Essential Biochemistry, 3rd Edition Pratt, Cornely Solution Manual.pdf (12)

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Type:

Solution Manual

Resource:

Essential Biochemistry

Edition: Author(s):

3rd Edition Charlotte W. Pratt Kathleen Cornely

SOLUTIONS

(b) O

Carboxyl

Amino

C

Carboxylate

O

NH Amido

C

CH3

2

H

C

OH

C

NH

C

H

HO

C C

H

H

C

OH

Amido

CH3

H3N

O

CH2

Hydroxyl

(CH 214 )

O

P

Amino

OH

OH

CH C CH2 SH

Carboxylate

Sulfhydryl

O

CH

HO

C

O

O

OH

O

CH2

Hydroxyl

CH2 OH

(d)

O

Ester

Carbonyl

O

H

(c)

O

CH2

added to some pet foods and milk products from China so that they would appear to contain more protein. Melamine is toxic to pets and children.) (c) Compound C is an amino acid, so it would already be present in protein-containing food.

8. A diet high in protein results in a high urea concentration, since urea is the body’s method of ridding itself of extra nitrogen. Nitrogen is found in proteins but is not found in signifi cant amounts in lipids or carbohydrates. A low-protein diet provides the patient with just enough protein for tissue repair and growth. In the absence of excess protein consumption, urea production decreases, and this puts less strain on the patient’s weakened kidneys.

9. All amino acids have carboxylate groups. All have primary amino groups except for proline, which has a secondary amino group.

10. The carbon marked with an asterisk is chiral. This means that alanine has two possible enantiomers, or mirror-image isomers.

Phosphoester O

H

(f)

Carbonyl

C

H

C

OH

C Secondary amino group

Hydroxyl

CH2 OH

HN

CH3

O Carboxylate

2. The functional groups are identifi ed below.

Ether Carbonyl

OH HO

Hydroxyl

CH O O C CH C H2 C HO C OH

O O

Carbonyl

Amido group

O C

Carboxyl OH

O 3C Ether H H3C

11. Asn has an amido group and Cys has a sulfhydryl group. 12. Two hydrogen atoms and one oxygen atom are lost when Asn and Cys form a dipeptide. This is an example of a condensation reaction. The carboxy late functional group on the Asn is lost and the amino functional group on the Cys is lost. An amido group is formed.

H 3N

N Nicotinic acid (niacin)

Vitamin C

O

Carbonyl CH3 C H H C 2

CH C CH2 C

O

NH

CH C O CH2 SH

NH2

CH2 H C C H 3 n

Coenzyme Q

3. Amino acids, monosaccharides, nucleotides, and lipids are the four types of biological small molecules. Amino acids, monosaccharides, and nucleotides can form polymers of proteins, polysaccharides, and nucleic acids, respectively.

4. (a) N-acetylglucosamine is a monosaccharide. (b) CMP is a nucleotide. (c) Homocysteine is an amino acid. (d) Cholesteryl ester is a lipid.

C* C

HN 3

O

O

(e)

H O

CH C CH2

O

H3N

COOH

O

Chapter 1 1. (a)

13.

O

Carbonyl

C H H

O H

HO

H

H

OH

H

Hydroxyl

OH CH2 OH

14. (a) Fructose has the same molecular formula, C6H12O6, as glucose. (b) Fructose is a ketone, whereas glucose is an aldehyde.

15. Uracil has a carbonyl functional group, whereas cytosine has an amino 5. (a) Mainly C and H plus some O functional group. (b) C, H, and O (c) C, H, O, and N plus small amounts of S 16. Nucleotides consist of a fi ve-carbon sugar, a nitrogenous ring, and one or more phosphate groups linked covalently together.

6. It is a lipid (it is actually lecithin). It is mostly C and H, with relatively little O and only one N and one P. It has too little O to be a carbohydrate, 17. As described in the text, palmitate and cholesterol are highly nonpolar too little N to be a protein, and too little P to be a nucleic acid. and are therefore insoluble in water. Both are highly aliphatic. Alanine is 7. (a) You should measure the nitrogen content, since this would indicate water soluble because its amino group and carboxylate group are ionized, the presence of protein (neither lipids nor carbohydrates contain nitrogen).which render the molecule “saltlike.” Glucose is also water soluble because (b) You could add the compound that contains the most nitrogen, com- its aldehyde group and many hydroxyl groups are able to form hydrogen bonds with water. pound B, which is melamine. (Melamine is a substance that has been

1

2

Ch 1 Solutions

18. Glucose has several hydroxyl groups and is a polar molecule. As such, it will have diffi culty crossing the nonpolar membrane. The 2,4-dinitrophenol The reaction is not favorable at 4 C. molecule consists of a substituted benzene ring and has greater nonpolar (b) DG5(6000 J?mol21)2(371273 K)(21 J?K21?mol21) char- acter. Of the two molecules, the 2,4-dinitrophenol will traverse the DG52510 J?mol21 membrane more easily.

The reaction is favorable at 37 C. 19. DNA forms a more regular structure because DNA consists of only four ? ? mol21 ) different nucleotides, whereas proteins are made up of as many as 20 30. 0.15,000 J?mol212(T)(51 JK21 different 215,000.2(T)(51 K21 ) amino acids. In addition, the 20 amino acids have much more individual 15,000,(T)(51 K21 ) 294 K,T variaThe reaction is favorable at temperatures of 21 C and higher. tion in their structures than do the four nucleotides. Both of these factors result 31.

0.214.3 kJ?mol212(273125 K)( DS) in a more regular structure for DNA. The cellular role of DNA relies on the 14.3 kJ?mol21.2(273125 K)( DS) sequence of the nucleotides that make up the DNA, not on the overall shape of 248 J?mol21.DS the DNA molecule itself. Proteins, on the other hand, fold into unique S could be any positive value, or it could have a negative value smaller than 48 J mol 1. shapes, as illustrated by endothelin in Figure 1-4. The ability of proteins to fold into 32. 263 kJ?mol215DH 2 (273 1 25 K)(190 JK21 ? ? mol21 ) a DH5263 kJ?mol21156.6 kJ?mol21 wide variety of shapes means that proteins can also serve a wide variety of DH526.4 kJ?mol21 bioThe reaction is exothermic. chemical roles in the cell. According to Table 1-2, the major roles of proteins in 33. Process (d) is never spontaneous. the cell are to carry out metabolic reactions and to support cellular 34. (a) Because the rubber band becomes cooler, it is absorbing heat. Therestructures. fore, its change in enthalpy is positive.

20. Polysaccharides serve as fuel-storage molecules and can also serve as (b) In order for G to be negative (a spontaneous process), the entropy structural support for the cell. change must negative so that it can compensate for an unfavorable (posi21. The pancreatic amylase is unable to digest the glycosidic bonds that link tive) change in enthalpy. the glucose residues together in cellulose. Figure 1-6 shows the structural 35. The dissolution of urea in water is an endothermic process and has a differences between starch and cellulose. Pancreatic amylase binds to starch positive H value. In order to be spontaneous, the process must also have prior to catalyzing the hydrolysis of the glycosidic bond; thus, the enzyme and the starch must have shapes that are complementary. The enzyme would a positive S value in order for the free energy change of the process to be be unable to bind to the cellulose, whose structure is much different from negative. Solutions have a higher order of entropy than the solvent and solute alone. that of starch.

22. Cellulose cannot be digested in mammals and therefore the energy yield is 36. (a) The reaction is exothermic because the value of H is negative.

0 kilocalories per gram. Although both starch and glycogen are polymers of (b) DG5DH2T DS 217200 J?mol21529500 J?mol212(310 K)(DS) glucose, the glucose residues are linked together differently in the two moleDS525 J?K21?mol21 cules and pancreatic amylase is unable to hydrolyze the glycosidic bonds (see (a) negative Solution 21). Cellulose provides (b) no positive energy to the diet but is an important The positive value of S indicates that the reaction proceeds with an (c) positive (d) positive component of the diet as fi ber. increase in entropy. (e) negative 23. A positive entropy change indicates that the system has become more (c) The H term makes a greater contribution to the G value. This 24. (a) decrease disordered; a negative entropy change indicates that the system has become indicates that the reaction is spontaneous largely because the reaction is increase more(b) ordered. exothermic.

25. The polymeric molecule is more ordered and thus has less entropy. A 37. (a) The conversion of glucose to glucose-6-phosphate is not favorable mixture of constituent monomers has a large number of different arrangebecause the G value for the reaction is positive, indicating an enderments (like the balls scattered on a pool table) and thus has greater entropy. gonic process.

26. Entropy increases as the reactants (7 molecules) are converted to products (b) If the two reactions are coupled, the overall reaction would be the sum of the two individual reactions. The G value would be the sum of (12 molecules). the G values for the two individual reactions.

27. The dissolution of ammonium nitrate in water is a highly endothermic process, as indicated by the positive value of H. This means that when ammonium nitrate dissolves in water, the system absorbs heat from the sur- ATP1glucoseΔADP1glucose-6-phosphate DG5216.7 kJ?mol21 roundings and the surroundings become cold. The plastic bag containing the ammonium nitrate becomes cold and can be used as a cold pack to treat Coupling the conversion of glucose to glucose-6-phosphate with the an injury. hydrolysis of ATP has converted an unfavorable reaction to a favorable 28. The dissolution of calcium chloride in water is a highly exothermic proreaction. The G value of the coupled reaction is negative, which indicess, as indicated by the negative value of H. This means that when calcium cates that the reaction as written is favorable. chloride dissolves in water, the system loses heat to the surroundings and the 38. (a) The reaction is not favorable because the G value for the reaction is surroundings become warm. The plastic bag holding the calcium chloride positive, indicating an endergonic process. solution becomes warm and can be used as a hot pack by the hiker at cold (b) temperatures.

29. First, calculate H and S, as described in Sample Calculation 1-1: DH5H2H DH560 kJ?mol21254 kJ?mol21 DH56 kJ?mol21 DS 5SB2SA DH543 J?K21?mol21222 J?K21?mol21 DH521 J?K21?mol21 (a) DG5(6000 J?mol21)2(41273 K)(21 J?K21?mol21) DG5180 J?mol21

GAP1P1NAD1Δ1,3BPG1NADH i 1,3BPG1ADPΔ3PG1ATP GAP1P1NAD1 i

1ADPΔ3PG1NADH1ATP

DG516.7 kJ? mol2 1 DG5218.8 kJ?mol21 DG5212.1 kJ?mol21

The coupled reaction is spontaneous because the G value is negative.

39. C (most oxidized), A, B (most reduced) 40. (a) reduction (b) oxidation

Ch 2 Solutions 41. (a) oxidized (b) oxidized (c) oxidized (d) reduced 42. (a) oxidizing agent (c) oxidizing agent

(b) oxidizing agent (d) reducing agent

43. (a) Palmitate’s carbon atoms, which have the formula OCH2O, are more reduced than CO2, so their reoxidation to CO2 can release free energy. (b) Because the OCH2O groups of palmitate are more reduced than those of glucose (OHCOHO), their conversion to the fully oxidized CO2 would be even more thermodynamically favorable (have a larger negative value of G ) than the conversion of glucose carbons to CO2. Therefore, palmitate carbons could provide more free energy than glucose carbons.

44. The complete oxidation of stearate to CO2 yields more energy because

3

4. Water has the highest melting point because each water molecule forms hydrogen bonds with four neighboring water molecules, and hydrogen bonds are among the strongest intermolecular forces. Ammonia is also capable of forming hydrogen bonds, but they are not as strong (due to the electronegativity difference between hydrogen and nitrogen). Methane cannot form hydrogen bonds; the molecules are attracted to their neighbors only via weak London dispersion forces.

5. The arrows point toward hydrogen acceptors and away from hydrogen donors.

O

CH2 O

17 of the 18 carbons of stearate are fully reduced. The conversion of these C C H3N CH N CH O CH3 carbons to CO2 provides more free energy than some of the carbons of CH 2 H Aspartame -linolenate, which participate in double bonds and are therefore already partially oxidized.

COO 45. The experiment was signifi cant because it demonstrated that it was posH O sible to synthesize the building blocks of biological macromolecules (amino H N acids, carbohydrates, and nucleic acids) using only inorganic gases as starting N O N materials and lightning as an energy source, that is, the conditions that most Uric acid O likely existed in the prebiotic world.

N H 46. The fi rst biological molecules would have had to polymerize, and they H O would have had to fi nd some way to make copies of themselves.

H2N 47. Morphological differences, which are useful for classifying large S NH2 organisms, are not useful for bacteria, which often look alike. Furthermore, O Sulfanilamide microscopic organisms do not leave an easily interpreted imprint in the fossil record, as vertebrates do. Thus, molecular information is often the only means for tracing the evolutionary history of bacteria.

6. Compound A does not form hydrogen bonds (the molecule has a hy48. It is diffi cult to envision how a single engulfment event could have given drogen bond acceptor but no hydrogen bond donor). Compounds B and rise to a stable and heritable association of the eukaryotic host and the bacte- C form hydrogen bonds as shown because each molecule contains at least rial dependent within a single generation. It is much more likely that natural one hydrogen bond donor and a hydrogen bond acceptor. The molecules in selection gradually promoted the interdependence of the cells. Over many D do not form hydrogen bonds with each other, whereas the molecules in E do because ammonia has a hydrogen bond donor and diethyl ether has a generations, genetic information supporting the association would have become widespread. 49. (a) H15 and H7 are closely related, as are H4 and hydrogen bond acceptor.

H14.

(b) H4 and H14 are most closely related to H3

50. A

B C

(B)N N H N N H

Chapter 2 1. The water molecule is not perfectly tetrahedral because the electrons in the nonbonding orbitals repel the electrons in the bonding orbitals more than the bonding electrons repel each other. The angle between the bonding orbitals is therefore slightly less than 109 .

2. Because the partial negative charges are arranged symmetrically (and the shape of the molecule is linear), the molecule as a whole is not polar.

(C) H3C

CH2 O H3C CH2 O

(E)

H N H H O H2C

H H CH2 CH3

H3C

7. (a) van der Waals forces (dipole–dipole interactions) (b) Hydrogen bonding (c) van der Waals forces (London dispersion forces) 3. Ammonia is polar because it has one unshared electron pair. Its shape is (d) Ionic interactions trigonal pyramidal, and the molecule is not symmetrical. Nitrogen is more elec-

O C

O

8. (a) H C N O F tronegative than hydrogen, so partial negative charges reside on the nitrogen and (b) The greater an atom’s electronegativity, the more polar its bond with partial positive charges on the hydrogens.

H and the greater its ability to act as a hydrogen bond acceptor. Thus, N, O, and F, which have relatively high electronegativity, can form hydrogen bonds, whereas C, whose electronegativity is only slightly greater N than hydrogen’s, cannot.

H

H H

9. From the highest melting point to the lowest melting point: C, B, E, A, D. Compound C (urea, melting point 133 C) has three functional groups that can serve as hydrogen donors and/or acceptors. Compound B (acetamide, melting

4

Ch 2 Solutions

point 80.16 C) has one less —NH group than Compound C and therefore 16. The waxed car is a hydrophobic surface. To minimize its interaction with forms fewer hydrogen bonds. Compound E (propionaldehyde, melting point the hydrophobic molecules (wax), each water drop minimizes its surface area 80 C) has one functional group that can serve as a hydrogen bond acceptor, by becoming a sphere (the geometrical shape with the lowest possible ratio of but it has no donors, so dipole–dipole forces are the strongest surface to volume). Water does not bead on glass, because the glass presents a intermolecular forces in a sample of this compound. Compound A (methyl hydrophilic surface with which the water molecules can interact. This allows ethyl ether, melt- ing point 113 C) also has one functional group that can the water to spread out. serve as a hydrogen bond acceptor (it has no donors), but it has hydrocarbon 17. Methanol, which has the highest dielectric constant, would be the best portions that interact with one another via London dispersion forces. solvent for the cationic NH 4. The polarity of the alcohols, which all contain Compound D (pentane, melting point 139.67 C) is nonpolar and experiences a primary OOH group, varies with the size of the hydrocarbon portion. only London dispersion forces, so it has the lowest melting point in the 1-Butanol, with the largest hydrophobic group, is the least polar and theregroup. fore has the highest dielectric constant.

10. Compound D is completely nonpolar and cannot form hydrogen bonds, 18. Isopropyl alcohol has intermolecular forces that are weaker and less so it will most likely be insoluble in water. Compounds E, B, and A can form numerous than those found in water. The isopropyl alcohol used to sponge hydrogen bonds with water, but these molecules also have hydrocarbon por- off sweat would evaporate more quickly from the body than water, since tions that will not interact with water favorably, so they will be only slightly less energy is required to convert the alcohol from the liquid to the gaseous soluble in water. Compound C forms many hydrogen bonds with water and phase. is likely to be very soluble in water.

19. (a) First, calculate the number of moles of protein using Avogadro’s 11. Aquatic organisms that live in the pond are able to survive the winter. number: Since the water at the bottom of the pond remains in the liquid form instead 1 mole 221 of freezing, the organisms are able to move around. The ice on top of the 1000 molecules3 6.0231023 molecules 5 3 pond also serves as an insulating layer from the cold winter air.

Next, calculate the volume of the cell, expressing r in centimeters: 12. Water is unique in that its liquid form is more dense than its solid form.

4p(531025 cm)3 4pr3 The weight of the skater puts pressure on the thin blade of the ice skate. The 55.2310213 cm3 3 volume5 3 5 ice melts under the blade because of this increased pressure. A higher Since 1 cm3 1 mL, the volume is 5.2 10 13 mL, or 5.2 10 16 L. pressure Therefore, the concentration of the protein is favors the liquid form of water over the solid form because the liquid form is more dense and takes up less volume.

1.66310221 moles 13. The positively charged ammonium ion is surrounded by a shell of water 53.331026 M, or 3.2 mM molecules that are oriented so that their partially negatively charged oxygen 5.2310216 L atoms interact with the positive charge on the ammonium ion. Similarly, the 531023 moles 6.0231023 molecules negatively charged sulfate ion is hydrated with water molecules oriented so (b) 3 L that the partially positively charged hydrogen atoms interact with the negamole 35.2310216 L51.63106 molecules tive charge on the sulfate anion. (Not shown in the diagram is the fact that the ammonium ions outnumber the sulfate ions by a 2:1 ratio. Also note that 20. (a) A 30X dilution is equivalent to multiplying one-tenth (10 1) by the exact number of water molecules shown isH unimportant.) itself 30 times: (10 1)30 10 30. The concentration would be 10 30 M. H (b) Use Avogadro’s number and multiply the concentration by the volO O ume to show that there is much less than one molecule present in 1 mL: H H (0.001 L)(10230 molesyL)(6.0231023 moleculesymole) O H H 56.02310210 molecules H H (c) The ability of water molecules to form a hydrogen-bonded coating, or O NH 4 O O O SO24 cage, around a solute molecule, particularly a hydrophobic one, might supH H H H O port the idea of water’s memory. However, water molecules are constantly H H in motion, so a group of water molecules that have been in contact with a H H solute do not retain an imprint of it.

H H O O 21. Compound A is amphiphilic and has a polar head and a nonpolar tail as indicated and can form a micelle (see Fig. 2-10). Compound B is H H nonpolar and cannot form a micelle or a bilayer. Compound C is polar (ionic) and cannot form a micelle or a bilayer. Compound D is amphiphilic and has a polar head and two nonpolar tails as indicated and can form a 14. Structure A depicts a polar compound, while structure B depicts an bilayer (see Fig. 2-11). Compound E is polar and forms neither a micelle ionic compound similar to a salt like sodium chloride. This is more consistent with glycine’s physical properties as a white crystalline solid with a high nor a bilayer. melting point. While structure A could be water soluble because of its ability to form hydrogen bonds, the high solubility of glycine in water is more CH3 consistent with an ionic compound whose positively and negatively charged 3 Polar head HC (C H2)11 N CH2COO (A) groups are hydrated in aqueous solution by water molecules.

15. (a) Surface tension is defi ned as the force that must be applied to surface CH3 molecules in a liquid so that they may experience the same forces as the Nonpolar tail molecules in the interior of the liquid. Water’s surface tension is greater than (D) ethanol’s because the strength and number of water’s intermolecular forces O (hydrogen bonds) are both greater. Ethanol’s OOH group also forms hyC CH2 (CH2)11 CH3 O Nonpolar drogen bonds, but the hydrocarbon portion of the molecule cannot interact tails favorably with water, and weaker London dispersion forces form instead.

(CH2)11 CH3 HC O C (b) The kinetic energy of the water molecules increases when temperaO HO CH2 ture increases. Intermolecular forces are weaker in strength as a consePolar head quence of the increased molecular motion. Because surface tension increases when the strength of intermolecular forces increases, as described 22. (a) In the nonpolar solvent, AOT’s polar head group faces the interior of in part (a), surface tension decreases when temperature increases. the micelle, and its nonpolar tails face the solvent.

Ch 2 Solutions CH2CH3 Nonpolar tails

H3C

(CH2)3

H3C

(CH2)3

CH

O

CH2 O

C

CH2O

O

C

CH

CH CH2

O

CH2CH3

S

Polar head

O

O AOT

(b) The protein, which contains numerous polar groups, interacts with the polar AOT groups in the micelle interior.

Isooctane

Water

Protein

23.

O

(a) Nonpolar tail

H3C (CH)211

O

ONa

S

Polar head

O O Na O O

S

O Na S O

CH 3 CH3 H3C

O Na O

O 21 ) O 1

)11 H2 (C O

O

O

(C H

S

(b)

S (CH)O211

Na O

O(H2C) 11

CH3

O

H3C

O 211 (CH)O

S O

O (C H S

Na O

)11O H2 (C

O

(CH)O211

21) 1

CH3 CHHC 3 3

O O

S

O O

O Na

O S

O Na

ONa

5

26. In reverse osmosis, water moves from an area of low concentration (high solute concentration) to an area of high concentration (low solute concentration). This movement is opposite that described for osmosis in Problem 25.

This is a nonspontaneous process that requires an input of energy in order to proceed, unlike osmosis, which occurs spontaneously without input of energy. (b)CO2 Glucose is polarand andwould wouldbe not be to able to pass through a bilayer be27. (a) is nonpolar able cross a bilayer. cause the presence of the hydroxyl groups means glucose is highly hydrated and would not be able to pass through the nonpolar tails of the molecules forming the bilayer. (c) DNP is nonpolar and would be able to cross a bilayer. (d) Calcium ions are charged and are, like glucose, highly hydrated and would not be able to cross a lipid bilayer.

28. Vesicles consist of a lipid bilayer that closes up to enclose an aqueous compartment. The polar drug readily dissolves in this aqueous compartment.

Delivery to the cell is accomplished when the vesicle membrane fuses with the cell membrane, releasing the drug into the cytosol.

29. Substances present at high concentration move to an area of low concentration spontaneously, or “down” a concentration gradient in a process that increases their entropy. The export of Na ions out of the cell requires that the sodium ions be transported from an area of low concentration to an area of high concentration. The same is true for potassium transport. Thus, these processes are not spontaneous, and an input of cellular energy is required to accomplish the transport.

30. The amount of Na (atomic weight 23 g mol 1) lost in 15 minutes, assuming a fl uid loss rate of 2 L per hour and a sweat Na concentration of 50 mM, is 23 g 1000 mg Na 1 2L 0.05 mol mol h 3 L 0.25 h3 3 3 g Na1 1 oz chips 3 200 mg Na1 52.9 oz chips It would take 2.9 ounces of potato chips (about a handful) to replace the lost sodium ions.

31. In a high-solute medium, the cytoplasm loses water; therefore its volume decreases. In a low-solute medium, the cytoplasm gains water and therefore its volume increases.

32. E. coli accumulates water when grown in a low-osmolarity medium.

However, regulation of water content only would cause a large increase in cytoplasmic volume. To avoid this large increase in volume, E. coli also exports K ions. The opposite occurs when E. coli is grown in a high-osmolarity medium—the cytoplasmic water content is decreased, but cytoplasmic osmolarity increases as E. coli imports K ions. [From Record, M. T., et al., Trends Biochem. Sci. 23, 143–148 (1998).]

(c) The hydrophobic grease can move into the hydrophobic core of the 33. Since the molecular mass of H 2O is 18.0 g mol 1 , a given volume (for ? water-soluble soap micelle. The “dissolved” grease can then be washed example, 1 L or 1000 g) has a molar concentration of 1000 g ? L 1 18.0 g ? away with the micelle. mol 1 55.5 M. By defi nition, a liter of water at pH 7.0 has a hydrogen 24. (a) It is doubtful that the contents of the ball could infl uence the behav- ion concentration of 1.0 10 7 M. Therefore, the ratio of [H 2O] to [H] is 55.5 M/(1.0 10 7 M) 5.55 108. ior of external water molecules separated by layers of rubber and plastic. (b) Even if water clusters were disrupted (which they are not), the re34. [H1]3[OH2]51.47310214 and [H1]5[OH2] moval of dirt requires more than one individual water molecule. In [H1]521.47310214 order for a dirt molecule to be washed away, it must be surrounded [H1]51.2131027 M (solubilized) by many water molecules. (c) Hot water, because of the higher energy of its water molecules, has pH52log(1.2131027 M) intrinsically better dirt-solubilizing power than cold water, regardless of pH56.92 the presence or absence of detergent. In the absence of detergent, hot 35.

The HCl is a strong acid and dissociates completely. This means that the water has signifi cant cleaning power on its own, which could be attribconcentration of hydrogen ions contributed by the HCl is 1.0 10 9 M. But uted to the presence of a laundry ball. the concentration of the hydrogen ions contributed by the dissociation of 25. (a) The nonpolar core of the lipid bilayer helps prevent the passage of water is 100-fold greater than this: 1.0 10 7 M. The concentration of the water since the polar water molecules cannot easily penetrate the hydro- hydrogen ions contributed by the HCl is negligible in comparison. Therephobic core of the bilayer. (b) Most human cells are surrounded by a fl fore, the pH of the solution is equal to 7.0. uid containing about 150 mM Na and slightly less Cl (see Fig. 2-13). A 36. The pH of this solution is 7.0 (see Solution 35). The concentration of solution containing 150 mM NaCl mimics the extracellular fl uid and hydroxide ions contributed by the dissociation of water is 100-fold greater therefore helps main- tain the isolated cells in near-normal conditions. If than that contributed by the dissociation of the dilute NaOH. the cells were placed in pure water, water would tend to enter the cells 37. In aqueous solution, where virtually all biochemical reactions take place, by osmosis; this might cause the cells to burst. an extremely strong acid such as HCl dissociates completely, so that all its

6

Ch 2 Solutions

protons are donated to water: HCl H2O S HO 3 H 3O as the only acidic species remaining.

38.

Cl . This leaves

−

[H1]52.2310212 M pH52log[H1] pH52log(2.2310212) pH511.6 43. The stomach contents have a low pH due to the contribution of gastric juice (pH 1.5–3.0). When the partially digested material enters the small intestine, the addition of pancreatic juice (pH 7.8–8.0) neutralizes the acid and increases the pH. 44. The carbonate ions accept protons from water and form hydroxide ions (as shown in the equation below), resulting in basic urine.

2 2 CO223(aq)1HO 2(l )SHCO3(aq)1OH(aq)

39. Since pH log[H ], [H ] 10 pH For saliva, [H ] 10 6.6 2.5 10 7 M For urine, [H ] 10 5.5 3.2 10 6 M

45. (a) CO22 (b) SO22 (c)3HPO22 (d) CO22

A B C D

3

(e) AsO32 (f ) PO32 (g) O22

40.

Acid, base, or neutral?

pH

[H1] (M)

[OH2] (M)

acid base neutral acid

5.60 7.65 7.00 2.68

2.5 3 1026 2.2 3 1028 1.0 3 1027 2.1 3 1023

4.0 3 1029 4.5 3 1027 1.0 3 1027 4.8 3 10212

46. (a) H2C2O4 (b) H2SO3 (c) H3PO4 (d) H2CO3 (f) 2HPO 4 (g) H2O2 2 (e) HAsO 4 47.

pH52log[H1] pH52log(0.0045) pH52.3 (b) Since NaOH dissociates completely, the added [OH ] is equal to the [NaOH]. The fi nal concentration of NaOH is (0.0015 L)(3.0 M)/ 1 L 0.0045 M. (The existing hydroxide ion concentration in the water itself, 1.0 10 7 M, can be ignored because it is much smaller than the hydroxide ion concentration contributed by the NaOH.) 214

1 5[H][OH] K w51.0310 1.0 3 10 214 [H1 ]5 [OH2] 214

[H1 ]5

1.0 3 10 (0.0045 M)

H Piperidine O

O

OH

H2N CH C

H

HO

CH2

O

C C OH Oxalic acid O

CH2 CH2

(b) The fi nal concentration of KOH is (0.015 L)( 1.0 M) 50.029 M 0.515 L Since KOH dissociates completely, the added [OH ] is equal to the [KOH]. (The existing hydroxide ion concentration in the water itself, 1.0 10 7 M, can be ignored because it is much smaller than the hy- droxide ion concentration contributed by the KOH.) 1 2 5[H][OH] w51.0310214 K 214 10 1.0 3 [H1 ]5 [OH2]

is equal to [HCl]. The fi nal concentration of HCl is (0.0015 L)(3.0 M)/ 1 L 0.0045 M. (The existing hydrogen ion concentration in the water itself, 1.0 10 7 M, can be ignored because it is much smaller than the hydrogen ion concentration contributed by the hydrochloric acid.)

H N

CH2 COOH Citric acid

(0.020 L)( 1.0 M) 50.038 M 0.520 L 3Since is a strong HNO acid and dissociates completely, the added [H] is equal to [HNO3]. (The existing hydrogen ion concentration in the water itself, 1.0 10 7 M, can be ignored because it is much smaller than the hydrogen ion concentration contributed by the nitric acid.) pH52log[H1] pH52log(0.038) pH51.4

1.0 3 10 214 (0.029 M) [H1]53.4310213 M pH52log[H1] pH52log(3.4310213) pH512.5 42. (a) Since HCl is a strong acid and dissociates completely, the added [H ]

COOH

C

41. (a) The fi nal concentration of HNO3 is

[H1 ]5

COOH

CH 2 HO

NH

CH2

O N H Barbituric acid

O

H NH 2 Lysine O O

NH

CH 2

CH2

S

O

O 4-Morphine ethanesulfonic acid (MES)

48. (a) CH3 C

O

COO

Pyruvate

49.

(b) The structure of pyruvate will predominate in the cell at pH 7.4.

The pK values for carboxylic acid groups are typically in the 2–3 range; therefore, the carboxylate group will be unprotonated at physiological pH. O H3N

CH

C

O

CH2 CH2 CH2 CH2 NH 3

2

50. Convert all the data to either Ka or pK values to evaluate (pK log Ka).

The greater the Ka value, the stronger the acid—that is, the greater the tendency for the proton to be donated. (The lower the pK value, the stronger the acid.) From strongest to weakest acid: E, D, B, A, C. Note that the stronger the acid, the weaker its conjugate base. For example, citric

Ch 2 Solutions acid is a stronger acid than citrate, and succinic acid is a stronger acid than succinate.

Acid citrate succinic acid succinate formic acid citric acid

A B C D E

Ka

pK

1.74 3 1025 6.17 3 1025 2.29 3 1026 1.78 3 1024 7.41 3 1024

4.76 4.21 5.64 3.75 3.13

51. The pK of the fl uorinated compound would be lower (it is 9.0); that is, the compound becomes less basic and more acidic. This occurs because the F atom, which is highly electronegative, pulls on the nitrogen’s electrons, loosening its hold on the proton.

52. pH 2 1H3NOCH2OCOOH pH 7 1H3NOCH2OCOO2 pH 10 H2NOCH2OCOO2

7

(b) The dissociation of the second proton has a pK of 6.82, which is closest to the pH of blood. Therefore, the weak acid present in blood is H 2PO4and the weak acid is HPO4.2 (c) The dissociation of the third proton has a pK of 12.38. Therefore, a buffer solution at pH 11 would consist of the weak acid HPO224 and its conjugate base, PO324 (supplied as the sodium salts Na2HPO4 and Na3PO4).

58. The aspirin is more likely to be absorbed in the stomach at pH 2. At this pH, the carboxylate group is mostly protonated and uncharged. This allows the aspirin to pass more easily through the nonpolar lipid bilayer. At the pH of the small intestine, the carboxylate group is mostly in the ionized form and will be negatively charged. Charged species are more polar than uncharged species (and are likely to be hydrated) and will have diffi culty traversing a lipid bilayer.

59. Calculate the fi nal concentrations of the weak acid (H 2PO4) 2 and conjugate base (HPO22). Note that K is a spectator ion.

(0.025 L)(2.0 M) The carboxylic acid group has a pK of 2.35, and the amino group has a pK 50.25 M [HPO2 4]5 2 0.200 L of 9.78. The Henderson–Hasselbalch equation can be used to calculate the exact percentage of protonated/unprotonated forms of each functional group, (0.050 L)( 2.0 M) 22 but that really isn’t necessary. Instead, the pK values for each group should be 50.50 M [HPO4]5 0.200 L compared to the pH. At pH 2, the pH is below both pK values, so both functional groups are mostly protonated. At pH 7, the pH is well above Next, substitute these values into the Henderson–Hasselbalch equation the pK for the carboxylic acid group but below the pK for the amino group. using the pK values in Table 2-4: Therefore the carboxylic acid group is unprotonated and the amino group is [A2] protonated. At pH 10, the pH is above the pK values of both functional pH5pK1log groups. Thus, both groups are mostly unprotonated.

[HA] pH56.821log(0.50 M)y(0.25 M) 53. (a) 10 mM glycinamide buffer because its pK is closer to the desired pH. pH56.8210.30 pH57.12 (b) 20 mM Tris buffer because the higher the concentration of the buffer60.

Use the pK value in Table 2-4 and the Henderson–Hasselbalch equation ing species, the more acid or base it can neutralize. (c) Neither; each solution will contain an equilibrium mixture of the to calculate the ratio of imidazole (A ) and the imidazolium ion (HA): [A2] pH5pK1log a boric acid and its conjugate base (borate).

[HA] b 54. (a) 10 mM acetic acid buffer because its pK is closer to the desired pH. (b) 20 mM acetic acid buffer because the higher concentration of buffer loga[A2] b5pH2pK [HA] species will allow it to neutralize a greater amount of acid or base. (c) Neither. Both a weak acid and a conjugate base are required buf[A2] 510(pH2pK ) fer constituents. The weak acid alone (acetic acid) or the conjugate base [HA] alone (sodium acetate) cannot serve as an effective buffer.

[A2] 510(7.427.0) 55. Because it is small and nonpolar (see Solution 2), CO2 can quickly dif[HA] fuse across cell membranes to exit the tissues and enter red blood cells.

[A2] 2.5 5 56. (a) H2CO3(aq)SH1(aq)1HCO23(aq) 1 [HA] 22 HCO2(aq)SH1(aq)1 CO3(aq) 3 61. First, determine the ratio of [A ] to [HA]: (b) The pK of the fi rst dissociation is closer to the pH; therefore the [A2] weak acid present in blood is H2CO3 and the conjugate base is HCO3. pH5pK1log [HA] 57. (a) The three ionizable protons of phosphoric acid have pK values of [A2] 2.15, 6.82, and 12.38 (Table 2-4). The pK values are the midpoints of 5pH2pK log the titration curve.

[HA] [A2]

pK3 Midpoint three 2 [PO4] 3 [HPO4]

510(pH2pK ) [HA] Substitute the values for the desired pH (5.0) and the pK (4.76):

14 12 10 8 pH

[PO34]

pK2 Midpoint two 2 [H2PO4] [HPO4 ]

Calculate the number of moles of acetate (A ) already present: (0.50 L)(0.20 mol?L21)50.10 moles acetate Calculate the moles of acetic acid needed, based on the calculated ratio:

6 4 2

pK1 Midpoint one 4] [H2PO4] [HPO 3

[HPO24]

[A2]

51.74 [HA] [HA] 5 0.10 moles

0 [HP 2O4]

[H3PO4]

0.5

1.0 1.5 2.0 H+ ions dissociated

[A2] [HA] 510(5.024.76)5100.2451.74

2.5

3.0

1.74 [HA]50.057 moles

8

Ch 2 Solutions

Finally, calculate the volume of glacial acetic acid needed: 0.057 moles 17.4 mol?L152

64. (a) HOHC 2

0.0033 L, or 3.3 mL

than 1%, which doesn’t introduce signifi cant error.

62. Adding NaOH to the acetic acid will convert some of the acetic acid (HA) to acetate (A ): 1H2O

3

For every mole of NaOH added, one mole of CH3COOH will be consumed and one mole of CHCOO will be generated. If x is the number of moles 3 of NaOH added, then x will also be the number of moles of A generated.

The initial amount of acetic acid is 0.10 mol, so the fi nal amount of acetic acid will be 0.10 mol x.

63. (a)

NH3

HOH2C

HO

NH

N

Weak acid

Conjugate base

(b) The pK of Tris is 8.30; therefore, its effective buffering range is 7.30–9.30. (c) Rearranging the Henderson–Hasselbalch equation gives [A2] 510(pH2pK )510(8.228.3)51020.150.79 [HA] Since [A2]1[HA]50.10 M, [A2]50.10 M2[HA], and (0.10[HA] M2[HA]) 5 0.79 0.79[HA]50.10 M2[HA]

(d) When HCl is added, an equivalent amount of Tris base (A ) is converted to Tris acid (HA). Let x moles of H added (0.0015 L) (3.0 mol ? L 1) 0.0045 moles 4.5 mmol.

The fi nal amount of A is 44 mmol 4.5 mmol 39.5 mmol.

The fi nal amount of HA is 56 mmol 4.5 mmol 60.5 mmol.

Use the Henderson–Hasselbalch equation to calculate the new pH:

1.58 g

(CH2)2

[A2 ]b [HA]

SO3 H2O

Weak acid (HA) HO

(H 2C)2

N

pH 5 8.3 1 N

(CH2)2

SO3

H3O

Conjugate base (A)

(b) The pK for HEPES is 7.55; therefore, its effective buffering range is 6.55–8.55. (c) 1.0 L3 260.3 g 0.10 mole 3 mol 526 g L Weigh 26 g of the HEPES salt and add to a beaker. Dissolve in slightly less than 1.0 liter of water (leave “room” for the HCl solution that will be added in the next step). (d) At the fi nal pH, [A2] 510(pH2pK )510(8.027.55)5100.4552.82 [HA] For each mole of HCl added, x, one mole of HEPES salt (A ) will be converted to a mole of HEPES acid (HA). The starting amount of A is (1.0 L)(0.10 mol ? L 1) 0.10 mole. After the HCl is added, the amount of A will be 0.10 mole x, and the amount of HA will be x.

Consequently, [A2] 52.825 [HA]

0.10 mole2x

x 2.82x50.10 mol2x 3.82x50.10 mol x50.10 moly3.8250.0262 mol Calculate how much 6.0 M HCl to add: 0.0262 mol ?

21

NH2

CH2OH

pH5pK1log a 2 (HC) 2

C

1.79[HA]50.10 M 0.10 M 50.056 M556 mM [HA]5 1.79 [A2]1[HA]50.10 M5100 mM, so [A2]544 mM

[A2] x 51.745 [HA] 0.10 mol2x x51.74(0.10 mol2x)50.174 mol21.74x 2.74x50.174 mol x50.174 moly2.7450.0635 mol Calculate the mass of NaOH to add: 0.0635 mol 0 g?mol152 4

C

CH2OH

CH2OH

The addition of 3.3 mL to a 500-mL solution dilutes the solution by less

NaOH1CH3COOHΔNa11CHCOO2

CH2OH

50.0044 L, or 4.4 mL

log a 39.5 mmol?1001.5 mL 60.5 mmol?1001.5 mL21

21

b

pH58.31(20.2) pH58.1 The buffer has been effective: The pH has declined about 0.1 unit (from pH 8.2 to pH 8.1) with the addition of the strong acid. In comparison, the addition of the same amount of acid to water, which is not buffered, resulted in a pH change from approximately 7.0 to 2.35 (see Problem 42a). (e) When NaOH is added, an equivalent amount of Tris acid (HA) is converted to Tris base (A ). Let x moles of OH added (0.0015 L) (3.0 mol ? L 1) 0.0045 moles 4.5 mmol.

The fi nal amount of A is 44 mmol 4.5 mmol 48.5 mmol.

The fi nal amount of HA is 56 mmol 4.5 mmol 51.5 mmol.

Use the Henderson–Hasselbalch equation to calculate the new pH: [A2 ]b [HA]

pH5pK1log a

pH 5 8.3 1

log a 48.5 mmol?1001.5 mL 51.5 mmol?1001.5 mL21

21

b

pH58.31(20.026) pH58.27 The buffer has been effective: The pH has increased only 0.07 unit (from pH 8.2 to pH 8.27) with the addition of the strong base. In comparison, the addition of the same amount of base to water, which is not buffered, resulted in a pH change from approximately 7.0 to 11.6 (see Problem 42b).

65. (a) First, calculate the ratio of [A ] to [HA]. Rearranging the Henderson–Hasselbalch equation gives [A2] [HA]

510(pH2pK )510(2.028.3)51026.35531027

To make the buffer, dissolve 26 g of HEPES salt [see part (c)] in less

Virtually all of the Tris is in the weak acid form. Therefore, the concen-

than 1.0 L. Add 4.4 mL of 6.0 M HCl, then add water to bring the fi nal volume to 1.0 L.

tration of the weak acid, HA, is 0.10 M and the concentration of the conjugate base, A , is 5.0 10 8 M.

Ch 3 Solutions 9 (b) The added HCl dissociates completely, so the amount of H added is Carbonic acid and bicarbonate ions are in equilibrium, as represented by the (0.0015 L)(3.0 mol ? L 1) 0.0045 mol. In an effective buffer, the acid following equation: would convert some of the conjugate Tris base to weak acid. But the H2COΔH11HCO2 3 3 concentration of conjugate base is already negligible. Therefore, the moles of additional H should be added to the concentration of hydro- Phosphate ions are in equilibrium, according to the following equation: gen ions already present (1.0 10 2 M), for a total concentration of 0.0145 H PO2 Δ H1 1 HPO22 M. In metabolic acidosis, the concentration of protons increases, so the equilib4, carbonic acid and ammonium ions. In order to rium shifts to form HPO 2 bring the pH back to normal, the kidney will excrete HPO 4 and ammo2 The buffer has not functioned effectively. There was not enough conju- nium ions and bicarbonate ions will be reabsorbed. The result is a decrease in gate base to react with the additional hydrogen ions added. The result is the concentration of protons and an increase in blood pH. a decrease in pH from 2.0 to 1.84.

68. The relevant equations are shown in Solution 67. In metabolic alka(c) When NaOH is added, an equivalent amount of Tris acid (HA) is losis there is an excess of hydroxide ions, which react with protons to form converted to Tris base (A ). Let x moles of OH added (0.0015 L) water. This causes the equilibria to shift to form HPO24 , NH and HCO 3.

3 (3.0 mol ? L 1) 0.0045 moles 4.5 mmol.

In order to bring the pH back to normal, the kidney reabsorbs NH 4 and The fi nal amount of A is 5.0 10 8 mol 4.5 mmol 4.5 mmol.

HPO 2 4and excretes HCO3.

The fi nal amount of HA is 100 mmol 4.5 mmol 95.5 mmol.

The new pH is determined by substituting the new concentrations of H and HA into the Henderson–Hasselbalch equation: Chapter 3 [A2] 1. The heat treatment destroys the polysaccharide capsule of the wild-type pH5pK1log Pneumococcus, but the DNA survives the heat treatment. The DNA then [HA] “invades” the mutant Pneumococcus and supplies the genes encoding the en(4.5 mmol) zymes needed for the capsule synthetic pathway that the mutant lacks. The pH58.31log (95.5 mmol) mutant is now able to synthesize a capsule and has the capacity to cause pH52log[H1

]5log (0.0145 M)51.84

pH 5 8.3 1 (21.3) 5 7.0

disease, which results in the death of the mice and the appearance of encapsulated Pneumococcus in the mouse tissue.

Tris is not an effective buffer at pH 2.0, a pH more than 6 units lower 2. These experiments showed that the transforming factor was neither a than its pK value. Virtually all of the Tris is in the weak acid form at this protein nor RNA. pH. If acid is added, there is not enough base to absorb the excess added hydrogen ions, and the pH decreases. If base is added, some of the weak 3. Some of the labeled “parent” DNA appears in the progeny, but none of acid is converted to the conjugate base and the pH approaches the value the labeled protein appears in the progeny. This indicates that the bacteriophage DNA is involved in the production of progeny bacteriophages, but of the pK. bacteriophage protein is not required.

66. The ratio of bicarbonate to carbonic acid in the patient’s blood can be 4. The triple-helical model is not consistent with the hydrophobic efdetermined using the Henderson–Hasselbalch equation: fect, which suggests that the nonpolar nitrogenous bases would reside in [A2] pH5pK1log a the center of the DNA structure and the hydrophilic phosphates would b reside on the surface. The triple-helical model also assumes that the phos[HA] phate groups are protonated and form stabilizing hydrogen bonds in the [HCO23] 7.5556.351log DNA interior. But the pK value for phosphate is well below 7, so the [H2CO3] phosphate groups would not be protonated at physiological pH. In the 2 H [CO3] absence of hydrogen bonds, there are no additional forces that would 101.2 5 [H2CO3] hold the strands of the triple helix together.

15.8 [HCO23] 1 5.

