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<html><body>CHAPTER 1 The Light Reactions of Photosynthesis 9

To this point, the authors have dealt with the mechanisms by which organisms

obtain energy from their environment by oxidizing fuels to generate ATP and reducing power. In this chapter, they describe how light energy is transduced into

the same forms of chemical energy, leading to conversion of CO2 into carbohydrate by photosynthetic organisms. Carbon fixation and sugar synthesis (the Dark Reaction) will be covered in Chapter 20 of the text.

The authors begin with the basic equation of photosynthesis and an overview of

the process. Next comes a description of the chloroplast, of chlorophyll, and of the relatively simple reaction center from a photosynthetic bacterium. They then describe the overall structures, components, and reactions of photosystems II and I, and the cytochrome bf complex, including the absorption of light, charge separation, electrontransport events, and the evolution of O2. They explain how these light reactions lead to the formation of proton gradients and the synthesis of ATP and NADPH.

A review of the basic concepts of metabolism in Chapter 14 and mitochondrial

structure, redox potentials, the proton-motive force, and free-energy changes in Chapter 18 will help you to understand this chapter.

When you have mastered this chapter, you should be able to complete the fol

lowing objectives. 331 332

CHAPTER 19 LEARNING OBJECTIVES Introduction

1. Distinguish between the light and dark reactions of photosynthesis. Photosynthesis Takes Place in Chloroplasts (Text Section 19.1)

2. Describe the structure of the chloroplast. Locate the outer, inner, and thylakoid membranes;

the thylakoid space; the granum; and the stroma. Associate these structures with the functions they perform.

3. Describe the properties of the thylakoid membrane.

4. Discuss the probable origin of the chloroplast and compare it to theories of the origin of

mitochondria. Light Absorption by Chlorophyll Induces Electron Transfer (Text Section 19.2)

5. List the structural components of chlorophyll a, and explain why chlorophylls are ef

fective photoreceptors.

6. Summarize the common features of diverse photosynthetic reaction centers, including

bacterial, photosystem II, and photosystem I.

7. Distinguish between bacteriochlorophyll and chlorophyll, also bacteriopheophytin and pheo

phytin.

8. Explain the significance of the two plastoquinone binding sites, QA and QB, in the bac

terial reaction center. Two Photosystems Generate a Proton Gradient and NADPH in Oxygenic Photosynthesis (Text Section 19.3)

9. Diagram photosystem II and identify its major components. Describe the roles of P680, pheophytin, and plastoquinone in the absorption of light, separation of charge, and electron transfer in photosystem II.

10. Explain the function of the manganese center in the extraction of electrons from water.

11. Describe the composition and function of the cytochrome bf complex, and outline the roles

of plastocyanin, Cu2+, and Fe-S clusters in the formation of a transmembrane proton gradient.

12. Compare and contrast the roles of plastocyanin in chloroplasts and cytochrome c in

mitochondria.

13. Diagram photosystem I. Indicate the components and reactions of photosystem I, in

cluding the roles of P700, A0, ferredoxin, FAD, NADPH, and plastocyanin(Cu+) in these processes. A Proton Gradient Across the Thylakoid Membrane Drives ATP Synthesis (Text Section 19.4)

14. Discuss the similarities and differences between the CF1–CF0 ATP synthase of chloro

plasts and the F1–F0 synthase of mitochondria.

15. Explain how photosystem I can synthesize ATP without forming NADPH or O2.

THE LIGHT REACTIONS OF PHOTOSYNTHESIS 333

16. Contrast the formation of ATP by cyclic photophosphorylation and by oxidative phos- phorylation.

17. Write the net reaction carried out by the combined actions of photosystem II, the cy

tochrome bf complex, and photosystem I. Accessory Pigments Funnel Energy into Reaction Centers (Text Section 19.5)

18. Explain how the components of the light-harvesting complexes interact to funnel light to

the reaction centers.

19. Relate the structure and color coding of Figure 19.28 on page 543 of the text to Figure

19.30 on page 544. Explain what is missing from Figure 19.30.

20. Explain the molecular and ecological functions of phycobilisomes in cyanobacteria and red algae.

21. Rationalize the differences in the photosynthetic assemblies in the stacked and unstacked re

gions of the thylakoid membranes.

22. Explain the mechanisms of common herbicides that work by inhibiting the light reaction. The Ability to Convert Light into Chemical Energy Is Ancient (Text Section 19.6)

23. List electron donors utilized by photosynthetic bacteria. Write the overall photosynthetic

reaction when H2S is the electron donor. SELF-TEST

1. Write the basic reaction for photosynthesis in green plants. Introduction

2. Assign each function or product from the right column to the appropriate structure or

pathway in the left column.

(a) chlorophyll

(1) O2 generation

(b) light-harvesting complex

(2) ATP synthesis

(3) light collection

(d) photosystem II

(4) NADPH synthesis (5) separation of charge (6) light absorption (7) transmembrane proton gradient Photosynthesis Takes Place in Chloroplasts

3. Thylakoid membranes contain which of the following?

(a) light-harvesting complexes

(e) galactolipids

(b) reaction centers

(f)

sulfolipids

(g) phospholipids

(d) electron-transport chains

4. The similarities between mitochondria and chloroplasts are obvious. In what ways are

they opposite? 334

CHAPTER 19 Light Absorption by Chlorophyll Induces Electron Transfer

5. Which of the following are constituents of chlorophylls?

(a) substituted tetrapyrrole

(d) Fe2+

(b) plastoquinone

(e) phytol

(f)

iron porphyrin

6. Why do chlorophylls absorb and transfer visible light efficiently?

7. Carefully read the description of the L, M, and H subunits of the bacterial reaction cen

ter and subunits D1 and D2 in photosystem II (pp. 532–535). How would you mark the locations of L, M, and H on the “box” structure of Figure 19.10? Where are D1 and D2 in Figure 19.13? Note that D1 contains the “loose” plastoquinone. Two Photosystems Generate a Proton Gradient and NADPH in Oxygenic Photosynthesis

8. Which of the following statements about photosystem II are correct?

(a) It is a multimolecular transmembrane assembly containing several polypeptides,

several chlorophyll molecules, a special chlorophyll (P680), pheophytin, and plastoquinones.

(b) It transfers electrons to photosystem I via the cytochrome bf complex. (c) It uses light energy to create a separation of charge whose potential energy can be

used to oxidize H2O and to produce a reductant, plastoquinol.

(d) It uses an Fe2+-Cu+ center as a charge accumulator to form O2 without generating

potentially harmful hydroxyl radicals, superoxide anions, or H2O2.

9. Which statement about the Mn center of photosystem II is INCORRECT?

(a) The Mn center has four possible oxidation states. (b) Electrons are transferred from the Mn center to P680+. (c) A tyrosine residue on the D1 protein is an intermediate in electron transfer. (d) The O2 released by the Mn center comes from the oxidation of water. (e) Each photon absorbed by the reaction center leads to the removal of an electron

from the Mn cluster.

10. Using the diagram of photosystem II (Figure 19.1), identify the appropriate compo

nents, sites, and functions listed below. The figure here is a greatly simplified version of Figure 19.12 on page 534 of the text, and a more complex version of Figure 19.13. Note that D1 contains the “loose” plastoquinone. FIGURE 19.1

Stroma

G

H

E

A

F

D1

D2

B C

Lumen

THE LIGHT REACTIONS OF PHOTOSYNTHESIS 335

(a) QA plastoquinone site (b) QB plastoquinone site (c) chlorophyll P680 (d) Mn2+ site (e) antennae chlorophylls (f)

photon absorption

(g) extraction of electrons from H2O (h) reaction center (i)

electron transfer chain

(j)

tyrosyl radical

11. Match the photosystems of the purple sulfur bacterium or of green plants with the ap

propriate properties listed in the right column.

(a) reaction center of

(1) contains an Mn center Rhodopseudomonas viridis

(2) contains two binding sites

(b) photosystem II of green plants

for plastoquinones

(3) absorbs light of >900 nm (4) energy conserving event is separation

of charge from a chlorophyll+ to pheophytin

(5) transfers electrons from QH2

to a cytochrome

(6) special-pair chlorophyll+ reduced by

electrons from H2O

(7) special-pair chlorophyll+ reduced

through a cytochrome with four covalently attached hemes

12. Explain how plastocyanin and plastoquinol are involved in ATP synthesis.

13. Write the net equation of the reaction catalyzed by photosystem I, and describe how

NADPH is formed. What is the role of FAD in this process? A Proton Gradient Across the Thylakoid Membrane Drives ATP Synthesis

14. Describe the experiment by which Jagendorf showed that chloroplasts could synthesize

ATP in the dark when an artificial pH gradient was created across the thylakoid membrane.

15. Which of the following statements about cyclic photophosphorylation are correct?

(a) It doesn’t involve NADPH formation. (b) It uses electrons supplied by photosystem II. (c) It is activated when NADP+ is limiting. (d) It does not generate O2. (e) It leads to ATP production via the cytochrome bf complex. (f)

It involves a substrate-level phosphorylation.

16. What is the overall stoichiometry of photosynthesis in chloroplasts? If eight photons are

absorbed the net yield is _____ O2 _____ NADPH _____ ATP 336

CHAPTER 19 Accessory Pigments Funnel Energy into Reaction Centers

17. Which of the following statements about the light-harvesting complex are true?

(a) It is a chlorophyll molecule. (b) It collects light energy through the absorption of light by chlorophyll molecules. (c) It surrounds a reaction center with a specialized chlorophyll pair that contributes

to the transduction of light energy into chemical energy.

(d) It contains chlorophyll molecules that transfer energy from one to another by di

rect electromagnetic interactions.

(e) It is the product of Planck’s constant h and the frequency of the incident light n.

18. Cyanobacteria and red algae are photosynthetic algae that live more than a meter un

derwater, where little blue or red light reaches them. Explain how they can carry out photosynthesis.

19. Match the descriptions with the pigments

A.

tetrapyrrole

a.

chlorophyll a

B.

polyene

b.

b-carotene

C.

contains Mg

c.

pheophytin

d.

phycocyanobilin

20. Which of the following statements about the thylakoid membrane are correct?

(a) It contains photosystem I and ATP synthase in the unstacked regions. (b) It contains the cytochrome bf complex in the unstacked regions only. (c) It contains photosystem II mostly in the stacked regions. (d) It facilitates communication between photosystems I and II by the circulation of

plastoquinones and plastocyanins in the thylakoid space.

(e) It allows direct interaction between P680* and P700* reaction centers through its

differentiation into stacked and unstacked regions. The Ability to Convert Light into Chemical Energy Is Ancient

21. Chlorobium thiosulfatophilum uses hydrogen sulfide as a source of electrons for photo

synthesis. Write the basic equation for H2S-based photosynthesis. Given the fact that the standard reduction potential for S + 2 H+

H2S is +0.14 V, does it seem likely that

two photosystems would be required for this process? Why or why not? ANSWERS TO SELF-TEST

1. The basic reaction for photosynthesis in green plants is

H O + CO

Light

 →

 (CH O) + O

2

where (CH2O) represents carbohydrate.

2. (a) 3, 5, 6 (b) 3, 6 (c) 2, 3, 4, 5, 6, 7 (d) 1, 2, 3, 5, 6, 7. Chlorophylls are involved in

light absorption, light collection in the antennae, and reaction center chemistry. Photosystems I and II cooperate to generate a transmembrane proton-motive force that can synthesize ATP.

THE LIGHT REACTIONS OF PHOTOSYNTHESIS 337

3. a, b, c, d, e, f, g

4. Chloroplasts produce oxygen from water; mitochondria use oxygen and produce water.

The direction of the proton gradient and ATPase are reversed in the two organelles. Electrons travel only from higher to lower energy in mitochondria, but with the aid of photons, can travel “uphill” in chloroplasts. Other differences (iron in heme vs. magnesium in chlorophyll; cytochrome c vs. plastocyanin) aren’t “opposites.”

5. a, c, e

6. The polyene structure (alternating single and double bonds) of chlorophylls causes them

to have strong absorption bands in the visible region of the spectrum. Their peak molar absorption coefficients are higher than 105 cm−1 M−1. Also, while this is not emphasized in the text, iron porphyrins (heme groups) return to the ground state much more rapidly than excited magnesium tetrapyrroles. Thus chlorophyll has more time to transfer a high-energy electron before the excitation is dissipated as heat.

7. The H subunit would be outside the box, mainly underneath in Figure 19.10. To delin

eate the L and M subunits, draw a vertical line dividing the box in half. We know that steps 1 and 2 show electron transfers in the L subunit, so it is the half on the left. The M subunit, on the right, contains QB, the loosely bound quinone. Figure 19.13 is already divided into vertical halves. Notice that it is upside-down compared to Figure 19.10 because the special pair is shown at the bottom. The text doesn’t emphasize the fact, but it is known that D1 contains the exchangeable plastoquinone QB. That means that D1 is parallel to M, and is represented by the blue rectangle on the left of Figure 19.13. That means that D2, where the first electron transfers occur, is parallel to L. Note that D1 is shown in red, not blue, in Figure 19.12.

8. a, b, c. Answer (d) is incorrect because a cluster of four manganese ions serves as a charge

accumulator by interactions with the strong oxidant P680+ and H2O to form O2.

9. a. Answer (a) is incorrect. The Mn center contains four Mn atoms, and can adopt five

oxidation states (S0–S4). Each manganese ion can exist in four oxidation states.

10. (a) H (b) G (c) A (d) B (e) E, F (f) A, E, F (g) B (h) D1, D2 (i) D1, D2 (j) D1

11. (a) 2, 3, 4, 5, 7 (b) 1, 2, 4, 5, 6

12. Two electrons from plastoquinol (QH2) are transferred to two molecules plastocyanin

(PC) in a reaction catalyzed by the transmembrane cytochrome bf complex; in the process, two protons are pumped across the thylakoid membrane to acidify the thylakoid space with respect to the stroma, and two more protons are contributed by QH2 (Figure 19.18). The transmembrane proton gradient is used to synthesize ATP. This process closely resembles the mitochondrial Q cycle except that plastoquinone replaces ubiquinone (CoQ), plastocyanin replaces cytochrome c, and “inside” and “outside” are reversed.

13. The net reaction catalyzed by photosystem I is

PC(Cu+) + ferredoxinoxidized

PC(Cu2+) + ferredoxinreduced

where PC is plastocyanin. Reduced ferredoxin is a powerful reductant. Two reduced ferredoxins reduce NADP+ to form NADPH and two oxidized ferredoxins in a reaction catalyzed by ferredoxin-NADP+ reductase. FAD is a prosthetic group on the enzyme that serves as an adapter to collect two electrons from two reduced ferredoxin molecules for their subsequent transfer to a single NADP+ molecule. 338

CHAPTER 19

14. In Jagendorf’s experiment, chloroplasts were equilibrated with a buffer at pH 4 to acid

ify their thylakoid spaces. The suspension was then rapidly brought to pH 8, and ADP and Pi were added. The pH of the stroma suddenly increased to 8, whereas that of the thylakoid space remained at 4, resulting in a pH gradient across the thylakoid membrane. Jagendorf observed that ATP was synthesized as the pH gradient dissipated and that the synthesis occurred in the dark.

15. a, c, d, e. Answer (b) is incorrect because photosystem I provides the electrons for pho

tophosphorylation.

16. Eight photons would yield one O2, two NADPH, and three ATP. 17. b, c, d

18. Cyanobacteria and red algae contain protein assemblies called phycobilisomes, which ab

sorb some of the green and yellow light that reaches them to perform photosynthesis.

19. A: a,c,d; B: a,b,c,d; C: a. The “bilin” pigments are linear tetrapyrroles, structurally related

to bilirubin and biliverdin, the pigments found in bruises and largely responsible for the yellow color of jaundiced skin. See pages 688 and 689 in the text to compare these structures. It may be easier to remember the terms phycoerythrobilin and phycocyanobilin if you know that in Greek erythro means “red” and cyano means “blue.”

