Untitled Notes

Home Explore Exams

Daniel

@danisgreat

view

Rating

0.0(0)

Explore Top Notes

Chapter 21 - Reaction, Revolution, and Romanticism, 1815-1850

Note

Studied by 77 people

5.0 Stars(2)

Chapter 6 Language Communication and Belief

Note

Studied by 5 people

4.0 Stars(1)

Chapter 20 - Collapse at the Center

Note

Studied by 80 people

5.0 Stars(1)

Chemistry of Life, Biology

Note

Studied by 45 people

3.3 Stars(3)

European Revolutions- 1830 & 1848

Note

Studied by 44 people

5.0 Stars(3)

Dev - Lecture 13B: Paleozoic Life

Note

Studied by 8 people

5.0 Stars(1)

Untitled

<html><body>2

CHAPTER 8 LEARNING OBJECTIVES

When you have mastered this chapter, you should be able to complete the following objectives. Enzymes Are Powerful and Highly Specific Catalysts (Text Section 8.1)

1. Explain why enzymes are versatile biological catalysts.

2. Appreciate that catalytic power and specificity are critical characteristics of enzymes. Give

examples of the rate enhancements of enzymes and the substrate selectivity they display.

3. Realize that both protein and RNA molecules are enzymes.

4. Define substrate, cofactor, prosthetic group, apoenzyme, and holoenzyme.

5. Provide examples of proteases with diverse substrate specificity, and explain how substrate

specificity arises from precise interactions of the enzyme with the substrate.

6. Provide examples of enzymes that transduce one form of energy into another. Free Energy is Useful Thermodynamic Function for Understanding Enzymes (Text Section 8.2)

7. Describe how DG can be used to predict whether a reaction can occur spontaneously. 8. Write the equation for the DG of a chemical reaction. Define the standard free-energy change (DG∞); define DG„ and DG∞ „. Interconvert kilocalories and kilojoules.

9. Derive the relationship between DG∞ „ and the equilibrium constant (K„eq) of a reaction.

Relate each tenfold change in K„eq to the change in DG∞„ in kilocalories per mole (kcal/mol) or kJ/mol.

10. Relate the concentrations of reactants and products to DG„. Define endergonic and exergonic. 11. Explain why enzymes do not alter the equilibrium of chemical reactions but change only

their rates. Enzymes Accelerate Reactions by Facilitating the Formation of the Transition State (Text Section 8.3)

12. Define the transition state and the free energy of activation (DG‡), and describe the effect

of enzymes on DG‡.

13. Describe the formation of enzyme-substrate (ES) complexes and discuss their properties.

14. Summarize the key features of the active sites of enzymes, and relate them to the speci

ficity of binding of the substrate. The Michaelis-Menten Model Accounts for the Kinetic Properties of Many Enzymes (Text Section 8.4)

15. Outline the Michaelis-Menten model of enzyme kinetics and describe the molecular nature

of each of its components.

16. Reproduce the derivation of the Michaelis-Menten equation in the text. Relate the

Michaelis-Menten equation to experimentally derived plots of velocity (V) versus substrate concentration [S]. List the assumptions underlying the derivation.

ENZYMES: BASIC CONCEPTS AND KINETICS 3

17. Define Vmax and KM, and explain how these parameters can be obtained from a plot of V versus [S] or a plot of 1/V versus 1/[S] (a Lineweaver-Burk plot).

18. Explain the significance of Vmax, KM, k2, kcat, and kcat/KM. Define kinetic perfection as it per

tains to enzyme catalysis.

19. Distinguish sequential displacement and double displacement in reactions involving multi

ple substrates. Provide examples of enzymes using each mechanism.

20. Contrast the kinetics of allosteric enzymes with those displaying simple Michaelis-Menten

kinetics. Describe the molecular basis of allostery. Enzymes Can Be Inhibited by Specific Molecules (Text Section 8.5)

21. Describe the functions and uses of enzyme inhibitors. Contrast reversible and irreversible

inhibitors.

22. Describe the effects of competitive and noncompetitive inhibitors on the kinetics of enzyme

reactions. Apply kinetic measurements and analysis to determine the nature of an inhibitor.

23. Explain how irreversible inhibitors are used to learn about the active sites of enzymes.

Provide examples of group-specific, substrate-analog, suicide, and transition-state inhibitors.

24. Contrast the properties of substrates and transition-state analogues.

25. Describe the formation of catalytic antibodies and recognize their uses.

26. Outline the mechanism of action of the antibiotic penicillin. Vitamins and Coenzymes (Text Section 8.6)

27. Explain the relationship of vitamins to coenzymes.

29. Relate the molecular function of each of the coenzymes.

30. List the water-soluble and fat-soluble vitamins and relate the deficiency of each to a patho

logical condition.

31. Explain the role of ascorbate (Vitamin C) in collagen formation. Outline the post-transla- tional modification of proline to hydroxyproline. SELF-TEST Enzymes Are Powerful and Highly Specific Catalysts

1. Which of the following are not true of enzymes?

(a) Enzymes are proteins. (b) Enzymes have great catalytic power. (c) Enzymes bind substrates with high specificity. (d) Enzymes use hydrophobic interactions exclusively in binding substrates. (e) The catalytic activity of enzymes is often regulated.

2. Enzymes catalyze reactions by

(a) binding regulatory proteins. (b) covalently modifying active-site residues. (c) binding substrates with great affinity. (d) selectively binding the transition state of a reaction with high affinity. 4

CHAPTER 8

3. The combination of an apoenzyme with a cofactor forms what? What are the two types

of cofactors? What distinguishes a prosthetic group from a cosubstrate?

4. Name a process that converts the energy of light into the energy of chemical bonds. Free Energy is Useful Thermodynamic Function for Understanding Enzymes

5. Which of the following statements is correct? The free energy change of a reaction

(a) if negative, enables the reaction to occur spontaneously. (b) if positive, enables the reaction to occur spontaneously. (c) is greater than zero when the reaction is at equilibrium. (d) determines the rate at which a reaction will attain equilibrium.

6. Explain why the thermodynamic parameter DS cannot be used to predict the direction

in which a reaction will proceed.

7. If the standard free-energy change (DG∞) for a reaction is zero, which of the following

statements about the reaction are true?

(a) The entropy (DS∞) of the reaction is zero. (b) The enthalpy (DH∞) of the reaction is zero. (c) The equilibrium constant for the reaction is 1.0. (d) The reaction is at equilibrium. (e) The concentrations of the reactants and products are all 1 M at equilibrium.

8. The enzyme triose phosphate isomerase catalyzes the following reaction: k1

Dihydroxyacetone phosphate G glyceraldehyde 3-phosphate k-1

The DG∞ „ for this reaction is 1.83 kcal/mol. In light of this information, which of the following statements are correct?

(a) The reaction would proceed spontaneously from left to right under standard con

ditions.

(b) The rate of the reaction in the reverse direction is higher than the rate in the for

ward direction at equilibrium.

pound on the left, dihydroxyacetone phosphate.

(d) The data given are sufficient to calculate the equilibrium constant of the reaction. (e) The data given are sufficient to calculate the left-to-right rate constant (k1).

9. Glycogen phosphorylase, an enzyme involved in the metabolism of the carbohydrate

polymer glycogen, catalyzes the reaction:

Glycogen +

n

phosphateGglucose 1-phosphate+glycogenn-1

glu

[

cose 1

phosphate glycogen

][

]

K eq =

=

n 1 = .

0 088

[phosphate] glycogen

[

n ]

Based on these data, which of the following statements are correct?

(a) Because glycogen phosphorylase normally degrades glycogen in cellular metabolism,

there is a paradox in that the equilibrium constant favors synthesis.

(b) The DG∞„ for this reaction at 25∞C is 1.43 kcal/mol.

ENZYMES: BASIC CONCEPTS AND KINETICS 5

(c) The phosphorolytic cleavage of glycogen consumes energy, that is, it is endergonic. (d) If the ratio of phosphate to glucose 1-phosphate in cells is high enough, phospho

rylase will degrade glycogen.

10. The reaction of the hydrolysis of glucose 6-phosphate to give glucose and phosphate has

a DG∞ „=-3.3 kcal/mol. The reaction takes place at 25∞C. Initially, the concentration of glucose 6-phosphate is 10-5 M, that of glucose is 10-1 M, and that of phosphate is 10-1 M. Which of the following statements pertaining to this reaction are correct?

(a) The equilibrium constant for the reaction is 267. (b) The equilibrium constant cannot be calculated because standard conditions do not

prevail initially.

(c) The DG„ for this reaction under the initial conditions is-0.78 kcal/mol. (d) Under the initial conditions, the synthesis of glucose 6-phosphate will take place

rather than hydrolysis.

(e) Under standard conditions, the hydrolysis of glucose 6-phosphate will proceed

spontaneously. Enzymes Accelerate Reactions by Facilitating the Formation of the Transition State

11. The transition state of an enzyme-catalyzed reaction that converts a substrate to a product

(a) is a transient intermediate formed along the reaction coordinate of the reaction. (b) has higher free energy than either the substrates or products. (c) is the most populated species along the reaction coordinate. (d) is increased in concentration because the enzyme binds tightly to it. (e) determines the velocity of the reaction.

12. Explain briefly how enzymes accelerate the rate of reactions.

13. Which of the following statements is true? Enzyme catalysis of a chemical reaction

(a) decreases DG„ so that the reaction can proceed spontaneously. (b) increases the energy of the transition state. (c) does not change DG∞ „, but rather changes the ratio of products to reactants at

equilibrium.

(d) decreases the entropy of the reaction. (e) increases the forward and reverse reaction rates.

14. Which of the following statements regarding an enzyme-substrate complex (ES) is true?

(a) The heat stability of an enzyme frequently changes upon the binding of a substrate. (b) At sufficiently high concentrations of substrate, the catalytic sites of the enzyme be

come filled and the reaction rate reaches a maximum.

(c) An enzyme-substrate complex can usually be isolated. (d) Enzyme-substrate complexes can usually be visualized by x-ray crystallography. (e) Spectroscopic changes in the substrate or the enzyme can be used to detect the for

mation of an enzyme-substrate complex.

15. Why is there a high degree of stereospecificity in the interaction of enzymes with their

substrates?

16. Explain why the forces that bind a substrate at the active site of an enzyme are usually weak. 6

CHAPTER 8 The Michaelis-Menten Model Accounts for the Kinetic Properties of Many Enzymes

17. Which of the following statements regarding simple Michaelis-Menten enzyme kinetics

are correct?

(a) The maximal velocity Vmax is related to the maximal number of substrate molecules

that can be “turned over” in unit time by a molecule of enzyme.

(b) KM is expressed in terms of a reaction velocity (e.g., mol S-1). (c) KM is the dissociation constant of the enzyme-substrate complex. (d) KM is the concentration of substrate required to achieve half of Vmax. (e) KM is the concentration of substrate required to convert half the total enzyme into

the enzyme-substrate complex.

18. Explain the relationship between KM and the dissociation constant of the enzyme-sub

strate complex KES.

19. Myoglobin binds and releases O

2 in muscle cells; myoglobin+O2

myoglobin·O2.

The fraction of myoglobin saturated with O2 (Y) is given by the equation pO Y =

2 pO2 + P50

where p is the partial pressure of the O2 and P50 is the pressure of O2 at which 50% of the myoglobin is saturated with O2. (This value reflects the equilibrium constant for the reaction.) Note the similarity between this equation and the Michaelis-Menten equation V

[S]

=

. V

[S] + K

max

m

Explain the relationships between the two equations.

20. From the plot of velocity versus substrate concentration shown in Figure 8-1, obtain the

following parameters. (The amount of enzyme in the reaction mixture is 10-3 mmol.) (a) KM (b) Vmax (c) k2/KM (d) Turnover number FIGURE 8.1 Plot of reaction velocity versus substrate concentration.

8

6

4

mmol/min) V(

2

0

1.0

2.0

3.0

4.0

[S]¥103 (M)

21. What is the significance of kcat/KM? 22. Which of the following statements is correct? The turnover number for chymotrypsin is

100 S-1, and for DNA polymerase it is 15 S-1. This means that

(a) chymotrypsin binds its substrate with higher affinity than does DNA polymerase. (b) the velocity of the chymotrypsin reaction is always greater than that of the DNA

polymerase reaction.

saturating substrate levels is lower than that of the DNA polymerase reaction under the same concentration conditions.

(d) the velocities of the reactions catalyzed by both enzymes at saturating substrate

levels could be made equal if 6.7 times more DNA polymerase than chymotrypsin were used. Enzymes Can Be Inhibited by Specific Molecules

23. Which of the following statements about the different types of enzyme inhibition are

correct?

(a) Competitive inhibition occurs when a substrate competes with an enzyme for bind

ing to an inhibitor protein.

(b) Competitive inhibition occurs when the substrate and the inhibitor compete for the

same active site on the enzyme.

amounts of substrate.

(d) Competitive inhibitors are often similar in chemical structure to the substrates of

the inhibited enzyme.

(e) Noncompetitive inhibitors often bind to the enzyme irreversibly.

24. If the KM of an enzyme for its substrate remains constant as the concentration of the in

hibitor increases, what can be said about the mode of inhibition?

25. The kinetic data for an enzymatic reaction in the presence and absence of inhibitors are

plotted in Figure 8-2. Identify the curve that corresponds to each of the following:

(a) No inhibitor (b) Noncompetitive inhibitor (c) Competitive inhibitor (d) Mixed inhibitor FIGURE 8.2 Effects of inhibitors on a plot of V versus [S].

1

2

3

V

4

[S] 8

CHAPTER 8

26. Draw approximate Lineweaver-Burk plots for each of the inhibitor types in Question 25.

1/V

0

1/[S]

27. Which statements are not true about a transition state analog?

(a) It fits better in the active site than the substrate. (b) It increases the rate of product formation. (c) It can be used as a hapten to produce catalytic antibodies. (d) It is usually a distorted or strained molecule. (e) It is a potent inhibitor of the enzyme.

28. The inhibition of bacterial cell wall synthesis by penicillin is a classic example of a med

ically significant inhibition of an enzymatic reaction. Which of the following statements about the inhibition of glycopeptide transpeptidase by penicillin is true?

(a) The inhibition is noncompetitive. (b) Penicillin binds irreversibly to an allosteric site of the enzyme. (c) Penicillin inhibits bacterial cell wall synthesis by incorrectly cross-linking the pep

tides of the proteoglycan.

(d) The penicilloyl-enzyme intermediate may be dissociated by high concentrations of

D-alanine.

(e) Penicillin resembles acyl-D-Ala-D-Ala, one of the substrates of the transpeptidase. Vitamins and Coenzymes

29. Which of the following correctly pairs a coenzyme with the group transferred by that

coenzyme?

(a) CoA, electrons (b) Biotin, CO2 (c) ATP, one-carbon unit (d) NADPH, phosphoryl group (e) Thiamine pyrophosphate, acyl group

30. Which of the following water-soluble vitamins forms part of the structure of CoA?

(a) Pantothenate (b) Thiamine (c) Riboflavin (d) Pyridoxine (e) Folate

31. Which of the vitamins in Question 30 is referred to as vitamin B1?

ENZYMES: BASIC CONCEPTS AND KINETICS 9

32. Match the lipid-soluble vitamins in the left column with the appropriate biological func

tions they or their derivatives serve in the right column.

(a) Vitamin A

(1) Protection of unsaturated membrane lipids from

(b) Vitamin D

oxidation

(2) Carboxylation of glutamate residues of clotting factors

(d) Vitamin K

(3) Participation in Ca2+and phosphorus metabolism (4) Precursor of retinal, the light-absorbing group in visual

pigments

(5) Related to fertility in rats (6) Activates transcription of some growth and development

genes

33. The hydroxylation of proline in nascent collagen polypeptide chains does not require

which of the following?

(a) O2

(d) Pyridoxal phosphate

(b) Dioxygenase

(e) a-Ketoglutarate

34. Why does hydroxylation increase the stability of the collagen triple helix?

(a) It promotes hydrogen bonding with water. (b) It increases hydrogen bonding between polypeptide chains. (c) It expands the helix and allows the glycine residues to better fit in the interior. (d) It decreases the melting temperature of nascent collagen. (e) It helps neutralize the charge on lysine residues. ANSWERS TO SELF-TEST

1. a, d. a is incorrect because some enzymes are RNA.

2. d. c is incorrect because, although tight binding to the substrates helps confer specificity

on the reaction, it increases the activation barrier to reaction. Tight substrate binding makes binding to the transition state of the reaction more energetically costly, that is, it increases the free energy of activation of the reaction.

3. Holoenzyme. Cofactors may be metal ions or low molecular weight organic molecules.

A prosthetic group is a tightly bound cofactor that seldom dissociates from the enzyme. Cofactors that are loosely bound behave like cosubstrates; they are easily bound and released from the enzyme.

4. Photosynthesis. The sun provides light energy that photosynthesis converts into chem

ical bond energy in the form of ATP. Other examples of energy transduction include the use of an ion gradient in mitochondria to drive the synthesis of chemical bonds, and the use of the energy in ATP to cause the movement of muscles.

5. a

6. The thermodynamic parameter DS for a chemical reaction is not easily measured. Even

if it were easily determined, its value depends on changes that occur not only in the system under study but also in the surroundings (see Chapter 1, Section 1.3.3). Intrinsically unfavorable reactions (DG∞ „>0) can take place if a change in the surroundings compensates for a decrease in the entropy (negative DS) of the reaction.

7. c, e. DG∞=-2.303 RT log

10 Keq. When Keq

1, DG∞=0 because the log of 1=0.

e is correct by definition. 10

CHAPTER 8

8. c, d

9. All of the statements are correct.

(a) The paradox is that although glycogen normally degrades glycogen to form glucose

1-phosphate, the standard free energy change of the reaction is positive, that is, the reaction is endergonic. See the answer to (d) for a resolution of the paradox.

(b) Using K„eq, one can calculate the DG∞ „ for the phosphorylase reaction: G∞ = 2 303

. RT log K

10

eq

=

cal

2 303

.

1 9

. 8

298∞

K

og

l

(0.088)

∞

10

mol K

= 1360 cal / mol

1 055

.

= 1430 cal / mol = 1 4

. 3 kcal / mol

lated; therefore, energy is consumed rather than released by this reaction.

(d) In cells, the ratio of phosphate to glucose 1-phosphate is so large that phosphory

lase is mainly involved with glycogen degradation.

10. a, d, e

(a) G∞ = 2 303

. RT log K

10

eq

3 3

. kcal / m =

ol

1 3

. 6 kcal / mol

log K

10

eq log K

10

eq = 2 4

. 3 K eq = 267

(b) Incorrect. K„eq is a constant; it is independent of the initial concentrations. (c) Incorrect.

[glucose][phosphate] G = G∞ +2 303

. RT log10 [glucose 6

]

phosphate

1

=

10

3 3

. kcal / mol + 1 3

. 6 kcal / mol

log10

10 5

2

=

10

3 3

. kcal / mol + 1 3

. 6 kcal / mol

log10 10 5

= 3 3

. kcal / mol + (1.36 kcal / mol

= +0 7

. 8 kcal / mol

(d) Correct. Under the initial conditions, DG„ is positive; therefore, the reaction will

proceed toward the formation of glucose 6-phosphate.

(e) Correct. The negative DG∞ „ value (at standard conditions) indicates that the reac

tion will proceed spontaneously toward the hydrolysis of glucose 6-phosphate.

11. a, b, d, e. c is incorrect because it has the most energy and is therefore hardest to form. The

velocity of the reaction is directly proportional to the concentration of the transition state.

ENZYMES: BASIC CONCEPTS AND KINETICS 11

12. Enzymes have evolved to bind tightly the transition state of the reaction they catalyze.

By binding the transition state with high affinity, they facilitate its formation. Hydrogen bonds and ionic and hydrophobic interactions can be involved in binding the transition state. The more transition state formed, the faster the reaction.

13. e. The enzyme speeds up the rate of attainment of equilibrium.

14. a, b, e. Turnover of ES to form P usually makes isolating ES difficult. In reactions re

quiring two substrates, an enzyme-substrate complex of one of the substrates can be isolated in the absence of the other substrate if the complex is very stable. The absence of the cosubstrate precludes turnover of ES. The same consideration applies to ES complexes formed by x-ray crystallography.

15. The formation of an enzyme-substrate complex involves a close, complementary fitting

of the atoms of the amino-acid-residue side chains that make up the active site of the enzyme with the atoms of the substrate. Since stereoisomers have different spatial arrangements of their atoms, only a single stereoisomer of the substrate usually fits into the active site in a form capable of being acted upon by the enzyme.

16. The enzyme-substrate and enzyme-product complexes must be reversible for catalysis

to proceed; therefore, weak forces are involved in the binding of substrates to enzymes.

17. a, d. Answer e is correct only when K =

M KES. See Question 18.

18. K

«

= +

M can be equal to KES when the rate constant k2 k-1. Since KM

(k2 k-1)/k1,

when k2 is negligible relative to k-1, KM becomes equal to k-1/k1, which is the dissociation constant of the enzyme-substrate complex.

19. These equations are related because they express the occupancy of saturable binding sites

as a function of either O2 or substrate concentration. The fraction of active sites filled, as reflected in V/Vmax, is analogous to Y, the degree of myoglobin saturation with oxygen; [S] and pO2 are the concentrations of substrate and O2, respectively; and KM and P50 are substrate or O2 concentrations at half-maximal saturation.

20. (a) K =

5¥10-4 M. KM is the value of the asymptote in Figure 8-1; it is equal to [S]

at 1/2 Vmax. Note that the units of [S] are mM. The factor 103 is used to multiply the actual concentrations. For example,

[S]

103=

.

2 0 M

[S] = .

2 0

10 3 M

(b) V

=

max

6 mmol/min. Vmax is obtained from Figure 8-1; it is the maximum velocity.

=

3/KM

2¥105 S-1 M-1. In order to calculate this ratio, k2 must be known. Since Vmax=k2[ET], k2=Vmax/[ET]. Thus

6 mol / min k =

2

3

10

mol

3

1

= 6 1

0 min

= 100 1

S

Using KM from part (a), K

1

2

100 S

5

=

= 2 10

1

S

M

K

4

5 10 M

(d) The turnover number is 100 S-1, equal to k2, which was calculated in part (c). 12

CHAPTER 8

21. Since V =

0

(kcat/KM) [S] [ET], kcat/KM represents the second-order rate constant for the

encounter of S with E. The ratio kcat/KM thus allows one to estimate the catalytic efficiency of an enzyme. The upper limit for kcat/KM, 108 to 109 M-1 S-1, is set by the rate of diffusion of the substrate in the solution, which limits the rate at which it encounters the enzyme. If an enzyme has a kcat/KM in this range, its catalytic velocity is restricted only by the rate at which the substrate can reach the enzyme, which means that the enzymatic catalysis has attained kinetic perfection.

22. d. V

=

max k2[ET]; thus, if 6.7 times more DNA polymerase than chymotrypsin is used, Vmax for both enzymes is the same:

100 S-1=6.7¥15 S-1

Answer (a) is incorrect because the affinity of substrate for the enzyme is given by K =

ES k-1/k1. Answer (b) is incorrect because the velocity of the enzymatic reactions is a

function of KM, Vmax, and substrate concentration. Answer (c) is incorrect because for the same enzyme concentration, V

=

max k2[ET] is greater for chymotrypsin than for DNA

polymerase.

23. b, c, d

24. The inhibition is noncompetitive because the proportion of bound substrate remains the

same as the concentration of the inhibitor increases.

25. (a) 1 (b) 3 (c) 2 (d) 4

26. See Figure 8-3. Plots 1 and 2 have the same 1/V intercept; plots 1 and 3 have the same

1/[S] intercept; and plots 1 and 4 have different 1/V and 1/[S] intercepts. FIGURE 8.3 Lineweaver-Burk plots for competitive (2), noncompetitive (3), and mixed (4)

inhibition, relative to the enzymatic reaction in the absence of inhibitors (1).

1/V

4

3

2

1

0

1/[S]

27. b, d. Answer (b) is incorrect because transition state analogs are inhibitors of the corre

sponding enzymes. Therefore, they decrease rather than increase enzyme reaction rates. Answer (d) is incorrect because transition-state analogs are not necessarily strained or distorted; rather, they mimic the shape of the transition state, which may be strained or distorted.

28. e

29. b

30. a

31. Thiamine

32. (a) 4, 6; (b) 3; (c) 1, 5; (d) 2

33. d

34. b PROBLEMS

1. Calculate the values for DG∞ „ that correspond to the following values of K„eq. Assume

that the temperature is 25∞C. (a) 1.5¥104 (b) 1.5 (c) 0.15 (d) 1.5¥10-4

2. Calculate the values for K„eq that correspond to the following values of DG∞ „. Assume

that the temperature is 25∞C. (a) -10 kcal/mol (b) -1 kcal/mol (c) +1 kcal/mol (d) +10 kcal/mol

3. The enzyme hexokinase catalyzes the following reaction:

Glucose+ATPGglucose 6-phosphate+ADP

For this reaction, DG∞ „=-4.0 kcal/mol.

(a) Calculate the change in free energy DG„ for this reaction under typical intracellular

conditions using the following concentrations: glucose, 55 mM; ATP, 5.0 mM; ADP, 1.0 mM; and glucose 6-phosphate, 0.1 mM. Assume that the temperature is 25∞C.

(b) In the typical cell, is the reaction catalyzed by hexokinase close to equilibrium or

far from equilibrium? Explain.

4. The enzyme aldolase catalyzes the following reaction:

Fructose 1,6-bisphosphateG

dihydroxyacetone phosphate+glyceraldehyde 3-phosphate

For this reaction, DG∞ „=+5.7 kcal/mol.

(a) Calculate the change in free energy DG„ for this reaction under typical intracellular

conditions using the following concentrations: fructose 1,6-bisphosphate, 0.15 mM; dihydroxyacetone phosphate, 4.3¥10-6 M; and glyceraldehyde 3-phosphate, 9.6¥10-5 M. Assume that the temperature is 25∞C.

(b) Explain why the aldolase reaction occurs in cells in the direction written despite the

fact that it has a positive free-energy change under standard conditions.

5. The text states (p. 197) that a decrease of 1.36 kcal/mol in the free energy of activation

of an enzyme-catalyzed reaction has the effect of increasing the rate of conversion of substrate to product by a factor of 10. What effect would this decrease of 1.36 kcal/mol in the free energy of activation have on the reverse reaction, the conversion of product to substrate? Explain. 14

CHAPTER 8

6. What is the ratio of [S] to KM when the velocity of an enzyme-catalyzed reaction is 80%

of Vmax?

7. The simple Michaelis-Menten model (equation 9 in the text, p. 201) applies only to the

initial velocity of an enzyme-catalyzed reaction, that is, to the velocity when no appreciable amount of product has accumulated. What feature of the model is consistent with this constraint? Explain.

8. Two first-order rate constants, k-1 and k2, and one second-order rate constant, k1, de

fine KM by the relationship k k

K

= 1 + 2

M k1

By substituting the appropriate units for the rate constants in this expression, show that KM must be expressed in terms of concentration.

9. Suppose that two tissues, tissue A and tissue B, are assayed for the activity of enzyme X.

The activity of enzyme X, expressed as the number of moles of substrate converted to product per gram of tissue, is found to be five times greater in tissue A than in tissue B under a variety of circumstances. What is the simplest explanation for this observation?

10. Sketch the appropriate plots on the following axes. Assume that simple Michaelis-Menten

kinetics apply, and that the pre-steady state occurs so rapidly that it need not be considered (see Section 8.4). FIGURE 8.10 (a)

V (b)

V

[S]

[E ]

T (c)

[ES] (d)

[S]

Time

Time (e)

[P]

Time

ENZYMES: BASIC CONCEPTS AND KINETICS 15

11. Suppose that the data shown below are obtained for an enzyme-catalyzed reaction. V S (mmol ml-1 min-1)

0.1

3.33

0.2

5.00

0.5

7.14

0.8

8.00

1.0

8.33

2.0

9.09

(a) From a double-reciprocal plot of the data, determine KM and Vmax. (b) Assuming that the enzyme present in the system had a concentration of 10-6 M,

calculate its turnover number.

12. Suppose that the data shown below are obtained for an enzyme-catalyzed reaction in the

presence and absence of inhibitor X. V (mmol ml-1 min-1) S Without X With X

0.2

5.0

3.0

0.4

7.5

5.0

0.8

10.0

7.5

1.0

10.7

8.3

2.0

12.5

10.7

4.0

13.6

12.5

(a) Using double-reciprocal plots of the data, determine the type of inhibition that has

occurred.

(b) Does inhibitor X combine with E, with ES, or with both? Explain. (c) Calculate the inhibitor constant Ki for substance X, assuming that the final con

centration of X in the reaction mixture was 0.2 mM.

13. Suppose that the data shown below are obtained for an enzyme-catalyzed reaction in the

presence and absence of inhibitor Y. V (mmol ml-1 min-1) S Without Y With Y

0.2

5.0

2.0

0.4

7.5

3.0

0.8

10.0

4.0

1.0

10.7

4.3

2.0

12.5

5.0

4.0

13.6

5.5

(a) Using double-reciprocal plots of the data, determine the type of inhibition that has

occurred. 16

CHAPTER 8

(b) Does inhibitor Y combine with E, with ES, or with both? Explain. (c) Calculate the inhibitor constant Ki for substance Y, assuming that the final concen

tration of Y in the reaction mixture was 0.3 mM.