NH 2 [H2CO3] 5 Similarly, the ratio of bicarbonate to carbonic acid in a normal person’s blood can be determined: [A2] pH5pK1log a b [HA] 7.456.351log a[HCO]b

[HCO23] [H2CO3]

5

N

H 5-Methylcytosine

6. (a)

H

[HCO23] [H2CO3] 11.2

O

N

2 3

[H2CO3] 101.055

H3C

CH3 N N

N N

N

1 H N 6-Methyladenine In order to serve as an effective buffer (i.e., absorb both added H and OH ), both a conjugate base and a weak acid must be present. In the patient, the (b) The N 6-DNA methyltransferase might be a good drug target. If ratio of conjugate base to weak acid does not lie within an effective buffering methylation of certain adenine residues is required for virulence, then it range. The bicarbonate concentration (conjugate base) is too high relative is possible that inhibition of the bacterial transferase enzyme might to the carbonic acid (weak acid) concentration; thus the relative amount of prevent adenine methylation and thus prevent disease caused by weak acid is insuffi cient. pathogenic bacteria.

67. Ammonia and ammonium ions are in equilibrium, as represented by the 7. The base, 5-chlorouracil, is a substitute for thymine (5-methyluracil). following equation: 8. A chlorine is substituted for a hydrogen in 5-chlorouracil, which closely 1NH3 resembles thymine (see Solution 7). Therefore, to real chlorouracil the culture NH1ΔH1 4

10

Ch 3 Solutions

containing 5-chlorouracil will incorporate this base in place of thymine as 16. The genome contains 28.7% T, 21.5% G, 29.2% A, and 20.6% C. the DNA replicates. The 5-chlorouracil has a greater mass than thymine, so Chargaff’s rules do not apply because the viral genome is composed of singleDNA isolated from this culture will have a greater mass than DNA isolated stranded DNA. from the control culture.

17. It is a G:C base pair.

9. Thymine (5-methyluracil) contains a methyl group attached to C5 of

18.

the pyrimidine ring of uracil.

10. (a) A diphosphate bridge links the ribose groups in each dinucleotide.

N H

N

This linkage is a variation of the monophosphate bridge (phosphodiester linkage) in DNA and RNA. (b) The adenosine group in CoA bears a phosphoryl group on C3 .

H

N

O

P

O H

O

O

N

O

NH2

O

O CH2 H

N N H

O

H

H

OH

OH

N N

If the dinucleotide were DNA, it would lack OH groups at each ribose C2 position.

O

H2 C OH

O P O

H

H

O

OH

N

NH

H

N

N Hypoxanthine

N

H

O

O

H

H Uracil

OH

P

H

N N

H

13. The total amount of purines (A G) in DNA must equal the total amount of pyrimidines (C T) because each base pair in the double-stranded DNA 23. molecule consists of a purine and a pyrimidine. This is not true for RNA, which is single-stranded.

14. The organism must also contain 19% A (since [A] [T] according to Chargaff’s rules) and 62% C G (or 31% C and 31% G, since [C] [G]). Each cell is a diploid, containing 60,000 kb, or 6 107 bases.

Therefore, [A] [T] (0.19)(6 107 bases) 1.14 107 bases [C] [G] (0.31)(6 107 bases) 1.86 107 bases 15. (a) Using Chargaff’s rules (see Solution 14), the number of C residues must also be 24,182. Subtracting 2 3 24,182) from 97,004 yields 48,640 (A T) residues. Dividing this number by 2 yields 24,320 residues each of A and T. (b) GenBank reports only the sequence of a single strand of DNA, since the sequence of the complementary strand can easily be deduced using Chargaff’s rules.

NH N

N Hypoxanthine

19. The statement is false because the greater stability of GC-rich DNA is due to the stronger stacking interactions involving G:C base pairs and does not depend on the number of hydrogen bonds in the base pairs.

20. It is certainly the case that hydrogen bonds hold A:T and G:C base pairs together and that these interactions are very favorable. But upon denaturation of the DNA, each nitrogenous base has the opportunity to form equally favorable hydrogen bonds with water. Therefore, forces other than hydrogen bonds must contribute to the overall stability of the DNA molecule.

21. The sugar–phosphate backbone is found on the outside of the molecule.

The polar sugar molecules can form hydrogen bonds with the surrounding water molecules. The negatively charged phosphate groups interact favorably with positively charged ions. The nonpolar nitrogen bases are found on the inside of the molecule and interact favorably via stacking interactions. In this way, contact with the aqueous solution is minimized, as described by the hydrophobic effect.

22. (a) Proteins are more likely to bind to the major groove, which can easily accommodate proteins. The larger surface area of the major groove allows multiple favorable interactions between the DNA and protein. (b) The positively charged side chains of the Lys and Arg residues form ion pairs with the negatively charged phosphate groups on the DNA backbone. These are strong interactions, so the histones have a high affi nity for DNA. 1.4 Relative absorbance at 260 nm

H

N

N

O

O

N H

H

O

N H Adenine

N

CH2

12.

Hypoxanthine

NH2

O

Phosphodiester bond

N

N

H Cytosine

O

H

N

O

N

11.

NH

O

H

1.3 D. discoideum

S. albus

1.2 Tm

1.1

1.0

30

Tm

50 70 90 Temperature (°C)

24. The DNA contains 50% G approximately 90 C.

C, so its melting point would be

Ch 3 Solutions 11 31. (a) TGTGGTACCACGTAGACTGA (b) ACACCAUGGUGCAUCUGACU

110

32. (a) The top strand is the coding strand and the bottom strand is the noncoding strand. (b) Only coding strands are published because the mRNA sequence is identical to the sequence of the coding strand, with the exception that U replaces T in the mRNA.

m

T (°C)

100 90

33. (a) A poly(Phe) polypeptide was produced. (b) Poly(A) produces poly(Lys); poly(C) yields poly(Pro); and poly(G) yields poly(Gly).

80

34. (a) The cell-free system produces polypeptides consisting of alternating Val and Cys residues. Some of the polypeptides begin with Val; others GC content (%) begin with Cys, depending on which reading frame is used. (b) The production of this peptide does not allow one to unambiguously 25. The DNA from the organisms that thrive in hot environments would assign the GUG and UGU codons. Additional experimental data are recontain more G and C than DNA from species living in a more temperate quired to make this assignment. environment. The higher GC content increases the stability of DNA at high 35. temperatures.

First reading frame: 26. The positively charged sodium ions can form ion pairs with the negaAGG TCT TCA GGG AAT GCC TGG CGA GAG GGG AGC AGC tively charged phosphate groups on the DNA backbone and “shield” the negative charges from one another. This increases the overall stability of Ser-Ser-Ser-Gly-Asn-Ala-Trp-Arg-Glu-Gly-Ser-SerDNA and makes it more diffi cult to melt.

TGG TAT CGC TGG GCC CAA AGG C 27. You should increase the temperature to melt out imperfect matches Trp-Tyr-Arg-Trp-Ala-Gln-Arg between the probe and the DNA. Second reading frame: 28. Heating denatures the target DNA (separates its two strands) so that A GGT CTT CAG GGA ATG CCT GGC GAG AGG GGA GCA a single-stranded probe can more easily form sequence-specifi c hydrogen Gly-Leu-Gln-Gly-Met-Pro-Gly-Glu-Arg-Gly-Alabonds with it.

GCT GGT ATC GCT GGG CCC AAA GGC 29. (a) An inherited characteristic could be determined by more than one Ala-Gly-Ile-Ala-Gly-Pro-Lys-Gly gene. (b) Some sequences of DNA encode RNA molecules that are not trans- Third reading frame: lated into protein (for example, rRNA and tRNA).

AG GTC TTC AGG GAA TGC CTG GCG AGA GGG GAG CAG (c) Some genes are not transcribed during a cell’s lifetime. This can ocVal-Phe-Arg-Glu-Cys-Leu-Ala-Arg-Gly-Glu-Glncur if the gene is expressed only under certain environmental conditions CTG GTA TCG CTG GGC CCA AAG GC or in certain specialized cells in a multicellular organism.

Leu-Val-Ser-Leu-Gly-Pro-Lys 30. The DNA isolated after one generation is a homogeneous sample of 70

0

10

20

30

40

50

60

70

80

DNA with a density intermediate between DNA containing all 14N and all 15 N. The DNA isolated after the second generation is heterogeneous. Half of the DNA has the same density as the fi rst generation; half of the DNA consists of all 14N DNA and has a lower density.

15N 14N

First-generation daughter molecules

36. The second reading frame, which produces a protein in which every third amino acid is Gly, is the correct reading frame.

37. Asparagine has two codons, AAU and AAC (see Table 3-3). An A S

G mutation at the second position could generate a codon for serine (AGU or AGC).

38. The CGA codon codes for the amino acid arginine; the mutation (C S T in the DNA) converts the codon to a stop codon. When the mRNA for the gene is translated, the protein terminates prematurely and is nonfunctional.

39. The genetic code (shown in Table 3-3) is redundant. Since there are 64 different possibilities for 3-base codons and only 20 amino acids, some amino acids have more than one codon. If a mutation just happens to occur in the third position (3 end), the mutation might not alter the protein sequence. For example, GUU, GUC, GUA, and GUG all code for valine.

A mutation in the third position of a valine codon would still result in the selection of valine and would have no effect on the amino acid sequence of the protein.

40. The same segment of DNA can encode two different proteins if each strand is a coding strand.

41. First, identify the translation start site, the Met residue whose codon is AUG in the mRNA (see Table 3-3) or ATG in the DNA. Translation stops at the DNA sequence TAA, which corresponds to the stop codon UAA in the mRNA. Use Table 3-3 to decode the intervening codons, substituting U for T.

Second-generation daughter molecules

CTCAGAGTTCACC ATG GGC TCC ATC GGT GCA GCA AGC ATG GAA Met Gly Ser lle Gly Ala Ala Ser Met Glu …1104 bp … UUC UUU GGC AGA UGU GUU UCC CCU UAA AAAGAA

……………. Phe Phe Gly Arg Cys Val Ser Pro *

12

Ch 3 Solutions

42. (a) The siRNA is an “antisense” mRNA, so its sequence should be complementary to that of the mRNA. The solution below shows an siRNA that corresponds to the 5 end of the mRNA: mRNA 5 AUG GGC UCC AUC GGA GCA GCA AGC AUG GAA 3 siRNA 3 UAC CCG AGG UAG CCA CGU CGU UCG UAC CUU 5

52. Chromosomes 10, 11, and 13 have genes carrying SNPs that are cor- related with the colon disease [From Garcia-Barcelo, M., et al., Proc. Natl. Acad. Sci. 106, 2694–2699 (2009).]

53. Polymerization occurs in the 5 S 3 direction and a 3 OH group must be available, so the primer must be complementary to the sequence as shown.

5 -AGTCGATCCCTGATCGTACGCTACGGTAACGT-3 (b) When the antisense RNA binds to the mRNA, the mRNA can no longer serve as a template for protein synthesis. (In the cell, the 3 -TGCCATTGCA-5 ATTGTTCCCACAGACCG binding of antisense RNA to mRNA involves an enzyme that degrades 54.

CGGCGAAGCATTGTTCC ACCGTGTTTCCGACCG the mRNA into smaller fragments.) (c) The siRNA molecules must be able to cross the cell membrane to enter the cell and fi nd the target TTGTTCCCACAGACCGTG 55. A restriction endonuclease is often used to prepare fragments of DNA mRNA. for insertion into a cloning vector. Since the cloned DNA contains the recog43. C. ruddii, with such a small genome and only 182 genes, must be some nition site (whose sequence is known), this sequence can be used as a starting sort of parasite rather than a free-living bacterium. (In fact, C. ruddii is an point to sequence the unknown DNA segment. 56. The polymerization insect symbiont.) reaction must be carried out at high temperatures 44. Prokaryotes tend to have smaller genomes than eukaryotes, so evolution (hence the need for a heat-stable DNA polymerase) in order to ensure that has shaped the prokaryote genome to pack in genes more effi ciently. the Eukaryotes, with larger genomes and more noncoding DNA, have more space template DNA remains an unknotted single strand. The high temperature is to arrange genes. necessary to melt GC-rich DNA, which is more stable than AT-rich DNA. 45. The 35 million differences out of 3.2 billion total nucleotides represent 57. You can use a DNA polymerase that is not heat-stable. You would have approximately 1%, or a bit less than the original claim. (This number refl ects to cool the reaction mixture to a temperature at which the polymerase works single-base differences and does not account for insertions and deletions of best, and you would have to add the enzyme at each reaction cycle because multiple bases.) it would be destroyed every time the temperature was raised to melt the 46. Questions neednetoorganismal be addressed in order to solve the are C-value • How doesthat one defi complexity?

Perhaps humans not double- stranded DNA. 58. To amplify the protein-coding DNA sequence, the primers should correthe most complex organisms. paradox: spond to the fi rst three and last three residues of the protein (each amino • How much of the organism’s genome codes for RNA? For protein? acid What are the biological roles of these gene products? represents three nucleotides, so the primers would each be nine bases long).

• How many copies of each gene are in the organism’s genome?

Use Table 3-3 to fi ndMet the codons that to the fi rst three residues: • What is the amount of noncoding DNA in the genome? Is the nonGlycorrespondSer AUG GGU UCU coding DNA really “junk DNA,” or does it have some biological role?

GGC UCC • How many transposable elements are in the noncoding DNA? GGA UCA 47. (a) The fi rst reading frame is the longest ORF. GGG UCG First reading frame: AGU TAT GGG ATG GCT GAG TAC AGC ACG TTG AAT GAG GCG AGC Tyr-Gly-Met-Ala-Glu-Tyr-Ser-Ser-Leu-Tyr-Glu-AlaUsing just the topmost set of codons, a possible DNA primer would therefore ATG GCC GCT GGT GAT G have the sequence 5 -ATGGGTTCT-3 . This primer could base pair with Met-Ala-Ala-Gly-Aspthe gene’s noncoding strand, and its extension from its 3 end would yield a copy of the coding strand of the gene (see Fig. 3-17). The other primer must Second reading frame: correspond to the last three amino acids of the protein: T ATG GGA TGG CTG AGT ACA GCA CGT TGA ATG AGG Val Ser Pro Met-Gly-Trp-Leu-Ser-Thr-Ala-Arg-stop-Met-ArgGUU UCU CCU CGA TGG CCG CTG GTG ATG GUC UCC CCC Arg-Trp-Pro-Leu-Val-Met GUA UCA CCA GUG UCG CCG Third reading frame: AGU TA TGG GAT GGC TGA GTA CAG CAC GTT GAA TGA GGC AGC -Tyr-Asp-Gly-stop-Leu-Gln-His-Val-Glu-stop-GlyAgain, considering just the topmost set of codons, a probable DNA coding GAT GGC CGC TGG TGA TG sequence would be 5 -GTTTCTCCT-3 . This sequence cannot be used as a Gly-Arg-Trp-stopprimer. However, a suitable primer would be the complementary sequence (b) Assuming the reading frame has been correctly identifi ed, the most 5 -AGGAGAAAC-3 , which can then be extended from its 3 end to yield a copy of the noncoding strand of the gene. The number of possible primer likely start site is the fi rst Met residue in the fi rst ORF. pairs is quite large because all but one of the amino acids has more than one 48. Like the bacterium described in Problem 43, the bacteriophage is codon. For the fi rst primer, there are 1 4 6 24 possibilities; for the second, 4 6 4 96 possibilities. There are 24 96 2304 different pairs of parasitic primers that could be used to amplify the gene by PCR. and makes use of the host tRNAs (and likely other protein synthesizing 59. MspI, AsuI, EcoRI, PstI, SauI, and NotI generate sticky ends. AluI and machinEcoRV generate blunt ends. ery as well) in order to replicate itself.

60. The enzyme MspI generates sticky ends. The single- stranded regions are 49. If a SNP occurs every 300 nucleotides or so, and if there are about then removed by the action of the exonuclease, releasing the free nucleotides 3 million kb in the human genome (see Table 3-4), then a SNP occurs every C and G. (3 3 106/300) 5 10,000 kb or so. [Source: http://ghr.nlm.nih.gov/handbook/ Exonuclease genomicresearch/snp] 50. Because ,1.4% of the human genome consists of cleavage protein-coding genes, about 1.4% of the 3 million SNPs, or 42,000, are likely to affect these genes.

Sinceand 67,450,000. (b) Gene B contains associated the disease genesdiffer.

A and TGCTTAGC CGGAACGA the total number of genesSNPs is only ,21,000,with in theory, everywhereas gene would B do not.

[From Duerr, R. H., et al., Science 314, 1461–1463 (2006).]

A C G A A T C G G C CTTGCT 51. (a) The strongest associations are located between positions 67,400,000

Ch 4 Solutions 13 61. The restriction enzyme with the longer recognition sequence would be a 2. Threonine has two chiral carbons; therefore, four stereoisomers are rare cutter because it is likely to encounter this sequence less often and possible. therefore will cleave the DNA less frequently than a restriction enzyme with a shorter recognition sequence.

COO

COO

PstI

H

2 kb

8 kb

7 kb

H OH

H

C

NH

HO

C

H

CH3

3 kb

COO 63. If you were to use plasmids, you would need at least 150,000 clones to accommodate the 3 106–kb genome in fragments of 20 kb each (see Table H3N C H 3-6). This is an almost unmanageable number. However, if you were to use HO C H yeast artifi cial chromosomes, you would need only a minimum of 3000 clones to accommodate the 3 106–kb genome in fragments of 1000 kb.

CH3 64. To design an oligonucleotide probe for a gene, the researcher must apply the genetic code in reverse, that is, select codons that correspond to the amino acids in the protein. Most amino acids can be encoded by more 3. (a) His, Phe, Pro, Tyr, Trp than one codon, so the researcher would have to choose one of them and (b) His, Phe, Tyr, Trp hope that it matched the DNA well enough for the probe to successfully (c) His, Cys, Ser, Thr, Tyr hybridize with the DNA. Met and Trp, however, are encoded by only one (d) Gly codon each, so by using these codons as part of the probe, the researcher (e) Arg, Lys can be assured of a perfect match with the DNA, at least for these three (f) Asp, Glu nucleotides. (g) Cys, Met 65. A stop codon would need to follow the Leu residue (instead of the Ser 4. that is present in the wild-type protein). A mismatched primer could substitute a UCA (Ser) codon for a UGA stop codon. There are several correct answers due to the redundancy of the genetic code. One possible sequence is H3N CH C GTTTTCGCTGTTCTTTCAUGA. [From Yue, L., J. Virol. 83, 11588– 11598 HO 3 (2009).]

CH2 66. (a) The primers are highlighted in gray below.

C O

3

CH3

COO H

C

NH 3

H

C

OH

CH3

O

O

Chapter 4 1. (a) The chiral carbon is marked with an asterisk. D -Alanine, the mirror image isomer, is shown.

COO

COO H3N

*C

H

CH3 L-Alanine

H

*C

NH2 Asparagine

H3N CH C (CH2)2 C O NH2 Glutamine

H3N CH C HO 3

NH 4

(CH2)2 C O O Glutamate

5. From least soluble to most soluble: Trp, Val, Thr, Ser, Arg. You can use Table 4-3 as a guide, but you should also be able to do this type of problem without using the table.

6. The tetrapeptide would be less soluble than free His because the amino and lost carboxylate groups that interact with water and make His soluble are in forming the peptide bonds in the tetrapeptide. 7. This combination cannot occur in signifi cant amounts at any pH. An unprotonated amino group cannot exist with a protonated carboxylate group because the amino group’s pK is much greater than the carboxyl group’s pK (therefore the carboxyl group ionizes at a lower pH than the amino group).

8. If amino acids existed in the nonionic form described in Problem 7, they

NH

CH3 D-Alanine

O

O Aspartate

O

(b) Add the sequence recognized by the EcoRI endonuclease (GAATTC; see Table 3-5) to the 5 end of each primer. So the “left” primer would have a sequence of GAATTCATGGGCTCCATCGGTGCA and the “right” primer would have a sequence of 5 -GAATTCTTCTTTTTAAGGGGAAAC-3 . In practice, however, you’d want to embed the recognition sequence rather than having the recognition sequence on the ends; the reason for this will be explained in Chapter 20.

C

O

AAGAAACCGTCTACACAAAGGGGAATTTTTCTT-5

CH2

O

5 -ATGGGCTCCATCGGTGCA - 3 3-TACCCGAGGTAGCCACGTCGTTCGTACCTT . . .

H3N CH C

H

TTCTTTGGCAGATGTGTTTCCCCTTAAAAAGAA-3 3 -CAAAGGGGAATTTTTCTT-5

O

5 -ATGGGCTCCATCGGTGCAGCAAGCATGGAA . . .

NH 4

O

Eco RI

C C

H

PstI

H3N

O

62.

3

would have lower melting points and would be soluble in nonpolar organic solvents. High melting points and water solubility are characteristics of ionic substances. These observations support the zwitterionic (doubly ionic) form of the amino acid. 9. In a free amino acid, the charged amino and

(b) Since the majority of proteins contain L-amino acids, the presence carboxylate groups, which of D-amino acids in the bacterial cell wall renders the cell wall less sus- are separated only by the alpha carbon, electronically infl uence each other. ceptible to digestion by proteases (enzymes produced by certain organ- When the amino acid forms a peptide bond, one of these groups is neutralisms to destroy bacteria). ized, thereby altering the electronic properties of the remaining group.

Ch 4 Solutions

14

10. Histones contain an abundance of the positively charged amino acids lysine and arginine. The positive charges of these amino acid side chains interact electrostatically to form ion pairs with the negatively charged phosphate groups on the backbone of the DNA molecule and to minimize charge–charge repulsion of the negatively charged phosphate groups.

17. (a)

O C

CH CH2

HN 3

O Glutathione (GSH)

CH2

11. At pH 6.0, the N-terminus is protonated ( 1 charge), and the Cterminus is unprotonated ( 1 charge). The four His side chains (pK 6.0) are half-protonated (see Fig. 2-17) so that each has a charge of 0.5. The tetrapeptide therefore has a net charge of 2.

12. (a) The net charge is 0, since the Pro side chains are neutral, the Tyr side chains are protonated and neutral, and there is no free N-terminus or C-terminus in the cyclic molecule. (b) If the molecule were linear, it would contain a free amino group ( 1 charge) and a free carboxylate group ( 1 charge), but its net charge would still be 0.

C

O

O

HN

CH

C

O HN

C

CH

CH2

O

H

SH

(b) HN 3

13. (a) The three amino acids are Ser, Tyr, and Gly. (b) Cyclization of the polypeptide backbone occurs between the carbonyl carbon of Ser and the amide nitrogen of Gly. (c) Oxidation results in a double bond in the Tyr side chain between C and C (the second carbon of the side chain).

CH CH2

O C

O

CH2

OH

C

O

O

HN

CH

C

O HN

CH2

C

CH

O

H

S CH

OH

SOxidized glutathione (GSSG)

(c) O

CH 2N

O

O

C

NH CH C N CH2 Ser

C O

Gly

H CH

NH

C O

CH2 CH NH O C

(b)

CH2

Tyr

14.

Peptide bond O

O H 3N

N-Terminus

CH C

-Amino group

-Amino group

15.

O H3N CH

CN

CH2

CH 2

CH 2

CH2

C

CH 2

O

H

CH2 C

CH

HN

CH2

NH 3

O

C

O

CH CH2

18. A polypeptide is a single polymer of amino acids. A protein may consist of one or more polypeptide chains.

-Carboxylate group O

19.

-Carboxylate group

3 HN

O C

CH CH2

H N

CH

O C

COO

OCH3

OH

O

Aspartame (Asp–Phe–OMe)

H3N

O CH C N CH2

H

O CH2

CN

H

CH2

O C

H N

CH CH2

O C

H N

CH CH2

O C

H N

CH2

O C

H N

CH CH2

CH2 S OH

CH3

H N

CH CH2 2CH CH2

O

16.

NH 3

C-Terminus

Lys-Glu

O C

C O

CH2 CH

OH

O C

O

O C

H N

CH CH2

O C

H N

CH CH2 CH2

COO 4

CH2

CH2

CH2

NH

NH 3

3

O C

O

Ch 4 Solutions 15 20. The FLAG epitope is highly polar. Seven of the eight residues are charged. 31.

The FLAG epitope will interact favorably with water; therefore, the claim made by the biotech company is supported.

Q

23. (a) tertiary

R

W

H

G

R

(b) quaternary (c) primary (d) secondary

W

Y V

W

L

The polar amino acid residues are shown in red; the nonpolar residues are shown in blue. The polar residues are mainly on one side of the helix, while the nonpolar residues are on the other side. Quite a few of the polar side chains are positively charged. [This is an example of an amphipathic helix. From Martoglio, B., Graf, R., and Dobberstein, B., EMBO J. 16, 6636–6645 (1997).]

O H O C

K

G

24. The primary structure of GFP is discussed in Problem 13.

C

E

K

22. As in Problem 21, there are 6 possible tripeptides; adding a fourth residue gives 6 4 24 possibilities.

H

R

W

21. There are six possible sequences: HPR, HRP, PHR, PRH, RHP, RPH.

25.

M

C N H

H

O

32.

G

H

F

L

S

26. The cis confi guration (shown here) is less likely because of the steric hindrance that would result from the close approach of the polypeptide chains attached to the C carbons.

O

A

L

K

A

S

W I

K

C C

L

N C

H

A

P G

L

F

27. Both the DNA helix and the helix turn in the right-handed direction. Both helices have tightly packed interiors: In DNA, the interior of The polar amino acid residues are shown in red; the nonpolar residues the helix is occupied by nitrogenous bases; in the helix, the atoms of the are shown in blue. This is another example of an amphipathic helix polypeptide backbone contact one another. In the helix, the side chains (see Problem 31). The hydrophobic side of the helix will interact favorextend outward from the helix; no such structure exists in the DNA helix. ably with the nonpolar membrane and disrupt its structure, resulting in cell lysis. [The investigators of the study found that the peptide was able to form a pore in the membrane, disrupting the ionic balance and killing the cell. From Corzo, G., Escoubas, P., Villegas, E., Barnham, K. J., 28.

Weilan, H. E., Norton, R. S., and Nakajima, T., Biochem. J. 359, 35–45 R2O R3O (2001).]

HN

C

C H O

HC

R1

N

H

C

NH

C H

C

NH

HC R4 C O H N HC R5 C O

33. Triose phosphate isomerase is an example of an / protein.

34. (a) Leu

(b) Ala (c) Phe

(d) Trp (e) Ile

35. It’s possible that the ligand has a positive charge and forms an ion pair with the negatively charged Glu on the receptor. When the Glu is mutated to an Ala, the negative charge on the receptor is lost and the ion pair between the receptor and the ligand can no longer form.

36. The mutant in which the Lys is converted to an Arg would most likely

29. The amino group of Pro is linked to its side chain (see Fig. 4-2), which have the least effect because Lys and Arg side chains are both positively limits the conformational fl exibility of a peptide bond involving the amino charged at physiological pH. The mutation that would have the greatest group. The geometry of this peptide bond is incompatible with the bond effect is the substitution of the Lys by the negatively charged Glu. [From Dong, L.-Y., Ueno, Y., Hata, S., and Isui, K., Plant Cell Physiol. 38, 1340– angles required for a polypeptide to form an helix.

1345 (1997).]

30. Whereas a backbone chain containing Pro has limited fl exibility (see 37. There are many possible answers for this question. An example is shown Solution 29), the backbone chain containing Gly is too fl exible to conform for each (next page). to the structure of the helix.

16

(a)

Ch 4 Solutions H N

CH CH2 CH2 CH2

A Lys-Glu CH 2 ion pair NH 3 COO CH2 CH2O

O C

(b)

H N

Hydrogen bonding between Tyr and Ser

CH CH2

O C

(c)

H N

Van der Waals forces between Leu and Val O H O

H

O C

CH CH3 CH CH2 CH3 CH 3

H3C CH O H N CH C

CH2O

groups in the reduced (¬SH form). [From Winter, J., Lilie, H., and Rudolph, R., Anal. Biochem. 310, 148–155 (2002).]

43. When the temperature increases, the vibrational and rotational energy of the atoms making up the protein molecules also increases, which increases the chance that the proteins will denature. Increasing the synthesis of chaperones under these conditions allows the cell to renature, or refold, proteins that have been denatured by heat.

44. Proteins with a higher ¬SH molar content have a greater opportunity to form disulfi de bonds, which play a role in stabilizing the protein.

Proteins strengthened with disulfi de bonds would require higher temperatures to denature and would therefore have higher Tm values. [From Lacy, E. R., Baker, M., and Brigham-Burke, M., Anal. Biochem. 382, 66–68 (2008).]

45. Proline does not fi t well into the structure of the helix, because of both its geometry (see Problem 29) and the absence of a peptide ¬NH to contribute to hydrogen bonding (see Problem 37). This amino acid substitu38. Substitution of a histidine for an arginine evidently causes a change in tion would produce a protein with decreased stability, which would affect the three-dimensional structure of the protein, which adversely affects its the function. This could occur for a variety of reasons. Histidine’s side chain is ability of red blood cells to deform their shape in order to squeeze through composed of a fi ve-membered ring, whereas arginine’s side chain is a capillaries. The cells would become damaged and would be removed from straight-chain structure. The change in shape of the side chain could lead to circulation, causing anemia. [From Johnson, C. P., Gaetani, M., Ortiz, V., an overall change in the shape of the protein. Because of the difference in Bhasin, N., Harper, S., Gallagher, P. G., Speicher, D. W., and Discher, D. E., Blood 109, 3538–3543 (2007).] their structures (although both amino acids can form hydrogen bonds and ion pairs), the substituted histidine might not form an interaction that is 46. Protein loops are often at the protein surface, whereas regular secondcrucial to the proper functioning of the protein. A “permanent” positive ary structure predominates in the protein core. The loops are better able to charge might also be necessary. Arginine has a pK of 12.5 and thus is always accommodate changes in size and amino acid composition than the core protonated at physiological pH. Even though the pK value of an amino acid segments, where a change in size or composition may disrupt an helix or in a protein is not necessarily the same as the pK value of the free amino sheet that is an essential part of the protein’s structure. acid, the great difference in the pK values indicates that the arginine is far 47. If the proteins were homodimers, they would be more likely to have more likely to be protonated than the histidine. If a full strong charge at two identical sites to interact with their palindromic recognition sites in the this site is necessary for protein function, its replacement could result in a DNA. Heterodimeric proteins would likely lack the necessary symmetry. defective protein.

(In fact, these enzymes are homodimeric.)

39. A polypeptide synthesized in a living cell has a sequence that has been 48. Proteins and their component amino acids (and many other organic optimized by natural selection so that it folds properly (with hydrophobic compounds) have inherent “handedness” and therefore cannot be interconresidues on the inside and polar residues on the outside). The random severted through mirroring. Consider a right-handed helix: Its mirror image quence of the synthetic peptide cannot direct a coherent folding process, would be a left-handed helix, which is not found in nature. so hydrophobic side chains on different molecules aggregate, causing the 49. The arginine residues, with their positively charged side chains, and the polypeptide to precipitate from solution. aspartate residues, with their negatively charged side chains, are likely to be 40. (a) Heating causes a protein to “melt,” or unfold, because heating infound on the surface of the monomer. These residues likely form ion pairs creases the vibrational and rotational energy of atoms in the protein, that stabilize the dimeric form. When these residues were mutated to neutral which disrupts the weak interactions that keep the protein in its propamino acid side chains, the ion pairs could not form, the dimer could not erly folded state. form, and the equilibrium shifted in favor of the monomers. [From Huang, (b) pH changes alter the ionization states of amino acid side chains. This Y., Misquitta, S., Blond, S. Y., Adams, E., and Colman, R. F., J. Biol. Chem., affects the ability of side chains to form ion pairs. Hydrogen bonds may 283, 32800–32888 (2008).] also be broken as protonation or deprotonation renders amino acid side 50. Because the tetramers dissociate into dimers in the presence of the amchains unable to serve as hydrogen bond donors or acceptors. phipathic SDS, it is possible that hydrophobic interactions are important in (c) Detergents have a nonpolar domain that allows them to penetrate stabilizing the dimer. The hydrophobic surface of the dimer could interact into the interior of the protein, thus interfering with the hydrophobic with the nonpolar tail of SDS when the tetramer dissociates into dimers. But interactions responsible for the protein’s tertiary structure. the monomers might be stabilized by a different type of interaction, either (d) Reducing agents, such as 2-mercaptoethanol, break disulfi de bonds polar or ionic. The nonpolar tail of SDS would not interact favorably with (converting them to the ¬SH form) and destabilize proteins that rea polar surface, and the negatively charged head group of SDS would be quire disulfi de bonds in order to assume the correct conformation. repelled by negatively charged amino acid side chains involved in ionic inter41. Anfi nsen’s ribonuclease experiment demonstrated that a protein’s actions. The association of the monomers with one another is more favorable primary structure dictates its three-dimensional structure. Although some than their association with SDS would be, so the dimers resist dissociation proteins, like ribonuclease, can renature spontaneously in vitro, most pro- into monomers in the presence of detergent. [From Gentile, F., Amodeo, teins require the assistance of molecular chaperones to fold properly in vivo. P., Febbraio, F., Picaro, F., Motta, A., Formisano, S., and Nucci, R., J. Biol.

42. (a) The loss of the C chain means the loss of information that is essential Chem., 277, 44050–44060 (2002).] to the proper folding of insulin. Removal of the C chain leaves two separate 51. The only ionizable groups in the dipeptide are the C-terminus (pK 3.5) chains (A and B), and it is much more diffi cult for two chains to resume their and the N-terminus (pK 9.0). pI ½ (3.5 9.0) 6.25. native conformation than for one chain to do so. Proinsulin has no diffi culty 52. In order for the tripeptide to have no net charge, its C-terminal group resuming its native conformation in a denaturation/renaturation experiment must be unprotonated (negatively charged), its Tyr side chain must be probecause proinsulin consists of only one peptide chain. When proinsulin is tonated (neutral), and its N-terminal group must be protonated (positively converted to insulin, two peptide bonds are cleaved to produce the correctcharged). Because deprotonation of the side chain (pK 10.5) or deprotonly cross-linked A and B chains. [From Dixon, G. H., and Wardlaw, A. C., ation of the N-terminal group (pK 9.0) would change the net charge, the Nature, 188, 721 (1960).] pK values of these groups (10.5 and 9.0) should be used with Equation 4-1: (b) At the higher pH, the sulfhydryl group is ionized (¬S ). These pI ½ (10.5 9.0) 9.75. sulfhydryl groups more readily form disulfi de bonds than sulfhydryl HN

CH

C

HN

CH

C

Ch 4 Solutions 53. The protein must contain groups that undergo protonation/deprotonation at pH values near 4.3. The only amino acids with side chain pK values in this range are Asp and Glu (Table 4-1), so the protein likely contains an abundance of these residues.

54. The protein will likely have a net charge of 0 or greater. The pK value of the C-terminal carboxylate group is 3.5 (Table 4-1), so at pH values below 3.5, the group will be predominantly neutral. All other ionizable groups in proteins have higher pK values and therefore will also be predominantly neutral or positively charged below pH 3.5.

55. At pH 7.0, the peptide likely has a net positive charge since Arg (R) and Lys (K) outnumber Asp (D) and Glu (E). Therefore, the peptide is likely to bind to CM groups but not to DEAE groups.

58. The cleavage site for each fragment is highlighted. A fragment not ending in Phe, Tyr, or Trp (for chymotrypsin cleavage) or Lys or Arg (for trypsin cleavage) must be the carboxyl terminal fragment. Use “overlap” to work backward from the C-terminus, beginning with the chymotrypsin fragment, to determine the sequence of the polypeptide.

Chymotrypsin fragments AGKQLEDGRTLSDY IPPDQQRLIF

LR AK

NIQKESTLHLVLRLRGG MQIF

IQDK LIFAGK

CH

O C

CH2

H N

CH

O C

CH2

DTT

S

HN

Trypsin fragments

VKTLTGKTITLDVEPSDTI ENVKAKIQDKEGI EGI

56. (a) H N

17

H N

CH

IPPDQQR

S

CH2COO

S

SH

S

CH2COO

CH2

CH2

CH2

C

HN

CH

C

HN

CH

GGMQIFVK TLSDYNIQK ESTLHLVLR TITLDVEPSDTIENVK

C O

O

O

TLTGK

CH2

ICH2COOH

SH

CH

QLEDGR

O C

(b) A simplifi ed version of the anion exchange chromatogram is shown.

IPPDQQRLIFAGKQLEDGRTLSDYNIQKESTLHLVLRLRGGMQIFVKTLTGKTITLDVEPSDTIENVKAKIQDKEGI

Absorbance @ 280 nm

59. Edman degradation of a polypeptide with a disulfi de cross-link would

Ara h8

Ara h6 high salt

not work properly when the fi rst Cys became exposed at the N-terminus of the polypeptide (the Cys would not be released in the next reaction since it would still be covalently linked to a Cys residue farther along the polypeptide chain). Reduction before sequencing breaks the disulfi de bond, and alkylation of the two free Cys groups prevents re-formation of the bond.

60. Thermolysin would yield the most fragments (9) and chymotrypsin would yield the fewest (3).

61. (a) If only one proteolytic cleavage is carried out, the sequences of the

fragments could be determined but it would not be possible to place the fragments in the proper order. If different enzymes are used to generate two sets of fragments, overlapping peptides would allow ordering of all Solvent volume the sequences. (c) Treatment of the proteins with a reducing agent followed by iodo(b) Trypsin cleaves after Lys and Arg residues. The fragments resulting acetic acid converts fi ve disulfi de bridges (which are neutral) in Ara from digestion of the heavy and light chains are shown below, identifi ed h6 to ten negatively charged side chains. The treatment produces a by residue number. modifi ed Ara h6 protein with a more acidic pI. The two proteins were Light chain Heavy chain successfully separated because the Ara h8 protein is less attracted to the positively charged anion exchange beads and elutes sooner. The Ara 6 1–7 h6 binds more strongly to the anion exchange beads and elutes only 7–9 8–18 after salt has been added to the elution buffer. [From Riecken, S., 10–11 19–31 Lindner, B., Petersen, A., Jappe, U., and Becker, W.-M., J. Biol. Chem., 12 32–35 389, 415–423 (2008).]

13–20 36–38 21–28 39–42 57. The amino terminal residue is Ala. The carboxyl terminal residue must 29–38 43–57 be Met, since the dodecapeptide was not cleaved when CNBr was added.

58–61 Chymotrypsin cleaves after Phe. Fragment II contains the Asp, so it appears 62–71 in the sequence fi rst, and Phe must be the cleavage site. Trypsin cleaves after 78–85 Lys. Since fragment III contains Asp, Lys must be the cleavage site. Elastase 86–91 cleaves after Gly, Val and Ser. Val must occupy the second position followed by a Pro, and this Val was not cleaved. (c) Chymotrypsin would be a good choice for the second enzyme beTrypsin

Chymotr ypsin

Asp–Val–Pro–Lys–Ser–Asp–Gln–Phe–Val–Gly–Leu–Met Elastase

Elastase

[Based on Anastasi, A., Montecucchi, P., Erspamer, V., and Visser, J., Experientia 33, 857–858 (1977).]

cause it cleaves after Phe, Tyr, and Trp. The following fragments would be obtained: Light chain 6–13 14–25 26–38

Heavy chain 1–63 64–83 84–91

62. Sequencing this particular peptide using chemical or enzymatic cleavage methods would be diffi cult. There is only one N-terminal Met, so CNBr

18

Ch 5 Solutions

would not cleave the peptide. Cleavage with trypsin would not occur because Chapter 5 the peptide lacks Lys and Arg residues. Cleavage with elastase would yield 1. Globin lacks an oxygen-binding group and therefore cannot bind O2. single amino acids, so locating areas of overlap would be diffi cult. Cleav- age Heme alone is easily oxidized and therefore cannot bind O2. The bound with chymotrypsin would produce three fragments, but the lack of a second heme gives a protein such as myoglobin the ability to bind O2. In turn, the cleavage agent would make it diffi cult to put the peptide fragments in the protein helps prevent oxidation of the heme Fe atom. correct order. [From Burstein, I., and Schechter, Y., Biochemistry 17, 2392– 2. The mitochondrial proteins accept oxygen from hemoglobin and thus 2400 (1978).] must be able to bind oxygen. Therefore these proteins, like myoglobin and hemoglobin, must have a heme group as part of their structure.

63. Since each codon corresponds to an amino acid, the error rate is 3. Myoglobin facilitates oxygen diffusion in the cell by acting as a “molecular bucket brigade.” It accepts oxygen delivered to the cell by hemoglobin 531024 error 3500 residues50.25 and then transfers the oxygen to proteins in the mitochondrion. Myoglobin residue is 50% saturated with oxygen when the intracellular oxygen concentration is equal to its p50 value, so it functions most effectively under these conditions.

About one-quarter of the polypeptides would contain a substitution.

At intracellular oxygen concentrations greater than p50, oxygen remains 64. Virtually all the 2000-residue polypeptides would contain a substibound to hemoglobin and is not transferred; at oxygen concentrations less tution: than p50, myoglobin doesn’t bind suffi cient oxygen. In either case, the transfer of oxygen from hemoglobin to mitochondrial proteins is compromised.

531024 error 4. Equation 5-4 can be used to calculate the fractional saturation (Y ) at 32000 residues51.0 residue 1.5 torr and 3.5 torr, letting K 2.8 torr: 65. Leu and Ile are isomers and have the same mass; therefore, mass spectrometry cannot distinguish them.

66. (a) A sequence of two Gly residues (mass 57 D 57 D) could be erroneously identifi ed as Asn (mass 114 D). (b) A sequence of a Ser residue and a Val residue (masses 87.0 D and 99.1 D, respectively) would be erroneously identifi ed as Trp (mass 186 D).

67.

Phe

Val H3C

Ala

Asp

O

O

O

H N

p O2

At 1.5 torr, Y5

K1p O2 1.5 torr 2.8 torr11.5 torr 50.35

At 3.5 torr, Y5

2.8 torr13.5 torr

3.5 torr

50.55

At 1.5 torr, 35% of Mb has oxygen bound, but at 3.5 torr, 55% of myoglobin

O

CH3

Y5

has oxygen bound. Thus a small change in oxygen partial pressure can lead to a large change in the percentage of oxygenated myoglobin in the muscle cell.

5. (a) The interior of the roast is purple because it does not have access to oxygen and thus the myoglobin in the meat is primarily in the deoxygenated form, which is purple. When sliced, the myoglobin is exposed O CH 3 O to oxygen and becomes oxygenated, turning red. (b) When meat is cooked, the globin chains are denatured and the iron is oxidized to Fe3 , converting the myoglobin to met-myoglobin, which is brown. (c) In a vacuum package, the cells rapidly use up any available oxygen in metabolism, which results in mostly deoxygenated myoglobin, which is Dashed lines indicate broken bonds. The smallest charged fragment is the purplish. Meat packaged in oxygen-permeable packaging continues to N-terminal residue (Phe), which has a mass of approximately 149 D (9 C 1 be metabolically active so that most of the myoglobin is oxygenated N 1 O 11 H). and red, which is more appealing to consumers. [Source: “Color 68. The next-smallest fragment is the Phe–Val dipeptide. The difference in Changes in Cooked Beef”, James Claus, National Cattlemen’s Association.] mass between the smallest and next-smallest fragments is the mass of the Val residue, or approximately 99 D. 6. (a) The p50 values from the oxygen binding curve are 0.6, 1, and 1.4 mm 69. In a protein crystal, the residues at the end of a polypeptide chain may Hg for the tuna, bonito, and mackerel, respectively. experience fewer intramolecular contacts and therefore tend to be less (b) The tuna has the lowest p50 value and the highest oxygen-binding ordered affi nity; the mackerel has the highest p50 value and the lowest oxygen (more mobile in the crystal). If their disorder prevents them from generating affi nity. Interestingly, the p50 values are the same when adjusted for the a coherent diffraction pattern, it may be impossible to map their electron body temperatures of the fi sh, which average 25 C, 20 C, and 13 C density. for the tuna, bonito, and mackerel, respectively. [From Marcinek, D. J., 70.

Bonaventura, J., Wittenberg, J. B. and Block, B. A. (2001). Am. J. Physiol.

H Regul. Integr. Comp. Physiol. 280, R1123–R1133 (2001).]

7. Isoleucine has a larger side chain than valine and will decrease the size N H O of the pocket. This steric hindrance results in decreased binding of oxygen.

H H C C C [From Olson, J. S., and Phillips, Jr., G. N. J. Biol. Chem. 271, 17593–17596 H H H H (1996).]

H C H H3N

H

N H

C

H

C N H C H O

H

O H N C C H H

C S

O

N H

H

C

H

C N H C H O

8. Because oxygen is nonpolar, it is advantageous for the heme pocket to be nonpolar as well. Therefore, substituting the nonpolar valine with the polar serine would result in decreased oxygen-binding affi nity.

9. In the arteries, nearly all the hemoglobin is oxygenated and therefore takes on the color of the Fe(II), in which the sixth coordination site is occupied by O2. Blood that has passed through the capillaries and given up some of its oxygen contains a mixture of oxy- and deoxyhemoglobin. Deoxyhemoglobin, in which the Fe(II) has only fi ve ligands, imparts a bluish tinge to venous blood.

Ch 5 Solutions

19

10. A patient with a light complexion would have fl ushed, reddened skin due dramatically, as shown in Solution 15. Under these conditions, hemogloto the color of the HbCO complex. The patient would also suffer from bin is nearly saturated with oxygen when pO2 120 torr (Y 0.96). dizziness, headaches, and shortness of breath due to lack of oxygen.

But when pO2 is 25 torr in the tissues, Y decreases to 0.20. So 76% of the oxygen is delivered to tissues. In general, higher CO2 concentrations 11. (a) Position 6 (Gly) and position 9 (Val) appear to be invariant. (b) Conservative substitutions occur at position 1 (Asp and Lys, both assist hemoglobin in delivering oxygen from the lungs to the tissues.