20. a, c, d. Answer (b) is incorrect because the cytochrome bf complex is uniformly distrib

uted throughout the thylakoid membrane. Answer (e) is incorrect because the differentiation into stacked and unstacked regions probably prevents direct interaction between the excited reaction center chlorophylls P680* and P700*.

21. The basic equation for this process is given on page 547 of the text:

CO +

2

2 H2S

(CH2O) + 2 S + H2O

Compare the standard reaction potential of S + 2 H+

H2S = +0.14 V to that of

water, given on page 495 of the text: 1/2 O +

2

2 H+

H2O = +0.82 V. Hydrogen sul

fide’s electrons are much higher in energy than those from water, so it is easier to promote them to the level of NADPH (standard reduction potential = −0.32, same as NADH). If we reverse the calculation on page 496, we see that the free energy change to move two electrons from oxygen to NADH would cost +52.6 kcal of free energy. Starting with hydrogen sulfide cuts this value roughly in half, so a single photosystem suffices. PROBLEMS

1. In Figure 19.10, the “special pair” is shown at the top of the photosystem, while in

Figures 19.13 and 19.20 the “special pair” is shown at the bottom. What does this difference imply? Why aren’t these homologous systems all shown in the same orientation?

2. NADP+ and A0 are two components in the electron-transport chain associated with pho

tosystem I (see text, Figure 19.22, p. 538). A −

0 is a chlorophyll that carries a single elec

tron, whereas NADPH carries two electrons. Write the overall reaction that occurs, and calculate DE′

−

0 and DGº′ for the reduction of NADP+ by A0

using the fact that the stan

dard reduction potential for A

= −

0 is DE′0

1.1 V. Other useful data is presented on pages

495 and 496 of the text.

THE LIGHT REACTIONS OF PHOTOSYNTHESIS 339

3. Calculate the maximum free-energy change DGº′ that occurs as a pair of electrons is

transferred from photosystem II to photosystem I, that is from P680* (excited) to P700 (unexcited). Estimate the E′0 values from Figure 19.22 on page 538 of the text. Then compare your answer with the free-energy change that occurs in mitochondria as a pair of electrons is transferred from NADH + H+ to oxygen. (See text, p. 496.)

4. Explain the defect or defects in the hypothetical scheme for the light reactions of pho

tosynthesis depicted in Figure 19.2 below. FIGURE 19.2 A hypothetical scheme for photosynthesis.

D

Y

D

:

D

1.0

D

) V

:0.5

D

Y

NADP;

D

XD

0

x potential (

Redo

;0.5

H O

2 DD

;1.0

X

5. Explain why ATP synthesis requires a larger pH gradient across the thylakoid membrane

of a chloroplast than across the inner membrane of a mitochondrion.

6. Would you expect oxygen to be evolved when NADP+ is added to an illuminated sus

pension of isolated chloroplasts? Explain briefly.

7. Would your answer to problem 6 change if the chloroplasts were illuminated with ex

tremely monochromatic light of 700 nm? Explain the basis for your answer.

8. Suppose you were designing spectrophotometric assays for chlorophyll a and chloro

phyll b. What wavelengths would you use for the detection of each? (See text, Figure 19.29, p. 544.) Explain your answer very briefly.

9. Green light has a wavelength of approximately 520 nm. Explain why solutions of chloro

phyll appear to be green. (See text, Figure 19.29, p. 544.)

10. If you were going to extract chlorophylls a and b from crushed spinach leaves, would

you prefer to use acetone or water as a solvent? Explain your answer briefly.

11. The spectrophotometric absorbance A of a species is given by the Beer-Lambert law: A = E × l × [c], where [c] is the molar concentration of the absorbing species, l is the length of the light path in centimeters, and E is the absorbance of a 1 M solution of the species in a 1-cm cell. Suppose that a mixture of chlorophylls a and b in a 1-cm cell gives

340

CHAPTER 19

an absorbance of 0.2345 at 430 nm and an absorbance of 0.161 at 455 nm. Calculate the molar concentration of each chlorophyll in the mixture. Use the values in the margin for E. (Hint: The absorbance of a mixture is equal to the sum of the absorbances of its constituents.) E430 E455

Chlorophyll a

1.21 × 105

0.037 × 105

Chlorophyll b

0.53 × 105

1.55 × 105

12. Section 19.5.5 in the text describes inhibitors of the light reaction that are used as her

bicides. Inhibiting photosynthesis is a good way to produce compounds that can kill plants while keeping toxicity toward animals to a minimum. Can you think of another way to produce herbicides that would be relatively safe for animals?

13. Why is it considered likely that the photosystems found in chloroplasts evolved from

earlier photosynthetic organisms? What is the minimum age for water-based, oxygenproducing photosynthesis? ANSWERS TO PROBLEMS

1. In all of the figures mentioned, “down” corresponds to “inside” the membrane, and

“up” is “outside.” In other words, in Rhodopseudomonas viridis the special pair of chlorophylls is near the cytoplasmic side of the membrane (inside the cell), whereas in both photosystems I and II, the special pairs are near the stroma and away from the thylakoid lumen, that is, facing “outward.” This is interesting because other structures in the thylakoid membrane are also “upside down” compared to bacterial and mitochondrial systems.

2. A −

0 is the stronger reductant in the system because it has the more negative standard re

ducing potential. A −

0 will therefore reduce NADP+ to NADPH under standard conditions.

Following the convention for redox problems presented in the text on page 496, we first write the partial reaction for the reduction involving the weaker reductant (the half-cell with the more positive standard reducing potential):

NADP+ + H+ + 2 e−

NADPH E′ = −

0

0.32 V (1)

Next, we write, again as a reduction, the partial reaction involving the stronger reductant (the half-cell with the more negative standard reducing potential):

A +

−

= −

0 e−

A0 E′0

1.1 V (2)

To get the overall reaction that occurs, we must equalize the number of electrons by multiplying equation 2 by 2. (We do not, however, multiply the half-cell potential by 2.)

2 A +

−

= −

0

2 e−

2 A0 E′0

1.1 V (3)

Then we subtract equation 3 from equation 1. This yields

NADP+ + H+ + 2 A −

= +

0

NADPH + 2 A0

DE′0

0.78 V

THE LIGHT REACTIONS OF PHOTOSYNTHESIS 341

In calculating DE′0 values, do not make the mistake of multiplying half-cell reduction potentials by factors used to equalize the number of electrons. Remember that DE′0 is a potential difference and hence, at least for our purposes, is independent of the amount of electron flow. For example, in a house with an adequate electrical power supply, the potential difference measured at the fuse box is approximately 117 V regardless of whether the house is in total darkness or all the lights are turned on.

To get the free-energy change for the overall reaction, we start with the relationship

given on page 495 of the text:

DGº′ = −nF DE′0

Substitution yields

DGº′ = −2 × 23.06 × 0.78 = −36 kcal/mol

3. In this process, electrons are transferred down an electron-transport chain from P680*

to P700. The E′0 value for P680* is approximately −0.8 V, and that for P700 is approximately 0.4 V; DE′0 is therefore +1.2 V. The free-energy change is calculated from the relationship given on page 495 of the text:

DGº′ = −nF DE′0

= −2 × 23.06 × 1.2 = −55 kcal/mol

In the mitochondrial electron-transport chain, the free-energy change as a pair of electrons is transferred from NADH + H+ to oxygen is −52.6 kcal/mol (see p. 496 of the text). In both cases the large “span” of free energy is used to drive the formation of ATP.

4. In the scheme in Figure 19.1, electrons are shown flowing “uphill” from X* to Y as

ATP is being formed. This is a thermodynamic impossibility. For electrons to flow spontaneously from X* to Y, the redox potential of X* must be more negative than that of Y. For ATP to be formed as electron transfer occurs, the free-energy change must be of sufficient magnitude to allow for ATP biosynthesis. In order to make electrons flow from X* to Y as depicted in the hypothetical scheme, ATP would be consumed, not generated.

5. The synthesis of ATP in both the chloroplast and the mitochondrion is driven by the

proton-motive force across the membrane. In mitochondria, a membrane potential of 0.14 V is established during electron transport. In chloroplasts, the light-induced potential is close to 0. Therefore, there must be a greater pH gradient in the chloroplast to give the same free-energy yield (see text, pp. 508, 540).

6. Oxygen would be evolved. NADP+ is the final electron acceptor for photosynthesis; see

the summary in Figure 19.25, page 541 of the text. Adding NADP+ will drive the process to the right.

7. Yes. Little oxygen would be evolved when 700-nm light is used. Oxygen is evolved

by photosystem II, which contains P680 and is therefore not maximally excited by 700-nm light. 342

CHAPTER 19

8. You would use 430 nm for chlorophyll a and 455 nm for chlorophyll b. These are the

wavelengths of maximum absorbance, so they would provide the most sensitive spectrophotometric assays.

9. Chlorophyll appears to be green because it has no significant absorption in the green re

gion of the spectrum and therefore transmits green light.

10. Acetone is the preferred solvent. Because of the hydrophobic porphyrin ring and the very

hydrophobic phytol tail of the chlorophylls, they are soluble in organic solvents like acetone but are insoluble in water.

11. The total absorbance at each wavelength is the sum of the absorbances due to chloro

phylls a and b separately. At 430 nm,

0.2345 = (1.21 × 105)[ch1 a] + (0.53 × 105)[ch1 b]

At 455 nm,

0.161 = (0.037 × 105)[ch1 a] + (1.55 × 105)[ch1 b]

Solving these two equations for the two unknowns yields

[ch1 a] = 1.5 × 10−6 M and [ch1 b] = 1.0 × 10−6 M

Beer’s law is discussed on page 46 of the text.

12. Higher animals tend to have limited biosynthetic abilities because their diet contains

plants and sometimes other animals. Thus there are many amino acids which animals can’t synthesize. One of the most popular herbicides in use today is glyphosate or Roundup, which inhibits the synthesis of phenylalanine. See page 679 in the text (Section 24.2.10) for more discussion. There are several herbicides that block amino acid biosynthesis. They are not very toxic to animals because they block enzymes we lack.

13. The fact that it requires two separate photosystems to promote electrons from water to

NADPH implies that the system evolved from something simpler with a single photosystem, which is why the apparatus from Rhodopseudomonas viridis is presented first in the chapter. Also, the 4-manganese center which interacts with water (Figure 19.15) is quite complex, and might relate to a 2 manganese center found in catalase. There is evidence for some O2 appearing in the atmosphere a little more than two billion years ago (see text, Figure 2.27, page 37). Water-based photosynthesis cannot be younger than this, and could be significantly older assuming the oxygen evolved was scavenged up locally. Stromatolites resembling those found today in Shark Bay, Australia, (which use oxygen-producing photosynthesis) can be found in layers dated 3.2 billion years old. One interesting theory about early photosynthesis postulates that a major source of electrons could have been the Ferrous (Fe+2) ions which were abundant in the Earth’s oceans before oxygen precipitated most of the iron as banded iron formations (Trends in Biochem. Sci. 23[1998]:94).

THE LIGHT REACTIONS OF PHOTOSYNTHESIS 343 EXPANDED SOLUTIONS TO TEXT PROBLEMS

1. DE′ = −

0

0.32 − (−0.43) = +0.11 V. DGº′ (to reduce 1 mol of NADP+) =

−2 × 23.06 × 0.11 = −5.08 kcal/mol.

2. (a) Some process for trapping energy from an external source is necessary for life as we

know it. Because radiation is an efficient mechanism for transferring the energy, it is likely that photons would be involved in bringing the energy to the places where life exists. (Alternatively, a local and stable long-term source of energy would be needed. Nevertheless, time as well as energy is a critical factor because the evolution of life is a slow process. Therefore, the hypothetical energy source would need to be reliable for a long time.)

(b) No. Electron donors other than water can be used for photosynthesis, for example,

H2, H2S, or other small organic molelcules (see Table 19.1).

3. DCMU inhibits electron transfer between Q and plastoquinone in the link between pho

tosystems II and I. O2 evolution can occur in the presence of DCMU if an artificial electron acceptor such as ferricyanide can accept electrons from Q.

4. Cyclic photophosphorylation could occur. Electrons would go from P700* to ferrodoxin to

cytochrome b6/f (generating a proton gradient for ATP synthesis), and finally to plastocyanin and back to P700. The site of DCMU inhibition (Q to plastoquinone) is outside this cycle.

5. (a) The energy of photons is inversely proportional to the wavelength. Since 600-nm

photons have an energy content of 47.6 kcal/einstein, 1000-nm light will have an energy content of 600/1000 × 47.6 kcal/einstein. Since 1 cal = 4.184 joule, 28.7 kcal × 4.184 = 120 kJ/einstein.

(b) −28.7 kcal/mol (DGº′) = −1 × 23.06 × V. Therefore, V = −28.7/−23.06 = 1.24 volts. (c) If 1000-nm photons have a free-energy content of 28.7 kcal/einstein and ATP has

a free-energy content of 12 kcal/mol, then 1000-nm photon has the free-energy content of 28.7/12, or 2.39 ATP. Therefore, a minimum of 0.42 (1/2.39) photon is needed to drive the synthesis of an ATP.

6. From Section 18.2.3, the electron transfer rate is about 1013 s−1 for groups in contact and

falls by a factor of 10 for every 1.7 Å through a protein environment. Dividing 22 Å by 1.7 Å gives an estimated decrease of about 12.94 orders of magnitude. The estimated rate is therefore 1013/1012.94 = 100.06 1 event per second. (This is much too slow for photosynthesis! Charge recombination at the chlorophyll would dominate.)

7. Phycoerythrin, the most peripheral protein in the phycobilisome.

8. From Section 18.2.3, the rate will decrease by a factor of 10 for about every 1.7 Å in

crease in the separation distance. (20 Å − 10 Å) / 1.7 Å = 5.88, so the rate will decrease by a factor of about 105.88. The time for electron transfer therefore will increase to (10 ps)*(105.88) = 106.88 ps = 7.6 * 106 ps, or about 7.6 ms.

9. The Hill reaction uses photosystem I. Electrons from P680 are excited and are replenished

by electrons from water (leading to evolution of O2). The excited electrons in P680* pass to pheophytin and then to Q and finally to the artificial acceptor such as ferricyanide. 344

CHAPTER 19

10. (a) Thioredoxin is the natural regulator in vivo.

(b) There is no effect on the control mitochondrial enzyme, but increasing the reduc

ing power increases the activity of the modified (chimeric) enzyme.

factor of approximately two. Since the DTT alone, especially at the higher concentrations, should provide sufficient reducing power, the additional enhancement with thioredoxin could be due to another effect. For example, thioredoxin could bind to the enzyme and induce a conformational change to a more active state. (In vivo— without DTT—the thioredoxin also would serve a reducing role.)

(d) Yes. The segment that was removed and replaced is responsible for the redox reg

ulation that is observed in chloroplasts but not in mitochondria.

(e) For chloroplasts, the redox potential of the stroma provides a way to link the ac

tivities of key enzymes to the level of illumination. Enzymes that do not respond to light directly are thereby able to respond to the levels of reducing agents and have their activities coordinated with the extent of ongoing photosynthesis.

(f)

The sulfhydryl groups of Cys are likely to be influenced. The Cys side chains can exist in −SH (reduced) and disulfide (−S−S−; oxidized) forms.