14. Although the double-reciprocal plot is the most widely used plotting form for enzyme ki

netic data, it suffers from a major disadvantage. If linear increments of substrate concentration are used, thereby minimizing measurement errors in the laboratory, data points will be obtained that cluster near the vertical axis. Thus the intercept on the ordinate can be determined with great accuracy, but the slope of the line will be subject to considerable error, because the least reliable data points, those obtained at low substrate concentrations, have greater weight in establishing the slope. (Remember that many enzymes are protected against denaturation by the presence of their substrates at high concentrations.)

Because of the limitation of double-reciprocal plots described above, other linear plot

ting forms have been devised. One of these, the Eadie plot, graphs V versus V/[S]. Another, the Hanes-Woolf plot, ([S]/V versus [S]) is perhaps the most useful in minimizing the difficulties of the double-reciprocal plot.

(a) Rearrange the Michaelis-Menten equation to give [S]/V as a function of [S]. (b) What is the significance of the slope, the vertical intercept, and the horizontal in

tercept in a plot of [S]/V versus [S]?

toside (ONPG) by E. coli b-galactosidase. Use both double-reciprocal and HanesWoolf plots to analyze these data, and calculate values for KM and Vmax from both plots. (We suggest that you use a graphing program to generate a scatter plot, and then fit the data using a linear curve-fitting algorithm.) V S (mmol ml-1 min-1)

0.5

8.93

1.0

14.29

1.5

16.52

2.0

19.20

2.5

19.64

(d) Make a sketch of a plot of [S]/V versus [S] in the absence of an inhibitor and in the

presence of a competitive inhibitor and in the presence of a noncompetitive inhibitor.

15. Suppose that a modifier Q is added to an enzyme-catalyzed reaction with the results de

picted in Figure 8-4. What role does Q have? Does it combine with E, with ES, or with both E and ES? FIGURE 8.4 Effects of modifier Q on a plot of 1/V versus 1/[S].

1/V

Lower [Q]

Higher [Q]

0

1/[S]

16. The enzyme DNA ligase catalyzes the formation of a phosphodiester bond at a break

(nick) in the phosphodiester backbone of a duplex DNA molecule. The enzyme from bacteriophage T4 uses the free energy of hydrolysis ATP as the energy source for the formation of the phosphodiester bond. A covalently modified form of the enzyme in which AMP is bound to a lysine side chain is an intermediate in the reaction. The intermediate is formed by the reaction of E+ATP to form E-AMP+PPi. In the next step, the AMP is transferred from the enzyme to a phosphate on the DNA to form a pyrophosphate-linked DNA-AMP. In the last step of the reaction, the phosphodiester bond is formed by the free enzyme to seal the nick in the DNA and AMP is released.

(a) Write chemical equations that show the individual steps that occur over the course

of the overall reaction.

(b) Does this enzyme catalyze a double-displacement reaction? (c) Do you think that if DNA were omitted from the reaction mixture, the enzyme

would catalyze a partial reaction? If so, what reaction might it catalyze?

17. If you were studying an enzyme that catalyzed the reaction of ATP and fructose 1-phos

phate to form fructose 1,6-bisphosphate and ADP and discovered that a plot of the initial velocity of formation of fructose 1,6-bisphosphate versus ATP concentration was not hyperbolic, but rather sigmoid, what would you suspect? ANSWERS TO PROBLEMS

1. The values for DG∞„ are found by substituting the values for K„eq into equation 6 on page

195 of the text.

(a) G∞ = 2 303

. RT log K

10

e

= 2 303

.

1

( 9

. 8

10 3)

298 log

1

( 5

.

104)

10

= 5 7

. kcal / mol

(b) -0.24 kcal/mol (c) +1.1 kcal/mol (d) +5.2 kcal/mol

2. Equation 8 in Section 8.2.2 is used to find the answers.

(a) K

1

/ 3

. 6

e = 10 G∞

= 10 ( 10/1 3.6)

= 2 3

.

107

(b) 5.4 (c) 0.18 (d) 4.4¥10-8

3. (a) The applicable relationship is equation 1 in Section 8.2.2 of the text:

[C][D] G = G∞ +RT 1n [A][B]

=

[glucose 6 - phosphate][ADP] G∞ +RT 1n

[glucose][ATP] 18

CHAPTER 8

= 4 0

. kcal / mol

3

+ (

3

1

][ 0

.

]

1 9

. 8 10 )

[0.1 10

10

298

1n 55

[

10 3] 5

[

10 3]

= 4 0

. kcal / mol 4 7

. kcal / mol

= 8 7

. kcal / mol

(b) The large negative value DG„ means that the reaction is far from equilibrium in the

typical cell and thus has a strong thermodynamic drive to go in the direction of product formation. (Remember that at equilibrium, DG„=0.)

4. (a) The applicable relationship is the same equation used in 3a:

[C][D] G = G∞ +RT 1n [A][B]

=

[DHAP][G3P] G∞ +RT 1n

[FBP]

= +5 7

. kcal / mol + (1.98 10 3

)

298

(4.3 10 6) (96. 10 5)

1n

0 1

. 5

10 3

= +5 7

. kcal / mol 7 6

. kcal / mol

= 1 9

. kcal / mol

(b) The reaction occurs in the direction written because of the effects of the concen

trations on the free-energy change. The concentration term in the equation is much smaller than 1.0, which is its value under standard conditions. Removal of G3P by a subsequent reaction keeps its concentration low.

5. The rate of the reverse reaction must also increase by a factor of 10. Enzymes do not

alter the equilibria of processes; they affect the rate at which equilibrium is attained. Since the equilibrium constant Keq is the quotient of the rate constants for the forward and reverse reactions, both rate constants must be altered by the same factor. If the rate of the forward reaction is increased by a factor of 10, the rate of the reverse reaction must also increase by the same factor.

6. Start with the Michaelis-Menten equation, equation 23 on page 203 of the text:

[S] V = Vmax [S]+KM

ENZYMES: BASIC CONCEPTS AND KINETICS 19

Substituting 0.8 Vmax for V yields

[S]

0 8

.

=

max V

max V

[S]+ KM

0 8

. [S]+ 0 8

. KM = [S]

0 8

. KM = 0 2

. [S]

[S] = 4KM

[S] = 4 KM

Thus, a substrate concentration four times greater than the Michaelis constant yields a velocity that is 80% of maximal velocity.

7. Equation 9 on page 201 of Stryer shows the k2 step as being irreversible. This is true in

practice at the initial stage of the reaction because P and E cannot recombine to give ES at an appreciable rate if negligible P is present. Note that the equation reveals nothing about the relative magnitudes of k2 and the reverse rate constant for this step, k-2: k k

1

2

E + S

ES + E P k k

1

2

The reverse constant k-2 may actually be quite large compared with k2; nevertheless, the reverse reaction will not occur when little product is present, since the rate of the k-2 step depends on the concentrations of P and E as well as on the magnitude of its rate constant.

8. The first-order rate constants have the dimensions t-1, whereas the second-order con

stant has the dimension conc-1 t-1. Thus, we can carry out the following dimensional analysis: k k K

= 1 + 2

M k1 t 1 t 1

=

+ conc t

1 1

= conc

9. For the activity of enzyme X to be five times greater in tissue A than in tissue B, tissue

A must have five times the amount of enzyme X as does tissue B. Enzyme activity is directly proportional to enzyme concentration. 20

CHAPTER 8

10. The sketches should resemble the following:

V max (a)

V (b)

V

[S]

[E ]

Steady state: d[ES]/dt=0 (c)

[ES] (d)

[S]

Equilibrium concentration

Time

Equilibrium concentration (e)

[P]

Time

11. (a) See the graph, Figure 8-5. V

=

max

1/0.1=10 mmol ml-1 min-1.

0 3

.

0 1

.

Slope =

= 0 0

. 2

10 K

Slope =

M

max V K =

M

0.02¥10=0.2 mM FIGURE 8.5 A double-reciprocal plot of data for problem 11.

0.3

0.2

1/V

0.1

V max

0

2

4

6

8

10

1/[S]

ENZYMES: BASIC CONCEPTS AND KINETICS 21

(b) The turnover number is equal to the rate constant k3 in equation 10 on page 201

of the text. Rearrangement of the equation gives V k = max

2

[E ]

T

1

10

=

mmol

ml min

6

1

10 mol liter

1

10

=

mol

liter min

6

1

10 mol liter

7

1

5

1

= 10 min or 1.7 10 S

12. (a) See Figure 8-6. The double-reciprocal plots intersect on the y-axis, so the inhibi

tion is competitive. FIGURE 8.6 A double-reciprocal plot of data for problem 13 showing the effects

of an inhibitor X.

0.3

With X

0.2

1/V

Without X

0.1

0

1

2

3

4

5

6

1/S

(b) The inhibitor combines only with E, the free enzyme. A competitive inhibitor can

not combine with ES because the inhibitor and the substrate compete for the same binding site on the enzyme.

I

[ ]

Slope

= slope

1 +

inhib

uninhib Ki

The slope with X is

0 333

.

0 067

.

Slope

=

inhib

0 0532

.

5

The slope without X is

0 200

.

0 067

.

Slope

=

uninhib

0 0266

.

5 22

CHAPTER 8

Substituting in these values yields

0 2

. mM

0 0532

.

= 0 0266

.

1 + Ki Ki = 0 2

. mM

13. (a) See Figure 8-7. The inhibition was noncompetitive, as indicated by the fact that the

double-reciprocal plots intersect to the left of the y-axis. FIGURE 8.7 A double-reciprocal plot of data for problem 13 showing the effects

of an inhibitor Y.

0.5

With Y

0.4

0.3

1/V

0.2

0.1

Without Y

-3 -2

-1

0

1

2

3

4

5

6

1/S

(b) A noncompetitive inhibitor combines at a site other than the substrate binding site.

Thus, it may combine with both E and ES. In the case illustrated, the inhibitor has equal affinity for E and ES, which is shown by the fact that the plots intersect on the x-axis.

I

[ ]

Slope

= Slope

1 +

inhib

uninhib Ki

The slope with Y is

0 500

.

Slope

=

inhib

0 0667

.

5 0

.

( 2 5

. )

The slope without Y is

0 200

.

Slope

=

uninhib

0 0267

.

5 0

.

( 2 5

. )

Substituting in these values yields

0 3

. mM

0 0667

.

= 0 0267

.

1 + Ki Ki = 0 2

. mM

ENZYMES: BASIC CONCEPTS AND KINETICS 23

14. (a) We start with the Michaelis-Menten equation: V=V

+

max [S]/(KM

[S])

Cross multiplying yields V(K +

[S])=Vmax[S]

Division of both sides by V/Vmax gives

[S]/V=(K +

M

[S])/Vmax

[S]/V=K

+

M/Vmax

[S]/Vmax

[S]/V=(1/Vmax)[S] +KM/Vmax

(b) The linear equation above is in the form, y=mx+b, where m is the slope, and

b the y-intercept. Therefore, the slope of a Hanes-Woolf plot is (1/Vmax), and the intercept on the y-axis is KM/Vmax. The plot will intercept the x-axis when [S]/V is zero. Then

0=(1/Vmax)[S]+KM/Vmax

-K

=

M/Vmax

(1/Vmax)[S]

[S]=-KM

(c) See Figure 8-8. The y-intercept of the double-reciprocal plot is 1/Vmax. Therefore V

=

max

1/0.034=29.4 mmol l-1 min-1. The slope of the double-reciprocal plot is KM/Vmax. Therefore,

0.039=KM/29.4 K =

M

1.15 mM FIGURE 8.8 A double-reciprocal plot of data for problem 14.

0.15

34

0.0

0.01

x+

39

0.0

1/V

y=

0.05

0

0.5

1

1.5

2

2.5

1/[S] 24

CHAPTER 8

See Figure 8-9. The slope of the Hanes-Woolf plot is 1/Vmax. Therefore V

=

max

1/0.035=28.6 mmol 1-1 min-1. The y-intercept of the Hanes-Woolf plot

is KM/Vmax. Therefore,

0.037=KM/28.6 K =

1.06 mM. FIGURE 8.9 Hanes-Woolf plot of data for problem 14.

0.15

0.037

0.01

x+

0.035

[S]/V

y=

0.05

0

0.5

1

1.5

2

2.5

3

[S]

In this instance, both plots give good fits of the data, and the values derived from each for KM and Vmax do not differ significantly. We can conclude that the measurements at low substrate concentration are reliable.

(d) See Figure 8-10. FIGURE 8.10 Hanes-Woolf plots depicting effects of competitive

and non-competitive inhibitors.

With non-competitive inhibitor

With competitive inhibitor

[S]/V

No inhibitor

1/[S]

15. Q increases the rate of reaction, so it is an activator, or perhaps a second substrate. It

combines with both E and ES.

ENZYMES: BASIC CONCEPTS AND KINETICS 25

16. (a) The overall reaction proceeds as follows

(1) E+ATPFE-AMP+PPi (2) E-AMP+nicked DNAFnicked DNA-AMP+E (3) nicked DNA-AMP+EFsealed DNA+E Â ATP+nicked DNAFsealed DNA+AMP+PPi

(b) Yes, a substituted enzyme intermediate (E-AMP) is formed. (c) In the absence of DNA, the enzyme catalyzes the partial reaction of the formation

of the E-AMP with the release of PPi. DNA is not involved in the first part of the double-displacement reaction. (This problem is derived from Weiss, B., and Richardson, C.C. (1964). Enzymatic breakage and joining of deoxyribonucleic acid. 3. An enzyme-adenylate intermediate in the polynucleotide ligase reaction. J. Biol. Chem. 243:4556–4563. See also Lehman, I.R. (1974). DNA ligase: Structure, mechanism, and function. Science 186:790–797, for a complete review.)

17. In the absence of additional information, you would suspect that the enzyme had al

losteric properties; its initial velocity was being influenced by binding of one of the substrates to a site different from the active site. EXPANDED SOLUTIONS TO TEXT PROBLEMS

1. (a) Note that a unit of activity is 10-5 mol/15 min and when the substrate concentra

tion is much greater than KM the enzyme operates at Vmax. Then (2800 units¥10-5 mole)/(15 min¥60s)=31.1¥10-6 mol/s.

(b) Remember grams/gram molecular weight=moles, thus 10-3 g/ (20¥103 g/mol

subunit)=5¥10-8 mol active site.

substrate converted to product by one molecule of enzyme per unit time, in this case, seconds. Hence, the turnover number=31.1¥10-6 mol S-1

s /5¥10-8 mol

E=622 molecules S per second per molecule E. Turnover number is discussed on page 204 of the text.

2. For (a) and (b), proper graphing of the data given will provide the correct answers: K =

=

5.2¥10-6 M; Vmax

6.84¥10-10 mol/min

3. Penicillinase, like glycopeptide transpeptidase, forms an acyl-enzyme intermediate

with its substrate but transfers it to water rather than to the terminal glycine of the pentaglycine bridge.

4. For (a) and (b), proper graphing of the data given will provide the correct answers:

(a) In the absence of inhibitor, Vmax is 47.6 mmol/min and KM is 1.1¥10-5 M. In the

presence of inhibitor, Vmax is the same, and the apparent KM is 3.1¥10-5 M.

(b) Since Vmax is not altered by the inhibitor, this is competitive inhibition. (c) Since this is competitive inhibition, the equation on page 222 of the text applies. The

only difference between this equation and equation 31 (text p. 221) is the factor

26

CHAPTER 8

(1+[I]/Ki). Since in competitive inhibition Vmax does not change, this factor describes the relationship between K

=

M and apparent KM. Hence, the apparent KM KM

(1+[I]/Ki). Therefore, using the data in (a) and (b), 3.1¥10-5 M=(1.1¥10-5 M) (1+2¥10-3 M/K

=

i) and Ki

1.1¥10-3 M.

(d) The [S]/(K +

[S]) term in the Michaelis-Menten equation tells us the fraction of en

zyme molecules bound to substrate. Thus, 1¥10-5/(1+3.1)10-5=ƒ =

ES

0.243.

Since K =

=

i

[E] [I]/[EI], [EI]/[E]=[I]/Ki

2¥10-3/1.1¥10-3=1.82. However,

the sum of [EI] +[E] is only 0.757 of the total enzyme because the remaining 0.243 is bound to substrate. Therefore, 1.82=ƒEI/ (0.757-ƒEI). Solving this equation gives ƒ =

EI

0.488.

(e) Using [S]/(K +

[S]), 3¥10-5/1.1+3) ¥10-5=0.73=ƒES in the absence of

inhibitor. 3¥10-5/(3.1+3) ¥10-5=0.49=ƒES in the presence of inhibitor. This ratio, 0.73/0.49 and 33.8/22.6 (the velocity ratio) are equal.

5. For (a) and (b), proper graphing of the data given will provide the correct answers:

(a) Vmax is 9.5 mmol/min. KM is 1.1¥10-5 M, the same as without inhibitor. (b) Since KM does not change, this is noncompetitive inhibition. (c) To answer this question you need to obtain a value for Vmax (47.6 mmol/min) from

the graphs used in question 4(a) above. Because this is noncompetitive inhibition, use the equation on text page 222 as follows: 9.5 mmol/min=47.6 mmol/min/ (1+10-4 M/Ki). Solving for Ki one obtains the answer 2.5¥10-5 M.

(d) Since an inhibitor does not affect KM, the fraction of enzyme molecules binding sub

strate=[S]/(K +

[S]), with or without inhibitor. For solution, see 4(e).

6. (a) V

[S] V =

max

(KM +[S] V(KM +[S] = Vmax

[S] VK V +

M = Vmax

[S] VK V = V

M

max

[S] K V V = V

M

max

[S]

(b) The slope of a straight line is the constant that the x-coordinate is multiplied by in

the equation for the straight line. Thus, in the Lineweaver-Burk plot, KM/Vmax is the slope; in the Eadie-Hofstee plot the slope is-KM because V/[S] is plotted on the x axis; see (a). By inspection, the y-intercept is Vmax. The x-intercept is Vmax/KM because one is extrapolating to [S]=0.

(the slope of 2 is greater than the slope of 1). In contrast, with a noncompetitive inhibitor, KM does not change; 1 and 3 have the same slope (while Vmax decreases).

7. Potential hydrogen-bond donors at pH 7 are the side chains of the following residues:

arginine, asparagine, glutamine, histidine, lysine, serine, threonine, tryptophan, and tyrosine. For a more detailed discussion of hydrogen-bond donors and acceptors, see problem 2-8.

ENZYMES: BASIC CONCEPTS AND KINETICS 27

8. The rates of utilization of A and B are given by equation 25 (text p. 204):

1 k3 2 V =

1–—2 [E] [A]

1 KM 2 A

and

1 k3 2 V =

1–—2 [E] [B]

1 KM 2 B

Hence, the ratio of these rates is

1 k3 2

1 k3 2 V

=

A/VB

1–—2 [A] / 1–—2 [B]

1 KM 2 A

1 KM 2 B

Thus, an enzyme discriminates between competing substrates on the basis of their values of k3/KM rather than of KM alone. Note that the velocity is dependent on the constants (k3/KM) and the concentrations of enzyme and substrate.

9. A tenfold change in the equilibrium constant corresponds to a standard free-energy

change (DG∞„) of 1.36 kcal/mol (1.36 comes from 2.303 RT). If a mutant enzyme binds a substrate, S, 100-fold as tightly as does the native enzyme, more Gibbs free energy of activation (DG‡) is needed to convert S to S‡ (transition state). In fact, the DG‡ is increased by 2.72 kcal/mol (2.303 RT log 100) and the velocity of the reaction will be slowed down by a factor of 100.

10. The uncompetitive inhibitor binds to the enzyme-substrate complex, but not to the free

enzyme. Both KM and Vmax are affected, but the ratio KM/Vmax remains constant. The reaction velocity obeys: V

S

V =

max[ ] KM + [jS]

where j=1+[I] / KI , and [I] is the inhibitor concentration. a. Standard Michaelis-Menten graph of reaction velocity, v, versus substrate concentration, [S]:

Increasing [I]

o

V

[S] 28

CHAPTER 8

Double-reciprocal plot for increasing concentrations of an uncompetitive inhibitor:

1/Vo

Increasing [I]

1/[S]

b. The double reciprocal plot shows lines of constant slope KM/Vmax. The rate equation can be derived using:

[E] = [E]+[ES]+

T

[

]

ESI .

[E][S]

K

~

M

= [ES]

[ES ][I]

KI = [ESI]

Therefore, [E] =

ES.

The rate equation is: v=k2[ES]. If we let j=(1+[I]/K

=

I), and Vmax

k2[E]T, then:

V

S

[ ]

v =

max

, or v =

max

j + K / S

[ ]

j S

[ ]+ K

m

For the double-reciprocal plot, 1/v=(1/[S])(KM/Vmax)+j/Vmax. Therefore, the slope of the double reciprocal plot remains the same (KM/Vmax) in the presence of an uncompetitive inhibitor.

11. By substituting [S]=0.1*KM into the Michaelis-Menten equation,

v=(Vmax)([S]) / ([S]+KM), we can show that: v=(Vmax)(0.1)(KM) / ((0.1+1.0)KM), or v=(1/11)Vmax. So with v=1.0 mmol min-1, V

=

max

11.0 mmol min-1.

12. KM will remain the same (center graph), and the apparent Vmax will change with the dif

ferent amounts of enzyme (y-intercept in center graph). Therefore, the correct answer is the center graph.

ENZYMES: BASIC CONCEPTS AND KINETICS 29

13. a.

The double-reciprocal plot will turn up to form a second line near the 1/v axis, giving an approximately “V-shaped” graph. b. The decrease in reaction velocity at high substrate concentration could be due to an allosteric inhibition by substrate at a second binding site. The binding affinity of the second (allosteric) site for substrate could be lower than the affinity of the catalytic site for substrate.

14. The step catalyzed by EA will be rate-limiting because the actual substrate concentration

(10-4 M) is much less than that needed to achieve half-maximal reaction velocity (10-2 M) for this step.

15. The mechanism suggests that H+ is behaving as a competitive inhibitor. Therefore at

sufficiently high substrate concentration, the substrate will overcome the inhibition, and the velocity, vo, will equal Vmax, independent of pH (part a). At a low (constant) substrate concentration, the observed vo will follow a titration curve with a pK of 6.0 (parts b, c).

a.

e activity

Relativ

0

3

6

9

12

pH

b.

e activity

Relativ

0

3

6

9

12

pH

c.

At pH 6.0 half of the enzyme will be in the E: form and the reaction velocity therefore will be 1/2 of Vmax.

16. a.

The enzyme is unstable at 37∞C. It unfolds or denatures as a function of time of storage at 37∞C.

b.

The PLP coenzyme partially protects the enzyme against the thermal unfolding. When PLP is bound to the enzyme, the rate of denaturation is slower. This page intentionally left blank. CHAPTER Catalytic Strategies 9

In the previous chapter, you learned that the catalytic activity of enzymes is based

on their ability to stabilize the transition states of chemical reactions and thereby decrease the energy-activation barrier to reactivity. In Chapter 9, the authors de

scribe in detail the structures, active-site configurations, binding of substrates, and catalytic mechanisms of four well-understood enzymes: chymotrypsin, carbonic anhydrase, restriction endonuclease EcoRV, and nucleoside monophosphate kinase. Using these specific enzymes as models, fundamental principles of enzyme catalysis are exemplified: specific binding of substrates, induced fit of enzyme-substrate complexes, covalent catalysis, general acid-base catalysis by active-site residues, catalysis by propinquity and by metal ions, formation and stabilization of transition states, and reversibility of catalytic steps. The principles employed by these enzymes illustrate how enzymes use basic chemistry to perform reactions at rapid rates and with high fidelity. Because the interactions of enzymes with substrates depend on the chemical properties of amino acid residues and on protein structure in general, a review of Chapter 3 would be helpful before reading this chapter. In addition, refresh your understanding of the basic concepts of enzyme action, thermodynamics, and kinetics presented in Chapter 8. 1 2

CHAPTER 9 LEARNING OBJECTIVES

When you have mastered this chapter, you should be able to complete the following objectives. Basic Catalytic Principles (Text Section 9.0.1)

1. Define binding energy as it relates to enzyme-substrate interactions and explain how it is

used in enzyme catalysis.

2. List four strategies commonly employed by enzymes to effect catalysis. Proteases and Chymotrypsin (Text Section 9.1)

3. Define proteolysis. Draw the reaction for peptide-bond hydrolysis, and explain why pep

tide bonds are resistant to spontaneous hydrolysis.

4. Indicate the amino acid sequence specificity of the cleavage catalyzed by chymotrypsin

and explain its molecular basis.

5. List the evidence that indicates a serine hydroxyl serves as a nucleophile in the reaction

catalyzed by chymotrypsin.

6. Explain why a “burst” of product appears when chymotrypsin reacts with a chromogenic ester substrate, and relate this phenomenon to covalent catalysis.

7. Indicate how X-ray crystallography was used to learn about the mechanism of the chy

motrypsin reaction.

8. Describe the formation and stabilization of the transient tetrahedral intermediate produced

from the scissile, planar peptide bond during hydrolysis.

9. Summarize the roles of the catalytic triad in the mechanism of chymotrypsin and the re

lationship of the oxyanion hole to the tetrahedral intermediate of the reaction. Appreciate that these features are present in other proteases, esterases, and lipases.

10. Describe how site-directed mutagenesis was used to prove the role of the catalytic triad in subtilisin catalysis.

11. List other catalytic mechanisms by which peptide bonds can be hydrolyzed and provide

examples for each.

12. Provide examples of protease inhibitors that serve as therapeutic agents. Carbonic Anhydrases (Text Section 9.2)

13. Outline the relationship of CO2 to aerobic metabolism and indicate how most of the CO2

generated by peripheral tissues is transported to the lungs.

14. Write the chemical equation for the hydration of carbon dioxide, and explain why bicar- bonate is formed at physiologic pH values.

15. Indicate the physiologic requirement for catalysis of the reaction that hydrates CO2. 16. Explain how carbonic anhydrase uses Zn21to activate a water molecule to attack CO2. 17. Describe why a buffer must be present at high concentrations to allow carbonic anhy

drase to function rapidly, and explain how a proton shuttle is involved in buffer action.

18. Using the carbonic anhydrases as examples, describe why convergent evolution is thought

to have selected a common active-site structure.

CATALYTIC STRATEGIES 3 Restriction Enzymes (Text Section 9.3)

19. Write the reaction catalyzed by restriction endonucleases and explain why these enzymes

must show very high substrate specificity to achieve their biological function.

20. Write the reaction and explain the biological role of methylases (DNA methyltransferases)

in restriction-modification systems.

21. Draw a phosphodiester bond. Deduce how phosphorothioates could be used to differentiate

the achiral oxygens of a phosphodiester bond and to distinguish between direct hydrolysis of the bond and a mechanism involving a covalent enzyme-DNA intermediate.

22. Draw the pentacoordinate, trigonal bipyramidal structure of the transition state of a phos

phodiester bond undergoing an in-line displacement reaction.

23. Compare the primary role of Mg21 in the mechanism of restriction endonucleases with

that of Zn21 in the carbonic anhydrases.

24. Summarize the ways in which restriction enzymes use binding energy to attain high sub

strate specificity. Consider the role of DNA distortion in achieving catalytic fidelity.

25. Provide evidence that restriction enzymes employed horizontal gene transfer to spread

among bacteria. Nucleoside Monophosphate Kinases (Text Section 9.4)

26. Write the general reaction of phosphoryl transfer for a kinase.

27. Indicate how induced fit is used to preclude hydrolysis of ATP during phosphoryl transfer.

28. Relate the P-loop of kinases to the phosphoryl group of enzyme-bound ATP, and appre

ciate the ubiquity of P-loop NTPase domains.

29. Rationalize the chemical functions of the Mg21-nucleotide complex in a kinase reaction.

30. Appreciate the ubiquity of P-loops and rationalize their wide distribution. SELF-TEST Basic Catalytic Principles

1. The free energy released when an enzyme binds a substrate

(a) arises from many weak intermolecular interactions. (b) contributes to the catalytic efficiency of the enzyme. (c) is more negative when an incorrect substrate is bound. (d) becomes more positive as the transition state of the reaction develops. (e) becomes more negative the more tightly the enzyme binds the substrate.

2. Which of the following are used by enzymes to catalyze specific reactions?

(a) Metal-ions

(d) General acid-base reactions

(b) Temperature changes

(e) Covalent enzyme-substrate complexes

3. Why is the peptide bond, which is thermodynamically unstable, resistant to spontaneous

hydrolysis? 4

CHAPTER 9 Chymotrypsin and Other Proteolytic Enzymes

4. The alkoxide group on chymotrypsin that attacks the carbonyl oxygen of the peptide

bond of the substrate arises from which amino acid side chain?

(a) an aspartate

(d) a threonine

(b) a histidine

(e) a tyrosine

5. Which of the following experimental observations provide evidence for the formation of

an acyl-enzyme intermediate during the chymotrypsin reaction?

(a) A biphasic release of p-nitrophenol occurs during the hydrolysis of the p-nitro

phenyl ester of N-acetyl-phenylalanine.

(b) The active serine can be specifically labeled with organic fluorophosphates. (c) The pH dependence of the catalytic rate is bell shaped, with a maximum at pH 8. (d) A deep pocket on the enzyme can accommodate a large hydrophobic side chain

of the recognized substrate.

6. Three essential amino acid residues in the active site of chymotrypsin form a catalytic

triad. Which of the following are roles for these residues in catalysis?