[From Bohr, C., Hasselbalch, K., and Krogh, A., Skand. Arch. Physiol. 16, charged), position 10 (Ile and Leu, similar in structure and hydrophobicity), and position 2 (all uncharged bulky side chains). Positions 5 and 401–412 (1904).]

17. At the high altitude where the bar-headed goose resides, less oxygen is 8 appear to tolerate some substitution. available to bind to hemoglobin in the lungs. The bar-headed goose hemo(c) The most variable positions are 3, 4, and 7, where a variety of residues appear. globin has a lower p50 value and a higher oxygen affi nity than the plainsdwelling grelag goose hemoglobin, so the bar-headed goose hemoglobin 12. His F8 is the most likely to be invariant. The nitrogen in the imidazole can more easily bind oxygen in order to deliver it to the tissues. [From Jessen, ring of His F8 serves as one of the ligands to the iron in the heme group. This T.-H., Weber, R. E., Fermi, G., Tame, J., and Braunitzer, G., Proc. Natl. residue plays a critical role in the structures of myoglobin and hemoglobin and Acad. Sci. 88, 6519–6522 (1991).] is essential for the proper binding of oxygen; substitution of this amino acid would likely interfere with the ability of the protein to bind and release oxygen18. (a) Vitamin O is useless because the body’s capacity to absorb oxygen is not limited by the amount of oxygen available but by the ability of effectively. hemoglobin to bind and transport O2. Furthermore, oxygen is nor13. Use Equation 5-4 to calculate the fractional saturation (Y ) for hypermally introduced into the body via the lungs, so it is unlikely that the bolic binding, letting K 26 torr: gastrointestinal tract would have an effi cient mechanism for extracting oxygen. p O2 (b) The fact that oxygen delivery in vertebrates requires a dedicated Y5 K1p O2 O2-binding protein (hemoglobin) indicates that dissolved oxygen by itself cannot attain the high concentrations required. (In fact, the 30 torr 50.54 solubility of oxygen in pure water is only about 0.1 mM; the hemoAt 30 torr, Y5 26 torr130 torr globin in blood boosts the solubility to about 10 mM.) In addition, a few drops of vitamin O would make an insignifi cant contribution 100 torr 50.79 to the amount of oxygen already present in a much larger volume of At 100 torr, Y5 26 torr1100 torr blood.

Therefore, if hemoglobin exhibited hyperbolic oxygen-binding behavior, 19. The increased O2 release is the result of the Bohr effect. The increase it would be only 79% saturated in the lungs (where pO2 < 100 torr) and would exhibit a loss of saturation of only 25% (79% 54%) in the tissues (where pO2 < 30 torr). Hemoglobin’s sigmoidal binding behavior allows it to bind more oxygen in the lungs so that it can deliver relatively more oxygen to the tissues (for an overall change in saturation of about 40%; see Fig. 5-7).

14. Y5

(120 torr)3

in [H ] promotes the shift from the oxy to the deoxy conformation of hemoglobin. The decrease in oxygen affi nity improves oxygen delivery to the muscle, where it is needed.

20. A decrease in pH diminishes hemoglobin’s affi nity for oxygen (the Bohr effect), thereby favoring deoxyhemoglobin. Since only deoxyhemoglobin S polymerizes, sickling of cells is most likely to occur when the parasiteinduced drop in pH promotes the formation of deoxyhemoglobin.

21. (a) Asp 94 and His 146, which is protonated, form an ion pair in the deoxy form of hemoglobin. The protons are produced by cellular respiration. As described by the Bohr effect, an increase in hydrogen ion concentration favors the deoxy form of hemoglobin so that oxygen can be delivered to the tissues. (b) The presence of the negatively charged Asp 94 increases the pK value of the imidazole ring of His and promotes the formation of the ion pair between the two side chains. The increased pK value means that the imidazole ring’s affi nity for protons is increased. [From Berenbrink, M., Resp. Physiol. Neurobiol. 154, 165–184 (2006).]

(15 torr)31(120 torr)3

22. The p50 for hemoglobin when pCO2 5 torr is given as 15 torr, whereas

(pO)n 2 (p)n50 1(pOn2) (25 torr)3

Y5

(15 torr)31(25 torr)3

Y50.82 Y5

(pO)n 2 (p)n50 1(pOn2)

Y5 Y51.00 15. Y5

(pO)n 2 (p n50)1(pOn 2) (25 torr)3

Y5

(40 torr)31(25 torr)3

Y50.20 Y5

(pO)n 2 (p)n50 1(pOn2)

Y5

(120 torr)3 (40 torr)31(120 torr)3

Y50.96 16. As shown in Solution 14, hemoglobin is completely saturated with oxygen (Y 1.00) when the pO2 120 torr. When the pO2 decreases to 25 torr, Y 0.82. This means that 18% of oxygen bound to hemoglobin is delivered to the tissues when pCO2 5 torr. But when the pCO2 increases to 40 torr, the amount of oxygen delivered to tissues increases

the p50 when pCO2 80 torr is given as 40 torr. When the partial pressure of CO2 increases, the p50 for hemoglobin increases as well. An increased p50 indicates a lower affi nity of hemoglobin for oxygen. Therefore, when the concentration of CO2 increases, as it does in metabolically active tissues, hemoglobin’s affi nity for oxygen decreases and hemoglobin unloads its oxygen to the cells that need it.

23. Negatively charged glutamate side chains on the surface of the oxygenated lamprey monomer would resist association due to the charge–charge repulsion. But when the pH decreases, excess protons would bind to the glutamate side chains, neutralizing them. The monomers would associate to form the deoxygenated tetramer. In this manner, oxygen is delivered to lamprey tissue when the pH decreases in metabolically active tissue. [From Qiu, Y., Maillett, D. H., Knapp, J., Olson, J. S., and Riggs, A. F., J. Biol. Chem.

275, 13517–13528 (2000).]

24. (a) Bicarbonate is formed when carbon dioxide reacts with water to form carbonic acid, a reaction in the blood catalyzed by carbonic anhydrase.

The carbonic acid dissociates to form protons and bicarbonate. (b) Both curves are sigmoidal. According to the investigators, the p50 for crocodile hemoglobin is 6.8 torr in the absence of bicarbonate and 44 torr in the presence of bicarbonate. The higher p50 value in

20

Ch 5 Solutions the presence of bicarbonate means that the hemoglobin has a lower affi nity for oxygen.

w/o HCO

H1(aq)1HCO23(aq)ΔH2CO3(aq)

3

Y w/ HCO 3

p O2

third equation to shift to the right. This depletes CO2(aq), which causes the second equation to shift right, thus depleting carbonic acid. This causes the fi rst equation to shift right as hydrogen ions and bicarbonate ions combine to form more carbonic acid. Depletion of hydrogen ions results in a basic pH.

H2CO3(aq)ΔCO2(aq)1H2O(l ) CO2(aq)ΔCO2(g) (b) The decrease in alveolar pCO2 concentration can be explained by the hyperventilation, as described in part (a). The concentration of 2,3-BPG increases in order to convert more of the hemoglobin molecules to the low-affi nity T form so that oxygen can be effectively delivered to the cells.

(c) The lysine side chain is positively charged and can interact with the 30. (a) The p50 for hemoglobin in these animals is lower (by about 10 torr) negative charge of the bicarbonate. The phenolate oxygens on the two than in their sea-level counterparts. The animals indigenous to high altityrosine side chains can act as hydrogen bond acceptors for the bicartudes require a hemoglobin with a higher affi nity (low p50 corresponds to bonate hydrogen. (d) All of the allosteric effectors are negatively charged. They bind to hemo- high affi nity, since this means that a lower oxygen partial pressure is required to half-saturate the hemoglobin with oxygen). Since these animals globin by forming ion pairs with positively charged amino acid side chains have higher oxygen-binding affi nity, their hemoglobin molecules are able (such as Lys or Arg) or with the positively charged -amino groups at the to bind oxygen effectively even at the low oxygen pressure that exists at amino termini of the four polypeptide chains. [From Komiyama, N. H., high altitudes.

Miyazaki, G., Tame, J, and Nagai, K., Nature 373, 244–246 (1995).] (b) Fetal hemoglobin has a higher oxygen affi nity than normal adult he25. (a) Curve A represents fetal hemoglobin, which has a higher oxygen moglobin because amino acid changes in the central cavity of hemogloaffi nity than adult hemoglobin (curve B). Curve A shows greater fracbin have replaced positively charged amino acid side chains with neutral tional saturation than curve B at any given oxygen concentration. (b) In amino acid side chains (see Solution 25). The neutral side chains cannot the chain, the positively charged His 21 in the chain is re- placed by form ion pairs with the negatively charged BPG, so BPG (which binds the neutral Ser, resulting in a central cavity with fewer positive charges. only to the T form) does not bind, the R (high affi nity) form of hemoThe allosteric effector BPG cannot bind to Hb F as effectively as it binds globin is favored, and the p50 decreases. The persistence of fetal hemogloto Hb A because the negatively charged BPG forms fewer ion pairs in bin in these animals endows the animals with yet another high-affi nity the central cavity of Hb F. BPG plays a role in decreasing oxygenoxygen-binding strategy so that the hemoglobin can bind oxygen, even binding affi nity, so the inability of BPG to bind to Hb F means that at the low oxygen partial pressures typical of high altitudes. fetal hemoglobin has a higher oxygen-binding affi nity. Oxygen transfer 31.

The substitution at the C-terminus could affect the position of the F8 from the maternal circulation to the fetal circulation can be accomplished more effi ciently if the fetal hemoglobin has a higher oxygen histidine in such a way that oxygen either binds more readily or dissociates less readily. The substitution is quite near the His 146 involved in binding affi nity than maternal hemoglobin. protons (see Problem 21) and may decrease this side chain’s proton-binding 26. Hydroxyurea increases the patient’s blood concentration of hemoglobin F, affi nity. Since His 146 is located in the central cavity, BPG would bind less which has two chains and two chains rather than two chains and two readily, which favors the oxygenated or R form of hemoglobin. chains. Since the chain has been mutated in the patient suffering from sickle cell anemia, the increased synthesis of hemoglobin F allows the patient to use 32. These patients are sometimes referred to as “human llamas” because, hemoglobin F to transport oxygen instead of the defective hemoglobin S. In like these animals, the patients are adapted for survival at high altitudes. Like the llamas (see Solution 30), these patients have mutant hemoglobins with a effect, the defective chains are replaced with healthy chains. higher oxygen-binding affi nity, which results in hemoglobin molecules with 27. (a) Normal hemoglobin has a sigmoidal curve, which means that the an enhanced ability to bind oxygen effectively, even at the low oxygen presbinding and release of oxygen from normal hemoglobin is cooperative. sure that exists at high altitudes.

The hyperbolic binding curve of Hb Great Lakes indicates that there is 33. Both types of molecules are proteins and consist of polymers of amino little cooperativity in oxygen binding and release. acids. Both contain elements of secondary structure. But globular proteins (b) Hb Great Lakes has a higher affi nity for oxygen. More than 60% are water soluble and nearly spherical in shape. Examples include proteins of the mutant hemoglobin has bound oxygen. Hb A is about 30% such as hemoglobin and myoglobin as well as enzymes. Their cellular role oxygenated. involves participating in the chemical reactions of the cell in some way. In (c) Both hemoglobins are essentially 100% oxygenated and therefore contrast, fi brous proteins tend to be water insoluble and have an elongated have equal affi nities. shape. Their cellular role is structural—as elements of the cytoskeleton of the (d) Normal hemoglobin is more effi cient at oxygen delivery. It delivers cell and the connective tissue matrix between cells. about 70% of its bound oxygen (since it is 100% oxygenated at 75 torr and 30% oxygenated at 20 torr). Hb Great Lakes is less effi cient since 34. (a) Keratin and collagen; (b) myosin and kinesin; (c) actin and tubulin; less than 40% of its bound oxygen is delivered to the tissues. [From (d) actin, myosin, tubulin, and kinesin.

Rahbar, S., Winkler, K., Louis, J., Rea, C., Blume, K., and Beutler, E., 35. Microfi laments and microtubules consist entirely of subunits that are Blood 58, 813–817 (1981).] assembled in a head-to-tail fashion, so the polarity of the subunits (actin 28. Hb Bruxelles is the most unstable, since Phe is deleted. Hb Hammermonomers in microfi laments and tubulin dimers in microtubules) is presmith is next, since the nonpolar Phe has been mutated to a polar Ser. Hb served in the fully assembled fi ber. In intermediate fi lament assembly, only Sendagi and Hb Bucuresti have nonpolar amino acid substitutions, but Hb the initial step (dimerization of parallel helices) maintains polarity. In subseBucuresti is more stable than Hb Sendagi because Leu is larger than Val and quent steps, subunits align in an antiparallel fashion, so in a fully assembled thus closer in size to the large nonpolar Phe. [From Griffon, N., Baden, C., intermediate fi lament, each end contains heads and tails.

Lena-Russo, D., Kister, J., Barkakdjian, J., Wajcman, H., Marden, M. C., 36. (a) Adding another protein was necessary because G-actin in solution and Poyart, C., J. Biol. Chem. 271, 25916–25920 (1996).] tends to polymerize rather than form crystals. The added protein bound 29. (a) A person hyperventilates in order to obtain more oxygen, and the to actin and prevented its polymerization. blood pH increases as a result, as shown by the equations below. Exces- (b) The crystal structure of the protein alone was needed so that it could sive removal of CO2(g) from the lungs during hyperventilation causes the be “subtracted” from the crystal structure of the actin–protein complex.

Ch 5 Solutions

21

therefore not fi t as well in the area of contact between the two polypep37. Because phalloidin binds to F-actin but not to G-actin, the addition of tides in a coiled coil (see Fig. 5-25). phalloidin fi xes actin in the fi lamentous form. This impairs cell motility because cell movement requires both actin polymerization at the leading edge 50. A fi brous protein such as keratin does not have a discrete globular core. of the cell and depolymerization at the trailing edge. In the presence of phalMost of the residues in its coiled-coil structure are exposed to the solvent. loidin, depolymerization does not occur and cell movement is not possible.

The exception is the strip of nonpolar side chains at the interface of the two 38. The use of fl uorescently tagged phalloidin allows the microscopist to coils. visualize “fi xed” cellular F-actin. G-actin is not visualized because phalloidin 51. The reducing agent breaks the disulfi de bonds (¬S¬S¬) between binds only to F-actin. In this way, only microfi laments are visualized. keratin molecules. Setting the hair brings the reduced Cys residues (with 39. During rapid microtubule growth, -tubulin subunits containing GTP their ¬SH groups) closer to new partners on other keratin chains. When the accumulate at the ( ) end because GTP hydrolysis occurs following subhair is then exposed to an oxidizing agent, new disulfi de bonds form between unit incorporation into the microtubule. In a slowly growing microtubule, the Cys residues and the hair retains the shape of the rollers. the ( ) end will contain relatively more GTP that has already been hydrolyzed to GDP. A protein that preferentially binds to ( ) ends that contain 52. The difference in sulfur content can be explained by the amount of disulfi de bonding in each type of keratin. Hard keratins are less pliable than GTP rather than GDP could thereby distinguish fast- and slow-growing soft keratins and are stabilized by a higher number of disulfi de bonds than microtubules. in the soft keratins.

40. Because each protofi lament contains many tubulin dimers, the separation of protofi laments during fraying represents a greater departure of tubu- 53. (a) Actin’s primary structure is its amino acid sequence. Its secondary lin subunits than if the tubulin subunits left the microtubule one dimer at structure includes its helices, sheets, and other conformations of a time. the polypeptide backbone. Its tertiary structure is the arrangement of its 41. Polymers composed of -tubulin molecules allowed to polymerize in backbone and all its side chains in a globular structure. Monomeric actin the presence of a nonhydrolyzable analog of GTP are more stable. When by defi nition has no quaternary structure. However, when actin monomers associate to form a microfi lament, the arrangement of subunits the -tubulin subunits are exposed to GTP in solution, the GTP binds to becomes the fi lament’s quaternary structure. Thus, actin is an example the -tubulin and then is hydrolyzed to GDP, which remains bound to the of a protein that has quaternary structure under certain conditions.

-tubulin. Additional heterodimers are then added. The microtubule (b) Collagen’s primary structure is its amino acid sequence. Its secondends with GDP bound to the -tubulin are less stable than those bound ary structure is the left-handed helical conformation characteristic of to GTP because protofi laments with GDP bound are curved rather than the Gly–Pro–Hyp repeating sequence. Its tertiary structure is essentially straight and tend to fray. If a nonhydrolyzable analog is bound, it will the same as its secondary structure, since most of the protein consists of resemble GTP and the protofi lament will be straight rather than curved. one type of secondary structure. Collagen’s quaternary structure is the It is less likely to fray and the resulting protofi lament is more stable as a arrangement of its three chains in a triple helix. It is also possible to view result. the triple helix as a form of tertiary structure, with quaternary structure 42. Although the drugs have opposite effects on microtubule dynamics, they both interfere with the normal formation of the mitotic spindle, which referring to the association of collagen molecules. is required for cell division.

54. Consumption of animal protein high in collagen is not required for 43. Microtubules form the mitotic spindle during cell division. Because cancer cells are rapidly dividing cells, and hence undergo mitosis at a rate human collagen synthesis, and this diet is unlikely to cause wrinkles to disapmore rapid than in most other body cells, drugs that target tubulin and thus pear. A diet that includes high-quality protein and vitamin C–containing interfere with the formation of the mitotic spindle in some way will slow the fruits and vegetables is suffi cient to promote synthesis of collagen. growth of cancerous tumors.

55. The bacterial enzymes degrade collagen, the major protein in connective 44. Paclitaxel increases the stability of the microtubule if it is able to override tissue. Treatment of the tissue with these enzymes degrades the collagen in the effects of GTP hydrolysis, since it is GTP hydrolysis to GDP that results the extracellular matrix without harming the cells themselves and thus faciliin a curved protofi lament that causes the microtubule to have a “frayed” tates the preparation of cells for culturing. [Source: Worthington Biochemical appearance and increases the likelihood that the tubulin dimers in the pro- Corporation.] tofi lament will dissociate. [From Amos, L. A., and Löwe, J., Chem. Biol. 6, 56. The enzyme is a collagenase. Cleaving the bond indicated disrupts the R65–R69 (1999).] triple-helical structure of the collagen molecule and causes it to be denatured, 45. Colchicine, which promotes microtubule depolymerization, inhibits which in effect destroys the collagen. This enzyme is necessary in the develthe mobility of the neutrophils because cell mobility results from polymer- oping tadpole because the collagen that forms the structure of the tadpole’s ization and depolymerization of microtubules. tail will need to be degraded when the tadpole develops into a frog (which 46. Microtubules in the cell spindle are dynamic structures since the mihas no tail). [From Stolow, M. A., Bauzon, D. D., Li, J., Sedgwick, T., Liang, totic spindle must form during mitosis and disassemble after cell division has V. C., Sang, Q. A., and Shi, Y. B., Mol. Biol. Cell 7, 1471–1483 (1996).] occurred. Therefore, spindle microtubules are less stable than the microtu- 57. (a) Collagen B is from rat, and collagen A is from the sea urchin. bules that defi ne the elongated structures of the axons of nerve cells. (b) The stability of each of these collagens is correlated with their hy47. As shown in Figure 5-22, microtubules link replicated chromosomes to droxyproline content. The higher the percentage of hydroxyproline, two points at opposite sides of the cell. Vinblastine’s ability to stabilize the mithe more regular the structure and the more diffi cult it is to melt, crotubules at the ( ) end while destabilizing the ( ) end disrupts this linkage. resulting in more stable collagen. The rat has a more stable collagen, Mitosis slows down or completely halts as a result. [From Panda, D., Jordan, and the sea urchin, which lives in cold water, has a less stable collagen.

M. A., Chu, K. C., and Wilson, L. J. Biol. Chem., 271, 29807–29812 (1996).]

It is impor- tant to note that the melting temperatures of each 48. Mitotic spindles are composed of microtubules and are required for cell collagen molecule are higher than the temperature at which each division. The formation of the mitotic spindle is a highly dynamic process. organism lives. Thus, each organism has stable collagen at the The presence of a drug that affected the formation of the mitotic spindle temperature of its environment. [From Mayne, J., and Robinson, J. J., J. would therefore have a much greater effect on a population of rapidly dividCell. Biochem. 84, 567–574 (2001).] ing cells.

58. (a) The melting temperature, or Tm, can be used to rank the stability 49. (a) The fi rst and fourth side chains are buried in the coiled coil, but the of the triple helices formed by the three peptides. The higher the melting temperature, the more stable the triple helix. Therefore, peptide 1 remaining side chains are exposed to the solvent and therefore tend to has the highest stability, followed by peptide 2 and then peptide 3. The be polar or charged. stability is correlated with Pro and Hyp content. Proline and hydroxy(b) Although the residues at positions 1 and 4 in both sequences are proline are responsible for the formation of the correct geometry of the hydrophobic, Trp and Tyr are much larger than Ile and Val and would left-handed helices of the three chains.

22

Ch 5 Solutions

(b) Peptide 3 contains a great number of charged residues, both acidic 63. Because collagen has such an unusual amino acid composition (almost and basic. Its maximum Tm is at pH 7, when both the acidic and basic two-thirds consists of Gly and Pro or Pro derivatives), it contains relatively residues are charged. (At the low pH, the acidic residues are protonated fewer of the other amino acids and is therefore not as good a source of and neutral, whereas at the high pH the basic residues are deprotonated amino acids as proteins containing a greater variety of amino acids. In and neutral.) The negatively charged acidic residues may form ion pairs particular, gelatin lacks tryptophan and contains only small amounts of with the positively charged basic residues, enhancing the stability of the methionine. triple helix at pH 7.

64. (a) The collagenase hydrolyzes the peptide bonds of collagen, thereby (c) The Tm values of peptide 1 do not vary as much when the pH breaking down the meat to make it easier to chew and digest. changes because the only ionizable groups are the amino and carboxyl (b) The collagenase in the fresh fruit will degrade the gelatin (collagen) termini. so that it cannot gel. Cooked fruit contains no collagenase activity be(d) The peptide with the greatest overall Tm values (and therefore the cause the heat of cooking destroys the collagenase. greatest stability) is peptide 1, which cannot form ion pairs to as great an extent as can peptide 3. Therefore, Pro and Hyp content plays a 65. (a) The patients all suffer from scurvy, a disease resulting from the lack of vitamin C, or ascorbate, in the diet. (b) Ascorbic acid is necessary for greater role in stability than the ability of the amino acid side chains the formation of hydroxyproline resi- dues in newly synthesized collagen to form ion pairs. [From Venugopal, M. G., Ramshaw, J. A. M., chains. Underhydroxylated collagen is less stable, so tissues containing Braswell, E., Zhu, D., and Brodsky, B., Biochemistry 33, 7948–7956 the defective collagen are less sound, leading to bruising, joint swelling, (1994).] fatigue, and gum disease. (c) Patients with a gastrointestinal disease may 59. (a) (Pro–Pro–Gly)10 has a melting temperature of 41 C, while (Pro– actually be consuming foods with vitamin C, but the disease impairs Hyp–Gly)10 has a melting temperature of 60 C. (Pro–Hyp–Gly)10 and absorption. Patients suf- fering from poor dentition and alcoholism may (Pro–Pro–Gly)10 both have an imino acid content of 67%, but (Pro– have overall diffi culties with food intake. Patients following various Hyp–Gly)10 contains hydroxyproline, whereas (Pro–Pro–Gly)10 does food fads might consume diets that are so unusual or restrictive that not. Hydroxyproline therefore has a stabilizing effect relative to proline. their intake of vitamin C is insuffi cient to support healthy collagen (b) (Pro–Pro–Gly)10 and (Gly–Pro–Thr(Gal))10 have the same meltsynthesis. [From Olmedo, J. M., Yiannias, J. A., Windgassen, E. B., and ing point, indicating that they have equal stabilities. This is interesting Gornet, M. K. Int. J. Dermatol., 45, 909–913 (2006).] because (Pro–Pro–Gly)10 has an imino acid content of 67%, whereas (Gly–Pro–Thr(Gal))10 has an imino acid content of only 33%. The glycosylated threonine must have an effect similar to that of proline. It is 66. The individual collagen chains are synthesized on the ribosome from the possible that the galactose, which contains many hydroxyl groups, provides additional sites for hydrogen bonding and would thus contribute constituent amino acids. Proline is one of the 20 amino acids for which a codon exists on the DNA. Some of the prolines are modifi ed post-translationally (after to the stability of the triple helix. the protein is synthesized) to hydroxyproline. Individual hydroxyproline amino (c) The inclusion of (Gly–Pro–Thr)10 is important because the results acids are not incorporated into the protein during synthesis (there is no codon for show that this molecule doesn’t form a triple helix. This molecule is inhydroxyproline). So [14C]-hydroxyproline is not used in the synthesis of collagen cluded as a control to show that the increased stability of the (Gly–Pro– and no radioactivity appears in the collagen product.

Thr(Gal))10 is due to the galactose, not to the threonine residue itself.

[From Bann, J. G., Peyton, D. H., and Bächinger, H. P., FEBS Lett. 473, 67.

O 237–240 (2000).]

NH

60. (a) The investigators chose fl uorine because they were interested in looking at electronic effects on the pyrrolidine ring and fl uorine is the most electronegative element that is not capable of forming hydrogen bonds. (b) (Pro–Flp–Gly)10 is the most stable collagen, even though it doesn’t contain hydroxyproline. However, it does contain a modifi ed proline with a strongly electronegative element attached to it. This has a large inductive effect on the ring. The investigators concluded that hydrogen bonds with bridging water molecules are not responsible for the stabilization of the collagen molecule. Instead, it is the inductive effect of the hydroxyl group. (They have suggested that the inductive effect favors the trans confi guration of the hydroxyproline peptide bond and that this trans confi guration is responsible for conferring stability on the collagen molecule.) [From Holmgren, S., Taylor, K., Bretscher, L. E., and Raines, R., Nature 392, 666–667 (1998).]

CH

2

CH2

C

1

3

CH2

4

CH

5

OH

CH2

6

NH 3

68. Hydroxylysine most often takes the place of hydroxyproline, since it includes a hydroxyl group that can participate in hydrogen bonding with water molecules, as hydroxyproline does. Hydroxylysine would not be a good substitute for proline, which lacks hydrogen-bonding groups in its side chain, or for glycine, which lacks a side chain entirely.

69. (a) Minoxidil inhibits the lysyl hydroxylase enzyme. In the presence of minoxidil, fewer Lys residues are hydroxylated, as demonstrated by the decrease of [3H]-lysine incorporated into collagen.

61.

R H R (b) Since minoxidil inhibits lysyl hydroxylase, procollagen chains would be underhydroxylated. Lysines lacking hydroxyl groups deN crease the stability of the collagen and increase the likelihood that the O collagen will be degraded once it is secreted from the fi broblast cell.

H R O This would be effective in reducing collagen concentrations in patients H with fi brosis.

N (c) A similar explanation indicates that long-term minoxidil use in N patients without fi brosis has the potential to compromise collagen synH O H thesis in fi broblasts of the skin. The underhydroxylated collagen synthesized in the presence of minoxidil will be less stable, and skin structure 62. For many proteins, bacterial expression systems offer a convenient source might be affected as a result. The medical literature reports only scalp of protein for structural studies. However, even if collagen genes were irritation, dryness, scaling, itching, and redness as a side effect in some successfully introduced into bacterial cells, the cells would not be able to men who received topical minoxidil treatments for nearly two years, produce mature collagen molecules because collagen is processed after it is however. [From Murad, S., Walker, L. C., Tajima, S., and Pinnell, Sr. R., synthesized. Bacteria are unable to undertake some of the processing steps, Arch. Biochem. Biophys. 308, 42–47 (1994) and Price, V. H., N. Engl. J. such as cleavage by extracellular proteases and covalent modifi cation of Pro Med. 341, 964–973 (1999).] and Lys residues.

Ch 6 Solutions 23 70. (a) The open reading frame is shown below. Collagen has the structure (Gly–X–Y)n, where X is often proline and Y is often hydroxyproline. The codon for Gly is GGX, and the codon for Pro is CCX (where X represents any nucleotide). In the mutant polypeptide, a Gly residue has been replaced by Ala. Normal 1(II) collagen gene: T| AAC | GGC | GAG | AAG | GGA | GAA | GTT | GGA | CCT | CCT –Asn–Gly–Glu–Lys–Gly–Glu–Val–Gly–Pro–Pro– Mutant 1(II) collagen gene: T | AAC | GGC | GAG | AAG | GCA | GAA | GTT | GGA | CCT | CCT –Asn–Gly–Glu–Lys–Ala–Glu–Val–Gly–Pro–Pro– (b) The Tm value for the mutant collagen would be lower than the Tm for normal collagen. The substitution of an Ala for a Gly would disrupt the triple helix in this region of the molecule. In order for the three chains to pack together to form the triple helix, every third residue must be a Gly. There are three amino acids per turn; therefore, the side chain of the Gly ends up on the interior of the triple helix. There is not suffi cient room to accommodate a larger side chain of an amino acid such as Ala. (c) The patient’s bones could not form properly due to the defects in the structure of collagen, and the patient died as a result. [From Freisinger, P., Ala-Kokkon, L., LeGuelle, D., Franc, S., Bouvier, R., Ritvaniemi, P., Prockop, D. R., and Bonaventure, J. J. Biol. Chem. 269, 13663–13699 (1994).]

71. Myosin is both fi brous and globular. Its two heads are globular, with several layers of secondary structure. Its tail, however, consists of a single fi brous coiled coil.

72.

Cargo

ADP actin ATP

ATP binding to the trailing head induces a conformational change that causes it to release actin.

ATP ADP

Hydrolysis of ATP to ADP + Pi triggers a conformational change that swings the trailing head forward. This also increases the affinity of the head for actin.

ADPP i ADP

ADP

The new leading head binds to the actin filament. This causes the release of the ADP from the new trailing head.

ADPP i

Pi

Pi dissociates from the leading head, preparing the myosin for another reaction cycle.

[From Walker, M. L., Burgess, S. A., Sellers, J. R., Wang, F., Hammer, J. A., III, Trinick, J., and Knight, P. J., Nature 405, 804–807 (2000).]

73. (a) Diffusion is a random process. It tends to be slow (especially for large substances and over long distances). Because it is random, it operates in three dimensions (not linearly) and has no directionality. (b) An intracellular transport system must have some sort of track (for linear movement of cargo) and an engine that moves cargo along the track by converting chemical energy to mechanical energy. The engine must operate irreversibly to promote rapid movement in one direction.

Finally, some sort of addressing system is needed to direct cargo from its source to a certain destination.

74. Myosin’s two heads act independently, so only one binds to the actin fi lament at a time. Consequently, when myosin advances from one actin subunit to another, the protein dissociates completely from its track, much like a one-legged hop. In contrast, kinesin’s two heads work together, so one head remains bound to the microtubule track when the other is released.

This mechanism is similar to a two-legged walk.

75. When muscles contract, myosin heads bind and release actin in a process that requires ATP for the physical movement of myosin along the actin fi lament. At the time of death, cellular processes that generate ATP cease.

Myosin heads remain bound to actin, but in the absence of ATP, the conformational change that causes myosin to release the actin does not occur, and stiffened muscles are the consequence.

76. (a) If rigor mortis sets in shortly after butchering, the result is a tough meat product. Butchers therefore have developed strategies to prevent the onset of rigor mortis after butchering. (b) A forensic pathologist could examine the body to determine whether the body’s position matches the location where it was found. An inconsistency would suggest that the body had been moved shortly after death.

77. Normal bone development involves the formation of bone tissue in response to stresses placed on the bone. When muscle activity is impaired, as in muscular dystrophy, the forces that shape bone development are also abnormal, leading to abnormal bone growth. 78. The two heads are necessary for kinesin’s processivity. If the protein had just one head, the protein would dissociate from the microtubule track during each reaction cycle, which would prevent its sustained forward movement.

Chapter 6 1. A globular protein can bind substrates in a sheltered active site and can support an arrangement of functional groups that facilitates the reaction and stabilizes the transition state. Most fi brous proteins are rigid and extended and therefore cannot surround the substrate to sequester it or promote its chemical transformation.

2. Myosin and kinesin are enzymes because they catalyze the hydrolysis of ATP. For myosin, the reaction is ATP1HOSADP1P For kinesin, the reaction is 2 ATP12 H2OS2 ADP12 Pi Two cycles of ATP hydrolysis are required to restore the two-headed kinesin motor to its original position.

3. The rate enhancement is calculated as the ratio of the catalyzed rate to the uncatalyzed rate: 61 s21 1.3310210 s21 5 4.7 3 1011 [From Bryant, R. A. R., and Hansen, D. E., J. Am. Chem. Soc., 118, 5498– 5499 (1996).]

4. (a) The rate constant is calculated by converting the half-life to seconds and then using the relationship k 0.693/t1/2 : 6.63106 years3

365 days 1 year

3

24 hours 1 day

3

3600 s hour

52.131014 s

Ch 6 Solutions

24

k5

0.693 t1y2 0.693

k5

(c) Alcohol dehydrogenase is an oxidoreductase. Acetaldehyde is reduced to ethanol or ethanol is oxidized to acetaldehyde. (d) Hexokinase is a transferase. The phosphate group is transferred from ATP to glucose to form glucose-6-phosphate. (e) Chymotrypsin is a hydrolase. Chymotrypsin catalyzes the hydrolysis of peptide bonds.

53.3310215 s21

2.131014 s

(b) The rate enhancement is determined by calculating the ratio of the 10. (a) isomerase (b) lyase (c) oxidoreductase (d) hydrolase

catalyzed rate to the uncatalyzed rate: 2.63103 s21 7.9 3 10 17 3.3310215 s21 5 [From Wolfenden, R., and Yuan, Y. J. Am. Chem. Soc. 130, 7548–7549 (2008).]

11. Succinate dehydrogenase is an oxidoreductase.

5. For adenosine deaminase:

1.83

COO

COO

370 s21

152.131012

1010 2 s2

H

C

H

H

C

H

For triose phosphate isomerase: 4300 s21 4.331026 s21

5 1.0 3 109

The rate of the uncatalyzed reaction is slower for the adenosine deaminase reaction than for the triose phosphate isomerase reaction. But adenosine deaminase is able to catalyze its reaction so that it occurs more quickly than the reaction catalyzed by triose phosphate isomerase.

Therefore, the rate enhancement for the adenosine deaminase reaction is greater.

6. As shown in Table 6-1, the only relationship between the rate of

succinate dehydrogenase

CH CH

COO

COO

Succinate

Fumarate

12. Malate dehydrogenase is an oxidoreductase.

CH 2 CH

COO OH

COO Malate

CO O

CH 2

malate dehydrogenase

C

O

COO Oxaloacetate

catalyzed and uncatalyzed reactions is that the catalyzed reaction is faster than the uncatalyzed reaction. The absolute rate of an uncatalyzed reaction does not correlate with the degree to which it is accelerated by an 13. A kinase transfers a phosphate group from ATP to a substrate. enzyme.

7.

H3 N

O

H N

N O

N H

O

8. (a)

H N O

NH2

O O

C NH2 N CH3

ATP

ADP

NH C

PO2 3

CH 2COO

N NH2 CH3 CH 2COO

Creatine

Creatine phosphate

NO2

O

O

HN

H2N

14. (a) glucose-6-phosphate dehydrogenase (b) isocitrate lyase (c) phosphoglycerate kinase (d) pyruvate carboxylase 15. (a) Reaction 4; (b) Reaction 1; (c) Reaction 3; (d) Reaction 2

O

16. (a) Reaction 2; (b) Reaction 4; (c) Reactions 1 and 3

17. 2 H2O2ΔO212 H2O

HN NH2 H2N

18. The common name is argininosuccinate lyase.

19. Every tenfold increase in rate corresponds to a decrease of about 5.7 kJ ?

(b) One of the products of the reaction is p-nitrophenolate, which is mol 1 in G ‡. For the nuclease, with a rate enhancement on the order of bright yellow. The progress of the reaction can be monitored in a spec- 1014, G ‡ is lowered about 14 5.7 kJ ? mol 1, or about 80 kJ ? mol 1. trophotometer at 410 nm by measuring the increase in absorbance per 20. At first, the temperature increases the reaction rate because heat unit time. increases the proportion of reacting groups that can achieve the transi9. (a) Pyruvate decarboxylase is a lyase. During the elimination of the tion state in a given time. However, when the temperature rises above a certain point, the heat causes the enzyme, which is a protein, to denacarboxylate group (OCOO ) of pyruvate, a double bond is formed in ture. Because most proteins are only marginally stable (see Section 4-3), CO2 (OPCPO). denaturation occurs readily, accounting for the steep drop in enzymatic (b) Alanine aminotransferase is a transferase. The amino group is transactivity. ferred from alanine to -ketoglutarate.

Ch 6 Solutions 25 21. (a)

Fast

(b) Amino acid side chains with hydroxyl or phenol groups (such as Ser, Thr, or—rarely—Tyr), sulfhydryl groups (Cys), amino groups (Lys), and imidazole groups (His) can serve as nucleophiles (see Figure 6-8).

Slow

G

G

27. (a) In order for any molecule to act as an enzyme, it must be able to rec-

(b)

One-step G

(c)

Two-step G

Endergonic

G

(d)

ognize and bind a substrate specifi cally, it must have the appropriate functional groups to effect a chemical reaction, and it must be able to position those groups for reaction. (b) Functional groups on the nitrogen bases can participate in chemical reactions in much the same way as amino acid side chains on proteins. For example, the amino groups on adenine, guanine, and cytosine bases could act as nucleophiles and could also act as proton donors. (c) DNA, as a double-stranded molecule, has limited conformational freedom. RNA, which is single-stranded, is able to assume a greater range of conformations. This fl exibility allows it to bind to substrates and carry out chemical transformations.

Exergonic

28. (a) DNA is not susceptible to base hydrolysis because it lacks a 2 OH group. (b) DNA’s resistance to base hydrolysis makes it a more stable molecule and a better candidate for the genetic material than RNA, which is susceptible to base-catalyzed hydrolysis.

G

Initial slow step

G

29. His 57 abstracts a proton from Ser 195, thus rendering the serine oxygen a better nucleophile. When Ser 195 is modifi ed by formation of a covalent bond with DIP, the proton is no longer available and Ser 195 is unable to function as a nucleophile.

30. Chymotrypsin can degrade neighboring molecules by catalyzing the

Initial fast step

G

22.

1st tetrahedral intermediate

hydrolysis of peptide bonds on the carboxyl side of Phe, Tyr, and Trp. If the chymotrypsin is stored in a solution of weak acid, His 57 would be protonated and would be unable to accept a proton from Ser 195 to begin hydrolysis.

31.

2nd tetrahedral intermediate

Acetylcholinesterase

F

O

CH2OH

E

CH

O P CH3

S

CH3 CH 3 Sarin

HF

G

CH2

ES covalent intermediate

H3C H3C

E

CH

O

P

O P

Modified enzyme O

CH3

Reaction coordinate 23. Yes. An enzyme decreases the activation energy barrier for both the for- 32. Acetylcholinesterase ward and the reverse directions of a reaction.

24. The nucleophilicity of a side chain is inversely related to its acidity. The O2N less acidic (the more basic or electron-rich) a group is, the more nucleophilic CH2 OH it is and the more likely it is to react with an electrophilic group.

25. (a) Gly, Ala, and Val have side chains that lack the functional groups required for acid–base or covalent catalysis. (b) Mutating one of these residues may alter the conformation at the active site enough to disrupt the arrangement of other groups that are involved in catalysis.

26. (a) The His side chain has a pK value close to neutrality and often serves as a proton donor in enzymatic reactions. But pK values of amino acid side chains in proteins often vary depending on their environment; thus, Asp, Glu, Lys, Cys and Tyr side chains can also serve as proton donors (see Figure 6-6).

Paraoxon OCH2CH3 O

P

O

OCH2CH3

OCH2CH3 CH2

O

P

O

O2N

OH

OCH2CH3 Modified enzyme

p-Nitrophenol

Ch 6 Solutions

26

33. His residues are often involved in proton transfer. A carboxymethylated His would be unable to donate or accept protons.

38. (a)

His 124

HN

O

CH

O

C

HN

CH2

HN

HBr

CH

BrCHCOO 2

HN

HN N H

N CHCOO 2

[From Shapiro, R., Weremowicz, S., Riordan, J. F., and Vallee, B., Proc. Natl.

Acad. Sci. 84, 8783–8787 (1987).]

(a)

His

Lys

Ser

(CH2)4

CH2

N

O

CH2 H

CH

C

RNA Substrate

O

H

C O H

O

O

O H

O

P

O

O

H

N

N

34.

O

CH CH2

CH2

C CH2

HN

Asp 70

O C

N H

N

H

H

H

(b) The experiment indicates that a Lys residue may be part of the active site. It is also possible that a modifi ed Lys elsewhere on the enzyme affects catalytic activity by altering the conformation of the enzyme.

(b) His 124 most likely has a pK of less than 5.0 because the His shown in the relay is unprotonated. It must be unprotonated in order to accept a proton from the water molecule. If His 124 had a higher pK value, it would be more likely to be partially protonated at physiological pH and less able to accept a proton. (c) Alanine has an aliphatic side chain and is unable to accept a pro- ton in the relay, as shown in part (a). [From Oda, Y., Yoshida, M., and Kanaya, S., J. Biol. Chem. 268, 88–92 (1993).]

39. (a) Glu 35 has a pK of 5.9 and Asp 52 has a pK of 4.5. (b) Lysozyme is inactive at pH 2.0 because both the Glu and the Asp are protonated. The Asp is no longer negatively charged and cannot nucleophilically attack the carbocation intermediate. Lysozyme is inactive at pH 8.0 because both the Glu and the Asp are unprotonated. The Glu would be unable to donate a hydrogen to cleave the bond between the sugar residues. (c) Glu 35

Asp 52

OOC

COO

Glu 35 Asp 52

O

COO

O

35. (a) HO

F NO2

HF

H Glu 35 COOH

OH

NO2

H2O

CH2

O

Asp 52

OH OOC

Glu 35 COO

H O

CH2

Asp 52

O O

NO2

NO2

40. (a) RNase A proceeds via acid–base catalysis. His 12 acts as a base and abstracts a proton from the 2 OH group on the RNA, allowing the oxygen to nucleophilically attack the phosphorous atom. His 119, acting as an acid, protonates the leaving group and promotes the breaking of the (b) Since benzoate resembles the substrate, it is likely that benzobond. (b) His 12 has a pK value of 5.4 and His 119 has a pK value of 6.4. ate binds to the active site of the enzyme. Under these conditions, His 12 is unprotonated and has the lower pK, while His 119 is protonated FDNP does not have access to the active site and will be unable to and has the higher pK. (c) The pH optimum is 6 because at this pH His react with the tyrosine (which is also assumed to be part of the en12 is unprotonated (the pH is greater than the pK ) and His 119 is zyme’s active site because of its unusual reactivity). [From protonated (the pH is less than the pK ). At a pH less than 6, His 12 Nishino, T., Massey, V., and Williams, C. H., J. Biol. Chem., 255, would be protonated and would not serve as a nucleophile to abstract 3610–3616 (1979).] the 2 -hydroxyl hydrogen. At pH greater than 6, His 119 would be unprotonated and would not be able to donate a hydrogen to the scissile 36. X could be Glu, which is similar to Asp, and Y could be Thr, which is bond. (d) Ribonuclease does not hydrolyze DNA because DNA contains similar to Ser. For example, Glu–His–Ser or Asp–His–Thr. de- oxyribose and is missing the 2 hydroxyl group, which serves as the 37. At very low pH values, His would be protonated and unable to form a at- tacking nucleophile for the phosphorus (once the His 12 has removed the hydrogen). hydrogen bond with Ser. Asp would also be protonated and unable to form a hydrogen bond with His. At very high pH values, His would be unprotonated an unable to form a hydrogen bond with Asp.

Ch 6 Solutions 27 44. (a) 41. (a) In the fi rst part of the reaction, the ester bond is cleaved and the chymotrypsin is acetylated. The p-nitrophenolate ion is quickly released, which accounts for the rapid increase in absorbance seen at 410 Acid catalyst Nucleophile nm. The enzyme must be regenerated before a second round of catalysis CH2 can begin, which requires a deacetylation step. This step is much slower Enzyme CH2 than the fi rst step. Once the acetate is released, the en- zyme is S NH regenerated and another molecule of substrate can bind and react. Thus C H N R1 HN a steady state is reached and the absorbance increases at a uniform rate Peptide R2 until the substrate is depleted.

Acid catalyst CH2

O

H3C

H

NO2

O

C

H

N

R2

Amine product

p -Nitrophenylacetate Acid catalyst

fast

O

NO2 p -Nitrophenolate

O

S C OH R1

Base catalyst

CH2

CH2

H

(yellow)

CH2

NH

R1 Acyl–enzyme intermediate

O

R1

CH3 Nucleophile

H2O

Chymotr ypsin

Enzyme

CHCOO 3

H

CH2 S

O

Acid catalyst CH2 NH

42. (a)

Asp 32

O R

OH C

O Asp 215

H N

O

H

Water

(b) The mechanism employs acid–base catalysis as well as covalent catalysis. (c) The reaction coordinate diagram would look like Figure 67, since it is a two-step reaction. (d) Bromelain acts as a meat tenderizer because it is a protease and can hydrolyze the peptide bonds in the structural proteins of the meat. This makes the meat easier to chew and digest. (e) Cys has a pK of ,3 since it must be unprotonated to act as a nucleophile, and His has a pK ,8 since it must be protonated to be active; see part (a).