(g) Directed mutagenesis experiments to change selected cysteines to alanine or serine

could confirm their importance in the regulatory mechanism. CHAPTER 2 The Calvin Cycle and the Pentose Phosphate Pathway 0

Chapter 16 introduced the glycolytic and gluconeogenic pathways in which glucose

was either broken down into or synthesized from pyruvate. These pathways were in many ways mirror images of each other in which many of the same enzymes

were used in both pathways. This chapter introduces two pathways that, like the glycolytic and gluconeogenic pathways, are mirror images of each other. The Calvin cycle (sometimes referred to as the reductive pentose phosphate pathway) uses NADPH to convert carbon dioxide into hexoses, and the pentose phosphate pathway breaks down glucose into carbon dioxide to produce NADPH. The Calvin cycle constitutes the dark reactions of photosynthesis. The light reactions were discussed in Chapter 19 and transform light energy into ATP and biosynthetic reducing power, nicotinamide adenine dinucleotide phosphate (NADPH). While the dark reactions do not directly require light, they do depend on the ATP and NADPH that are produced by the light reactions. The Calvin cycle synthesizes hexoses from carbon dioxide and water in three stages: (1) fixation of CO2 by ribulose-5-phosphate to form two molecules of 3-phosphoglycerate, (2) reduction of 3-phosphoglycerate to form hexose sugars, and (3) regeneration of ribulose-5-phosphate so that more CO2 can be fixed. After a discussion of the reactions of the Calvin cycle, the authors proceed to the regulation of the cycle. Carbon dioxide assimilation by the Calvin cycle operates during the day, while carbohydrate degradation to yield energy occurs at night. The discussion of the Calvin cycle concludes with two environmentally dependent modifications to the pathway used by tropical plants and succulents to respond to high temperatures and drought.

The authors next turn their attention to the pentose phosphate pathway, which

is common to all organisms. The role of the pentose phosphate pathway is to produce NADPH, which is the currency of reducing power utilized for most reductive biosyntheses. In addition, this pathway generates ribose 5-phosphate needed for DNA synthesis and can produce various three-, four-, five-, six-, and seven-carbon sugars. 345 346

CHAPTER 20

The pathway can be separated into oxidative steps in which glucose-6-phosphate and NADP+ are converted into ribulose-5-phosphate, CO2, and NADPH; and nonoxidative steps in which ribulose-5-phosphate is converted into three 7-carbon sugars. The pentose phosphate pathway is linked to glycolysis (Chapter 16) by the common intermediates glucose 6-phosphate, fructose 6-phosphate, and glyceraldehyde 3-phosphate. The authors discuss the mechanisms of the two enzymes that catalyze the conversion of ribose-5-phosphate into glyceraldehyde 3-phosphate and fructose 6-phosphate, transketolase and transaldolase, respectively. The regulation of the pentose phosphate pathway and the ways in which its activity is coordinated with glycolysis are discussed. The chapter concludes with the role of glucose-6-phosphate dehydrogenase in protection against reactive oxygen species and the physiological consequences of deficiencies in the enzyme. LEARNING OBJECTIVES

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

1. Distinguish between the dark and light reactions of photosynthesis.

2. Explain the functions of the Calvin cycle and the pentose phosphate pathway.

3. Discuss the mirror image nature of the Calvin Cycle and the pentose phosphate pathway. The Calvin Cycle Synthesizes Hexoses from Carbon Dioxide and Water (Text Section 20.1)

4. Differentiate between heterotrophs and autotrophs.

5. Outline the three stages of the Calvin cycle.

6. Describe the formation of 3-phosphoglycerate by ribulose 1,5-bisphosphate carboxylase (ru- bisco). Note the activator and substrate roles of CO2 and the role of Mg2+ in the reaction.

7. Outline the formation of phosphoglycolate by the oxygenase reaction of rubisco, and fol

low its subsequent metabolism. Define photorespiration.

8. Outline the conversion of 3-phosphoglycerate into fructose 6-phosphate and the regen- eration of ribulose 1,5-bisphosphate.

9. Write a balanced equation for the Calvin cycle, and account for the ATP and NADPH ex

pended to form a hexose molecule.

10. Explain the formation of starch and sucrose. The Activity of the Calvin Cycle Depends on Environmental Conditions (Text Section 20.2)

11. List the four light-dependent changes in the stroma that regulate the Calvin cycle.

12. Outline the role of rubisco and thioredoxin in coordinating the light and dark reactions of

photosynthesis.

13. Describe the C4 pathway and its adaptive value to tropical plants. Explain how CO2 trans

port suppresses the oxygenase reaction of rubisco.

THE CALVIN CYCLE AND THE PENTOSE PHOSPHATE PATHWAY 347 The Pentose Phosphate Pathway Generates NADPH and Synthesizes Five-Carbon Sugars (Text Section 20.3)

14. List the two phases of the pentose phosphate pathway (oxidative generation of NADPH

and nonoxidative interconversion of sugars). List the biochemical pathways that require NADPH from the pentose phosphate pathway.

15. Describe the reactions of the oxidative branch of the pentose phosphate pathway and the

regulation of glucose 6-phosphate dehydrogenase by NADP+ levels.

16. Explain how the pentose phosphate pathway and the glycolytic pathway are linked

through reactions catalyzed by transaldolase and transketolase.

17. Outline the sugar interconversions of the nonoxidative branch of the pentose phosphate

pathway.

18. Compare the role of thiamine pyrophosphate (TPP) in transketolase with its role in pyru- vate dehydrogenase and a-ketoglutarate dehydrogenase. Outline the enzymatic mechanisms of transketolase and transaldolase. The Metabolism of Glucose-6-Phosphate by the Pentose Phosphate Pathway Is Coordinated with Glycolysis (Text Section 20.4)

19. State the different product stoichiometries obtained from the pentose phosphate path

way under conditions in which (1) more ribose 5-phosphate than NADPH is needed, (2) there is a balanced requirement for both, (3) more NADPH than ribose 5-phosphate is needed, and (4) both NADPH and ATP are required. Glucose-6-Phosphate Dehydrogenase Plays a Key Role in Protection Against Reactive Oxygen Species (Text Section 20.5)

20. Discuss the effects of glucose 6-phosphate dehydrogenase deficiency on red cells in drug

induced hemolytic anemia, and relate them to the biological roles of glutathione.

21. Discuss the reduction of glutathione by glutathione reductase. SELF-TEST The Calvin Cycle Synthesizes Hexoses from Carbon Dioxide and Water

1. Which of the following statements about ribulose 1,5-bisphosphate carboxylase (rubisco)

are correct?

(a) It is present at low concentrations in the chloroplast. (b) It is activated by the addition of CO2 to the e-amino group of a specific lysine to

form a carbamate that then binds a divalent metal cation.

the cleavage of a six-carbon diol derivative of arabinitol to form two three-carbon compounds.

(d) It catalyzes a reaction between ribulose 1,5-bisphosphate and O2 that decreases the

efficiency of photosynthesis.

(e) It catalyzes the carboxylase reaction more efficiently and the oxygenase reaction less

efficiently as the temperature increases. 348

CHAPTER 20

2. The rubisco-catalyzed reaction of O2 with ribulose 1,5-bisphosphate forms which of

the following?

(a) 3-phosphoglycerate

(d) glycolate

(b) 2-phosphoacetate

(e) glyoxylate

3. Which of the following statements about 3-phosphoglycerate (3-PG) produced in the

Calvin cycle is NOT true?

(a) It can be used to produce glucose-1-phosphate, glucose-6-phosphate, and fructose

6-phosphate.

(b) It is converted to hexose phosphates in a series of reactions that are identical to

those in the gluconeogenic pathway.

for glucose synthesis.

(d) The conversion of 3-PG into hexose phosphates produces energy and reducing

equivalents.

(e) While both glyceraldehyde 3-phosphate (GAP) and dihydroxyacetone phosphate

(DHAP) can be produced from 3-PG, only GAP can be used in further sugar producing reactions.

4. Place the following sugar conversions in the correct order used to regenerate starting ma

terial for the Calvin cycle, and name the enzyme that catalyzes each reaction:

(a) C7-ketose + C3-aldose

C5-ketose + C5-aldose

(b) C6-ketose + C3-aldose

C4-aldose + C5-ketose

C7-ketose

5. Match the two major storage forms of carbohydrates, starch and sucrose, with the ap

propriate items listed in the right column.

(a) starch (1) contains

glucose

(b) sucrose

(2) contains fructose (3) is a polymer (4) is synthesized in the cytosol (5) is synthesized in chloroplasts (6) is synthesized from UDP-glucose The Activity of the Calvin Cycle Depends on Environmental Conditions

6. Which of the following statements about the Calvin cycle are true?

(a) It regenerates the ribulose 1,5-bisphosphate consumed by the rubisco reaction. (b) It forms glyceraldehyde 3-phosphate, which can be converted to fructose 6-phosphate. (c) It requires ATP and NADPH. (d) It is exergonic because light energy absorbed by the chlorophylls is transferred

to rubisco.

(e) It consists of enzymes, several of which can be activated through reduction of disul

fide bridges by reduced thioredoxin.

(f)

It is controlled, in part, by the rate of the rubisco reaction.

(g) Its rate decreases as the level of illumination increases because both the pH and the

level of Mg2+ of the stroma decrease.

THE CALVIN CYCLE AND THE PENTOSE PHOSPHATE PATHWAY 349

7. Which of these statements about thioredoxin is correct?

(a) It contains a heme that cycles between two oxidation states. (b) Its oxidized form predominates while light absorption is taking place. (c) It activates some biosynthetic enzymes by reducing disulfide bridges. (d) It activates some degradative enzymes by reducing disulfide bridges. (e) Oxidized thioredoxin is reduced by plastoquinol.

8. Answer the following questions about the C4 pathway in tropical plants.

(a) What is the three-carbon CO2 acceptor in mesophyll cells? (b) What is the four-carbon CO2 donor in bundle-sheath cells? (c) What is the net reaction for the C4 pathway? (d) Is the C4 pathway a type of active or passive transport?

9. Both the productive carboxylase and the wasteful oxygenase reactions of rubisco require

the presence of CO2. Why then would increasing the concentration of CO2 in the chloroplasts of tropical plants via the C4 pathway shift rubisco toward the carboxylase reaction and away from the oxygenase reaction? (Hint: Look at the reactions on pages 554–555 in text.) The Pentose Phosphate Pathway Generates NADPH and Synthesizes Five-Carbon Sugars

10. Which of the following compounds is not a product of the pentose phosphate pathway?

(a) NADPH

(d) ribulose 5-phosphate

(b) glycerate 3-phosphate

(e) sedoheptulose 7-phosphate

11. Figure 20.1 shows the first four reactions of the pentose phosphate pathway. Use it to

answer the following questions.

(a) Which reactions produce NADPH? (b) Which reaction produces CO2? (c) Which compound is ribose 5-phosphate? (d) Which compound is 6-phosphoglucono d-lactone? (e) Which compound is 6-phosphogluconate? (f)

Which reaction is catalyzed by phosphopentose isomerase?

(g) Which enzyme is deficient in drug-induced hemolytic anemia? (h) Which compound can be a group acceptor in the transketolase reaction? FIGURE 20.1 Reactions of the pentose phosphate pathway.

O:

K

J

CH OPO 2:

2:

HJC

CH OPO

2

3

2

3

J

O

HJCJOH

CH OH

CHO

H

J

H

2

H

J

H

J

H

D

K

D

O

HOJCJH

D

CKO

D

HJCJOH

J

J

OH

H

J

OH

H

HO

OH B D

J F

J H

J

OH

HJCJOH

J

J

J

O H

O H

HJCJOH

HJCJOH A

J

J C

CH OPO 2:

CH OPO 2:

2

3

2

3

2

3 E G I 350

CHAPTER 20

12. Which of the following statements about glucose 6-phosphate dehydrogenase are correct?

(a) It catalyzes the committed step in the pentose phosphate pathway. (b) It is regulated by the availability of NAD+. (c) One of its products is 6-phosphogluconate. (d) It contains thiamine pyrophosphate as a cofactor. (e) It is important in the metabolism of glutathione in erythrocytes.

13. The nonoxidative branch of the pentose phosphate pathway does NOT include which

of the following reactions?

(a) Ribulose 5-P

ribose 5-P

(b) Xylulose 5-P + ribose 5-P

sedoheptulose 7-P + glyceraldehyde 3-P

sedoheptulose 7-P

(d) Sedoheptulose 7-P + glyceraldehyde 3-P

fructose 6-P + erythrose 4-P

(e) Ribulose 5-P

xylulose 5-P

14. Liver synthesizes fatty acids and lipids for export to other tissues. Would you expect

the pentose phosphate pathway to have a low or a high activity in this organ? Explain your answer.

15. Transaldolase and transketolase have which of the following similarities?

(a) Both require thiamine pyrophosphate. (b) Both form a Schiff base with the substrate. (c) Both use an aldose as a group donor. (d) Both use a ketose as a group donor. (e) Both form a covalent addition compound with the donor substrate. The Metabolism of Glucose-6-Phosphate by the Pentose Phosphate Pathway Is Coordinated with Glycolysis

16. Which of the following conversions take place in a metabolic situation that requires

much more NADPH than ribose 5-phosphate, as well as complete oxidation of glucose 6-phosphate to CO2? The arrows represent one or more enzymatic steps. (a) Glucose 6-phosphate

ribulose 5-phosphate

(b) Fructose 6-phosphate

glyceraldehyde 3-phosphate

ribose 5-phosphate

fructose 6-phosphate

glyceraldehyde 3-phosphate

(d) Glyceraldehyde 3-phosphate

pyruvate

(e) Fructose 6-phosphate

glucose 6-phosphate Glucose-6-Phosphate Dehydrogenase Plays a Key Role in Protection Against Reactive Oxygen Species

17. Which of the following statements regarding reduced glutathione is NOT true?

(a) It contains one g-carboxyglutamate, one cysteine, and one glycine residue. (b) It keeps the cysteine residues of proteins in their reduced states. (c) It is regenerated from oxidized glutathione by glutathione reductase. (d) It reacts with hydrogen peroxide and organic peroxides. (e) It is decreased relative to oxidized glutathione in glucose 6-phosphate dehydroge

nase deficiency.

18. Suggest reasons why glucose 6-phosphate dehydrogenase deficiency may be manifested

in red blood cells but not in adipocytes, which also require NADPH for their metabolism.

THE CALVIN CYCLE AND THE PENTOSE PHOSPHATE PATHWAY 351 ANSWERS TO SELF-TEST

1. b, c, d. Answer (e) is incorrect because the rate of the oxygenase reaction increases relative

to that of the carboxylase reaction as the temperature increases; the altered ratio of the two reaction rates decreases the efficiency of photosynthesis as the temperature increases.

2. a, c

3. The incorrect statements are b, d, and e. Statement (b) is incorrect because the gluco

neogenic pathway uses NADH and not NADPH; (d) is incorrect because the conversion requires both ATP and NADPH, and e is incorrect because both GAP and DHAP can be used to produce larger sugars.

4. The correct order is b, c, a. Transketolase catalyzes reactions (a) and (b) while aldolase

catalyzes reaction (c).

5. (a) 1, 3, 5 (b) 1, 2, 4, 6

6. a, b, c, e, f

7. c. Thioredoxin contains cysteine residues that cycle between two oxidation states. It is

reduced by ferredoxin while the light reactions are proceeding. It activates biosynthetic enzymes and inhibits degradative enzymes by reducing their disulfide bridges.

8. (a) Phosphoenolpyruvate is the three- carbon CO2 acceptor in mesophyll cells.

(b) Malate is the four-carbon CO2 donor in bundle-sheath cells. (c) CO2 (in mesophyll cell) + ATP + H2O

CO2 (in bundle-sheath cell)

+ AMP + 2P +

i

H+

(d) It is a type of active transport because it requires ATP to function.

9. Although both reactions require CO2 to form a lysine carbamate on rubisco, the car

boxylase reaction requires an additional CO2 to proceed. Increasing the concentration of CO2 would therefore be expected to accelerate the carboxylase reaction while not affecting the oxygenase pathway.

10. b

11. (a) B, F (b) F (c) I (d) C (e) E (f) H (g) B (h) I

12. a, e

13. c

14. The activity of the pentose phosphate pathway in the liver is high. The biosynthesis of

fatty acids and lipids requires reducing equivalents in the form of NADPH. In all organs that carry out reductive biosyntheses, the pentose phosphate pathway supplies a large proportion of the required NADPH.

