686$ $left-bracket upper E Subscript t Baseline right-bracket$ at 1 nm and [HAPPY] at 30 μm, the researchers find that $mathml alt text687$ $upper V 0 equals 300 nM s Superscript negative 1$ . What is the measured $mathml alt text688$ $upper K Subscript m$ of happyase* for its substrate HAPPY? (Include appropriate units.)
3. Further research shows that the purified happyase* used in the first two experiments was actually contaminated with a reversible inhibitor called ANGER. When ANGER is carefully removed from the happyase* preparation and the two experiments are repeated, the measured $mathml alt text689$ $upper V Subscript max$ in (a) is increased to $mathml alt text690$ $4.8 mu upper M s Superscript negative 1$ , and the measured $mathml alt text691$ $upper K Subscript m$ in (b) is now 15 μm. Calculate the values of α and $mathml alt text692$ $alpha prime$ for ANGER.
4. Based on the information given, what type of inhibitor is ANGER?
11. Applying the Michaelis-Menten Equation IV Researchers discover an enzyme that catalyzes the reaction $X ⇌ Y$ $upper X right harpoon over left harpoon upper Y$ . They find that the $mathml alt text694$ $upper K Subscript m$ for the substrate X is 4 μm, and the $mathml alt text695$ $k Subscript cat$ is $20 min Overscript negative 1 Endscripts$ $20 min Overscript negative 1 Endscripts$ .
1. In an experiment, $mathml alt text697$ $left-bracket upper X right-bracket equals 6 mM comma$ and $mathml alt text698$ $upper V 0 equals 480 nM min Superscript negative 1$ . What was the $mathml alt text699$ $left-bracket upper E Subscript t Baseline right-bracket$ used in the experiment?
2. In another experiment, $mathml alt text700$ $left-bracket upper E Subscript t Baseline right-bracket equals 0.5 mu upper M$ , and the measured $mathml alt text701$ $upper V 0 equals 5 mu upper M min Overscript negative 1 Endscripts$ . What was the [X] used in the experiment?
3. The researchers discover that compound Z is a very strong competitive inhibitor of the enzyme. In an experiment with the same $mathml alt text702$ $left-bracket upper E Subscript t Baseline right-bracket$ as in (a), but a different [X], they add an amount of Z that produces an α of 10 and reduces $mathml alt text703$ $upper V 0$ to $mathml alt text704$ $240 nM min Superscript negative 1$ . What is the [X] in this experiment?
4. Based on the kinetic parameters given, has this enzyme evolved to achieve catalytic perfection? Explain your answer briefly, using the kinetic parameter(s) that define catalytic perfection.
12. Estimation of $mathml alt text705$ $bold-italic upper V Subscript bold max$ and $mathml alt text706$ $bold-italic upper K Subscript bold m$ by Inspection Graphical methods are available for accurate determination of the $mathml alt text707$ $upper V Subscript max$ and $mathml alt text708$ $upper K Subscript m$ of an enzyme-catalyzed reaction. However, these quantities can sometimes be estimated by inspecting values of $mathml alt text709$ $upper V 0$ at increasing [S]. Estimate the $mathml alt text710$ $upper V Subscript max$ and $mathml alt text711$ $upper K Subscript m$ of the enzyme-catalyzed reaction for which the data in the table were obtained.

[S] (m) $mathml alt text712$ $bold-italic upper V bold 0 bold left-parenthesis bold-italic mu bold upper M bold slash min bold right-parenthesis$

$mathml alt text713$ $2.5 times 10 Superscript negative 6$ 28

$mathml alt text714$ $4.0 times 10 Superscript negative 6$ 40

$mathml alt text715$ $1 times 10 Superscript negative 5$ 70

$mathml alt text716$ $2 times 10 Superscript negative 5$ 95

$mathml alt text717$ $4 times 10 Superscript negative 5$ 112

$mathml alt text718$ $1 times 10 Superscript negative 4$ 128

$mathml alt text719$ $2 times 10 Superscript negative 3$ 139

$mathml alt text720$ $1 times 10 Superscript negative 2$ 140

[S] (m)	$mathml alt text712$ $bold-italic upper V bold 0 bold left-parenthesis bold-italic mu bold upper M bold slash min bold right-parenthesis$
$mathml alt text713$ $2.5 times 10 Superscript negative 6$	28
$mathml alt text714$ $4.0 times 10 Superscript negative 6$	40
$mathml alt text715$ $1 times 10 Superscript negative 5$	70
$mathml alt text716$ $2 times 10 Superscript negative 5$	95
$mathml alt text717$ $4 times 10 Superscript negative 5$	112
$mathml alt text718$ $1 times 10 Superscript negative 4$	128
$mathml alt text719$ $2 times 10 Superscript negative 3$	139
$mathml alt text720$ $1 times 10 Superscript negative 2$	140