(a) The histidine residue facilitates the reaction by acting as an acid-base catalyst. (b) The aspartate residue orients the histine properly for reaction. (c) The serine residue acts as a nucleophile during the reaction with the substrate. (d) The aspartate residue acts as an electrophile during the reaction with the substrate. (e) The aspartate residue initiates the deacylation step by a nucleophilic attack on the

carbonyl carbon of the acyl intermediate.

(f)

They comprise the oxyanion hole.

7. Which of the following enzymes can be irreversibly inactivated with diisopropylphos

phofluoridate (DIPF)?

(a) Carboxypeptidase II (b) Trypsin (c) Lysozyme (d) Subtilisin (e) Thrombin

8. The three enzymes trypsin, elastase, and chymotrypsin

(a) likely evolved from a common ancestor. (b) have major similarities in their amino acid sequences and three-dimensional

structures.

(c) catalyze the same general reaction: the cleavage of a peptide bond. (d) catalyze reactions that proceed through a covalent intermediate. (e) have structural differences at their active sites.

9. Match the enzyme in the right column with the proteolytic-enzyme class to which it be

longs in the left column.

(a) Metalloprotease

(1) Papain

(b) Serine protease

(2) Pepsin

(3) Elastase

(d) Acid (aspartyl) proteases

(4) Thermolysin

10. Why might inhibitors of specific proteases be useful therapeutic agents? Provide a

specific example.

CATALYTIC STRATEGIES 5 Carbonic Anhydrases

11. Match the molecule in the first column with the appropriate item in the second column.

(a) Water

(1) pKa~3.5

(b) Bicarbonate

(2) pKa~7

(3) pKa~14

(d) Water bound to Zn21

(4) pKa~10.3

in carbonic anhydrase II

12. Given the pKa values of the compounds shown in the previous question (Number 11),

what is the significance of the water that is bound to the zinc ion in carbonic anhydrase?

13. Several different carbonic anhydrases coordinate Zn21 in their active sites using the

amino acid side chains of His exclusively or of His and of Cys. Rationalize how the binding of water to the coordinated Zn21 lowers the pKa value of the water. Restriction Enzymes

14. Is the DNA sequence 5¢-GAATTC-3¢ palindromic when it is in a duplex? Why? 15. Bacteria use restriction enzymes to destroy invading, exogenous DNA; for instance, DNA

injected during bacteriophage infection. How can the restriction enzyme hydrolyze the foreign DNA and not destroy the DNA of the bacterium in which it resides?

16. List all the substrates and cofactors used by type II restriction endonucleases and type II

DNA methylases?

17. Which of the following DNA sequences is likely to be cut by a restriction enzyme? Only

one strand, written in the 5¢ to 3¢ orientation is shown, but you should assume the opposite strand is present to form a duplex.

(a) TAGCAT (b) CTGCAG (c) CAGGAC (d) GAATTC (e) TCGA

18. Which of the following amino acids in the active site of a typical restriction enzyme

would you expect to be involved in binding Mg2+?

(a) D

(d) N

(b) Y

(e) E

19. Which of the following are roles for Mg21in reactions that use ATP as a phosphoryl

donor, e.g., the NMP kinases?

Mg21

(a) binds to the enzyme and activates a water molecule. (b) neutralizes partially the negative charge on the triphosphate group of the ATP. (c) forms a stable conformation of ATP by binding to its phosphoanhydride “tail.” (d) provides potential binding points on the ATP for the enzyme to recognize. 6

CHAPTER 9

20. The P-loops of ATP-using kinases

(a) are formed mostly of P (proline) residues. (b) are held rigidly in place by intramolecular interactions with adjacent a helices. (c) interact with the phosphates of the substrate nucleotides. (d) move extensively upon ATP binding. (e) help promote phosphoryl transfer and not hydrolysis. PROBLEMS

1. Why is histidine a particularly versatile amino acid residue in terms of its involvement

in enzymatic reaction mechanisms?

2. Although chymotrypsin is a proteolytic enzyme, it is quite resistant to digesting itself.

How would you explain its resistance to self-proteolysis?

3. For each enzyme in the left column, indicate the appropriate transition state or chemical

entity in the right column that has a postulated involvement in its catalytic mechanism.

(a) Carbonic Anhydrase

(1) Mixed anhydride

(b) Nucleoside Monophosphate Kinase

(2) Oxyanion hole

(4) Carbonium ion (5) Tetrahedral carbon intermediate

4. A pH-enzyme activity curve is shown in Figure 9-1. Which of the following pairs of amino

acids would be likely candidates as catalytic groups? (See Primary Text, Table X-X for the pKa values of amino acid residues.) (a) Glutamic acid and lysine (b) Aspartic acid and histidine (c) Histidine and cysteine (d) Histidine and histidine (e) Histidine and lysine FIGURE 9.1

pH

5. Consider the fact that DNA methylases put methyl groups only onto preformed DNA, i.e., after DNA has been synthesized from unmethylated dNTPs. Because DNA replication is semiconservative, what would be the methylation state of a restriction site immediately after replication in a bacterium containing a restriction-modification system? What can you conclude about the number of methyl groups needed per restriction site to render the DNA refractory to cleavage by the cognate restriction endonuclease?

CATALYTIC STRATEGIES 7

6. On average, how many EcoRV restriction sites would you expect in the genome of E. coli?

The genome is 4.6¥106 base pairs and its composition is approximately 50% G+C.

7. Examine the reaction of nucleoside monophosphate kinase shown on page 252 of the

primary text. Estimate what the equilibrium constant would be for the reaction under standard biochemical conditions.

8. If you introduced a mutation into adenylate kinase that prevented P-loop movement and

subsequent lid closure, what reaction would you expect the enzyme to catalyze when incubated with AMP and ATP?

9. Trypsin, chymotrypsin, and carboxypeptidase A fail to cleave peptide bonds involving

proline. Trypsin, for example, will not cleave a peptide at a Lys-Pro junction. Why do you think this is the case?

10. Place slash marks at the sites where you would expect chymotrypsin to cleave the fol

lowing peptide:

Lys-Gly-Phe-Thr-Tyr-Pro-Asn-Trp-Ser-Tyr-Phe

11. Many enzymes can be protected against thermal denaturation during purification pro

cedures by the addition of substrate. Propose an explanation for this phenomenon.

12. What are the main structural features of an enzyme that determine its substrate speci

ficity?

13. DNA methyltransferases (DNA methylases) use S-adenosylmethionine (AdoMet) as the

methyl donor in a reaction that methylates a specific base at a specific sequence in DNA and releases the AdoMet remnant S-adenosylhomocysteine (AdoHcy). The DNA methyltransferase RsrI catalyzes the reaction DNA+AdoMet Æ methylated DNA+AdoHcy where the methyl group is deposited on the exocyclic amino group of the second A in the recognition sequence GAATTC. A burst of incorporation of methyl groups into DNA occurred in an experiment in which the enzyme was saturated with AdoMet radiolabeled with 14C in its activated methyl group. The enzyme was first saturated with an excess of [14C]AdoMet and then a saturating excess of unmethylated DNA containing the target sequence was added along with more radiolabeled AdoMet to maintain its original concentration and specific activity. The incorporation of isotope into the DNA was monitored. A rapid incorporation of methyl groups occurred (burst) upon addition of the DNA+[14C]AdoMet. The burst was followed by a slower, steady-state rate of DNA methylation. A plot of the formation of labeled DNA as a function of time is shown in Figure 9-2. The mol of methyl groups incorporated into DNA/mol of enzyme is plotted on the ordinate, and time in seconds is plotted on the abcissa. When the steady-state phase of the reaction curve was extrapolated back to the ordinate (Y-axis), the value obtained was 0.94 mol methyl group deposited on DNA/mol of enzyme.

a.

What can you conclude from this experiment about: 1) the mechanism of the reaction, and 2) the proportion of molecules of enzyme that were active? FIGURE 9.2

Steady-state phase

roup/

yl g

1

mol enzyme

Burst phase

mol Meth

Seconds 8

CHAPTER 9

b.

In a second experiment with RsrI methylase, the same protocol described for the burst experiment was followed except that when the excess DNA was added to the enzyme that had been saturated with radiolabeled AdoMet, the solution containing the DNA also contained the initial, 50-fold concentration excess of unlabelled AdoMet. The incorporation of the radiolabeled methyl groups of the pre-bound AdoMet into DNA was again followed. In this case, a smaller burst, approximately 10% of that observed in the first experiment, was detected before the steady state rate of reaction began. Explain what the smaller burst implies about the order of addition of the substrates, DNA and AdoMet, to the enzyme.

c.

At physiological pH values, the dissociation of water cannot supply a sufficient concentration of protons to support the full catalytic potential of carbonic anhydrase. As a result, the enzyme has evolved to use buffers as acid-base catalysts to increase local proton concentrations in the active site. Some of these buffers have molecular dimensions too large to allow them to penetrate into the active site and gain proximity to the protein-bound Zn21. Despite their exclusion by steric factors from the active site, the buffers support efficient catalysis by the enzyme. In addition, ionizable groups in the active site of the enzyme are involved in the delivery and removal of the required protons. Site-specific mutagenesis that substituted some of these residues with other ionizable amino acids having different pKa values failed to inactivate the enzyme. What can you conclude about the molecular mechanisms by which protons are shuttled into the active site of carbonic anhydrase? ANSWERS TO SELF-TEST

1. a, b, e. The DG∞¢ of the reaction becomes more negative as the binding affinity of the en

zyme for the substrate increases. Interactions between the substrate and enzyme promote the reaction when they are fully formed during the development of the transition state of the reaction. Favorable interactions between the enzyme and the substrate in its ground state before development of the transition state can hinder the reaction by lowering the valley preceding the activation barrier in the reaction coordinate diagram if they do not also contribute to binding the transition state. For instance a substrate analog that is a good competitive inhibitor forms strong interactions with the enzyme, but cannot develop a transition state.

2. a, c, d, e

3. A peptide bond is stabilized by resonance, which gives the carbonyl-carbon–to–amide

nitrogen link partial double-bond character, making it more stable to hydrolysis. In addition, the carbonyl carbon of the peptide bond is linked to a partially negatively charged carbonyl oxygen that decreases the susceptibility of the carbon atom to nucleophilic attack by a hydroxyl ion.

4. c

5. a, b. The pH versus activity curve indicates only that some step in the mechanism is sen

sitive to the state of dissociation of a proton donor on the protein.

6. a, b, c. d is incorrect, because the aspartic acid carboxylate is ionized, and bearing a neg

ative charge, is not an electrophile.

7. a, b, d, e. The mechanism of lysozyme does not involve the nucleophilic attack on

the substrate by an activated hydroxyl of the enzyme. The other three enzymes do have such an activated serine hydroxyl and react to form a covalent, inactive complex with DIPF.

CATALYTIC STRATEGIES 9

8. a, b, c, d, e. The enzymes differ in structure at the sites at which they interact with the

amino acid side chains to determine their substrate specificity.

9. (a) 4 (b) 3 (c) 1 (d) 2

10. Proteases that are specific for particular amino acid sequences play important roles in

normal and pathological physiology in humans. For example, a protease, the angiotensinconverting enzyme (ACE), is involved in blood-pressure regulation. A specific inhibitor would prevent the hypertension that arises from overactivity of ACE. Similarly, a specific protease is necessary for human immunodeficiency virus maturation after infection. Inhibition of this protease could limit HIV infection. Carbonic Anhydrases

11. (a) 3, (b) 4, (c) 1, (d) 2. The CO2 buffering system is unusual because one of its compo

nents, dissolved CO2,, is a volatile gas in equilibrium with atmospheric CO2. By convention, [H

+

2CO3] is used to represent the total concentration of dissolved CO2

H2CO3.

Carbonic acid that dissociates to form H1+bicarbonate is immediately replaced by the reaction of CO2 with water. The observed pKa value of carbonic acid in a solution in equilibrium with gaseous CO2 in the lungs is approximately 6.1, not 3.5, because of its equilibrium with dissolved CO2, which exceeds it in concentration by approximately 1000-fold. Thus, although the “true” pKa value of carbonic acid is 3.5, it behaves in gas transport in mammals as if the value were approximately 6.1. The pKa value for the dissociation of bicarbonate is ~10.3.

12. With the pKa value of water lowered to near physiological pH values, an appreciable

amount of zinc-bound hydroxyl ion will be formed by dissociation of a proton from the zinc-bound water. The hydroxyl ion is the nucleophile that attacks the carbonyl carbon of CO2 to form the bicarbonate ion. Thus, the enzyme generates a reactive substrate by binding water to Zn21, thereby facilitating its dissociation to form the reactive substrate.

13. The positive charge on the zinc ion withdraws electrons from the oxygen of the bound

water, weakens the bonds to its hydrogen atoms, and promotes the dissociation of a proton to form an enzyme-bound hydroxyl. Restriction Enzymes

14. Yes, because the complementary strand is identical, namely, 5¢-GAATTC-3¢. Remember,

the strands of duplex DNA have opposite polarity. A palindromic sequence has two-fold rotational symmetry. If you rotate the duplex molecule 180° about an axis located perpendicular to its long axis and piercing between the two strands between the AT sequences in each strand, you will generate the starting configuration of atoms.

15. A restriction enzyme recognizes and hydrolyzes a particular DNA sequence. The same

sequence is recognized and methylated by the partner DNA methylase of the restriction enzyme. A methylated restriction site is immune to cleavage by the restriction enzyme. The methylase keeps the host DNA methylated and thus protected. The invading DNA, if unmethylated itself, will be cleaved by the restriction enzyme and subsequently destroyed by less specific nucleases.

16. Restriction enzymes require only target DNA, Mg21, and water. DNA methylases require

only unmethylated target DNA and S-adenosylmethionine.

17. b, d, e. Each of these sequences has an identical complementary strand.

18. a, e. The carboxyl groups of Asp and Glu can bind Mg21 effectively. 10

CHAPTER 9 Nucleoside Monophosphate Kinases

19. b, c, d

20. c, d, e. The P-loops are highly mobile. ANSWERS TO PROBLEMS

1. The imidazole ring of histidine can act as an acid-base catalyst, a nucleophile, or a chela

tor (coordinator) of metal ions. The first and third functions were illustrated by chymotrypsin and carbonic anhydrase, respectively.

2. Chymotrypsin specifically cleaves peptide bonds whose C-terminal amino acid is adja

cent to non-polar aromatic amino acid residues or the bulky, hydrophobic methionine. Because these residues are often buried in the interior of proteins, including chymotrypsin, the self-hydrolysis of native, folded chymotrypsin is very inefficient. In fact, during digestion, chymotrypsin acts most effectively on partially degraded and denatured (unfolded) proteins.

3. (a) 1; (b) 1, 3; (c) 3; (d) 2, 5

4. c

5. A parent DNA molecule in a cell with a restriction/modification (R/M) system would have

both strands of its restriction sites methylated. Upon semiconservative replication, the newly synthesized daughter strand would be transiently unmethylated for a short time. Because the cell survives during this time, you can conclude that only one methyl group on a restriction site can stop the endonuclease from cutting.

6. For DNA that has equal proportions of A, C, G, and T, each base has a 0.25 probability

of appearing at any position in the sequence. Since the EcoRV site, GATATC, is six bases long, (0.25)6¥4.6¥106=976. We would, thus, expectª1000 EcoRI sites in the genome of E. coli.

7. The Keq value would be near one because a phosphoanhydride bond between the b and

l phosphorous atoms is broken in ATP and the same bond is formed to link the a and

b phosphorous atoms of the product ADP.

8. The enzyme would likely hydrolyze ATP to ADP by transferring its l phosphoryl group

to water rather than to AMP. The inability of the mutant to close the lid would allow water into the active site where it would react with the ATP to form ADP and Pi.

9. Because of its ring structure, the imino acid proline cannot be accommodated in the sub

strate binding sites of trypsin, chymotrypsin, or carboxypeptidase A. Therefore these proteases fail to cleave peptide bonds involving proline.

10. Chymotrypsin would produce the following four fragments:

Lys-Gly-Phe

Thr-Tyr-Pro-Asn-Trp

Ser-Tyr Phe

11. When the substrate occupies the active site in the enzyme, the weak bonds that it forms

with groups on the enzyme help to stabilize the tertiary structure of the enzyme and protect it against thermal denaturation.

12. The enzyme must have functional groups in the active site that can interact specifically

with the substrate to distinguish it from other similar molecules and position it properly for a productive reaction. Usually, the enzyme must also have catalytic residues

that react with a specific chemical bond of the substrate during the development of the transition state. Both ground-state interactions with the substrate by specific binding and the ability to catalyze the chemistry of the reaction determine the ability of an enzyme to convert a substrate to a product.

13. a.

Unlike chymotrypsin, the existence of the burst with RsrI DNA methylase is not due to a covalent enzyme-substrate complex. The appearance of a burst in an enzyme reaction reveals only that some step past the chemistry occurring during bond making and breaking is limiting the overall rate. The RsrI methylase reaction is known to proceed without a covalent enzyme-substrate intermediate. With this methylase, the burst indicates that some step subsequent to the addition of the methyl group onto the DNA is the rate-limiting step of the reaction. Likely, the ratelimiting step is the release of product from the enzyme. The fact that 0.94 mol of DNA was methylated per mol of enzyme indicates that at least 94% of the enzyme molecules were active. The experiment measured the initial reaction of all the enzyme molecules present because the enzyme was preloaded with AdoMet and then given DNA at a concentration that would also saturate it with the methyl acceptor. No excess, free enzyme existed in the solution, and the initial reaction observed (the burst) measured a single turnover.

b.

The protocol in problem 2 is an isotope-partitioning experiment. That a burst was detected when the enzyme was preloaded with labeled AdoMet before being mixed with excess DNA and a 50-fold excess of unlabeled AdoMet means that AdoMet bound to the enzyme before the DNA binds can be catalytically competent. If the radiolabeled AdoMet would have dissociated from the enzyme before reacting, its specific activity would have been decreased 50-fold by the unlabeled AdoMet in the solution, and the maximum incorporation would have been 2% of that seen in the burst experiment (Question 1). This result does not prove that the reaction is ordered with the order of binding being AdoMet first and DNA second. It only shows that AdoMet can be bound first and be used in the reaction after DNA binds. The order of addition of the substrates to the enzyme might be random with either AdoMet or DNA binding first. Further experiments would be needed to resolve this question (Both of these questions were derived from Szegedi, S.S., Reich, N.O., and Gumport, R. I. [2000]. Substrate binding in vitro and kinetics of RsrI [N6-adenine] DNA methyltransferase. Nucleic Acids Res. 28: 3962–3971.)

c.

The buffers must donate and accept protons at some distance from the active center of the enzyme because they are too large to access it. The protons supplied by these buffers reach the reaction center by being transported or shuttled through a network of proton carriers that comprises ionizable groups on the protein and water molecules. The fact that active site residues with pKa values different from those of the wild-type enzyme function in the reaction suggests that the precise location and strength of the ionizable groups are not critical to the functioning of the shuttle. The malleability of the positioning of the active site, ionizing amino acid side chains probably results from the formation of different networks of variable numbers of hydrogen-bonded water molecules. These networks form in various shapes to accommodate the altered positions of the variant amino acid side chains. (This question was derived from Qian, M., Earnhardt, J. N., Qian, M., Tu, C., Laipis, P. J., and Silverman D. N. [1998]. Intramolecular proton transfer from multiple sites in catalysis by murine carbonic anhydrase V. Biochemistry 37: 7649–7655.) 12

CHAPTER 9 EXPANDED SOLUTIONS TO TEXT PROBLEMS

1. The answer concerns the different kinetic behavior of chymotrypsin toward amide and

ester substrates. Substrate A is N-acetyl-L-phenylalanine p-nitrophenyl amide, rather than N-acetyl-L-phenylalanine p-nitrophenyl ester for which the initial “burst” activity was described in the text. The burst is observed if the first step of a reaction (in this case, acylchymotrypsin formation, together with release of p-nitrophenyl amine) is much faster than the second step (release of N-acetyl-phenylalanine and free chymotrypsin). With the amide substrate, however, the relative rates of the two steps are more nearly equal; therefore no burst is observed.

2. The Ala-64 subtilisin lacks the critical histidine in the catalytic triad of the active site and

therefore cleaves most substrates much more slowly than does normal subtilisin. However, the histidine in substrate B can act as a general base; thereby the substrate itself partially compensates for the missing histidine on the mutant enzyme.

3. The statement is false (incorrect). Either mutation alone is such a serious impairment for

the enzyme that the second mutation will be of little additional consequence.

4. A reasonable prediction is that the substrate specificity of the mutant protease would re

semble that of trypsin. The mutant enzyme would be predicted to hydrolyze peptide bonds that follow either lysine or arginine in the sequence (i.e., peptide bonds whose carbonyl groups are from either lysine or arginine).

5. Small molecular buffers such as imidazole can diffuse into the active site of carbonic an

hydrase and substitute for the proton shuttle function of His 64 near the zinc ion. Large molecular buffers cannot fit into the active site because of their steric bulk and therefore cannot compensate for the loss of the side chain of His 64.

6. The enzyme would not be useful. The probability of finding a particular unique restric

tion site of length 10 is 1/(410)=1/1,048,576, i.e., about once per million base pairs of DNA. Therefore most viral genomes that contain only 50,000 base pairs have little statistical chance of having a site that would be recognized by this enzyme.

7. The increased rate would not be beneficial. Because only a small number of cuts (or

even a single cut) of an invading foreign DNA molecule will be sufficient to inactivate the foreign DNA, the host cell would realize no practical benefit from a faster rate of hydrolysis. Specificity is more important than turnover number for restriction endonucleases.

8. In the absence of the gene for the corresponding methylase, there would be no benefit.

The restriction endonuclease from the newly acquired gene would digest the host cell’s own DNA.

9. a. (Assuming that magnesium also is present), ATP and AMP will be generated from two

molecules of ADP in a “reverse” adenylate kinase reaction. Enzymes catalyze both forward and reverse reactions.

b. The answer will require knowledge of an equilibrium constant for the reaction ATP+AMPG2 ADP. In this reaction, the two substrates together are approximately isoenergetic with the products. If one therefore assumes an equilibrium constant of one, then: [ADP]2/([ATP] [AMP])=1.

CATALYTIC STRATEGIES 13

Let [ATP]=x at equilibrium. Then [AMP]=x, and [ADP]=((1 mM)-(2x)). (1-2x)2/(x2)=1. (1-4x+4x2)=x2. 3x2-4x+1=0.

Solve for x.

Two answers emerge. Either x=1, or x=1⁄3. However, x=1 is physically unreasonable (impossible). Therefore x=1⁄3. The concentrations of [ATP], [ADP], and [AMP] therefore are all 0.3333 mM.

10. The chelator will remove the zinc from the active site of the enzyme. Without zinc, the

carbonic anhydrase is inactive.

11. Molecule A is an analogue of N-acetyl-lysine that is likely to inhibit trypsin. The posi

tively charged e-ammonium group will bind in trypsin’s substrate specificity pocket. Additionally, the B-O group is likely to bind in the oxyanion hole (Figure 9.9). Because there is no peptide bond to be cleaved, the inhibitor will remain bound to the enzyme and will interfere competitively with the binding of natural substrates.

12. Aldehydes can react with one molecule of an alcohol to form a hemiacetal (see Chapter

11). Because the catalytic site of elastase contains an active serine hydroxyl group, it is reasonable that an aldehyde derivative of a peptide substrate of elastase would react with the serine -OH group to form a hemiacetal, which is a tetrahedral analogue of the transition state of the peptide hydrolysis reaction. (See also Robert C. Thompson and Carl A. Bauer. [1979]. Biochemistry 18, 1552–1558.)

13. a.

O

X

K

X

HX

K

–

H

–

H

–

H

N

N- - - -H–S

R

+ N–H- - - -S

R

N

S

R

2

3

O

4

HX

X

K

1

R

O

K

–

H

–

H

N

N- - - -H–S

N

S

H O

O

2

5

K

8

R

OH

O

HO

H

O

HO

K

|

–

H

–

H

K

–

| O

N

H

N- - - -H–S

R

+ N–H- - - -S

R

N

S

R

7

6 14

CHAPTER 9

b.

R

H—

HO

K - - -

O

O- - -

O

–

H

O

H

X

—

X

—

O

K

2

OH

O

OK—

3

O

X

K

1

R

K - - -

O

H—

O

HX

H

O - - - -

HO

—

O

H

–

—

O

H

OK—

O

OK—

4

HX

R

5

KO

K - - -

O

H

HO

O

HO

—

H O

OK—

2

c.

O

X

K

R

H—

BA

R

HO

—

BA - - -

O

|

X

—

|

2

—

|

H

|

Zn2+

—

Zn2+

—

O

3

X

K

1

R

K

HX

OH

BA - - -

O

H

—

|

BA

|

—

O

Zn2+

—

|

—

H

| |

Zn2+

—

4

HX

HO

R

5

R

K

BA

KO

—

O

| |

HO

|

H O

Zn2+

—

2

— This page intentionally left blank. CHAPTER 1 Regulatory Strategies: Enzymes and Hemoglobin 0

The theme of Chapter 10 is the regulation of protein function. Four major types

of regulatory mechanisms are discussed in detail: allosteric control, isozymes, reversible covalent modification, and proteolytic activation. The authors use spe

cific examples to illustrate the general structure-function relationships involved in these control mechanisms. To illuminate allosteric control, the authors discuss E. coli aspartate transcarbamoylase (ATCase) and hemoglobin (Hb), two well-understood allosterically regulated proteins. ATCase is the enzyme that catalyzes the condensation of carbamoyl phosphate and aspartate in the first step of pyrimidine biosynthesis. Its activity is regulated both positively and negatively and provides a classic example of feedback inhibition of enzymes in multistep biosynthetic pathways. Coorperative binding of oxygen to hemoglobin is critical to its ability to efficiently transport oxygen in blood and release it to myoglobin in tissues. Because it is also regulated by H+, CO2, and 2,3-biphosphoglycerate (2,3-BPG), Hb provides an excellent example of both homotropic and heterotropic allosteric regulation of proteins.

After the sections on allosteric control, the authors illustrate the use of isozymes

to regulate enzymes in a developmental and/or tissue-specific manner using lactate dehydrogenase as an example. Next the authors discuss the regulation of enzymes by covalent modifications such as phosphorylation, acetylation, lipidation, and ubiquination. The authors focus on reversible phosphorylation as a control mechanism and use cAMPdependent protein kinase (PKA) as an example of how phosphorylation of target proteins can be regulated. The authors then turn to the activation of enzymes by proteolytic cleavage. They describe the proteolytic steps and conformational rearrangements that produce the active forms of chymotrypsin, trypsin, and pepsin from their inactive zymogens. The mechanisms of action of the digestive enzymes were presented in Chapter 9. The authors conclude Chapter 10 with a discussion of the blood clotting cascade— the series of proteolytic activations of clotting factors that lead to the formation of fibrin clots. Several specific stimulating and inhibiting proteins are described in connection with the proteolytic enzymes. 159 160

CHAPTER 10 LEARNING OBJECTIVES

When you have mastered this chapter, you should be able to complete the following objectives. Introduction

1. List the four major regulatory mechanisms that control enzyme activity and give exam

ples of each. Aspartate Transcarbamoylase Is Allosterically Inhibited by the End Product of Its Pathway (Text Section 10.1)

2. Describe the reaction catalyzed by aspartate transcarbamoylase (ATCase), the regulation

of ATCase by CTP and ATP, and the biological significance of this regulation.

3. Describe the composition and arrangement of the subunits of ATCase and the major fea

tures of its active site as revealed by the binding of N-(phosphonacetyl)-L-aspartate (PALA). Explain the effects of subunit dissociation on the allosteric behavior of the enzyme.

4. Outline the structural effects of binding of CTP and PALA to ATCase.

5. Describe the experimental evidence for a concerted allosteric transition during the bind

ing of substrate analogs to ATCase.

6. Outline the effects of heterotropic and homotropic allosteric interactions on the equilib

rium between the T and R forms of ATCase.

7. Differentiate between concerted and sequential mechanisms of allosteric regulation. Hemoglobin Transports Oxygen Efficiently by Binding Oxygen Cooperatively (Text Section 10.2)

8. Contrast the oxygen binding properties of myoglobin and hemoglobin. Define the coop- erative binding of oxygen by hemoglobin and summarize how it makes hemoglobin a better oxygen transporter.

9. Explain the significance of the differences in oxygen dissociation curves, in which the frac- tional saturation (Y) of the oxygen-binding sites is plotted as a function of the partial pressure of oxygen (pO2), for myoglobin and hemoglobin.

10. State the major structural differences between the oxygenated and deoxygenated forms of

hemoglobin.

11. Explain the effects of CO2 and H+ (the Bohr effect) and 2,3-bisphosphoglycerate (BPG) on

the binding of oxygen by hemoglobin. Describe the structural bases for the effects of these molecules on the binding of oxygen by hemoglobin. Explain the consequences of the metabolic production of CO2 and H+ on the oxygen affinity of hemoglobin.

12. Rationalize the existence of fetal hemoglobin. Isozymes Provide a Means of Regulation Specific to Distinct Tissues and Developmental Stages (Text Section 10.3)

13. Define isozyme. Give examples of ways in which isozymes of a given enzyme can be dif

ferentiated from each other.