15. d, e

16. a, c, e. Glucose 6-phosphate is converted to ribulose 5-phosphate, producing CO2 and

NADPH in the process. Then ribulose 5-phosphate, via ribose 5-phosphate, is transformed into fructose 6-phosphate and glyceraldehyde 3-phosphate. These two glycolytic intermediates are converted back to glucose 6-phosphate, and the cycle is repeated until the equivalent of six carbon atoms from glucose 6-phosphate are converted to CO2.

17. a. Answer (a) is incorrect because the glutamate residue in glutathione is not g-car

boxyglutamate; rather, the glutamate in glutathione forms a peptide bond with the adjacent cysteine residue via its g-carboxyl group.

18. The glucose 6-phosphate dehydrogenase in erythrocytes and that in adipocytes are spec

ified by distinct genes; they have the same function but different structures—that is, they

352

CHAPTER 20

are isozymes. Furthermore, NADPH synthesis by the pentose phosphate pathway may not be as critical in the cells of other tissues as it is in erythrocytes because other tissues have other sources of NADPH. PROBLEMS

1. Outline the synthesis of fructose 6-phosphate from 3-phosphoglycerate.

2. How many moles of ATP and NADPH are required to convert 6 moles of CO2 to fruc

tose 6-phosphate?

3. Describe photorespiration, and explain why it decreases the efficiency of photosynthesis.

4. It is said that the C4 pathway increases the efficiency of photosynthesis. What is the jus

tification for this statement when more than 1.6 times as much ATP is required to convert 6 moles of CO2 to a hexose when this pathway is used in contrast with the pathway used by plants lacking the C4 apparatus? Account for the extra ATP molecules used in the C4 pathway.

5. In addition to the well-understood ferredoxin-thioredoxin couple, NADPH can regulate

Calvin cycle enzymes. The text gives the example of a recently discovered assembly protein CP12 (p. 561) which binds to and inhibits phosphoribulose kinase (PRK) and glyceraldehyde 3-phosphate dehydrogenase (GAPDH) in the dark and releases them in the light (Wedel et al. PNAS 94[1997]:10479–10484). The authors in the PNAS paper found that NADPH triggers the release of PRK and GADPH from CP12 and is also necessary for PRK activity after its release. They also noted that PRK is rapidly oxidized in the absence of reduced thioredoxin, while it remains reduced when bound to CP12.

(a) Why would PRK require NADPH for full activity given that is does not catalyze a

reduction reaction?

(b) Given the above information, what is a possible role of PRK binding to CP12?

6. The conversion of glucose 6-phosphate to ribose 5-phosphate via the enzymes of the

pentose phosphate pathway and glycolysis can be summarized as follows:

5 Glucose 6-phosphate + ATP

6 ribose 5-phosphate + ADP + H+

Which enzyme uses the molecule of ATP shown in the equation?

7. Liver and other organ tissues contain relatively large quantities of nucleic acids. During di

gestion, nucleases hydrolyze RNA and DNA, and among the products is ribose 5-phosphate.

(a) How can this molecule be used as a metabolic fuel? (b) Another product formed by the degradation of nucleic acids is 2-deoxyribose 5

phosphate. Can this molecule be converted to glycolytic intermediates through the action of the pentose phosphate pathway? Explain your answer.

8. You have glucose that is radioactively labeled with 14C at C-1, and you have an extract

that contains the enzymes that catalyze the reactions of the glycolytic and the pentose phosphate pathways, along with all the intermediates of the pathways.

(a) If the enzymes of the oxidative branch of the pentose phosphate pathway are not ac

tive in your extract, is it possible to obtain labeled sedoheptulose 7-phosphate using glucose labeled with 14C at C-1? Explain.

THE CALVIN CYCLE AND THE PENTOSE PHOSPHATE PATHWAY 353

(b) Suppose that in a second experiment all the enzymes of both the oxidative branch

and the nonoxidative branch of the pentose phosphate pathway are active. Will the labeling pattern of sedoheptulose 7-phosphate be different? Explain.

9. Why is the pentose phosphate pathway more active in cells that are dividing than in cells

that are not?

10. A bacterium isolated from a soil culture can utilize ribose as a sole source of carbon when

grown anaerobically. Experiments show that in the anaerobic pathways leading to ATP production, three molecules of ribose are converted to five molecules of CO2 and five molecules of ethanol. These organisms also use ribose for the production of NADPH. The assimilation of ribose begins with its conversion to ribose 5-phosphate, with ATP serving as a phosphoryl donor.

(a) Explain how ribose can be converted to CO2 and ethanol under anaerobic condi

tions. Write the overall reaction, showing how much ATP can be produced per pentose utilized.

(b) Write an equation for the generation of NADPH using ribose as a sole source of

carbon.

11. Mature erythrocytes, which lack mitochondria, metabolize glucose at a high rate. In re

sponse to the increased availability of glucose, erythrocytes generate lactate and also evolve carbon dioxide.

(a) Why is generation of lactate necessary to ensure the continued utilization of glucose? (b) In erythrocytes, what pathway is likely to be used for the generation of carbon diox

ide from glucose? Can glucose be completely oxidized to CO2 in erythrocytes? Explain.

12. A biochemist needs to determine whether a particular tissue homogenate has a high level

of pentose phosphate pathway activity. She incubates one sample with 14C-1 glucose, and another with 14C-6 glucose. Then she measures the specific activity of radioactive CO2 generated by each sample. Her measurements show that the specific activity of CO2 from the experiment using glucose labeled at C-1 is much higher than that from the sample in which glucose labeled at C-6 was employed. What is her conclusion?

13. Even if glucose 6-phosphate dehydrogenase is deficient, the synthesis of ribose 5-phos

phate from glucose 6-phosphate can proceed normally. Explain how this is possible. ANSWERS TO PROBLEMS

1. Phosphoglycerate kinase converts 3-phosphoglycerate, the initial product of photosyn

thesis, to the glycolytic intermediate 1,3-bisphosphoglycerate, which is then converted to glyceraldehyde 3-phosphate (G-3-P) by an NADPH-dependent G-3-P dehydrogenase in the chloroplast. Triosephosphate isomerase converts G-3-P to dihydroxyacetone phosphate, which aldolase can condense with another G-3-P to form fructose 1,6-bisphosphate. The phosphate ester at C-1 is hydrolyzed to give fructose 6-phosphate. The result of this pathway, which is functionally equivalent to the gluconeogenic pathway, is the conversion of the CO2 fixed by photosynthesis into a hexose.

2. Eighteen moles of ATP and twelve moles of NADPH are required to fix six moles of

CO2. Two moles of ATP are used by phosphoglycerate kinase to form two moles of

354

CHAPTER 20

1,3-bisphosphoglycerate, and one mole of ATP is used by ribulose 5-phosphate kinase to form one mole of ribulose 1,5-bisphosphate per mole of CO2 fixed. Two moles of NADPH are used by G-3-P dehydrogenase to form two moles of G-3-P per mole of CO2 incorporated. Therefore, three moles of ATP and two moles of NADPH are used for each mole of CO2 fixed.

3. The oxygenase reaction of rubisco and the salvage reactions that convert two resulting

phosphoglycolate molecules into serine are called photorespiration because CO2 is released and O2 is consumed in the process. Unlike genuine respiration, no ATP or NADPH is produced by photorespiration. Ordinarily, no CO2 is released during photosynthesis, and all the fixed CO2 can be used to form hexoses. During photorespiration, no CO2 is fixed, and the products into which ribulose 1,5-bisphosphate is converted by the oxygenase reaction of rubisco cannot be completely recycled into carbohydrate because of the loss of CO2 in the phosphoglycolate salvage reactions.

4. Plants lacking the C4 pathway cannot compensate for the relative increase in the rate of

the oxygenase reaction of rubisco with respect to the rate of the carboxylase reaction that occurs as the temperature rises. Plants with the C4 pathway increase the concentration of CO2 in the bundle-sheath cell, where the Calvin cycle occurs, thereby increasing the ability of CO2 to compete with O2 as a substrate for rubisco. As a result, more CO2 is fixed and less ribulose 1,5-bisphosphate is degraded into phosphoglycolate, which cannot be efficiently converted into carbohydrate. Thus, the Calvin cycle functions more efficiently in these specialized plants under conditions of high illumination and at higher temperatures than it would otherwise.

The concentration of CO2 is increased by an expenditure of ATP. The collection of one

CO2 molecule and its transport on C4 compounds from the mesophyll cell into the bundlesheath cell is brought about by the conversion of one ATP to AMP and PPi in a reaction in which pyruvate is phosphorylated to PEP. The PPi is hydrolyzed, and two ATP are required to resynthesize ATP from AMP. Thus, an extra ATP/CO ×

2

6 CO2/hexose = 12 ATP/hexose

are used by the C4 pathway.

5. (a) Since the purpose of PRK is to regenerate ribulose 1,5-bisphosphate for use in the

Calvin cycle, it does not make sense to have it active when there is not enough NADPH to run the cycle. The PRK reaction requires ATP and would be wasteful if ribulose 1,5-bisphosphate were not needed.

(b) In the absence of CP12 complex formation, PRK is rapidly oxidized and becomes

inactive. In conditions of low NADPH, if complex formation did not occur, PRK would reoxidze and become inactive before producing ribulose 1,5-bisphosphate. The light energy used to reduce thioredoxin would therefore be wasted. By keeping thioredoxin-reduced PRK bound to CP12 until enough NADPH is present, the light energy is not wasted.

6. Phosphofructokinase uses ATP to convert fructose 6-phosphate to fructose 1,6-bisphosphate,

which is then cleaved by aldolase to yield dihydroxyacetone phosphate (DHAP) and glyceraldehyde 3-phosphate. The conversion of DHAP to a second molecule of glyceraldehyde 3-phosphate provides the molecules that are needed for the synthesis of ribose 5-phosphate.

7. (a) The most direct route for the oxidative degradation of ribose 5-phosphate is its con

version to glycolytic intermediates by the nonoxidative enzymes of the pentose phosphate pathway. The overall reaction is

3 ribose 5-phosphate

2 fructose 6-phosphate + glyceraldehyde 3-phosphate

(b) The formation of glycolytic intermediates from 2-deoxyribose 5-phosphate is not

possible, because unlike ribose 5-phosphate, 2-deoxyribose 5-phosphate lacks a hydroxyl group at C-2. It is therefore not a substrate for phosphopentose isomerase, whose action is required to convert ketopentose phosphates to substrates that can be utilized by other enzymes of the pentose phosphate pathway. Most deoxyribose phosphate molecules are used in salvage pathways to form deoxynucleotides.

8. (a) Yes. The most direct route would be the conversion of glucose to fructose 6-phos

phate, followed by the condensation of fructose 6-phosphate with erythrose 4-phosphate to form sedoheptulose 7-phosphate and glyceraldehyde 3-phosphate. The labeled carbon of glucose becomes the C-1 of fructose 6-phosphate and C-1 of sedoheptulose 7-phosphate.

(b) The labeling pattern will be the same, although the amount of labeled carbon in

corporated into the heptose will be reduced. In the oxidative branch of the pentose phosphate pathway, the labeled glucose is converted to glucose 6-phosphate with the 14C label on C-1. Glucose 6-phosphate then undergoes successive oxidations and decarboxylation to form ribulose 5-phosphate. The label is lost when the C-1 carbon is removed during decarboxylation.

hemiketal linkage. The most likely link would be between the keto group at C-2 and the hydroxyl group at C-6.

9. Cells have a high rate of nucleic acid biosynthesis when they grow and divide. Among

the precursors needed is ribose 5-phosphate, which is synthesized through the action of the enzymes of the glycolytic and the pentose phosphate pathways. Biosynthetic reactions requiring NADPH occur at a high rate in growing and dividing cells. For these reasons, the enzymes of the pentose phosphate pathway will be extremely active in dividing cells.

10. (a) To generate ATP, ethanol, and CO2, ribose must first be converted to ribose 5-phos

phate, with ATP serving as a phosphate donor. Then, in the nonoxidative branch of the pentose phosphate pathway, three molecules of ribose 5-phosphate are converted to two molecules of fructose 6-phosphate and one molecule of glyceraldehyde 3-phosphate. Two molecules of ATP are required for the production of fructose 1,6-bisphosphate from fructose 6-phosphate. The formation of a total of five molecules of glyceraldehyde 3-phosphate is achieved through the action of aldolase and triose phosphate isomerase. These five molecules are converted to five molecules of pyruvate, yielding ten ATP molecules and five NADH molecules. To keep the anaerobic cell in redox balance, the pyruvate molecules are converted to five molecules of ethanol, with the production of five CO2 molecules and five NAD+. The overall reaction is 3 ribose + 5 ADP + P

+

5 ethanol + 5 CO2

5 ADP.

(b) Ribose 5-phosphate molecules must first be converted to glucose 6-phosphate for

the oxidative enzymes of the pentose pathway to generate NADPH. The stoichiometry of the reactions is

6 Ribose 5-phosphate

4 fructose 6-phosphate + 2 glyceraldehyde

3-phosphate

4 Fructose 6-phosphate

4 glucose 6-phosphate

2 Glyceraldehyde 3-phosphate

glucose 6-phosphate + Pi

5 Glucose 6-phosphate + 10 NADP+ + 5 H2O

5 ribose 5-phosphate

+ 10 NADPH + 10 H+ + 5 CO2

11. (a) Because erythrocytes lack mitochondria, they cannot use the citric acid cycle to re

generate the NAD+ needed to sustain glycolysis. Instead, they regenerate NAD+ by reducing pyruvate through the action of lactate dehydrogenase; NAD+ is then reduced in the reaction catalyzed by glyceraldehyde 3-phosphate dehydrogenase during glycolysis. Failure to oxidize the NADH generated in the glycolytic pathway will cause a reduction in the rate of glucose breakdown.

(b) In erythrocytes, the pentose phosphate pathway is the only route available to yield

CO2 from glucose. Glucose can be completely oxidized by first entering the oxidative branch of the pathway, generating NADPH and ribose 5-phosphate. Transaldolase and transketolase then convert the pentose phosphates to fructose 6phosphate and glyceraldehyde 3-phosphate. Part of the gluconeogenic pathway is used to convert both the products to glucose 6-phosphate. The net reaction is

Glucose 6-P + 12 NADP+ + 7 H

+

2O

6 CO2

12 NADPH + 12 H+ + Pi

12. The experiments show that the activity of the pentose phosphate pathway is high. In the

pentose phosphate pathway, glucose labeled at C-1 is decarboxylated, while glucose labeled at C-6 is not. On the other hand, both C-1- and C-6-labeled glucose are decarboxylated to the same extent by the combined action of the glycolytic pathway and the citric acid cycle. Because in these experiments, the specific activity (ratio of labeled CO2 to total CO2) is higher for C-1-labeled glucose, much of the glucose in the experiment must be moving through the pentose phosphate pathway.

13. Ribose 5-phosphate can be synthesized from fructose 6-phosphate and glyceraldehyde

3-phosphate, both of which are glycolytic products of glucose 6-phosphate. These reactions are carried out by transketolase and transaldolase in a reversal of the nonoxidative branch of the pentose phosphate pathway and do not involve glucose 6-phosphate dehydrogenase. EXPANDED SOLUTIONS TO TEXT PROBLEMS

1. Aldolase participates in the Calvin cycle, whereas transaldolase participates in the pen

tose phosphate pathway.

2. The conversion of ribulose 1,5-bisphosphate to 3-phosphoglycerate does not require

ATP, so it will continue until the ribulose 1,5-bisphosphate is largely depleted.

3. When the concentration of CO2 is drastically decreased, the rate of conversion of ribu

lose 1,5-bisphosphate to 3-phosphoglycerate will greatly decrease, whereas the rate of utilization of 3-phosphoglycerate will not be diminished.

THE CALVIN CYCLE AND THE PENTOSE PHOSPHATE PATHWAY 357

4. (a)

O

CH OPJO:

2

O:

HJCJCOO:

J

HJCJOH

J

HJCJOH

O

J

CH OPJO:

2

O: 2-Carboxyarabinitol 1,5 bisphosphate (CABP)

(b) CABP resembles the addition compound formed in the reaction of CO2 and ribu

lose 1,5-bisphosphate.