13. Properties of an Enzyme of Prostaglandin Synthesis Prostaglandins are one class of the fatty acid derivatives called eicosanoids. Prostaglandins produce fever and inflammation, as well as the pain associated with inflammation. The enzyme prostaglandin endoperoxide synthase, a cyclooxygenase, uses oxygen to convert arachidonic acid to $mathml alt text721$ $PGG Subscript 2$ , the immediate precursor of many different prostaglandins (prostaglandin synthesis is described in Chapter 21).

Ibuprofen inhibits prostaglandin endoperoxide synthase, thereby reducing inflammation and pain. The kinetic data given in the table are for the reaction catalyzed by prostaglandin endoperoxide synthase in the absence and presence of ibuprofen.

Based on the data, determine the

mathml alt text722

$upper V Subscript max$ and

mathml alt text723

$upper K Subscript m$ of the enzyme.

[Arachidonic acid] (mm)	Rate of formation of $bold PGG Subscript bold 2$ $bold PGG Subscript bold 2$ $mathml alt text724$ $bold left-parenthesis bold m bold upper M bold min Superscript negative bold 1 Baseline bold right-parenthesis$	Rate of formation of $mathml alt text725$ $bold PGG Subscript bold 2$ with 10 mg/mL ibuprofen $mathml alt text726$ $bold left-parenthesis bold m bold upper M bold min Superscript negative bold 1 Baseline bold right-parenthesis$
0.5	23.5	16.67
1.0	32.2	25.25
1.5	36.9	30.49
2.5	41.8	37.04
3.5	44.0	38.91

Based on the data, determine the type of inhibition that ibuprofen exerts on prostaglandin endoperoxide synthase.

14. Graphical Analysis of $mathml alt text727$ $bold-italic upper V Subscript bold max$ and $mathml alt text728$ $bold-italic upper K Subscript bold m$ A kinetic study of an intestinal peptidase using glycylglycine as the substrate produced the experimental data shown in the table. The peptidase catalyzes this reaction:

$Glycylglycine + H_{2} O \to 2 glycine$ $Glycylglycine plus upper H Subscript 2 Baseline upper O right-arrow 2 glycine$

[S] (mm) Product formed (μmol/min)

1.5 0.21

2.0 0.24

3.0 0.28

4.0 0.33

8.0 0.40

16.0 0.45

Use the Lineweaver-Burk equation to determine the $mathml alt text730$ $upper V Subscript max$ and $mathml alt text731$ $upper K Subscript m$ for this enzyme preparation and substrate.
15. The Eadie-Hofstee Equation There are several ways to transform the Michaelis-Menten equation so as to plot data and derive kinetic parameters, each with different advantages depending on the data set being analyzed. One transformation of the Michaelis-Menten equation is the Lineweaver-Burk, or double-reciprocal, equation. Multiplying both sides of the Lineweaver-Burk equation by $mathml alt text732$ $upper V Subscript max$ and rearranging gives the Eadie-Hofstee equation:

$V_{0} = (- K_{m}) \frac{V_{0}}{[S]} + V_{max}$ $upper V 0 equals left-parenthesis minus upper K Subscript m Baseline right-parenthesis StartFraction upper V 0 Over left-bracket upper S right-bracket EndFraction plus upper V Subscript max$

Consider the plot of $mathml alt text734$ $upper V 0$ versus $mathml alt text735$ $upper V 0 slash left-bracket upper S right-bracket$ for an enzyme-catalyzed reaction. The slope of the line is $mathml alt text736$ $minus upper K Subscript m$ . The x intercept is $mathml alt text737$ $upper V Subscript max Baseline slash upper K Subscript m$ . The control reactions (the blue line in the plot) did not contain any inhibitor.
1. Which of the other lines (A, B, or C) depicts this enzyme’s activity in the presence of a competitive inhibitor? Hint: See Equation 6-33.
2. Which line (A, B, or C) depicts this enzyme’s activity in the presence of an uncompetitive inhibitor?
  