(b) The reaction coordinate diagram will look like the one in Figure 6-7, since this is a two-step reaction. Each step has a characteristic activation energy. The acetylated chymotrypsin is the intermediate. (c) Yes, chymotrypsin and trypsin use the same catalytic mechanism, so trypsin can act as an esterase as well as a protease.

O

H

HN

CH2OH

C

O H

N

Carboxylate product O C H

O

slow

NH

C

O

Second tetrahedral intermediate

CH2

CH2

S

N H

Chymotr ypsin

C

N

First tetrahedral intermediate

O

CH2OH

NH R2 H

O

Chymotr ypsin

CH2

S C R1 NH

C O

H

Asp 32

Asp 32

C

C

O

OH

H N

R

R

O C

O Asp 215

(b) Renin uses an acid–base catalytic strategy. (c) The pK value of Asp 32 is lower than that of Asp 215 since the Asp 32 is unprotonated and the Asp 215 residue is unprotonated.

43. This is an example of convergent evolution, in which unrelated proteins evolve similar characteristics.

OH C R OH

O

O

OH C R

OH C O Asp 215

O R NH 2

28

Ch 6 Solutions

49. (a) The deamidation reaction for asparagine is shown. The deamidation 45. Cys 278 is highly exposed and unusually reactive compared to other cysteines in creatine kinase. Cys 278, because of its high reactivity, is probreaction for glutamine is similar. ably one of the catalytic residues in the enzyme. The other cysteine residues are not as reactive because they are not directly involved in catalysis and/ or because they are shielded in some way that prevents them from reacting O O with NEM. NH CH C NH CH C NH 4 46. The enzyme would be inactive because Cys often functions as a nucleoH2O phile and would be unable to do so if a carboxymethyl group were covalently CH2 CH2 attached to the sulfur.

C O C O NH2

O

Asparagine

Aspartate

HI I

CH2

COO

CH2

CH2

SH

S CHCOO 2

47.

H OH N

O (CH2)2

C

Glu 224

CH2

CH2

CH2

CH2

C

1

O

H

NH2

C O

OH

(b)

C

A

HA 2

CH2 CH

O

HO

2

O

NH3

NH 3 A

O H

H

Zn2

C

3

His His Glu

CH2 CH2 C O C

O (CH 2)2 Glu 224

C O

CH2

4

CH2

NH 4

NH 3

O

O

OH H2N

Zn2

O

O

H

(c) Ser and Thr residues could stabilize the transition state. They could also serve as bases (if unprotonated) and accept a proton from water to form a hydroxide ion that would act as the attacking nucleophile. Ser and Thr (in the unprotonated form) could also act as attacking nucleophiles themselves. (d) The mechanism for the deamidation of an amino terminal Gln resi- due is shown. Amino terminal Asn residues are not deamidated because a four-membered ring, which is unstable, would result.

His His Glu

[From Li, L., Binz, T., Niemann, H., and Singh, B. R., Biochemistry 39, 2399–2405 (2000).]

48.

C

O HN

CH C

O NH2

CH2

H

C

H

Im

NH2

O

Zn2

CH C H

Im Zn2

O

HN N

H

O C

Im O

Im

C

H Im Zn2 Im

O

CH

C

CH2 CH 2

C

NH

NH3 HA H

A

Im H2O

H2O

O

NH

Im

O O

NH

CH2

N

O

CH2

C

CH2O

HN HN

CH

Im HCO 3

(e) Water is a substrate in the reaction. The Asn and Gln residues on the surface of the protein have much greater access to water molecules than interior Asn and Gln residues. [From Wright, H. T., Crit. Rev. Bio- chem.

Mol. Biol. 26, 1–52 (1991).]

Ch 6 Solutions 29 50.

CH2 O R2

C

CH2 HO

H N

CH2

(CH2)4

(CH2)4

CH2

NH 3

NH3 O H O C H N R1 O First tetrahedral intermediate

R2 R1

O H2N R1 Amino product CH2

CH2

(CH2)4

CH2

CH2

(CH2)4

O

O

NH 3

O C O

O

NH3

(b) The imidazole ring is already tethered to the nitrophenyl group, so the reaction is unimolecular rather than bimolecular. In the bimolecular reaction, the reactants must fi rst encounter each other via diffusion in solution. The unimolecular reaction proceeds faster because the reacting groups are already in close proximity. (c) Enzymes speed up reaction rates in part by proximity and orienta- tion effects.

By binding to the enzyme, the reactants are in close proxim- ity to the one another. The enzyme also assists in binding the reactants in the proper orientation so that the reaction can occur with less added free energy, which results in a more rapid reaction.

59. A mutation can increase or decrease an enzyme’s catalytic activity, depending on how it affects the structure and activity of groups in the active site.

60. A water molecule is able to enter the active site after the fi rst product, the C-terminal portion of the substrate, diffuses away.

61. (a) Trypsin cleaves peptide bonds on the carboxyl side of Lys and Arg residues, which are positively charged at physiological pH. These residues Acyl–enzyme intermediate O O fi t into the specifi city pocket and interact electrostatically with Asp 189. (b) A mutant trypsin with a positively charged Lys residue in its speciH fi city pocket would no longer prefer basic side chains because the like charges would repel one another. The mutant trypsin might instead preCH2 CH2 (CH2)4 CH2 CH2 (CH2)4 fer to cleave peptide bonds on the carboxyl side of negatively charged residues such as Glu and Asp, whose side chains could interact electroO OH NH 3 NH3 HO O statically with the positively charged Lys residue.

C R2 OH O (c) If the substrate specifi city pocket does not include a positively C H charged Lys residue, then there would be no reason to expect the mutant O Second tetrahedral intermediate enzyme to prefer substrates with acidic side chains. Instead, the mutant O R2 enzyme is more likely to prefer substrates with nonpolar side chains such [From Shin, S., Yun, Y. S., Koo, H. M., Kim, Y. S., Choi, K. Y., and Oh, B.- as Leu or lIe. [From Graf, L., Craik, C. S., Patthy, A., Roczniak, S., Fletterick, R. J., and Rutter, W. J. Biochemistry 26, 2616–2623 (1987).]

H., J. Biol. Chem. 278, 24937–24943 (2003).]

R2

C

H2O

H

R2

62. (a) The pocket that holds the P1 side chain is nonpolar, fairly large, and able to hold side chains of varying sizes. The Asp–Ala–Phe–Leu peptide is hydrolyzed fastest, followed by Asp–Phe–Ala–Leu. Thus, the pocket can accommodate small aliphatic side chains such as Ala as well as larger aromatic side chains such as Phe. A positively charged amino acid does not fi t well into this pocket, since Asp–Lys–Ala–Leu is hydrolyzed more slowly than most of the other artifi cial peptides. The P2 pocket most likely accommodates a large hydrophobic side chain, since access to the active site, to stabilize the changing electronic structure of the the Asp–Ala–Phe–Leu peptide was hydrolyzed faster than the peptides with Ala at the P2 position. A substrate with a charged residue such as transition state, and to accommodate the reaction products.

Lys or Asp is hydrolyzed relatively slowly.

53. In a serine protease, there is no need to exclude water from the active (b) Aspartame is a good candidate for hydrolysis by aspartyl aminopepsite, since it is a reactant for the hydrolysis reaction catalyzed by the enzyme. tidase, since the enzyme catalyzes hydrolysis of peptides on the carboxyl 54. Yes. The smaller substrate (xylose) is unable to trigger the same conside of Asp residues (that is, Asp is in the P1 position). The amino acid formational changes elicited by glucose binding, so the enzyme is not able in the P1 position would be Phe, a large hydrophobic residue that should fi t in the P1 pocket, as suggested by the studies of the artifi cial to fully prevent the ATP hydrolysis reaction. substrates.

55. The zinc ion participates in catalysis by polarizing the water molecule so (c) that its proton is more easily abstracted by Glu 224. The positively charged zinc ion stabilizes the negatively charged oxygen in the transition state.

51. The ability of an enzyme to accelerate a reaction depends on the free energy difference between the enzyme-bound substrate and the enzymebound transition state. As long as this free energy difference is less than the free energy difference between the unbound substrate and the uncatalyzed transition state, the enzyme-mediated reaction proceeds more quickly.

52. The enzyme’s conformation must be fl exible enough to allow substrates

56. Like serine proteases, a tetrahedral intermediate forms. The intermediate might be stabilized by an “oxyanion hole” similar to that of the serine proteases.

57. The transition state structure is likely tetrahedral at position 6 on the purine ring, since adenosine is planar whereas 1,6-dihydropurine is tetrahedral at this position. Enzymes bind the transition state much more tightly than the substrate.

58. (a)

HN 3

H3N

CH CH2 C

O

O C

H N

O

O

C

CH CH2

O C

O

CH3

aspartyl aminopeptidase

O C

CH CH2 C

O

O

O

H2O H3N CH

O C

OCH3

CH2 Aspartame

N NH

O C

NO2

O N NH

O

NO2

p -Nitrophenolate

(d) The enzyme likely consists of eight identical subunits with a subunit molecular mass of 55 kD. The subunits are associated with one another noncovalently, so the entire enzyme complex has a molecular mass of 440 kD. [From Wilk, S., Wilk, E., and Magnusson, R. P., J. Biol. Chem.

273, 15961–15970 (1998).]

30

Ch 7 Solutions

2531026 M 63. During chymotrypsin activation, chymotrypsin cleaves other chymod[P] 50 d324 h?day2133600 s?h21 dt 5 trypsin molecules at a Leu, a Tyr, and an Asn residue. Only one of these (Tyr) 5. v5 fi ts the standard description of chymotrypsin’s specifi city. Clearly, chymov55.8310212 M?s21 trypsin has wider substrate specifi city, probably determined in part by the identities of residues near the scissile bond.

[From Bryant, R. A. R., and Hansen, D. E., J. Am. Chem. Soc., 118, 5498–

Chymotr ypsinogen

Trypsin Chymotr ypsin

65. No, the compound shown in Problem 8 would not be hydrolyzed by chymotrypsin. The side chain on the carboxyl side of the amide bond is an arginine side chain which would not fi t into the chymotrypsin’s specifi city pocket.

[Hippurylphenylalanine] (mM)

phenylalanine (M s) Rate of appearance of

7. (5.8 66. Yes, DIPF inactivates both trypsin and elastase. All three of these enzymes 8. have the same catalytic triad and have identical mechanisms and would thus react with DIPF in a similar manner.

67. (a) Persistent activation of trypsinogen to trypsin also results in the activation of chymotrypsinogen to chymotrypsin (see Problem 64) and causes proteolytic destruction of the pancreatic tissue. (b) Since trypsin is at the “top of the cascade,” it makes sense to inactivate it by using a trypsin inhibitor. [From Hirota, M., Ohmuraya, M., and Baba, H., Postgrad. Med. J. 82, 775–778 (2006).]

. 1

Trypsinogen

6.

3 102

12 M?s21

)(4.7 3 10 11 ) 5 2.7 M?s21

. 1

Enteropeptidase

5499 (1996).]

phenylalanine (M s) Rate of appearance of

64. Chymotrypsin activation is a cascade mechanism, since chymotrypsinogen is activated by trypsin, which is in turn activated by enteropeptidase.

68. Serine protease inhibitors would interfere with the digestion of proteins in the small intestine. Undigested proteins could not be absorbed by the cells of the small intestine, and thus their nutritional value would be lost. Gastric [Hippurylphenylalanine] (mM) upset is also a possibility.

69. A protease with extremely narrow substrate specifi city (that is, a protease The graph in Problem 6 is drawn as a straight line because in the absence with a single target) would pose no threat to nearby proteins because these of enzyme, the reaction rate is linearly proportional to the concentration of substrate. Here, the graph is drawn as a hyperbolic curve that reaches a proteins would not be recognized as substrates for hydrolysis. maxi- mum value. In the presence of enzyme, a maximum value is reached 70. Chymotrypsin prefers substrates with large hydrophobic side chains, so when all of the binding sites on the enzyme are occupied. it is surprising to fi nd asparagine as the “bait” for the chymotrypsin inhibitor. d [S] [From Chen, C., Hsu, C.-H., Su, N.-Y., Lin, Y.-C., Chiou, S.-H., and Wu, 9. v52 dt S.-H., J. Biol. Chem. 276, 45079–45087 (2001).]

0.065 M v52 60 s Chapter 7 v521.131023 M?s21 1. The hyperbolic shape of the velocity versus substrate curve suggests that the enzyme and substrate physically combine so that the enzyme becomes d [S] 2 d [P] 5 sat- urated at high concentrations of substrate. The lock-and-key model 10. v 52 dt dt describes the interaction between an enzyme and its substrate in terms of a 2 mol glucose 1.131023 M maltose highly spe- cifi c physical association between the enzyme (lock) and the v5 3 1 mol maltose s substrate (key).

21 23 ?

5 3 11.

2. Enzyme activity is measured as an initial reaction velocity, which is the Reaction velocity before much substrate has been depleted and before much product has Molecularity Rate equation been generated. It is easier to measure the appearance of a small amount of A S B 1 C Unimolecular Rate 5 k [A] A1BSC prodBimolecular Rate 5 k [A][B] 4.0 s21 k cat uct product s21?M21 than to measure the disappearance of a 2 A S B 3. from 5a baseline of zero 52.83104 small Bimolecular Rate 5 k [A]2 K M 3 1024against amountM of 1.4 substrate a background of a high concentration of 2 A S B 1 C Bimolecular Rate 5 k [A]2 d[S] substrate.

4. v 5 2 Reaction velocity dt Units of k Order proportional to . . .

0.025 M v52 s21 First [A] 6.9 3 1025 s M21 ? s21 Second [A] and [B] v 5 2 360 M ? s21 M21 ? s21 Second [A] squared [From Wolfenden, R., and Yuan, Y., J. Am. Chem. Soc. 13, 7548–7549 M21 ? s21 Second [A] squared (2008).]

Ch 7 Solutions 31 18. See Figure 7-4 for the answers to parts (a) through (d).

12. rate 5 k [sucrose] The reaction is fi rst-order overall.

(a)

13. rate5k [sucrose] rate5(5.0310211 s21)(0.050 M) rate52.5310212 M?s21

(b)

[P]

[S]

14. rate5k [sucrose4 rate5(1.03104 s21

)(0.050 M)

Time

rate55.03102 M?s21 15. (a) The reaction is a second-order reaction because the units of the rate constant k are M 1 s 1. (b) Convert the partial pressure of CO2 into units of molar concentration by using the ideal gas law:

Time

(c)

(d)

[ES]

[E]

PV5nRT n V5

P RT Time

1 atm 40 torr3 n 760 torr 5 0.0821 L?atm V 3310 K ? n 50.0021 M V

Time

(e)

(f) v0

v0

Next, substitute values for concentration and k into the rate law: rate5k[RNH2][CO2]

[E]

54950 M21?s2130.631023 M30.0021 M 56.231023 M?s21 (c) The rate constant k would increase with increasing pH because an amino group is more likely to be unprotonated (and thus able to react) at a higher pH. [From Gros, G., Forster, E., and Lin, L., J. Biol. Chem. 251, 4398–4407 (1976).]

rate k [RNH2]

[CO2] 5

0.045 M ? s21 4950 M21? s2130.631023

M

[CO2] 5 0.015 M P5 P 5

21. (a) Velocity measurements can be made using any convenient units. KM is by defi nition a substrate concentration, so its value does not refl ect how the velocity is measured. (b) It’s not necessary to know the enzyme concentration in order to determine KM or V max; only [S] and v0 must be known so that a Lineweaver–Burk plot can be constructed. The value of [E]T is required to calculate kcat since kcat V max /[E]T, according to Equation 7-23.

Vmax[S] 22. v0 5 KM 1 [S]

16. ratěˇ 5 k [RNH][CO] [CO2] 5

[S]

19. The apparent KM would be greater than the true KM because the experimental substrate concentration would be less than expected if some of the substrate has precipitated out of solution during the reaction.

20. The enzyme concentration is comparable to the lowest substrate concentration and therefore does not meet the requirement that [E]V[S]. You could fi x this problem by decreasing the amount of enzyme used for each measurement.

(65 mmol?min21

nRT

v0 5

0.015 mol

0.0821 L ?atm 3 3

760 torr 1 atm

?

P 5 290 torr

v50 33 10 K 23. v05

)(1.0 mM)

(0.135 mM) 1 (1.0 mM) 57 mmol?min21 .

V max[S] K M1[S]

7.5 mmol?min21?mg2130.15 mM

17.

zero order

v05

0.5 mM10.15 mM 7.5 mmol?min21?mg2130.15 mM 0.65 mM

v05

v0

v057.5 mmo? lmin1 first order

v50 [S]

1.7 mo m l?min21?

2?mg2130.23 mg 2 1

[From Phillips, R. S., Parniak, M. A., and Kaufman, S., J. Biol. Chem. 259, 271–277 (1984).]

32

Ch 7 Solutions

24.

v0 5 v0 5 Vmax 0.95

Vmax[S] KM 1 [S]

v05

Vmax320 KM 21 KM

v050.95 Vmax

[S] KM1 [S] 1.0 mM

K M11.0 mM 0.9 K M10.9 mM51.0 mM K M50.11 mM [From Zhu, X., Byrnes, M., Nelson, J. W., and Chang, S. H., Biochemistry 34, 2560–2565 (1995).]

When [S] 20 K M, the velocity is 95% of the maximum velocity. Therefore, a fourfold increase in substrate concentration causes a smaller proportional increase in velocity (from 83% to 95%). Estimating V max from a plot of v0 versus [S] is diffi cult because the substrate concentration must be quite high in order to achieve a maximal velocity of close to 100%. (And it is possible that these points on the hyperbolic curve cannot be experimentally measured since at high concentration the substrate may not be soluble in the reaction medium.) It is better to obtain V max by fi tting experimental data to the equation of a hyperbola or, alternatively, from a Lineweaver–Burk plot.

25. The V max is app roximately 30 M s and1 the KM is approximately Vmax 5 M. 29. kcat5 [E]T 26. The K M of enzyme 1 is about 2 mM and the K M for enzyme 2 is about 4.031027 M?s21 1 Vmax for enzyme 2 5 mM. The V max for enzyme 1 is about 3.5 mM s; the kcat5 1.031027 M is about 5.5 mM s 1. (a) The [S] is about half as large as the KM when the [S] is 1 mM for k cat 54.021 s enzyme 1. For enzyme 2, the [S] is about one-fi fth of the K M. (b) The [S] is fi ve times as great as the K M for enzyme 1. For enzyme 2,The kcat is the turnover number, which is the number of catalytic cycles per unit time. Each molecule of the enzyme therefore undergoes 4 catalytic the [S] is about twice as much as the K M. cycles per second.

27. (a) v0 0.75 V max, so substitute in: k cat 4.0 s 2 1 52.83104 s21?M21 30.

5 24 Vmax [S] KM 1.4310 M 31.

0.75 V max5 [S]1K M Reaction 1/K M 1/V max V max KM (M 1) (s M 1) (M s 1) Vmax cancels out on both sides.

(M) 4 4 0.25 1 0.25 [S] 1 2 0.50 0.755 2 1.0 [S]1K M 2 2 0.50 3 0.5 0.75 ([S]1K M)5[S] Reaction 1 has the lowest K M; Reactions 2 and 3 are tied for the highest V max.

0.75 K M50.25[S] 32. The simultaneous collision of three molecules (E, A, and B) is an un3 K M5[S] likely event. It is much more likely that the enzyme binds fi rst one and then the other substrate. For example, the fi rst bimolecular reaction might be Thus, the substrate concentration is three times as high as the K M. E AΔEA, and the second would be EA BΔEAB. (b) v0 0.9 V max, so substitute in: 33. (a) N-Acetyltyrosine ethyl ester, with its lower K M value, has a higher af0.9 Vmax5

Vmax [S] [S] 1 KM

Vmax cancels out on both sides.

0.95

[S] [S]1K M

0.9 ([S]1K M)5[S] 0.9 K M50.1[S] 9 K M5[S] Thus, the substrate concentration is nine times as high as the KM. 28.

V max35 K M v05 5 K M1K M V max35 K M v05 6 K M v050.83 V max

When [S] 5 KM, the velocity is 83% of the maximum velocity. v05

Vmax320 KM 20 KM1KM

fi nity for the chymotrypsin enzyme. The aromatic tyrosine residue more easily fi ts into the nonpolar “pocket” on the enzyme than does the smaller aliphatic valine residue. (b) The value of V max is not related to the value of K M, so no conclusion can be drawn.

34. The higher K M indicates that hexokinase has a lower affi nity for fructose than for glucose. But once the substrate binds to the enzyme, the fructose is converted to product more rapidly than glucose.

35. The maximum rate at which two molecules can collide with one another in solution is 108 to 109 M 1 s 1. Enzymes with kcat/K M values in this range can be considered to be diffusion-controlled, which means the reaction is catalyzed as rapidly as the two reactants can encounter each other in solution.

Thus, enzymes B and C are diffusion-controlled but enzyme A is not.

Enzyme A B C

KM 0.3 mM 1 nM 2 mM

kcat 5000 s21 2 s21 850 s21

kcat/K M 1.7 3 107 M21 ? s21 2 3 109 M21 ? s21 4.2 3 108 M21 ? s21

36. Enzyme B has the greater catalytic effi ciency (see Solution 35), so Product Q will be more abundant.

37. (a) The reaction is a trisubstrate reaction and therefore does not obey Michaelis–Menten kinetics. (b) The K M value for one substrate is obtained by varying its concentration while holding the concentrations of the other two substrates constant at saturating levels.

Ch 7 Solutions 33 (c) V max is achieved by saturating the enzyme with each substrate.

Therefore, the concentration of each substrate must be much greater than its K M value. [From Brekken, D. L., and Phillips, M. A., J. Biol.

Chem. 273, 26317–26322 (1998).]

38. The V max can be calculated by taking the reciprocal of the y intercept: 1 Vmax 5 y int 1

Vmax5

4.4131024 mM21?h

3 mM?h21 2 Vmax5.27310

The K M can be determined by fi rst calculating the x intercept and then by taking its reciprocal: x int5 2 x int52

b Enzyme

m 4.4131024 mM21?h 0.26 mM21?h?mM

x int521.7031023 mM21 1 K M 5 2 x int KM52

18-fold increase in K M when the Arg 386 is mutated to a lysine and an 80-fold increase in K M when Arg 292 is mutated to a lysine (which is similar to the double mutant). The Arg 292 must play a critical role in binding the aspartate substrate to the enzyme. (b) Although both Arg and Lys have positive charges that would attract the carboxylate group on the substrate, there must be other interactions that are important in addition to ionic interactions. For example, the ability of Arg and Lys side chains to form hydrogen bonds differs. Also, steric considerations may play a role, since the Lys side chain has a different shape than the Arg side chain. (c) Substitution of Arg with Lys greatly decreases the catalytic effi ciency of the mutant enzyme. Changing only one arginine decreases the catalytic effi ciency by four orders of magnitude, but the catalytic effi ciency decreases by six orders of magnitude when both arginines are replaced.

1 21.7031023 mM21

K M Aspartate (mM)

21

kcat (s )

kcat/K M (mM21 ? s21) 1.3 3 105

Wild-type Asp AT (Arg 292 Arg 386)

4

Mutant Asp AT (Lys 292 Arg 386)

326

4.5

1.4 3 101

Mutant Asp AT (Arg 292 Lys 386)

72

9.6

1.33 3 102

Mutant Asp AT (Lys 292 Lys 386)

300

0.055

1.8 3 1021

530

K M5590 mM 39. (a) The enzyme catalyzes the hydrolysis of the peptide bond on the carboxyl side of the Phe residue. One of the products, p-nitrophenolate, is bright yellow, and the rate of its appearance was monitored spectrophotometrically. (b) The K M values are nearly identical, which means that each enzyme has the same affi nity for its substrate. The kcat value for the Leu 31 enzyme is nearly six times greater than the kcat value for the wildtype enzyme, which means that the mutant enzyme has a greater catalytic effi ciency and a higher turnover rate of substrate converted to product per minute. The k cat/K M ratio refl ects the specifi c reactivity of the substrate AAPF with the enzyme, and this ratio is larger in the mutant enzyme than in the wild type. (c) The mutant enzyme is nearly threefold more active toward the casein substrate than the wild-type enzyme. Thus the improvement in catalytic activity seen in the mutant is not just toward an artifi cial substrate but toward a natural substrate as well. (d) Ile 31 is near the Asp–His–Ser catalytic triad of the subtilisin E enzyme. Because Leu 31 improves the catalytic activity, this residue must somehow improve the function of the catalytic triad. Residue 31 is especially close to Asp, so it's possible that this residue plays a role in assisting Asp in its function in the catalytic triad. Histidine acts as a base catalyst and abstracts a proton from serine. The imidazole ring of histidine is positively charged as a result, and the role of Asp is to stabilize the positively charged imidazole ring. Thus leucine must somehow enable the Asp to fulfi ll this function better than isoleucine.

Another possibility is that the substitution of leucine for isoleucine has altered the three-dimensional protein structure such that the catalytic triad residues are closer to one another and thus proton transfer is facilitated. (e) Subtilisin would remove protein stains by hydrolyzing the peptide bonds of the protein and producing amino acids or short peptides as products. These products could be easily washed away from the clothing. [From Takagi, H., Morinaga, Y., Ikemura, H., and Inouye, M., J. Biol. Chem. 263, 19592–19596 (1988).]

(d) Changing the amino acid from Arg to Lys may cause a three-dimensional conformational change that would interfere with the ability of the catalytic apparatus to function properly. If proton donation is important to the mechanism, this would be affected as well, since the lysine side chain donates a proton more readily than the resonancestabilized gua- nidino group in the side chain of arginine. [From Vacca, R. A., Giannat- tasio, S., Graber, R., Sandmeier, E., Marra, E., and Christen, P., J. Biol. Chem. 272, 21932–21937 (1997).]

41. (a) If an irreversible inhibitor is present, the enzyme's activity would be exactly 100 times lower when the sample is diluted 100-fold. Dilution would not change the degree of inhibition. (b) If a reversible inhibitor is present, dilution would lower the concentrations of both the enzyme and the inhibitor enough that some inhibitor would dissociate from the enzyme. The enzyme’s activity would therefore not be exactly 100 times less than the diluted sample; it would be slightly greater because the proportion of uninhibited enzyme would be greater at the lower concentration.

42. By irreversibly reacting with chymotrypsin’s active site, DIPF would decrease [E]T. The apparent Vmax would decrease since Vmax kcat[E]T (Equation 7-23). K M would not be affected since the unmodifi ed enzyme would bind substrate normally.

43. (a) Since the structures are similar (both have choline groups), the inhibitor is competitive. Competitive inhibitors compete with the substrate for binding to the active site, so the structures of the inhibitor and the substrate must be similar. (b) Yes, the inhibition can be overcome. If large amounts of substrate are added, the substrate will be able to effectively compete with the inhibitor such that very little inhibitor will be bound to the active site. The substrate “wins” the competition when it is in excess. (c) Like all competitive inhibitors, the inhibitor binds reversibly.

44. Indole is a competitive inhibitor of chymotrypsin because its structure resembles the side chain of tryptophan, which fi ts into the specifi city pocket of chymotrypsin. Thus, indole and tryptophan side chains compete with 40. (a) The wild-type enzyme has the lowest KM, indicating that it has each other for binding to the active site. Indole does not fi t into the specifi cthe highest affi nity for aspartate. The aspartate substrate is not able ity pocket of either trypsin or elastase and therefore is not an effective comto bind to the mutant enzymes with as high an affi nity. There is an petitive inhibitor of these enzymes.

34

Ch 7 Solutions

45. It is diffi cult to envision how an inhibitor that interferes with the cata- [From Gross, R. W., and Sobel, B. E., J. Biol. Chem. 258, 5221–5226 lytic function (represented by kcat or Vmax) of amino acid side chains at the (1983).] active site would not also interfere with the binding (represented by KM) of 54. (a) The inhibitor is a competitive inhibitor because the KM increases in a the presence of the inhibitor while the Vmax stays the same. substrate to a site at or near those same amino acid side chains. (b) The values are shown in the table below. The K M is calculated by 46. Since competitive inhibitors compete with substrate for binding to the taking the negative reciprocal of the x intercept (which is calculated active site of the enzyme, it is possible (especially if the concentration of the from the equation of the straight line and is equal to b/m). The V max is substrate increases in vivo) that this inhibition can be overcome. There might the reciprocal of the y intercept. also be cellular complications if these competitive inhibitors, which resemble substrates, accumulate. Noncompetitive and uncompetitive inhibitors do x Intercept KM Vmax 1 1 not resemble the substrate, and inhibition cannot be overcome by increasing ( mol min 1) (mL mg ) (mg mL ) the concentration ubstrate.

If these inhibitors of s bind to their targets very 1.16 tightly, small amounts can be used to inhibit the enzyme effectively. [From Without 0.86 0.44 Westley, A. M., and Westley, J. J. Biol. Chem. 271, 5347–5352 (1996).] inhibitor 0.48 2.1 0.43 47. (a) NADPH is structurally similar to NADP and is likely to be a comWith inhibitor P56 petitive inhibitor. (b) The Vmax is the same in the presence and absence of the inhibitor since inhibition can be overcome at high substrate concentrations. The K M (withI) K Mincreases bec ause a higher concentration of substrate is needed to (c) a 5 K M(no I) achieve half-maximal activity in the presence of an inhibitor. , indicating that the enzyme 2.1 mg?mL21 (c) The K is M 400 times greater for NAD prefers NADP as a cofactor. The differences in Vmax are not as great. a 5 0.86 mg mL 2 1 5 2.44 ?

[From Hansen, T ., Schicting, B., and Schonheit, P., FEMS Microbiol.

Lett. 216, 249–253 (2002).]

[I] a5 1 1 K 48. (a) NADPH is the most effective inhibitor because it has the lowest 0.22 mM K I value.

KI (b) Inorganic phosphate is likely to be completely ineffective because 2.44 5 1 1 its K Ivalue is so much greater than the K values of either of the two M K I 5 0.15 mM substrates. This indicates that the enzyme has a much greater affi nity for its substrates and thus will bind the substrates rather than the inhibitor.

Thus, the inhibitor will be ineffective.

[From Amlabu, E., et al., J. Cell Molec. Biol. 9, 43–49 (2011).]

49. The compound is a transition state analog (it mimics the planar transi55. The inhibitor is a mixed inhibitor. The V max is decreased and the K M tion state of the reaction) and therefore acts as a competitive inhibitor. value is increased in the presence of the inhibitor.

50. Compound A has a K value of 1.2 10 12 M. The lower K I value I The V max in the absence of inhibitor can be calculated by taking the reciproindicates that it is a more effective inhibitor. Compound A is more effective cal of the y intercept: because its structure most closely resembles the structure of the transition state. Compound B lacks the large polar OH group that mimics the transi1 Vmax5y int tion state of the reaction.

51. The structure of coformycin structurally resembles the proposed transi1 tion state for adenosine deaminase (see Section 7-3), and this supports the Vmax5 1.51 (OD 2 1 ?min) proposed structure. However, 1,6-dihydroinosine has a K of 1.5 10 13 M I

whereas coformycin’s K I is about 0.25 M; thus, 1,6-dihydroinosine more closely resembles the transition state than does coformycin.

21

Vmax5 0.66 OD ? min The Vmax in the presence of inhibitor can be calculated similarly:

52. (a) Zanamivir, with a lower K I (indicating tighter binding of the inhibitor to the enzyme) would work better.

1 Vmax5 y int (b) Vmax is about the same, but K M increases, so the mutant enzyme probably binds the substrate more poorly than the wild-type enzyme but 1 does not exhibit any decrease in turnover number (kcat, refl ected in Vmax).

Vmax5 21 ?min) 4.27 (OD (c) The mutation increases the K forI oseltamivir by 265 times but increases the K I for zanamivir by only 2 times. Therefore, zanamivir Vmax 50.23 OD?min21 would be a better inhibitor of neuraminidase in the mutant virus. [From Collins, P. J., Haire, L. F., Lin, Y. P., Liu, J., Russell, R. J., Walker, P. A., The K M value in the absence of inhibitor and can be determined by fi rst calculating the x intercept and then by taking its reciprocal: Skehel, J. J., Martin, S. R., Hay, A. J., and Gamblin, S. J., Nature 453, 1258–1261 (2008).] b m xint52 K M(with I) 53. a5 K M(no I) OD21?min 1.521.51 min?OD21?mM xint52 40 M m 5 4 a5 10 mM xint520.99 mM21 [I] a 5 1 1 K 4511 KI510 mM

I

30 mM KI

KM52

1 x int

KM52

1 20.99 mM2

KM51.0 mM

1

Ch 7 Solutions 35 The KM value in the presence of inhibitor and can be similarly determined: b m

xint52

57. (a) The Lineweaver–Burk plot is shown. The KM is calculated from the x intercept, the Vmax from the y intercept.

Inhibition of PTP1B by vanadate

4.27 OD21?min

xint52

0.4

1.58 min?OD21?mM x int522.70 mM21

y 9.1369x 0.0352

1

KM52 K

0

1 K M 5 2 x int

1/v (s/nM)

0.3 0.2 0.1

22.70 mM21 50.37 mM

Dodecyl gallate is an uncompetitive inhibitor. In the presence of the inhibitor, the Vmax and the K M values decreased to a similar extent. The slopes of the lines in the Lineweaver–Burk plot are nearly the same. [From Kubo, I., Chen, Q.-X., and Nihei, K.-I., Food Chem. 81, 241–247 (2003).]

0.05

y 0.05 1/[FDP] ( M 1)

0

0.4

0

1/v (min/nmol)

y 3.4337x 0.038 0.3 0.2 y

0

0.02

0.04

1.8985x

0.06

0.08

0.0349

0.1

With vanadate

x intercept ( M 1)

0.038

0.0039

KM ( M) y intercept (s nM 1) Vmax (nM s 1 )

26

260

0.035

0.035

28.5

28.5

presence and absence of the inhibitor, but the KM has increased tenfold, indicating that the vanadate is competing with the substrate for binding to the active site of the enzyme.

58. To calculate KI, rearrange Equation 7-28 and solve for :

a 5

Without inhibitor With inhibitor

Vmax [S] 2[S] v0 K

Substitute in values for Vmax and K M (in the absence of vanadate). The sub-

Without p6*

With p6*

x intercept (mM21)

20.0185

20.011

[Vanadate] (mM)

v0 (nM ? s21)

K M (mM)

54

89

0.035

0.0385

28.5

26.0

0.0 0.2 0.4 0.7 1.0 2.0 4.0

5.70 3.83 3.07 2.35 2.04 1.18 0.71

y intercept (min ? nmol21) V max (nmol? min

0.15

Without vanadate

1/[S] ( M 1)

21

0.1

(b) The inhibitor is a competitive inhibitor. The Vmax is the same in the

0.1

0.02

0.035

Without inhibitor With inhibitor

56. (a) Lineweaver–Burk plots are shown below. The K M is calculated from the x intercept, the Vmax from the y intercept.

Inhibition of HIV-1 protease by p6*

0.9122x

)

strate concentration is 6.67 mM. The values for a are presented in the table: a 1.0 1.7 2.1 2.9 3.3 6.0 10.1

(b) The inhibitor is a competitive inhibitor. The Vmax is the same in the presence and absence of the inhibitor (within experimental error), but the KM has increased nearly twofold, indicating that the p6* is competA plot of a versus inhibitor concentration, shown below, gives a slope of ing with the substrate for binding to the active site of the enzyme.

21yaK M 2.26 mM21. The value of K I is the reciprocal of the slope, or 0.44 mM. 1 (c) 5 21yK M a Determination of K I 2 (20.011M2ym 120.0185 mM1)50.595 12

a51.68 a 5 1 1 1.68511

(I)

10

KI

8

10 mM

6

KI 0.68510 mMyKI KI515 mM [From Paulus, C., Hellebrand, S., Tessmer, U., Wolf, H., Kräusslich, H.-G., and Wagner, R., J. Biol. Chem. 274, 21539–21543 (1999).]

y

2.2599x

1.1918

4 2 0

0

1

2 [Vanadate] ( M)

3

4

36

Ch 7 Solutions

[From Huyer, G., Liu, S., Kelly, J., Moffat, J., Payette, P., Kennedy, B., Tsaprailis, G., Gresser, M., and Ramachandran, C., J. Biol. Chem. 272, 843–851 (1997).]

59. (a) The Lineweaver–Burk plot is shown. The KM is calculated from the x intercept, the Vmax from the y intercept.

Without inhibitor

With inhibitor

y intercept, (mM/min)21

0.704

1.90

Vmax (mM/min) x intercept (mM)21

1.42 –0.949

0.52 –2.54

K M (mM)

1.05

0.39

1/v (mMmin) 1.

0

y

0.751x

2.0

3.0

2.5

2.0

1.5 1.0 0.5 0.0 0.5 1/[S] (mM 1)

1.0

0.742x

0.704

2.0

2.5

1.5

2.38

x intercept (mL ? mg–1)

26.56

K M(mg ? mL21

0.102

0.152 15.6

min21 Vmax (nmol mL21 ? ?

4.19 0.239

) y intercept (min ? mL ? nmol21) )

0.064

61. (a) ATCase is an allosteric enzyme because its activity versus [S] curve is sigmoidally shaped. (b) CTP is a negative effector, or inhibitor, because when CTP is added, the KM increases and thus the affi nity of the enzyme for the substrate decreases. CTP is the eventual product of the pyrimidine biosynthesis pathway; thus, when the concentration of CTP is suffi cient for the needs of the cell, CTP inhibits an early enzyme in the synthetic pathway, ATCase, by feedback inhibition. (c) ATP is a positive effector, or activator, because when ATP is added, the KM decreases and thus the affi nity of the enzyme for its substrate increases. ATP is a reactant in the reaction sequence, so it serves as an activator. ATP is also a purine nucleotide, whereas CTP is a pyrimidine nucleotide. Stimulation of ATCase by ATP encourages CTP synthesis when ATP synthesis is high, thus balancing the cellular pool of purine and pyrimidine nucleotides.

1.905

y

0.427 29.81

Slope (min ? mL ? nmol–1)/(mL ? mg–1)

Hannun, Y. A., J. Biol. Chem. 274, 21335–21341 (1999).]

3.0

1.0

With inhibitor

[From Kishikawa, K., Chalfant, C. E., Perry, D. K., Bielawska, A., and

5.0

4.0

Without inhibitor

3.0

62. (a) The combination that gives the most effective inhibition is

CTP UTP. More than 90% of the ATCase activity is inhibited when CTP and UTP are both present. In contrast, CTP alone inhibNo inhibitor With homoarginine its only 60% of ATCase activity, and UTP alone inhibits only 5% (b) Homoarginine is an uncompetitive inhibitor. The slopes of the lines of ATCase activity. in the Lineweaver–Burk plot are nearly identical. A proportional de(b) UTP and CTP are both pyrimidines. Regulation of the ATCase in this manner makes metabolic sense because CTP is synthesized from crease in Vmax and KM occurs in the presence of the inhibitor. (c) Because homoarginine is an uncompetitive inhibitor, it does not UTP and thus is produced in larger quantities fi rst. The CTP produced bind to the active site of the alkaline phosphatase enzyme but to another can then inhibit the ATCase enzyme, but not completely. Only when site that interferes with the activity of the enzyme in some way. The UTP levels rise do UTP and CTP act synergistically to completely inintestinal alkaline phosphatase catalyzes the same reaction as the bone hibit the ATCase enzyme. alkaline phosphatase, so the active sites of the two enzymes are likely to (c) The graph below shows a hypothetical curve for the effect of CTP be similar, but the structures of the enzymes may be suffi ciently differ- and UTP combined on ATCase activity. The KM value is increased, inent that the intestinal enzyme lacks the binding site for homoarginine. dicating that a higher concentration of aspartate is required to achieve [From Lin, C.-W., and Fishman, W. H., J. Biol. Chem. 247, 3082–3097 half-maximal activity. Although there isn’t enough information to make (1972).] a defi nite conclusion, the Vmax value is probably unchanged, as the Vmax values are unchanged in the presence of ATP alone and CTP alone.

60. (a) This is an example of mixed inhibition.

5 Inhibition of PP1 by PA 4

300

40

20

v (U mg) .1

15.637

0

200

2.3826x

. . −1

1/v (min mL nmol)

0

3 y

2 1

100 0

y 0.4268x 4.1878 0

20

40

60

80

100

1/[MBP] (mL/mg) Without inhibitor With inhibitor

(b) The KM has increased by 50% in the presence of PA, but there has been a fourfold decrease in Vmax as shown in the table.

0

5

10

15

20

25

[Aspartate] (mM) Control With ATP

With CTP With CTP and UTP

63. The formation of a disulfi de bond under oxidizing conditions, or its cleavage under reducing conditions, could act as an allosteric signal by altering the conformation of the enzyme in a way that affects the groups at the active site.

Ch 8 Solutions 37 64. (a) The M1 isoform of pyruvate kinase is not an allosteric enzyme, as shown by the hyperbolic shape of the curve. F16BP does not infl uence the activity of the enzyme. (b) Substitution of Ala by Arg at an intersubunit contact converted the pyruvate kinase from a nonallosteric form to an allosteric form. F16BP is an allosteric activator of the mutant enzyme, as shown by the curve, which shifts left. It’s possible that substitution of the neutral Ala by the positively charged Arg produced intersubunit contacts, perhaps in the form of ion pairs, that allowed the subunits to behave in a cooperative manner. [From Ikeda, Y., Tanaka, T., and Noguchi, T. (1997) J. Biol.

Chem. 272, 20495–20501.]

6.

OH H3C

(CH)2 7

CH

CH

C H

(CH2)6

COO

Miner val 7.

H H3C

C

(CH)2 7

C

(CH2 )7

COOH

H Elaidic acid Chapter 8 1. (a)

The melting point of elaidic acid is higher than the melting point of oleic acid, since the trans double bond of elaidic acid gives it an elongated shape, whereas the cis double bond oleic acid gives it a bent shape.

COO

(CH2)12

HC

Myristate (14:0)

(b)

H3C

CH

CH

(CH)25

COO

(CH2)7

8.

Palmitoleate (16:1 n-7)

(c) H3C

(d)

CH2

H3C

(CH CH CH 2)3 (CH2)6 -Linolenate (18:3 n-3) CH

CH

(CH)27

O CH2

COO

HO

C

C

(CH)2 7

CH3

C

C

(CH2 )7

CH3

H

CH2

COO

(CH2)13

O

H

OH

Nervonate (24:1 n-9) 2. (a) H3C

CH2

CH

(CH

CH 2)5

(CH2)2

[From Chao-Mei Y., Curtis, J. M., Wright, J. L. C., Ayer, S. W., and FathiAfshar, Z. R., Can. J. Chem. 74, 730–735 (1996).]

COO

EPA (20:5 n-3)

(b) H3C

CH2

(CH CH CH2)6 DHA (22:6 n-3)

9. (a) Because it has a head group with just one anionic charge, SQDG is likely to substitute for phosphatidylglycerol (net charge 1) rather than phosphatidylethanolamine (with a positive and a negative charge). (b) When phosphorus is limited, the organism produces relatively more of the sulfur-containing SQDG to substitute for phospholipids in the cell membrane.

COO

CH2

3.

H3C

(CH)2 4

CH

CH

CH2

CH

CH

(CH2)4

CH

5, 11, 14

Sciadonate (20:3

CH

(CH)2 3

COO

10.

)

O

[From Sayanova, O., Haslam, R., Venegas Caleron, M., and Napier, J. A., Plant Physiol. 144, 455–467 (2007).]

4. (a)

H 3C

(CH2)7

CH

CH

(CH2)11

O H3C

COOH

(H2C) 14

CH2

C

O

H

(b) H3C

(CH2)4

C

C

C

H

H

H

C

C

(CH2)6

COOH

H

H3C

H CH2 CH CH (CH2)7 C OH Ricinoleic acid (12-hydroxy-18:1 9)

(CH2)5

COOH

5.

H 3C

CH

(CH)213

CH

(CH2 )2 24:2

CH

CH

(CH)214

CH3

(CH)214

CH3

O O

C

R1 R2 C

C

O

O

H2C

C

CH2

H

O

O

O

[From Cahoon, E. B., and Kinney, A. J., Eur. J. Lipid Sci. Technol. 107, 239– 243 (2005).] (a)

11.

O

Calendic acid (18:3 8trans, 10trans, 12cis)

(c)

C

Tripalmitin

H C

CH CH2

Erucic acid (22:1 13)

O

(CH)2 3

COO

5, 9

P O

HO H

O H

H HO OH H H

OH H

OH

(b) H3C

(CH)2 4

(CH

CH

CH2)2

(CH2)3

CH

CH

24:2 5, 9, 15, 18

(CH)2 2

CH

CH

(CH)2 3

COO

12. The inositol head group has fi ve hydroxyl groups that could potentially undergo phosphorylation.

38

Ch 8 Solutions

13.

CH3

O O

P

O

(CH2)2

O H2C

CH

O

O

(CH)214 CH 3

CH3

CH2

19. Both DNA and phospholipids have exposed phosphate groups that are recognized by the antibodies.

20. (a)

C

O C

N CH3

18. Phosphatidylcholine and phosphatidylethanolamine are neutral. Phosphatidylglycerol and phosphatidylserine are negatively charged.

O

O

DPPC

O

(CH2)14

O

CH3

CH2

CH

H2C OH

14.