REGULATORY STRATEGIES: ENZYMES AND HEMOGLOBIN 161

14. Explain the purpose of isozymes in metabolism.

15. Explain the role of isozymes in the tissue-specific regulation of lactate dehydrogenase. Covalent Modification Is a Means of Regulating Enzyme Activity (Text Section 10.4)

16. List the common covalent modifications used to regulate protein activity.

17. Write the basic reactions catalyzed by protein kinases and protein phosphatases.

18. List the reasons why phosphorylation is such an effective control mechanism.

19. Describe the activation of protein kinase A (PKA) by cyclic AMP (cAMP) and the mode of

interaction of PKA with its pseudosubstrate. Many Enzymes Are Activated by Specific Proteolytic Cleavage (Text Section 10.5)

20. Define zymogen. Give examples of enzymes and proteins that are derived from zymogens

and the biological processes they mediate.

21. Summarize the enzymes and conditions required for the activation of all the digestive enzymes.

22. Explain how trypsin is inhibited by the pancreatic trypsin inhibitor.

23. Describe the stabilization of the fibrin clot by Factor XIIIα, an active transamidase. Explain

the role of thrombin in the activation of fibrinogen and Factor XIII.

24. Compare the cleavage specificity of thrombin with those of the pancreatic serine proteases.

25. Discuss the requirement for vitamin K in the synthesis of prothrombin. Outline the mech

anism of prothrombin activation.

26. Explain the genetic defect in hemophilia and discuss how recombinant DNA technology

has been used to produce human Factor VIII (antihemophilic factor).

27. State the general mechanisms for the control of clotting and explain the specific role of antithrombin III in the clotting cascade. Note the effect of heparin on antithrombin III.

28. Describe the lysis of fibrin clots by plasmin and the activation of plasminogen by tissue- type plasminogen activator (TPA). SELF-TEST Aspartate Transcarbamoylase Is Allosterically Inhibited by the End Product of Its Pathway

1. The dependence of the reaction velocity on the substrate concentration for an allosteric

enzyme is shown in Figure 10.1 as curve A. A shift to curve B could be caused by the

(a) addition of an irreversible inhibitor. (b) addition of an allosteric activator. (c) addition of an allosteric inhibitor. (d) dissociation of the enzyme into subunits. 162

CHAPTER 10 FIGURE 10.1 Reaction velocity versus substrate concentration for an allosteric enzyme.

V

A

B

[S]

2. In E. coli, ATCase is inhibited by CTP and is activated by ATP. Explain the biological sig

nificance of these effects.

3. Which of the following statements regarding the structure of ATCase in E. coli are in

correct?

(a) ATCase consists of two kinds of subunits and a total of 12 polypeptide chains. (b) Reaction with mercurials dissociates each ATCase into three r2 and two c3 subunits. (c) ATCase has a threefold axis of symmetry and a large inner cavity. (d) The active sites of ATCase are located at the interface between c and r subunits. (e) The separate subunits r2 and c3 retain their respective ligand-binding capacities.

4. Which of the following methods can provide information about the subunit dissociation

of ATCase or the structural changes that occur when ATCase binds a substrate analog?

(a) x-ray crystallography (b) Western blotting (c) sedimentation-velocity ultracentrifugation (d) SDS-polyacrylamide gel electrophoresis (e) gel-filtration chromatography

5. The allosteric effect of CTP on ATCase is called

(a) homotropic activation. (b) homotropic inhibition. (c) heterotropic activation. (d) heterotropic inhibition. Hemoglobin Transports Oxygen Efficiently by Binding Oxygen Cooperatively

6. Which of the following statements are false?

(a) The oxygen dissociation curve of myoglobin is sigmoidal, whereas that of hemo

globin is hyperbolic.

(b) The affinity of hemoglobin for O2 is regulated by organic phosphates, whereas the

affinity of myoglobin for O2 is not.

(c) Hemoglobin has a higher affinity for O2 than does myoglobin. (d) The affinity of both myoglobin and hemoglobin for O2 is independent of pH.

REGULATORY STRATEGIES: ENZYMES AND HEMOGLOBIN 163

7. Several oxygen dissociation curves are shown in Figure 10.2. Assuming that curve 3 cor

responds to isolated hemoglobin placed in a solution containing physiologic concentrations of CO2 and BPG at a pH of 7.0, indicate which of the curves reflects the following changes in conditions:

(a) decreased CO2 concentration (b) increased BPG concentration (c) increased pH (d) dissociation of hemoglobin into subunits FIGURE 10.2 Oxygen dissociation curves.

1

2

3

4

ation (Y)

Satur

pO2

8. Which of the following statements concerning the Bohr effect are true?

(a) Lowering the pH shifts the oxygen dissociation curve of hemoglobin to the right. (b) The acidic environment of an exercising muscle allows hemoglobin to bind O2 more

strongly.

(c) The affinity of hemoglobin for O2 is diminished by high concentrations of CO2. (d) In the lung, the presence of higher concentrations of H+ and CO2 allows hemoglo

bin to become oxygenated.

(e) In the lung, the presence of higher concentrations of O2 promotes the release of

CO2 and H+.

9. Explain why fetal hemoglobin has a higher affinity for oxygen than does maternal he

moglobin, and why this is a necessary adaptation.

10. The oxygen dissociation curve for hemoglobin reflects allosteric effects that result from

the interaction of hemoglobin with O2, CO2, H+, and BPG. Which of the following structural changes occur in the hemoglobin molecule when O2, CO2, H+, or BPG bind? (a) The binding of O2 pulls the iron into the plane of the heme and causes a change in

the interaction of all four globin subunits, mediated through His F8.

(b) BPG binds at a single site between the four globin subunits in deoxyhemoglobin

and stabilizes the deoxyhemoglobin form by cross-linking the b subunits.

environment of His and the a-NH2 groups of the a chains changes, rendering these groups less acidic when O2 is released.

(d) The binding of CO2 stabilizes the oxy form of hemoglobin. 164

CHAPTER 10

11. The structure of deoxyhemoglobin is stabilized by each of the following interactions except for

(a) BPG binding. (b) salt bridges between acidic and basic side chains. (c) coordination of the hemes with the distal histidine. (d) hydrophobic interactions. (e) salt bridges involving N-terminal carbamates.

12. In the transition of hemoglobin from the oxy to the deoxy form, an aspartate residue is

brought to the vicinity of His 146. This increases the affinity of this histidine for protons. Explain why. Isozymes Provide a Means of Regulation Specific to Distinct Tissues and Developmental Stages

13. Which of the following would not be useful in distinguishing one isozyme from another?

(a) electrophoretic mobility (b) gene sequence (c) kinetic rate constant (d) allosteric regulators Covalent Modification Is a Means of Regulating Enzyme Activity

14. Protein kinases

(a) transfer a phosphoryl group from one protein to another. (b) use AMP as a substrate. (c) use Thr, Ser, or Tyr as the acceptor groups for phosphoryl transfer. (d) transfer the a phosphorus atom of ATP. (e) are located on the external surface of cells.

15. Explain how a phosphoryl group can change the conformation of a protein.

16. Protein kinase A

(a) is activated by ATP. (b) consists of two catalytic (c) and two regulatory (r) subunits in the absence of ac

tivator.

(c) upon binding the activator dissociates into one c2 and two r subunits. (d) contains a pseudosubstrate sequence in the c subunits. Many Enzymes Are Activated by Specific Proteolytic Cleavage

17. The pancreas is the source of the proteolytic enzyme trypsin. Which of the following are

reasons trypsin does not digest the tissue in which it is produced?

(a) It is synthesized in the form of an inactive precursor that requires activation. (b) It is stored in zymogen granules that are enclosed by a membrane. (c) It is active only at the pH of the intestine, not at the pH of the pancreatic cells. (d) It requires a specific noncatalytic modifier protein in order to become active.

REGULATORY STRATEGIES: ENZYMES AND HEMOGLOBIN 165

18. Activation of chymotrypsinogen requires

(a) the cleavage of at least two peptide bonds by trypsin. (b) structural rearrangements that complete the formation of the substrate cavity and

the oxyanion hole.

(c) major structural rearrangements of the entire protein molecule. (d) the concerted proteolytic action of trypsin and pepsin to give a-chymotrypsin.

19. Match the zymogens in the left column with the enzymes that participate directly in their

activation which are listed in the right column.

(a) chymotrypsinogen

(1) trypsin

(b) trypsinogen

(2) enteropeptidase

(3) carboxypeptidase

(d) procarboxy-peptidase

20. Explain why the new carboxyl-terminal residues of the polypeptide chains produced

during the activation of pancreatic zymogens are usually Arg or Lys.

21. The inactivation of trypsin by pancreatic trypsin inhibitor involves

(a) an allosteric inhibition. (b) the covalent binding of a phosphate to the active site serine. (c) the facilitated self-digestion of the enzyme. (d) denaturation at the alkaline pH of the duodenum. (e) the nearly irreversible binding of the protein inhibitor at the active site.

22. Match fibrinogen and fibrin with the appropriate properties in the right column.

(a) fibrinogen

(1) is soluble in blood

(b) fibrin

(2) is insoluble in blood (3) forms ordered fibrous arrays (4) contains a-helical coiled coils (5) may be cross-linked by transamidase

23. Which of the following statements about prothrombin are incorrect?

(a) It requires vitamin K for its synthesis. (b) It can be converted to thrombin by the decarboxylation of g-carboxyglutamate

residues.

(c) It is activated by Factor IXa and Factor VIII. (d) It is anchored to platelet phospholipid membranes through Ca2+ bridges. (e) It is part of the common pathway of clotting.

24. Explain the role of the g-carboxyglutamate residues found in clotting factors.

25. Which of the following mechanisms is not involved in the control of the clotting process?

(a) the specific inhibition of fibrin formation by antielastase (b) the degradation of Factors Va and VIIIa by protein C, which is in turn switched on

by thrombin

(c) the dilution of clotting factors in the blood and their removal by the liver (d) the specific inhibition of thrombin by antithrombin III

26. Explain the effects of each of the following substances on blood coagulation or clot

dissolution:

(a) heparin (b) dicumarol (c) tissue-type plasminogen activator

27. Which of the following statements about plasmin are true?

(a) It is a serine protease. (b) It diffuses into clots. (c) It cleaves fibrin at connector rod regions. (d) It is inactivated by a1-antitrypsin. (e) It contains a “kringle” region in its structure for binding to clots. ANSWERS TO SELF-TEST

1. c

2. The activation of ATCase by ATP occurs when metabolic energy is available for DNA

replication and the synthesis of pyrimidine nucleotides. Feedback inhibition by CTP prevents the overproduction of pyrimidine nucleotides and the waste of precursors.

3. d

4. a, c, e

5. d

6. a, c, d

7. (a) 2 (b) 4 (c) 2 (d) 1

8. a, c, e

9. Fetal hemoglobin is composed of different subunits than adult hemoglobin and binds

BPG less strongly. As a result, the affinity of fetal hemoglobin for oxygen is higher, and the fetus can extract the O2 that is transported in maternal blood.

10. a, b, c

11. c

12. The pK values of ionizable groups are sensitive to their environment. The change in

the environment of His 146 in deoxyhemoglobin increases its affinity for protons as a result of the electrostatic attraction between the negative charge of the aspartate and the proton.

13. b

14. c

15. A phosphoryl group introduces two negative charges that can affect the electrostatic

interactions within the protein. In addition, a phosphoryl group can form three highly directional hydrogen bonds to adjacent H-bond partners in the protein. These local effects can be transmitted to more distant parts of the protein in a manner similar to allosteric effects.

16. b

17. a, b

18. b

19. (a) 2 (b) 3, 2 (c) 2 (d) 2

20. Because trypsin is the common activator of the pancreatic zymogens, its specificity for

Arg-X and Lys-X peptide bonds will produce Arg and Lys carboxyl-terminal residues.

21. e

22. (a) 1, 4 (b) 2, 3, 4, 5

23. b, c

24. The g-carboxyglutamate residues are effective chelators of Ca2+. This Ca2+ is the elec

trostatic anchor that binds the protein to a phospholipid membrane, thereby bringing interdependent clotting factors into close proximity.

25. a

26. (a) Heparin enhances the inhibitory action of antithrombin III.

(b) Dicumarol is a vitamin K analog, and, as such, it interferes with the synthesis of the

factors that contain g-carboxyglutamate residues.

minogen into plasmin directly on the clot.

27. a, b, c PROBLEMS

1. What would be the kinetic consequences if a substrate were to have exactly equal affini

ties for the R form and the T form of an allosteric enzyme?

2. In a spectroscopic study designed to elucidate the mechanism of the allosteric transi

tion in ATCase, hybrid enzyme molecules were formed containing (in addition to regulatory subunits) three native catalytic subunits, and three modified catalytic subunits. The modified catalytic subunits contained the nitrotyrosine reporter group and a modified lysine that precluded substrate binding. Why was this modified lysine necessary to the experiment?

3. Aspartate transcarbamoylase catalyzes the first step in the biosynthetic pathway leading

to the synthesis of cytidine triphosphate (CTP). CTP serves as an allosteric inhibitor of aspartate transcarbamoylase that shuts off the biosynthetic pathway when the cell has an ample supply of CTP. Although the first step in a pathway may often be the principal regulatory step, such is not always the case. Figure 10.3 shows a hypothetical degradative metabolic pathway in which step 3 is the principal regulatory step. In this pathway, what advantage does regulation at step 3 have over regulation at step 1 or 2? FIGURE 10.3 A hypothetical metabolic pathway.

A

1

2

C B

3

D

4

E

5

F

6

X 168

CHAPTER 10

4. Explain why the reagent N-(phosphonacetyl)-L-aspartate (PALA) has been especially use

ful in the investigation of the properties of ATCase.

5. Explain how PALA can act as both an activator and an inhibitor of ATCase.

6. One molecule of 2,3-bisphosphoglycerate binds to one molecule of hemoglobin in a cen

tral cavity of the hemoglobin molecule. Is the interaction between BPG and hemoglobin stronger or weaker than it would be if BPG bound to the surface of the protein instead? Explain your answer.

7. An effective respiratory carrier must be able to pick up oxygen from the lungs and de

liver it to peripheral tissues. Oxygen dissociation curves for substances A and B are shown in Figure 10.4. What would be the disadvantage of each of these substances as a respiratory carrier? Where would the curve for an effective carrier appear in the figure? FIGURE 10.4 Oxygen dissociation curves for substances A and B.

1.0

A

B

0.5

ation (Y)

Satur

0

pO

2

in tissues in lungs

8. Patients suffering from pneumonia have a portion of their lungs filled with fluid, and

therefore have reduced lung surface area available for oxygen exchange. Standard hospital treatment of these patients involves placing them on a ventilating machine set to deliver enough oxygen to keep their hemoglobin approximately 92% saturated. Why is this value selected rather than one lower or higher?

9. What major differences exist between the sequential and concerted models for allostery

in accounting for hemoglobin that is partially saturated with oxygen?

10. Predict whether each of the following peptide sequences is likely to be phosphorylated

by protein kinase A. Briefly explain your answers, and indicate which residue would be phosphorylated.

(a) Ala-Arg-Arg-Ala-Ser-Leu (b) Ala-Arg-Arg-Ala-His-Leu (c) Val-Arg-Arg-Trp-Thr-Leu (d) Ala-Arg-Arg-Gly-Ser-Asp (e) Gly-Arg-Arg-Ala-Thr-Ile

11. Consider the hypothetical metabolic sequence shown in Figure 10.5. Suppose it is

known that protein kinase A phosphorylates both enzyme 1 and enzyme 2, and that an increase in intracellular cAMP levels increases the steady-state [B]/[A] ratio. In

REGULATORY STRATEGIES: ENZYMES AND HEMOGLOBIN 169

order for a given increment in cAMP concentration to result in the largest change in the steady-state [B]/[A] ratio, what should be the effect of phosphorylation on the activities of enzymes 1 and 2? FIGURE 10.5 Hypothetical metabolic sequence.

F

B

C

Enz 2

Enz 1

E

A

D

12. Figure 10.6 shows time courses for the activation of two zymogens, I and II. Which of

the time courses more resembles that of the activation of trypsinogen, and which corresponds to the activation of chymotrypsinogen? Explain. FIGURE 10.6 Time courses for the activation of two zymogens, I and II.

ation

e enzyme concentr

II

Activ

Time

13. Trypsin has 13 lysine and 2 arginine residues in its primary structure. Why does trypsin

not cleave itself into 16 smaller peptides?

14. Although thrombin has many properties in common with trypsin, the conversion of

prothrombin to thrombin is not autocatalytic whereas the conversion of trypsinogen to trypsin is autocatalytic. Why is the conversion of prothrombin to thrombin not autocatalytic?

15. Because many clotting factors are present in blood in small concentrations, direct chem

ical measurements often cannot be used to determine whether the factors are within normal concentration ranges or are deficient. Once a deficiency has been established, however, plasma from the affected person can be used to screen for the presence of the deficiency in other people. A rare deficiency in Factor XII leads to a prolongation of clotting time. Assuming that you have plasma from someone in which this deficiency has been established, design a test that might help determine whether another person has a Factor XII deficiency. 170

CHAPTER 10

16. In general, regulatory enzymes catalyze reactions that are irreversible in cells, that is,

reactions that are far from equilibrium. Why must this be the case?

17. Amplification cascades, such as the one involved in blood clotting, are important in a

number of regulatory processes. Figure 10.7 shows a hypothetical cascade involving conversions between inactive and active forms of enzymes. Active enzyme A serves as a catalyst for the activation of enzyme B. Active B in turn activates C, and so forth. Assume that each enzyme in the pathway has a turnover number of 103. How many molecules of enzyme D will be activated per unit time when one molecule of active enzyme A is produced per unit time? FIGURE 10.7 A hypothetical regulatory cascade.

Aactive

B

inactive

active

C

inactive

active

D

inactive

active

18. Thrombin and trypsin are both serine proteases that are capable of cleaving the peptide

bond on the carboxyl side of arginine; thrombin, however, is specific for Arg-Gly bonds. Describe briefly the similarities and differences in the active sites of these two enzymes. ANSWERS TO PROBLEMS

1. If a substrate were to have equal affinities for the R and T forms, the forms would be in

distinguishable kinetically and the system would behave as if all the enzyme were present in a single form. Thus, Michaelis-Menten kinetics would apply, and a plot of the reaction velocity versus the substrate concentration would be hyperbolic.

2. The point of the experiment was to show that the subunits containing the reporter group

undergo conformational change because substrate is bound to a neighboring subunit of the enzyme. Thus binding of substrate to the subunit containing the reporter group had to be precluded if the experiment was to give meaningful results.

3. The pathway shown in Figure 10.3 is branched. If regulation were to occur at step 1

only, there would be no control over the production of X from B. If only step 2 were regulated, there would be no regulation over the production of X from A. Regulation at step 3 provides control of the amount of X produced from both A and B. In branched pathways, the principal regulatory step is usually after the branch point.

4. PALA is a bisubstrate analog; that is, it resembles a combination of both substrates, and

it is a transition state analog for the carbamoyl phosphate-aspartate complex during catalysis by ATCase. X-ray diffraction analysis of ATCase with bound PALA has revealed the location of the active site and interactions that occur within it. In addition, comparisons of structures with and without PALA have indicated the large structural changes that ATCase undergoes upon binding substrates.

REGULATORY STRATEGIES: ENZYMES AND HEMOGLOBIN 171

5. PALA is an analog of both substrates of ATCase; therefore, it binds to the active site and

acts as a potent inhibitor. At low concentrations, however, binding of PALA shifts the distribution of ATCase molecules to the R conformation. This increases binding of substrates and enzymatic activity.

6. BPG binds to hemoglobin by electrostatic interactions. These interactions between the

negatively charged phosphates of BPG and the positively charged residues of hemoglobin are much stronger in the interior hydrophobic environment than they would be on the surface, where water would compete and weaken the interaction by binding both to BPG and to the positively charged residues. Remember that the force of electrostatic interactions, given by Coulomb’s law, is inversely proportional to the dielectric constant of the medium. The dielectric constant in the interior of a protein may be as low as 2. Hence, electrostatic interactions there are much more stable than those on the surface, where the dielectric constant is approximately 80.

7. Substance A would never unload oxygen to peripheral tissues. Substance B would never

load oxygen in the lungs. An effective carrier would have an oxygen dissociation curve between those depicted for substance A and substance B. It would be relatively saturated with oxygen in the lungs and relatively unsaturated in the peripheral tissues.

8. Look at the oxygen saturation curve for hemoglobin given in Figure 10.17 of the text

book. Administering enough oxygen to give saturation levels greater than approximately 92% would be wasteful of oxygen, because one reaches the point of diminishing returns. Administering oxygen in amounts less than that required for 92% saturation runs the risk of compromising oxygen delivery to the tissues.

9. According to the concerted model, hemoglobin partially saturated with oxygen is com

posed of a mixture of fully oxygenated molecules with all subunits in the R form and fully deoxygenated molecules, with all subunits in the T form. According to the sequential model, individual molecules would have some subunits that are oxygenated (in the R form) and some that are deoxygenated (in the T form).

10. The consensus motif recognized by protein kinase A is Arg-Arg-X-Ser-Z, where X is a

small residue and Z is a large hydrophobic residue. The site of phosphorylation is either Ser or Thr.

(a) Ser would be phosphorylated. (b) There would be no phosphorylation because neither Ser nor Thr is present. (c) There would be no phosphorylation because Trp is a bulky group and residue X

must be small.

(d) There would be no phosphorylation because Asp is polar and charged and residue

Z must be hydrophobic.

(e) Thr would be phosphorylated.

11. Phosphorylation should increase the activity of enzyme 2 and decrease the activity of en

zyme 1. That being the case, an increase in intracellular cAMP levels could greatly increase the steady-state ratio of [B]/[A]. Such coordinated, reciprocal control of opposing metabolic sequences is observed frequently in cells.

12. Curve I corresponds to the activation of trypsinogen, a process that is autocatalytic. As the

process occurs, trypsin is produced, which can then cleave yet more trypsinogen. Curve II corresponds to the activation of chymotrypsinogen. The activation of chymotrypsinogen is not autocatalytic. Rather, tryspin catalyzes the conversion of chymotrypsinogen to active p-chymotrypsin. Therefore, its time course is initially linear. 172

CHAPTER 10

13. The lysine and arginine residues must be partially buried and inaccessible to the active

site of trypsin.

14. Thrombin specifically cleaves Arg-Gly bonds. The two bonds that are broken when pro

thrombin is converted to thrombin are Arg-Thr and Arg-Ile. Therefore, the conversion cannot be autocatalytic.

15. Prepare two samples of blood from the person to be tested. Add normal plasma to one

sample and Factor XII-deficient plasma to another. If clotting time is restored to normal in both samples, Factor XII deficiency is probably not involved. If the addition of normal plasma restores normal clotting time but the addition of Factor XII-deficient plasma does not, then a Factor XII deficiency must be suspected.

16. Suppose that a reaction is at equilibrium. If the enzyme catalyzing that reaction were

made more active, nothing would happen. The reaction would still be at equilibrium. If, on the other hand, the reaction is displaced far from equilibrium and the enzyme catalyzing the reaction is made more active, more product will be produced. Thus, a regulatory enzyme must catalyze an irreversible step if it is to increase the flux rate through a pathway when it is allosterically activated.

17. One molecule of active A will lead to the activation of 109 molecules of enzyme D per

unit time. Active A will produce 103 molecules of active B. Each of the 103 molecules of active B will activate 103 molecules of C per unit time. Since there are 103 molecules of B, this gives a total of 106 molecules of active C. Similar reasoning leads to the answer of 109 molecules of active D.

18.

Because both thrombin and trypsin are serine proteases, they both have an oxyanion

hole and a catalytic triad at the active site. Also, the substrate-specificity sites of both have a similar, negatively charged pocket capable of binding Arg. However, thrombin probably has just enough space to accommodate a Gly residue next to the Arg binding site in contrast to trypsin, which has no restrictions as to the amino acid residue that can be accommodated at the corresponding position. EXPANDED SOLUTIONS TO TEXT PROBLEMS

1. Since histidine 134 is thought to stabilize the negative charge on the carbonyl oxygen in

the transition state (Figure 10.7), the protonated imidazole ring (which carries a positive charge) must be the active species. That being the case, the enzyme velocity V should be half of Vmax at a pH of 6.5 (the pK of an unperturbed histidine side chain in a protein). Raising the pH above 6.5 will remove protons from the imidazole ring, thus causing a decrease in V; lowering the pH below 6.5 will have a reverse effect.

2. (a) One can show that the change in [R]/[T] is the same as the ratio of the substrate

affinities of the two forms. For example, the mathematical constant for the conversion of R to T

→

S is the same whether one proceeds R → T → TS

or R → RS

TS.

Let us assume that the constant for the conversion of R to T and R to RS is 103. Since the affinity of R for S is 100 times that of T for S, it follows that the constant for the conversion of T to TS is 10. The constant for the conversion of RS to TS is therefore equal to 103 × 10/103, or 10. Note that the binding of substrate with a hundredfold tighter binding to R changes the R to T ratio from 1/1000 to 1/10.

REGULATORY STRATEGIES: ENZYMES AND HEMOGLOBIN 173

(b) Since the binding of one substrate molecule changes the [R]/[T] by a factor of 100,

the binding of four substrate molecules will change the [R]/[T] by a factor of 1004 = 108. If [R]/[T] in the absence of substrate = 10−7, the ratio in the fully liganded molecule will be 108 × 10−7 = 10.

3. Following the nomenclature in section 10.1.5, L is [T]/[R], the ratio of T to R in the ab

sence of ligand, and L = 105. With j ligands bound, a new Lj will equal (KR/KT)j*L. Then we have [R]L =

j

[T]. The fraction of molecules in the R state therefore is [R]/([R] + [T]),

or [R]/([R] + [R]Lj), or 1/(1 + Lj). Now we can set up a table, using L = 105, K =

=

5 × 10−6 M, KT

2 × 10−3 M, and j from 0 to 4: Ligands bound (j) Lj = (KR/KT)j*L Fraction R = 1/(1 + Lj)

0

1.0 * 105

10−5

1

250

0.004

2

0.625

0.615

3

1.6 * 10−3

0.998

4

3.9 * 10−6

1.000

4. The concerted model, in contrast with the induced-fit (sequential) model, cannot ac

count for negative cooperativity because, according to this model, the binding of substrate promotes a conformational transition of all subunits to the high-affinity R state. Hence, homotropic allosteric interactions must be cooperative if the concerted model holds. In the sequential (induced-fit) model, the binding of ligand changes the conformation of the subunit to which it is bound but not that of its neighbors. This conformational change in one subunit can increase or decrease the binding affinity of other subunits in the same molecule and can, therefore, account for negative cooperativity.

5. The binding of PALA switches ATCase from the T to the R state because it acts as a

substrate analog. An enzyme molecule containing bound PALA has fewer free catalytic sites than does an unoccupied enzyme molecule. However, the PALA-containing enzyme will be in the R state and hence have higher affinity for the substrates. The dependence of the degree of activation on the concentration of PALA is a complex function of the allosteric constant L0, and of the binding affinities of the R and T states for the analog and the substrates. For an account of this experiment, see J. Foote and H. K. Schachman, J. Mol. Biol. 186(1985):175.

6. The mutant enzyme would be in the R state essentially all of the time, independent

of whether or not substrate was present. The reaction rate would therefore depend only on the fraction of active sites occupied by substrate, according to a classical saturation curve, and would follow approximately Michaelis-Menten kinetics (with a classical apparent KM).

7. (a) A higher pH corresponds to a lower [H+]. By the Bohr effect, the oxygen affinity

will be higher at pH 7.4

(b) Increasing the partial pressure of CO2 will lower the oxygen affinity (again by means

of the Bohr effect).

[2,3-BPG] will therefore lower the oxygen affinity and favor the release of oxygen.

(d) The monomeric subunits will lack cooperativity and will behave approximately as

myoglobin. The isolated subunits therefore will have higher affinity for oxygen. 174

CHAPTER 10

8. To replace the function of 2,3-BPG, a molecule should have a high density of negative

charge. Among the molecules listed, the best candidate is (b) inositol hexaphosphate, an analogue of a major natural hemoglobin effector, inositol pentaphosphate in avian or turtle erythrocytes. (See J. Biol. Chem., 274[10][March 5, 1999]:6411–6420, and references therein.)

9. (a) A nearby positively charged lysine side chain will stabilize the negatively charged

carboxylate form of the Glu side chain. This effect will favor the release of H+ from the side chain, thereby lowering the pK.

(b) As a converse to part (a), a nearby negatively charged carboxyl group will favor the

retention of H+ on the Glu side chain, and will therefore raise the pK.

the neutral form of the side chain over the negatively charged carboxylate form, and will therefore promote retention of the H+ and raise the pK.

10. The activation of zymogens involves the cleavage of one or more peptide bonds. In the

case of pepsinogen, when the catalytic site is exposed by lowering the pH, it hydrolyzes the peptide bond between the percursor and pepsin moities. Note that this activation is autocatalytic. Therefore, the time required for activation of half the pepsinogen molecules is independent of the total number of the molecules present.

11. If both patients have a Factor VIII deficiency, a mixture of the two bloods will not clot.

However, if the second patient’s bleeding disorder is due to the deficiency of another factor, a mixture of the two bloods should clot. This type of assay is called a complementation test.