  A dotted horizontal line labeled V subscript max runs horizontally just below the top of the graph. A highlighted line begins where the dotted horizontal V subscript max line meets the vertical axis and runs to meet the horizontal axis near the right end of the graph. This point is labeled V subscript max over K subscript lowercase M. Text reads, control slope equals negative K subscript lowercase M. Line A begins where the dotted horizontal V subscript max line meets the vertical axis and forms a diagonal line that decreases to meet the horizontal axis at a point about one-third of the distance across the horizontal axis. Line B begins just below the midpoint of the horizontal axis at the left side of the graph and forms a diagonal line down to the point where the blue line ends that is labeled V subscript max over K subscript lowercase M. Line C begins at the same point as line A and the blue line, where V subscript max meets the vertical axis, and descends diagonally to end at the right side of the horizontal axis at about half the height of the vertical axis. All data are approximate.
16. The Turnover Number of Carbonic Anhydrase Carbonic anhydrase of erythrocytes $mathml alt text738$ $left-parenthesis upper M Subscript r Baseline 30,000 right-parenthesis$ has one of the highest turnover numbers known. It catalyzes the reversible hydration of $mathml alt text739$ $CO Subscript 2$ :

$H_{2} O + {CO}_{2} ⇌ H_{2} {CO}_{3}$ $upper H Subscript 2 Baseline upper O plus CO Subscript 2 Baseline right harpoon over left harpoon upper H Subscript 2 Baseline CO Subscript 3 Baseline$

This is an important process in the transport of $mathml alt text741$ $CO Subscript 2$ from the tissues to the lungs. If $mathml alt text742$ $10.0 mu g$ of pure carbonic anhydrase catalyzes the hydration of 0.30 g of $mathml alt text743$ $CO Subscript 2$ in 1 min at $mathml alt text744$ $37 Superscript degree Baseline upper C$ at $mathml alt text745$ $upper V Subscript max$ , what is the turnover number $mathml alt text746$ $left-parenthesis k Subscript cat Baseline right-parenthesis$ of carbonic anhydrase (in units of $mathml alt text747$ $min Overscript negative 1 Endscripts$ )?
17. Describing Reactions with the Cleland Shorthand The chymotrypsin-catalyzed reaction is diagrammed using the Cleland shorthand. Match the letters in the drawing with each description:
1. The product that includes the amino group from the cleaved peptide bond
2. The product that includes the carbonyl group from the cleaved peptide bond
3. Free chymotrypsin (nothing bound to it)
4. Water
5. The peptide substrate
6. The acyl-enzyme intermediate
  
  A horizontal line begins with the label E beneath the first section, then an arrow points down from A to its right. E A dash F P is beneath the next section, then an arrow labeled P points upward. F is beneath the next region, then an arrow labeled B points down. F B dash E Q is beneath the next region, then an arrow labeled Q points upward. E is beneath the end of the line.
18. Kinetic Inhibition Patterns Indicate how the observed $mathml alt text748$ $upper K Subscript m$ of an enzyme would change in the presence of inhibitors having the given effect on α and $mathml alt text749$ $alpha prime$ :
1. $mathml alt text750$ $alpha greater-than alpha prime$ ; $mathml alt text751$ $alpha prime equals 1.0$
2. $mathml alt text752$ $alpha prime greater-than alpha$
3. $mathml alt text753$ $alpha equals alpha prime$ ; $mathml alt text754$ $alpha prime greater-than 1.0$
4. $mathml alt text755$ $alpha equals alpha prime$ ; $mathml alt text756$ $alpha prime equals 1.0$
19. Deriving a Rate Equation for Competitive Inhibition The Michaelis-Menten rate equation for an enzyme subject to competitive inhibition is

$V_{0} = \frac{V_{max} [S]}{α K_{m} + [S]}$ $upper V 0 equals StartFraction upper V Subscript max Baseline left-bracket upper S right-bracket Over alpha upper K Subscript m Baseline plus left-bracket upper S right-bracket EndFraction$

Beginning with a new definition of total enzyme as

$[E_{t}] = [E] + [ES] + [EI]$ $left-bracket upper E Subscript t Baseline right-bracket equals left-bracket upper E right-bracket plus left-bracket ES right-bracket plus left-bracket EI right-bracket$

and the definitions of α and $mathml alt text759$ $upper K Subscript upper I$ provided in the text, derive the first rate equation. Use the derivation of the Michaelis-Menten equation as a guide.
20. Irreversible Inhibition of an Enzyme Many enzymes are inhibited irreversibly by heavy metal ions such as $mathml alt text760$ $Hg Superscript 2 plus Baseline comma Cu Superscript 2 plus$ , or $mathml alt text761$ $Ag Superscript plus$ , which can react with essential sulfhydryl groups to form mercaptides:

$Enz-SH + {Ag}^{+} \to Enz–S–Ag + H^{+}$ $Enz hyphen SH plus Ag Superscript plus Baseline right-arrow Enz en-dash upper S en-dash Ag plus upper H Superscript plus$