OH CH

CH

(b) H 2C

CH

CH

O

O

(CH2)2

C

C

R

R

CH

O C O H

O O

CH2

P

CH3 O

(CH2)2

O

CH3

N

H

CH3

O

(c)

CH3

COO

NH 3

OH

O

CH

O

NH 3CH

C

CH 2

O

R

OC R

OH

H2C

P

O P

O

CH

O

O

CH (CH2)6 CH3

H2C

CH

O

O

C

C

R

R

O

CH2 HO

CH2 CH

CH2 OH

OH O

H

[From Ohasi, Y., Tanaka, T., Akashi, S., Morimoto, S., Kishimoto, Y., and Nagai, Y., J. Lipid Res. 41, 1118–1124 (2000).]

15. Glucose

O

CH2

HO

CH

CH

CH

NH

CH

C

21. The spicy ingredient in the food is a powder made from peppers that contains the hydrophobic compound capsaicin (see Section 8-1). Yogurt containing whole milk also contains hydrophobic ingredients that can cleanse the palate of the irritating capsaicin. Water is polar, so it does not dissolve the capsaicin and cannot cleanse the palate.

22. Vitamins A, D, and K are isoprenoids and are nonpolar. These dietary

O

vitamins are soluble in the synthetic lipid Olestra® and pass out of the intestinal tract along with the Olestra® without being absorbed. Adding these vitamins to the product helps saturate the synthetic lipid with vitamins so that dietary vitamins are not excreted.

(H2C)12 (CH2)26 CH3 CH2 O

C

(CH2)6(CH2CH

CH)2(CH2)4CH3

O

16.

HO

HO

CH2

CH

CH

CH

NH

CH

C

tissue. Vitamin C is water soluble and any excess is excreted in the urine.

25. (b) is polar, (d) is nonpolar, and (a), (c), and (e) are amphipathic.

O

26. Only (e) will form a bilayer (see Fig. 8-4). The molecules shown in (a) and (c), though amphipathic, have one “head” and only one “tail” and will therefore preferentially form micelles (see Fig. 2-10).

(H2C)12 (CH2)26 CH3CH2 O

23. Vitamin A, and the compound from which it is derived, -carotene, are lipid-soluble molecules. The vegetables in a typical salad do not contain large amounts of lipid. The addition of the lipid-rich avocado provided a means to solubilize the -carotene and thus increase its absorption. [From Unlu, N. Z., Bohn, T., Clinton, S. K., and Schwartz, S. J., J. Nutr. 135, 431–436 (2005).]

24. Vitamins A and D are lipid soluble and therefore accumulate in the fatty

C O

(CH2)2

NH

27. (a) A hydrocarbon chain is attached to the glycerol backbone at position

CH

1 by a vinyl ether linkage. In a glycerophospholipid, an acyl group is attached by an ester linkage. (b) The presence of this plasmalogen would not have a great effect since it has the same head group and same overall shape as phosphatidylcholine.

C

O

17. All except phosphatidylcholine have hydrogen-bonding head groups.

Ch 8 Solutions 39 28. A glycerophospholipid with two saturated acyl chains has a cylindrical shape, whereas a glycerophospholipid with two unsaturated, kinked acyl chains would be more cone-shaped:

36. When phytanic acid is incorporated into membrane phospholipids, the resulting membrane is more fl uid. The presence of the methyl groups on the phytanic acid results in an acyl chain that has a decreased ability to interact with neighboring acyl chains. This decreases the number of van der Waals interactions and decreases the melting point, which increases membrane fl uidity. [From van den Brink, D. M., van Miert, J. N. I., Dacremont, G., Rontani, J.-F., and Wanders, R. J. A., J. Biol. Chem. 280, 26838–26844 (2005).]

37. The cyclopropane ring in lactobacillic acid produces a bend in the aliphatic chain and thus its melting point should be closer to the melting point of oleate, which also has a bend due to the double bond. The presence of bends decreases the opportunity for van der Waals forces to act among neighboring molecules. Less heat is required to disrupt the intermolecular forces, resulting in a melting point that is lower than that of a saturated fatty Saturated acyl chains Unsaturated acyl chains acid with a similar number of carbons. Therefore stearate has the highest melting point (69.6 C), followed by lactobacillic acid (28 C) and oleate 29. Lipids that form bilayers are amphiphilic, whereas triacylglycerols are (13.4 C). nonpolar. Amphiphilic molecules orient themselves so that their polar head 38. Increasing the temperature would make the membrane more fl uid. To groups face the aqueous medium on the inside and outside of the cell. Also, maintain constant fl uidity, the bacteria synthesize fatty acids with more triacylglycerols are cone-shaped rather than cylindrical and thus would not carbons and with fewer double bonds. fi t well in a bilayer structure, as shown in Figure 8-4.

39. Cholesterol’s planar ring system interferes with the movement of acyl 30. Phospholipase A1 catalyzes the hydrolysis of one of the acyl chains on a phospholipid. The resulting product, a lysophospholipid (see Problem 20a), has a cone-shaped structure (see Fig. 8-4). Lysophospholipids do not assemble to form bilayers as the cylindrical-shaped glycerophospholipids do. The conversion of a signifi cant portion of glycerophospholipids to lysophospholipids results in the destruction of the red blood cell membrane.

31. Two factors that infl uence the melting point of a fatty acid are the number of carbons and the number of double bonds. Double bonds are a more important factor than the number of carbons, since a signifi cant change in structure (a “kink”) occurs when a double bond is introduced. An increase in the number of carbons increases the melting point, but the change is not nearly as dramatic. For example, the melting point of palmitate (16:0) is 63.1 C, whereas the melting point of stearate (18:0) is only slightly higher at 69.1 C. However, the melting point of oleate (18:1) is 13.2 C, a dramatic decrease with the introduction of a double bond.

32. (b) (a) (c). Trans-oleate has a melting point similar to that of stearate (18:0) because the trans double bond does not produce a kink in the molecule. Its geometry more closely resembles that of a single bond.

chains and thus tends to decrease membrane fl uidity. At the same time, cholesterol prevents close packing of the acyl chains, which tends to prevent their crystallization. The net result is that cholesterol helps the membrane resist melting at high temperatures and resist crystallization at low temperatures.

Therefore, in a membrane containing cholesterol, the shift from the crystalline form to the fl uid form is more gradual than it would be if cholesterol were absent.

40. In a bilayer, one end of each lipid acyl chain is fi xed by its attachment to a head group. The methylene groups closer to the head group have the least conformational freedom, whereas the methylene groups farthest from the head group, near the bilayer center, have the greatest freedom.

41. No. Higher temperatures increase fatty acid fl uidity. To counter the effect of temperature, the plants make relatively more fatty acids with higher melting points. Dienoic acids have higher melting points than trienoic acids because they are more saturated. Therefore, the plants convert fewer dienoic acids into trienoic acids.

42. The cold temperatures might induce the activation of desaturase enzymes in the plant that convert the 18:0, 18:1, and 18:2 fatty acids to 18:3 fatty 33. In general, animal triacylglycerols must contain longer and/or more satacids. Of these four fatty acids, 18:3 is the most unsaturated and has the urated acyl chains than plant triacylglycerols, since these chains have higher lowest melting point. Membranes composed of phospholipids containing melting points and are more likely to be in the crystalline phase at room unsaturated fatty acids maintain their fl uidity in cold temperatures. [From temperature. The plant triacylglycerols must contain shorter and/or less Shi, Y., An, L., Zhang, M., Huang, C., Zhang, H., and Xu, S., Protoplasma 232, satu173–181 (2008).] rated acyl chains in order to remain fl uid at room temperature.

43. (a) PS and PE both contain amino groups.

34. Peanut oil has a higher melting point because the fatty acids that com(b) PC and SM both contain choline groups. pose the monounsaturated triacylglycerols have a higher melting point than (c) PE, PC, and SM are all neutral, but PS carries an overall negative the more highly unsaturated fatty acids of the vegetable oil. Each double charge. Since PS is exclusively found on the cytosolic-facing leafl et, this bond introduces a “kink” in the molecule, which means that the fatty acids side of the membrane is more negatively charged than the other side. don’t pack together as well. The number of double bonds corresponds to the 44.

The fl ippase enzyme has a strong preference for PS, which is consistent number of “bends” in the acyl chain. Fatty acids that do not pack together well have fewer London dispersion forces among the chains, and less heat with the observation that PS is located exclusively in the cytosolic-facing energy is required to disrupt the forces and melt the solid. Therefore, the leafl et of the membrane. Translocation of PE also explains the vegetable oil, with a higher percentage of polyunsaturated triacylglycerols, predominance of PE in the cytosolic leafl et. PC and SM, both cholinecontaining lipids, are not translocated and remain in the extracellular leafl et has a lower melting point and does not freeze. of the membrane. The fl ippase has a preference for amino-containing 35. The lipids from the meat of the reindeer slaughtered in February phospholipid head groups and eschews choline-containing phospholipids. contained fewer unsaturated acyl chains than their healthy counterparts. The fl ippase requires ATP, probably because translocation occurs against a Unsaturated fatty acids have a lower melting point due to the presence concentration gradient, and magnesium ions. A cysteine amino acid side of double bonds that prevent them from packing together tightly. These chain must be essential for translocation activity since chemical modifi lipids therefore help membranes remain fl uid even at low temperatures cation of this side chain abolishes translocation activity. [From Daleke, D. L., as the reindeer walks through the snow. Saturated fatty acids, which pack and Heustis, W. H., Biochemistry 24, 5406–5416 (1985).] together more effi ciently and have higher melting points, decrease mem45. (a) Detergents are required to solubilize a transmembrane protein bebrane fl uidity at low temperatures. The decreased percentage of lipids cause the protein domains that interact with the nonpolar acyl chains with unsaturated fatty acyl chains result in decreased membrane fl uidity are highly hydrophobic and would not form favorable interactions with and may compromise the ability of the animal to survive a cold winwater. ter. [From Suppela, P., and Nieminen, M., Comp. Biochem. Physiol. 128, 53–72 (2001).]

40

Ch 8 Solutions (b) A schematic diagram of the detergent SDS interacting with a trans(c) A barrel could contain some parallel strands, but these could membrane protein is shown here. The polar head group of SDS is repnot be consecutive. A barrel with consecutive parallel strands could resented by a circle and the nonpolar tail as a wavy line. The nonpolar occur only if the strands were linked by additional membranetail of the SDS interacts with the nonpolar regions of the protein, efspanning segments (such as a transmembrane helix or a structure fectively masking these regions from the polar solvent. The polar head that passed through the center of the barrel). group of the SDS interacts favorably with water. The presence of the detergent effectively solubilizes the transmembrane protein so that it 51. A steroid is a hydrophobic lipid that can easily cross a membrane to enter the cell. It does not require a cell-surface receptor, as does a polar molcan be purifi ed. ecule such as a peptide. 52. Melittin is more conformationally restricted in

SDS

the membrane when it is associated with glycerophospholipids containing more saturated fatty acids (like oleate, 18:1) and less restricted when associated with lipids containing more unsaturated fatty acids (like arachidonate, 20:4). The presence of double bonds in the acyl chains produces “bends” that prevent the acyl chains from packing closely together. The unsaturated fatty acids thus provide a less conformationally restricted environment for the melittin peptide.

In contrast, when there are fewer double bonds, such as in oleate, the fatty acyl chains pack more closely together and interact via van der Waals forces.

This environment restricts the conformation of melittin. [From Raghuraman, H., and Chattopadhyay, A., Biochim. Biophys. Acta 1665, 29–39 (2004).]

53. (a) Glycosphingolipids pack together loosely because their very large head groups do not allow tight association. (b) The lipid raft is less fl uid because of both the presence of cholesterol and the saturated fatty acyl chains, which pack together more tightly than unsaturated acyl chains. [From Pike, L., J. Lipid Res. 44, 655–667 (2003).]

54. (a) Cholesterol and cholesteryl esters are highly nonpolar and must be

solubilized via their association with protein and phospholipids in order to be transported through the aqueous bloodstream. (b) An LDL particle is similar to a membrane in that the surrounding 46. (a) A or D (b) A or C (c) D layer consists of phospholipids (however, this is a single layer, rather (d) D (e) B (f) B or E than a bilayer). Cholesterol is associated with this layer of phospholipids in a manner similar to a membrane. However, unlike a membrane, the 47. A. Fatty acyl–anchored protein (the acyl group is myristate) interior of the LDL is completely nonpolar.

B. Prenyl-anchored protein (c) Apolipoprotein B, like membrane proteins, associates with the choC. Glycosylphosphatidylinositol (GPI)-anchored protein lesteryl esters via its nonpolar domains and with the aqueous medium 48. Cytochrome c is a peripheral membrane protein that is only loosely as- via its polar domains. A large fraction of the protein contains nonpolar sociated with the membrane and can therefore be removed by gentle means domains, which makes it diffi cult to solubilize and purify using tradisuch as a salt solution. Cytochrome oxidase is an integral membrane protein tional column chromatography. that completely spans the bilayer and has large hydrophobic portions where the protein spans the bilayer. Thus, it is diffi cult to remove unless nonpolar 55. (a) Alcohols, ether, and chloroform are nonpolar molecules and can easily pass through the nonpolar portion of the lipid bilayer, the aliphatorganic solvents or amphipathic detergents are used to dissociate it from the ic acyl chains of the phospholipids. Salts, sugars, and amino acids are membrane. highly polar and would not be able to traverse the nonpolar portion of the membrane. (b) Cells contain proteins that serve as transporters.

Proteins that transport water, known as aquaporins, have been identifi Peripheral ed. [From Kleinzeller, A., News Physiol. Sci. 12, 49–54 (1997).]

Integral

49. The membrane-spanning segment is a stretch of 19 residues that are all uncharged and mostly hydrophobic.

LSTTEVAMHTTTSSSVSKSYISSQTNDTHKRDTYAATPRAHEVSEISVRTVYPPEEETGERVQLAHHFSEPEITLIIFGVMAGVIGTILLISYGIRRLIKKSPSDVKPLPSPDTDVPLSSVEIENPETSDQ 50. (a) A fully hydrogen-bonded barrel can form only if the number of strands is even. A sheet with an odd number of strands could not close up on itself to form a barrel. (b) The strands are antiparallel because adjacent strands can be easily linked by loops on the solvent-exposed portions of the protein.

56. The sandwich model assumes that all membranes are the same. However, it has been noted that different membranes have a variety of functions.

Multiple functions would not be possible with the same structure. The model also does not explain how membrane transport could occur, since it would be diffi cult for transported molecules to get past the protein “caps” on either side of the membrane.

57. After fusion, the green and red markers were segregated because they represent cell-surface proteins derived from two different sets of cells. Over time, the cell-surface proteins that could diffuse in the lipid bilayer became distributed randomly over the surface of the hybrid cell, so the green and red markers were intermingled. At 15 C, the lipid bilayer was in a gel-like rather than a fl uid state, which prevented membrane protein diffusion. Edidin’s experiment supported the fl uid mosaic model by demonstrating the ability of proteins to diffuse through a fl uid membrane.

58. The bleached area “recovers” its fl uorescence as fl uorophore-labeled molecules diffuse out of the small area and unbleached fl uorophore-labeled molecules diffuse in. These experiments are useful for measuring diffusion rates of the target molecules.

Ch 9 Solutions 41 DG5(8.314531023 kJ?K21?mol21)(201273 K)

Chapter 9 1.

Dc50.058 log 20.07050.058 log 21.205log

[Na1]in [Na1]out

ln

[Na1]in

DG5225.8 kJ?mol21213.5 kJ?mol21

[Na1]out

DG5239.3 kJ?mol21

[Na1]in

Since the concentrations of both ions are greater outside the cell than in-

[Na1]out

side and the cell potential is negative, the passive movement of the ions will be from outside the cell to inside. In order to maintain the ion concentrations given in the problem, energy-consuming active transport processes are required. 6. The extracellular Na concentration is 150 mM and the

1 [Na]in 1021.20 5 [Na1]out

[Na1]in

0.063

2.

intracellular concentration is about 12 mM, whereas the extracellular K concentration is about 4 mM and the intracellular concentration is 140 mM.

5 [Na1]out

1

Dc50.058 log 10.05050.058 log 0.865log 100.86 5 7.3

5

1

[Na1]in [Na1]out

DG5RT ln

[Na1]out [Na1]in

12 mM 1(1)(96,48531023 kJ?V21?mol21)(20.070 V) 150 mM DG526.15 kJ?mol2126.75 kJ?mol21 ln

[Na1]out [N1a]in

DG5212.90 kJ?mol21

[Na1]out DG5RT ln

[Na1]out

[Na1]in [Na1]out

0.063 1

1 (11)(96,48531023 kJ?V21?mol21)(20.070 V) DG527.12 kJ?mol2126.75 kJ?mol21 DG5213.9 kJ?mol21 At the resting potential, the movement of Na ions into the cell is a favorable process.

[Na1]out

1ZF Dc

DG5(8.314531023 kJ?K21?mol21)(310 K)ln

7.3 1

1(1)(96,48531023 kJ?V21?mol21)(10.050 V) DG515.12 kJ?mol2114.82 kJ?mol21 DG519.94 kJ?mol21 The free energy change for this process is positive, so there is no longer a driving force to move additional Na ions into the cell.

[Na1]in

1ZF Dc [Na1]out DG5(8.314531023 kJ?K21?mol21)(201273 K)

5. DG5RT ln

40 mM

1(1)(96,48531023 kJ?V21?mol21)(20.070 V) 450 mM DG525.90 kJ?mol2126.75 kJ?mol21 ln

DG5212.64 kJ?mol21 DG5RT ln

[Ca21]in [C2a1]

140 mM

1(1)(96,48531023 kJ?V21?mol21)(20.070 V) 4 mM DG58.66 kJ?mol2126.75 kJ?mol21 ln

DG51.91 kJ?mol21

1ZF Dc

DG5(8.314531023 kJ?K21?mol21)(310 K)ln

4. DG5RT ln

[K1]in

1ZF Dc [K1]out DG5(8.314531023 kJ?K21?mol21)(201273 K)

[Na1]in

[Na1]in

[Na1]in

1ZF Dc [Na1]out DG5(8.314531023 kJ?K21?mol21)(201273 K)

[Na1]in

When a nerve cell is depolarized, sodium ions enter the cell; therefore, the ]out ratio is more than 100-fold greater in the depolarized cell [Na ]/[Na in than in the resting cell.

3. DG5RT ln

0.0001 mM 1(2)(96,48531023 kJ?V21?mol21)(20.070 V) 4 mM

The free energy change for the movement of Na ions into the cell is negative, indicating that movement of these ions in this direction occurs passively.

In contrast, the free energy change for the movement of K ions into the cell is positive, indicating that K ions passively move from the inside of the cell to the outside. An active transport process is required to transport K ions into the cell.

7. Use Equation 9-4 and let Z 2 and T 310 K: (a)

DG5RT ln

[Ca21]in 2

[Ca 1]out

1Z FDc

DG5(8.3145 J?K21?mol21)(310 K)ln

1027 1023

1 (2)(96,485 J?V21?mol21)(20.05 V) DG5223,700 J?mol2129600 J?mol21 DG5233,300 J ?mol215233.3 kJ?mol21 The negative value of G indicates a thermodynamically favorable process.

[Ca21]in (b) DG5RT ln 1ZFDc 2 [Ca 1]out 1027 DG5(8.3145 J?K21?mol21)(310 K)ln 1023 1(2)(96,485 J?V21?mol21)(10.05 V) DG5223,700 J?mol2119600 J?mol21 DG5214,100 J ?mol215214.1 kJ?mol21

1 Z F Dc

The negative value of G indicates a thermodynamically favorable process, but not as favorable as in part (a).

42

Ch 9 Solutions [Ca21]in Z [Ca2 1]out 1 F D c

8. DG5RT ln

DG5(8.3145 J?K21?mol21)(310 K)ln

13. The less polar a substance, the faster it can diffuse through the lipid bilayer. From slowest to fastest: C, A, B. 14. The permeability coeffi cients are positively correlated with membrane solubility.

The greater the permeability coeffi cient, the more easily the solute

1024 mM 2 mM

passes through the membrane. This observation is consistent with the ranking shown in Solution 13 in that butyramide has the highest permeability coeffi cient and urea the lowest.

1(2)(96,485 J?V21?mol21)(20.05 V) DG5225,500 J?mol2129600 J?mol21 DG5235,100 J ?mol215235.1 kJ?mol21 Ca2 ions move spontaneously from the outside of the cell to the cytosol.

9. (a) Since all the terms on the right side of Equation 9-1 are constant, except for T, the following proportion for the two temperatures (310 K and 313 K) applies: 270 mV Dc 5 310 K 313 K Dc5270.7 mV The difference in membrane potential at the higher temperature would not signifi cantly affect the neuron’s activity. (b) It is more likely that an increased temperature would increase the fl uidity of cell membranes. This in turn might alter the activity of membrane proteins, including ion channels and pumps, which would have a more dramatic effect on membrane potential than temperature alone.

[H1]in

10. (a) Dc50.058 log

[H1]out

[H1]in

1ZF Dc [H1]out DG5(8.3145 J?K21?mol21)(371273 K) 1026.88

1(1)(96485 J?V21?mol21)(20.052 V) 1027.78 DG515344 J?mol21125017 J?mol21 DG5327 J?mol21 ln

[From Porcelli, A. M., Ghelli, A., Zanna, C., Pinton, P., Rizzuto, R., and Rugolo, M., Biochem. Biophys. Res. Commun. 326, 799–804 (2005).]

11. (a) DG5RT ln

[glucose]in [glucose]out

DG5(8.314531023 kJ?K21?mol21)(310 K)ln DG528.8 kJ?mol21

(b) DG5RT ln

0.5 mM 15 mM

[glucose]in [glucose]out

DG5(8.314531023 kJ?K21?mol21)(310 K)ln DG525.4 kJ?mol21

0.5 mM 4 mM

[glutamate]in [glutamate]out 12. DG5RT ln

1ZF Dc

DG5(8.314531023 kJ?K21?mol21)(310 K) ln

10 mM

1(21)(96,48531023 kJ?V21?mol21)(20.070 V)

DG511.9 kJ?mol2116.75 kJ?mol21 DG518.6 kJ?mol21

ic barrier to diffusing across the lipid bilayer.

17. (a) Phosphate ions are negatively charged, and lysine side chains most likely carry a full positive charge at physiological pH. It is possible that an ion pair forms between the phosphate and the lysine side chains and that the lysine side chains serve to funnel the phosphate ions through the porin. (b) If the hypothesis described in part(a) is correct, the replacement of lysines with the negatively charged glutamates would abolish phosphate transport by the porin, due to charge–charge repulsion. Possibly the mutated porin might even transport positively charged ions instead of phosphate. [From Sukhan, A., and Hancock, R. E. W., J. Biol. Chem.

271, 21239–21242 (1996).]

18. The D-E-K-A sequence already contains two negatively charged side chains, so the lysine (K, which is positively charged) and the alanine (A, which is neutral) can be mutated to negatively charged amino acid side chains, either Asp (D) or Glu (E). [From Miedema, H., MeterArkema, A., Wierenga, J., Tang, J., Eisenberg, B., Nonner, W., Hektor, H., Gillespie, D., and Meijberg, W., Biophys. J. 87, 3137–3147 (2004).] 19. (a) Acetylcholine binding triggers the opening of the channel, an example of a ligand-gated transport protein. (b) Na ions fl ow into the muscle cell, where their concentration is low. (c) The infl ux of positive charges causes the membrane potential to increase.

(1027.78) (1026.88) Dc520.052 V5252 mV Dc50.058 log

(b) DG5RT ln

15. (a) Glucose has a slightly larger permeability coeffi cient than mannitol and therefore moves across the synthetic bilayer more easily. (b) Both solutes have higher permeability coeffi cients for the red blood cell membrane, indicating that transport is occurring via a protein transporter rather than diffusion through the membrane. The transporter binds glucose specifi cally and transports it rapidly across the membrane, whereas it is less specifi c for mannitol and transports it less effectively.

16. CO2 is small and nonpolar and therefore experiences no thermodynam-

20. At neutral pH, the Asp and Glu side chains are unprotonated and can participate in ion pairing that holds the protein in its closed conformation.

In the presence of acid (low pH, high [H ]), the Asp and Glu side chains become protonated. This disrupts ion pairing, and the conformation of the protein shifts to an open state. 21. The transfer to pure water increases the infl ux of water by osmosis, and the cell begins to swell. Swelling, which puts pressure on the cell membrane, causes mechanosensitive channels to open. As soon as the cell’s contents fl ow out, the pressure is relieved and the cell can return to its normal size.

Without these channels, the cell would swell up and burst. 22. (a) Rhcg appears to encode an ammonia channel, since its absence decreases the amount of NH3 that crosses the cell membrane. (b) Because the large-scale transport of water (a small polar molecule) requires aquaporin, it is not surprising that the transport of ammonia (also a small nonpolar molecule) requires a transport protein, especially in the kidneys, which are responsible for ammonia excretion. [From Biver, S., Belge, H., Bourgeois, S., Van Vooren, P., Nowik, M., Scohy, S., Houillier, P., Szpirer, J., Szpirer, C., Wagner, C. A., Devuyst, O., and Marini, A. M., Nature 456, 339–343 (2008).]

23. The hydroxyl and amido groups act as proton donors to coordinate the negatively charged chloride ion. Cations could not interact with the protons and so would be excluded.

24. (a) Because this sequence is so similar to the KcsA and NaK sequences, it probably comes from a channel that is specifi c for K or for Na and K . (b) This sequence is most similar to the Ca2 II channel sequence and so probably comes from a Ca2 channel.

25. (a) A transport protein, like an enzyme, carries out a chemical reaction (in this case, the transmembrane movement of glucose) but is not

Ch 9 Solutions 43 permanently altered in the process. Because the transport protein binds 29. As the glutamate (charge 2) enters the cell, four positive charges also glucose, its rate does not increase in direct proportion to increases in enter (3 Na , 1 H ) for a total of two positive charges. Since 1 K exits the glucose concentration, and it becomes saturated at high glucose con- cell at the same time, a total of one positive charge is added to the cell for centrations. (b) The transport protein has a maximum rate at which it each glutamate transported inside. can operate 30. (a) The oxalate (charge 2) enters the cell and the formate (charge 1) (corresponding to Vmax, the upper limit of the curve). It also binds exits the cell, so one negative charge is added to the cell for each oxalate glucose with a characteristic affi nity (corresponding to KM, the glucose transported inside the cell. concentration at half-maximal velocity). The estimated Vmax for this (b) The antiporter illustrated here is a hydrogen pump because one intratransporter is about 0.8 106 mM ? cm ? s 1, and the KM is about cellular H is consumed for every oxalate transported inside the cell. [From 0.5 mM. Anantharam, V., Allison, M. J., and Maloney, P. C., J. Biol. Chem. 264, 7244–7250 (1989).]

1.0 H+

. . −1

GuoeumMms× 6lcs flx ( c 10)

Vmax

oxalate2−

formate−

formate−

decarboxylase CO2

0.5

outside

KM 0

0

2

4 6 [Glucose] (mM)

8

10

26.

No inhibitor

. . −1

1.0 GuoeumMms× 6lcs flx ( c 10)

oxalate2−

0.5

With inhibitor

31. CO2 produced by respiring tissues enters the red blood cell and combines with water to form carbonic acid, which then dissociates to form H and HCO 3 ion s. The HCO3 ions are transported out of the cell by Band 3 in exchange for Cl ions, which enter the cell. The HCO 3 ions travel through the circulation to the lungs, where they recombine with H ions to form carbonic acid, which subsequently dissociates to form water and CO2.

The CO2 is then exhaled in the lungs.

32. When red blood cells pass through tissues, HCO 3 ions are transported out of the cell by Band 3 (see Solution 31). However, in the lungs, HCO 3 ions are transported into red blood cells by Band 3 (in exchange for Cl ions, which exit the cell). The intracellular HCO ions combine with H 3 ions that dissociate from hemoglobin (promoting conversion of hemoglobin from the deoxy T form to the oxy R form), forming carbonic acid. The carbonic acid dissociates into water and CO2, which diffuses out of the cell and is exhaled.

33.

H N 0

0

2

4 6 [Glucose] (mM)

8

10

inside

O C

CH CH2 C O

O

The curve shown in Problem 25b would be shifted to the right. In the presP ence of the inhibitor, higher concentrations of glucose are required to achieve the same rate of glucose transport. [From Barnett, J. E. G., Holman, O O O G. D., Chalkley, R. A., and Munday, K. A. Biochem J. 145, 412–429 (1975).]

27. Intracellular exposure of the glucose transporter to trypsin indicates that there is at least one cytosolic domain of the transport protein that is essenAspartyl phosphate tial for glucose transport. Hydrolysis of one or more peptide bonds in this domain(s) abolishes glucose transport. But extracellular exposure of the ghost 34. Treatment with ouabain “freezes” the Na,K-ATPase in the phosphorytransporter to trypsin has no effect, so there is no trypsin-sensitive lated conformation, and the reaction cycle cannot be completed. As a result, the pump is nonfunctional and the concentration of sodium ions inside the extracellular domain that is essential for transport. This experiment also shows that the cell increases. Water enters the cell along with the sodium ions, and the cells glucose transporter is asymmetrically arranged in the erythrocyte membrane. will eventually lyse due to the increased osmotic pressure.

28. The experiments with the derivatized glucose molecules indicate that 35. (a) A transporter similar to a porin would be inadequate since even there is an extracellular contact between the C1 hydroxyl group of glucose a large barrel would be far too small to accommodate the massive and the glucose transporter, but the C6 hydroxyl group does not contact the ribosome. Likewise, a transport protein with alternating conformatransporter. When glucose binds to the transporter on the outside of the cell, tions would not be up to the task due to its small size relative to a conformational change occurs that brings the glucose inside the cell. The the ribosome. In addition, neither type of protein would be suited mode of glucose–transporter binding changes somehow in that the fi rst for transporting a particle across two membranes. In fact, ribosomes binding site is closed and a new intracellular binding site opens up that has and other large particles move between the nucleus and cytoplasm via an essential contact with glucose at the C6 hydroxyl group but not the C1 nuclear pores, which are constructed from many different proteins hydroxyl group. Glucose is delivered to the cytosol, and the transporter re- and form a structure, even larger than the ribosome, that spans both sumes its initial confi guration, ready to bind another molecule of glucose on nuclear membranes. the extracellular surface.

44

Ch 9 Solutions

(b) Initially, one might expect ribosomal transport to be a thermody- [From Wakisaka, M., Yoshinari, M., Yamamoto, M., Nakamura, S., Asano, T., namically favorable process, since the concentration of ribosomes is Himeno, T., Ichikawa, K., Doi, Y., and Fujishima, M., Biochim. Biophys. Acta greater in the nucleus, where they are synthesized. However, free 1362, 87–96 (1997).] energy would ultimately be required to establish a pore (which would 39. (a) The maximal velocity is estimated to be between 6 and 7 pmol span two membrane thicknesses) for the ribosome to pass through. In choline ? mg 1 ? mL 1. The KM is the substrate concentration at halffact, the nucleocytoplasmic transport of all but very small substances maximal velocity and is about 40 mM. requires the activity of GTPases that escort particles through the (b) The choline transporter responds to increased choline nuclear pore assem- bly and help ensure that transport proceeds in concentrations one direction. by increasing its rate of transport so that effi cient uptake occurs over a 36. (a) range of choline concentrations near the KM. At low concentrations of OH choline (10 M), the transporter operates at 20% of its maximal velocCA II ity, whereas at high concentrations of choline (80 M), the transporter H H C OH O C O operates at nearly 100% of its maximal velocity. The KM value of 40 M O is between the low and high physiological concentrations of choline. (c) C a r b o n dWater i o x i d e OCarbonic acid It is possible that the choline transporter cotransports hydrogen ions and choline. A hydrogen ion might be exported when choline is imported. This is an example of antiport transport. (d) TEA is structurally H2CO3 HCO H similar to choline and acts as a competitive in- hibitor. TEA might bind Bicarbonate to the choline transporter, preventing choline from binding. In this manner, TEA is brought into the cell and choline transport is inhibited.

[From Sinclair, C. J., Chi, K. D., Subramanian, V., Ward, K. L., and (b) Bone-resorbing Green, R. M., J. Lipid Res. 41, 1841–1847 (2000).]

40. (a) Vmax is calculated from the y intercept: compartment + Na

pH = 5.5

1 Vmax5 y int

K+

H+ Na+

Vmax5 CA II H2O + CO2 H2CO3 H+ + HCO− HCO 2 3 3

Vmax 5.6 1mmol?g21 VI max5 VI

CI− HCO−

pH = 7

3

Hydrogen ions are produced from carbonic acid, which is produced from water and carbon dioxide via the carbonic anhydrase reaction.

The hydrogen ions exit the cell via the Na /H exchanger. These hydrogen ions acidify the bone-resorbing compartment as sodium ions enter the cell. The sodium ions leave the cell via the Na,KATPase, while potas- sium ions enter the cell. Bicarbonate, the other product of the dissociation of carbonic acid, leaves the cell via the Cl /HCO 3exchanger as chloride enters the cell.

37. (a) Glucose uptake increases as sodium concentration increases in pericytes. In endothelial cells, glucose uptake is constant regardless of sodium ion concentration. (b) The shape of the curve for the pericytes indicates that a protein transporter is involved. Glucose uptake initially increases as sodium ion concentration increases, then reaches a plateau at high sodium ion concentration, indicating that the transporter is saturated and is operating at its maximal capacity. (c) It is likely that the pericytes use secondary active transport to import glucose. Sodium ions and glucose molecules enter the cell in symport.

The sodium ions are then ejected from the cell by the Na,K-ATPase transporter.

1 38.

KM52 x int KM52 Vmax5 Vmax5

1 20.25 mM

54 mM

1 y int 1 0.030 mg?h?nmol21

Vmax 533 nmolmg21 ?

?h 2 1

1 0.64 g?min?mmol21

max

?min21

1 0.65 g?min?mmol21

5 1.5 mmol?g21

? min21

KM is calculated from the x intercept (without inhibitor): x int52bym x int 5

0.64 g?min?mmol21 5.4 g?min?mmol21?mM

xint520.12 mM21 1 K M 5 2 x int 1

KM52

20.12 mM21 KM58.3 mM KappM (the apparent KM in the presence of inhibitor) is calculated from the slope (with inhibitor): x int52bym x int5

0.65 g?min?mmol21 13.8 g?min?mmol21?mM

xint520.047 mM21 Kapp M52 Kapp M52

1 1

20.047 mM21 Kapp M521.2 mM (b) Phlorizin is a competitive inhibitor. The Vmax has remained nearly unchanged while the KM increased nearly threefold in the presence of the inhibitor. Phlorizin competes with glucose for binding to the transporter.

[From Betz, A. L., Drewes, L. R., and Gilboe, D. D., Biochim. Biophys.

Acta 406, 505–515 (1975).]

Ch 9 Solutions 45 41. The ABC transporters bind ATP, then undergo a conformational change 49.

CH(CH3)2 as the ATP is hydrolyzed and Pi is released, leaving ADP. Vanadate, O a phosphate analog, might serve as a competitive inhibitor by binding to Acetylcholinesterase the phosphate portion of the ATP binding site. With ATP unable to bind, F P O DIPF the necessary conformational change cannot occur and the transporter is CH2OH O inhibited.

42. The fi rst vinblastine binds to a low-affi nity binding site, the second CH(CH3)2 vinblastine to a high-affi nity site. This indicates that when vinblastine binds to LmrA, it does so cooperatively. A conformational change occurs that allows the second molecule of vinblastine to bind with greater affi nity than HF the fi rst.

43. Both transporters are examples of secondary active transport. The H /Na exchanger uses the free energy of the Na gradient (established by CH2 the Na,K-ATPase) to remove H from the cell as Na enters. Similarly, a O preexisting Cl gradient (see Fig. 2-13) allows the cell to export HCO23 as Cl Modified enzyme enters.

P (H3C)2HC O O 44. Ammonium ions exiting the cell could be exchanged for entering Na O ions. The free energy for NH14 transport would be provided directly by the movement of Na ions down their gradient and indirectly by ATP, which is CH(CH3)2 used by the Na,K-ATPase to establish the Na gradient.

45. Di- and tripeptides enter the cell in symport with H ions. The H ions leave in exchange for Na via the antiport protein. The Na ions are ejected via the Na,K-ATPase. This is an example of secondary active transport in 50. If the serine in the active site of acetylcholinesterase is covalently modi- fi which the expenditure of ATP by the Na,K-ATPase pump is the driving ed, the enzyme will not be able to function. Acetylcholine will build up in force the synaptic cleft and will not be hydrolyzed. The postsynaptic cell remains for peptide entry into the cell. polarized and does not return to its resting state, so it cannot receive the next nerve impulse from the presynaptic cell.

Peptide

Peptide

H+

H+

Na+

+

+

Na+

H

H

3 Na+

51. The tetanus toxin cleaves the SNAREs, which are required for the fusion of synaptic vesicles with the neuronal plasma membrane. This prevents the release of acetylcholine, interrupting communication between nerves and muscles and causing paralysis.

52. Botulinum toxin destroys SNAREs, which are required for the fusion of synaptic vesicles with the neuronal plasma membrane. This interrupts communication between facial nerves and muscles. The result is paralysis of the muscles whose contraction accentuates wrinkles.

53. By adding a phosphate group, the kinase increases the size and negative 2 K+

ATP

3 Na+

2 K+

charge of the lipid head group, which then occupies a larger volume and more strongly repels neighboring negatively charged lipid head groups. The phosphatidylinositol would become more cone-shaped, thereby increasing bilayer curvature, which is a necessary step in the formation of a new vesicle by budding.

54. Diacylglycerol is a lipid without a phosphate-derivative head group. Because it consists mostly of lipid “tails,” it could promote the inward curvature of the bilayer, which occurs during membrane fusion (see Fig. 9-22).

55. Two lipid bilayers separate the damaged organelle from the rest of the 46. AMPPCP is a nonhydrolyzable ATP analog (due to the substitution of cell. the methylene group for the oxygen linking the and phosphate groups). Its structural similarity to ATP allows it to bind to the Ca2 -ATPase, but since hydrolysis cannot occur, the transport cycle cannot be completed. This “freezes” the ATPase in one conformational state and allows the crystallographers to obtain high-resolution images of the protein.

47. Acetylcholinesterase inhibitors will prevent the enzyme from breaking down acetylcholine (see Section 9-4). This increases the concentration of acetylcholine in the synaptic cleft and increases the chances that acetylcholine will bind to a dwindling number of receptors in the postsynap- 56. Lysosome containing tic cell. [From Thanvi, B. R., and Lo, T. C. N., Postgrad. Med. J., 80, hydrolytic enzymes 690–700 (2004).]

48. In the presynaptic cell, when an action potential reaches the axon terminus, voltage-gated Ca2 channels open, fl ooding the cell with Ca2 ions and triggering exocytosis of the acetylcholine-containing synaptic vesicles.

Acetylcholine diffuses across the synaptic cleft and binds to receptors on the postsynaptic cell, triggering a series of events that results in muscle contraction (see Fig. 9-19). If antibodies block the Ca2 channels, acetylcholine is not released from the presynaptic cell, receptors on the postsynaptic cell are not activated, and muscle contraction does not occur. Thus, patients with this autoimmune disorder suffer from muscle weakness.

Ch 10 Solutions

46

Chapter 10 7.

1. Signal molecules that are lipids (cortisol and thromboxane) or very small (nitric oxide) can diffuse through lipid bilayers and do not need a receptor on the cell surface.

2. Prednisone is a steroid, a hydrophobic molecule that can diffuse through the lipid bilayer without the need for a cell-surface receptor. The receptor for prednisone is located either in the cytosol or in the nucleus. Upon binding to its receptor, the ligand–receptor complex binds to DNA and induces transcription of specifi c genes.

3. Let [R ? L] x and [R] 0.010 x Kd5 [R?L]5x5 x5 x5

[R]T5[R]1[RL] [R]5[R]T2[RL] Kd5

[RL](Kd1[L])5[R]T[L]

[R][L]

[RL] [L] [R]T 5 Kd 1 [L]

[R ?L] [R][L] 8. (a)

Kd (0.010 2 x)(0.0025) 0.0015 (0.000025 2 0.0025x)

5

[L]

[R]T Kd 1 [L] Kdy5 [RL] 5 Kd 1 Kdy5 [R]T [R]T

5

0.2 Kd Kd10.2 Kd

[RL] 5 0.2 Kd 50.17 (b)

The percentage of receptors occupied by ligand is 6.25 mM/10 mM, or 62.5%.

4. Let [R ? L] x and [R] 0.010 x [R][L] K d 5 [R ?L] [R][L]

(c)

Kd (0.010 2 x)(0.0025 ) 0.0003 (0.000025 2 0.0025x)

0.0003 0.0003x 5 0.000025 2 0.0025x 9.

0.0028x 5 0.000025 x5 0.00893 5 8.93 mM 5 [R?L] The percentage of receptors occupied by ligand is 8.93 mM/10 mM, or 89.3%. A higher percentage of the receptors are occupied by a ligand with a smaller Kd. 5.

[RL]

[RL]

x50.0062556.25 mM5[R?L]

x5

[RL]

Kd[RL]5[R]T[L]2[RL][L]

0.0040x50.000025

x5

([R]T 2 [RL])[L]

(Kd[RL])1([RL][L])5[R]T[L]

0.0015 0.0015x50.00002520.0025x

[R?L]̌ˇ 5 x 5

[R][L] [RL]

Kd5

1.2 Kd [R]T [L] [RL] 5 Kd 1 [L] [R]T Kd [RL] 5 Kd 1 Kd [R]T Kd 50.50 [RL] 5 2 Kd [R]T [L] [RL] 5 K 1 [ L ] [R] 5 Kd [RL] 5 Kd15 Kd [R]T [RL] 5

5 Kd 50.83

[R]T

6 Kd

[L] [RL] 5 [L] 1 Kd [R]T

[L] 100 5 [L]11.0310210 M 1000 0.10([L]11.0310210 M)5[L] 0.10[L]11.0310211 M5[L] 1.0310211 M50.9[L]

1.0

0.6 T

[RL]/[R]

[L]51.11310211 M 10. If the number of receptors decreases to 150, a greater proportion of these receptors must have bound ligand, since occupation of 100 receptors is required for a maximal response (see Problem 9). A 20-fold increase in the concentration of ligand is required in order to achieve a maximal response.

Kd 0.2

[RL] 0.1

0.2

0.3

0.4

0.5

0.6

0.7

[L] (mM)

The Kd estimated from the curve is about 0.1 mM. 6. The experimental results would produce a curve that is similar in shape but shifted to the right (similar to the plot shown in Figure 7-9). As an antagonist that competes with adenosine for binding the same receptors, caffeine acts like a competitive inhibitor.

[RT]

5

[L] [L] 1 Kd

100 5 150

[L] [L]11.0310210 M

0.67([L]11.0310210 M)5[L] 0.67[L]16.7310211 M5[L] 6.7310211 M50.33[L] [L]52.0310210 M

Ch 10 Solutions 47 11. (a) The binding site with a Kd of 0.35 M is the high-affi nity binding site, and the one with a Kd of 7.9 M is the low- affi nity binding site. Kd is the ligand concentration at which the receptor is half-saturated with ligand; therefore, the lower the Kd, the lower the concentration of ligand required to achieve half-saturation. (b) The high-affi nity binding site with a Kd of 0.35 M is most effective in the 0.1–0.5 M range because at the upper limit of this range the high-affi nity binding sites will be more than 50% occupied, whereas the low-affi nity sites will be less than 50% occupied. (c) Both of these agonists can compete at high concentrations, but the methylthio-ADP has a lower Kd and will be a more effective inhibitor at low concentrations. [From Jefferson, J. R., Harmon, J. T., and Jamieson, G. A. Blood 71, 110–116 (1988).]

12. Use the equation derived in Problem 7.

[RL] 45 [RT 136,000

[L] Kd 1 [L] 5

160,000 0.85 (0.35 mM 1 [L]) 5 [L]

[L] 0.35 mM 1 [L]

0.2975 mM 1 0.85[L] 5 [L] 0.2975 mM5 0.15[L] [L]5 1.98 mM 13. The Kd is obtained from the x intercept in the double-reciprocal plot.

10

be solubilized by adding a detergent. An affi nity column is constructed by covalently attaching a specifi c ligand to the chromatography matrix. The sol- ubilized membranes are then loaded onto the column. Cellular proteins that do not bind to the ligand will pass through the column. The receptor pro- teins are then eluted from the column using a concentrated salt solution to disrupt the intermolecular interactions between the receptor and the ligand.

17. The different types of G protein–linked receptors are found in different types of cells. The cellular response elicited when a ligand binds to a receptor depends on how that particular cell integrates and processes the signal. Different cells have different intracellular components, which results in different responses to what appears to be the same signal.

18. When glucagon and epinephrine bind to their respective G protein– coupled receptors, the result is the same—an enzyme is activated that results in the synthesis of the second messenger cAMP. Since the binding of both ligands to their receptors results in the activation of the same second messenger, the same signal transduction pathway is activated and the same cellular response is observed.

19. If receptors are removed from the cell surface, the ligand cannot bind and an intracellular response cannot occur. If a receptor is phosphorylated, it binds to arrestin, which blocks the ligand from binding.

20. The strategy shown in the fi gure results in amplifi cation of the signal, since a single molecule of A can activate many molecules of B, which can then go on to activate even more molecules of C, and fi nally D. Amplifi cation of a signal is an advantage to the cell because a small signal (i.e., a low concentration of A) can result in a large response (a high concentration of D). The strategy shown allows cells to respond to changing conditions by dramatically altering the activities of intracellular components.

21.

1/[R L] ( M) . 1

8

3

CH S C O

1/K d

O C

CH2 6 4

4

H N

(CH)214

CH3

2

2

1

1 0 1/[L] ( M 1 )

xint52 22 mM21

1 Kd

15

2

3

4

22. Both hormones lack tyrosine’s carboxylate group and have hydroxyl groups attached to the ring and to the carbon. In epinephrine, the amino nitrogen bears a methyl group.