5. Glyoxalate + glutamate

glycine + a-ketoglutarate

O

K

NH ;

J

3

J

D

:

H

COO: +

OOC

H COO:

NH ;

J

O

K

3

J

H

:

H

COO: +

OOC

COO:

6. Two high-energy bonds from ATP are used by pyruvate-Pi dikinase in forming phos

phoenolpyruvate and the side-product pyrophosphate (which subsequently hydrolyzes to 2 Pi ). Therefore, two ATP equivalents are used. The phosphoenolpyruvate contains sufficient energy to drive its carboxylation to oxaloacetate.

7. The crabgrass adapts better to the hot and dry conditions. One could speculate that

crabgrass may close the stomata of their leaves during the day and use CO2 that has been stored as malate in vacuoles the previous night (Crassulacean acid metabolism, Section 20.2.4).

8. C4 plants have the advantage in hotter environments and so may become more promi

nent at higher latitudes as well as lower latitudes, under the influence of global warming.

9. Since the C-1 of glucose is lost during the conversion to pentose, carbon atoms 2 through

6 of glucose become carbon atoms 1 through 5 of the pentose. That is, each pentose carbon is numerically 1 less than its counterpart in glucose.

10. Note that in the oxidative decarboxylation of 6-phosphogluconate, oxidation occurs at

the carbon b to the carboxyl group. A similar b-oxidation occurs during the decarboxylation of isocitrate in the citric acid cycle. In both cases a b-keto acid intermediate is formed. Since b-keto acids are relatively unstable, they are easily decarboxylated. 358

CHAPTER 20

11. Ribose 5-P is first converted to xylulose 5-P (labeled in C-1) via ribulose 5-P.

Transketolase can then catalyze the conversion of ribose 5-P + xylulose 5-P to sedoheptulose 7-P (labeled in C-1 and C-3) and glyceraldehyde 3-P. Transaldolase then transfers carbons 1 through 3 of sedoheptulose 7-P to glyceraldehyde 3-P, forming erythrose 4-P, which is unlabeled (from carbons 4 through 7 of sedoheptulose), and fructose 6-P, which is labeled in C-1 and C-3 (from C-1 and C-3 of sedoheptulose).

12. (a) To make six pentoses, four glucose 6-phosphates must be converted to fructose 6

phosphate (no ATP required), and one glucose 6-phosphate must be converted to two molecules of glyceraldehyde 3-phosphate (this requires one ATP). These are converted to pentoses by the following reactions (Table 20.3 in the text). 2 Fructose 6-phosphate + 2 glyceraldehyde 3-phosphate

2 erythrose

4-phosphate + 2 xylulose 5-phosphate 2 Fructose 6-phosphate + 2 erythrose 4-phosphate

2 glyceraldehyde

3-phosphate + 2 sedoheptulose 7-phosphate

2 Glyceraldehyde 3-phosphate + 2 sedoheptulose 7-phosphate 2 xylulose 5-phosphate + 2 ribose 5-phosphate

(b) What really happens is that six molecules of glucose 6-phosphate are converted to

6 CO +

2

6 ribulose 5-phosphates + 12 NADPH + 12 H+ (see Table 20.3 in the text).

The ribulose phosphates are then converted back to five molecules of glucose 6phosphate by the action of transketolase and transaldolase. By these reactions three pentoses are converted to two hexoses and one triose. Thus six pentoses can be converted to four hexoses plus two trioses, and the latter can be converted to the fifth hexose.

13. The double bond in Schiff bases can be reduced to stable molecules with sodium boro

hydride (NaBH4). Since transaldolase forms a Schiff base with a ketose substrate, this enzyme-substrate complex can be reduced with tritiated NaBH4 to yield a stable radioactive derivative of the active-site lysine. Fingerprinting the labeled enzyme will identify the lysine at the active site.

14. The DE′0 for the reduction of glutathione by NADPH is + 0.09 V. Then

DGº′ = −nFDE′ = −

=

0

2 × 23.06 × 0.09 = −4.15 kcal/mol. Also, K′eq

10−D Gº′/1.36

= 104.15/1.36 = 1.126 × 103. Thus,

GSH 2 NADP+

[

] [

] K

=

eq

1126

GSSG

[

][NADPH]

After substituting the given concentrations for GSH and GSSG,

[ .

0 01 ]

M [

2 NADP+] = .1126 × 103

[ .

0 001 ][

M

]

NADPH

Therefore,

[NADP+ ] = .1126 × 104

[

]

NADPH

THE CALVIN CYCLE AND THE PENTOSE PHOSPHATE PATHWAY 359

and

[

]

NADPH

.

8 9

105

=

↔

[NADP+ ]

.

1 126 ↔ 104

1

Remember, in equilibrium constants the molar concentrations of the reactants are raised to a power equal to the number of moles taking part in the reaction. Therefore, in this problem the [GSH] is squared because, for each mole of GSSG, NADP+, and NADPH, two moles of GSH are involved.

15. In similar fashion to the traditional mechanism, the enolate form of dihydroxyacetone

phosphate could be used. A metal ion instead of a protonated Schiff base could stabilize an enolate anion intermediate. The enolate anion could then add to the aldehyde of glyceraldehyde-3 phosphate:

CH OPO 2:

2

3

KOH

CH OPO 2:

CH OPO 2:

2

3

2

3

HOJJJH

KO

D

KO

D

J

HJJJOH

HJCJOH

J

:

HJJJ OH

H

M;

J

CH OPO 2:

2

3

D

H

O

J

K

HJJJOH

J

CH OPO 2:

2

3

16. The reaction is similar to the hexose phosphate isomerase and triose phosphate isomerase

reactions of glycolysis and probably proceeds through an enediol intermediate:

H

O

H

O

H

O

HJJJOH

JOH

KOH

HJJJOH

G

HJJJOH

G

HJJJOH

HJJJ OH

HJJJ OH

J

CH OPO :2

J

CH OPO :2

J

CH OPO :2

2

3

2

3

2

3 Aldose Enediol Ketose

17. Labels at C-1 and C-6 of glucose will behave identically in glycolysis (both emerging at

C-3 of pyruvate) and the citric acid cycle. Both labels will transfer to acetyl-CoA (methyl group) and will remain in the citric acid cycle for two rounds. Only with the third turn of the cycle will the C-1 and C-6 labels from glucose finally begin to be released as CO2 (50% of remaining C-1 and C-6 during the third and each subsequent turn). It is important to note that none of the C-1 or C-6 label will be released in the early stages of glycolysis or the citric acid cycle. By contrast, in the pentose phosphate pathway, all of the C-1 label (and none of the C-6 label) will be released very quickly as CO2 at the step where ribulose-5-phosphate is formed. We can put all of these facts together to propose our experiment: incubate a portion each tissue with each labeled glucose sample, and

360

CHAPTER 20

measure the specific activity of CO2 that is released as a function of time in each experiment. The extent by which release of C-1 label precedes the release of C-6 label will reflect the level of activity of the pentose phosphate pathway. If both labels are released at the same rate by a particular tissue, then the dominant pathway follows glycolysis and the citric acid cycle.

18. From the stoichiometry of the Calvin cycle, two moles of NADPH are needed for every

mole of CO2 that is incorporated into glucose:

6 CO +

2

18 ATP + 12 NADPH + 12 H2O

C

+

6H12O6

18 ADP + 18 Pi

12 NADP+ + 6 H+

The production of each molecule of NADPH requires illumination from two photons (to activate photosystems I and II, which also produce the necessary ATP). Therefore, the energy of four photons is needed for every CO2 that is reduced to the level of hexose. The efficiency is: (477 kJ)/(4 * 199 kJ) = 60%. Photosynthesis is remarkably efficient!

19. (a) The C4 plant is more efficient at higher temperature. Therefore, the curve (on the

right) that peaks sharply at about 39ºC represents the C4 plant.

(b) 1. The oxygenase activity of rubisco increases with temperature. Other key enzymes

may become inactive at high temperature.

sheath cells.

(d) The C3 activity depends on passive diffusion of CO2, whereas the C4 activity de

pends on the active transport of CO2 into the bundle-sheath cells. Once the transport system is saturated (working at maximum rate), then no further increase in photosynthetic activity is possible. By contrast, higher CO2 concentrations continue to enhance the rate of diffusion and cause increased availability of CO2 for the C3 plants. CHAPTER 2 Glycogen Metabolism 1

The topic of carbohydrate metabolism presented in Chapters 16 and 20 is further

developed in this chapter with a detailed discussion of the metabolism of glycogen, the intracellular storage form of glucose. Glycogen is important in the me

tabolism of higher animals because its glucose residues can be easily mobilized by the liver to maintain blood glucose levels and by muscle to satisfy the energy needs during bursts of contraction. The text first reviews the structure and the physiologic roles of glycogen and provides an overview of its metabolism. You were introduced briefly to the structure of glycogen, a polymer of glucose, in Section 11.2.2 of Chapter 11. Next, with glucose as the ending and starting points, the text presents the enzymatic reactions of glycogen degradation and synthesis. This is followed by a discussion of the control of these reactions by allosteric mechanisms and the phosphorylation and dephosphorylation of the key enzymes in response to hormonal signals. AMP, ATP, glucose, and glucose 6-phosphate act as allosteric effectors; and the hormones insulin, glucagon, and epinephrine function as signals in transduction pathways that control critical enzyme phosphorylations and dephosphorylations. The text describes relevant structures and control mechanisms for phosphorylase, phosphorylase kinase, glycogen synthase, the branching enzyme, and protein phosphatase 1. The differences in glycogen metabolism in muscle and liver are related to the distinct physiologic functions these tissues perform. The text concludes the chapter with a discussion of the biochemical bases of several glycogen storage diseases. 361 362

CHAPTER 21 LEARNING OBJECTIVES

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

1. Describe the structure of glycogen and its roles in the liver and muscle. Distinguish be

tween a-1,4 glycosidic linkages and their a-1,6 glycosidic isomers.

2. List the properties of glycogen granules and their location within cells.

3. Describe the three steps of glycogen catabolism and the three fates of its product, glucose 6-phosphate.

4. Describe the precursors of glycogen anabolism.

5. Explain the role of hormones in regulating glycogen metabolism. Glycogen Breakdown Requires the Interplay of Several Enzymes (Text Section 21.1)

6. Write the reaction catalyzed by glycogen phosphorylase.

7. Explain the advantage of the phosphorolytic cleavage of glycogen over its hydrolytic cleavage.

8. Outline the steps in the degradation of glycogen, and relate them to the action of phos

phorylase, transferase, and a-1,6-glucosidase, which is also known as the debranching enzyme. Explain why the glycogen molecule must be remodeled during its degradation.

9. Compare the reaction mechanisms of phosphoglucomutase and phosphoglyceromutase and

describe their common mechanistic features.

10. Explain the importance of glucose 6-phosphatase in the release of glucose by the liver. Note

the absence of this enzyme in the brain and muscle.

11. Describe the roles of pyridoxal 5a-phosphate, general acid-base catalysis, Schiff-base formation

and the carbonium ion intermediate in the mechanism of action of glycogen phosphorylase.

12. Define processivity as it relates to glycogen phosphorylase activity. Phosphorylase Is Regulated by Allosteric Interactions and Reversible Phosphorylation (Text Section 21.2)

13. Appreciate that the two primary mechanisms of glycogen phosphorylase are allosteric ef- fectors and reversible covalent modifications.

14. Describe the phosphorylation of phosphorylase by phosphorylase kinase and its dephos

phorylation by protein phosphatase 1 (PP1).

15. Explain the relationships between phosphorylase a and phosphorylase b, the T (tense) and R (relaxed) forms of each, and the allosteric effectors that mediate their interconversions in skeletal muscle. Outline the molecular bases for the relative inactivities of the T states.

16. Contrast the regulation of liver phosphorylase and muscle phosphorylase.

17. Contrast the important structural features of phosphorylase a and phosphorylase b. Note

the variety of binding sites, their functional roles, and the critical location of the phosphorylation and AMP binding sites near the subunit interface.

GLYCOGEN METABOLISM 363

18. Describe the major compositional features of phosphorylase kinase and its activation

by protein kinase A (PKA). Explain the effects of calmodulin and Ca2+ on glycogen metabolism in muscle and liver. Epinephrine and Glucagon Signal the Need for Glycogen Breakdown (Text Section 21.3)

19. Compare the effects of glucagon and epinephrine on glycogen metabolism in liver and in

muscle.

20. List the sequence of events from the binding of hormones by their receptors to the phos

phorylation of glycogen synthase and phosphorylase. Explain the roles of cAMP and PKA in these processes.

21. Explain the role of protein phosphatase 1 (PP1) in the control of the activities of glycogen

phosphorylase and synthase. Glycogen Is Synthesized and Degraded by Different Pathways (Text Section 21.4)

22. Explain the roles of UDP-glucose and inorganic pyrophosphatase in the synthesis of

glycogen.

23. Outline the steps in the synthesis of glycogen, name the pertinent enzymes, and note

the requirement for a primer. Describe the actions of glycogenin and its effect on glycogen synthase.

24. Explain why glycogen per se lacks glucose residues that can be reduced.

25. Explain the functional importance of branching in the glycogen molecule.

26. Discuss the efficiency of glycogen as a storage form of glucose. Glycogen Breakdown and Synthesis Are Reciprocally Regulated (Text Section 21.5)

27. Distinguish between the active and inactive forms of glycogen synthase in terms of their

states of phosphorylation and contrast the effects of phosphorylation on glycogen synthase and glycogen phosphorylase. Appreciate the reciprocal regulation strategies employed and the consequences of amplification cascades.

28. Describe the events that lead to the inactivation of phosphorylase and the activation

of glycogen synthase by glucose in the liver. Note the role of phosphorylase a as the glucose sensor in liver cells and the participation of phosphorylase a and PP1 in glucose sensing.

29. Outline the effects of insulin on glycogen. Rationalize the existence of distinct pathways

for the biosynthesis and degradation of glycogen.

30. Provide examples of glycogen storage diseases, and relate the biochemical defects with

the clinical observations. Use the disease discovered by von Gierke to show how a deficiency in any of several different enzymes can cause the same disease. 364

CHAPTER 21 SELF-TEST Introduction

1. Answer the following questions about the glycogen fragment in Figure 21.1. FIGURE 21.1 Fragment of glycogen. (R represents the rest of the glycogen molecule.)

HOCH2

O

H

J

H

H

OH

H

O

HO

J

CH2

O H

G

HOCH

2

O

H

J

H

J

H

J

H

J

H

J

H

HJ

H

J

OH

H

O

OH

H

O

OH

H

O

OH

H

O

OH

H

O

OH

H

R

HO

J

O H

O H

O H

O H

O H

O H

A B C D E F

(a) Which residues are at nonreducing ends? (b) An a-1,6 glycosidic linkage occurs between which residues? (c) An a-1,4 glycosidic linkage occurs between which residues? (d) Is the glycogen fragment a substrate for phosphorylase a? Explain. (e) Is the glycogen fragment a substrate for the debranching enzyme? Explain. (f)

Is the glycogen fragment a substrate for the branching enzyme? Explain.

2. Which of the following statements about glycogen storage are INCORRECT?

(a) Glycogen is stored in muscles and liver. (b) Glycogen is a major source of stored energy in the brain. (c) Glycogen reserves are less rapidly depleted than fat reserves during starvation. (d) Glycogen nearly fills the nucleus of cells that specialize in glycogen storage. (e) Glycogen storage occurs in the form of dense granules in the cytoplasm of cells.

3. Is the largest total mass of glycogen found in the liver or muscle? Glycogen Breakdown Requires the Interplay of Several Enzymes

4. Explain why the phosphorolytic cleavage of glycogen is more energetically advantageous

than its hydrolytic cleavage.

5. Which of the following statements about the role of pyridoxal phosphate in the mecha

nism of action of phosphorylase are correct?