The affinity of $mathml alt text763$ $Ag Superscript plus$ for sulfhydryl groups is so great that $mathml alt text764$ $Ag Superscript plus$ can be used to titrate —SH groups quantitatively. An investigator added just enough $mathml alt text765$ $AgNO Subscript 3$ to completely inactivate a 10.0 mL solution containing 1.0 mg/mL enzyme. A total of 0.342 μmol of $mathml alt text766$ $AgNO Subscript 3$ was required. Calculate the minimum molecular weight of the enzyme. Why does the value obtained in this way give only the minimum molecular weight?
21. Clinical Application of Differential Enzyme Inhibition Human blood serum contains a class of enzymes known as acid phosphatases, which hydrolyze biological phosphate esters under slightly acidic conditions (pH 5.0):

A molecule has P bonded to O minus above, bonded to O minus on the right, double bonded to O below, and bonded to O on the left that is further bonded to R. H 2 O is added. The reaction yields two molecules. These are R bonded to O H and P bonded to O minus above and on the right, double bonded to O below, and bonded to O H on the left.

Acid phosphatases are produced by erythrocytes and by the liver, kidney, spleen, and prostate gland. The enzyme of the prostate gland is clinically important, because its increased activity in the blood can be an indication of prostate cancer. The phosphatase from the prostate gland is strongly inhibited by tartrate ion, but acid phosphatases from other tissues are not. How can this information be used to develop a specific procedure for measuring the activity of prostatic acid phosphatase in human blood serum?
22. Inhibition of Carbonic Anhydrase by Acetazolamide Carbonic anhydrase is strongly inhibited by the drug acetazolamide, which is used as a diuretic (i.e., to increase the production of urine) and to lower excessively high pressure in the eye (due to accumulation of intraocular fluid) in glaucoma. Carbonic anhydrase plays an important role in these and other secretory processes because it participates in regulating the pH and bicarbonate content of several body fluids. Carbonic anhydrase activity can be analyzed using the initial reaction velocity (as percentage of $mathml alt text767$ $upper V Subscript max$ ) versus [S]. The black curve of the graph shows the uninhibited activity; the blue curve shows activity in the presence of acetazolamide. Based on the data provided, determine the nature of the inhibition by acetazolamide. Explain your reasoning.

The graph plots concentration of S in millimolar on the horizontal axis, ranging from 0 to 1 and labeled in increments of 0.2 It plots V as percentage of V subscript max on the vertical axis, ranging from 0 to 100 and labeled in increments of 50. Both curves begin at the origin. The first curve, labeled no inhibitor, rises rapidly at first, then begins to level off around 0.18 and 48 percent. It slows and becomes almost horizontal by the far right side of the graph, where it ends near the top of the vertical axis. The second curve, labeled acetazolamide, begins more slowly than the first curve and begins to level off at about 0.13 and 30 percent. It rises very slowly after 0.2 on the horizontal axis and about 35 percent on the vertical axis and is almost horizontal for the right half of the graph before ending at slightly under 50 percent on the vertical axis. All data are approximate.
23. The Effects of Reversible Inhibitors The Michaelis-Menten rate equation for reversible mixed inhibition is written as

$V_{0} = \frac{V_{max} [S]}{α K_{m} + α^{'} [S]}$ $upper V 0 equals StartFraction upper V Subscript max Baseline left-bracket upper S right-bracket Over alpha upper K Subscript m Baseline plus alpha prime left-bracket upper S right-bracket EndFraction$

Apparent, or observed, $mathml alt text769$ $upper K Subscript m$ is equivalent to the [S] at which

$V_{0} = \frac{V_{max}}{2 α^{'}}$ $upper V 0 equals StartFraction upper V Subscript max Baseline Over 2 alpha prime EndFraction$