12. The function of Factor X is to convert prothrombin to thrombin on phospholipid mem

branes derived from blood platelets. This proteolytic activation removes the aminoterminal fragment of prothrombin, which contains Ca+2-binding sites, and releases thrombin to activate fibrinogen. Meanwhile, Factor X remains bound to the platelet membrane, where it can activate other prothrombin molecules, because during activation it retains the Ca+2-binding g-carboxyglutamate residues.

13. Apparently antithrombin III is a very poor substrate for thrombin. Remember, many en

zyme inhibitors have high affinity for active sites. Thrombin, not prothrombin, can react with antithrombin III because it has available a fully formed active site.

14. One needs to remember a-helical coiled coils, introduced in Chapter 3 of your textbook

(p. 56). Examination of Figure 3.34 (p. 58) suggests that near the axis of the superhelix some amino acid residues are located in the interior (hydrophobic) portion of the molecule. Since this is a long molecule made up of repeating units, one would expect to have hydrophobic side chains at regular intervals in the molecule.

15. Methionine 358 has a side chain that not only is essential for the binding of elastase by

a1-antitrypsin but also is most susceptible to oxidation by cigarette smoke. What is needed is a side chain resistant to oxidation yet having a strong binding affinity for elastase. A likely choice would be leucine, the side chain of which is much more stable than that of methionine but which has nearly the same volume and is very hydrophobic.

16. The concerted model (in which all subunits change conformation in response to the first

instance of substrate binding) predicts that the change in fR should precede the change

in Y. By contrast, the sequential model predicts that the fraction of subunits in the R state (fR) should equal the fraction containing bound substrate (Y). The results in the figure therefore are best explained by the concerted model.

17. As in problem 16, this experiment also supports a concerted mechanism. The change in

the absorbance at 430 nm reports a conformational change in response to substrate binding at a distant site (on another trimer). (Substrate is prevented from binding to the same trimer that reports the 430 nm absorbance change.) Thus, the binding of succinate to the active sites of a native trimer alters the structure of a different trimer (that carries the reporter nitrotyrosine group).

18. The binding of ATP to the regulatory subunits produces the same absorbance change

at 430 nm as did substrate binding in problem 17. ATP therefore is an allosteric activator that drives the catalytic subunits into the active conformation (R state). CTP has a converse or opposite effect, driving the catalytic subunits into an inactive conformation (T state) and decreasing the absorbance at 430 nm.

19. The hydrophobic effect is at work here. The valine side chain on the surface seeks to

avoid water and finds that it can make favorable van der Waals interactions with the leucine and phenylalanine side chains on another deoxy molecule. The effect is to reduce the solubility of the deoxyhemoglogin and cause the crystallization of long fibers that distort the shapes of the red blood cells into the sickled motif.

20. In step 1, the aspartate smino group carries out a nucleophilic attack on the carbonyl

carbon of the carbamoyl phosphate to give a tetrahedral transition state. The histidine in the active site can stabilize the negatively charged oxyanion of this transition state. In step 2, phosphate is the leaving group to generate the N-carbamoylaspartate.

NH

:O

O

H;

J

K

P

:

J

O

NH

:

2

O

:

O

1

O

J

NH

J

K

P

2

:

J

O

+

J N

J

COO:

O

H

J

J

H

:

J

CH COO:

OOC-CH

J J

J

2

2

H N

¨

COO:

(; H;)

2

2

O

KJ NH

2

J JN

H

J

COO:

JJ

N

-carbamoylaspartate :OOC-CH H

2

+

O

K

J

P

phosphate

J

HO

O:

O: 176

CHAPTER 10

21. The reaction is equivalent to a “hydrolysis” (or transfer) of the g-phosphate of ATP, with

the serine-OH group taking the role of water and accepting the phosphate. The enzyme’s active site will need a group to accept the proton from the serine oxygen during the attack on the g-phosphate in step 1. (Histidine plays such a role in the serine proteases (e.g., trypsin and chymotrypsin) and could play a similar role here.) Another valuable functional group at the active site would be one that could stabilize the extra negative charge on the pentacoordinate phosphate intermediate between steps 1 and 2 (before ADP is lost as the leaving group in step 2).

J J

H C

2

J

OH

J

+H;

+

1

J

J

CH2

O

J O-

K

J

K

J J

P

J J

P

J J

P

J

J ribose-A

J J

P

J J

P

J J

P

J

J ribose-A

O

:

:

O

O

:

:

:

:

2

J J

H C

=

2

J

OPO3

+

O

K

J J

P

J J

P

J

J ribose-A

O

:

O

O

:

: CHAPTER 1 Carbohydrates 1

Carbohydrates are one of the four major classes of biomolecules; the others are

proteins, nucleic acids, and lipids. In Chapter 11, the authors describe the chemical nature of carbohydrates and summarize their principal biological roles.

First, they introduce monosaccharides, the simplest carbohydrates, and describe their chemical properties. Since these sections assume familiarity with the properties of aldehydes, ketones, alcohols, and stereoisomers, students with a limited background in organic chemistry should review these topics in any standard organic chemistry text. Next, the chapter discusses simple derivatives of monosaccharides, including sugar phosphates and disaccharides. Sugar is the common name for monosaccharides and their derivatives. You have already seen some monosaccharide derivatives in the structures of nucleic acids in Chapter 5 and nucleotides in Chapter 9. Then, the text discusses polysaccharides and oligosaccharides as storage and structural polymers and as components of proteoglycans and glycoproteins.

Glycoproteins are proteins with carbohydrates attached, generally as oligosac

charides. The attachment of sugars takes place either in the lumen of the endoplasmic reticulum or in the Golgi complex. One reason for attachment of sugars is the targeting of specific proteins to specific sites. For example, attachment of mannose 6phosphate sends proteins from the Golgi complex to the lysosomes. A eucaryotic cell has many different subcellular compartments, each of which has to have a certain array of enzymes and proteins. The Golgi complex functions as the “post office” for the cell, and the attached oligosaccharides function as the “ZIP codes.” Attached sugars can also function as signals for proper folding, or as sites of interaction between cells. Lectins and selectins are proteins that bind specific oligosaccharide clusters on the cell surface. The A, B, and O blood group antigens are examples of cell-surface oligosaccharides. Hemagglutinin allows the influenza virus to bind to sialic acid and thus attach to cells before invading them. 177 178

CHAPTER 11 LEARNING OBJECTIVES

When you have mastered this chapter, you should be able to complete the following objectives. Introduction

1. List the main roles of carbohydrates in nature. Monosaccharides Are Aldehydes or Ketones with Multiple Hydroxyl Groups (Text Section 11.1)

2. Define carbohydrate and monosaccharide in chemical terms.

3. Relate the absolute configuration of monosaccharide D or L stereoisomers to those of glyceraldehyde.

4. Associate the following monosaccharide class names with their corresponding structures: aldose and ketose; triose, tetrose, pentose, hexose, and heptose; pyranose and furanose.

5. Distinguish among enantiomers, diastereoisomers, and epimers of monosaccharides.

6. Draw the Fisher (open-chain) structures and the most common Haworth (ring) structures

of D-glucose, D-fructose, D-galactose, and D-ribose.

7. Explain how ring structures arise through the formation of hemiacetal and hemiketal

bonds. Draw a ring structure, given a Fisher formula.

8. Distinguish between a and b anomers of monosaccharides.

9. Compare the chair, boat, and envelope conformations of monosaccharides.

10. Define O-glycosidic and N-glycosidic bonds in terms of acetal and ketal bonds. Draw the

bonds indicated by such symbols as a-1,6 or b-1,4. Complex Carbohydrates Are Formed by Linkage of Monosaccharides (Text Section 11.2)

11. Explain the role of O-glycosidic bonds in the formation of monosaccharide derivatives, disaccharides, and polysaccharides.

12. Draw the structures of sucrose, lactose, and maltose. Give the natural sources of these com

mon disaccharides.

13. Describe the structures and biological roles of glycogen, starch, amylose, amylopectin, and cellulose.

14. Give examples of enzymes involved in the digestion of carbohydrates in humans.

15. List the major kinds of glycosaminoglycans and name their sugar components.

16. Explain the differences between the oligosaccharide antigens for A, B, and O blood types.

CARBOHYDRATES 179 Carbohydrates Can Be Attached to Proteins to Form Glycoproteins (Text Section 11.3)

17. Name the amino acid residues that are used for attachment of carbohydrates to glyco

proteins.

18. Describe the steps required for synthesis of the enzyme elastase and its preparation for

export from the cell.

19. Describe the structure of dolichol phosphate and outline its role in the synthesis of the pen- tasaccharide core of N-linked oligosaccharides. Relate the effects of bacitracin and tunicamycin to dolichol phosphate metabolism.

20. Describe the Golgi complex and list its major functions. Distinguish among the cis, me- dial, and trans compartments of the Golgi.

21. Distinguish between core and terminal glycosylation of glycoproteins and provide an

overview of the reactions that occur in the three compartments of the Golgi.

22. State the molecular basis of I-cell disease. Explain how this disorder revealed the molec

ular signal that directs hydrolytic enzymes to the lysosome.

23. Explain the functions of the repeated addition and removal of glucose from the

oligosaccharide of luminal ER proteins and of calnexin in selecting properly folded proteins for export.

24. Explain briefly how biological oligosaccharides can be “sequenced” using mass spec

trometry methods in conjunction with specific enzyme cleavage. Lectins Are Specific Carbohydrate-Binding Proteins (Text Section 11.4)

25. Give examples of lectins and outline their functions and uses.

26. Explain why the influenza virus would have two proteins, hemagglutinin and neu

raminidase, which perform diametrically opposite tasks. SELF-TEST Introduction

1. Which of the following are roles of carbohydrates in nature? Carbohydrates

(a) serve as energy stores in plants and animals. (b) are major structural components of mammalian tissues. (c) are constituents of nucleic acids. (d) are conjugated to many proteins and lipids. (e) are found in the structures of all the coenzymes.

2. In the human diet, carbohydrates constitute approximately half the total caloric intake,

yet only 1% of tissue weight is carbohydrate. Explain this fact. 180

CHAPTER 11 Monosaccharides Are Aldehydes or Ketones with Multiple Hydroxyl Groups

3. Examine the following five sugar structures: FIGURE 11.1

CH OH

CH OH

2

CH OH

H COH

2

2

O

O

H

O

CH OH

H

J

H

O

H

2

H

J

H

J

J

OH

H

O

H

HO

J

H

HO

H

OCH

HO

CH OH

H

OH

3

2

J

OH

O H

OH

H

OH

H

A B C

CH OH

CH OH

H

OH

2

O

H

J

H

J

H

HO

J

H

OH

H

J

OH

H

O

OH

H

H

HO

O H

O

J

O H

O H

CH OH

2 D E

Which of these sugars

(a) contain or are pentoses? (b) contain or are ketoses? (c) contain the same monosaccharides? Name those monosaccharides. (d) will yield different sugars after chemical or enzymatic hydrolysis of glycosidic bonds? (e) are reducing sugars? (f)

contain a b-anomeric carbon?

(g) is sucrose? (h) are released upon the digestion of starch?

4. Consider the aldopentoses shown below. FIGURE 11.2

CHO

J

HCOH

HOCH

J

J

HOCH

HCOH

J

J

HCOH

HOCH

J

J

CH OH

CH OH

2

2 A B C

Aldopentoses

(a) Name the types of stereoisomers represented by each pair.

A and B are B and C are A and C are

(b) Name sugar B. (c) Draw the a-anomeric form of the furanose Haworth ring structure for sugar A.

5. Identify the properties common to D-glucose and D-ribose. Both monosaccharides

(a) are reducing sugars. (b) form intramolecular hemiacetal bonds. (c) have functional groups that can form glycosidic linkages. (d) occur in hexose form. (e) are major constituents of glycoproteins.

CARBOHYDRATES 181

6. Referring to the structure of ATP as shown below, which of the following statements are

true? The structure of ATP FIGURE 11.3

NH

2

J C

N

C

CH

HC

C

N

O

K

:OJPJOJPJOJPJOCH

O

2

J

O:

J

H

H

J

H

H

J

HO

OH

Adenosine triphosphate (ATP)

(a) contains a b-N-glycosidic linkage. (b) contains a pyranose ring. (c) exists in equilibrium with the open Fischer structure of the sugar. (d) preferentially adopts a chair conformation. (e) contains a ketose sugar. Complex Carbohydrates Are Formed by Linkage of Monosaccharides

7. Draw the structure of the disaccharide glucose a-1,6-galactose in the b-anomeric form.

8. If one carries out the partial mild acid hydrolysis of glycogen or starch and then isolates

from the product oligosaccharides all the trisaccharides present, how many different kinds of trisaccharides would one expect to find? Disregard a or b anomers. (a) 1 (b) 2 (c) 3 (d) 4 (e) 5

9. A sample of bread gives a faint positive color with Nelson’s reagent for reducing sugars.

After an equivalent bread sample has been masticated, the test becomes markedly positive. Explain this result.

10. Why does cellulose form dense linear fibrils, whereas amylose forms open helices?

11. For the polysaccharides in the left column, indicate all the descriptions in the right col

umn that are appropriate.

(a) amylose

(1) contains a-1,6 glucosidic bonds

(b) cellulose

(2) is a storage polysaccharide in yeasts

and bacteria

(d) glycogen

(3) can be effectively digested

(e) starch

by humans

(4) contains b-1,4 glucosidic bonds (5) is a branched polysaccharide (6) is a storage polysaccharide

in humans

(7) is a component of starch

12. a-Amylase

(a) removes glucose residues sequentially from the reducing end of starch. (b) breaks the internal a-1,6 glycosidic bonds of starch. (c) breaks the internal a-1,4 glycosidic bonds of starch. (d) cleaves the a-1,4 glycosidic bond of lactose. (e) can hydrolyze cellulose in the presence of an isomerase.

13. Which of the following statements about glycosaminoglycans are true?

(a) They contain derivatives of either glucosamine or galactosamine. (b) They constitute 5% of the weight of proteoglycans. (c) They contain positively charged substituent groups. (d) They include heparin, chondroitin sulfate, and keratan sulfate. (e) They have repeating units of four sugar groups.

14. Look at Figure 11.17 in the text, which shows the structures of the A, B, and O blood

antigens. Based on the structures of the three antigens, can you suggest why type O blood is the “universal donor” and can be transfused into people with type A or type B without provoking an immune response? Carbohydrates Can Be Attached to Proteins to Form Glycoproteins

15. Glycoproteins

(a) contain oligosaccharides linked to the side chain of lysine or histidine residues. (b) contain oligosaccharides linked to the side chain of asparagine, serine, or threonine

residues.

(c) contain linear oligosaccharides with a terminal glucose residue. (d) bind to liver cell-surface receptors that recognize sialic acid residues. (e) are mostly cytoplasmic proteins.

16. Translocated proteins may undergo which of the following modifications in the lumen

of the ER?

(a) signal sequence cleavage (b) the attachment of dolichol phosphate to form a lipid anchor (c) folding, disulfide-bond formation and isomerization, and cis-trans isomerization of

X-Pro peptide bonds

(d) the addition of oligosaccharides to their asparagine residues to form N-linked

derivatives

(e) the addition of oligosaccharides to their tyrosine residues to form O-linked de

rivatives

17. Which of the following statements about dolichol phosphate are correct?

(a) It serves as an acceptor of monosaccharides. (b) It serves as a donor of both monosaccharides and oligosaccharides. (c) It acts as a lipid carrier to facilitate the transfer of sugar residues from the cytosol

to the lumen of the ER.

(d) It is converted to dolichol pyrophosphate by a kinase that uses ATP as a phos

phate source.

(e) It is produced from dolichol pyrophosphate by a phosphatase.

18. Which of the following statements about the Golgi complex are correct?

(a) It is a stack of flattened proteoglycan sacs. (b) It carries out core glycosylation of the proteins being transported.

CARBOHYDRATES 183

(c) It is the major protein-sorting center of the cell. (d) It receives proteins from the ER by fusion with transport vesicles. (e) It forms secretory granules in its trans compartment. (f)

The cisternae of the cis, medial, and trans compartments are connected by pores.

19. Which of the following statements about I-cell disease are correct?

(a) It results from the inability of lysosomes to hydrolyze glycosaminoglycans and gly

colipids.

(b) It results from a chromosomal deletion of the genes specifying at least eight acid

hydrolases ordinarily found in the lysosomes.

a core oligosaccharide that is normally found on lysosomal enzymes.

(d) It arises from the absence of a mannose 6-phosphate receptor in the trans Golgi

complex.

20. Matching:

____

membrane bound ER

A.

glucosidase

chaperone for protein folding

B.

calreticulin

____

enzyme that removes

C.

calnexin

glucose from oligosaccharide

D.

glucosyltransferase

____

enzyme that adds glucose to unfolded protein Lectins Are Specific Carbohydrate-Binding Proteins

21. Which of the following statements about lectins are true? Lectins

(a) are produced by plants and bacteria. (b) contain only a single binding site for carbohydrate. (c) are glycosaminoglycans. (d) recognize specific oligosaccharide patterns. (e) mediate cell-to-cell recognition.

22. Which of the following statements are true? Selectins

(a) circulate in blood as free proteins. (b) are cell-surface receptor proteins. (c) are carbohydrate-binding adhesive proteins. (d) recognize and bind collagen in the extracellular matrix. (e) mediate the binding of immune cells to sites of injury during the inflammation

process. ANSWERS TO SELF-TEST

1. a, c, d

2. Most of the carbohydrates in the human diet are used as fuel to supply the energy re

quirements of the organism. Although some carbohydrate is stored in the form of glycogen, the mass stored is relatively small compared with adipose tissue and muscle mass. The carbohydrate present in nucleic acids, glycoproteins, glycolipids, and cofactors, although functionally essential, contributes relatively little to the weight of the body. 184

CHAPTER 11

3. (a) A

(b) B, C (c) B and C contain fructose; B, D, and E contain or are glucose. Note that glucose is

in the a-anomer form in sugars B and D and is in the b-anomer form in sugar E.

(d) A, B, D (e) C, D, E (f)

B and E. In structure B, the fructose ring is flipped over.

(g) B (h) D and E. Although E is in the b-anomer form, recall that in solution it can “mutaro

tate” or change back to the a-anomer.

4. (a) A and B are 3-epimers. B and C are diastereoisomers. A and C are enantiomers.

(b) D-ribose (c) See Figure 11.4. FIGURE 11.4 HOCH 2

O

H

J

OH

H

H

OH

J

O H

Sugar A

(a-anomeric form)

5. a, b, c. Note that glycosidic refers to bonds involving any sugars; however, glucosidic and galactosidic refer specifically to bonds involving the anomeric (reducing) carbons of glucose and galactose, respectively.

6. a

7. See Figure 11.5. FIGURE 11.5

HOCH2

O

H

J

H

J

OH

H

O

CH

HO

2

O

OH

J

OH

O H

H OH

H

H

J

O H

Glucose a (1–6)-galactose

(b-anomeric form)

8. Both c and d are correct. Since there are two glucosidic bonds in each trisaccharide and

each bond can be a-1,4 or a-1,6, the total number of possible kinds of trisaccharides is four. However, two consecutive a-1,6 bonds would be very rare in glycogen or starch; therefore, one would be more likely to find three kinds.

9. The carbohydrate in bread is mostly starch, which is a polysaccharide mixture contain

ing D-glucose residues linked by glucosidic bonds. All the aldehyde groups in each poly

saccharide, except one at the free end, are involved in acetal bonds and do not react with Nelson’s reagent. During mastication, a-amylase in saliva breaks many of the internal a1,4 glucosidic bonds and exposes reactive aldehyde groups (reducing groups). Note: Nelson’s reagent consists of copper sulfate in a hot alkaline solution; a reducing sugar, such as glucose, reduces the copper, which in turn reduces the arsenomolybdate in the reagent, producing a blue complex.

10. Both cellulose and amylose are linear polymers of D-glucose, but the glucosidic linkages

of cellulose are b-1,4 whereas those of amylose are a-1,4. The different configuration at the anomeric carbons determines a different spatial orientation of consecutive glucose residues. Thus, cellulose is capable of forming a linear, hydrogen-bonded structure, whereas amylose forms an open helical structure (see Figure 11.14 in the text).

11. (a) 3, 7 (b) 4 (c) 1, 2, 5 (d) 1, 3, 5, 6 (e) 1, 3, 5, and, if you wish, 7.

12. c

13. a, d

14. The O antigen lacks the extra galactose or N-acetylgalactosamine that the other antigens

have. Antibodies will react to the presence of an unfamiliar “bump” in the shape of an oligosaccharide, but evidently not to the lack of a sugar. It is also possible that individuals with Type A or Type B blood have a small amount of O antigen because of inefficient transfer of the final galactose, or perhaps hydrolysis of the galactose. This would prevent the immune system from seeing the O antigen as “foreign.”

15. b

16. a, c, d. Answer (e) is incorrect because threonine and serine provide hydroxyls for the

formation of O-linked oligosaccharides. Answer (b) is incorrect because dolichol phosphate is attached to an oligosaccharide, not a protein.

17. a, b, c, e. Sugar-substituted dolichol phosphates serve both as acceptors of monosac

charides from nucleotide sugars and other dolichol phosphate sugars and as donors of monosaccharides and oligosaccharides to other dolichol phosphate sugar derivatives and proteins. As a result of glycosyl transfer by the dolichol oligosaccharide, dolichol pyrophosphate is formed. This compound must be hydrolyzed to dolichol phosphate by a phosphatase to regenerate the sugar carrier for continued use. (See Section 11.3.3 in the text.)

18. c, d, e. The Golgi complex carries out terminal glycosylation by modifying and adding

to the core oligosaccharides that were constructed in the ER. Answer (f) is incorrect because the compartments of the Golgi are distinct, and components are transferred between them by vesicles.

19. c. The disease results from a deficiency in a sugar phosphotransferase that initiates a two

step sequence leading to the formation of a mannose 6-phosphate terminus on an oligosaccharide substituent of the eight or more affected lysosomal hydrolases. The phosphotransferase attaches a GlcNAc phosphate to a mannose residue of the oligosaccharide. Removal of the GlcNAc leaves the phosphate on the mannose. The enzymes lacking this mannose 6-phosphate “address” label are erroneously exported from the cell rather than being directed to the lysosomes. (See Figure 11.25 in Section 11.3.5 in the text.)

20. C, A, D

21. a, d, e

22. b, c, e. Collagen is a fibrous protein that is bound by proteins called “integrins.” 186

CHAPTER 11 PROBLEMS

1. Glucose and other dietary monosaccharides like fructose and galactose are very soluble

in water at neutral pH. For example, over 150 g of glucose can be dissolved in 100 ml water at 25°C.

(a) What features of the chemical structure of glucose make it so soluble in water? (b) What features of the proteoglycans found in cartilage make them so highly hydrated

and contribute to their ability to spring back after deformation?

2. Indicate whether the following pairs of molecules are enantiomers, epimers, di

astereoisomers, or anomers.

(a) D-xylose and D-lyxose (b) a-D-galactose and b-D-galactose (c) D-allose and D-talose (d) L-arabinose and D-arabinose

3. What is the name of the compound that is the mirror image of a-D-glucose?

4. Compound X, an aldose, is enzymatically reduced using NADPH as an electron donor,

yielding D-sorbitol (Figure 11.6). This sugar alcohol is then oxidized at the C-2 position with NAD+ as the electron acceptor; the products are NADH and a ketose, compound Y.

(a) Name compound X and write its structure. (b) Will sorbitol form a furanose or pyranose ring? Why? (c) Name compound Y and write its structure. FIGURE 11.6

CH OH

2

J

HCOH

J

HOCH

J

HCOH

J

HCOH

J

CH OH

2 D-Sorbitol

5. In Section 11.1.3 of the text, reducing sugars are defined as those with a free alde

hyde or keto group that can reduce cupric ion to the cuprous form. The reactive species in the reducing sugar reaction is the open-chain form of the aldose or ketose. The reaction can be used to estimate the total amount of glucose in a solution such as blood plasma. An aqueous solution of glucose contains only a small amount of the open-chain form. How can the reaction be used to provide a quantitative estimate of glucose concentration?

6. Compare the number of dimers that can be prepared from a pair of alanine molecules

and from a pair of D-galactose molecules, each of which is present as a pyranose ring. For the galactose molecules, pairs may be made using the a or b anomers.

7. Storage polysaccharides, like starch and glycogen, often contain over a million glu

cose units. The energetic cost of synthesizing polysaccharides is high (about one high energy phosphate bond per sugar residue added). Suppose that in a liver cell, the glucosyl residues in large numbers of glycogen molecules were replaced with an equivalent number of molecules of free glucose. What problems would this cause for the liver cell?

CARBOHYDRATES 187

8. You have a sample of glycogen that you wish to analyze using exhaustive methylation and

acid hydrolysis. Using a sample of 0.4 g, you incubate the glycogen with methyl iodide, which methylates all free primary or secondary alcohol groups on sugars. Then you subject the sample to acid hydrolysis, which cleaves glycosidic linkages between adjacent glucose residues. You then determine the yield of 2,3-dimethylglucose in your sample.

(a) Why is a 2,3-dimethylglucose residue produced from a branch point in glycogen? (b) The yield of 2,3-dimethylglucose is 0.247 mmol. What fraction of the total residues

in each sample are branch points? The molecular weight of a glycosyl residue in glycogen is 162.

branch? Why?

9. Shown below (Figure 11.7) is one example of the storage oligosaccharides that account in

part for the flatulence caused by eating beans, peas, and other legumes. These oligosaccharides cannot be digested by enzymes in the small intestine, but they can be metabolized by anaerobic microorganisms in the large intestine. There, they undergo oxidation, with the production of large quantities of carbon dioxide, hydrogen sulfide, and other gases. Solutions are now on the market containing one or more enzymes that, when ingested with the offending legumes at mealtime, convert the oligosaccharides to digestible products. FIGURE 11.7

CH OH

2

O

H

J

H

J

OH

H

O

CH

HO

2

O

HO

J

J

O H

O

H

CH2

CH OH

O

2

OH

H

H

O

O

H

J

H

HO

H

H

O H

J

OH

H

OH

HO

H

CH OH

2

J

O H

(a) Name the oligosaccharide shown above. (b) Given that free hexoses can pass easily through intestinal cells into the blood, what

types of enzymes do you think are included in the commercial products that aid in legume oligosaccharide digestion?

sprouting. What happens to those oligosaccharides when cooking is employed? When the beans are sprouted before cooking or eating?

(d) When small amounts of cellulose are ingested purposely or accidentally (e.g., by

pets or young children), there is usually no gas production. In fact, the primary concern about paper ingestion by pets or small children is intestinal blockage. Why?

10. Explain the roles of (a) the phosphate group and (b) the long lipid chain of dolichol

phosphate in the transport of polysaccharides across membranes.

11. Suppose that glucose 1-phosphate labeled with 32P is added to a cellular system designed

to study the synthesis and processing of N-glycosylated proteins. When bacitracin is added to the system, a lipid-soluble intermediate labeled with 32P accumulates. In the absence of bacitracin, the label appears in inorganic phosphate. Explain these results, and identify the lipid-soluble intermediate that accumulates. (Refer to Section 11.3.3 in the text.) 188

CHAPTER 11

12. MALDI–TOF MS stands for Matrix Assisted Laser Desorption/Ionization–Time of Flight

Mass Spectrometry. It is a highly sophisticated technique (also used for proteins, see text Chapter 4, Section 4.1.7), but it can’t solve oligosaccharide structures without input from other techniques. Why not?

13. Why is the structural analysis of an oligosaccharide containing eight monosaccharide

residues more complicated than a similar analysis for an octanucleotide or an octapeptide? This is not a quantitative question, a qualitative description will do.

14. In the 1950s, Morgan and Watkins showed that N-acetylgalactosamine and its a-methyl

glycoside inhibit the agglutination of type A erythrocytes by type A–specific lectins, whereas other sugars had little effect. What did this information reveal about the structure of the glycoprotein on the surface of type A cells? ANSWERS TO PROBLEMS

1. (a) Glucose and other hexose monosaccharides have five hydroxyl groups and an oxy

gen in the heterocyclic ring that can all form hydrogen bonds with water. The ability to form these hydrogen bonds with water and other polar molecules enables hexoses and other carbohydrates to dissolve easily in aqueous solution.

(b) In addition to hydrogen bonding of water to hydroxyl groups and oxygen atoms in

the repeating disaccharide units of cartilaginous proteoglycans like keratan sulfate and chondroitin sulfate, these molecules also contain charged sulfate and carboxylate groups that can also interact with water. Compression of these large hydrated polyanions can drive some water out of the cavities between them, but the high degree of hydration of the molecules, as well as charge repulsion between the sulfate and carboxylate groups, contributes to the tendency of these compounds to resume their normal conformations after deformation.

2. (a) D-xylose and D-lyxose differ in configuration at a single asymmetric center; they are

epimers.

(b) a-D-galactose and b-D-galactose have differing configurations at the C-1, or

anomeric, carbon; they are anomers.

tions at one or more chiral centers, but they are not complete mirror images.

(d) L-arabinose and D-arabinose are mirror images of each other and are therefore enan

tiomers.

3. Although the mirror image of a D compound is an L compound, the mirror image of an

a compound is an a compound. (An a compound has a 1-hydroxyl group in the a position.) Thus, a-L-glucose is the compound that is the mirror image, or enantiomer, of a-D-glucose.