23. Epinephrine and norepinephrine are ligands that bind to 2-adrenergic receptors. When these ligands bind to their receptors, the signal transduction process produces a number of effects, including increases in heart rate, muscle contraction, and blood pressure. In addition, the smooth muscle in the bronchial tubes relaxes, making it easier to expand the lungs. These physiological effects are all necessary components of the fi ght-or-fl ight response but

are harmful to a person suffering from high blood pressure. -Blockers are antagonists that bind to the same 2-adrenergic receptors but do not elicit a response. By occupying the receptors, the antagonists prevent epinephrine and norepinephrine from doing so. The result is that the heart rate and 21 14. slope 5 K d blood pressure decrease and the heart contractions are less intense.

24. The inhibition of the intrinsic GTPase activity results in a continuously 1 2 active G protein. This increases the activity of adenylate cyclase, which renM21 0.067 5 2 Kd sults in an increase in the concentration of intracellular cAMP. In intestinal cells, the increased cAMP concentration leads to the loss of water and Kd 5 15 nM electrolytes from the cells and results in diarrhea that can potentially be fatal.

[From Hubbard, M. J., and Klee, C. B., J. Biol. Chem. 262, 15062–15070 25. Stimulation of GTPase activity by RGS accelerates the hydrolysis of GTP (1987).] to GDP, converting the receptor-associated G protein into its inactive form 15. Cell-surface receptors are diffi cult to purify because they are usually integral membrane proteins and require the addition of detergents to dis- more rapidly. This will shorten the duration of signaling. sociate them from the membrane. The receptor proteins constitute a very 26. GTPgS can bind to a G protein, but since it cannot be hydrolyzed, the small proportion of all of the proteins in the cell; this makes it diffi cult for G protein is in a persistently active state. If GTPgS binds to a stimulatory the experimenter to isolate the receptor protein from other cellular proteins. G protein, then adenylate cyclase is continually active, which has the effect of increasing cellular cAMP concentration. If GTPgS binds to an inhibi16. First the cells would be lysed and insoluble components removed by centrifugation. The membranes and the associated membrane proteins can tory G protein, adenylate cyclase is continually inhibited, and cellular cAMP concentration decreases.

Kd Kd50.5 mM

Ch 10 Solutions

48 27. (a)

OC HN

CH

OC

protein kinase A

CH3

ATP

CH

HN

CH

ADP

CH3

CH

OH

O P

O

O

O

(b)

OC HN

CH

O

phosphatase

CH3

H2O

CH

HN

Pi

C

CH

CH3

OH

O O

CH

P

O

O

28.

N

CH2 N

this occurs, inositol trisphosphate is no longer able to activate protein kinase B. In the absence of protein kinase B, cells are not stimulated to grow and proliferate and instead undergo apoptosis.

34. Yes, mutations in which the gene coding for PTEN is either nonfunctional or absent are commonly found in cancer cells. In the absence of PTEN, inositol trisphosphate is not dephosphorylated, and protein kinase B is highly active. As described in Problem 33, activation of protein kinase B promotes cell growth, so a highly active protein kinase B could account for the uncontrolled cell growth observed in cancer cells.

35. (a) Upon stimulation by an action potential, acetylcholine-containing synaptic vesicles in neurons fuse with the plasma membrane and release their contents into the synaptic cleft (see Section 9-4). Acetylcholine then diffuses across the synaptic cleft to the endothelial cell. (b) Acetylcholine binds to G protein–linked cell-surface receptors in the endothelial cell and activates phospholipase C, which hydrolyzes phosphatidylinositol bisphosphate to diacylglycerol and inositol trisphosphate. The inositol trisphosphate binds to calcium channels in the endoplasmic reticulum, which opens the channels and fl oods the cell with Ca2 . Calcium ions bind to calmodulin, changing its conformation and allowing it to bind to NO synthase to activate the enzyme. (c) Cyclic GMP is formed, along with pyrophosphate, from GTP. The enzyme that catalyzes the reaction is guanylate cyclase.

PO2 3

O

29. Because of their structural similarity to diacylglycerol, phorbol esters will stimulate protein kinase C, as diacylglycerol does. Increased protein kinase C activity will lead to an increase in the phosphorylation of the kinase’s cellular targets. Because protein kinase C phosphorylates proteins involved in cell division and growth, the addition of phorbol esters will have profound effects on the rates of cell division and growth if added to cells in culture.

30.

N

O

O P O

O P O

O P

O

O

O

O

R

C

HC HN

CH

CH

CH CH 2

(CH 212 )

CH 3

O O

P

CH2

CH2

O

Sphingomyelin

N

H

H

OH

OH

N N

R

C

HN

CH

(CH2)12

H

CH3

CH CH2

O

OH

Ceramide

O O

P O

O

CH 2

CH2

N CH3

Phosphocholine

CH3

N

NH2

H

H

O

OH

H

P O

CH 3

NH

CH2 O

OH CH

GTP

O

CH3

O

HC

H

CH3

sphingomyelinase

O

NH2

guanylate cyclase

CH 3 O

N

O

CH2 H

OH O

NH

N

Cyclic GMP

(d) It’s possible that cGMP activates protein kinase G in a manner analogous to that of cAMP activating protein kinase A; that is, cGMP binding could displace regulatory subunits from protein kinase G to release active catalytic subunits. The active protein kinase G would next phosphorylate proteins involved in the muscle contraction process, perhaps myosin or actin, resulting in smooth muscle relaxation.

31. The T cell is stimulated when an extracellular ligand binds to a G protein–linked receptor and activates phospholipase C. The activated phospholipase C catalyzes the hydrolysis of phosphatidylinositol bisphosphate, 36. (a) Acetylcholine in the synaptic cleft is rapidly hydrolyzed by acetylchoyielding diacylglycerol and inositol trisphosphate. The inositol linesterase (see Section 9-4). trisphosphate binds to channel proteins in the endoplasmic reticulum and (b) Various calcium pumps, using the energy of ATP hydrolysis, pump allows calcium ions to fl ow into the cytosol. Calcium ions then bind to calcium ions out of the cytosol. The Ca2 concentration in the cell calmodulin, causing a conformational change that allows it to bind and drops, the ion channels close, and Ca2 ions dissociate from calmoduactivate calcineurin. The activated calcineurin then activates NFAT, as lin, changing its conformation and rendering it unable to bind to NO described in the problem. synthase. Without bound calmodulin, the NO synthase is inactive.

32. If calcineurin is bound to cyclosporine A, then calcineurin will be in(c) The signaling molecule NO rapidly decomposes. In its absence, guaactive as a phosphatase and will be unable to catalyze the removal of the nylate cyclase is not activated. Any cGMP still present is hydrolyzed to phosphate group from NFAT. The NFAT remains in the cytosol, and the GMP by cGMP phosphodiesterase. specifi c genes required for T cell activation are not expressed. The T cell, an (d) In the absence of cGMP, protein kinase G reassociates with its reguimportant component of the immune response, is not activated, which aclatory subunits and is inactivated. Any phosphorylated proteins are acted upon by phosphatase enzymes, which remove the phosphate groups counts for the immunosuppressive properties of cyclosporine A. by hydrolysis.

33. Overexpression of PTEN in mammalian cells would promote apoptosis. PTEN removes a phosphate group from inositol trisphosphate; when

Ch 10 Solutions

49

synthase, so the cultured cells would be expected to show an increase in gly37. If the NO enzyme is missing, the signaling pathway described in Probcogen synthesis. Enhanced activation of protein kinase C would stimulate lem 35 cannot be completed. NO cannot be synthesized in the absence of translocation of glucose transporters to the plasma membrane; thus, an inthe NO synthase enzyme, and subsequent steps, including the production crease in glucose import would be observed in the cultured cells. of the second messenger cGMP and activation of protein kinase G, do not occur. Protein kinase G acts on muscle in such a way that the muscle relaxes. 49. As noted in Problem 29, phorbol esters are diacylglycerol analogs that If this does not occur, muscles lining the blood vessels are constricted, result- can activate protein kinase C. According to the information given in this ing in high blood pressure. This makes it more diffi cult for the heart to pump problem, protein kinase C activates the MAP kinase cascade, which leads to blood through the circulatory system, leading to an increased heart rate and the phosphorylation of proteins that infl uence gene expression. When these increased size of the ventricular heart chambers. genes are expressed, progression through the cell cycle is altered and cells are stimulated to grow and proliferate, a characteristic of tumor cells.

38. NO synthase is activated when calcium ions bind to calmodulin (see Solution 35). Clotrimazole is a calmodulin antagonist and prevents NO syn- 50. A ligand could bind to a G protein–coupled receptor and subsequently thase from binding to calmodulin. In the absence of active NO synthase, activate phospholipase C. This enzyme catalyzes the cleavage of phosphatidylcyclic GMP is not produced and protein kinase G is not activated. Celluinositol bisphosphate to diacylglycerol and inositol trisphosphate. Diacyllar targets of protein kinase G are not phosphorylated in the presence of glycerol activates protein kinase C, which then activates the MAP kinase clotrimazole. cascade, as shown in Problem 49. This is an example of cross-talk, in which 39. Nitroglycerin decomposes to form NO, which passes through cell memsignaling pathways share intracellular components. branes in tissues of the tongue to enter the bloodstream. NO activates gua- 51. In order to become activated, two inactive PKR proteins must come nylate cyclase in smooth muscle cells, as described in Problem 35, producclose enough to phosphorylate each other (autophosphorylation). A long ing cyclic GMP, which subsequently activates protein kinase G. The kinase RNA molecule can bind two PKR proteins simultaneously, holding them phosphorylates proteins involved in muscle contraction, which leads to the in close proximity so that they can activate each other. Short RNA molrelaxation of the smooth muscle cell. This increases blood fl ow to the heart ecules prevent PKR activation because when a short RNA molecule ocand relieves the pain associated with angina. cupies the PKR RNA-binding site, the PKR cannot bind to another RNA 40. Inhibition of cGMP phosphodiesterase, the enzyme that hydrolyzes where it might encounter a second PKR and get phosphorylated. [From cGMP, increases the intracellular cGMP concentration. The activity of pro- Nallagatla, S. R., Toroney, R., and Bevilacqua, P. C., Curr. Opin. Struct. Biol. tein kinase G is increased as a result, as is the phosphorylation of the enzyme’s 21, 119–127 (2011).] targets, which are proteins involved in smooth muscle cell relaxation. The 52. (a) The injection of YopH into a cell causes rapid dephosphorylation of overall result is increased blood fl ow. phosphorylated tyrosines and disrupts receptor tyrosine kinase signaling 41. (a) Adenylate cyclase generates the second messenger cAMP in response pathways by favoring the “off” position of the pathway, which would to activation of G proteins by G protein–coupled receptors. The EF decrease the likelihood of cell survival. toxin would generate large amounts of cAMP in the absence of any (b) The YopH protein does not affect the bacterium itself because bacspecifi c hormone signal. teria do not have proteins that are phosphorylated on tyrosine residues. (b) When Ca2 calmodulin is bound to EF, it is not available to activate any (c) Because YopH is a mammalian phosphatase homolog, care must be other Ca2 -sensitive proteins that might be involved in normal cell signaling. taken to develop an inhibitor that targets YopH specifi cally, not mammalian phosphatases.

42. By inactivating a signaling molecule required for cell proliferation, LF can 53. Substances cannot enter the nucleus unless they possess a nuclear loblock the production of white blood cells. This impairs the ability of the immune calization signal, a sequence that interacts with the nuclear pore and allows system to fi ght off the infection, so the bacteria are able to multiply in the entry into the nucleus. The nuclear localization signal on the progesterone receptor must be exposed, even when ligand is not bound. But the nuclear body. localization signal on the glucocorticoid receptor must be masked. When 43. Since the growth factor stimulates kinase activity, the H2O2 second ligand binds, a conformational change occurs that unmasks the nuclear lomessenger is likely to produce similar responses, so it must inactivate the calization signal, and the complex can pass through the nuclear pore and phosphatases. enter the nucleus.

44. PTEN is a phosphatase (see Problem 33); therefore, if H2O2 inacti54. (a) In order to treat cancer in these tissues, membrane-permeable drugs vates cellular phosphatases (see Solution 43), it likely inhibits PTEN as well. could be developed that would interact with intracellular steroid recep[From Gough, D. R., and Cotter, T. G., Cell Death and Disease (Nature tors to prevent binding by the steroid ligand.

Publishing Group), 2, e213 (2011).] (b) Because the activation of Ras is a common feature of many trans45. In the presence of GEF, the activity of the signaling pathway increases, formed cells (see Box 10-B), it’s possible that some steroids use a second signaling mechanism that involves activation of Ras, either since GEF promotes dissociation of bound GDP, and Ras ? GDP is inactive. through binding to G protein–coupled receptors or to receptor tyroOnce GDP has dissociated, GTP can bind and activate Ras. The opposite sine kinases. is true in the presence of GAP. Ras ? GTP is active, but when the GTP is hydrolyzed to GDP, Ras is converted from the active to the inactive form.

55. Arachidonate is the substrate for the production of prostaglandins, 46. In the presence of Ras with this particular mutation, the signaling path- many of which have infl ammatory properties. Stimulating the release of way would be constitutively active. Ras is active when bound to GTP and is arachidonate from the membrane by C1P increases the concentration of inactivated when GTP is hydrolyzed to GDP. If hydrolysis of GTP cannot substrate available for prostaglandin synthesis. One of the enzymes that cataoccur, Ras cannot be inactivated and the signaling pathway cannot be turned lyzes the fi rst step in the production of prostaglandins is COX-2, which is off. The cell is stimulated to grow and proliferate even in the absence of stimulated by S1P. Both C1P and S1P can potentially increase production of growth-signaling ligand. prostaglandins, which accounts for their infl ammatory properties, as shown 47. Phosphatases that remove phosphate groups from the insulin receptor in the diagram. would turn the insulin signaling pathway off and protein kinases B and C 56. Protein kinase B (Akt) is anti-apoptotic (see Problem 33). If S1P actiwould not be activated. Without active protein kinase B, glycogen synthase is vates protein kinase B, then the cell will not undergo programmed cell death inactive and glycogen cannot be synthesized from glucose. Without protein but will grow and proliferate. This explains S1P’s ability to promote cell surkinase C, glucose transporters are not translocated to the membrane and vival, as shown in the diagram in Problem 55. glucose is not brought into the cell but remains in the blood. Drugs that act 57. S1P might use a variety of mechanisms to activate Ras, either through as inhibitors of these phosphatases would potentiate the action of the insulin receptor, allowing the receptor to remain active with a lower concentration receptor tyrosine kinases or activation of protein kinase C. Ras then activates of the MAP kinase pathway (see Problem 49), which leads to the phosphorylaligand, and thus are potentially effective treatments for diabetes. tion of transcription factors that promote the expression of proteins involved 48. Overexpression of IRS-1 would stimulate the activity of the signaling in the cell cycle, ultimately leading to cell survival. pathway and would lead to enhanced activation of protein kinases B and C. The activation of protein kinase B would result in the activation of glycogen

50

Ch 11 Solutions

58. Both ceramide kinase and sphingosine kinase are potential drug targets, 64. In the mutant yeast the Ras-cAMP pathway could potentially still funcsince these enzymes catalyze the synthesis of C1P and S1P, respectively, and tion, but in the absence of a functioning HOG pathway (see Solution 63), both of these products promote cell survival. Inhibiting the enzymes that the activation of PFK2 would be nearly the same in isotonic and hypertonic catalyze production of these pro-survival signaling molecules may inhibit the media, depending on the extent of the HOG pathway’s contribution to the growth of cancer cells. But caution should be exercised, since inhibitors of phosphorylation of PFK2. [From Dihazi, H., Kessler, R., and Eschrich, K., COX-2, such as Vioxx, were determined to be unsuitable drugs because of J. Biol. Chem. 279, 23961–23968 (2004).] unexpected side effects.

59.

O

O

C O OH COCH3

C

Chapter 11 1. (a) aldose OH

OH Salicylic acid

Acetylsalicylic acid Enzyme

CH2O

C

(c) aldose

2. (a) aldotetrose (threose) (b) ketopentose (ribulose) 3. Chiral carbons are indicated with an asterisk:

Acetylated enzyme

CH2OH

(b) ketose

CH3

(a)

CHO

HO

C* C*H

H

OH CH2OH

(b)

HO

O

CH2OH

(c)

CHO

C

O

HO

C*

H

C*

H

HO

C*

H

OH

HO

C*

H

H

C*

OH

C*

H

CH2OH

Without knowing the mechanism of the enzyme, it is not possible to say CH2OH for certain why acetylating the serine inhibits cyclooxygenase activity.

But it’s possible that the acetylation alters the structure of the active site such that the arachidonic acid substrate is unable to bind. It’s also 4. Coenzyme A, NAD, and FAD all contain ribose residues. possible that serine participates in catalysis, possibly as a nucleophile, as (b) L (c) D (d) L in chymotrypsin. An acetylated serine would be unable to function as a 5. (a) D nucleophile, which would explain why the modifi ed enzyme is 6. (a) 22 4 (b) 23 8 (c) 24 16 catalytically inactive.

7. (a) D-Psicose and D-sorbose are epimers.

60. Aspirin inhibits cyclooxygenase (see Box 10-D) by acetylating an essen(b) D-Sorbose and D-fructose are structural isomers. tial Ser residue on the enzyme. In the presence of aspirin, thromboxanes are (c) D-Fructose and L-fructose are enantiomers. not synthesized. Thromboxanes promote platelet aggregation and vasocon(d) D-Ribose and D-ribulose are structural isomers. striction, which could promote clot formation and high blood pressure and 8. -D-Glucose and -D-glucose are anomers. could lead to a heart attack. 61. Phospholipase A2 catalyzes the release of 9. Fructose and galactose are isomers of glucose. arachidonate from membrane phospholipids. Blocking this reaction would prevent the COX-catalyzed conversion of arachidonate to proinfl ammatory prostaglandins. 62. The 10.

CHO

other product is arachidonate.

HOCH

High osmolarity (salt or glucose)

63.

HOCH HCOH HCOH

Ras

CH2OH

Raf

cAMP

MEK

PKA

11. (a)

CH2OH

MAP kinase

PFK2

Pi

Activation of glycolysis Accumulation of glycerol

(b)

CH2OH

C

O

C

O

HO

C

H

H

C

OH

HO

C

H

H

C

OH

H

C

OH

HO

C

H

CH2OH

CH2OH

(c) Tagatose is less effectively absorbed in the small intestine because the transport proteins in the epithelial cells lining the small intestine do not bind and transport tagatose as effi ciently as they do sugars that are more naturally and commonly present in the diet.

Ch 11 Solutions 51 12.

18.

CH2OH O C

O C

CH3

H

OH C

H

OH C

H

OH C

OH

HO

H

CH2OH

H

CH2OH

H

H

HO

OH

H

H OH -D-Galactose

19. All the sugar molecules will be converted to product because the and anomers are in equilibrium. Depletion of molecules in the form will cause more of the anomers to convert to anomers, which will then be converted to product.

because the phosphorylated glucose is negatively charged and cannot easily cross the nonpolar lipid bilayer by passive diffusion. Glucose-6-phosphate cannot exit the cell via transport proteins either, as these transporters are specifi c for glucose, not glucose-6-phosphate.

CH2OH

CH2OH O H H OH

O H

H OH

OH

22.

H

C H H

O H

H

H

OH

HO

OH OH -D-Ribose

O H

H H

OH H

16. A fi ve-membered ring results.

H

H

H H

HOCH2 O

OH

OH

H

H

OH OH -D-Ribose

H

H

O

CH2

O H

CH2

H

O H

CH2

CH2OH

NH 3

CH2OH

HO

H

OH

O HO

OH CH2OH

OH H -D-Fructose

OH H -D-Fructose

H C

COO

NH 3

HO H H OH H OH

OH

CH2OH

(b) no (c) no (d) yes (However, in the presence of the strong base in Benedict’s solution, fructose undergoes isomerization to an aldose, which does react with Cu2 .)

[From Benedict, S. R., J. Biol. Chem. 5, 485–487 (1908).]

25.

H

(CH2 )4

24. (a) yes

OH OH -D-Ribose

HOCH2

H H N C

O

H OH

H

H

17. (a) A fi ve-membered ring results.

HOCH2

O H

CH CH2

23. Glucose is a reducing sugar because the cyclization reaction can occur in reverse to form the straight-chain structure that contains an aldehyde group (see Fig. 11-1). Fructose contains a ketone functional group that cannot be oxidized by Cu2 , so it is not a reducing sugar. (However, in the presence of the strong base in Benedict’s solution, fructose undergoes isomerization to an aldose, which does react with Cu2 .)

OH OH -D-Ribose

HOCH2 O

H 3N

H

HO

H

H H

O C

O

H OH -D-Galactose

H

H

OH

21. Glucose-6-phosphate is more likely than glucose to remain in the cell

-Mannose

HO

R H C

(¬OH and ¬CH2OH groups) are in the equatorial position on the chair conformation of glucose. The anomer has one hydroxyl group in the axial position, so the equilibrium position favors formation of the anomer and the concentrations of the anomers are not equal.

OH

H

H

H

OH

20. The anomer is more stable because all of the bulky substituents O HO

H

HO

15.

C

N-Acetylneuraminic acid

13. The anomer is more stable because most of the bulky hydroxyl substituents are in the equatorial position.

H

H

COOH

H

OH

[From Chung, M.-Y., Oh, D.-K., and Lee, K. W., J. Agric. Food Chem., 60, 863–869 (2012).]

HO

O

H

CHOH 2

14.

H R H

H N

H HO

CHOH 2 O H H OH H

H

H

CH3I

OH

H3CO

OH

CH 2OCH 3 O H H OCH3H OCH3 H

-D-Glucose

OCH3

1,2,3,4,6-Penta-O-methyl-D-glucose

(b) A six-membered ring results.

H H HO

H O

H H

OH

OH H -D-Fructose

CH 2OH OH

H HO

O H H

OH OH CH2OH

H H3CO

CH2OCH3 O H 3 OCHH H

OH H -D-Fructose

CHO H OCH3

OCH3

1,2,3,4,6-Penta-O-methyl-D-glucose

H, H2O CH3OH

H H3CO

OCH3 H OCH3

H

OCH

H

3

CH2OCH3 2,3,4,6-Tetra-O-methyl-D-glucose

Ch 11 Solutions

52

26. There are several possibilities; one is shown here.

(b)

CH2OH O H H H OH OH NH2 H

HO

CH2OH HCOH

H

O

OH H

H

H H

OH

(c)

OH

CH2OH H HO

H

27.

COO H

OH

HO

H

H

OH

H

OH

HO

H

H

OH

H

OH

33.

CH2OH H HO

OH

H

OH

O

H H OH

H

KDG

HO

H OH H

2

CHOP O3 2 O H

H

OH

H

OH CH 2OPO32

CH2OH O H H OH H

30. For (b) and (c), only one of several possible structures is shown.

HO

H

O

OH

H H OH

ture of the and anomers.

CH2OH

(a)

O

34. Trehalose is not a reducing sugar because the anomeric carbons of both glucose residues are involved in the glycosidic bond.

35. Trehalase digestion produces glucose, which exists in solution as a mix-

O H

CH2OH

Cellobiose is a reducing sugar. The anomeric carbon of the glucose on the right side is free to reverse the cyclization reaction to re-form the aldehyde functional group, which can be reduced.

COO

H

H OH H

Sorbitol

H

CH3

Individuals whose ancestors domesticated cows and goats relied on the milk produced from these animals as a food source and became lactose tolerant as a result of natural selection. Those individuals who maintained the ability to produce lactase were more likely to survive to adulthood to produce offspring with the same metabolic capability.

CH2OH

29.

C

the glucose residue). Sucrose is not a reducing sugar because the anomeric carbons of both glucose and fructose are involved in the glycosidic bond. 32.

CH2OH H

HN

31. Lactose is a reducing sugar because it has a free anomeric carbon (C1 of

Gluconate

OH

H OH

O

CH2OH

28.

O H

H OH

H

H

OH

HO

CH2OH O H H OH H

OH

OH H

OH

36. The numerous hydroxyl groups on the trehalose molecule enable it to form a network of hydrogen bonds with numerous water molecules. In this manner, cellular structures are stabilized in a dehydrated form.

37.

CHOH 2 H HO

H OH H

CHOH 2 O H

H

H O

OH

H OH H

O H

OH

H OH

Ch 11 Solutions 53 38. The unknown sugar is isomaltose.

H HO

CH2OH O H H OH H

46. Potato starch consists of a mixture of amylose and amylopectin. The iodine can fi t into the center of the helical structure of the amylose and produce a blue color. Apples, on the other hand, contain mostly cellulose.

Cellulose consists of sheets of extended (1 S 4) glycan chains and does not form a helical structure. Consequently, iodine does not form a complex with the cellulose and a blue color is not produced.

H O

47. Pectin is a highly hydrated polysaccharide, so it thickens the fruit preparation and helps turn it into a gel.

OH H2C H HO

O H OH

H

H

OH

H

48.

H

OH

H

O HO

OH

H

O CH2 O

O

HO

OH

H

OH

OH

O

HOCH 2 H

H

OH

H

H HO

H

H O

H

H

OH

H

CH2OH O H H OCH3 OH H

H

HO

H

O

O

H

H

HO

52.

HOCH 2

O

CH2OH O H OH H OCH3 OH H

H

CH2OH

[From Jackson, M., and Brennan, P., J. Biol. Chem. 284, 1949–1953 (2009).] 53.

H

H

O

50. (a) Cotton clothing consists mostly of cellulose fi bers. Limited digestion by cellulase destroys some of the fi bers, creating the weathered look of stonewashing. (b) Prolonged treatment with cellulase would completely destroy the cotton fi bers and would ruin the garment.

H

H

H

H

49. The cellulose-based plant cell wall is strong and rigid, but it must be remodeled as the plant cell grows. The cell uses cellulase to weaken the cell wall so that the cell can expand.

44. There is one reducing end.

CH 2

H

H

H

42. The fungus must have contained an -galactosidase enzyme capable of digesting the (1 S 6) linkage and a sucrase-like enzyme that could hydrolyze the (1 S 2) glycosidic bond formed between the anomer of glucose and the anomer of fructose. [From Feng, S., Saw, C. L., Lee, Y. K., and Huang, D., J. Agric. Food Chem. 56, 10078–10084 (2008).]

45. (a)

O H

H OH

HO

41. Humans do not have one of the enzymes needed to hydrolyze the glycosidic bonds in order to convert the trisaccharide raffi nose to its constituent monosaccharides. The glycosidic linkages in raffi nose are an (1 S 6) link51. age between galactose and glucose and an (1 S 2) glycosidic linkage between glucose and fructose. Humans have the enzymes to digest the second glycosidic bond but not the fi rst.

43. Starch, glycogen, cellulose, and chitin are homopolymers. Peptidoglycan and chondroitin sulfate are heteropolymers.

OH

H OH

H

39. In order for sorbitol to be catabolized to yield energy, it would need to enter the same catabolic pathway as glucose. The enzymes that catalyze glucose catabolism are specifi c for glucose and do not bind to sorbitol, so the sugar alcohol is not metabolized and passes through the body undigested.

In this manner, sorbitol contributes no calories to the food containing it, but it is nearly identical to its parent monosaccharide in taste and sweetness.

40. Celery is mainly cellulose and water, neither of which provides nutritive calories. Humans do not have -glucosidase enzymes and cannot hydrolyze the -glycosidic bonds linking the glucose resides in cellulose.

Since cellulose is not digested, the body does not spend any energy to further process it. Foods like celery contribute roughage, or fi ber, to the diet, but these foods neither provide nor cost the body much in the way of energy.

O

H

CH2OH

The N-linked saccharide is N-acetylglucosamine, and the bond has the confi guration. The O-linked oligosaccharide is N-acetylgalactosamine, and the bond has the confi guration.

54.

(b) Humans do not have the enzymes to digest (2 S 1) glycosidic bonds (although the bacteria that inhabit the small intestine possess the necessary enzymes and do have this capability). Nondigestible carbohydrates are often classifi ed by food manufacturers as “fi ber”; thus, inulin extracted from chicory root is often added to processed foods to boost their fi ber content.

CH2OH OH CH2OH OH H

H OH

O

O H

OH

H H

H H

O H

H

H

NH H 2 CH O C C NHCOCH3 O

Ch 12 Solutions

54

55. The residues of the disaccharide are glucuronate linked by a (1 S 3) Chapter 12 glycosidic bond to N-acetylgalactosamine-4-sulfate. Disaccharides are linked 1. (a) chemoautotroph (b) photoautotroph to each other by (1 S 4) bonds. (d) heterotroph (c) chemoautotroph 56. Each disaccharide unit of chondroitin sulfate has two negatively charged (f) chemoautotroph (e) heterotroph (g) photoautotroph groups: a carboxylate group and a sulfate group. One hundred of these disaccharide units would yield a net charge of 200.

2. The purple nonsulfur bacteria are photoheterotrophs. They are similar to the photoautotrophs in that they can capture energy from sunlight but differ in that they are unable to fi x CO2. These bacteria are similar to the 57. heterotrophs in that an organic source of carbon is required. Thus the term CH2OH CH3 NH photoheterotroph accurately describes the trophic strategy of this organism.

O O C CH OH 3. The pH of the stomach is 2. At this pH, the salivary amylase is denaH H H tured and can no longer catalyze the hydrolysis of glycosidic bonds in dietary C O OH carbohydrates.

H

H

H

58.

4. (a) The main products of pancreatic amylase digestion of amylose are maltotriose (a glucose trisaccharide) and maltose (a glucose disaccharide). (b) Maltose and maltotriose are also products of amylopectin digestion; in addition, a “limit dextrin” of 5–6 glucoses that includes one (1 S 6) bond is also formed.

OH

CH2OH H HO

O H

H OH

H O

R

NH

C H

CH

NH

H

C

C

5. Maltase is required to hydrolyze the (1 S 4) glycosidic bonds in maltotriose and maltose (see Solution 4). Isomaltase is needed to hydro- lyze the (1 S 6) glycosidic bonds in the limit dextrins because -amylase only catalyzes the hydrolysis of (1 S 4) glycosidic bonds and cannot accommodate branch points (see Problem 4). These enzymes are required to completely hydrolyze starch to its component monosaccharides, since only monosaccharides can be absorbed.

O

O

6. Monosaccharides enter the cells lining the intestine via secondary active transport (see Fig. 9-18). Na ions and glucose enter the cell in symport, both along their concentration gradients. The Na ions are subsequently pumped out of the cell by the Na,K-ATPase transporter, which uses the free energy of ATP hydrolysis to eject the Na ions against their concentration gradient.

7. Sugar alcohols are not present in abundance naturally, which explains the [From Schirm, M., Schoenhofen, I. C., Logan, S. M., Waldron, K. C., and absence of transporters for these molecules. Passive diffusion is less effec- tive Thibault, P., Anal. Chem. 77, 7774–7782 (2005).] than passive transport (as shown in Problem 9-15).

59. The monosaccharide is N-acetylglucosamine.

8. Ethanol is water soluble but also is small and has suffi cient nonpolar character to pass through the membranes of the cells lining the stomach and 60. Lysozyme catalyzes hydrolysis of the (1 S 4) link between the monosaccharide residues in peptidoglycan, thereby destroying the bacterial cell the small intestine without requiring a transport protein. Ethanol absorption occurs quickly on an empty stomach but more slowly if ethanol is consumed wall. with food.

61. The amide bond forms between Ala and the carboxylate group on the 9. Because the products of nucleic acid digestion are relatively large, charged nucleotides, a transport protein is required to facilitate their move- ment C4 substituent in the disaccharide. across the cell membrane. The transport protein most likely uses the free energy of a Na gradient (active transport), as is the case for intestinal monosaccharide and amino acid transporters.

CH2OH CH2OH 10.

CH3

R H (Ser) R CH3 (Thr)

H

O H OH

H

H

NH

H O

H

O H

O

O

H

H C

C

O

H

CH3 O

CH 3CHCOO

NH

C O

CH3 n

62. (a) Sialic acid is negatively charged. The presence of sialic acid on their surfaces weakens the attachment of the tumor cells to each other and may promote the detachment process. (b) A drug could act as an inhibitor of one of the enzymes in the biochemical pathway for synthesizing sialic acid. Alternatively, foreign sialic acid precursors could be administered that would be taken up by the tumor cells and used for sialic acid synthesis. The use of a foreign precursor would result in the synthesis of a sialic acid derivative that is more immunogenic. [From Fuster, M. M., and Esko, Nature Rev. Cancer, 5, 526–542 (2005).]

NH

H 2O

HCl

C

O

H 3N

O

11. The low pH denatures the protein, unfolding it so peptide bonds are more accessible to proteolytic digestion by stomach enzymes.

12. The active site of pepsin is lined with Asp and Glu residues that are unprotonated at pH 7. These residues form ion pairs with the basic peptide, which contains positively charged Arg and Lys residues. When the pH drops to 2, the Asp and Glu residues become protonated and lose their negative charge. Ion pairs can no longer form, so the affi nity of the basic peptide for the active site decreases. The basic peptide dissociates, the enzyme cleaves the peptide and the enzyme is activated autocatalytically.

13. The pH optimum for pepsin is 2, which is the pH of the stomach. The pH optimum for trypsin and chymotrypsin is 7–8, as the small intestine is slightly basic (see Table 2-3). Each enzyme functions optimally in the conditions of its environment.

14. The interaction between the botulinum toxin and its “shield” protein appears to be pH dependent, with a strong interaction occurring at low pH and a weaker interaction at high pH. It’s possible that the interface between these two proteins involves ion pairing and hydrogen bonding.

Ch 12 Solutions

55

At the extremely low pH of the stomach, all basic amino acid side chains 20. Cholesteryl esters are completely nonpolar and are located in the center and many of the acidic side chains are protonated. When the complex of the chylomicron particle, where they will not interact with water. The enters the small intestine, the acidic residues become unprotonated and hydroxyl group of the cholesterol molecule can be found on the surface of negatively charged and exert repulsive forces on the shield protein, the chylomicron, where it forms hydrogen bonds with water; the rest of the causing it to dissociate and releasing the toxin. [From Gu, S., Rumpel, S., molecule resides in the chylomicron’s nonpolar core.

Zhou, J., Strotmeier, J., Bigalke, H., Perry, K., Shoemaker, C. B., Rummel, 21. (a) The polar glycogen molecule is fully hydrated, so its weight refl ects A., and Jin, R., Science 24, 977–981 (2012).] a large number of closely associated water molecules. Fat is stored in 15. Amino acids enter the cells lining the small intestine via secondary active anhydrous form. Therefore, a given weight of fat stores more free energy transport. This system is similar to the process for glucose absorption shown than the same weight of glycogen. in Figure 9-18. (b) Because it must be hydrated, a glycogen molecule occupies a large effective volume of the cytoplasm, which it shares with other glycogen + Amino acid Na molecules, enzymes, organelles, and so on. Because hydrophobic fat molecules are sequestered from the bulk of the cytoplasm, they do not Na-amino acid transporter INTESTINAL SPACE have the same potential for interfering with other cellular constituents, so their collective volume is virtually unlimited.

22. A branched glycogen chain has many nonreducing ends but only one reducing end. Enzymes specifi c for the nonreducing ends can act on many ends simultaneously, building up the glycogen molecule when glucose is plentiful Na+ Amino acid and degrading glycogen when glucose is in short supply. Enzymes that acted on the reducing end would accomplish these processes much more slowly.

K+ 23. The phosphorylated glucose molecule is not recognized by the glucose ATP transporter. Removal of the phosphate group allows the glucose to more easADP + Pi ily leave the cell.

24. If lysosomal hydrolytic enzymes function optimally at pH 5, they will Amino acid transporter Na,K-ATPase not function well at the cytosolic pH of 7. Amino acid side chains essential for catalytic activity will become deprotonated at the higher pH. If lysosomal Na+ K + Amino acid enzymes leak out of the lysosome into the cytosol, cellular components will 16. The entry of glucose and amino acids into cells lining the small intestine be protected from degradation by the destructive power of these enzymes. is accomplished by a secondary active transport process in which sodium ions 25. are transported into the cell along with the glucose or amino acid (see Fig. 9Citric acid Fatty acid 18 and Problems 6 and 15). Adding an electrolyte such as sodium chloride cycle metabolism Glycolysis provides the sodium ions required for this cotransport. As the intes- tinal cells import glucose and amino acids, water is also absorbed.

Acetyl-CoA ✓ ✓ 17. (a) Glyceraldehyde-3-P ✓ O O Pyruvate ✓ INTESTINAL CELL

O

R2

C

H2C

O

O

R1

O

C

O

Gastric lipase

O

CH

H2C

C

R3

H2O

Triacylglycerol

R2

C

H2C

O

R3

C

CH

H2C

O

O

OH

Diacylglycerol

O

C

R1

Triacylglycerol synthesis

Acetyl-CoA ✓ (b) Both diacylglycerol and fatty acids are amphipathic molecules— they Glyceraldehyde-3-P have both hydrophilic and hydrophobic domains. These molecules can Pyruvate form micelles that emulsify the dietary triacylglycerols, which are 26. (a) oxidized nonpolar and are unable to form micelles. (b) reduced (c) reduced 27. (a) NAD 18. Fatty acids, with one head and one tail, resemble membrane phospho(b) NADPH (c) NADH lipids (with one head and two tails) and are thus able to diffuse through the 2 4 SO4 S HCO 3 HS H2O 28. (a) CH membrane without the need for a transmembrane transport protein.

4 is oxidized to HCO3. (b) CH 19. (c) SO24 is reduced to HS. Fatty acid

O H3C

(H2C) 17

C

Cholesteryl ester

O O

H3C

(H 2C)17

C

Cholesteryl esterase O

Fatty acid

HO

Cholesterol

Photosynthesis

Transamination

✓ ✓ (d) oxidized (d) NADP

29. Individuals with gastrointestinal disorders might have a gastrointestinal tract that is not colonized by the appropriate vitamin B12–synthesizing bacteria. A defi ciency in haptocorrin or intrinsic factor would be manifested as a vitamin B12 defi ciency, since these proteins are essential for absorption of the vitamin. Vegetarians and vegans who consume no animal products would also be at risk for a defi ciency of vitamin B12.

30. (a) A high-protein diet, which contains a mixture of all 20 amino acids, including the essential amino acids (most of which are nonpolar; Table 12-1), may provide enough amino acids to compensate for the poor absorption of nonpolar amino acids due to the defective transporter. Small peptides containing the nonpolar amino acids can still enter cells via other transporters. (b) The amino acid transporter defect in Hartnup disease leads to a shortage of the amino acid tryptophan, which can be converted to niacin, the vitamin whose defi ciency causes pellagra.

Ch 12 Solutions

56 31. (a)

H N

K 5e2DG8¿yRT )(298 K) Keq 5210e kJ?mol21y(8.314531023 kJ?K21?mol21 eq Keq 524.04 e K eq50.018 K 5e2DG8¿yRT )(298 K) Keq 5220 e kJ?mol21y(8.314531023 kJ?K21?mol21 eq Keq 528.07 e

O C

CH CH2 CH

COO

OOC

-Carboxyglutamate (b) The additional carboxylate group on the glutamate residue confers a 2 charge on the side chain, generating a high-affi nity binding site for the Ca2 ions essential for blood clotting.

32. Vitamin A would be absorbed more effectively from cooked carrots because the cooking process breaks down the cell walls, releasing any vitamin bound to the cell walls and increasing the bioavailability of the vitamin.

33. (a) Vitamin C (b) biotin (c) pyridoxine (d) pantothenic acid 34. Vitamins are compounds that human beings need but cannot synthesize and thus must be obtained in the diet. Niacin is technically not a vitamin because it can be synthesized from tryptophan, although ingestion of this essential amino acid is required to maintain proper cellular niacin levels.

35. Because Keq is the ratio of the product concentration to the reactant con-

K eq50.000313 Small changes in G result in large changes in Keq. Doubling the G value (a positive, unfavorable value) leads to a 60-fold decrease in Keq. (b) K eq 5 e 2 D G 8 ¿ y R T )(298 K) K eq 52(210 e kJ?mol21)y(8.314531023 kJ?K21?mol21 eq K 54.04 e K eq556.6 K eq 5 e 2 D G 8 ¿ y R T )(298 K) K eq 52(220 e kJ?mol21)y(8.314531023 kJ?K21?mol21 eq5e 8.07 K K eq53200

The same conclusion can be made: Small changes in G lead to large changes in Keq. Doubling a (favorable) G results in a Keq value that is nearly 60 times as large. centration at equilibrium, the reaction with the larger Keq will have a higher concentration of product. Therefore, the concentration of B in Tube 1 will be 42. (a) greater than the concentration of D in Tube 2. glucose1Pi Sglucose-6-phosphate1H2O 36. Use Equation 12-2. For the AΔB reaction, DG8¿5113.8 kJ ?mol21 DG 8¿ 5 2RT ln K eq DG 8¿ 5 2(8.3145 31023 D G¿8 5 2

glucose-1-phosphate1H2OSglucose1Pi kJ ? K21 ?mol21)( 310 K)ln 10

5.9 kJ ?mol2 1

glucose-1-phosphateSglucose-6-phosphate DG8¿527.1 kJ ?mol21

For the CΔD reaction, DG8¿ 5 2(8.3145 3 DG8¿ 5 5.9 kJ ?mol21

DG8¿5220.9 kJ ?mol21

10 2 3 kJ ?K21?mol21)(310 K)ln 0.1

The reaction is spontaneous under standard conditions.

[G6P] [G1P] (b) DG 5 DG8¿1RT ln 37. (a) Since Keq 1, ln Keq 0 and G is also equal to zero (Equation 12-2).

2 1 1 (8.3145 3 10 2 3 kJ ? K2 1 ?mol 2 1 ) D G 5 2 7.1 kJ mol ? (b) Since Keq 1, the concentrations of reactants and products must 10 2 3 M be equal at equilibrium. If the reaction started with 1 mM F, the equilibb (310 K)ln a53 rium concentrations will be 0.5 mM E and 0.5 mM F. 0.13 10 2 3 M 38. Since Keq 1, the concentrations of reactants and products must be DG 5 3.0 kJ? mol21 the same at equilibrium. Therefore the reaction will proceed in the forward The reaction is not spontaneous under these conditions. direction with 1.5 mM E being converted to 1.5 mM F, resulting in equal concentrations (3.5 mM of each) at equilibrium.

43. The complete reaction is ATP H2O S ADP Pi. Use Equation 12-3 and the value of G from Table 12-4. The concentration of water is [B] assumed to be equal to 1.

39. DG5DG 8¿1RT ln [A] [ADP][Pi] DG5DG8¿1RT ln DG525.9 kJ?mol211(8.314531023 kJ?K21?mol21) [ATP] DG5230.5 kJ?mol211(8.314531023 kJ?K21?mol21) a0.131023 M b (310 K)ln 0.9 103 2 M 3 ( 0.001 )( 0.005 ) (0.003) (310 K)ln DG5211.6 kJ?mol21 The reaction will proceed as written, with A converted to B until the ratio

DG5230.5 kJ?mol21216.5 kJ?mol21

of 40.[B]/[A] DG 5 DG10/1.

8¿1RT ln

DG5247 kJ?mol21

[D] [C] DG 5 5.9 kJ ?mol211(8.3145 31023

DG 5 0.23 kJ ?mol21

kJ?K21 ?mol21) M (310 K)ln a 0.131023 Mb 0.9 3 1032

The reaction will proceed in the reverse direction as written, with D converted to C until the ratio of [C]/[D] 1/10.

41. (a) The equilibrium constant can be determined by rearranging Equation 12-2 (see Sample Calculation 12-2):

44. (a) See Sample Calculation 12-2. The equilibrium constant can be derived by rearranging Equation 12-2: 8¿y K e q 5 e 2 DGRT ?mol21 )(298 K) 527.9 kJ?mol21y(8.314531023kJ?K21 K eq e K eq 523.19 e Keq50.041 (b) At 37 C, T 310 K: DG5DG 8¿1RT ln

[dihydroxyacetone phosphate] [glyceraldehyde-3-phosphate]

Ch 12 Solutions DG57.9 kJ?mol211(8.314531023kJK1

? 2 ? mol12 ) (310 K)ln

(5 3 1024) (1 3 1024)

DG57.9 kJ?mol2114.1 kJ?mol21

(b) 100 moles ATP 3

505 g mol 3

57

1 lb 2200 g 5 23 lb

(c) ATP does not accumulate but instead is constantly recycled. As ATP is used, its hydrolysis products, ADP and Pi, serve as reactants for ATP synthesis in the process of oxidative phosphorylation.

DG512 kJ?mol21 (c) The reaction is not spontaneous as written. The reverse reaction, in which G 12 kJ ? mol 1, would be spontaneous.

45. First, convert Calories to joules: 72,000 cal 4.184 J/cal 300,000 J or 300 kJ. Since the ATP S ADP P reaction releases 30.5 kJ ? mol 1 , the i apple contains 300 kJ/30.5 kJ ? mol 1 or the equivalent of about 9.8 moles of ATP. 46. See Solution 45. A 760-Calorie drink contains 760,000 calories, which is equivalent to 3.18 106 J or 3180 kJ. The hot chocolate drink contains 3180 kJ/30.5 kJ ? mol 1 or the equivalent of about 104 moles of ATP. 47.

a

53. The apple provides 9.8 moles of ATP (see Solution 45). The moderately active female described in Problem 52 requires 100 moles ATP daily. Therefore, 100 moles ATP/9.8 apples mol 1 10.2 apples. Keeping the 33% effi ciency in mind, 10.2/0.33 31 apples would be required.