Derive an expression for the effect of a reversible inhibitor on apparent $mathml alt text771$ $upper K Subscript m$ from the previous equation.
24. Perturbed $mathml alt text772$ $bold p bold-italic upper K Subscript bold a Baseline$ Values in Enzyme Active Sites Alanine racemase is a bacterial enzyme that converts l-alanine to d-alanine, which is needed in small amounts to synthesize the bacterial cell wall. The active site of alanine racemase includes a Tyr residue with a $mathml alt text773$ $p upper K Subscript a Baseline$ value of 7.2. The $mathml alt text774$ $p upper K Subscript a Baseline$ of free tyrosine is 10. The altered $mathml alt text775$ $p upper K Subscript a Baseline$ of this residue is due largely to the presence of a nearby charged amino acid residue. Which amino acid(s) could lower the $mathml alt text776$ $p upper K Subscript a Baseline$ of the neighboring Tyr residue? Explain your reasoning.
25. pH Optimum of Lysozyme The enzyme lysozyme hydrolyzes glycosidic bonds in peptidoglycan, an oligosaccharide found in bacterial cell walls. The active site of lysozyme contains two amino acid residues essential for catalysis: $mathml alt text777$ $Glu Superscript 35$ and $mathml alt text778$ $Asp Superscript 52$ . The $mathml alt text779$ $p upper K Subscript a Baseline$ values of the carboxyl side chains of these residues are 5.9 and 4.5, respectively. What is the ionization state (protonated or deprotonated) of each residue at pH 5.2, the pH optimum of lysozyme? How can the ionization states of these residues explain the pH-activity profile of lysozyme shown?

The horizontal axis shows p H ranging from 2 to 10, labeled in increments of 2. The vertical axis shows activity as a percentage of maximal, ranging from 0 to 100 and labeled in increments of 50. The curve begins at (2, 1), increases in almost a bell-shaped curve to reach a maximum at (5.5, 100), and then descends slightly more gradually to have a longer tail on the right that ends at (10, 5). All data are approximate.

[S] (mm)	Product formed (μmol/min)
1.5	0.21
2.0	0.24
3.0	0.28
4.0	0.33
8.0	0.40
16.0	0.45

DATA ANALYSIS PROBLEM

26. Mirror-Image Enzymes As noted in Chapter 3, “The amino acid residues in protein molecules are almost all l stereoisomers.” It is not clear whether this selectivity is necessary for proper protein function or is an accident of evolution. To explore this question, Milton and colleagues (1992) studied an enzyme made entirely of d stereoisomers. The enzyme they chose was HIV protease, the proteolytic enzyme made by HIV that converts inactive viral pre-proteins to their active forms as described earlier in Figure 6-28.

Previously, Wlodawer and coworkers (1989) had reported the complete chemical synthesis of HIV protease from l-amino acids (the l-enzyme), using the process shown in Figure 3-30. Normal HIV protease contains two Cys residues at positions 67 and 95. Because chemical synthesis of proteins containing Cys is technically difficult, Wlodawer and colleagues substituted the synthetic amino acid l-α-amino-n-butyric acid (Aba) for the two Cys residues in the protein. They did this to “reduce synthetic difficulties associated with Cys deprotection and ease product handling.”
1. The structure of Aba is shown below.
  
  Why was this a suitable substitution for a Cys residue?
  
  In their study, Milton and coworkers synthesized HIV protease from d-amino acids, using the same protocol as the earlier study (Wlodawer et al.). Formally, there are three possible outcomes for the folding of the d-protease: (1) the same shape as the l-protease; (2) the mirror image of the l-protease, or (3) something else, possibly inactive.
2. For each possibility, decide whether or not it is a likely outcome and defend your position.
  In fact, the d-protease was active: it cleaved a particular synthetic substrate and was inhibited by specific inhibitors. To examine the structure of the d- and l-enzymes, Milton and coworkers tested both forms for activity with d and l forms of a chiral peptide substrate and for inhibition by d and l forms of a chiral peptide-analog inhibitor. Both forms were also tested for inhibition by the achiral inhibitor Evans blue. The findings are given in the table.
  
  Inhibition
  
  Substrate hydrolysis Peptide inhibitor
  
  HIV protease d form l form d form l form
  
  l form − + − +
  
  d form + − + −
3. Which of the three models proposed above is supported by these data? Explain your reasoning.
4. Would you expect chymotrypsin to digest the d-protease? Explain your reasoning.
5. Would you expect total synthesis from d-amino acids followed by renaturation to yield active enzyme for any enzyme? Explain your reasoning.

	Inhibition
	Substrate hydrolysis	Peptide inhibitor
HIV protease	d form	l form	d form	l form
l form	−	+	−	+
d form	+	−	+	−

References

Milton, R.C., Milton, S.C., and Kent, S.B. 1992. Total chemical synthesis of a d-enzyme: the enantiomers of HIV-1 protease show demonstration of reciprocal chiral substrate specificity. Science 256, 1445–1448.
Wlodawer, A., Miller, M., Jaskólski, M., Sathyanarayana, B.K., Baldwin, E., Weber, I.T., Selk, L.M., Clawson, L., Schneider, J., and Kent, S.B. 1989. Conserved folding in retroviral proteases: crystal structure of a synthetic HIV-1 protease. Science 245, 616–621.