4. (a) Compound X is D-gulose; it is the only D-aldose whose reduction will yield a hex

itol with the same conformation as that of D-sorbitol. The less common sugar Lgulose would also yield the same result. With sugar alcohols, there is no most-oxidized carbon, so it is hard to define which end is “carbon one” but the use of an enzyme greatly favors D-aldose as the starting material.

(b) Sorbitol cannot form a hemiacetal because it has no aldehyde or ketone group.

Therefore, neither type of ring can be formed by sorbitol.

bitol. Enzymes would be very unlikely to produce an L-ketose, so this is the only expected result.

CARBOHYDRATES 189

5. In water, an equilibrium exists among three forms of glucose. Two-thirds is present as

the b anomer, one-third as the a anomer, and less than 1% as the open-chain form. When excess cupric ion reacts with the open-chain form, glucose is oxidized to gluconic acid. Through the law of mass action, the a and b anomers of glucose are then converted to the open-chain aldose form. Continued production of gluconic acid from the openchain form leads to the ultimate conversion of all glucopyranoses to the open-chain form, which reacts quantitatively with cupric ion. Thus the total amount of glucose in a known volume of blood plasma or other solution can be determined.

6. Only one dimer, alanylalanine, can be made from two alanine molecules linked via a

peptide bond. However, the presence of several hydroxyl groups and the aldehydic function at the C-1 position of each D-galactose molecule provides an opportunity to make a larger number of dimers. Both the a and b forms of one molecule can form glycosidic linkages with the C-2, C-3, C-4, or C-6 hydroxyl groups of the other. Recall that the C-5 position is not available, because it participates in the formation of the pyranose ring. To these eight dimers can be added those dimers formed through glycosidic linkages involving the aa, ab, or bb configurations. Thus, 11 possible dimers exist. If one is allowed to use L forms, then the number of possible dimers increases greatly. This variety of linkages makes the sugars very versatile molecules and yields many different structures that may be useful in biology. However, this variety has also made the systematic study of the chemistry of polysaccharides very difficult.

7. The primary consequence of a high concentration of free glucose molecules in the cell

would be a dramatic and probably catastrophic increase in osmotic pressure. In aqueous solutions, colligative properties like boiling and freezing points, vapor pressure, and osmotic pressure depend primarily on the number of molecules in the solution. Thus a glycogen molecule containing a million glucose residues exerts one-millionth the osmotic pressure of a million molecules of free glucose. Osmotic pressure exerted by high glucose concentration would induce entry of water into the cell, in an attempt to equalize pressure inside and outside the cell. Unlike bacterial or plant cells, which have a rigid cell wall that can help resist high pressures, animal cells have a comparatively fragile plasma membrane, which will burst when osmotic pressures are too high.

8. (a) A glycosyl residue at a branch point has three of its five carbons linked to other

glucose residues; these are carbons 1, 4, and 6. Only C-2 and C-3 of a branch point residue will have alcohol or hydroxyl groups that are free and therefore available for methylation. Thus residues at a branch point are converted to 2,3-dimethylglucose after methylation and hydrolysis. Those glucosyl residues not at a branch point would be converted to 2,3,6-trimethylglucose by the same procedure, except for the single residue at the reducing end, which could be converted to 1,2,3,6-tetramethylglucose.

(b) The original sample of 0.4 g corresponds to 0.4 g ÷ 162 g/mole, or 2.47 × 10−3

mole, or 2.47 mmol glucose residues, which is 10% of the total sample. Thus 10% of the glucosyl residues are at branch points.

of the anomeric linkage. Acid hydrolysis cleaves both a- and b-anomeric linkages and does not allow distinctions between them.

9. (a) Glu a-1,6 Gal a-1,6 Fru b-1,4 Glu.

(b) The solution must contain enzymes that hydrolyze the glycosidic linkages between

the monosaccharides. For example, an activity that would be required for the oligosaccharide shown would be a type of a-1,6-glycosidase, which would cleave the a-1,6 linkage between glucose and galactose. Another would be the b-1,4-fructosidase, a different glycosidase. The glycosidases are needed to convert the

190

CHAPTER 11

oligosaccharides to free hexoses, which then pass easily into the circulation. The three common sugars found in the oligosaccharide shown in this problem are easily metabolized by the liver and other cells.

found in the oligosaccharides. Sprouting or germinating beans undergo a reduction in oligosaccharide concentration because hydrolase proteins induced during germination produce free hexoses, which can be used in the developing plant tissues as a source of carbon for biosynthesis.

(d) Because cellulose is an unbranched polymer of glucose residues joined by b-1,4

linkages, the molecule is resistant to hydrolysis, even by anaerobic bacteria in the human intestine. Small amounts of cellulose and other indigestible complex carbohydrates are virtually unaltered as they pass through the digestive system. Thus no gases from carbohydrate breakdown are generated in the large intestine. Intestinal blockage may result from ingestion of large quantities of cellulose because there are no enzymes available to cleave the glycosidic linkages. Organisms that use cellulose as an energy source (e.g., cows and termites) have gut flora which make cellulase, and can provide the service of breaking these b-1,4 bonds.

10. (a) The phosphate group serves as the site for the covalent attachment of sugar residues

to the carrier.

(b) The long lipid chain renders the carrier highly hydrophobic and thus membrane

permeable.

11. The lipid-soluble intermediate that accumulates is dolichol pyrophosphate, whose ter

minal phosphate comes from glucose 1-phosphate. (See Figure 11.23 in Section 11.3.3 in the text.) Bacitracin is an antibiotic that forms a 1:1 complex with dolichol pyrophosphate, preventing its hydrolysis to dolichol phosphate and inorganic phosphate. Thus, in the presence of bacitracin, the label will remain in dolichol pyrophosphate. In the absence of bacitracin, the terminal phosphate will be released as inorganic phosphate.

12. MALDI–TOF MS only provides a very accurate molecular weight for an oligosaccharide

or other complex molecule. If you have, say, ten sugars, they can be rearranged in many different isomeric forms that all would have the same molecular weight. Enzymes that can cleave only certain sugars in certain positions provide extra information that is critical to the “sequencing” of an oligosaccharide.

13. In oligosaccharides, there are a number of different types of potential glycosidic linkages

that can be formed among eight residues, because each free hydroxyl group as well as the anomeric carbon on a particular monosaccharide could be linked to similar groups on adjacent residues. An octooligosaccharide could be linear or branched, and could be composed of as many as eight different monosaccharides, each of which could require additional steps to analyze completely. Analysis of an oligonucleotide is somewhat less complicated, because usually only four different bases will be found during the analysis, and the linkage between adjacent nucleotides is almost always 3′

5′; in addition, the oligonucleotide

molecule is not likely to be branched. Although there may be as many as eight different amino acid residues in an octapeptide, all 20 different amino acids found in most proteins are relatively easy to characterize and the octapeptide is unlikely to be branched.

14. The observations of Morgan and Watkins suggested that the sugar N-acetylgalactosamine in

a linkage is the determinant of blood group A specificity. The galactose derivative binds to type A lectins, occupying the sites that would otherwise bind to glycoproteins, having Nacetylgalactosamine end groups, on the surfaces of type A cells. The papers establishing the structures of the blood group oligosaccharides were among the first of Winifred M. Watkins’ long and distinguished career. The fields of Biochemistry and Molecular Biology have provided several early female role models including such important scientists as Maud Menten (who collaborated with L. Michaelis to study enzymology) and Rosalind Franklin (who de

termined the structure of the A-form, and worked on the B-form, of double helical DNA). Dr. Watkins was elected as a Fellow of the Royal Society in 1998. [W. M. Watkins, & W. T. J. Morgan. Nature 178[1956]:1289, and other papers.] EXPANDED SOLUTIONS TO TEXT PROBLEMS

1. Although it can be risky business, chemists have always tried to gain some insight into

molecular structure from knowledge of the empirical formula. Since the empirical formula for carbohydrates is (CH2O)n, it is not surprising that in the latter half of the nineteenth century the name carbohydrate was coined.

2. To begin, there are six different ((3)(2)(1)) ways to specify the order of monosaccha

ride units. Then the first glycosidic bond can join the first two monomers in any of 25 or 32 ways, a or b from the C1 oxygen of the first sugar to OH #2, 3, 4, or 6 of the second sugar. Finally, the second glycosidic bond can join the second and third monomers in any of 26 or 64 ways, a or b from the C1 oxygen of the second sugar to OH #1 (nonreducing) or #2, 3, 4, or 6 of the third sugar. Putting this all together, one has (6)(32)(64) or 12,288 possible trisaccharides.

For tripeptides, there are only 6 different sequences that use exactly one each of three

different amino acids ((3)(2)(1) = 6).

3. To answer this problem, one must know the structures of the molecules in question and

a couple of definitions. By definition, epimers are a pair of molecules that differ from each other only in their configuration at a single asymmetric center. Anomers are special epimers that differ only in their configuration at a carbonyl carbon; hence, they are usually acetals or hemiacetals. An aldose-ketose pair is obvious. Inspection of Fischer representations of the molecular pairs leads to the conclusion that (a), (c), and (e) are aldose-ketose pairs; (b) and (f) are epimers; and (e) are anomers.

4. A mild oxidant, Tollens’ reagent converts aldoses to aldonic acids and free silver as fol

lows:

RCHO + 2 Ag(NH

+

− +

+ +

3)2

H2O

RCO2

2 Ag0 + 3 NH4

NH3

However, cyclic hemiacetals are oxidized directly to lactones, which are hydrolyzed to the corresponding aldonic acid under alkaline conditions. Thus, in the case of glucose, the major first reaction product is D-d-gluconolactone. To prepare aldonic acids, Br2 is usually used as the oxidant because it gives fewer side reactions than does Tollens’ reagent. FIGURE 11.8

CH OH

CH OH

2

HO

O

HO

O+2 Ag0 +2 NH +2 NH ;

3

4

2 Ag(NH ) ;

3 2

HO

HO

OH

OH

OH

K

H

O a-D-Glucose D-d-Gluconolactone (cyclic hemiacetal) (D-glucono-(1D5)-lactone)

5. The reason the specific rotation of a-D-glucopyranose changes after it is dissolved in

water is that the ring form is in equilibrium with a small amount of the straight-chain form of glucose. The straight-chain form then converts to either a-D-glucopyranose or b-D-glucopyranose. This process, called mutarotation, continues until after 1–2 hours a thermodynamically stable mixture of the a and b anomers is obtained. Its specific rotation is 52.7º. The difference in the specific rotations of the two anomers is 93.3º (112º − 18.7º),

192

CHAPTER 11

and the difference between the equilibrium value and that of the b anomer is 34º (52.7º − 18.7º). Since the optical rotation of the equilibrium mixture is closer to that of the b anomer than it is to that of the a anomer, obviously more than half the equilibrium mixture is in the b configuration. The fraction present in the a configuration is 34º ÷ 93.3º = 0.36. The fraction in the b configuration is 1 − 0.36 = 0.64.

6. Glucose reacts slowly because the predominant hemiacetal ring form (which is inactive)

is in equilibrium with the active straight-chain free aldehyde. The latter can react with terminal amino groups to form a Schiff base, which can then rearrange to the stable amino ketone, sometimes referred to as Hb AIc, which accounts for approximately 3% to 5% of the hemoglobin in normal adult human red cells. In the diabetic, its concentration may rise 6% to 15% owing to the elevated concentrations of glucose. FIGURE 11.9

HJNJHb

O

NJHb

J

K

K

HJC

HJCJH

J

H NJ(Val)Hb

HJCJOH

2

HJCJOH

Amadori

CKO

J

rearrangement

J

HOJCJH

J

J

HJCJOH

HJCJOH

J

J

HJCJOH

J

CH JOH

CH JOH

2

2 Glucose Schiff base Amino ketone (aldimine)

7. Whereas pyranosides have a series of three adjacent hydroxyls, furanosides have only

two. Therefore, oxidation of pyranosides uses two equivalents of periodate and yields one mole of formic acid, whereas oxidation of furanosides uses only one equivalent of periodate and yields no formic acid. FIGURE 11.10

OH

J

O

CH

H

CH OH

2

K

2

O

H

O

HO

+IO :

+IO :

3

J

3

K

H

OCH

O

HO

3

OCH

OH

3

H

OH

H b-D-Methylglucopyranoside

2nd equivalent of IO :

4

O

CH OH

O

K

2

H

O

K

IO :+HJCJOH+

3

H

Formic acid

K

O

OCH

3

J

H

CH OH

CH OH

2

2

O

OCH

O

OCH

3

+IO :

+IO :

4

3

J

H

HO

J

H

CH OH

H

H

CH OH

2

H

J

K

OH

H

O

O b-D-Methylfructofuranoside

CARBOHYDRATES 193

8. The formation of acetals (such as methylglucoside) is acid-catalyzed. In a mechanism

similar to that of the esterification of carboxylic acids (shown in most organic chemistry texts), the anomeric hydroxyl group is replaced. The resulting carbocation is susceptible to attack by the nucleophilic oxygen of methanol, leading to the incorporation of this oxygen into the methylglucoside molecule. FIGURE 11.11

HOCH

+

HOCH

2

O

2

O

Q

2

O

HO

HO

HO

H;

G

H O

HO

OH

+

2

HO

HO

OH

HO

H

HO

H D-Glucose (b-pyranose form)

Electron pair on ring oxygen

can stabilize carbocation at

anomeric position only

9. By inspection, A, B, and D are the pyranosyl forms of D-aldohexoses because the

CH2OH is above the plane of the ring. In Haworth projections, OH’s above the ring are to the left in Fischer projections, and those below the ring are to the right. Therefore, A is b-D-mannose, B is b-D-galactose, and D is b-D-glucosamine. By similar use of the Haworth projection, C can be identified as b-D-fructose. All these sugars are b because, in Haworth projections, when the CH2OH attached to the C-5 carbon (the carbon that determines whether the sugar is D or L) is above the ring, if the anomeric hydroxyl is also above the ring, the sugar is b.

10. The trisaccharide itself should be a competitive inhibitor of cell adhesion if the trisac

charide unit of the glycoprotein is critical for the interaction.

11. The nonreducing carbon-1 oxygens cannot be methylated, whereas the carbon-1 hy

droxyls at the reducing ends can be methylated. Conversely, most of the carbon-6 hydroxyls can be methylated, but not at the branch points. Therefore, the ratio of methylated to nonmethylated C-1 hydroxyls in the final digestion mixture will indicate the relative proportion of reducing ends. Likewise, the ratio of nonmethylated to methylated C-6 hydroxyls in the digestion mixture will indicate the relative proportion of branch points.

12. (a) No. There is no hemiacetal linkage in raffinose, but rather two acetal linkages.

(b) galactose, glucose, and fructose. (c) Galactose and sucrose. (After digestion, the released galactose—in water solution—

will establish an equilibrium among the a, b, and open-chain forms. See also the answer to Text Problem 13, below.)

13. The hemiacetal of the a anomer opens in water to give the open-chain aldehyde/alcohol

form. The open form then can reclose the ring with either the a or b configuration. In water solution, an equilibrium will be established among the b anomer, the a anomer, and a small amount of the open-chain form, through which the two pyranose ring forms interconvert. CHAPTER 1 Lipids and Cell Membranes 2

In this chapter, the authors describe the composition, structural organization, and

general functions of biological membranes. After outlining the common features of membranes, a new class of biomolecules, the lipids, are introduced in the context

of their role as membrane components. The authors focus on the three main kinds of membrane lipids—the phospholipids, glycolipids, and cholesterol. The amphipathic nature of membrane lipids and their ability to organize into bilayers in water are then described. An important functional feature of membranes is their selective permeability to molecules, in particular the inability of ions and most polar molecules to cross membrane bilayers. This aspect of membrane function is discussed next and will be revisited when the mechanisms for transport of ions and polar molecules across membranes is discussed in Chapter 13.

Next, the authors turn to membrane proteins, the major functional constituents

of biological membranes. The arrangement of proteins and lipids in membranes is described and the asymmetric, fluid nature of membranes is stressed. The important differentiation between integral and peripheral membrane proteins is discussed as well as the chemical forces that bind them to the membrane. The high-resolution analyses of the structures of selected membrane proteins are discussed, including structure prediction of membrane-spanning proteins. The chapter concludes with a discussion of internal membranes within eukaryotic cells and the mechanisms by which proteins are targeted to specific compartments within cells. 195 196

CHAPTER 12 LEARNING OBJECTIVES

When you have mastered this chapter, you should be able to complete the following objectives. Introduction

1. List the functions of biological membranes. Many Common Features Underlie the Diversity of Biological Membranes (Text Section 12.1)

2. Describe the common features of biological membranes. Fatty Acids Are Key Constituents of Lipids (Text Section 12.2)

3. Draw the general chemical formula of a fatty acid and be able to use standard notation

for representing the number of carbons and double bonds in a fatty acid chain.

4. Distinguish between saturated and unsaturated fatty acids.

5. Explain the relationship between fatty acid chain length and degree of saturation and the

physical property of melting point. There Are Three Common Types of Membrane Lipids (Text Section 12.3)

6. Define lipid and list the major kinds of membrane lipids.

7. Recognize the structures and the constituent parts of phospholipids (phosphoglycerides and sphingomyelin), glycolipids, and cholesterol.

8. Describe the general properties of the fatty acid chains found in phospholipids and

glycolipids.

9. Draw the general chemical formula of a phosphoglyceride, and recognize the most common alcohol moieties of phosphoglycerides (e.g., choline, ethanolamine, and glycerol).

10. Distinguish between membranes of archaea and those of eukaryotes and bacteria.

11. Describe the composition of glycolipids. Note the location of the carbohydrate compo

nents of membranes.

12. Recognize the structure of cholesterol.

13. Describe the properties of an amphipathic molecule. Phospholipids and Glycolipids Readily Form Bimolecular Sheets in Aqueous Media (Text Section 12.4)

14. Distinguish among oriented monolayers, micelles, and lipid bilayers.

15. Describe the self-assembly process for the formation of lipid bilayers. Note the stabilizing

intermolecular forces.

16. Outline the methods used to prepare lipid vesicles (liposomes) and planar bilayer mem- branes. Point out some applications of these systems.

17. Explain the relationship between the permeability coefficients of small molecules and ions

and their solubility in a nonpolar solvent relative to their solubility in water.

LIPIDS AND CELL MEMBRANES 197 Proteins Carry Out Most Membrane Processes (Text Section 12.5)

18. Distinguish between peripheral and integral membrane proteins.

19. Describe the structure of glycophorin. Explain how transmembrane a helices can be pre

dicted from hydropathy plots. Lipids and Many Membrane Proteins Diffuse Rapidly in the Plane of the Membrane (Text Section 12.6)

20. Describe the evidence for the lateral diffusion of membrane lipids and proteins. Contrast

the rates for lateral diffusion with those for transverse diffusion.

21. Describe the features of the fluid mosaic model of biological membranes.

22. Explain the roles of the fatty acid chains of membrane lipids and cholesterol in control

ling the fluidity of membranes.

23. Discuss the origin and the significance of membrane asymmetry. Eukaryotic Cells Contain Compartments Bounded by Internal Membranes (Text Section 12.7)

24. Give examples of the compositional and functional varieties of biological membranes.

25. Discuss the role of targeting sequences in eukaryotic proteins.

26. Describe the recognition of a nuclear localization signal by a-karyopherin.

27. Describe the process of receptor-mediated endocytosis of low-density lipoproteins (LDL). SELF-TEST Introduction

1. Which of the following statements about biological membranes are true?

(a) They constitute selectively permeable boundaries between cells and their environ

ment and between intracellular compartments.

(b) They are formed primarily of lipid and carbohydrate. (c) They are involved in information transduction. (d) Targeting across them requires specific systems. (e) They are dynamic structures. Many Common Features Underlie the Diversity of Biological Membranes

2. Which of the following statements about biological membranes is not true?

(a) They contain carbohydrates that are covalently bound to proteins and lipids. (b) They are very large, sheetlike structures with closed boundaries. (c) They are symmetric because of the symmetric nature of lipid bilayers. (d) They can be regarded as two-dimensional solutions of oriented proteins and lipids. (e) They contain specific proteins that mediate their distinctive functions. 198

CHAPTER 12 Fatty Acids Are Key Constituents of Lipids

3. Which of the following fatty acids is polyunsaturated?

(a) arachididic (b) arachidonic (c) oleic (d) palmitic (e) stearic There Are Three Common Types of Membrane Lipids

4. Which of the following are membrane lipids?

(a) cholesterol (b) glycerol (c) phosphoglycerides (d) choline (e) cerebrosides

5. The phosphoinositol portion of the phosphatidyl inositol molecule is called which of

the following?

(a) the amphipathic moiety (b) the hydrophobic moiety (c) the hydrophilic moiety (d) the micelle (e) the polar head group

6. Acid hydrolysis will break all ester, amide, and acetal chemical linkages. Which of the

following statements is incorrect about the acid hydrolysis of various lipids?

(a) A cerebroside will release two fatty acids and one monosaccharide per mole of

cerebroside.

(b) Phosphatidylcholine will release two fatty acids and one glycerol molecule per mole

of phosphatidylcholine.

(d) Sphingomyelin and phosphatidylcholine will release equivalent molar amounts of

choline and phosphoric acid.

(e) Cerebrosides and sphingomyelin will each release one mole of sphingosine.

7. After examining the structural formulas of the four lipids in Figure 12.1, answer the

following questions.

(a) Which are phosphoglycerides? (b) Which is a glycolipid? (c) Which contain sphingosine? (d) Which contain choline? (e) Which contain glycerol? (f)

Name the lipids.

LIPIDS AND CELL MEMBRANES 199 FIGURE 12.1 Membrane lipids R1 and R2 represent hydrocarbon chains.

O:

J

H

H

H

H

2

;

J

2

;

OKPJOJCJCJN(CH

)

OKPJOJCJCJN(CH

)

3 3

J

J

O

OH

O

H

J

H

H CJCJ

CH

HCJCJ

CH

2

2

J

J

J

O

HC

NH

J

J

K

J

OKC

CKO

CH

CKO

J

R R

(CH )

R

1

2

2 12

1

J

CH3 A B

O: J

H

2

Glucose

OKPJOJCJCJC

J

J

OH

O

HO

OH

J

H

J

H

J

HCJCJ

CH

H CJCJ

CH

2

J

J

J

HC

NH

O

K

J

CH

CKO

OKC

CKO

J

J

J

(CH )

R R

2 12

1

2

J

CH3 C D Phospholipids and Glycolipids Readily Form Bimolecular Sheets in Aqueous Media

8. Which of the following statements are NOT true of both a micelle and a lipid bilayer?

(a) Both assemble spontaneously in water. (b) Both are made up of amphipathic molecules. (c) Both are very large, sheetlike structures. (d) Both have the thickness of two constituent molecules in one of their dimensions. (e) Both are stabilized by hydrophobic interactions, van der Waals forces, hydrogen

bonds, and electrostatic interactions.

9. A triglyceride (triacylglycerol) is a glycerol derivative that is similar to a phosphoglyc

eride, except that all three of its glycerol hydroxyl groups are esterified to fatty acid chains. Would you expect a triglyceride to form a lipid bilayer? Explain. 200

CHAPTER 12

10. What is the volume of the inner water compartment of a liposome that has a diameter

of 500 Å and a bilayer that is 40 Å thick?

(a) 5.6 × 105 Å3

(d) 7.3 × 107 Å3

(b) 7.3 × 106 Å3

(e) 3.9 × 108 Å3

11. Arrange the following in the order of decreasing permeability through a lipid bilayer.

(a) urea

(d) Na+

(b) tryptophan

(e) glucose

12. Why is an a helix the preferred structure for transmembrane protein segments?

13. Show which of the properties listed on the right are characteristics of peripheral mem

brane proteins and which are characteristics of integral membrane proteins.

(a) peripheral

(1) require detergents or organic

(b) integral

solvent treatment for dissociation from the membrane

(2) require mild salt or pH treatment for

dissociation from the membrane

(3) bind to the surface of membranes (4) have transmembrane domains Lipids and Many Membrane Proteins Diffuse Rapidly in the Plane of the Membrane

14. Which of the following statements about the diffusion of lipids and proteins in mem

branes is NOT true?

(a) Many membrane proteins can diffuse rapidly in the plane of the membrane. (b) In general, lipids show a faster lateral diffusion than do proteins. (c) Membrane proteins do not diffuse across membranes at measurable rates. (d) Lipids diffuse across and in the plane of the membrane at equal rates.

15. Which of the following statements about the asymmetry of membranes are true?

(a) It is absolute for glycoproteins. (b) It is absolute for phospholipids, but only partial for glycolipids. (c) It arises during biosynthesis. (d) It is structural but not functional.

16. If phosphoglyceride A has a higher Tm than phosphoglyceride B, which of the following

differences between A and B may exist? (In each case only one parameter—either chain length or double bonds—is compared.)

(a) A has shorter fatty acid chains than B. (b) A has longer fatty acid chains than B. (c) A has more unsaturated fatty acid chains than B. (d) A has more saturated fatty acid chains than B. (e) A has trans unsaturated fatty acid chains, whereas B has cis unsaturated fatty

acid chains.

17. Explain how the mobility of a protein in a membrane might be restricted far beyond

what a simple consideration of its native molecular weight would lead you to conclude.

LIPIDS AND CELL MEMBRANES 201 Eukaryotic Cells Contain Compartments Bounded by Internal Membranes

18. Which of the following sequences would target a protein to the nucleus?

(a) -SKL-COO− (b) -KKLK(c) -KDEL-COO− (d) -KKLK-COO− ANSWERS TO SELF-TEST

1. a, c, d, e

2. c

3. b

4. a, c, e

5. c, e

6. a

7. (a) A, D (b) C (c) B, C (d) A, B (e) A, D (f) A is phosphatidyl choline, B is sphingomyelin,

C is cerebroside, and D is phosphatidyl glycerol.

8. c

9. No. Although a triglyceride has hydrophobic fatty acyl chains attached to a glycerol back

bone, it lacks a polar head group; therefore, it is not an amphipathic molecule and is incapable of forming a bilayer.

10. The correct answer is (c). The volume of a sphere is 4/3 pr3, so we just need the radius of

the inner compartment to do the calculation. Using Figure 12.2 to represent the liposome, we can calculate the diameter of the inner water compartment by subtracting the width of the bilayer from the left and right sides of the liposome from the diameter of the outer compartment. FIGURE 12.2

40 Å

500 Å

Diameter of inner water compartment = 500 Å − (2 × 40 Å) = 420 Å

Since the radius of a circle is half the diameter, the radius of the inner compartment r = 1/2(420 Å) = 210 Å.

Therefore the volume of inner water compartment in the liposome 4/3 p (210 Å)3

= 3.9 × 107 Å3.

11. c, a, b, e, d

18. b PROBLEMS

1. The ability of bacteria, yeasts, and fungi to convert aliphatic hydrocarbons to carbon diox

ide and water has been studied intensively over the past decade because of concerns about the effects of crude oil spills on the environment. Microorganisms cannot survive when they are placed in high concentrations of crude oil or any of its components. However, they can utilize hydrocarbons very efficiently when they are placed in a medium in which an extensive lipid-water interface is created by agitation and aeration. Why?

2. Phytol, a long-chain alcohol, appears as an ester in plant chlorophyll. When consumed

as part of the diet, phytol is converted to phytanic acid (see Figure 12.3). FIGURE 12.3

( CH

CH

3

J

) J

H CJ CHJCH JCH JCH

JCKCJCH OH

3

2

2 3

2

J

H Phytol

( CH

CH

3

O

J

) J

J

H CJ CHJCH JCH JCH

JCHJCH JC

3

2

2 3

2

K

O: Phytanic Acid

People who cannot oxidize phytanic acid suffer from a number of neurological disorders that together are known as Refsum’s disease. The symptoms may be related to the fact that phytanic acid accumulates in the membranes of nerve cells. What general effects of phytanic acid on these membranes would be observed?

3. Bacterial mutants that are unable to synthesize fatty acids will incorporate them into their

membranes when fatty acids are supplied in their growth medium. Suppose that each of two cultures contains a mixture of several types of straight-chain fatty acids, some saturated and some unsaturated, ranging in chain length from 10 to 20 carbon atoms. If one culture is maintained at 18ºC and the other is maintained at 40ºC over several generations, what differences in the composition of the cell membranes of the two cultures would you expect to observe?

LIPIDS AND CELL MEMBRANES 203

4. Given two bilayer systems, one composed of phospholipids having saturated acyl chains

20 carbons in length and the other having acyl chains of the same length but with cis double bonds at C-5, C-8, C-11, and C-14, compare the effect of the acyl chains on Tm for each system.

5. Hopanoids are pentacyclic molecules that are found in bacteria and in some plants. A

typical bacterial hopanoid, bacteriohopanetetrol, is shown in Figure 12.4. Compare the structure of this compound with that of cholesterol. What effect would you expect a hopanoid to have on a bacterial membrane? FIGURE 12.4 H C

3

H C

3

JCH3

JCH3

H CJ

3

H C

3

H CJCH

3

J

CH

2

J CH

2

J

HCJOH

J

HOJCH

J

HCJOH

J

CH

2

J CH OH

2 Bacteriohopanetetrol

6. As early as 1972, it was known that many biological membranes are asymmetric with

respect to distribution of phospholipids between the inner and outer leaflets of the bilayer. Once such asymmetry is established, what factors act to preserve it?

7. As discussed in the text, the length and degree of saturation of the fatty acyl chains in

membrane bilayers can affect the melting temperature Tm. (a) The value of Tm for a pure sample of phosphatidyl choline that contains two

12-carbon fatty acyl chains is −1ºC. Values for phosphatidyl choline species with longer acyl chains increase by about 20ºC for each two-carbon unit added. Why?