54. The hot chocolate drink provides 104 moles of ATP (see Solution 46).

The moderately active female described in Problem 52 requires 100 moles ATP daily. Therefore 100 moles ATP/104 hot chocolate drinks mol 1 0.96 drinks. Keeping the 33% effi ciency in mind, 0.96/0.33 3 drinks would be required. In other words, one of these drinks provides one-third of the daily energy needs for the moderately active female.

55. Reactions involving phosphoenolpyruvate, 1,3-bisphosphoglycerate, and phosphocreatine could drive the synthesis of ATP because transfer of a phosphoryl group from one of these compounds occurs with a greater change in free energy than the transfer of a phosphoryl group to ADP. 56. Reactions involving the conversion of phosphate to pyrophosphate

G

and the phosphorylation of glucose (to glucose-1-phosphate and glucose-6phosphate) and glycerol (to glycerol-3-phosphate) would require the hydrolysis of ATP to drive the reaction, because transfer of a phosphoryl group from ATP occurs with a greater change in free energy than the transfer of a phosphoryl group to one of these compounds.

57. (a) The equilibrium constant can be determined by rearranging Equa-

c b Reaction coordinate

48. Creatine, which undergoes phosphorylation to serve as a reservoir to phosphorylate ADP, becomes depleted after a few seconds of muscle activity.

Therefore, it would not be expected to play a role in maintaining ATP concentrations for longer periods.

49. (a) The phosphate groups on the ATP molecule would be less negative at a lower pH. Therefore, there would be less charge–charge repulsion and therefore less energy released upon hydrolysis. The G would be less negative at a lower pH. (b) Magnesium ions are positively charged and form ion pairs with the negatively charged phosphate groups. Thus, magnesium ions serve to decrease the charge–charge repulsion associated with the phosphate groups. In the absence of magnesium ions, the charge–charge repulsion is greater; thus, more free energy is released upon the removal of one of the phosphate groups. This results in a G value that is more negative.

50. The hydrolysis of pyrophosphate releases 19.2 kJ ? mol 1 (see Table 12-4). These two reactions are coupled, and the overall G value for the coupled reactions is negative; thus UDP–glucose formation occurs spontaneously.

51. (a) The synthesis of ATP from ADP requires 30.5 kJ ? mol 1 of energy: ADP Pi S ATP HO G2 30.5 kJ ? mol 1 0.33 2850 kJ?mol21 5 30.8 ATP 2 1 3 30.5 kJ? mol 9781 kJ?mol21 (b) 30.5 kJ?mol213

0.33 5 105.8 ATP

(c) For glucose, 30.8 ATP/6 carbons 5.1 ATP/carbon. For palmitate, 105.8 ATP/16 carbons 6.6 ATP/carbon. Most of the carbon atoms of fatty acids are fully reduced ¬CH2¬ groups. Most of the carbon atoms of glucose have hydroxyl groups attached to them (¬CHOH¬) and are therefore already partly oxidized. Consequently, more free energy is available from a carbon in a triacylglycerol than from a carbon in a glycogen molecule.

G 30.5 kJ mol 1 ?

52. (a) ADP P S ATP HO 2 i 2200 Calories day

3

1 mol ATP 30.5 kJ

3

4.184 kJ 1 Calorie

3 0.33 5 100 moles ATP

tion 12-2 (see Sample Calculation 12-2): Keq5e2DG8¿yRT 25 kJ?mol21y(8.314531023 kJ?K21?mol21

Keq5e

)(298 K)

Keq5e22.02 Keq50.133 (b) Since Keq5

[isocitrate] [citrate] 50.133

[isocitrate]50.133 [citrate] The total concentration of isocitrate and citrate is 2 M, so [isocitrate]52 M2[citrate] Combining the two equations gives 0.133 [citrate]52 M2[citrate] 1.133 [citrate]52 M [citrate]51.77 M [isocitrate]52 M21.77 M50.23 M (c) The preferred direction under standard conditions is toward the formation of citrate. (d) The reaction occurs in the direction of isocitrate synthesis because standard conditions do not exist in the cell. Also, the reaction is the second step of an eight-step pathway, so isocitrate is removed as soon as it is produced in order to serve as the reactant for the next step of the pathway.

58. (a) Use Equation 12-2 to solve for G : DG8¿52RT ln K eq DG8¿52(8.314531023

1

1

DG8¿52.2 kJ?mol21 The reaction will proceed in the opposite direction as written.

58

Ch 12 Solutions [fructose-6-phosphate] ( b ) D G 5 D G 8 ¿ 1 R T l n [glucose-6-phosphate] DG52.2 kJ?mol211(8.314531023 kJ?K21?mol21) (310 K)ln

(5 331024) (2.0 1023)

DG52.2 kJ?mol2123.57 kJ?mol21 DG521.37 kJ?mol21

(f) The reaction can be accomplished at a much lower glucose concentration when the phosphorylation of glucose is coupled to ATP hydrolysis (7.4 10 8 M instead of 13 M). This can be done because the second reaction couples the phosphorylation of glucose with the exergonic hydrolysis of ATP. Thus, an unfavorable reaction is converted to a favorable reaction.

61. The equilibrium constant can be determined by rearranging Equa-

59. Under these conditions, the reaction will proceed as written.

Mechanism 1: glutamate1NH3Sglutamine

tion 12-2 (see Sample Calculation 12-2): (a) K eq5 e2DG8¿yRT )(298 K) K 5247.7 e kJ?mol21y(8.314531023 kJ?K21?mol21 eq

K eq5e 219.2

ATP1H2OSADP1Pi

glutamate 1 NH3 1 ATP 1 H2O S glutamine 1 ADP 1 Pi

60. (a) The equilibrium constant can be determined by rearranging Equation 12-2 (see Sample Calculation 12-2): K eq5 e2DG8¿yRT y(8.314531023 K eq5 e213.8 kJ ?mol21

kJ ?K21 ?mol21 )(298 K)

0.00385

[glucose-6-phosphate] (2.5 3 1026) [glucose-6-phosphate]59.531028 M

(250 3 1026) [glucose](5.0 3 1023)

Driving the reaction to the right using this method is not feasible because it is impossible to achieve a concentration of 13 M glucose inside the cell. kJ ?K21 ?mol21 )(298 K)

K eq5 e6.74 K eq5850 K eq5 8505

52.2310211

(b) fructose-6-phosphate 1 Pi Δ fructose-1,6-bisphosphate ATP1HOΔADP1P 2

eq5 K eq5 K

[glucose]513 M

(e)

[fructose-6-phosphate]

DG8¿5230.5 kJ?mol21

i

DG8¿517.2 kJ?mol21

K eq5 e2DG8¿yRT 5e217.2 kJ ? mol21y(8.314531023 kJ ?K 21 ? mol21)(298 K) K eq

(5.0 3 1023)(5.0 3 1023)

K eq5 e2DG8¿yRT K eq 5e2(216.7 kJ ?mol21 )y(8.314531023

[fructose-1,6-bisphosphate]

(c)

Under the given conditions, the reaction would produce only 9.5 10 8 M glucose-6-phosphate and thus is not a feasible route to the production of this compound for the glycolytic pathway. [glucose-6-phosphate] [glucose][Pi] (c) K eq5

(d)

[fructose-6-phosphate](5.0 3 1023)

fructose-1,6-bisphosphate1ADP

[glucose-6-phosphate]

0.00385

[fructose-1,6-bisphosphate]

4.4310295

fructose-6-phosphate1ATP1H2OΔ

Keq50.0038 [glucose-6-phosphate] Keq5 [glucose][Pi] 0.00385

[fructose-6-phosphate][Pi]

DG8¿547.7 kJ?mol21

K eq5 e25.57 (b)

[fructose-1,6-bisphosphate]

K eq5

Mechanism 2: glutamate1ATPSg- glutamylphosphate1ADP g-glutamylphosphate 1 H2O 1 NH3 S glutamine 1 Pi Mechanism 2 is the more likely mechanism because it proceeds through a phosphorylated intermediate that “captures” the energy of the phosphoanhydride bond of ATP. In the fi rst mechanism, the two reactions are not linked to a common intermediate. ATP is hydrolyzed, but this energy is not “harnessed” in any way and is simply dissipated as heat.

29

K eq54.4310

glutamate1NH31ATP1H2OS glutamine1ADP1Pi

[glucose-6-phosphate][ADP] [glucose][ATP] (250 3 1026)(1.25 3 1023)

[glucose](5.0 3 1023) [glucose]57.431028 M

e26.94 1.0 3 1023

K eq5 [fructose-1,6-bisphosphate][ADP] [fructose-6-phosphate][ATP]

1.0310235

[fructose-1,6-bisphosphate][ADP] [fructose-6-phosphate][ATP]

(1.0 3 1023)[ATP] [ADP]

5

(1.0 3 1023)(3.0 3 1023) (1.0 3 1023) 3.0310235

[fructose-1,6-bisphosphate] [fructose-6-phosphate] 5

[fructose-1,6-bisphosphate] [fructose-6-phosphate]

[fructose-1,6-bisphosphate] [fructose-6-phosphate]

(d) The conversion of fructose-6-phosphate to fructose-1,6-bisphosphate is unfavorable. The ratio of products to reactants at equilibrium is 2.2 10 11 under standard conditions. But if the conversion of fructose-6-phosphate to fructose-1,6-bisphosphate is coupled with the hydrolysis of ATP, the reaction becomes more favorable and the ratio of fructose-1,6-bisphosphate to fructose-6-phosphate increases to 3 10 3, a change of eight orders of magnitude. (e) The second mechanism is biochemically feasible because the two steps are coupled via a common phosphorylated intermediate that “captures” the energy of ATP. In the fi rst mechanism, the two steps are not coupled. The ATP is hydrolyzed and the energy is lost as heat instead of being used to assist the conversion of fructose-6-phosphate to fructose-1,6-bisphosphate.

Ch 12 Solutions 62.

I. GAPΔ1,3-BPG

DG8¿56.7 kJ?mol21

1,3-BPG1H2OΔ3PG1Pi

DG8¿5249.3 kJ?mol21

GAP1H2OΔ3PG1Pi

DG8¿5242.6 kJ?mol21

II. GAPΔ1,3-BPG 1,3-BPG1ADPΔ3PG1ATP

DG8¿56.7 kJ?mol21 DG8¿5218.8 kJ?mol21

GAP1ADPΔ3PG1ATP

DG8¿5212.1 kJ?mol21

Coupling the synthesis of palmitoyl-CoA with the hydrolysis of ATP to AMP has improved the [product]/[reactant] ratio. The formation of products is now favored. (d) Coupling the synthesis of palmitoyl-CoA with ATP hydrolysis to AMP and PPi followed by PPi hydrolysis produces a standard free energy of –34.2 kJ ? mol 1 for the coupled process [–14.1 kJ ? mol 1 ( 19.2 kJ ? mol 1) 33.3 kJ ? mol 1]. palmitate1CoA1ATP1H2OSpalmitoyl-CoA1AMP12 Pi DG8¿5233.3 kJ?mol21 Kq5e2DG8¿yRT e

The second scenario is more likely. The fi rst coupled reaction is more exergonic, but the second coupled reaction “captures” some of this free energy in the form of ATP, which the cell can use.

21 y ?

2 kJ 2 m ol)(8.314510 Kq5e(33.3 kJ 3K e

eq K eq 5e212.7

26

K

5 3.0 3 10 [palmitoyl-CoA] Keq5 [palmitate][CoA] [palmitoyl-CoA] 3.0310265

23

21 ? 2 1 ?mol)(298 K)

13.4

K eq5e

5

K eq56.9310

63. (a) The equilibrium constant can be determined by rearranging Equation 12-2 (see Sample Calculation 12-2): K eq 5 e 2 D G 8 ¿ y R T ?m K) ol21 K eq 5231.5 e kJ ?m ol21y(8.314531023 kJ ?K 21 )(298

59

[palmitoyl-CoA][AMP][P i]2

6.9 3 105 5

[palmitate][CoA][ATP]

Coupling the activation of palmitate to palmitoyl-CoA with the hydrolysis of ATP to AMP, with subsequent hydrolysis of pyrophosphate, is a thermodynamically effective means of accomplishing the reaction. Coupling the reaction with hydrolysis of ATP to ADP is not effective.

64. (a) The value of C is determined from the slope of the plot. The slope is equal to 1.9 10 6 M 1.

[palmitate][CoA]

1.2

DG8¿51.0 kJ?mol21 K eq 5 e 2 D G 8 ¿ y R T )(298 K) K eq 521.0 e kJ?mol21y(8.314531023 kJ?K21?mol21 K

eq

0.8 −4

The reaction is not favorable. (b) Coupling the synthesis of palmitoyl-CoA with ATP hydrolysis to ADP produces a standard free energy change of 1.0 kJ ? mol 1 for the coupled process [31.5 kJ ? mol 1 (–30.5 kJ ? mol 1) 1.0 kJ ? mol 1]. palmitate CoA ATP S palmitoyl-CoA ADP Pi

[nick]/[phosphodiester] (10)

Therefore the ratio of products to reactants is 3.0 10 6:1.

y

0.4

0

520.40 e

10

0

(1.9

20

Keq50.67 K eq5

0.675

2.0

30

40

10 5

50

[AMP] (mM)

[palmitoyl-CoA][ADP][Pi] [palmitate][CoA][ATP] [palmitoyl-CoA][ADP][Pi]

(b) C5

[palmitate][CoA][ATP]

Keq5

Coupling the synthesis of palmitoyl-CoA with the hydrolysis of ATP to ADP for- has improved the [product]/[reactant] ratio considerably, but the mation of products is still not favored. (c) Coupling the synthesis of palmitoyl-CoA with ATP hydrolysis to AMP produces a standard free energy change of 14.1 kJ ? mol 1 for the coupled process [31.5 kJ ? mol 1 ( 45.6 kJ ? mol 1) 14.1 kJ ? mol 1]. palmitate 1 CoA 1 ATP S palmitoyl-CoA 1 AMP 1 PPi DG8¿5214.1 kJ?mol21 K eq 5 2eD G 8 ¿ y R T )(298ol21 K) kJ ?m ol21)y(8.314531023 kJ ?K 21 ?m K eq 52(214.1 e eq5e 5.7 K Keq5296 2965

10 6) x

[palmitoyl-CoA][AMP][PPi] [palmitate][CoA][ATP]

[PPi] K eq[ATP] [PPi] C[ATP] (1.0 3 1023)

Keq5

(1.9 3 1023)(14 3 1026)

Keq53.83

10 4

(c) Use Equation 12-2 to solve for G : DG852RT ln Keq ¿ DG852(8.314531023 kJ?K21 ¿

?mol2 1 )

(298 K) ln 3.83104 DG85226 kJ?mol21 ¿ (d) nick Δ

phosphodiester bond

ATP Δ

AMP1PPi

ATP 1 nick Δ AMP1PPi 1 phosphodiester bond

D G¿8 5 ?

21 D G ¿8 5 248.5 kJ?mol D G¿8 5 2 26 kJ?mol

21

Ch 13 Solutions

60

The G for the formation of the phosphodiester bond is 22.5 kJ ? mol 1. (e) The G value for the hydrolysis of a phosphodiester bond in DNA is –22.5 kJ · mol 1, whereas the G value for the hydrolysis of a typical phosphomonoester bond is 13.8 kJ ? mol 1. Therefore, the phosphodiester bond in DNA is less stable than the typical phosphomonoester bond. [From Dickson, K., Burns, C. M., and Richardson, J., J. Biol.

Chem. 275, 15828–15831 (2000).]

Chapter 13 1. (a) Reactions 1, 3, 7, and 10; (b) Reactions 2, 5, and 8; (c) Reaction 6; (d) Reaction 9; (e) Reaction 4.

In order to reverse the reaction, the ratio of glucose-6- phosphate to glucose would have to be 8.3 103:1.

4. (a)

DG8¿52RT ln

2.2 kJ? mol

23 2 1? mol 2 1 ) 5 2 (8.3145 3 10 kJ ? K [fructose-6-phosphate] (298 K) ln [glucose-6-phosphate]

2.2 kJ mol ?

21

522.48 kJ?m21

16.7 kJ?mol2152.48 kJ?mol21 ln

[glucose-6-phosphate]

[glucose-6-phosphate] [fructose-6-phosphate]

0.41 5 1

(b)

[fructose-6-phosphate]

[glucose-6-phosphate]

[glucose-6-phosphate]

216.7 kJ ? mol2152(8.314531023 kJ ? K21 ? mol21) [glucose-6-phosphate][ADP] [glucose][ATP] (298 K)ln

ln

[fructose-6-phosphate]

e20.89 5

[glucose-6-phosphate][ADP] [glucose][ATP]

ol

[fructose-6-phosphate]

20.89 5 ln

versible. Metabolically irreversible reactions are good “control points” for a pathway.

DG8¿52RTln

[glucose-6-phosphate]

21

2. Reactions 1, 3, and 10 are irreversible; the remaining reactions are re-

3. (a)

[fructose-6-phosphate]

DG5DG8¿1RT ln

[fructose-6-phosphate] [glucose-6-phosphate] 1 (8.3145 3 10 kJK?21? mol ) [fructose-6-phosphate] (310 K) ln [glucose-6-phosphate]

21.4 kJ?mol215 2.2 kJ?mol21

23

[glucose-6-phosphate][ADP] [glucose][ATP] [glucose-6-phosphate][ADP]

e6.735 8405 8405 8.431035 (b)

[glucose-6-phosphate][ADP] [glucose][ATP]

[glucose-6-phosphate](1)

[glucose][ADP] [glucose-6-phosphate][ADP]

[glucose][ATP]

216.7 kJ?mol2152.45 kJ?mol21 ln

1.2310235 1.2310245

[glucose-6-phosphate][ADP] [glucose][ATP]

[glucose-6-phosphate][ADP] [glucose][ATP] [glucose-6-phosphate][ADP] [glucose][ATP] [glucose-6-phosphate][ADP] [glucose](10) [glucose-6-phosphate](1) [glucose] [glucose-6-phosphate]

21

ln

[fructose-6-phosphate] [glucose-6-phosphate]

[fructose-6-phosphate] [glucose-6-phosphate] [fructose-6-phosphate] [glucose-6-phosphate] [fructose-6-phosphate] [glucose-6-phosphate]

The reaction will proceed in the direction of fructose-6-phosphate syn-

[glucose]

[glucose][ATP] (298 K)ln [glucose-6-phosphate][ADP]

1.2310235

0.25 5 1

[glucose](10)

16.7 kJ?mol2152(8.314531023 kJ?K21?mol21)

e26.735

e21.4 5

[glucose][ATP]

[glucose-6-phosphate]

5 2.6 kJ ? mol

21.4 5 ln

[glucose-6-phosphate][ADP]

DG8¿52RT ln

26.735ln

23.6 kJ?mol21

[glucose][ATP]

6.735ln

21

thesis since G 0.

5. (a) Because the brain relies on glucose from the blood, it stores very little glucose in the form of glycogen. Therefore, glucose rather than phosphorylated glucose is the substrate that enters the glycolytic pathway. The fi rst step of glucose catabolism in the brain is catalyzed by hexokinase, so this step is the rate-determining step of the pathway. In other tissues that break down glycogen for glycolysis, the hexokinase step is bypassed. (b) The low KM means that the enzyme will be saturated with glucose and will therefore operate at maximum velocity. Even if the concentration of glucose were to fl uctuate slightly, the brain’s ability to catabolize glucose would not be affected.

6. This would not be benefi cial to the patient. In order to enter the glycolytic pathway, the glucose-6-phosphate would fi rst have to enter the cells.

Glucose transporters recognize glucose, not glucose-6-phosphate; thus, glucose-6-phosphate would be unable to enter the cell for oxidation through the glycolytic pathway. 7. One might expect the product of a reaction to inhibit the enzyme that catalyzes the reaction, while the reactant would act as an activator. Although it is true that ADP is a direct product of the PFK reaction, PFK is sensitive to the ATP needs of the cell as a whole. Rising ADP concentrations are an indication that ATP is needed; the subsequent stimulation of PFK increases glycolytic fl ux and generates ATP as a fi nal pathway product.

8. (a) R

(b) T

(c) R

Ch 13 Solutions 9. In the presence of the inhibitor, the curve is sigmoidal and the KM increases dramatically (nearly 10-fold, to 200 M), indicating that a greater quantity of substrate is required to achieve ½ Vmax. PEP stabilizes the T

61

(b) The structure of GAPDH in cancer cells is probably different from the structure of GAPDH in normal cells. The structure of the active site in GAPDH from cancer cells might be altered in such a way that the binding of methylglyoxal is permitted, which then precludes the binding of the substrate. Or the altered GAPDH might have a binding site for methyl- glyoxal elsewhere on the protein, which causes a conformational change in the protein that alters the substrate binding site so that the substrate can no longer bind. [From Ray, M., Basu, N., and Ray, S., Mol. Cell. Biochem. 177, 21–26 (1997).]

form of PFK (see Solution 8).

10. The negatively charged fructose-1,6-bisphosphate is repelled by the negatively charged Glu 161 in the active site in the low-affi nity form of the enzyme, whereas the substrate is attracted to the positively charged Arg 162 positioned in the active site in the high-affi nity form.

11. Glycerol can serve as an energy source because glycerol can be converted 17. Arsenate is a metabolic poison, and the cells eventually die. In the presto glyceraldehyde-3-phosphate, which can then enter the glycolytic pathway ence of arsenate, 1,3-bisphosphoglycerate is not formed; instead, this step “below” the phosphofructokinase step. The mutants cannot grow on glucose is essentially skipped. Two moles of ATP per mole of glucose are normally because glucose enters the glycolytic pathway by fi rst being converted to generated at this step. If ATP is not generated at this step, the net ATP yield glufor the glycolytic pathway is zero, and the cells die because they are unable to cose-6-phosphate, then fructose-6-phosphate. The next step, conversion to meet their energy requirements. fructose-1,6-bisphosphate, requires the phosphofructokinase enzyme. Thus, 18. As the NADH/NAD ratio increases, the activity of GAPDH deglycerol is a suitable substrate for this mutant, but glucose is not. phate groups may interact with the serine in some manner, perhaps by creases and less 1,3-bisphosphoglycerate is produced from glyceraldehyde12. (a)hydrogen When F26BP bindsBut to the PFK, one of its bonding. if allosteric the serinesiteis on replaced with thephosnegatively 3-phosphate. NAD is a reactant and NADH is a product of the reaction, charged aspartate, the negative charges on the F26BP are repulsed and so as NAD becomes less available and NADH accumulates, the ratio of the F26BP cannot bind to the enzyme. (b) Apparently PFK cannot be [product]/[reactant] increases and the activity of the enzyme decreases. fully active in the absence of the F26BP allosteric activator, which plays an important role in stimulating glycoly- sis. In its absence, the 19. Phosphoglycerate kinase catalyzes the conversion of 1,3-bisphosphoglycto 3-phosphoglycerate with concomitant production of ATP from glycolytic pathway is less active, which accounts for the observed erate ADP. decrease in glucose consumption and ethanol produc- tion. [From The kinase can generate the ATP required by the ion pump, and the ADP Heinisch, J. J., Boles, E., and Timpel, C., J. Biol. Chem. 271, 15928–15933 produced when the pump is phosphorylated can serve as a substrate in the (1996).] kinase reaction.

13. In the presence of iodoacetate, fructose-l,6-bisphosphate accumulates, 20. Hexokinase-defi cient erythrocytes have low levels of all glycolytic inwhich suggests that iodoacetate inactivates the enzyme that uses fructose- termediates, since hexokinase catalyzes the fi rst step of glycolysis. Therefore, l,6-bisphosphate as a substrate. Because iodoacetate reacts with Cys residues, the concentration of 2,3-BPG in the erythrocyte will be decreased as well, the inactivation of the enzyme by the reagent suggests that a Cys residue is in favoring the oxygenated form of hemoglobin and decreasing its p50 value. the active site. Later studies showed that Cys was not part of the active site; Pyruvate kinase–defi cient erythrocytes have high levels of 2,3-BPG since instead, the reaction of the Cys with iodoacetate probably caused a conforma- pyruvate kinase catalyzes the last step of glycolysis. This blockade at the last tional change that rendered the enzyme inactive. step causes the concentrations of all of the intermediates “ahead” of the block 14.

to be increased. Thus, the oxygen affi nity of hemoglobin is decreased with increased 2,3-BPG concentration, and the p50 value increases as a result.

OH H

C

O C CH OPO23

100

2

90

Possible enediolate intermediate

[GAP] [DHAP]

4.4 kJ?mol2157.9?mol21 1(8.314531023 kJ?K21?mol21)(310 K) ln 23.5 kJ?mol21

52.58 kJ?mol21 ln

21.365ln

[GAP] [DHAP]

[GAP] [DHAP]

70 60

40 30

[DHAP]

0

[DHAP]

The ratio of [GAP] to [DHAP] is 0.26:1, which seems to indicate that the formation of DHAP, not the formation of GAP, is favored. However, GAP, the product of the triose phosphate isomerase reaction, is the substrate for the glyceraldehyde-3-phosphate dehydrogenase reaction. The continuous removal of the product GAP by the action of the dehydrogenase shifts the equilibrium toward formation of GAP from DHAP. 16. (a) The cancer cells may express the GAPDH protein at higher levels (i.e., transcription of the GAPDH gene and translation of its mRNA may occur at a higher rate).

Pyruvate kinase deficient

20 10

[GAP]

Normal erythrocytes

50

[GAP]

[G AP] e21.36 5 [DHAP] 0.265

Oxygen saturation (%)

DG5DG8¿1RT ln

15.

Hexokinase deficient

80

0

10

20

30

40

50

60

pO2 (torr)

21. Phosphate levels increase because phosphate is a reactant for the GAPDH enzyme. ATP levels decrease, since ATP is a product of the GAPDH reaction. Levels of 2,3-BPG decrease as well, since levels of 1,3-BPG decrease as a result of GAPDH inhibition.

22. (a) In hepatocytes, the phospho-His on the phosphoglycerate mutase transfers its phosphate to the C2 position of 3PG to form 2,3-BPG. The [32P]-labeled phosphate on the C3 position is transferred back to the enzyme to form the 2PG product, so initially the enzyme would be labeled. In the next round of catalysis, the labeled phosphate on the enzyme is transferred to the C2 position of the next molecule of 3PG substrate, so 2PG becomes labeled. Eventually, this phosphate is transferred to ADP to form ATP, so ATP is labeled.

62

Ch 13 Solutions

(b) In the plant, the labeled phosphate is transferred to C2 to form 2PG, B., Walsh, M. C., van Dam, K., and Westerhoff, H. V., Trends Biochem. Sci. so 2PG is labeled and then eventually ATP. The plant enzyme is not 23, 162–169 (1998).] labeled.

31. (a) One mole of ATP is invested when KDG is converted to KDPG. One mole of ATP is produced when 1,3-BPG is converted to 3PG. One 23. Fluoride inhibits the enzyme enolase. If the enzyme is inactive, its mole of ATP is produced when phosphoenolpyruvate is converted to substrate, 2-phosphoglycerate, will accumulate. The previous reaction is at pyruvate. Therefore, the net yield of this pathway (per mole of glucose) equilibrium, so 3-phosphoglycerate will accumulate as 2-phosphoglycerate is one mole of ATP. accumulates. (b) In order to keep the pathway going, subsequent reactions would 24. (a) The glucose–lactate pathway releases 196 kJ ? mol 1 of free energy, need to reoxidize the NADPH that is produced when glucose is convertenough theoretically to drive the synthesis of 196/30.5, or about 6, ATP. ed to gluconate and the NADH that is produced by GAPDH. [From (b) The complete oxidation of glucose releases 2850 kJ ? mol 1 of free Johnsen, U., Selig, M., Xavier, K. B., Santos, H., and Schönheit, P., energy, enough for 2850/30.5, or about 93, ATP. Arch. Microbiol. 175, 52–61 (2001).]

25. In cells with a pyruvate kinase defi ciency the [ADP]/[ATP] ratio increases and the [NAD ]/[NADH] ratio decreases. Pyruvate kinase catalyzes the second ATP-generating step in glycolysis; in the absence of this reaction, ATP levels decrease and the [ADP]/[ATP] ratio increases. The substrate PEP accumulates in a pyruvate kinase defi ciency, stimulating PFK and increasing the concentration of F16BP. This ultimately leads to an increase in glyceraldehyde-3-phosphate, which reacts with NAD to form NADH. In the absence of pyruvate kinase, pyruvate is not produced, so lactate cannot subsequently re-oxidize NADH to NAD ; therefore, the [NAD ]/[NADH] decreases.

26. Cells swell and lyse in the absence of ATP (see Solution 25) because the Na,K-ATPase pump (see Fig. 9-15) is unable to function. Consequently, Na ions enter the cell, along with water, which eventually causes the cells to swell and lyse, resulting in anemia.

27. The alcohol dehydrogenase enzyme catalyzes the reduction of acetaldehyde to ethanol. Concomitantly, NADH is oxidized to NAD . The NADH reactant is produced by glycolysis in the GAPDH reaction. The NAD produced in the alcohol dehydrogenase reaction can serve as a reactant for the glycolytic GAPDH reaction, allowing glycolysis to continue.

28. (a) If PFK is the only enzyme affected by aluminum treatment, then the production of pyruvate from perfused livers of control and aluminumtreated rats should be the same. In the liver, fructose is fi rst converted to fructose-1-phosphate by fructokinase, then fructose-1-phosphate aldolase converts it to glyceraldehyde and dihydroxyacetone phosphate.

The PFK reaction is bypassed, so inhibition of this enzyme should have no effect. (b) If the same experiment were carried out using glucose as an energy source, pyruvate production would be much lower in the livers from the aluminum-treated rats than in livers from the control rats. This result is due to the inhibition of the PFK enzyme, which is essential in the production of pyruvate from glucose. [From Xu, Z.-X., Fox, L., Melethil, S., Winberg, L., and Badr, M., J., Pharmacol. Exp. Ther. 254, 301–305 (1990).]

29. (a) Methanol reacts with alcohol dehydrogenase (as ethanol does) to produce formaldehyde (Section 13-1).

NAD H3 C

OH

Methanol

O

NADH, H H

alcohol dehydrogenase

C

H

Formaldehyde

32. (a) When glucose is converted to pyruvate, NAD is reduced to NADH in the glyceraldehyde-3-phosphate dehydrogenase reaction. Without a mechanism for regenerating NAD , glycolysis could not continue. (b) If the trypanosome reduced pyruvate to lactate, this step would regenerate NAD and no other pathway for regenerating NAD would be needed. (c) A drug that interfered with one of the enzymes in the trypanosome’s NADH-oxidizing pathway would likely have no effect on the host since mammals lack the enzymes that would be inactivated by the drug.

33. (a) Acetyl-CoA produced from pyruvate is a substrate for the citric acid cycle, an energy-producing pathway. When the cell's need for energy is low, acetyl-CoA accumulates and activates pyruvate carboxylase, which catalyzes the fi rst step of gluconeogenesis. As a result, the cell can synthesize glucose when the need to catabolize fuel is low. (b) Following its deamination, alanine is a substrate for gluconeogenesis. By inhibiting pyruvate kinase, alanine suppresses glycolysis so that fl ux through the shared steps of glycolysis and gluconeogenesis will favor gluconeogenesis.

34. Fructose-1,6-bisphosphatase catalyzes the conversion of fructose1,6-bisphosphate to fructose-6-phosphate in gluconeogenesis. When the enzyme is defi cient, the gluconeogenic pathway is severely impaired and glucose synthesis from amino acids, glycerol, and lactate precursors occurs at a low level. At the beginning of a fast, blood glucose levels are maintained at normal levels because the source of blood glucose is glycogenolysis in the liver, which occurs normally. Once liver glycogen has been depleted, a normal individual would rely on gluconeogenesis for endogenous glucose production, but this pathway occurs at a low level in this patient because of the defi cient enzyme; therefore, blood glucose levels decrease. Pyruvate and alanine, both gluconeogenic precursors, are elevated because they cannot be converted to glucose via gluconeogenesis. The defi ciency of fructose-1,6-bisphosphatase results in the buildup of the substrate, fructose-1,6-bisphosphate. This would promote the formation of GAP and DHAP, since the aldolase reaction is reversible.

Because glycolysis does not occur during a fast, GAP is not consumed by the GAPDH reaction, so the ratio of GAP to DHAP, which is normally low, increases.

35. Insulin, the hormone of the fed state, might be expected to suppress the transcription of the gluconeogenic enzymes pyruvate carboxylase, PEPCK, fructose-1,6-bisphosphatase, and glucose-6-phosphatase. [In fact, insulin has been shown to suppress the transcription of PEPCK and glucose-6-phosphatase.]

(b) Administering ethanol is a good antidote because ethanol will 36. In type 2 diabetes, insulin is unable to suppress the transcription of compete with methanol for binding to alcohol dehydrogenase and will PEPCK and glucose-6-phosphatase. This leads to an increase in gluconeoproduce the less harmful acetaldehyde. This allows time for methanol to genenic activity, resulting in the release of glucose from the liver and an be eliminated from the system. [From Cooper, J. A., and Kini, M., increase in blood glucose levels.

Biochem. Pharmacol. 11, 405–416 (1962).]

37. (a) The phosphatase activity is active under fasting conditions. The 30. If hexokinase cannot be inhibited, all glucose that enters the yeast phosphatase removes the phosphate group from F26BP, forming cell will be phosphorylated at the expense of ATP to produce glucose-6fructose-6-phosphate. Thus F26BP is not present to stimulate glycolysis phosphate. Glucose-6-phosphate is isomerized to fructose-6-phosphate, which is then phosphorylated to fructose-1,6-bisphosphate, again at the ex- (or to inhibit gluconeogenesis); therefore, gluconeogenesis is active. pense of ATP. If concentrations of glucose are high, and if there is no mecha- (b) The hormone of the fasted state is glucagon. nism to inhibit hexokinase, then cellular ATP will be depleted in these early (c) When glucagon binds to its receptors, cellular cAMP levels rise as steps. The early reactions of glycolysis proceed at a rate greater than the rate described in Section 10-2. This activates protein kinase A, which phosof the later ATP-generating steps of glycolysis. ATP is therefore used faster phorylates the bifunctional enzyme, resulting in the activation of the than it is regenerated, and the yeast mutants die as a result. [From Teusink, phosphatase activity and the inhibition of the kinase activity.

Ch 13 Solutions 38. Deamination of alanine produces pyruvate, a gluconeogenic substrate, and deamination of aspartate produces oxaloacetate, an intermediate of gluconeogenesis.

39. (a) Increasing the activity of the enzyme that produces F26BP would increase the concentration of this metabolite, which stimulates PFK and inhibits fructose-1,6-bisphosphatase. This would have the effect of stimulating glycolysis and inhibiting gluconeogenesis. Stimulation of PFK would increase the concentration of fructose-1,6-bisphosphate, which activates pyruvate kinase via feed-forward activation. Additional stimulation of pyruvate kinase by brazilin would lead to an increase in fl ux through the glycolytic pathway. (b) If brazilin is able to act on the liver to increase glycolysis and decrease gluconeogenesis, this could help alleviate the high blood glucose concentrations that occur in diabetes. An active gluconeogenic pathway in the liver would result in effl ux of glucose from the liver, which is not desirable in the diabetic patient. [From You, E.-J., Khill, L.-Y., Kwak, W.-J., Won, H.-S., Chae, S.-H., Lee, B.-H., and Moon, C.-K., J. Ethnopharmacol. 102, 53–57 (2005).]

63

47. This observation revealed that the pathways for glycogen degradation and synthesis must be different, since a defect in the degradative pathway has no effect on the synthetic pathway.

48. McArdle’s disease is also known as type V glycogen storage disease, as it results from a defi ciency of muscle (not liver) glycogen phosphorylase.

49. Normally, muscle glycogen is degraded to glucose-6-phosphate, which enters glycolysis to be oxidized to yield ATP for the active muscle. In anaerobic conditions, pyruvate, the end product of glycolysis, is converted to lactate, which is released from the muscle into the blood and enters the liver to be converted back to glucose via gluconeogenesis. The patient’s muscle cells are unable to degrade glycogen to glucose-6-phosphate; thus, there is no glucose-6-phosphate to enter glycolysis and lactate formation does not occur. [From Stanbury, J. B., Wyngaarden, J. B., and Fredrickson, D. S., The Metabolic Basis of Inherited Disease, pp. 151–153, McGraw-Hill, New York (1978).]

50. Blood glucose concentrations are regulated by pancreatic hormones act-

ing on the liver to stimulate glycogen synthesis or degradation, whatever is 40. Phosphoenolpyruvate carboxykinase catalyzes an essential step of glu- appropriate. Since the patient’s liver enzymes appear to function normally coneogenesis. Lower expression of this enzyme decreases the gluconeogenic (see Problem 48), the blood glucose concentration is properly regulated and output of the liver, which helps decrease the level of circulating glucose in the patient is neither hypo- nor hyperglycemic. patients with diabetes.

51. Glucose-6-phosphatase catalyzes the last reaction in gluconeogenesis (and glycogenolysis) in the liver. Glucose-6-phosphate is converted to glucose, 41. A diagram of this pathway, referred to as the Cori cycle, is shown in Figure 19-3. Lactate is released from the muscle as a result of anaerobic gly- and the glucose transporters export glucose to the circulation, where it is colytic activity. It travels via the bloodstream to the liver, where it is taken available to other body tissues that do not carry out gluconeogenesis and that do not store glycogen. In the absence of this enzyme, glucose-6-phosphate up, cannot be converted to glucose and instead accumulates in the liver and is converted to pyruvate, and then transformed back to glucose via gluconeo- converted to glucose-1-phosphate, which is used for glycogen synthesis. Glygenesis. The cost of running this cycle is 4 ATP, since 2 ATP are generated in cogen synthesis is therefore elevated in the livers of patients with this disease. glycolysis and the cost of running gluconeogenesis is 6 ATP. The accumulation of glycogen enlarges the liver and causes the abdomen to 42. The transformations described comprise the glucose–alanine cycle, protrude. which is shown in Figure 19-4. Prolonged operation of this cycle results in 52. As described in Solution 51, patients with von Gierke’s disease cannot muscle breakdown, since muscle protein is the source of the -amino acids convert glucose-6-phosphate to glucose in the absence of glucose-6-phospharequired in the transamination of pyruvate to alanine. tase; thus, these patients are hypoglycemic.

43. The starch in the grains must be converted to glucose because the yeast 53. In order for the enzyme to be active, the serine in the active site must that carry out fermentation use glucose as their starting material. be phosphorylated. This phosphate group is donated to C1 of the glucose44. Amylase hydrolyzes starch into oligosaccharides (see Section 12-1), 6-phosphate in the fi rst step of the conversion of glucose-6-phosphate to which are subsequently hydrolyzed to monosaccharides. Although amylases glucose-1-phosphate. If glucose-1,6-bisphosphate dissociates prematurely, are naturally present in bread fl our, adding additional enzyme speeds up the the serine is not phosphorylated, the enzyme is not regenerated, and further starch degradation process to provide monosaccharides for glycolysis, which rounds of catalysis cannot occur. ultimately results in the conversion of pyruvate to ethanol, yielding CO2 that 54. An increase in the activity of glycogen phosphorylase in the fat body causes the bread to rise. will result in the increased degradation of glycogen to glucose. Because 45. Production of glucose-1-phosphate requires only an isomerization reacfructose-2,6-bisphosphate concentrations are low, glycolysis will not be tion catalyzed by phosphoglucomutase to convert it to glucose-6-phosphate, stimulated (F26BP is a potent activator of the glycolytic enzyme PFK). which can enter glycolysis. This skips the hexokinase step and saves a mol- Instead, glucose can be used to synthesize trehalose, which leaves the fat ecule of ATP. Hydrolysis, which produces glucose, would require body and enters the hemolymph. In this way, the fat body produces sugars expenditure for use by other tissues in the fasting insect. [From Meyer-Fernandes, J. R., of aning ATP to phosphorylate glucose to glucose-6-phosphate.

Equation 12-2: Clark, C. P., Gondim, K. C., and Wells, M. A., Insect Biochem. Mol. Biol.

46. (a) See Sample Calculation 12-2. The Keq can be calculated by rearrang31, 165–170 (2001).]

K eq 5 e2DG8¿yRT kJ?K21?mol21 55. The ATP yield for the pathway is 2 ATP. T. tenax stores energy in the )(298 K) K eq 5 e2(3.1kJ?mol21)y(8.314531023 form of glycogen, which is phosphorolyzed to glucose-1-phosphate. The e1.25 K eq 52 glucose-1-phosphate is converted to glucose-6-phosphate, an isomerization reaction that does not require an investment of ATP. Glucose-6-phosphate [glycogen, n21][G1P] 0.29 [glycogen, n][Pi] is then converted to fructose-6-phosphate. The next step, in which frucKeq 5 1 5 tose-6-phosphate is converted to fructose-1,6-bisphosphate, also does not require an investment of ATP, since the phosphofructokinase reaction in 1 [G1P] 0.29 T. tenax is reversible and uses pyrophosphate rather than ATP. Since the 5 GAPDH reaction does not produce 1,3-bisphosphoglycerate, there is no 1 [Pi] 1 ATP produced until the last step, when phosphoenolpyruvate is converted 3.5 [Pi] to pyruvate. Since this reaction occurs twice for each glucose molecule, 5 1 the pathway yields two moles ATP per mole of glucose. [From Brunner, [G1P] N. A., Brinkmann, H., Siebers, B., and Hensel, R., J. Biol. Chem. 273, [glycogen, n21][G1P] (b) DG 5 DG8¿ 1 RT ln 6149–6156 (1998).] [glycogen][Pi] 56. (a) Inhibition of glycogen phosphorylase blocks glycogenolysis, and in(1)(1) 1 1 2 mol1 ?m2 DG53.1kJ 5 kJ?K2 )(298 K)ln ol (8.314 1 ?

(1)(50) 2 DG53.1 kJ mol2 1 ) ? mol1 1 27( kJ 9.? hibition of the bisphosphatase blocks gluconeogenesis. As a result, the DG526.6 kJ? mol2 1 liver is unable to produce glucose and the blood glucose level is low.

Ch 14 Solutions

64

(b) The inhibition of fructose-1,6-bisphosphatase prevents gluconeo- 62. If NADPH is used as a coenzyme for nitrate reduction, it will begenesis even when substrates such as glycerol and dihydroxyacetone come oxidized to NADP and will need to be regenerated. The pen- tose phosphate are supplied. Galactose relieves the hypoglycemia because it phosphate pathway produces NADPH, so it is possible that the can be phosphorylated, converted to glucose-6-phosphate, and NADPH-producing enzymes of this pathway, glucose-6-phosphate dehydrogenase and 6-phosphogluconate dehydrogenase, are stimulated by dephos- phorylated to yield glucose that enters the circulation. nitrate. [From Hankinson, O., and Cove, D. J., J. Biol. Chem. 249, 2344– 57. (a) The fi rst committed step of the pentose phosphate pathway is the 2353 (1974).] fi rst reaction, which is catalyzed by glucose-6-phosphate dehydrogenase 63. G16BP inhibits hexokinase but stimulates PFK and pyruvate kiand is irreversible. Once glucose-6-phosphate has passed this point, it has no other fate than conversion to a pentose phosphate. nase. This means that glycolysis will be active, but only if the substrate (b) The hexokinase reaction does not commit glucose to the glycolytic is glucose-6-phosphate, since glucose cannot be phosphorylated in the pathway, since the product of the reaction, glucose-6-phosphate, can also absence of hexokinase activity. The pentose phosphate pathway is inacenter the pentose phosphate pathway. tive, since 6-phosphogluconate dehydrogenase is inhibited. Phosphoglucomutase is activated, which converts glucose-1-phosphate (the prod58. (a) In a liver cell, glucose-6-phosphate has four possible fates: It can be uct of glycogenolysis) to glucose-6-phosphate. Thus, in the presence of used to synthesize glycogen, it can be catabolized via glycolysis, it can be G16BP, glycogenolysis is active and produces substrate for glycolysis but catabolized via the pentose phosphate pathway, and it can be converted not the pentose phosphate pathway. This is a more effi cient process than to glucose and released from the cell. using glucose taken up from the blood, which would need to be phos(b) In a muscle cell, only the fi rst three processes occur. Muscle cells phorylated at the expense of ATP. [From Beitner, R., Trends Biol. Sci. 4, lack glucose-6-phosphatase and therefore cannot release glucose from 228–230 (1979).] the cell.

64. The concentration of xylulose-5-phosphate, an intermediate of the 59. The pentose phosphate pathway in the red blood cell generates NADPH, which is used to regenerate oxidized glutathione. Glucose-6-phosphate dehy- pentose phosphate pathway, increases following a meal, when there is drogenase is the enzyme that catalyzes the fi rst step of the oxidative branch of plenty of glucose being catabolized by the pentose phosphate pathway. the pathway. Its defi ciency results in a decreased output of NADPH from the The increase in fructose-2,6-bisphosphate increases the fl ux through glypathway. As a result, glutathione remains in the oxidized form and cannot colysis while inhibiting fl ux through gluconeogenesis, so glycolysis profulfi ll its roles of decreasing the concentrations of organic peroxides, main- duces large amounts of pyruvate, which can be converted to acetyl-CoA. taining red blood cell shape, and keeping the iron ion of hemoglobin in the At the same time, the production of lipid-synthesizing enzymes increases, so the net result is that glucose is converted to acetyl-CoA and then to fat 2 form. Hemolytic anemia is the likely result. for storage.