(a) It interacts with orthophosphate. (b) It acts as a general acid-base catalyst.

GLYCOGEN METABOLISM 365

(c) It orients the glycogen substrate in the active site. (d) It donates a proton directly to the O-4 of the departing glycogen chain. (e) It binds water at the active site.

6. Match the enzymes that degrade glycogen in the left column with the appropriate prop

erties from the right column.

(a) phosphorylase

(1) is part of a single polypeptide chain

(b) a-1,6-glucosidase

with two activities

a-1,4 glucosidic bonds

(3) releases glucose (4) releases glucose 1-phosphate (5) moves three sugar residues from one

chain to another

(6) requires ATP

7. The phosphoglucomutase reaction is similar to the phosphoglyceromutase reaction of

the glycolytic pathway. Which of the following properties are common to both enzymes?

(a) Both have a phosphoenzyme intermediate. (b) Both use a glucose 1,6-bisphosphate intermediate. (c) Both contain pyridoxal phosphate, which donates its phosphate group to the substrate. (d) Both transfer the phosphate group from one position to another on the same molecule.

8. The activity of which of the following enzymes is NOT required for the release of large

amounts of glucose from liver glycogen?

(a) glucose 6-phosphatase (b) fructose 1,6-bisphosphatase (c) a-1,6-glucosidase (d) phosphoglucomutase (e) glycogen phosphorylase

9. Answer the following questions about the enzymatic degradation of amylose, a linear

a-1,4 polymer of glucose that is a storage form of glucose in plants. (a) Would phosphorylase act on amylose? Explain. (b) Would the rates of glucose 1-phosphate release from an amylose molecule by phos

phorylase relative to that from a glycogen molecule having an equivalent number of glucose monomers be equal? Explain.

internal glycosidic bonds, how might the rate of production of glucose 1-phosphate be affected?

10. Starting from a glucose residue in glycogen, how many net ATP molecules will be formed

in the glycolysis of the residue to pyruvate?

(a) 1 (b) 2 (c) 3 (d) 4 (e) 5 366

CHAPTER 21 Phosphorylase Is Regulated by Allosteric Interactions and Reversible Phosphorylation

11. Consider the diagram of the different conformational states of muscle glycogen phos

phorylase in Figure 21.2. Answer the following questions. FIGURE 21.2 Conformational states of phosphorylase in muscle.

2 ATP

2 ADP

G

P

2 P

2 H O

i

2 A B C D

(a) Which are the active forms of phosphorylase? (b) Which form requires high levels of AMP to become activated? (c) Which conversion is antagonized by ATP and glucose 6-phosphate? (d) What enzyme catalyzes the conversion of C to B?

12. How does the regulation of phosphorylase in the liver differ from the scheme for phos

phorylase regulation in muscle shown in Figure 21.2?

13. Indicate which of the following substances have binding sites on phosphorylase. For

those that do, give their major roles or effects.

(a) calmodulin (e) AMP (b) glycogen (f)

Pi

(g) ATP

(d) Ca2+

(h) glucose

14. Explain the roles of protein kinase A and calmodulin in the control of phosphorylase

kinase in muscle. Epinephrine and Glucagon Signal the Need for Glycogen Breakdown

15. What would increased epinephrine do to protein phosphatase 1 (PP1) in muscle, and

how would muscle glycogen metabolism be affected?

16. Place the following steps of the reaction cascade of glycogen metabolism in the proper

sequence.

(a) phosphorylation of protein kinase (b) formation of cyclic AMP by adenylate cyclase (c) phosphorylation of phosphorylase b (d) hormone binding to target cell receptors (e) phosphorylation of glycogen synthase a and phosphorylase kinase

17. Explain the effect of insulin on the activity of protein phosphatase 1 and the subsequent

effects on glycogen metabolism.

18. Why are enzymatic cascades, such as those that control glycogen metabolism and the

clotting of blood, of particular importance in metabolism?

GLYCOGEN METABOLISM 367 Glycogen Is Synthesized and Degraded by Different Pathways

19. Which of the following are common features of both glycogen synthesis and glycogen

breakdown?

(a) Both require UDP-glucose. (b) Both involve glucose 1-phosphate. (c) Both are driven in part by the hydrolysis of pyrophosphate. (d) Both occur on cytoplasmic glycogen granules. (e) Both use the same enzyme for branching and debranching.

20. If glycogen synthase can add a glucose residue to a growing glycogen molecule only if

the glucose chain is at least four units long, how does a new glycogen molecule start?

21. Why is the existence of distinct biosynthetic and catabolic pathways for glycogen im

portant for the metabolism of liver and muscle cells?

22. Is it true or false that branching in the structure of glycogen increases the rates of its syn

thesis and degradation? Explain.

23. Which of the following statements about glycogen synthase are correct?

(a) It is activated when it is dephosphorylated. (b) It is activated when it is phosphorylated. (c) It is activated when it is phosphorylated and in the presence of high levels of glu

cose 6-phosphate.

(d) It is activated when it is phosphorylated and in the presence of high levels of AMP.

24. Which of the following statements about the hormonal regulation of glycogen synthesis

and degradation are correct?

(a) Insulin increases the capacity of the liver to synthesize glycogen. (b) Insulin is secreted in response to low levels of blood glucose. (c) Glucagon and epinephrine have opposing effects on glycogen metabolism. (d) Glucagon stimulates the breakdown of glycogen, particularly in the liver. (e) The effects of all three of the regulating hormones are mediated by cyclic AMP.

25. Which of the following are effects of glucose on the metabolism of glycogen in the liver?

(a) The binding of glucose to phosphorylase a converts this enzyme to the inactive

T form.

(b) The T form of phosphorylase a becomes susceptible to the action of phosphatase. (c) The R form of phosphorylase b becomes susceptible to the action of phospho

rylase kinase.

(d) When phosphorylase a is converted to phosphorylase b, the bound phosphatase is

released.

(e) The free phosphatase dephosphorylates and activates glycogen synthase. Glycogen Breakdown and Synthesis Are Reciprocally Regulated

26. Explain how a defect in phosphofructokinase in muscle can lead to increased amounts

of glycogen having a normal structure. Patients with this defect are normal except for having a limited ability to perform strenuous exercise.

27. For the defect in question 26, explain why there is not a massive accumulation of glycogen. 368

CHAPTER 21 ANSWERS TO SELF-TEST

1. (a) A and G

(b) G and E (c) All the bonds are a-1,4 glycosidic linkages except for the one between residues

G and E.

(d) No. The two branches are too short for phosphorylase cleavage. Phosphorylase stops

cleaving four residues away from a branch point.

(e) Yes. Residue G can be hydrolyzed by a-1,6-glucosidase (the debranching enzyme). (f)

No. The branching enzyme transfers a block of about seven residues from a nonreducing end of a chain at least 11 residues long. Furthermore, the new a-1,6 glycosidic linkage must be at least four residues away from a preexisting branch point at a more internal site. The fragment of glycogen in Figure 21.1 does not fulfill these requirements.

2. b, c, d

3. Although the concentration of glycogen is higher in liver, the larger mass of muscle stores

more glycogen in toto.

4. The phosphorolytic cleavage of glycogen produces glucose 1-phosphate, which can enter

into the glycolytic pathway after conversion to glucose 6-phosphate. These reactions do not require ATP. On the other hand, the hydrolysis of glycogen would produce glucose, which would have to be converted to glucose 6-phosphate by hexokinase, requiring the expenditure of an ATP. Therefore, harvesting the free energy stored in glycogen by phosphorolytic cleavage rather than a hydrolytic one is more efficient because it decreases the ATP investment.

5. a, b

6. (a) 2, 4 (b) 1, 3 (c) 1, 2, 5. None of these enzymes requires ATP.

7. a. Answer (d) is incorrect because the phosphate group at one position on the small sub

strate molecule is transferred to, and remains for one cycle of reaction on, phosphoglucomutase and the phosphate at the other position on the small molecule product comes from a pre-existing phosphate on enzyme. For a given glucose 1-phosphate substrate, the phosphate on the product, glucose 6-phosphate, is not the same one that was present on the substrate; it came from the enzyme.

8. b

9. (a) Yes, phosphorylase would act on amylose by removing one glucose residue at a time

from the nonreducing end.

(b) No, the rate of degradation of amylose would be much slower than that of glyco

gen because amylose would have only a single nonreducing end available for reaction, whereas glycogen has many ends.

ing the chain into pieces with the endosaccharidase would allow a more rapid production of glucose 1-phosphate by phosphorylase.

10. c. A glucose molecule that is degraded in the glycolytic pathway to two pyruvate mole

cules yields two ATP; however, the formation of glucose-1-P from glycogen does not consume the ATP that would be required for the formation of glucose-6-P from glucose. Thus, the net yield of ATP for a glucose residue derived from glycogen is three ATP.

11. (a) A and D

(b) B (c) B to A (d) protein phosphatase 1

GLYCOGEN METABOLISM 369

The phosphorylated form of glycogen phosphorylase is phosphorylase a, which is mostly present in the active conformation designated as D in Figure 21.2. In the presence of high levels of glucose, phosphorylase a adopts a strained, inactive conformation, designated C in the figure. The dephosphorylated form of the enzyme is called phosphorylase b. Phosphorylase b is mostly present in an inactive conformation, labeled B in the figure. When AMP binds to the inactive phosphorylase b, the enzyme changes to an active conformation, designated A in the figure. The effects of AMP can be reversed by ATP or glucose 6-phosphate.

12. AMP does not activate liver phosphorylase (the B to A conversion shown in Figure 21.2),

and glucose shifts the equilibrium between the activated phosphorylase a toward the inactivated form (the D to C conversion).

13. (b) Glycogen, as the substrate, binds to the active site; there is also a glycogen particle

binding site that keeps the enzyme attached to the glycogen granule.

phorolysis and acts as a general acid-base catalyst.

(e) AMP binds to an allosteric site and activates phosphorylase b in muscle. (f)

Pi binds to the pyridoxal phosphate at the active site and attacks the a-1,4 glycosidic bond. Another Pi is covalently bound to serine 14 by phosphorylase kinase. This phosphorylation converts phosphorylase b into active phosphorylase a.

(g) ATP binds to the same site as AMP and blocks its effects in muscle; therefore,

energy charge affects phosphorylase activity.

(h) Glucose inhibits phosphorylase a in the liver by changing the conformation of the

enzyme to the inactive T form.

Answers (a) and (d) are incorrect because calmodulin and Ca2+ bind to phosphorylase kinase rather than to phosphorylase.

14. Protein kinase A, which is itself activated by cAMP, phosphorylates phosphorylase kinase

to activate it. Phosphorylase kinase can also be activated by the binding of Ca2+ to its calmodulin subunit. Upon binding Ca2+, calmodulin undergoes conformational changes that activate the phosphorylase kinase. The activated kinase in turn activates glycogen phosphorylase. These effects lead to glycogen degradation in contracting muscle.

15. Increased epinephrine activates PKA, which phosphorylates a subunit of PP1 and thus

reduces the ability of PP1 to act on its protein targets. Furthermore, inhibitor 1 is also phosphorylated by PKA so that it too decreases PP1 activity, albeit by a different mechanism. Inactivated PP1 leads to increased levels of activated (phosphorylated) phosphorylase and inactivated (phosphorylated) glycogen synthase. Glycogen breakdown would be stimulated under these conditions.

16. d, b, a, e, c

17. Insulin results in the activation of PP1. The hormone activates an insulin-sensitive pro

tein kinase that phosphorylates a subunit of PP1, rendering the phosphatase more active. The activated phosphatase dephosphorylates phosphorylase, protein kinase, and glycogen synthase. These changes result in a decrease in glycogen degradation and the stimulation of glycogen synthesis.

18. Enzymatic cascades lead from a small signal, caused by a few molecules, to a large sub

sequent enzymatic response. Thus, small chemical signals can be amplified in a short time to yield large biological effects. In addition, their effects can be regulated at various levels of the cascade.

19. b, d

20. The primer required to start a new glycogen chain is formed by the enzyme glycogenin,

which has a glucose residue covalently attached to one of its tyrosine residues. 370

CHAPTER 21

Glycogenin uses UDP-glucose to add approximately eight glucose residues to itself to generate a primer that glycogen synthase can extend.

21. The separate pathways for the synthesis and degradation of glycogen allow the synthe

sis of glycogen to proceed despite a high ratio of orthophosphate to glucose 1-phosphate, which energetically favors the degradation of glycogen. In addition, the separate pathways allow the coordinated reciprocal control of glycogen synthesis and degradation by hormonal and metabolic signals.

22. True. Since degradation and synthesis occur at the nonreducing ends of glycogen, the

branched structure allows simultaneous reactions to occur at many nonreducing ends, thereby increasing the overall rates of degradation or biosynthesis.

23. a, c

24. a, d

25. a, b, d, e

26. Since a defect in phosphofructokinase does not impair the ability of muscle to synthe

size and degrade glycogen normally, the structure of glycogen will be normal. However, the utilization of glucose 6-phosphate in the glycolytic pathway is impaired, and it equilibrates with glucose 1-phosphate; therefore, some net accumulation of glycogen will occur. The inability to perform strenuous exercise is likely due to the impaired glycolytic pathway in muscle and the diminished production of ATP.

27. Although the impaired use of glucose 6-phosphate in glycolysis will lead to the storage of

extra glycogen, it will not become excessive, because the increased concentration of glucose 6-phosphate will inhibit hexokinase and hence the sequestering of glucose in muscle. PROBLEMS

1. A patient can perform nonstrenuous tasks but becomes fatigued upon physical exertion.

Assays from a muscle biopsy reveal that glycogen levels are slightly elevated relative to normal. Crude extracts from muscle are used to determine the activity of glycogen phosphorylase at various levels of calcium ion for the patient and for a normal person. The results of those assays are shown in Figure 21.3. Briefly explain the clinical and biochemical findings for the patient. FIGURE 21.3 Response of glycogen phosphorylase to calcium ion in a patient

and in a normal person.

100

,

Normal

ylase

Patient

ram

50

uscle protein

Glycogen phosphor

activity per g

of m

0

10:7

10:6

10:5

[Ca2;]

2. A strain of mutant mice is characterized by limited ability to engage in prolonged exer

cise. After a high carbohydrate meal, these mice can exercise on a treadmill for only about 30 percent of the time a normal mouse can. At exhaustion, blood glucose levels in a mutant mouse are quite low, and they increase only marginally after rest. When liver glycogen in fed mutant mice is examined before exercise, the polymers have chains that are highly branched, with average branch lengths of about 10 glucose residues in either a-1,4 or a-1,6 linkage. Glycogen from exhausted normal mice has the same type of structure. Glycogen from exhausted mutant mice is still highly branched, but the polymer has an unusually large number of single glucose residues with a-1,6 linkages. Practically all the chains with a-1,4 linkages are still about 10 residues in length. Explain the metabolic and molecular observations for the mutant mice.

3. Your colleague discovers a fungal enzyme that can liberate glucose residues from cellulose.

The enzyme is similar to glycogen phosphorylase in that it utilizes inorganic phosphate for the phosphorolytic cleavage of glucose residues from the nonreducing ends of cellulose.

(a) Why would you suspect that other types of cellulases may be important in the rapid

degradation of cellulose?

(b) Do you think an enzyme that phosphorylizes cellulose external to the cell would

be useful to a fungal cell? Explain.

4. Consider a patient with the following clinical findings: fasting blood glucose level is 25 mg

per 100 ml (normal values are from 80 to 100 mg per 100 ml); feeding the patient glucose results in a rapid elevation of blood glucose level, followed by a normal return to fasting levels; feeding the patient galactose or fructose results in the elevation of blood glucose to normal levels; the administration of glucagon fails to generate hyperglycemia; biochemical examination of liver glycogen reveals a normal glycogen structure.

(a) Which of the enzyme deficiencies described in Table 21.1 of the text could account

for these clinical findings?

(b) What additional experiments would you conduct to provide a specific diagnosis for

the patient?