(b) Suppose you have a phosphatidyl choline species that has one palmitoyl group es

terified to C-1 of the glycerol moiety, as well as an oleoyl group esterified at C-2 of glycerol. How would Tm for this species compare with that of dipalmitoylphosphatidyl choline, which contains two esterified palmitoyl groups?

sphingosine backbone. Compare the Tm for this phospholipid with that of dipalmitoylphosphatidyl choline.

(d) The transition temperature for dipalmitoylphosphatidyl ethanolamine is 63ºC.

Suppose you have a sample of this phospholipid in excess water at 50ºC, and you add cholesterol until it constitutes about 50% of the total lipid, by weight, in the sample. What would you expect when you attempt to determine the transition temperature for the mixture? 204

CHAPTER 12

8. At least two segments of the polypeptide chain of a particular glycoprotein span the

membrane of an erythrocyte. All the sugars in the glycoprotein are O-linked.

(a) Which amino acids might be found in the portion of the chain that is buried in the

lipid bilayer?

(b) Why would you expect to find serine or threonine residues in the glycoprotein?

9. (a) Many integral membrane proteins are composed of a number of membrane-spanning

segments, which form bundles of a helices packed closely together, often forming a membrane channel or pore. Each membrane-spanning sequence of most integral membrane proteins is an a helix composed of 18 to 20 amino acids. What is the width of the hydrocarbon core of the membrane?

(b) The sequence of one of the a helices in a particular integral membrane protein is

shown in Figure 12.5, and the 19 residues in this helix are plotted in a helical wheel plot. Such a plot projects the side chains of the amino acid residues along the axis of the a helix (z-axis) onto an x-y plane. In an a helix, a full turn occurs every 3.6 residues, so that each successive residue is 100° apart on the helix wheel. Compare the location of hydrophobic side chains on the helix surface with those that are polar or hydrophilic. Where are the hydrophobic side chains, and how are they accommodated in the membrane? Where are the polar side chains? How are they accommodated in the protein-membrane complex?

Helix sequence:

Ser Val Tyr Asp Ile Leu Glu Arg Phe Asn Glu Thr Met Asn His Ala Val Ser Gly FIGURE 12.5

Gly

19

Ser

Arg

1

Thr

8

12

Ile

His

5

15

Asp 4

16 Ala

Glu 11

9 Phe

Ser 18

2 Val

13 Met

Glu 7

6 Leu

14

Asn

3

17

Tyr

10

Val

Asn

LIPIDS AND CELL MEMBRANES 205

10. A series of experiments that shed some light on the movement of lipids in membranes

were conducted by Rothman and Kennedy, using a gram-positive bacterium. They used 2,4,6-trinitrobenzenesulfonic acid (TNBS), which reacts with amino groups in phosphoethanolamine residues. Note that TNBS, shown in Figure 12.6, is charged at physiologic pH and cannot penetrate intact membrane vesicles. Incubation of TNBS with intact bacterial cells and with disrupted cells revealed that about two-thirds of the phosphoethanolamine molecules are located on the outside of the membrane, with the remaining residues on the inside. Rothman and Kennedy then incubated growing cells with a pulse of radioactive inorganic phosphate, to label newly synthesized phosphoethanolamine molecules in the membrane. Using TNBS once again to distinguish between residues on the two sides of the membrane, they determined that immediately after the radioactive pulse, all newly synthesized phosphoethanolamine residues were located on the inner face of the membrane. After 30 minutes, however, the original distribution of phosphoethanolamine residues on the inner and outer faces of the bacterial cell membrane was restored. What do these observations suggest about the movement of phospholipids in membranes? FIGURE 12.6

NO

2

:

+H NJR

+SO2 +H;

2

3

J

O N

NO

O N

NO

2

JJ

2

J

2

SO:

NH

3

JJ

R 2,4,6-Trinitrobenzenesulfonic acid (TBNS)

11. Mycoplasma cells can be grown under conditions so that their plasma membrane con

tains one type of glycolipid, such as mono- or diglucosylated sphingosine molecules.

(a) Membranes prepared from Mycoplasma cells undergo a phase transition when

heated. Suppose that sample A is isolated from cells whose glycolipids contain a very high percentage of unsaturated fatty acyl chains, whereas sample B is isolated from cells whose glycolipids contain a high percentage of saturated fatty acyl chains of the same length. When heated, which sample will exhibit a higher melting temperature? Why?

(b) Glycolipids from samples A and B are analyzed for the carbohydrate content of their

polar head groups. Those from sample A have a higher percentage of diglucosyl residues than those from sample B, which have mostly monoglucosyl residues. Explain how this observation is consistent with the lipid content of the two samples.

12. In mammals, lysophosphoglycerides (1-monoacylglycerol-3-phosphates) are generated

in small quantities in order to trigger physiologic responses. Hydrolysis of a fatty acyl group from the C-2 position of a glycerophospholipid yields a lysophosphoglyceride. The reaction is catalyzed by phospholipase A2, whose activity is strictly regulated. However, large quantities of phospholipase A2 are found in snake venom, and the active venom enzyme can generate high concentrations of lysophosphoglycerides from membranes of snakebite victims. Lysophosphoglycerides are so named because in high concentrations they can disrupt membrane structure. Why?

13. What features of liposomes make them potentially useful as a delivery system for trans

porting water-soluble drugs to target cells? Suggest how one could prepare a liposome that is specific for a particular type of cell. 206

CHAPTER 12

14. Explain the role of cholesterol in cell membranes.

15. During the solubilization of membranes, the purification of integral membrane proteins,

and the reconstitution of membranes, gentler detergents, such as octyl glucoside, are used in preference to sodium dodecyl sulfate (SDS). Explain why.

16. Why do membrane proteins not diffuse, that is, flip-flop, across membranes?

17. In a membrane, an integral membrane protein diffuses laterally an average distance of

4 × 10−6 m in 1 minute, whereas a phospholipid molecule diffuses an average distance of 2 mm in 1 second. (a) Calculate the ratio of the diffusion rates for these two membrane components. (b) Provide reasons for the difference between the two rates. ANSWERS TO PROBLEMS

1. All organisms require water for many biochemical reactions, and thus organisms can

live only in an aqueous environment. While a bacterial cell placed in a solution of crude oil might at least survive, it would not have enough water for growth and division. Microorganisms can best utilize crude oil or its hydrocarbon components when the microorganisms are present at a boundary layer between water and lipid. Aeration and agitation increase the effective area of such a layer. Some microorganisms that degrade hydrocarbons have a glycolipid-rich cell wall in which those compounds are soluble. After being solubilized, the compounds are transferred to the cytoplasmic membrane, where water-requiring reactions that initiate hydrocarbon degradation occur.

2. The four methyl side chains of each phytanic acid molecule interfere with the ordered

association of fatty acyl chains; thus, they increase the fluidity of nerve cell membranes. This increase in fluidity could interfere with myelin function or ion transport, but the actual molecular basis for the symptoms is not yet known. Many of the symptoms of Refsum’s disease can be eliminated by adopting diets that are free of phytol. The primary source of phytol in the human diet is from dairy products and other fats from ruminants. Cows, for example, consume large quantities of chlorophyll as they ingest grasses and plant materials. The symbiotic bacteria that inhabit the bovine rumen readily degrade chlorophyll, releasing free phytol, which is then converted to phytanic acid. Up to 10% of the fatty acids in bovine blood plasma are found as phytanic acid, which can then be incorporated into cell membranes and milk. For those who have Refsum’s disease, it is therefore necessary to restrict consumption of beef as well as dairy products like milk and butter. Because humans do not degrade chlorophyll extensively during digestion, restriction of green plants in the diet is usually unnecessary.

3. You would expect to find that the bacteria grown at the higher temperature will have

incorporated a higher number of the longer fatty acids and a greater proportion of the saturated fatty acids. The membranes of bacteria grown at 18°C will have more shortchain fatty acids and more that are unsaturated. These cells select fatty acids that will remain fluid at a lower temperature in order to prevent their membranes from becoming too rigid. The cells grown at the higher temperatures can select fatty acids that pack more closely. Cells in both cultures thus employ strategies designed to achieve optimal membrane fluidity.

4. The higher the number of cis double bonds, the less ordered the bilayer structure will

be and the more fluid the membrane system will be. You would therefore expect Tm for

the bilayer system containing the acyl chains with four unsaturated bonds to be much lower than that for the system containing the saturated fatty acid chains.

5. Like cholesterol, bacteriohopanetetrol is a pentacyclic molecule with a rigid, platelike,

hydrophobic ring structure; it has a hydrophilic region as well, although that region is on the opposite end of the molecule when compared with cholesterol. In bacterial membranes hopanoids may have a function similar to that of cholesterol in mammalian membranes; that is, they may moderate bacterial membrane fluidity by blocking the motion of fatty acyl chains and by preventing their crystallization.

6. Phospholipids have polar head groups, so their transfer across the hydrophobic interior

of the bilayer as well as their dissociation from water at the bilayer surface would require a positive change in free energy. Without the input of free energy to make the process a spontaneous one, the transfer of the polar head group is very unlikely, so the asymmetric distribution of the phospholipids is preserved.

7. (a) The longer the acyl groups, the larger the number of noncovalent interactions that

can form among the hydrocarbon chains. Higher temperatures are therefore required to disrupt the interactions of phospholipid species that have longer fatty acyl groups.

(b) The cis double bond in oleate produces a bend in the hydrocarbon chain, interfer

ing with the formation of noncovalent bonds between the acyl chains. Less heat energy is therefore required to cause a phase transition; in fact, the melting temperature for phosphatidyl choline with a palmitoyl and an oleoyl unit is −5ºC, while Tm for dipalmitoylphosphatidyl choline is 41ºC.

choline and of sphingomyelin are very similar to each other; both contain phosphoryl choline and both have a pair of hydrocarbon chains. Given similar chain lengths in palmitoylspingomyelin and in dipalmitoylphosphatidyl choline, you would expect that values of Tm for the two molecules are similar. Both species in fact exhibit a phase transition at 43ºC.

(d) At 50ºC, you should expect cholesterol to diminish or even to abolish the transition,

by preventing the close packing of the fatty acyl chains that impart rigidity to the molecular assembly. At higher temperatures, cholesterol in the mixture also prevents larger motions of fatty acyl chains, making the assembly less fluid. Studies show that in mixtures containing 30 to 35 mol % cholesterol, phase transitions are extinguished.

8. (a) You should expect to find nonpolar amino acid residues in the portions of the gly

coprotein chain that are buried in the membrane. Because the core portion of the membrane is 30 Å wide, up to 20 amino acids could be included in the buried segments, assuming that the amino acids are part of an a helix, in which the translation distance for each amino acid is 1.5 Å.

(b) The glycoprotein has O-linked carbohydrate residues, and in most such proteins

the sugars are attached to the side chains of serine or threonine residues. Were the sugars N-linked instead, you would expect to find one or more asparagine residues in the glycoprotein.

9. (a) In an a helix, each amino acid residue extends 1.5 Å (1.5 × 10−1 nm) along the

helix axis. Therefore a span of 20 amino acids in an a helix will be about 30 Å in length, corresponding to the width of the hydrophobic core of the membrane.

(b) The plot clearly shows that hydrophobic amino acids are concentrated along one

side of the surface of the helix. The side chains of those residues are likely to face the hydrophobic core of the membrane. Polar side chains are located on the opposite side of the helical surface; they are likely to be on the side of the chain that faces other a-helical bundles. They could form hydrogen or ionic bonds with polar residues in other bundles. 208

CHAPTER 12

10. In model membrane systems, the transfer of phospholipid head groups from one side

of the bilayer to the other is very slow, presumably because of the energy required to move the polar head group through the hydrophobic bilayer. The experiments carried out by Rothman and Kennedy indicate that a process that mediates the flip-flop of membrane lipids is operating in bacterial cells. Phospholipid synthesis takes place on the cytosolic face (the inner leaflet) of the membrane, and some of the newly synthesized lipids are moved through the bilayer to the outside surface of the membrane bilayer. While aminophospholipid translocases that can move polar lipids across membranes have been found in eukaryotes, it is not yet known how such a process occurs in bacteria.

11. (a) You would expect sample B, with a higher percentage of saturated fatty acids, to

have a higher melting point. The saturated chains will aggregate more closely with each other, requiring more thermal energy to disrupt that aggregation. The acyl chains of unsaturated fatty acids are kinked, and therefore cannot aggregate in regular arrays like saturated acyl chains of the same length. They are therefore disrupted at a lower temperature.

(b) In the membrane, the cross-sectional area occupied by unsaturated fatty acids is

larger than that occupied by saturated chains, because of kinks in the hydrocarbon chains due to double bonds. A diglucosyl head group is larger (i.e., has a larger cross-section size) than that of a monoglucosyl derivative, so that the larger head group would match the increase in cross-sectional area in the interior of the bilayer composed of unsaturated fatty acyl chains.

12. Glycerophospholipids contain two fatty acyl groups esterified to glycerol, to which a

polar head group is also attached at the C-3 carbon. When phospholipase A2 removes one of the fatty acyl chains, the polar head group is too large in relation to the single hydrocarbon chain to allow optimal packing in the bilayer. The regular association of the hydrocarbon tails is disrupted, and the plasma membrane dissolves.

13. Liposomes are essentially impermeable to water-soluble molecules. Therefore, water

soluble drugs could be trapped inside the liposomes and then be delivered into the target cells by fusing the liposomes with the cell membrane. To make a liposome specific for a particular type of cell, antibodies that have been prepared against a surface protein of the target cell could be attached to the liposome via a covalent bond with a bilayer lipid, for example, phosphatidyl ethanolamine. This would enable the liposome to recognize the target cells. Of course, strategies would also have to be devised to prevent the premature, nonspecific fusion of the liposome with other cells.

14. Cholesterol modulates the fluidity of membranes. By inserting itself between the fatty

acid chains, cholesterol prevents their “crystallization” at temperatures below Tm and sterically blocks large motions of the fatty acid chains at temperatures above Tm. In fact, high concentrations of cholesterol abolish phase transitions of bilayers. This modulating effect of cholesterol maintains the fluidity of membranes in the range required for biological function.

15. Although sodium dodecyl sulfate (SDS) is a very effective detergent for solubilizing

membrane components, the strong electrostatic interactions of its polar head groups with charged groups on the membrane proteins disrupt protein structure. A detergent such as octyl glucoside, which has an uncharged head group, allows the proteins to retain their three-dimensional structures while it interacts with their hydrophobic domains.

16. Membrane proteins are very bulky molecules that contain numerous charged amino acid

residues and polar sugar groups (in the case of glycoproteins) that are highly hydrated. Such molecules do not diffuse through the hydrophobic interior of the lipid bilayer.

LIPIDS AND CELL MEMBRANES 209

17. (a) Rate of protein diffusion:

6

−

1 min

4

10

6 7

.

10 8

×

=

×

−

m/ min

m /s

60 s

Rate of phospholipid diffusion:

2 mm/s = 2 × 10−6 m/s Ratio of phospholipid diffusion rate to protein diffusion rate:

2 × 1

06 m / s

= 30

6 7

. × 1

0 8

− m/s

(b) The difference in diffusion rates is due primarily to the difference in mass between

phospholipids, which have a molecular weight of approximately 800, and proteins, which have a molecular weight greater than 10,000. In addition, integral membrane proteins may associate with peripheral proteins, which would further decrease their lateral diffusion. EXPANDED SOLUTIONS TO TEXT PROBLEMS

1. 1 mm2 = (10−6 m)2 = 10−12 m2 70 Å2 = 70 (10−10 m)2 = 70 × 10−20 m2. Since the bi

layer has two sides, 2 × 10−12/(70 × 10−20) = 2.86 × 106 molecules.

2. Using the diffusion coefficient equation, s = (4Dt)1/2, one gets s = (4 × 10−8 × 10−6)1/2 or

(4 × 10−8 × 10−3)1/2 or (4 × 10−8 × 1)1/2. Solving for s gives 2 × 10−7 cm, 6.32 × 10−6 cm, and 2 × 10−4 cm, respectively.

3. The gram molecular weight of the protein divided by Avogadro’s number = 105 g/(6.02 ×

1023) = 1.66 × 10−19 g/molecule. 1.66 × 10−19 g/1.35 (density) = 1.23 × 10−19 cm3/molecule. The volume of a sphere equals 4/3 pr3 = 1.23 × 10−19 cm3. Solving for r, one gets 3.08 × 10−7 cm. By substituting this value into the equation given,

−

1 3

. 8

16

10

310 D =

×

6 × 3.14 × 1 × 3.08 ×

−7

10

= 7 3

. 7 ×

−9

10 cm2 / s

By substituting this value for D and the times given in the problem into the equation shown in the answer to problem 2, one obtains the distances traversed: 1.72 × 10−7 cm in 1 ms, 5.42 × 10−6 cm in 1 ms, and 1.72 × 10−4 cm in 1 s.

4. As its name implies, a carrier antibiotic must move from side to side when it shuttles ions

across a membrane. By contrast, channel formers allow ions to pass through their pores much like water through a pipe. Lowering the temperature caused a phase transition from a fluid to a nearly frozen membrane. In the nearly frozen state the carrier is immobilized, whereas the pore of the channel former remains intact, allowing ions to pass through it.

5. The initial decrease in the amplitude of the paramagnetic resonance spectrum results

from the reduction of spin-labeled phosphatidyl cholines in the outer leaflet of the bilayer. Ascorbate does not traverse the membrane under these experimental conditions, and so it does not reduce the phospholipids in the inner leaflet. The slow decay of the residual spectrum is due to the reduction of phospholipids that have flipped over to the outer leaflet of the bilayer. 210

CHAPTER 12

6. The large polar carbohydrate moieties of glycolipids are always on the extracellular side

of a cell membrane. Because the sugars are large and polar, there is a significant energy barrier to passing them through the interior of the lipid bilayer. Therefore, glycolipids do not spontaneously undergo flip-flop.

7. The cis double bonds interrupt the packing of fatty acid chains and make phospholipid

bilayers more fluid. By contrast, trans double bonds would pack in quite similar fashion as saturated chains (see Figure 12.7). FIGURE 12.7 saturated:

O

trans:

O

O:

O

cis:

K

O:

8. The helix formation would be more likely in the hydrophobic medium. Water molecules

would compete with peptide backbone NH and C=O groups for hydrogen bond formation; this competition would reduce the relative helix propensity in water. Additionally, the isolated NH and C=O groups would be quite unstable if not hydrogen bonded in a hydrophobic medium, and so would be driven to maximize their participation in hydrogen bonds.

9. Double bonds (cis) in the lipid acyl chains will increase the membrane fluidity. To maintain

a similar membrane fluidity at the lower temperature of 25ºC, the bacteria would incorporate more of the unsaturated fatty acids in their membrane phospholipids than at 37ºC.

10. (a) The main effect is to broaden the phase transition of the lipid bilayer. The relative

change in fluidity near Tm is much less dramatic when cholesterol is present. The effect could be biologically important in maintaining the functions of proteins that may be sensitive to membrane fluidity. In particular, such proteins will be less sensitive to small local fluctuations in temperature when cholesterol is present in the membrane.

11. We will presume that the hydropathy plots were constructed using 20-residue windows.

Plot c shows several peaks that surpass the criterion level of 20 kcal/mol−1 (84 kJ/mol− 1) for the hydropathy index, indicating possible regions for membrane-spanning a helices. Therefore plot c is likely to predict a membrane protein with about four (possibly five) membrane-spanning a helices. However, there are ambiguities: Membrane-spanning b-strands will escape detection by these hydropathy plots. (Plots a and b, in fact, are somewhat similar to Figure 12.18.) Additionally, a highly nonpolar segment of a protein sequence is not necessarily a transmembrane segment, but may simply be a hydrophobic segment that is buried in the core of the folded protein.

12. Membrane proteins are not soluble in water, and they require lipids for folding into their

proper functional states. Lipid/protein complexes are difficult to crystallize. In some cases, the lipids may be replaced by detergents that may solubilize particular membrane proteins (with retention of their biological functions), but some detergents may alter the folded state of the membrane protein. Furthermore, detergent/protein complexes also are difficult to crystallize (though easier than lipid/protein complexes). Key advances in the development of synthetic detergents and of methods for crystallization have led to several dozen crystal structures of important membrane proteins. CHAPTER 1 Membrane Channels and Pumps 3

The intrinsic impermeability of the lipid bilayer to polar molecules and ions can be

circumvented by two classes of transmembrane proteins—pumps and channels. This chapter describes some of the structural and functional features of these pro

teins. The authors first differentiate between active transport (used by pumps) and passive transport (used by channels) of molecules across a membrane and discuss how to quantitate the free energy stored in concentration gradients. The authors then discuss three types of active transport systems: the P-type ATPases and the ATP-binding cassette (ABC) pumps, both of which use ATP hydrolysis to drive the transport of ions across the membrane; and the secondary transporters, which couple the thermodynamically uphill flow of one molecule with the downhill flow of another. The wellstudied Na+-K+ ATPase and sarcoplasmic reticulum Ca2+-ATPase are used as examples of P-type ATPases, which have many common structural and mechanistic features. The authors then look at the more recently identified family of ABC pumps including the multidrug resistance protein and cystic fibrosis transmembrane conductance regulator. The discussion of active transport is concluded with an examination of the mechanism of secondary transporters, including the bacterial lactose permease, which uses the proton-motif force to drive the uptake of lactose against a concentration gradient.

In addition to active transport, ions can be transported across membranes by pas

sive methods such as through ion channels. The authors differentiate between voltagegated and ligand-gated channels and discuss the key properties of all channels. The powerful patch-clamp technique is described, which allows researchers to measure the activity of a single ion channel. Two types of ion channels important in the propagation of nerve impulses are examined in detail—the ligand-gated acetylcholine receptor and the voltage-gated sodium and potassium ion channels. The chapter concludes with a discussion of gap junctions, which act as cell-to-cell channels and allow all polar molecules with a molecular mass of less than 1 kDa to pass through. 211 212

CHAPTER 13 LEARNING OBJECTIVES

When you master this chapter, you should be able to complete the following objectives. Introduction

1. Distinguish between channels and pumps. List the forms of energy that can drive active transport.

2. Distinguish between ligand-gated and voltage-gated channels The Transport of Molecules Across a Membrane May Be Active or Passive (Text Section 13.1)

3. List the two factors determining whether a molecule will cross a membrane.

4. Distinguish between simple and facilitated diffusion.

5. Use the concepts of free-energy change (DG) and electrochemical potential to predict

active or passive transport. A Family of Membrane Proteins Uses ATP Hydrolysis to Pump Ions Across Membranes (Text Section 13.2)

6. Describe the defining features of P-type ATPases.

7. Describe the functions of the Na+-K+ ATPase or Na+-K+ pump.

8. Describe the structure and the functional sites of the sarcoplasmic Ca2+-ATPase.

9. Discuss the inhibition of the Na+ -K+ pump by cardiotonic steroids.

10. Outline the reaction cycle of the Ca2+-ATPase.

11. Compare the sarcoplasmic Ca2+-ATPase and the Na+-K+ pump in terms of functional sites

and reaction cycles. Multidrug Resistance and Cystic Fibrosis Highlight a Family of Membrane Proteins with ATP-Binding Cassette Domains (Text Section 13.3)

12. Define multidrug resistance.

13. Compare the functions of the multidrug resistance protein and the cystic fibrosis trans- membrane conductance regulator.

14. Describe the architecture of the ABC transporter family of proteins. Secondary Transporters Use One Concentration Gradient to Power Formation of Another (Text Section 13.4)

15. Define symporter and antiporter.

16. Describe the role of the sodium-calcium exchanger and compare its capacity to extrude

Ca2+ with the Ca2+-ATPase.

17. Using lactose permease as the example, explain how a proton gradient or proton-motive force promotes the accumulation of lactose in bacteria.

MEMBRANE CHANNELS AND PUMPS 213 Specific Channels Can Rapidly Transport Ions Across Membranes (Text Section 13.5)

18. List the key properties of ion channels. Compare the rates of active transport with rates

of transport through channels.

19. Outline the patch-clamp technique and note its use in electrical measurements of membranes.

20. Explain the function of neurotransmitters and ligand-gated channels in the transmission

of nerve impulses across synapses. Outline the effects of acetylcholine on the postsynaptic membrane.

21. Describe the subunit structure, ligand binding sites, and channel architecture of the acetylcholine receptor from Torpedo marmorata.

22. Define action potential and explain its mechanism in terms of the transient changes in

Na+ and K+ permeability of the plasma membrane of a neuron.

23. Explain the effects of tetrodotoxin on the sodium channel.

24. Outline the possible role of the S4 segments of sodium channels as voltage sensors. List

the sequence of steps in the cycling of sodium channels during an action potential.

25. Outline the possible role of the S5 and S6 segments of sodium, potassium, and calcium

channels as a key region of the ion channel pore.

26. Describe the selectivity filter of potassium channels.

27. Compare the structure, ion selectivity, and inactivation mechanism of the potassium

channel with the sodium channel.

28. Relate the structure of the potassium channel to its rapid rate of transport. Gap Junctions Allow Ions and Small Molecules to Flow Between Communicating Cells (Text Section 13.6)

29. Distinguish between gap junctions and other membrane channels. Give examples of mol

ecules that can pass through gap junctions.

30. Describe the structure of a gap junction and the role of connexin in the formation of the

structure. SELF-TEST Introduction

1. Which of these statements about membrane channels and pumps are true?

(a) Both are integral, transmembrane proteins. (b) Both can be ligand- or voltage-gated. (c) Both contain multiple subunits or domains. (d) Both carry out active transport of ions and polar molecules. (e) Both allow bidirectional flux of the transported molecule. The Transport of Molecules Across a Membrane May Be Active or Passive

2. What will be the free-energy change generated by transport of one mole of Na+ from a

concentration of 10 mM to 150 mM with a membrane potential of −25 mV at 37º C? Would the transport need to be active or passive? 214

CHAPTER 13 A Family of Membrane Proteins Uses ATP Hydrolysis to Pump Ions Across Membranes

3. The orientation of the Na+-K+ pump in cell membranes determines the side of the mem

brane where the various processes involved in the transport of Na+ and K+ will take place. Assign each of the steps or processes in the right column to the intracellular or extracellular side of the pump.

(a) intracellular side

(1) binding of cardiotonic steroids

(b) extracellular side

(2) hydrolysis of ATP and phosphorylation

of the pump

(3) binding of K+ (4) binding of Na+

4. The proposed model for the mechanism of the Ca2+-ATPase is based on the existence of

four conformational states of this enzyme. Match each conformational state in the left column with the appropriate descriptions from the right column.

(a) E1

(1) low affinity for Ca2+

(b) E1-P

(2) high affinity for Ca2+

(3) is phosphorylated by ATP upon Ca2+

(d) E2-P

binding

(4) ion-binding sites open to the cytosol (5) ion-binding sites open to the luminal side

of the membrane

(6) is dephosphorylated upon the release of Ca2+

5. Explain why an electric current is generated during the transport of Na+ and K+ by the

Na+-K+ pump.

6. Which of the following statements describe properties that are common to the Ca2+-ATPase

of the sarcoplasmic reticulum and the Na+-K+ pump?

(a) Both are very abundant membrane proteins in the sarcoplasmic reticulum. (b) Both have homologous N-terminal subunits containing numerous transmem

brane helices.

(c) Both contain an aspartate residue that is phosphorylated by ATP. (d) Both translocate the same number of ions per transport cycle. (e) Both probably have four major conformational states. Multidrug Resistance and Cystic Fibrosis Highlight a Family of Membrane Proteins with ATP-Binding Cassette Domains

7. Which of the following statements about proteins containing ABC domains is INCORRECT?

(a) They are members of the P-loop NTPase superfamily. (b) They usually consist of two membrane spanning domains and two ATP-binding

domains.

rate polypeptide chains.

(d) The ABC domains undergo conformational changes upon ATP binding and hy

drolysis.

MEMBRANE CHANNELS AND PUMPS 215 Secondary Transporters Use One Concentration Gradient to Power Formation of Another

8. Which of the following statements about the sodium-calcium exchanger is incorrect? The

sodium-calcium exchanger

(a) is a symporter for sodium and calcium transport. (b) is an antiporter for sodium and calcium transport. (c) is driven by the Na+ gradient generated by the Na+-K+ pump. (d) has a lower affinity for Ca2+ than the Ca2+-ATPase. (e) has a higher transport rate for Ca2+ than the Ca2+-ATPase.

9. Ascribe the characteristics in the right column either to active transport or to transport

of ions or molecules through channels.

(a) active transport

(1) flux ~107 s−1

(b) transport through channels

(2) flux 3 × 10 to 2 × 103 s−1 (3) Ions can flow from either side

of the membrane.

(4) flux in a specific direction

10. Which of the following statements about the lactose permease of E. coli under physio

logic conditions are correct?

(a) It derives energy for transport from an Na+ gradient. (b) It derives energy for transport from an H+ gradient. (c) It derives energy for transport from a Ca2+ gradient. (d) It is an antiporter for lactose and H+. (e) It is a symporter for lactose and Na+. Specific Channels Can Rapidly Transport Ions Across Membranes

11. List the events in the transmission of nerve impulses in synapses in their proper

sequence.

(a) binding of acetylcholine to acetylcholine receptor (b) depolarization of the postsynaptic membrane (c) release of acetylcholine from synaptic vesicles into the synaptic cleft (d) increase in postsynaptic membrane permeability to Na+ and K+ (e) increase of acetylcholine concentration in the synaptic cleft from ~10 nM to 500 mM

12. Explain the use of cobratoxin in the purification of acetylcholine receptor.

13. Which of the following is not a characteristic of the structure of the acetylcholine

receptor of Torpedo?

(a) It has five subunits of four different kinds. (b) Acetylcholine binds between the a–g and a–d interfaces. (c) It has a uniform channel 20 Å in diameter. (d) The genes for the subunits arose by the duplication and divergence of a common

ancestral gene.

(e) It has pentagonal symmetry. 216

CHAPTER 13

14. Which of the following statements about the plasma membrane of a neuron are correct?

(a) In the resting state, the membrane is more permeable to Na+ than to K+. (b) In the resting state, the membrane potential is approximately +30 mV. (c) The Na+ and K+ gradients across the membrane are maintained by the Na+-K+

pump.

(d) The equilibrium potential for K+ across the membrane is near −75 mV. (e) During the action potential, the membrane potential varies between the limits of

+30 mV and −75 mV.

15. Place the following events of the action potential in their correct sequence.

(a) the spontaneous closing of sodium channels (b) a membrane potential of −75 mV (c) the depolarization of the plasma membrane to approximately −40 mV (d) the opening of the potassium channels (e) the opening of sodium channels (f)

a membrane potential of +30 mV

(g) a membrane potential of −60 mV

16. Which of the following statements about the sodium channel, purified and reconstituted

in lipid bilayers, are correct?

(a) It is about 10 times more permeable to Na+ than to K+. (b) It is sensitive to voltage. (c) It is inhibited by cobratoxin. (d) It becomes inactivated spontaneously. (e) It consists of seven hydrophobic transmembrane segments.

17. Match the sodium (eel electric organ) and potassium (Shaker) channels with the corre

sponding properties listed in the right column.

(a) sodium channel

(1) four 70-kd subunits

(b) potassium channel

(2) single 260-kd polypeptide chain (3) fourfold symmetry (4) tetrodotoxin binding site (5) positively charged, voltage-sensing S4

helical segment

(6) ball-and-chain inactivation mechanism (7) 3-Å–diameter channel Gap Junctions Allow Ions and Small Molecules to Flow Between Communicating Cells

18. Which of the following statements are NOT true of gap junctions between cells?

(a) They allow the exchange of ions and metabolites between cells. (b) They allow the exchange of cytoplasmic proteins. (c) They are essential for the nourishment of cells that are distant from blood vessels. (d) They are made up of 10 molecules of connexin. (e) They are controlled by Ca2+ and H+ concentrations in cells.

MEMBRANE CHANNELS AND PUMPS 217 ANSWERS TO SELF-TEST

1. a, c

2. Using the equation from page 347 of the text: c

DG = RT

ln 2 +

D ZF V c 1

150mM

DG = 1.99 × 310 × ln

+ 1

( ) × (23 062

.

) × −

( 0 025

.

V)

5mM

= 1 6

. 7 kcal / mol + − .577 kcal

0

/ mol = + 1.09 kcal / mol

Since the sign of DG is positive, the transport is unfavorable in the direction indicated and therefore active transport must be used.

3. (a) 2, 4 (b) 1, 3

4. (a) 2, 3, 4 (b) 2, 4 (c) 1, 5, 6 (d) 1, 5

5. Since three Na+ ions are transported out for every two K+ ions that are transported in,

there is a net efflux of one positively charged ion. The net movement of ions sets up an electric current.

6. b , c, e

7. The answer is (c). While prokaryotic ABC proteins are often multisubunit proteins,

eukaryotic ABC proteins usually contain both membrane spanning and ATP-binding domains on the same polypeptide.

8. Answer (a) is incorrect. The sodium-calcium exchanger transports these cations in the

opposite direction; therefore, it is an antiporter, not a symporter.

9. (a) 2, 4 (b) 1, 3

10. b

11. c, e, a, d, b

12. Cobratoxin binds specifically and with very high affinity to the acetylcholine receptor;

therefore, a column with covalently attached cobratoxin can be used in the affinity purification of the receptor from a mixture of macromolecules in the postsynaptic membrane that has been solubilized by adding nonionic detergents.

13. Answer (c) is incorrect. The channel is not of uniform diameter.

14. c, d, e

15. g, c, e, f, a, d, b, g

16. a, b, d

17. (a) 2, 3, 4, 5, 6

(b) 1, 3, 5, 6, 7

18. b, d

1. Calculate the free-energy change for the transport of an uncharged species from a con

centration of 5 mM outside a cell to a concentration of 150 mM inside. Assume that the temperature is 25ºC. Now repeat the calculation for an ion with a charge of +1 that is crossing a membrane with a potential of −60 mV, with the interior negative with respect to the exterior. Would the transport of an ion with a −1 charge be more or less favorable?

2. In dog skeletal muscle, the extracellular and intracellular concentrations of Na+ are

150 mM and 12 mM, and those of K+ are 2.7 mM and 140 mM, respectively.

(a) Calculate the free-energy change as three Na+ are transported out and two K+ are

transported in by the Na+-K+ pump. Assume that the temperature is 25ºC and that the membrane potential is −60 mV.

(b) Does the hydrolysis of a single ATP provide sufficient energy for the process in

part (a)? Explain.

3. An uncharged molecule is transported from side 1 to side 2 of a membrane.

(a) If its concentration is 10−3 M on side 1 and 10−6 M on side 2, will the transport be

an active or a passive process? Explain your answer.

(b) If the concentration is 10−1 M on side 1 and 10−4 M on side 2, how will the free

energy change compare with that in part (a)? Explain.

4. In addition to the Na+-K+ ATPase, eukaryotic cells contain other ATP-driven pumps. One

such pump is the H+-K+ ATPase, in which a hydrogen ion is extruded from the cytoplasm in exchange for a potassium ion at the expense of ATP hydrolysis. Given that the interior of most animal cells is electrically negative with respect to the exterior, explain why the Na+-K+ ATPase can contribute to the membrane potential but the H+-K+ ATPase cannot.

5. What is the molecular basis for the phenomenon of multidrug resistance?

6. In experiments to investigate the mechanism of transport of two substances, X and Y,

across cell membranes, cells were incubated in media containing various concentrations of X and Y, and the initial rate of transport of each of the substances into the cell was determined. The results that were obtained are depicted in Figure 13.1. What conclusion is suggested by the results? Explain. It may be helpful to refer to the discussion of enzyme kinetics in the text. FIGURE 13.1 Initial velocity of transport versus concentration for substances X and Y.

Y

X

t

anspor

elocity of tr

Initial v

Concentration

7. Design an experiment using ATP labeled with 32P in the g position that would suggest

that the Na+-K+ ATPase reaction involves a stable enzyme-phosphate intermediate.

8. Figure 13.2 depicts a typical action potential that might be measured in an isolated

axon, such as the giant axon of a squid. Give the events that are responsible for (a) the rising phase of the action potential, and (b) the falling phase. Specify in each case whether ion flow occurs with or against concentration gradients, electrical gradients, or both. FIGURE 13.2 An action potential.

0

Rising

Falling

phase

phase

ane potential (mV) :60

TimeD

Membr

9. When the sciatic nerve is removed from a frog, placed in an isotonic salt solution,

and stimulated electrically, it will generate action potentials that can be measured by an electrode placed at some distance from the site of stimulation. When metabolic poisons are added to the preparation, the nerve retains the capability of generating action potentials even though the supply of ATP to drive its Na+-K+ pump has been depleted and it is thus incapable of carrying out active transport. Explain how this can be the case.

10. Suppose that a Glu residue is present in the narrow region of the sodium channel. A mu

tant protein is found in which that Glu is replaced by Val.

(a) Compare the Na+ conductance of the mutant as opposed to the normal channel.

Explain.

(b) Compare the sodium permeability as a function of pH in each case. (c) Compare the sensitivity of the normal and the mutant channels to tetrodotoxin. (d) Compare the magnitude of an action potential in nerves containing sodium chan

nels of the mutant type, as opposed to the normal type.

11. How could you produce a synthetic vesicle in which the uptake of lactose from the

medium into the vesicle against a concentration gradient could be driven by light?

12. Acetylcholine opens a single kind of cation channel that has a very similar permeabil

ity to Na+ and K+, yet the influx of Na+ is much larger than the efflux of K+. Explain this fact.

13. The K+ channel is over 100 times more permeable to K+ than to Na+. Explain the mo

lecular mechanism for this selectivity.

14. Explain the experimental evidence supporting the ball-and-chain model for channel in

activation. 220

CHAPTER 13 ANSWERS TO PROBLEMS

1. We use the following equation from page 347 in the text: c

DG = RT ln

2 c 1

=

150

1 9

. 9 × 298 × ln 5

= +2 0

. 2 kcal /mol

Now for the ion with a +1 charge: c

DG = RT ln 2 +

D ZF V c 1

=

150

1.99 × 298 × ln

+ [(+1) × 23 062

.

× −

( 0 060

.

)]

5

= +2 0

. 2 − 1 3

. 8

= +0.64 kcal / mol

Note that the membrane potential favors the entry of a positively charged ion and overcomes the unfavorable concentration gradient. If the calculation is repeated for a negatively charged molecule, the free energy change will be more positive and therefore less favorable.

2. (a) To solve this problem, we first calculate the free-energy change for transporting three

Na+ and then that for two K+. The total free-energy change will be the sum of the two. Again we use the following equation from the text: c

DG = RT ln 2 +

D ZF V c 1

Substituting in the values for Na+ yields

150

DG

= 1 9

. 9 × 298 × ln

+ [(+1) × 23 062

.

× +

0 060

.

]

Na

5

= +1 5

. 0 + 1 3

. 8

= +2 8

. 8 kcal / mol

Note that when Na+ is transported out of the cell, work must be done against both a concentration gradient and an electrical gradient.

Now we carry out the corresponding calculation for the K+ ion.

140

DG

= 1 9

. 9 × 298 × ln

+ [(+1) × 23 062

.

× − +

( 0 060

.

)]

2 7

.

= +2 3

. 4 − 1 3

. 8

= +0 9

. 6 kcal / mol

MEMBRANE CHANNELS AND PUMPS 221

Note that potassium ion is being transported against a concentration gradient but with an electrical gradient. Accordingly, the sign for the electrical term in the equation is negative.

To get the total energy expenditure, we must account for the stoichiometry of transport by summing the energy required for the transport of Na+ and that required for the transport of K+.

DG = 3DG + + 2 DG +

Na

= (3 × 2 88

.

) + (2 × 0 96

)

= +10 6

. kcal / mol

(b) Although the free-energy change for the hydrolysis of ATP under standard condi

tions is −7.3 kcal/mol, the free-energy change for the ATP concentrations that exist in typical cells is approximately −12 kcal/mol. Thus, the energy furnished by the hydrolysis of a single ATP is sufficient.

3. An uncharged molecule is transported from side 1 to side 2 of a membrane.

(a) The transport will be a passive process. Because the concentration on side 1 is

higher than that on side 2, the molecule will move spontaneously down its concentration gradient. One can use the expression c

DG = RT ln 2 + ZF D

V c1

−6

=

10 RT ln

−3

10

to show that DG has a negative value. Assuming a temperature of 25ºC

cal

DG = 1 9

. 9

× 298 × K × (−3)

mol ⋅K

= −4 1

. 0 kcal / mol

A negative DG value in the direction of movement is the hallmark of passive transport.

(b) Since the ratios of the concentrations in (a) and (b) are equal, the free-energy change

for the transport is the same in both cases.

−

cal

10 4

DG = 1 9

. 9

× 298 × K × 1n

mol ⋅K

10 − 1

= −4 1

. 0 kcal / mol

concentration of the chemical species undergoing the change. Thus, the rate of transport will be equal to k(c1). Since the concentration of c1 is greater in (b) than in (a), the rate of transport will be greater in (b). 222

CHAPTER 13

4. The difference between the two ATPase systems lies in the stoichiometry of their ex

change of ions. In the case of the Na+-K+ ATPase, three Na+ ions are extruded and two K+ ions are taken up during each pump cycle, making the interior of the cell more negative (less positive) for each pump cycle. The H+-K+ ATPase, in contrast, extrudes one H+ ion for each K+_taken up, so its operation is electrically neutral.

5. Multidrug resistance is said to occur when resistance to one drug makes cells less sensi

tive to a range of other drugs. The development of multidrug resistance is correlated with expression and activity of a 170 kDa protein called MultiDrug Resistance protein (MDR). MDR contains an ATP-binding cassette (ABC) domain and pumps drugs out of cells before the drugs can exert their effects.

6. The curve for X shows saturation, which would be expected if some membrane car

rier is involved in the transport of substance X. The curve for Y shows no saturation, which is consistent with the notion that substance Y diffuses through the membrane without a carrier. Such behavior is shown by lipid soluble substances, which dissolve in the hydrophobic tails of membrane phospholipids and can thus enter cells without a carrier.

7. If the Na+-K+ ATPase reaction involved a stable enzyme-phosphate intermediate, the

mechanism would be a two-step process involving two independent half-reactions:

E + ATP

Na



, Mg 2+ →

 E − P + ADP (1)

E − P + H O

K

 →

 E + Pi (2)

2

In the first half-reaction, ATP and unmodified enzyme interact in the presence of Na+ and Mg2+ to give phosphorylated enzyme and the first product of the overall reaction, ADP. In the second half-reaction, the phosphorylated enzyme is hydrolyzed by water in the presence of K+ to give unmodified enzyme and inorganic phosphate. An overall reaction that consists of two independent half-reactions is known as a double displacement.

The following experiment would suggest that such an overall reaction occurs.

Incubate a fragmented membrane preparation with K+ ion to hydrolyze any phosphate that might be bound to the enzyme. Then wash the membrane preparation to remove all K+, and transfer the membrane to a medium containing g-labeled ATP, Na+, and Mg2+. After a suitable incubation period, wash the membrane preparation to remove any unreacted labeled ATP. Then carry out scintillation counting on the membrane preparation to detect the presence of labeled phosphate. The presence of radioactivity in the membrane fraction would suggest that a stable enzyme-phosphate intermediate had been formed. In fact, a covalent aspartyl phosphate derivative is formed at the active site of the ATPase.

8. Nerve cells, like most animal cells, have a higher concentration of K+ inside than out

side and a higher concentration of Na+ outside than inside. In addition, there is a membrane potential; that is, the inside of the cell is negative (in this case −60 mV) with respect to the outside.

(a) The rising phase of the action potential is due to the influx of Na+ ions down a con

centration gradient and an electrical gradient.

(b) The falling phase is due to the efflux of K+ ions down a concentration gradient but

against an electrical gradient.

MEMBRANE CHANNELS AND PUMPS 223

9. For each action potential that is generated in an axon, only a very few Na+ ions enter

and a very few K+ ions depart the cell. Thus, in a poisoned nerve cell, many tens of thousands of impulses may be conducted before ionic equilibrium across the membrane is achieved. The active transport of Na+ and K+ across the membrane may be best viewed as necessary in the long run but not in the short run.

10. If a Glu residue is present in the narrow region of the sodium channel and a mutant pro

tein is found in which that Glu is replaced by Val:

(a) The mutant sodium channel would have decreased Na+ conductance. Charge attrac

tion between Na+ ions and carboxylate anions is important in drawing Na+ into the channel. In the mutant channel, Val, with its uncharged side chain, will not attract Na+ ions.

(b) In the normal channel, Na+ conductance decreases as pH is lowered below 5.4 and

the negatively charged carboxylate side chains are titrated to uncharged carboxyl groups. The mutant channel would show no such sensitivity to pH, because no charged groups are present in the side chain of Val.

nel would be less sensitive. Tetrodotoxin contains a positively charged guanido group (p. 358 in the text) that presumably interacts electrostatically with a negatively charged group in the sodium channel.

(d) The magnitude of the action potential would be reduced in nerves containing the

mutant channel. The action potential is generated by sodium flowing from the outside of the cell to the inside. Decreased sodium conductance in the mutant would reduce the amount of sodium influx, and hence reduce the magnitude of the action potential.

11. Form reconstituted vesicles containing lactose permease and bacteriorhodopsin, with

each oriented in the membrane so that its cytoplasmic face is toward the inside of the vesicle. Illumination of such vesicles will cause hydrogen ions to be extruded by bacteriorhodopsin. The resulting gradient of hydrogen ions will then drive the entry of lactose by its permease.

12. The electrochemical gradient for Na+ influx is steeper than that for K+efflux. The con

centration gradients across the membrane are similar for both ions, but Na+ moves from the positive to the negative side of the membrane, whereas K+ moves from the negative to the positive side; that is, the membrane potential favors Na+ influx.

13. At the entrance of the pore of the potassium channel there is a glutamate residue that

binds cations. The pore is about 3 Å in diameter at its narrowest point, so that only dehydrated small cations can fit; however, the energy required to dehydrate Na+ and smaller cations is too large and is not compensated by favorable polar interactions that occur in the case of K+.

14. There are two pieces of experimental evidence given in the text in support of the ball

and-chain model of channel inactivation. The first is that treatment of the cytoplasmic side of either the Na+ or K+ channel with trypsin yields a trimmed channel that stays open after depolarization. The second is that N-terminal splice variants of the potassium channel have altered inactivation kinetics. A deletion of 42 amino acids at the N-terminus of the Shaker channel causes the channel to open upon depolarization but not inactivate. Addition of a synthetic peptide corresponding to the deleted amino acids restores inactivation to the channel. 224

CHAPTER 13 EXPANDED SOLUTIONS TO TEXT PROBLEMS

1. From the data given, we get L =

0

[T0]/[R0] = 105 and c = 1/20 = 5 × 10−2. The ratio of

closed to open channels = [T]/[R]. When no ligands are bound, [T]/[R] = 105. For one ligand, [T]/[R] = L ×

0 c = 105 × 5 × 10−2 = 5 × 103.

For two ligands, [T]/[R] is 105 × (5 × 10−2)2 = 2.5 × 102. For three ligands, [T]/[R] is

105 × (5 × 10−2)3 = 1.25 × 10. For four ligands, it is 105 × (5 × 10−2)4 = 0.625.

From these ratios of closed/open, one can calculate that the fractions of open chan

nels are respectively 10−5, 2 × 10−4, 3.98 × 10−3, 7.41 × 10−2, and 0.615.

2. All three of these molecules contain highly reactive phosphoryl groups that readily react

with the active-site serine of acetylcholinesterase to form a stable derivative. Without active acetylcholinesterase, synaptic transmission at the cholinergic synapses is impossible, resulting in respiratory paralysis.

3. (a) The binding of the first acetylcholine increases the open/closed channels by a fac

tor of 1.2 × 10−3/(5 × 10−6) = 240, whereas the binding of the second acetylcholine increases this ratio by a factor of 14/(1.2 × 10−3) = 11.7 × 103.

(b) For the free-energy calculation, refer to Table 8.4 and the accompanying discussion

on pages 194–195 in your textbook. Note that for a tenfold change in the equilibrium constant, there is a standard free-energy (DGº′) change of 1.36 kcal/mol. Also note that the DGº′ varies with the log of K′eq. Therefore, the free-energy change during the binding of the first acetylcholine is the log (240) × 1.36 = 3.24 kcal/mol. For the second binding, DGº′ = log (11.7 × 103) × 1.36 = 5.54 kcal/mol.

effect on the closed/open ratio. The acetylcholine receptor channel is not perfectly symmetric. The two a chains are not in identical environments. Also, the presence of desensitized states in addition to the open and closed ones indicates that a more complex model is required.

4. (a) Since 37% of the channels are open at −25 mV and 59% are open at −20 mV, 50%

of the channels will be open at a voltage somewhere between −20 and −25 mV. Then we can say that

5 mV

= x

0 2

. 2 5

( 9% − 37%)

0 09 .

59

(

% − 50%)

Solving this equation gives x = 2 mV, where x is the difference in the voltage causing 59% of the channels to open and that causing 50% of the channels to open. Therefore the voltage required to open half the channels is −20 mV −2 mV = −22 mV.

(b)

) K

c

o zF(V V )/RT

= ƒ = − ƒ = −

−

e

O ,

ƒo

where f =

=

fraction closed and fo

fraction open.

Then,

(1 − ƒ )

log

o

= −zF(V − V ) / e RT

0

ƒo

ƒ

log

o

= −zF(V − V ) / e RT

(1 − ƒ )

0

o

MEMBRANE CHANNELS AND PUMPS 225

From the information given we calculate the information shown in the table below. Volts ƒo 1 ƒo loge(ƒo/(1 ƒo)

−0.045

0.02

0.98

−3.89

−0.04

0.04

0.96

−3.18

−0.035

0.09

0.91

−2.31

−0.03

0.19

0.81

−1.45

−0.025

0.37

0.63

−0.53

−0.02

0.59

0.41

0.36

−0.015

0.78

0.22

1.27

−0.01

0.89

0.11

2.09

−0.005

0.95

0.05

2.94

0

0.98

0.02

3.89

0.005

0.99

0.01

4.60

If we plot these data, with or without the aid of a computer program, we obtain the following graph FIGURE 13.3

5.0

3.0

)) 0ƒ

3.81

1.0

x+

/(1 0 :1.0

(ƒ

173.4

e

y=

log

:3.0

:5.0

:0.06

:0.03

0

0.03

Volts

Since the slope of the line = 173.4, and = zF/RT Slope = 173.4 = zF/RT RT

1 987

.

298 z = slope ×

= 173 4

. ×

×

= 4 4

. 5 F

23060

The answer to (a) can be obtained from the equation for this straight line because when f =

o

0.5, y = 0. Then, 0 = 173.4x + 3.81 and x = −3.81/173.4 V = −22 mV.

by DG = RT log

e c2/c1 ZFDV, where Z is the electrical charge of the transported

species, V is the potential in volts, and F is the faraday. To calculate the DG contributed by the movement of the gating charge, we need only consider the ZFDV portion of this equation. Substituting, we get 4.5 × 23 kcal V−1 mol−1 × 0.05 V = 5.2 kcal/mol.

5. The channels provide no energy and permit only passive transport of ions. When a

sodium channel is open, sodium ions flow from the region of high [Na+] outside a cell to the region of low [Na+] inside. Conversely, potassium ions flow in the opposite direction. The ion gradients are established by active pumps that require energy. 226

CHAPTER 13

6. The guanidino group of tetrodotoxin has a single positive charge, as does Na+. The pos

itively charged guanidino group probably binds to the entrance of a sodium channel, but the remainder of the tetrodotoxin molecule is too large to pass through the channel. Consequently, the channel is blocked.

7. Ion-channel blocking molecules will disrupt or halt electrical activity in the nervous sys

tem, leading to paralysis. As with tetrodotoxin, these snail toxins can be poisonous at quite low concentrations. For biochemical studies, such toxins could be useful for identifying and labeling new types of ion channels, and for investigations of the channels’ mechanisms of action.

8. This is a difficult yet crucial concept. Let us take the viewpoint of the sodium channels.

(Similar arguments would apply to potassium, on a slightly different time scale.) When sodium channels are closed, the extracellular/intracellular imbalance in Na+ concentration cannot be productively “sensed” by the cell, that is, it is not operative or, in thermodynamic terms, there is no pathway by which Na+ ions could diffuse to adjust their concentrations toward equilibrium values. An action potential change of only about 20 mV causes a small number of the voltage-gated sodium channels to open. Following this initial event, the process becomes autocatalytic, but with only a very small proportion of available Na+ ions actually flowing through the channels. The small initial increase in the sodium permeability (due to the opening of only a few channels) causes a further positive increase in the membrane potential (beyond the initial 20 mV change), and more channels are induced to open. The further change in sodium permeability due to these additional channel openings changes the membrane potential still further, and so very rapidly more and more channels are induced to open until the signal due to sodium ions peaks within about 1 ms. Each sodium channel is open for only a very short time (1 ms or less). Within 1.5 ms from the initial triggering event (beginning of the action potential), the change in potassium permeability will have caused all of the sodium channels to close (Figure 13.19B). During the short time that it is open, each channel permits the passage of only a few thousand Na+ ions (a very small fraction of total available Na+). The ability to generate a signal using the flux of only a small number of ions has two important physiological consequences: (a) an initial triggering event is efficiently amplified as many channels become involved, and (b) the nerve cell can recover quickly and transmit a new impulse every few milliseconds.

9. Normally the open state of sodium channels lasts for only about 1 ms because it spontaneously converts to an inactive state. Its return to a closed but activatable state requires repolarization. Since BTX keeps the sodium channels open after depolarization, it apparently blocks the transition from the open to the inactivated state.

10. (a) Open channels enable ions to flow rapidly through membranes in a thermodynam

ically downhill direction, that is, from higher to lower concentration. Therefore, chloride ions will flow into the cell.

(b) This flow of chloride ions increases the membrane polarization. Since depolariza

tion triggers an action potential, the chloride flux is inhibitory.

sist of five subunits.

MEMBRANE CHANNELS AND PUMPS 227

11. Since DG = 2.3 RT log c

+

2/c1 ZFDV, we can substitute and get the following:

−

1 5

.

10 3

DG =

×

RT log

+ 2 × 23 × 6 ×

−

2 3

10 2

.

4 ×

−

10 7

= 4.86 kcal / mol (chemical work) + 2.76 kcal / mol (electrical work)

12. (a) i

Conductance = g = V −Er

=

5pA

= 100 pS (picosiemens)

5 × −

10 2 V − 0V

(See text, p. 295.)

(b) Since a picoampere is 10−12 amperes, it is the flow of 6.24 × 106 charges per sec

ond. Therefore, 5 × 6.24 × 106 × 10−3 s = 3.12 × 104 charges per ms.

1 ms

(time)

= t

3 1

. 2 × 104 molecules

1 molecule

Solving this equation gives t = 32 × 10−6 ms = 32 ns (nanoseconds).

13. Membrane vesicles containing a high concentration of lactose in their inner volume could

be formed. The binding of lactose to the inner face of the permease would be followed by the binding of a proton. Both sides would then evert. Because the lactose concentration on the outside is low, lactose and the proton will dissociate from the permease. The downhill flux of lactose will drive the uphill flux of protons in this in vitro system.

14. For proper nerve activity, the change in membrane permeability caused by the opening

of acetylcholine receptor channels must be short-lived. In order to close the receptor channels, it is important to remove the source of the stimulation, the acetylcholine. Once initiated, the nerve impulse moves on and the postsynaptic membrane must return to its resting state in order to be ready to receive and propagate another signal. (Some notable nerve poisons such as DIPF—see problem 15—act by inhibiting the acetylcholinesterase.)

15.

O

;

K

N

K

+ H OD

+

H C

O (CH )

2

H C O:

(CH )

3

3 3

3

3 3

Acetylcholinesterase is a serine esterase whose catalytic mechanism is similar to that of the serine proteases. As with chymotrypsin and trypsin, the active site of acetylcholinesterase has serine as part of a Ser-His-Asp catalytic triad. The mechanism will involve covalent tetrahedral and acyl enzyme intermediates in which the substrate is bonded covalently to the active-site Ser. The reaction starts with nucleophilic attack on

228

CHAPTER 13

the substrate carbonyl group by the Ser OH group to give a tetrahderal intermediate (His acts as a base and accepts H+). Next, choline will be the leaving group and the acetyl group will be left bonded to the Ser in the acyl-enzyme intermediate. Then in the second half of the reaction, water will act as the nucleophile to attack the acyl enzyme, giving a second tetrahedral intermediate. Finally, the free enzyme will be regenerated when acetate leaves as the second product. The process then can repeat.

16. (a) The ASIC1a channels are most sensitive to the toxin (first set of recordings in part

(A) of the problem figure). The currents from these channels are completely inhibited for about 60 s following the application of the toxin; then there is a recovery.

(b) Yes, the effect of the toxin is reversible. Toward the end of the first set of record

ings in part (A) the ASIC1a channels are recovering as the PcTX1 is washed from the system.

slightly less than 1 nM. A good estimate is approximately 0.7 nM (reading from the logarithmic scale).

17. The channels with the bV266M mutation remain open for longer times. There are sev

eral possible explanations for the slower channel closing rate. For example, a tighter binding of acetylcholine (slower release) could keep the channels open longer. Alternatively, acetylcholine could be released at the normal rate, but the mutation could slow the conformational transition from the open state to the closed state. (Other explanations are possible.)

18. With fast channel syndrome, the recordings would show channel events that are very

brief, that is, with open channel lifetimes that are shorter than those of the control channels in problem 17. Possible explanations could include the converse of those in problem 17: quicker release of acetylcholine, a more rapid conformational transition from the open state to the closed state, and/or other factors.

19. The rate of indole transport is proportional to the indole concentration. This finding sug

gests that indole may diffuse freely across the cell membrane, without a need for a specific facilitated transport mechanism. By constrast, the rate of glucose transport is saturable and reaches a plateau (with no further rate increase) at high glucose concentrations. The finding for glucose is consistent with a specific protein-mediated uptake of glucose, in which glucose would bind to a specific membrane protein and then be transported across the membrane. The glucose transport rate would saturate when all of the protein-binding sites are occupied by glucose molecules. Effect of ouabain: ouabain is an inhibitor of the Na+-K+ ATPase. The inhibition of glucose transport by ouabain indicates that glucose transport requires energy and further suggests that glucose transport may be linked to the transport of Na+ or K+. </body></html>

Home