5. Vigorously contracting muscle often becomes anaerobic when the demand for oxygen

exceeds the amount supplied through the circulation. Under such conditions, lactate may accumulate in muscle. Under anaerobic conditions a certain percentage of lactate can be converted to glycogen in muscle. One line of evidence for this synthesis involves the demonstration of activity for malic enzyme, which can use CO2 to convert pyruvate to malate, using NADPH as an electron donor.

(a) Why is lactate produced in muscle when the supply of oxygen is insufficient? (b) In muscle, pyruvate carboxylase activity is very low. How could malic enzyme ac

tivity facilitate the synthesis of glycogen from lactate?

orous muscle contraction ceases?

(d) Is there an energetic advantage to converting lactate to glycogen in muscle rather

than using the Cori cycle for sending the lactate to the liver, where it can be reconverted to glucose and then returned to muscle for glycogen synthesis?

6. Cyclic nucleotide phosphatases are inhibited by caffeine. What effect would drinking a

strong cup of coffee have on glycogen metabolism when epinephrine levels are dropping in the blood?

7. During the degradation of branched chains of glycogen, a transferase shifts a chain of

three glycosyl residues from one branch to another, exposing a single remaining glycosyl residue to a-1,6-glucosidase activity. Free glucose is released, and the now unbranched chain can be further degraded by glycogen phosphorylase. 372

CHAPTER 21

(a) Estimate the free-energy change of the transfer of glycosyl residues from one branch

to another.

(b) About 10 percent of the glycosyl residues of normal glycogen are released as glu

cose, whereas the remainder are released as glucose 1-phosphate. Give two reasons why it is desirable for cells to convert most of the glycosyl residues in glycogen to glucose 1-phosphate.

pect liver extracts from such people to perhaps also lack a-1,6-glucosidase activity?

8. You are studying a patient with McArdle’s disease, which is described on page 595 of the

text. Explain what you would expect to find when you carry out each of the following analyses.

(a) fasting level of blood glucose (b) structure and amount of liver glycogen (c) structure and amount of muscle glycogen (d) change in blood glucose levels upon feeding the patient galactose (e) change in blood lactate levels after vigorous exercise (f)

change in blood glucose levels after administration of glucagon

(g) change in blood glucose levels after administration of epinephrine

9. An investigator has a sample of purified muscle phosphorylase b that she knows is rel

atively inactive. (a) Suggest two methods in vitro that could be employed to generate active phospho

rylase from the inactive phosphorylase b.

(b) After the phosphorylase is activated, the investigator incubates the enzyme with a

sample of unbranched glycogen in a buffered solution. She finds that no glycosyl residues are cleaved. What else is needed for the cleavage of glycosyl residues by active phosphorylase?

10. Arsenate can substitute in many reactions for which phosphate is the normal substrate.

However, arsenate esters are far less stable than phosphate esters, and they decompose spontaneously to arsenate and an alcohol:

R–OAsO 2− +

2−

3

H2O

R–OH + AsO4

(a) In which of the steps of glycogen metabolism might arsenate be used as a substrate? (b) What are the energetic consequences of utilizing arsenate as a substrate in glycogen

degradation?

11. As described on page 594 of your textbook, the ratio of glycogen phosphorylase to pro

tein phosphatase 1 is approximately 10 to 1. Suppose that in some liver cells the overproduction of the phosphatase results in a ratio of 1 to 1. How will such a ratio affect the cell’s response to an infusion of glucose?

12. A young woman cannot exercise vigorously on the treadmill without leg pains and stiff

ness. During exercise, lactate levels do not increase in her serum, in contrast to results of exercise in normal subjects. As is the case with normal subjects, no significant hypoglycemia is observed when the patient exercises or fasts. Analyses of muscle biopsy samples show that glycogen content is about 10 times greater than normal in the young woman, but the level of muscle phosphorylase activity is normal. Other experiments with biopsy samples show that rapid incorporation of 14C from radioactive glucose into fructose 6-phosphate and glycogen is observed, but very little incorporation of radioisotope into lactate is seen. When 14C-pyruvate is incubated with another sample of the homogenate, the radioisotope is readily incorporated into glycogen.

GLYCOGEN METABOLISM 373

What specific deficiency in a metabolic pathway could contribute to the observations

described above? Propose two additional studies that could confirm your conclusion.

13. In 1952, Dr. D. H. Andersen described a seriously ill infant with an enlarged liver as well

as cirrhosis. When epinephrine was administered, a relatively low elevation in the patient’s blood glucose levels was noted. Several days later when the infant was fed galactose, normal elevation of glucose was observed in the circulation. The infant died at the age of 17 months, and at autopsy Dr. Andersen found that glycogen from the liver, while present in unusually high concentration, was relatively insoluble, making it difficult to extract. She sent a sample of the liver glycogen to Dr. Gerty Cori. In an experiment designed to characterize the glycogen, Dr. Cori incubated a sample with orthophosphate (Pi) and two normal liver enzymes, active glycogen phosphorylase and debranching enzyme. She found that the ratio of glucose 1-phosphate to glucose released from the glycogen sample was 100:1, while the ratio from normal glycogen is 10:1.

(a) What enzyme of glycogen metabolism is most likely to be deficient in the liver tis

sue of the infant? Write a concise explanation for your answer, and relate it to the relative insolubility of the glycogen in the autopsy sample.

(b) Dr. Andersen, aware that a number of enzyme deficiencies might cause a glycogen

storage disease, sought to rule out a deficiency of a particular enzyme in the infant by studying the elevation of glucose levels after feeding galactose. What is that enzyme, and how does normal elevation of blood glucose after galactose feeding rule out a deficiency of that enzyme in the infant?

14. While muscle cells in tissue culture can be stimulated to break down glycogen only min

imally when incubated in a solution containing cyclic AMP, they are more readily stimulated by compounds like dibutyryl cyclic AMP, whose structure is shown in Figure 21.4. Explain the difference in response of cells to these two substances. FIGURE 21.4

O

NHJCJCH JCH JCH

2

3

J

C

N

C

CH

HC

C

N

OJCH

2

O

J

H

H

J

H

H

J

OKPJO

OJCJCH JCH JCH

2

3

J

O:

O Dibutyryl-cyclic AMP

15. One method for the analysis of glycogen involves incubating a sample with methyl io

dide, which methylates all free hydroxyl groups. Acid hydrolysis of exhaustively methylated glycogen yields a mixture of methyl glucosides, which can be separated and analyzed. Considering the various types of glycogen-storage diseases listed in Table 21.1 of the text, which of them could be diagnosed using exhaustive methylation and acid hydrolysis of glycogen?

16. Propose a scheme to identify the specific lysine residue in glycogen phosphorylase that

is in Schiff base linkage with pyridoxal phosphate. 374

CHAPTER 21

17. Patients with Cori’s disease lack the debranching enzyme, and therefore the structure of

liver and muscle glycogen is unusual, with short outer branches. Design an assay that would enable you to demonstrate the presence of short branches in glycogen from one of these patients. Also explain how you would demonstrate that the debranching enzyme is deficient in these patients.

18. Table 21.1 in the text lists eight diseases of glycogen metabolism, all of which affect the

level of glycogen in muscle and liver or the structure of the polysaccharide in one or both of those tissues. Another rare disease of glycogen metabolism is caused by a deficiency in liver glycogen synthase. After fasting, affected subjects have low blood glucose. Hyperglycemia and high blood lactate are observed after a meal.

(a) Briefly explain how these symptoms could be caused by glycogen synthase deficiency. (b) Under normal nutritional conditions, glycogen constitutes about 4% of the wet

weight of liver tissue in normal subjects. What proportion of glycogen in liver would you expect in a patient who lacks liver glycogen synthase?

19. Phosphoglucomutase converts the product of glycogen phosphorylase, glucose 1-phos

phate, to the glycolytic pathway component glucose 6-phosphate. The reaction catalyzed by phosphoglucomutase proceeds by way of a glucose 1,6-bisphosphate intermediate.

(a) What would happen to phophoglucomutase activity if the glucose 1,6-bisphosphate

intermediate were to dissociate from the enzyme before completion of the reaction?

(b) Would glucose 1,6-bisphosphate dissociation be equivalent to the hydrolysis of the

serine phosphate on the enzyme? Explain why.

might the enzyme be reactivated? ANSWERS TO PROBLEMS

1. Calcium ion normally activates muscle phosphorylase kinase, which in turn phospho

rylates muscle phosphorylase. In the patient, glycogen phosphorylase activity is less responsive to Ca2+ than it is in the normal subject. It is likely that Ca2+ cannot activate phosphorylase kinase in the patient, perhaps because the d subunit (calmodulin) of the enzyme is altered in some way. As a result, there are too few molecules of enzymatically active glycogen phosphorylase to provide the rate of glycogen breakdown that is needed to sustain vigorous muscle contraction. Elevated levels of muscle glycogen should be expected when glycogen phosphorylase activity is lower than normal.

2. The longer chains of glucose residues in a-1,4 linkage and the unusually high number

of single glucose residues in a-1,6 linkage suggest that while transferase activity is present, a-1,6-glucosidase activity is deficient in the mutant strain. For such chains, far fewer ends with glucose residues are available as substrates for glycogen phosphorylase. (Recall that glycogen phosphorylase cannot cleave a-1,6 linkages.) This limited ability to mobilize glucose residues means that less energy is available for prolonged exercise.

3. (a) Cellulose is an unbranched polymer of glucose residues with b-1,4 linkages.

Therefore, each chain has only one nonreducing end that is available for phosphorolysis by the fungal enzyme. Compared with the rate of breakdown of molecules of glycogen, whose branched chains provide more sites for the action of glycogen phosphorylase, the generation of glucose phosphate molecules from cellulose by means of the fungal enzyme alone could be quite slow. Therefore, you

would expect to find endocellulases that generate additional nonreducing ends in cellulose chains.

(b) No. In order for fungal cells to degrade cellulose, they must secrete the relevant en

zymes to the extracellular space, because cellulose is a large, insoluble macromolecule. It seems unlikely that a phosphorylase would be the enzyme of choice for cellulose degradation, because glucose phosphate molecules, which are negatively charged at neutral pH, would be unable to cross cellular membranes and enter the cytosol. In addition, concentrations of phosphate outside the cell could be too low to drive phosphorolysis of cellulose.

4. (a) A low fasting-blood-glucose level indicates a failure either to mobilize glucose pro

duction from glycogen or to release glucose from the liver. However, the elevations in blood glucose levels after feeding the patient glucose, galactose, or fructose indicate that the liver can release glucose derived from the diet or formed from other monosaccharides. The lack of response to glucagon indicates that the enzymatic cascade for glycogen breakdown is defective. Therefore, you would suspect a deficiency of liver glycogen phosphorylase or phosphorylase kinase. Of the diseases described in Table 21.1 of the text, both type VI and type VIII could account for the findings, which include increased amounts of glycogen with normal structure.

(b) The direct assay of the activities of glycogen phosphorylase and phosphorylase ki

nase would enable you to make a specific diagnosis. For these purposes, a liver biopsy would be necessary.

5. (a) Muscle cells produce lactate from pyruvate under anaerobic conditions in order to

generate NAD+, which is required to sustain the activity of glyceraldehyde 3-phosphate dehydrogenase in the glycolytic pathway.

(b) The low activity of muscle pyruvate carboxylase means that other pathways for the

synthesis of oxaloacetate must be available. The formation of malate, which is then converted to oxaloacetate, enables the muscle cell to carry out the synthesis of glucose 6-phosphate via gluconeogenesis. Glycogen can then be synthesized through the conversion of glucose 6-phosphate to glucose 1-phosphate, the formation of UDP-glucose, and the transfer of the glucose residue to a glycogen primer chain.

marily from the conversion of glycogen and glucose to lactate. The simultaneous conversion of lactate to glycogen would simply result in the unnecessary hydrolysis of ATP.

(d) In the liver, the conversion of two lactate molecules to a glucose residue in glyco

gen through gluconeogenesis requires seven high-energy phosphate bonds; six are required for the formation of glucose 6-phosphate from two molecules of lactate, and one is needed for the synthesis of UDP-glucose from glucose 1-phosphate. The conversion of lactate to glycogen in muscle requires two fewer high-energy bonds because the formation of oxaloacetate through the action of malic enzyme does not require ATP. Recall that pyruvate carboxylase requires ATP for the synthesis of oxaloacetate from pyruvate.

6. When epinephrine levels in the blood decrease, the synthesis of cyclic AMP decreases.

Existing cyclic AMP is degraded by cyclic nucleotide phosphatases. The inhibition of these enzymes by caffeine prolongs the degradation of glycogen because the remaining cyclic AMP continues to activate protein kinase, which in turn activates phosphorylase kinase. Glycogen phosphorylase is, in turn, activated by phosphorylase kinase. Sustained activation of phosphorylase results in continued mobilization of glucose residues from glycogen stores in liver. 376

CHAPTER 21

7. (a) Because the bonds broken and formed during the transferase action are both a-1,4

glycosidic bonds, the free-energy change is likely to be close to zero.

(b) The generation of glucose 1-phosphate rather than glucose means that one fewer

ATP equivalent is required for the conversion of a glucose residue to two molecules of lactate. The glucose residue released does not have to be phosphorylated for subsequent metabolism when phophorolysis produces it. In addition, the phosphorylation of glucose ensures that the molecule cannot diffuse across the cell membrane before it is utilized in the glycolytic pathway.

polypeptide chain. A significant alteration in the structure of the domain for glucosidase in the bifunctional enzyme could also impair the functioning of the transferase domain.

8. (a) In McArdle’s disease, muscle phosphorylase is deficient, but liver phosphorylase is

normal. Therefore, you would expect glucose and glycogen metabolism in the liver to be normal and the control of blood glucose by the liver also to be normal.

(b) Normal glyogen metabolism in the liver means that both the amount of liver glyco

gen and its structure would be the same as in unaffected people.

paired. Moderately increased concentrations of muscle glycogen could be expected, although the structure of the glycogen should be similar to that in unaffected people.

(d) Galactose can be converted to glucose 6-phosphate in the liver, which can then ex

port glucose to the blood. Because defective muscle phosphorylase has no effect on galactose metabolism, you would expect similar elevations in blood glucose after the ingestion of galactose in normal and affected people.

(e) During vigorous exercise, blood lactate levels normally rise as muscle tissue exports

the lactate generated through glycogen breakdown. The defect in muscle phosphorylase limits the extent to which glycogen is degraded in the muscle. This in turn reduces the amount of lactate exported during exercise, so the rise in blood lactate levels would not be as great in the affected person.

(f)

Glucagon exerts its effects primarily on liver, not muscle. In patients with McArdle’s disease, blood glucose levels increase normally in response to glucagon.

(g) A slight increase in blood glucose concentration may occur after epinephrine ad

ministration, because liver is somewhat responsive to this hormone. Epinephrine does have a greater glycogenolytic effect on muscle, but you would not expect to see any change in blood glucose concentration when it is administered. The reason is that, even if glycogen breakdown is accelerated (which is unlikely to occur in patients with McArdle’s disease), the glucose 6-phosphate produced cannot be converted to glucose for export into the circulation, because muscle lacks the enzyme glucose 6-phosphatase.

9. (a) The investigator could activate the phosphorylase by adding AMP to the sample or

by using active phosphorylase kinase and ATP to phosphorylate the enzyme.

(b) Inorganic phosphate is also required for the conversion of glycosyl residues in glyco

gen to glucose 1-phosphate molecules.

10. (a) Arsenate can substitute for inorganic phosphate in the glycogen phosphorylase re

action, generating glucose arsenate esters.

(b) When Pi is used as a substrate for glycogen phosphorylase, glucose 1-phosphate is

generated. The glucose 1-arsenate esters that are generated when arsenate is used as a substrate spontaneously hydrolyze to yield glucose and arsenate. The conver

sion of glucose to pyruvate requires one more ATP equivalent than does the conversion of glucose 1-phosphate to pyruvate.

11. The normal 10-to-1 ratio means that glycogen synthase molecules are activated only

after most of the phosphorylase a molecules are converted to the inactive b form, which ensures that the simultaneous degradation and synthesis of glycogen does not occur. A phosphorylase to phosphatase ratio of one to one means that, as soon as a few phosphorylase molecules are inactivated, phosphatase molecules that are no longer bound to phosphorylase begin to convert glycogen synthase molecules to the active form. Glycogen degradation and synthesis then occur simultaneously, resulting in the wasteful hydrolysis of ATP.

12. From the clinical observations, it appears that the pathway from pyruvate to glucose 6

phosphate and on to glycogen is functional and that gluconeogenesis is working normally in liver (there is no hypoglycemia during fasting or exercise, when demands for glucose increase). Although muscle glycogen content is higher, normal phosphorylase activity indicates that glycogen could be phosphorylized normally. You should then consider whether there is a deficiency in the glycolytic pathway, because lactate does not accumulate during exercise and it is not labeled when 14C-glucose is administered. Labeled fructose 6-phosphate can be made from radioactive glucose in the biopsy sample, but knowledge about subsequent glycolytic reactions is not available. There could be a significant block at the level of phosphofructokinase or beyond. Such a deficiency would mean that while normal demands for glucose can be taken care of, a high rate of glycolytic activity during vigorous exercise cannot be accommodated. You should consider analyzing for additional radioactive glycolytic intermediates when glucose is administered, then testing for deficiency of one or more glycolytic enzymes using biopsy tissues. The description of the disorder corresponds most closely to a known condition for a deficiency in muscle phosphofructokinase (Type VII glycogen-storage disease). One might also argue that lactate dehydrogenase could be absent, explaining why no lactate is generated during exercise. However, in cases in which muscle lactate dehydrogenase is defective, affected subjects cannot exercise vigorously, but they have no accumulation of glycogen in their muscle tissue.

13. (a) The branching enzyme was deficient in the infant. This enzyme removes blocks of

glucosyl residues from a chain of a-1

4-linked residues and transfers them

internally to form a branch with an a-1

6 link to a polymer chain. The most

important clue to the deficiency is found in the ratio of glucose 1-phosphate to glucose, which is 10 times higher in glycogen from the affected infant than from a normal polymer sample. Recall that glucose 1-phosphate is produced through the action of phosphorylase, which phosphorylizes a-1

4 linkages, while

glucose is produced when the glycogen debranching enzyme hydrolyzes a glucose in a-1

6 linkage at a branch point. Normal glycogen has a branch at every

10 or so glycosyl residues, so that treatment with a mixture of normal phosphorylase and debranching enzyme will yield a 10:1 ratio of glucose 1-phosphate to glucose. The autopsy sample yielded a ratio of 100:1, suggesting that there are far fewer branches in the sample. This conclusion is consistent with the relative insolubility of the infant’s glycogen, which, with fewer branches, is more like amylopectin, a linear glucosyl polymer which has limited solubility in water.

(b) The pathway for galactose metabolism includes its conversion, through steps that

include epimerization, to glucose 6-phosphate. Thus, feeding galactose should result in an increased concentration of glucose 6-phosphate in the liver cell. If

378

CHAPTER 21

glucose 6-phosphatase were deficient, glucose 6-phosphate would not be converted to glucose, so that the levels of blood glucose would not be elevated after galactose feeding. Dr. Andersen considered a glucose 6-phosphatase deficiency because of the limited increase in blood glucose levels after administration of epinephrine, so she used galactose feeding to increase glucose 6-phosphate levels in liver cells. When glucose levels rose in the blood, she concluded that glucose 6-phosphatase levels were normal. She subsequently considered other deficiencies that would result in storage of abnormal amounts of liver glycogen.

14. Like other nucleotides, cyclic AMP is polar and negatively charged at neutral pH. It there

fore crosses plasma membranes at a relatively low rate. The presence of two hydrophobic acyl chains on the molecule make it much more hydrophobic, so that it can more easily dissolve in the bilayer and more readily enter the cytosol.

15. Type IV glycogen-storage disease, in which glycogen with a much lower number of

a-1,6 glycosidic linkages is produced, could be analyzed using methylation and hydrolysis. Any glucose residue derived from a branch point will have methyl groups at C-2 and C-3, while all other residues (with one exception) will emerge from hydrolysis as 2,3,6-O-trimethyl glucose molecules. The glucose at the reducing end of the glycogen molecule, if it exists, will be converted to a tetramethlyglucoside. In normal subjects, the ratio of trimethylglucose to dimethyl glucose should be about 10 to 1, while glycogen from a person with a deficiency in the branching enzyme will have a much higher ratio.

16. Treat the protein with sodium borohydride, which reduces the Schiff base linkage be

tween pyridoxal phosphate and the lysine residue. Then convert the protein to peptide fragments using various proteases or chemical methods. This is a time-consuming task because the protein is composed of two identical chains, each containing 842 amino acids. Every isolated fragment that is known to contain lysine can be subjected to acid hydrolysis. A lysine residue from one of the fragments will be covalently attached to pyridoxamine (the phosphate group is usually removed during acid hydrolysis). Analysis of phosphorylase by these methods shows that Lys 680 is in Schiff base linkage with pyridoxal phosphate. Nowadays, mass spectrometry of the fragments would be an easier way to identify the proteolytic fragment bearing the pyridoxal phosphate.

17. A short outer branch in a glycogen molecule has only a small number of a-1,4-glucosyl

residues on the nonreducing side of a branch or an a-1,6 link. Incubating such a glycogen molecule with active phosphorylase and Pi will liberate only limited amounts of glucose 1-phosphate, compared with the number liberated from normal glycogen. Recall that phosphorylase cannot free glucose molecules that are within four residues of a branch point in glycogen. To demonstrate that phosphorylase action is limited by short outer branches, you can incubate another sample with purified debranching enzyme and phosphorylase, and you would expect to see an increase in production of glucose 1-phosphate. To demonstrate a debranching enzyme deficiency in a patient, you could treat normal glycogen with active muscle phosphorylase and muscle extracts from a patient with Cori’s disease. If debranching enzyme activity is low, only limited amounts of glucose 1-phosphate will be produced. Larger numbers of glucose 1-phosphate molecules will be released from a normal glycogen sample treated with active phosphorylase and muscle-cell extracts from a normal person.

18. (a) Lack of glycogen synthase implies that the ability of the liver to store glucose as

glycogen is impaired. After fasting, when blood glucose concentrations are low, liver glycogen is normally converted to glucose 6-phosphate, which is converted

to glucose and exported to the blood. Low glycogen levels in liver tissue would make it impossible for liver to maintain proper glucose levels in the blood. After a meal containing carbohydrates, the liver would be unable to convert glucose to glycogen. Even though glucokinase may convert glucose to glucose 6-phosphate, the high concentration of that substrate may cause accelerated conversion back to glucose through the action of glucose 6-phosphatase. Glucose levels would then increase in the circulation. The elevation of lactate levels in blood suggests that any glucose metabolized in the liver is preferentially converted to lactate rather than to glycogen.

(b) As discussed above, liver cells deficient in glycogen synthase would be unable to

synthesize large amounts of glycogen. You would therefore expect the percentage of glycogen in affected people to be lower. In those few patients with the disorder, glycogen makes up less than 1% of liver tissue.

19. (a) If the glucose 1,6-bisphosphate were to dissociate from the enzyme, the enzyme

would not have a phosphate on the serine hydroxyl that is necessary for activity. The dephosphorylated enzyme would lack the phosphate need for transfer to the incoming glucose 1-phosphate to form the bisphosphate intermediate and could not catalyze the mutase reaction.

(b) Yes, both bisphosphate dissociation or phosphoenzyme hydrolysis would lead to

an inactive, unphosphorylated enzyme.

quired for activity, some means of producing the phosphoenzyme is required. A protein kinase could replace the covalently bound enzyme phosphate or a phosphoglucokinase enzyme that produced glucose 1,6-bisphospate, which would bind to and phosphorylate phosphoglucomutase, could also form the phosphorylated enzyme. The latter mechanism is known. EXPANDED SOLUTIONS TO TEXT PROBLEMS

1. Recall that galactose enters metabolism by reacting with ATP in the presence of galac

tokinase to yield galactose 1-phosphate and subsequently glucose 1-phosphate. On the way to glycogen the latter reacts with UTP to give UDP–glucose. Hence:

Galactose + ATP + UTP + H

+

2O + (glycogen)n

(glycogen)n+ 1

ADP + UDP +

2 P +

i

H+

2. Glucose is mobilized from the nonreducing ends of glycogen. The unbranched a-amy

lose has only one nonreducing end, whereas glycogen has many of them. Therefore, glucose monomers can be released much more quickly from glycogen than from a-amylose.

3. In normal glycogen, branches occur about once in 10 units. Therefore, degradation of

this glycogen is expected to give a ratio of glucose 1-phosphate to glucose of about 10:1. An increased ratio (100:1) indicates that the glycogen has a much lower degree of branching, suggesting a deficiency of the branching enzyme.

4. The enzymatic defect in von Gierke’s disease is the absence of liver glucose 6-phos

phatase. The resulting high concentrations of glucose 6-phosphate allosterically activate the inactive glycogen synthase b, causing a net increase in liver glycogen. 380

CHAPTER 21

5. Glucose is an allosteric inhibitor of phosphorylase a. Hence, crystals grown in its

presence are in the T state. The addition of glucose 1-phosphate, a substrate, shifts the R

T equilibrium toward the R state. The conformational differences between

these states are sufficiently large that the crystal shatters unless it is stabilized by chemical cross-links. The shattering of a crystal caused by an allosteric transition was first observed by Haurowitz in the oxygenation of crystals of deoxyhemoglobin.

6. Since glucose 1,6-bisphosphate is an intermediate in the reaction catalyzed by phos

phoglucomutase and phosphorylates this enzyme during catalysis, it is reasonable to expect that it can phosphorylate the dephosphoenzyme. Glucose 1,6-bisphosphate is formed from glucose 1-phosphate and ATP by phosphoglucokinase.

7. Water is excluded from the active site to prevent hydrolysis. The entry of water could

lead to the formation of glucose rather than glucose 1-phosphate. A site-specific mutagenesis experiment is revealing in this regard. In phosphorylase, Tyr 573 is hydrogenbonded to the 2′-OH of a glucose residue. The ratio of glucose 1-phosphate to glucose product is 9000:1 for the wild-type enzyme, and 500:1 for the Phe 573 mutant. Model building suggests that a water molecule occupies the site normally filled by the phenolic OH of tyrosine and occasionally attacks the oxocarbonium ion intermediate to form glucose. See D. Palm, H. W. Klein, R. Schinzel, M. Buehner, and E. J. M. Helmreich, Biochemistry 29(1990):1099.

8. Glycogenin performs the priming function for glycogen synthesis. Without a-amylase to

degrade pre-existing chains, the glycogenin activity would be masked by the more prominent activity of glycogen synthase. The a-amylase treatment halts the activity of glycogen synthase by shortening existing glucose chains below the threshold size required for them to be substrates of glycogen synthase.

9. When two soluble enzymes catalyze consecutive reactions, the product formed by the

first enzyme must leave and diffuse to the second enzyme. Catalytic efficiency is substantially increased if both active sites are in close proximity in the same enzyme molecule. A similar advantage is obtained when consecutive enzymes are held close to each other in multienzyme complexes.

10. The mice will be unable to generate phosphorylase a from phosphorylase b, but phos

phorylase b will still have a low level of activity and will degrade glycogen, especially during exercise. Although the T state of phosphorylase b is favored, accumulation of AMP during exercise will convert some of the phosphorylase b to the active R state.

11. (a) Muscle phosphorylase b will be inactive even when the AMP level is high. Hence,

glycogen will not be degraded unless phosphorylase is converted into the a form by hormone-induced or Ca2+-induced phosphorylation.

(b) Phosphorylase b cannot be converted into the much more active a form. Hence, the

mobilization of liver glycogen will be markedly impaired.

glycogen phosphorylase. Little glycogen will be present in the liver because it will be persistently degraded.

(d) Protein phosphatase 1 will be continually active. Hence, the level of phosphorylase b will be higher than normal, and glycogen will be less readily degraded.

GLYCOGEN METABOLISM 381

(e) Protein phosphatase 1 will be much less effective in dephosphorylating glycogen

synthase and glycogen phosphorylase. Consequently, the synthase will stay in the less active b form, and the phosphorylase will stay in the more active a form. Both changes will lead to increased degradation of glycogen.

(f)

The absence of glycogenin will block the initiation of glycogen synthesis. Very little glycogen will be synthesized in its absence.

12. (a) Glycogen breakdown will persist for too long a time in response to epinephrine (or

glucagon), so that too much glucose will be released. The a subunit of GS will remain active for too long and will stimulate too much production of cAMP, which ultimately will keep glycogen phosphorylase active for too long (Sections 21.3.1 and 21.3.2).

(b) The liver enzyme will be locked in its virtually inactive dephosphorylated form

(phosphorylase b) because the alanine mutant will not be able to accept a phosphate group. Also, liver phosphorylase b is not activated by AMP, and so the liver will have almost no ability to mobilize glucose from glycogen.

extra glycogen phosphorylase and lead to an increased concentration of free glucose. (However, under suitably tight regulation conditions, a cell could be normal if the extra supply of kinase were kept in an inhibited state.)

(d) Protein phosphatase will be more active, so that glycogen phosphorylase will

be easily inactivated. Glucose mobilization will be impaired, and glucose will be less available.

(e) This will lead to a serious loss of ability to synthesize glycogen, for glycogen syn

thase is catalytically efficient only when bound to glycogenin. Both the number and size of glycogen granules will be very small. The organism will lack much of the normal ability to regulate the blood glucose level.

(f)

cAMP will persist for longer than normal (in response to epinephrine or glucagon), and so extra glucose will be released from glycogen stores even when no longer needed.

13. The slow phosphorylation of the a subunits of phosphorylase kinase serves to prolong

the degradation of glycogen. The kinase cannot be deactivated until its a subunits are phosphorylated. The slow phosphorylation of a assures that the kinase and, in turn, phosphorylase stay active for a defined interval. See H. G. Hers, Ann. Rev. Biochem. 45(1976):167.

14. When the a subunit is phosphorylated (Section 21.3.2), the b subunit is more suscep

tible to dephosphorylation by protein phosphatase, which causes inactivation. Phosphorylase kinase (and consequently glycogen phosphorylase) therefore would be less active, and the release of glucose from glycogen would be slowed.

15. An enzyme-bound intermediate is likely for amylase, and for the transferase and a-1,6

glucosidase (debranching enzyme). A nucleophile on the enzyme would need to break the a-1,4 bond (transferase) or the a-1,6 bond (debranching enzyme) and form a bond to one part of the carbohydrate chain (at C1). A second nucleophile would then attack and release the enzyme-bound chain. For the case of transferase, the second nucleophile

382

CHAPTER 21

would be a terminal C4-OH group of glycogen to receive the transferred tri-glucose unit, whereas debranching enzyme would use a water molecule as the second nucleophile to release a free glucose monomer.

16. (a) The antibodies will detect only glycogenin, and the glycogenin will be bound to

glycogen. Without a-amylase treatment, the glycogen will have a high molecular weight and will remain at the top of the gel.

(b) The glycogen is digested into small pieces that remain bound to the glycogenin. The

glycogenin migration distance now will reflect approximately its true molecular weight plus that of a small bound carbohydrate oligomer.

be present, but they were not stained with specific antibodies in the Western blot.

17. (a) The pattern reflects glycogenin bound to carbohydrate chains of varying sizes.

(b) When starved for glucose, the cells use most of their glycogen and the supply is

depleted.

that lane 3 resembles lane 1.

(d) The glycogen supply is replenished within one hour and does not further increase

in three hours.

(e) Amylase digests the glycogen in all samples to small fragments that are bound to

the glycogenin, whose size is ~66 kD. </body></html>

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