60. (a) Since serum withdrawal decreases G6PDH activity, serum withdrawal should decrease the NADPH/NADP ratio. NADPH is a product of the G6PDH reaction and NADP is a reactant, so decreased enzyme activity leads to an increase in the oxidized form of the coenzyme and a decrease in the reduced form.

Chapter 14 (b) In the presence of DHEA, the NADPH/NADP ratio should decrease, since DHEA inhibits G6PDH activity, for the reasons outlined in part (a). 1. In mammalian cells, pyruvate can be converted to lactate by lactate (c) In the absence of any other factor, adding H2O2 should not affect the dehydrogenase. Pyruvate can also be transformed into oxaloacetate; this reratio, since G6PDH can increase its activity to produce more NADPH action is catalyzed by pyruvate carboxylase. Pyruvate can be converted to acetyl-CoA by the pyruvate dehydrogenase complex. Pyruvate can be conto react with the H2O2. (d) In the absence of the serum containing growth factors, G6PDH cannotverted to alanine by transamination. handle the increased load of H2O2, and the ratio would decrease. [From 2. The decarboxylation step is metabolically irreversible since the CO2 Tian, W.-N., Braunstein, L. D., Pang, J., Stuhlmeier, K. M., Xi, Q.-C., product diffuses away from the enzyme. The other four reactions are transTian, X., and Stanton, R., J. Biol. Chem., 273, 10609–10617 (1998).] fer reactions or oxidation–reduction reactions (transfer of electrons) that are more easily reversed.

61.

3. The purpose of steps 4 and 5 is to regenerate the enzyme. In step 3,

CH 2OPO32

CHOPO2 2 3

O H HO

H H OH OH H

H HO

O

H

O

H

H

OH

H

O O

H

H

OH

OH CHOPO2 2 3

O H

OH

COO H HO H

H OH H

H

H HO

CHOPO2 2 3 H O H H OH O OH O H H

the product acetyl-CoA is released, but the lipoamide prosthetic group of E2 is reduced. In step 4, the E3 reoxidizes the lipoamide group by accepting the protons and electrons from the reduced lipoamide. In step 5, the enzyme is reoxidized by NAD . The product NADH then diffuses away.

4. TPP is a cofactor in two of the enzymes associated with the citric acid cycle—the pyruvate dehydrogenase complex and the -ketoglutarate dehydrogenase complex. If thiamine were defi cient, TPP would be defi cient as well and the activities of both these enzymes would decrease. As a result, the substrates of these two reactions, pyruvate and -ketoglutarate, would accumulate.

5. Arsenite reacts with the reduced lipoamide group on E2 of the pyruvate dehydrogenase complex to form a compound with the structure shown in the fi gure. The enzyme cannot be regenerated and can no longer catalyze the conversion of acetyl-CoA to pyruvate. The -ketoglutarate dehydrogenase complex has a lipoamide group on its E2 subunits and will be inhibited as well. The entire citric acid cycle cannot function, glucose cannot be oxidized aerobically, and respiration comes to a halt, which explains why these compounds are so toxic.

Ch 14 Solutions 65 6.

O C O

O

CH3

N C

C

S

CH 3 Pyruvate

R TPP

CH2

Br

O H

13. In the citrate synthase mechanism, a proton is removed from the acetyl group of acetyl-CoA. The histidine then forms a hydrogen bond to stabilize the resulting enediolate intermediate. The alanine side chain cannot form this hydro- gen bond; thus, the reaction cannot continue. [From Pereira, D. S., Donald, L. J., Hosfi eld, D. J., and Duckworth, H. W., J. Biol. Chem. 269, 412–417 (1994).]

14. (a) O

R

C

H

H CH 3

C

CH3

O

S

3

S

CH 3

R

CO2

3

HO

2

R

R N C

CN

C

H

H3C

OR

O

HO

H

SH O C

CH 2

S

CoA

S-Acetonyl-CoA

H R CN

H C

CoA H3C

Acetaldehyde

O

CH3 HBr

1

4

C

CH3

HO

R

H C

C

C S

3

Resonance-stabilized carbanion

N

CH3

C S

R

CH3

R

(b) S-acetonyl CoA is a competitive inhibitor. The Vmax of the citrate synthase reaction is the same in the absence and in the presence of the inhibitor. The KM has increased in the presence of the inhibitor, indicating that the enzyme’s affi nity for its substrate has decreased in the presence of the inhibitor. The inhibitor competes with acetyl-CoA for binding to the citrate synthase active site. S-acetonyl-CoA can do this because its structure resembles that of acetyl-CoA. (c) Acetyl-CoA binds to pyruvate carboxylase and acts as an activator. A thioester functional group must be required for binding, since SacetonylCoA, which lacks a thioester functional group, cannot bind to this binding site. [From Rubenstein, P., and Dryer, R., J. Biol. Chem. 255, 15. 7858–7862 (1980).]

OH CoA

S

C CH 2

7. In both cases, the activity of the pyruvate dehydrogenase complex de- 16. During the resting state, citric acid cycle activity is low as a result of the creases, as both NADH and acetyl-CoA are products of the reaction. Rising low activities of the two enzymes described. Any available citrate will inhibit concentrations of NADH and acetyl-CoA decrease pyruvate dehydrogenase citrate synthase, while the low concentrations of Ca2 ions ensure that the activity by competing with NAD and CoASH for binding sites on the activity of isocitrate dehydrogenase is also low. Upon beginning exercise, the Ca2 concentration increases in muscle cells, which increases the activity of enzyme. isocitrate dehydrogenase. This depletes the cellular concentration of citrate, 8. Ca2 inhibits the kinase and activates the phosphatase. In this way, the depriving citrate synthase of its inhibitor, so the activity of citrate synthase pyruvate dehydrogenase complex is active and can funnel substrates from also increases. Thus, when the cell transitions from resting to exercise mode, glycolysis into the citric acid cycle to provide ATP for the contracting the activities of the enzymes involved in the citric acid cycle increase in order to meet the increased demand for ATP in the working muscle. muscle.

9. The E1 subunit of the pyruvate dehydrogenase complex requires TPP, the 17. (a) Aconitase is the enzyme that catalyzes the reversible isomerization of phosphorylated form of thiamine, as a cofactor. Administering large doses of citrate to isocitrate. Because this reaction is followed by and preceded by thiamine might be a successful treatment option if the E1 mutation happens irreversible reactions, the inhibition of aconitase leads to an accumulation of citrate. The concentrations of other citric acid cycle intermediates will to occur in the thiamine-binding site.

10. If there is residual pyruvate dehydrogenase complex activity, the activity be decreased. can be enhanced by inhibiting the kinase, which inhibits pyruvate dehydro- (b) If the citric acid cycle and mitochondrial respiration are not functioning, the cell turns to glycolysis to produce the ATP required for its genase itself by phosphorylation. [From Brown, G. K., Otero, L. J., LeGris, energy needs. Consequently, fl ux through glycolysis increases. The inM., and Brown, R. M., J. Med. Genet. 31, 875–879 (1994).] crease in the rate of the pentose phosphate pathway is required to meet 11. The phosphofructokinase reaction is the major rate-control point the increased demand for reducing equivalents during hyperoxia. [From for the pathway of glycolysis. Inhibiting phosphofructokinase slows the Allen, C. B., Guo, X. L., and White, C. W., Am. J. Physiol. 274 (3 Pt. 1), entire pathway, so the production of acetyl-CoA by glycolysis followed L320–L329 (1998).] by the pyruvate dehydrogenase complex can be decreased when the citric acid cycle is operating at maximum capacity and the citrate concentration 18. Citrate accumulates when aconitase is inhibited (see Solution 17). Aconis high. itase is also inhibited when fl uoroacetate is added (see Solution 12) because 12. Citrate synthase acts on oxaloacetate (OAA) and fl uoroacetyl-CoA, citrate synthase reacts with oxaloacetate and fl uoroacetyl-CoA to produce an acetyl-CoA analog, to produce fl uorocitrate. The fl uorocitrate then fl uorocitrate, which inhibits aconitase. This transformation can be skipped if serves as an inhibitor of aconitase. This leads to an accumulation of citrate, since fl uorocitrate is added directly. Therefore all three mechanisms either directly or indirectly inhibit aconitase and thus would be predicted to have similar the citric acid cycle can go no further if the aconitase reaction is inhibited.

O effects on cells in culture.

19. Cis-aconitate is an intermediate in the reaction when citrate is converted FH2C COO FH2C C S CoA to isocitrate by aconitase. Trans-aconitate structurally resembles cis-aconitate Fluoroacetate Fluoroacetyl-CoA and would be expected to compete with cis-aconitate for binding to the enzyme. But because trans-aconitate is a noncompetitive inhibitor when citrate F CH COO is used as the substrate, the citrate binding site must be distinct from the citrate synthase aconitate binding site. Citrate and trans-aconitate do not compete for bindHO C COO ing and can bind to the enzyme simultaneously, but when both substrate and inhibitor are bound, the substrate cannot be converted to product. [From Oxaloacetate CH2 COO Villafranca, J. J., J. Biol. Chem. 249, 6149–6155 (1974).]

Fluorocitrate

66

Ch 14 Solutions

20. The yeast mutant would be unable to carry out the citric acid cycle and so would be unable to carry out aerobic oxidation of fuel molecules in order to generate ATP. The yeast mutant would obtain all of its energy from the anaerobic oxidation of glucose, even in the presence of oxygen.

Step 2. The succinyl group is then transferred to the lipoamide prosthetic group of E2 of the -ketoglutarate dehydrogenase complex.

COO

COO

21. See Sample Calculation 12-2. Keq can be calculated by rearranging Equation 12-2:

TPP

CH2

K eq 5 e 2 D G 8 ¿ y R T

CH2

CH2

K eq 5e2(221 kJ?mol21)y(8.314531023 kJ?K21?mol21)(298 K) eq 5e8.5 K eq54.8310 3 K

HO

C

23. (a) Usually, phosphorylation of an enzyme causes a conformational change in the protein that subsequently alters its activity. For the bacterial isocitrate dehydrogenase, however, phosphorylation of an active site Ser residue introduces negative charges that repel the negatively charged isocitrate and prevent it from binding. (b) Construction of the mutant supports this hypothesis. The introduc- tion of the negatively charged Asp residue in place of the Ser residue similarly introduces a negative charge to the active site and prevents iso- citrate binding in the same manner. [From Dean, A. M., Lee, M. H. I., and Koshland, D. E., J. Biol.

Chem. 264, 20482–20486 (1989).]

24. (a) When yeast use glucose as a carbon source, ATP is obtained from the oxidation of glucose by glycolysis. Glucose is oxidized anaerobically, without the citric acid cycle. But when the food source is shifted from glucose to acetate, yeast are required to obtain all of their energy from the citric acid cycle and oxidative phosphorylation. Acetate is converted to acetyl-CoA, which enters the citric acid cycle. In the absence of glycolysis, fl ux through the citric acid cycle increases. Because isocitrate dehydrogenase is one of the main regulatory enzymes of the cycle, increasing its expression increases the rate of its reaction and increases the fl ux through the citric acid cycle. In this manner, yeast can effectively use acetate as a food source. (b) For the wild-type yeast, the [NAD ]/[NADH] ratio increases slightly as the cells switch from using glycolysis to using the citric acid cycle to generate ATP, as described in part (a). There is an increase in fl ux through the citric acid cycle, but perhaps the isocitrate dehydrogenase cannot keep up with the demand, so the ratio increases. There is a more dramatic change in the ratio for the mutant because the isocitrate dehydrogenase enzyme is nonfunctional. This means that the reaction, in which NAD is a reactant and NADH is a product, cannot occur, so the NAD reactant accumulates. These cells will eventually die if not given a carbon source other than acetate. [From Minard, K. I., and McAlisterHenn, L., Arch. Biochem. Biophys. 483, 136–143 (2009).]

25. Step 1. In the fi rst step, -ketoglutarate is decarboxylated, a process that requires TPP. The carbon of the carbonyl group becomes a carbanion, which forms a bond with TPP. COO CH2

COO CO2

CH2

CH2 C

O

COO -Ketoglutarate

CH2 HO

C TPP

C

S

TPP

O

S S

22. The substrate is isocitrate, a compound with three carboxylate groups that are negatively charged at physiological pH. Four of the fi ve conserved amino acid residues have positively charged side chains that could form ion pairs with isocitrate. The enzyme–substrate complex should not be too stable, however; it is the role of the enzyme to stabilize the transition state. So it is possible that the positively charged side chains in the binding pocket stabilize the transition state. This stabilization must be important in the conversion of substrate to product, which explains why these residues are highly conserved throughout evolution. [From Hurley, J. H., Thorsness, P. E., Ramalingam, V., Helmers, N. H., Koshland, D. E., and Stroud, R. M., Proc. Natl. Acad.

Sci. 86, 8635–8639 (1989).]

CH2

R

Lipoamide

HS R

Step 3. The succinyl group is transferred to coenzyme A, and the lipoamide group is reduced.

HS

COO reduced lipoamide

CH2

COO CH2

HS R

CH2 C

CH2 C

O

S

O

CoA

Coenzyme A

Succinyl-CoA

HS R

Steps 4 and 5. The last two steps are the same as for the pyruvate dehydro- genase complex. E3 reoxidizes the lipoamide when its disulfi de group ac- cepts two protons and two electrons. The NAD reoxidizes the enzyme, and the NADH and H products diffuse away.

26. TPP attacks the carbonyl carbon of the succinyl phosphonate, but the subsequent step (the departure of the leaving group) cannot occur because the C¬P bond cannot be broken. A covalent bond forms between the enzyme and the inhibitor (although the reaction is reversible, the reverse reaction occurs slowly) and the substrate cannot bind.

H3C

H3C N C

N S O C P C

S O O P C

O O O

CH2

CH2

COO

O OH

CH2 CH2 COO [From Bunik, V. I., Denton, T. T., Xu, H., Thompson, C. M., Cooper, A. J. L., and Gibson, G., Biochemistry 44, 10552–10561 (2005).]

27. Succinyl-CoA resembles acetyl-CoA suffi ciently that it is able to compete with acetyl-CoA for binding to the active site of citrate synthase. Likewise, succinyl-CoA competes with CoASH binding to the active site of -ketoglutarate dehydrogenase. Both cases are examples of feedback inhibition: Succinyl-CoA inhibits the enzyme that produces it and inhibits an earlier enzyme in the pathway.

28. Because of the enzyme defi ciency, the citric acid cycle cannot be completed; in the absence of a functional citric acid cycle, glucose must be oxidized anaerobically. Lactate is a product of anaerobic oxidation of glucose; thus, lactate levels are elevated in the patients. The pyruvate S lactate

Ch 14 Solutions 67 transformation is reversible, so if lactate levels rise due to increased glyco- than the mutant enzyme. [From Estévez, M., Skarda, J., Spencer, J., lytic activity, pyruvate levels rise as well. However, lactate levels rise more Banaszak, L., and Weaver, T. M., Protein Sci., 11, 1522–1557 (2002).] (because there are other options for pyruvate), so the [lactate]/[pyruvate] 35. (a) malate1NAD1Soxaloacetate1NADH1H1 ratio is increased in the patient. [From Bonnefont, J.-P., et al., J. Pediatrics, DG8¿529.7 kJ?mol21 121, 255–258 (1992)]. oxaloacetate1acetyl-CoAScitrate1coenzyme A 29. When operating in reverse, succinyl-CoA synthetase catalyzes a kinaseDG8¿5231.5 kJ?mol21 type reaction, the transfer of a phosphoryl group from a nucleotide triphosmalate 1 NAD1 1 acetyl-CoA S citrate 1 NADH 1 H1 1 CoA phate (GTP or ATP).

DG8¿521.8 kJ ? mol21 30. (a) Succinyl-CoA synthetase catalyzes the only substrate-level phos(b) The equilibrium constant for the coupled reaction is 3.4 105 phorylation reaction in the citric acid cycle. The enzyme with the ADPspecifi c subunit could produce ATP in the brain and muscle to meet times greater than the equilibrium constant for the uncoupled the energy needs of these tissues. The enzyme with the GDP-specifi c reaction. subunit could produce GTP needed by phosphoenolpyruvate carReactions 1 and 8 boxykinase for gluconeogenesis in the liver and kidneys.

G¿yRT K e q 5e2 D 8 (b) Individuals who lack a functioning subunit cannot carry out the )(298 K) succinyl-CoA synthetase reaction in any tissue, since this subunit is K 52(21.8 e kJ?mol21)y(8.314531023 kJ?K21?mol21 eq common to both forms of the enzyme. Because aerobic respiration is 0.73 K eq5e impossible, the individual uses glycolysis followed by lactate fermentation to obtain ATP. This doesn’t provide enough energy to sustain life, so the person dies shortly after birth. (c) When the gene for one of the subunits is mutated, the gene for the other subunit is normal. The individual would suffer from decreased succinyl-CoA synthetase activity, but operation of the citric acid cycle would allow the individual to obtain some energy from aerobic respiration. However, the individual’s ability to meet his or her energy needs is still somewhat compromised, as shown by the elevated levels of lactate and decreased life span.

Keq52.1

Uncoupled Reaction 8 K eq 5 e 2 D G 8 ¿ y R T )(298 K) K eq 52(29.7 e kJ?mol21)y(8.314531023 kJ?K21?mol21 eq

K eq 5212.0 e K 5.231026 6 36. When cells obtain energy in the absence of oxygen, glucose is oxidized

to pyruvate, which is subsequently reduced to lactate with concomitant re31. Succinate accumulates because it cannot be converted to fumarate.

Succinyl-CoA also accumulates because the succinyl-CoA synthetase reac- generation of NAD . The citric acid cycle is not active, and the cells respond tion is reversible. However, the succinyl-CoA ties up some of the cell’s CoA by down-regulating the activity of the citric acid cycle enzyme malate dehysupply, so the -ketoglutarate dehydrogenase reaction, which requires CoA, drogenase. In the presence of oxygen, pyruvate is converted to acetyl-CoA, which enters the citric acid cycle so that ATP can be produced by oxidative slows. As a result, -ketoglutarate accumulates. phosphorylation. Under aerobic conditions, malate dehydrogenase is abso32. Isocitrate lyase, a glyoxylate pathway enzyme, catalyzes the conversion of lutely essential for the regeneration of oxaloacetate as part of the citric acid isocitrate to succinate and glyoxylate. The glyoxylate product of this reac- cycle; therefore, activity levels are much higher. tion goes on to form malate, which is eventually converted to glucose via gluconeogenesis. The succinate product is not part of this pathway and is 37. (a) The isotopic label on C4 of oxaloacetate is released as 14CO2 in the -ketoglutarate dehydrogenase reaction. “disposed of” by mitochondrial succinate dehydrogenase, which regenerates (b) The isotopic label on C1 of acetyl-CoA is scrambled at the succinylthe oxaloacetate needed to keep the glyoxylate pathway operating.

CoA synthetase step. Because succinate is symmetrical, C1 and C4 are [malate] chemically equivalent, so in a population of molecules, both C1 and C4 33. DG5DG8¿1RT ln would appear to be labeled (half the label would appear to be at C1 and [fumarate] half at C4). Consequently, one round of the citric acid cycle would yield 0523.4 kJ?mol211(8.314531023 kJ?K21?mol21) oxaloacetate with half the labeled carbon at C1 and half at C4. Both of [malate] these labeled carbons would be lost as 14CO2 in a second round of the (310 K) ln [fumarate] citric acid cycle.

38.

The volume increase that occurs when the bread dough rises is due to [malate] 3.4 kJ?mol2152.58 kJ ? mol2 1 ln the CO2 produced by the anaerobic oxidation of glucose. Sugar (sucrose) [fumarate] in the bread dough is hydrolyzed to glucose and fructose by enzymes in the yeast. Then the fructose and glucose both enter glycolysis. The end product is [malate] 1.325ln pyruvate, which is converted to acetaldehyde. In this step, a CO2 is released. [fumarate] The acetaldehyde is then reduced to ethanol, which evaporates when the [malate] bread is baked. e1.325 39. (a) Substrate availability: Acetyl-CoA and oxaloacetate levels regulate [fumarate] citrate synthase activity. [malate] 3.75 (b) Product inhibition: Citrate inhibits citrate synthase, NADH inhib[fumarate] its isocitrate dehydrogenase and -ketoglutarate dehydrogenase, and succinyl-CoA inhibits -ketoglutarate dehydrogenase.

The ratio of malate to fumarate is 3.7 to 1, indicating that the reaction pro(c) Feedback inhibition: NADH and succinyl-CoA inhibit citrate ceeds in the direction of formation of malate. This is not a control point for synthase. the citric acid cycle because the G is close to zero, indicating it is a near40.

The alternate pathway bypasses the succinyl-CoA synthetase reaction equilibrium reaction. of the standard citric acid cycle, a step that is accompanied by the phos34. The Vmax for the native enzyme is more than 10-fold greater than the phorylation of a nucleoside diphosphate. The alternative pathway therefore Vmax for the mutant enzyme, indicating that the wild-type fumarase converts generates one less nucleoside triphosphate than the standard citric acid cycle. substrate to product at a higher rate than the mutant. The KM values are There is no difference in the number of reduced cofactors generated. similar, indicating that the mutation doesn’t affect substrate binding much.

41. Pyruvate carboxylase converts pyruvate to oxaloacetate, one of the reacThe turnover numbers are refl ected in the differences in the Vmax values. The tants for the fi rst reaction of the citric acid cycle. If the fi rst reaction of the catalytic effi ciency is 13-fold greater in the native enzyme, indicating that cycle cannot take place, the remaining reactions cannot proceed. the native enzyme successfully encounters its substrate much more frequently

68

Ch 14 Solutions

42. Both reactants for the fi rst reaction of the citric acid cycle can be produced from pyruvate. Pyruvate dehydrogenase converts pyruvate to acetylCoA, while pyruvate carboxylase converts pyruvate to oxaloacetate. Stimulation of pyruvate carboxylase by acetyl-CoA occurs when excess acetyl-CoA is present and more oxaloacetate is needed. This regulatory strategy ensures that there are equimolar amounts of both of these reactants to begin the citric acid cycle.

43. (a) The citric acid cycle is a multistep catalyst. Degrading an amino acid to a citric acid cycle intermediate boosts the catalytic activity of the cycle but does not alter the stoichiometry of the overall reaction (acetylCoA S 2 CO2). (b) Pyruvate derived from the degradation of an amino acid can be converted to acetyl-CoA by the pyruvate dehydrogenase complex; these amino acid carbons can then be completely oxidized by the citric acid cycle.

44. In the reversible reaction shown, the amino acid aspartate and the glycolytic product pyruvate can undergo a transamination in which aspartate’s amino group is transferred, leaving oxaloacetate, which is a citric acid cycle intermediate.

45. Any metabolite that can be converted to oxaloacetate can enter gluconeogenesis and serve as a precursor for glucose. Biological molecules that are degraded to acetyl-CoA cannot be used as glucose precursors because acetylCoA enters the citric acid cycle and its two carbons are oxidized to carbon dioxide. Thus, glyceraldehyde-3-phosphate, tryptophan, phenylalanine, and pentadecanoate can serve as gluconeogenic substrates because at least one of their breakdown products can be converted to oxaloacetate. Palmitate and leucine are not glucogenic because their breakdown products are acetyl-CoA or one of its derivatives. Acetyl-CoA cannot be converted to pyruvate in mammals.

46. An increase in the concentration of glucose increases fl ux through gly-

of acetyl-CoA and oxaloacetate are required for the fi rst step of the citric acid cycle, some of the pyruvate was converted to oxaloacetate via the pyruvate carboxylase reaction. This explains why oxaloacetate concentrations increased. The concentration of pyruvate did not increase because a steady state was reached: The rate of production of pyruvate from phosphoenolpyruvate was equal to the rate of consumption of pyruvate.

52. During operation of the citric acid cycle, NAD and ubiquinone collect the electrons released as acetyl carbons are oxidized to CO2. Oxygen is required as the fi nal acceptor in the electron transport chain that reoxidizes NADH and QH2.

53. (a) Isocitrate is a branch point between the citric acid cycle and the glyoxylate pathway in organisms such as E. coli, which have the glyoxylate pathway. When acetate is the only food source, isocitrate dehydrogenase is inactive and acetate enters the glyoxylate pathway (as acetyl-CoA). This pathway produces glucose (by gluconeogenesis) and other intermediates that can be used as precursors in other biosynthetic reactions.

Citrate

Isocitrate isocitrate lyase

Succinate

Glyoxylate Acetate

Acetyl-CoA

Malate

malate synthase

Oxaloacetate

malate dehydrogenase

(b) When glucose is the substrate for the cultured E. coli, the glyoxylate pathway, which leads to glucose, is no longer necessary. Isocitrate dehydrogenase then becomes active so that glucose enters glycolysis and then the citric acid cycle.

colysis, which produces more pyruvate. The pyruvate dehydrogenase complex activity increases in response to the increase in pyruvate so that the conversion 54. Ethanol is converted to acetaldehyde and then to acetate, and acetate of pyruvate to acetyl-CoA can increase proportionally. At the same time, more is converted to acetyl-CoA. Acetyl-CoA then enters the glyoxylate pathway. oxaloacetate is required, since equimolar amounts of acetyl-CoA and oxaloac- The fi rst step is the synthesis of citrate from acetyl-CoA and oxaloacetate. etate are required for the fi rst step of the citric acid cycle. Therefore, the activity Citrate is isomerized to isocitrate. Isocitrate lyase then splits isocitrate into of pyruvate carboxylase, the enzyme that catalyzes the conversion of pyruvate succinate and glyoxylate. Succinate leaves the glyoxysome and enters the mitoto oxaloacetate, also increases. chondrion, where it participates in the citric acid cycle. The glyoxylate fuses 47. Alanine can be converted to pyruvate via the transamination reaction with a second molecule of acetyl-CoA to yield malate. Malate leaves the glyshown in Problem 44. In gluconeogenesis, pyruvate is converted to oxalo- oxysome and enters the cytosol, where it is converted to oxaloacetate via the acetate via the pyruvate carboxylase reaction, then oxaloacetate is converted malate dehydrogenase reaction. Oxaloacetate can then enter gluconeogenesis to phosphoenolpyruvate, and so on to produce glucose. But if pyruvate car- to form glucose. boxylase is defi cient, alanine is converted to pyruvate, but the gluconeogenic 55. If the fatty acid is broken down, the resulting acetyl-CoA units can enter pathway can go no further. the citric acid cycle for further oxidation. Because the labeled carbons are not 48. Glutamine can be converted to glutamate, which can be converted to lost as CO2 in the fi rst round of the cycle (see Fig.14-10), some of the label -ketoglutarate. This citric acid cycle intermediate is eventually converted to will appear in oxaloacetate. If some of this oxaloacetate is siphoned from the oxaloacetate, so the glutamine is a source of the oxaloacetate needed for glu- citric acid cycle for gluconeogenesis, some of the label will appear in glucose. coneogenesis when pyruvate carboxylase is not functioning. The oxaloacetate This sequence of reactions does not violate the statement that fatty acids derived from glutamine can also replenish citric acid cycle intermediates that cannot be converted to carbohydrates because two carbons have already been are normally produced by the pyruvate carboxylase reaction. lost as CO2. Thus, there is no net conversion of fatty acid carbons to glucose 49. Pyruvate carboxylase requires biotin as a cofactor (see Table 12-2). If the carbons.

nature of the defect involves a mutated pyruvate carboxylase with a decreased 56. The mutant cannot convert isocitrate to -ketoglutarate, so the citric acid cycle is blocked at this point. Adding glutamate to the culture medium affi nity for biotin, administering large doses of the vitamin might help treat the disease. Biotin treatment is ineffective for more severe forms of the disease in bypasses the block, because glutamate can be converted to -ketoglutarate which the mutation does not occur in the biotin-binding region or in which and the citric acid cycle can continue.

57. The concentration of AMP in muscle cells rises during periods of high the enzyme is expressed at extremely low levels or not at all.

50. If pyruvate dehydrogenase is defi cient, pyruvate cannot be converted to muscle activity, indicating the need for ATP production via glycolysis and acetyl-CoA. If pyruvate carboxylase is defi cient, pyruvate cannot be convert- the citric acid cycle. AMP is converted to IMP by adenosine deaminase, as ed to oxaloacetate. So either enzyme block means that pyruvate will accumu- shown in the fi gure. Breakdown of muscle protein produces aspartate, which late, which accounts for the elevated pyruvate levels. Pyruvate is converted joins with IMP to produce adenylosuccinate. This substrate is lysed to form to lactate in anaerobic metabolism; so if pyruvate levels are elevated, lactate AMP and fumarate. Fumarate is a citric acid cycle intermediate, and increasing its concentration leads to greater citric acid cycle activity. Thus, the purine will be elevated as well. nucleotide cycle acts as an anaplerotic mechanism for the citric acid cycle (at 51. The exercising muscle required greater concentrations of ATP to power the expense of muscle protein). it, so the rates of glycolysis and the citric acid cycle increased. Phosphoenol58. (a) Hydroxycitrate differs by the addition of a hydroxyl group. pyruvate was converted to pyruvate more rapidly, so the concentration of (b) Based on its structural similarity to the substrate citrate, hydroxyphosphoenolpyruvate decreased. Some of the pyruvate was converted to acecitrate is most likely to act as a competitive inhibitor. tyl-CoA via the pyruvate dehydrogenase reaction. Since equimolar amounts

Ch 14 Solutions

linked to the citric acid cycle, so the citrate synthase and malate dehydro- genase enzymes are upregulated as well. The glyoxylate pathway allows the pathogen to use two-carbon sources to synthesize malate and, from there, glucose and larger biological macromolecules.

The phagosome is not likely to be a rich source of glucose and amino acids that the patho- gen can use for biosynthesis, so the glyoxylate pathway is essential to its sur vival. (b) The host does not have the enzymes found in the glyoxylate pathway, so inhibitors of isocitrate lyase and malate synthase would be good targets, assuming that there is a way to deliver these inhibitors to the phagosome. [From Lorenz, M. C., and Fink, G. R., Euk. Cell 1, 657– 662 (2002).]

(c) Acetyl-CoA synthesized in the mitochondria from pyruvate, the product of carbohydrate catabolism, is made available in the cytosol by the action of ATP-citrate lyase (see Fig. 14-17). Inhibiting this enzyme might decrease the amount of acetyl-CoA available for fatty acid synthesis. (d) Since cholesterol synthesis also begins with acetyl-CoA, hydroxy- citrate might reduce the production of cholesterol and the structurally related steroid hormones. (Although hydroxycitrate inhibits ATP-citrate lyase in vitro, controlled clinical trials have shown that the compound has no signifi cant effect on weight loss in humans.)

59. (a) NADH, citrate, and succinyl-CoA inhibit citrate synthase in mammals but do not inhibit citrate synthase in H. pylori. Isocitrate dehydrogenase is NADP -dependent rather than NAD -dependent and is regulated differently (by higher concentrations of its substrates NADP and isocitrate instead of NADH). H. pylori lacks -ketoglutarate dehydrogenase and instead has -ketoglutarate oxidase. The enzyme succinyl-CoA synthetase is missing in H. pylori. This enzyme catalyzes the only substrate-level phosphorylation reaction in the citric acid cycle; therefore, no GTP is produced in the citric acid cycle of this organism. Succinate dehydrogenase is missing. Fumarate reductase is present.

Mammals do not have the glyoxylate pathway, but most bacteria do.

H. pylori has only one step, catalyzed by malate synthase (isocitrate lyase is not present). (b) High KM values indicate a low affi nity of the enzyme for the substrate. The high values indicate that substrate concentrations must be relatively high for the enzyme to attain half-maximal velocity. This indicates that the pathway does not operate unless the concentrations of the citric acid cycle intermediates are relatively high, which occurs when the H. pylori is in a nutrient-rich environment with plentiful resources (which would occur when the human “host” ingested a meal). Since the primary purpose of the citric acid cycle in H. pylori is to provide biosynthetic intermediates, it makes sense that the pathway operates only when metabolic resources are plentiful. (c) H. pylori citrate synthase is inhibited by ATP but is not affected by NADH or any of the other citric acid cycle intermediates. Since the citric acid cycle in this organism is not used to produce metabolic energy in the form of reducing equivalents for oxidative phosphorylation, it makes sense that NADH would not serve as an inhibitor. (d) Citrate synthase, isocitrate dehydrogenase, and -ketoglutarate oxidase may serve as regulatory control points since these enzymes catalyze irreversible reactions and are subject to activation and inhibition by allosteric modulators. (e) Enzymes unique to H. pylori would be good therapeutic targets: ketoglutarate oxidase, fumarate reductase, and malate synthase. [From Pitson, S. M., Mendz, G. L., Srinivasan, S., and Hazell, S. L., Eur. J. Biochem. 260, 258–267 (1999).]

69

63. (a) This reaction is an anaplerotic reaction in bacteria and plants, analogous to the pyruvate carboxylase reaction in animals. PPC produces oxaloacetate for the citric acid cycle to ensure its continued operation as a pathway for oxidizing fuel molecules and for producing intermediates for biosynthetic reactions. (b) Acetyl-CoA and oxaloacetate are required in equimolar amounts as substrates for the citrate synthase reaction that begins the citric acid cycle. If the concentration of acetyl-CoA rises, the concentration of oxaloacetate will need to increase as well, so acetyl-CoA stimulates the enzyme that produces its cosubstrate. The activation by fructose-1,6-bisphosphate appears to be a feed-forward mechanism to ensure that suffi cient oxaloacetate is present to condense with the acetyl-CoA produced by glycolysis and the pyruvate dehydrogenase reaction.

64. (a)

60. (a) The acetyl-CoA enters the glyoxylate pathway to produce malate, which can then be converted to oxaloacetate and used to synthesize glucose. Acetyl-CoA can also condense with oxaloacetate to form citrate, which can be converted to isocitrate and then to -ketoglutarate. Glutamate dehydrogenase converts -ketoglutarate to glutamate by reductive amination. Alternatively, -ketoglutarate can be converted to glutamate by transamination. (b) Aspartate undergoes transamination to form oxaloacetate, which can condense with acetyl-CoA to form citrate. Citrate is then converted to isocitrate, which is converted to -ketoglutarate. -Ketoglutarate can then be converted to glutamate.

PEP CO2

PEP carboxylase OAA

NADH NAD

malate dehydrogenase malate

H2O

fumarase

fumarate QH2 succinate dehydrogenase Q succinate

(b) If the bacteria were allowed to grow aerobically, the carbon atoms of fuel molecules would be completely oxidized to carbon dioxide. The citric acid cycle would function in a forward, catabolic direction.

Depriving the bacteria of oxygen ensures that the fi nal three steps of the citric acid cycle will function in a reverse, anabolic direction to produce the desired product, succinate. [From Agarwal, L., Isar, J., Meghwanshi, G. K., and Saxena, R. K., Enzyme Microb. Tech. 40, 629– 636 (2007).]

65. (a) Glutamine is converted to glutamate by deamidation, and glutamate is 61. Enzymes of the glyoxylate pathway, particularly malate dehydrogenase converted to -ketoglutarate by transamination. and isocitrate lyase (which are unique to this pathway), would be inacti(b) -Ketoglutarate dehydrogenase, succinyl-CoA synthetase, succinate vated. The glyoxylate pathway produces glucose from noncarbohydrate dehydrogenase, fumarase, and malic enzyme. sources but is not required when glucose is available. Enzymes required for gluconeogenesis that are not involved in glycolysis would also be inactivated, mainly phosphoenolpyruvate carboxykinase and fructose- 66. Glutamate is transaminated to -ketoglutarate; then isocitrate dehydrogenase, working in reverse, adds CO2 to -ketoglutarate to generate 1,6-bisphosphatase. isocitrate, which is in equilibrium with citrate. ATP-citrate lyase can then 62. (a) Isocitrate lyase and malate synthase are found in the glyoxysome, so break down the citrate into oxaloacetate and the acetyl-CoA needed for the pathogen is using the glyoxylate pathway. The glyoxylate pathway is fatty acid synthesis.

70

Ch 15 Solutions

Chapter 15

Use Equation 15-4 to calculate DG 89 for this reaction:

1. Reverse the NADH half-reaction and the sign of its E value to indicate oxidation, then combine the half-reactions and theirE values.

E89 5 20.185 V E89 5 0.315 V

pyruvate 1 2 H1 1 2 e2 S lactate NADH S NAD1 1 H1 1 2 e2 NADH 1 pyruvate 1 H1 S NAD1 1 lactate

D89 E 5 0.130 V

Use Equation 15-4 to calculate DG89 for this reaction: DG 8¿52nFD E8¿ DG852(2)(96 ¿ ,485 J?V

21

?mol 2 1 )(0.130 V)

?mol 2 1

DG85225.1 kJ ¿

DG 8¿52nFD E8¿ D G852(2)(96,485 ¿ JV ?

D G85251.3 ¿ kJ

?mol21)(0.266 V)

21

21 ? mol

The oxidation of acetaldehyde by NAD is spontaneous, as shown by the negative G value.

6. Consult Table 15-1 for the relevant half-reactions involving acetoacetate and NADH. Reverse the NADH half-reaction and the sign of its E value to indicate oxidation, then combine the half-reactions and theirE values. acetoacetate 1 2 H1 1 2 e2

S 3-hydroxybutyrate

NADH S NAD1 1 H1 1 2 e2

E8 9 5 20.346 V E8 9 50.315 V

acetoacetate 1 NADH 1 H1 S NAD1 1 3-hydroxybutyrate DE8 9 5 20.031 V

The reduction of pyruvate by NADH (Section 13-1) is spontaneous under Use Equation 15-4 to calculate DG89 for this reaction: standard conditions.

DG8¿52nF DE8¿ 2. Reverse the cytochrome a3 half-reaction and the sign of its E value DG8¿52(2)(96,485 J?V21?mol21)(20.031 V) to indicate oxidation, multiply the coeffi cients by 2 so that the number of electrons transferred will be equal, then combine the half-reactions and their DG8¿56.0 kJ?mol21 E values.

Consult Table 15-1 for the relevant half-reactions involving acetoacetate and 2 cyt a (Fe21) S 2 cyt a (F31 FADH2. Reverse the FADH2 half-reaction and the sign of its E value to 2 E89 5 20.385 V 3e) 1 2 e 3 indicate oxidation, then combine the half-reactions and theirE values.

E89 5 0.815 V 2 ½ O 21 2 H 2 1e1 S H2O acetoacetate 1 2 H 11 2 e2 S 3-hydroxybutyrate E8 9 5 20.346 V DE89 5 0.430 V 2 cyt a (Fe21) 1 ½ O 1 2 H1 S 3 2 FADH2 S FAD 1 2 H1 1 2 e2 E8 9 50.0 V 2 cyt a (Fe31 ) 1 H2O 3 acetoacetate 1 FADH 2 S FAD 1 3-hydroxybutyrate

Use Equation 15-4 to calculate DG89 for this reaction:

Use Equation 15-4 to calculate DG89 for this reaction:

DG8¿52nFDE8¿ DG8¿52(2)(96,485 J?V21?mol21)(0.430 V)

DG8¿52nF DE8¿ DG8¿52(2)(96,485 J?V21?mol21)(20.346 V) DG8¿566.8 kJ?mol21

DG8¿5283 kJ?mol21 The reaction is spontaneous under standard conditions.

3. The relevant reactions and their E values are obtained from Table 15-1: Dihydrolipoic acid S lipoic acid 1 2 H1 1 2 e2

E89 5 0.29 V NAD1 1 H1 1V2 e2 S NADH E89 5 20.315 Dihydrolipoic acid 1 NAD1 S lipoic acid 1 NADH DE89 5 20.025 V Use Equation 15-4 to calculate DG89 for this reaction: DG 8¿52nF DE8¿ DG852(2)(96,4 ¿ 85 J?V DG854.8 kJmol21 ¿ ?

21

?mol 2 1 )(20.025 V)

4. The relevant reactions and their E values are obtained from Table 15-1 or given in the problem: Cu(Cu21)1e2 A

SCu(Cu1 A

E8¿520.235 V )

cyt c (Fe21)1Cu(Cu21)Scyt c (Fe31)1Cu(Cu1 A

The reduction of acetoacetate by NADH is more favorable than reduction by FADH2, as shown by the G values calculated above. Although neither reaction is spontaneous under standard conditions, reduction by FADH2 is far more unfavorable. The reduction of acetoacetate by NADH is, in fact, a favorable process under cellular conditions, which differ from standard conditions.

7. Consult Table 15-1 for the relevant half-reactions involving ubiquinol and cytochrome c1. Reverse the ubiquinol half-reaction and the sign of its E value to indicate oxidation, multiply the coeffi cients in the cytochrome c1 equation by 2 so that the number of electrons transferred will be equal, then combine the half-reactions and their E values.

QH S Q 1 2 H1 1 2 e2 E89 5 20.045 V 2

2 cyt c 1(Fe31) 1 2 e2 S 2 cy

cyt c (Fe21)Scyt c (Fe31)1e2

E8¿50.245 V ) DE8¿50.010 V

A

DE8 9 5 20.346 V

Use Equation 15-4 to calculate G 89 for this reaction:

QH 21 2 cy

t c1 (Fe)31S

t c1 (Fe)21 21

Q 1 2 cyt c1 (Fe)

E89 5 0.220 V DE89 5 0.175 V

Use Equation 15-4 to calculate DG 89 for this reaction: DG 8¿52nF DE8¿ DG852(2)(96 ¿ ,485 J?V 2 1 ?mol 2 1 )(0.175 V) DG85233.8 kJ ¿ ? mol21

The reaction is spontaneous under standard conditions.

8. Consult Table 15-1 for the relevant half-reactions involving succinate 21 and NAD . Reverse the succinate half-reaction and the sign of its E value ?mol 2 1 )(0.010 V) to indicate oxidation, then combine the half-reactions and their E values.

D G¿ 8 5 2 0.96 kJ?mol 2 1 E89 5 20.031 V succinate S fumarate 1 2 H1 1 2 e2 5. Consult Table 15-1 for the relevant half-reactions involving acetalde- hyde NAD1 1 H1 1 2 e2 S NADH E89 5 20.315 V and NAD . Reverse the acetaldehyde half-reaction and the sign of its value to DE89 5 20.346 V succinate 1 NAD1 1 H1 S NADH 1 fumarate indicate oxidation, then combine the half-reactions and their E E values. 1 HO S acetate acetaldehyde Use Equation 15-4 to calculate DG89 for this reaction: DG 8¿52nFD E8¿ D G¿ 8 5 2 (1)(96,485 J?V

2

0.581 V E8 9 5 E8 9 5 20.315 V

1 3 H 1 1 2 e2

NAD1 1 H1 1 2 e2 S NADH 1 2 1 NAD S NADH 1 acetate acetaldehyde 1 HO

1 2 H1

DE8 9 50.266 V

D G ¿8 5 2 n F D E 8 ¿ DG852(2)(96,485 J?V21?mol21)(20.346 V) ¿ DG8566.8 kJ?mol21 ¿

Ch 15 Solutions

71

Consult Table 15-1 for the relevant half-reactions involving succinate and gained by cytochrome c1 will be passed along to Complex IV, in effect couFAD. Reverse the succinate half-reaction and the sign of its E value to indi- pling the two reactions, which would also tend to make the process more favorable than the G indicates. cate oxidation, then combine the half-reactions and their E values.

11. Consult Table 15-1 for the relevant half-reactions involving O2 and E89 5 20.031 V succinate S fumarate 1 2 H1 1 2 e2 NADH. Reverse the NADH half-reaction and the sign of its E value to E89 5 0.0 V FAD 1 2 H1 1 2 e2 S FADH2 indicate oxidation, then combine the half-reactions and their E values. succinate 1 FAD S fumarate 1 FADH2 DE89 5 20.031 V E89 5 0.815 V ½ O2 1 2 H1 1 2 e2 S H2O E89 5 0.315 V NADH S NAD1 1 H1 1 2 e2 Use Equation 15-4 to calculate DG 89 for this reaction: DE89 5 1.13 V

NADH 1 ½ O 21 H1 S NAD1 1 H2O

DG8¿52nFDE8¿ DG8¿52(2)(96,485 J?V21?mol21)(20.031 V)

Use Equation 15-4 to calculate DG 89 for this reaction:

DG8¿56.0 kJ?mol21

DG 8¿52nFDE8¿ 1 1 DG 8¿52(2)(96,485 J?V2?mol2)(1.13 V) The oxidation of succinate by FAD is more favorable than oxidation by 1 NAD , as shown by the G values calculated above. Although neither D G 8 ¿ 5 2218 kJ ?mol2 reaction is spontaneous under standard conditions, oxidation by NAD is The synthesis of 2.5 ATP requires a free energy investment of 2.5 30.5 kJ far more unfavorable. The oxidation of succinate by FAD is, in fact, a favormol 1, or 76.3 kJ mol 1. The effi ciency of oxidative phosphorylation is able process under cellular conditions (citric acid cycle, Section 14-2), which therefore 76.3/218 0.35, or 35%. differ from standard conditions.

12. (a) Consult Table 15-1 for the relevant half-reactions involving NADH 9. (a) Use Equation 15-2 to determine the E values for these two halfand coenzyme Q. Reverse the NADH half-reaction and the sign of its reactions.

E value to indicate oxidation, then combine the half-reactions and QH 2SQ1 2 H 1 1 2 e 2E89 5 20.045 V their E values.

E89 5 0.045 V [QH]2 Q 1 2 H1 1 2 e2 S QH2 0.026 V n ln E5E8¿2 E89 5 0.315 V [Q] NADH S NAD1 1 H1 1 2 e2 DE89 5 0.360 V NADH 1 Q 1 H1 S NAD1 1 QH2 0.026 V E520.045 V2 ln 10 Use Equation 15-4 to calculate DG 89 for this reaction: E520.075 V 2 cyt c (Fe31) 1 2 e2 S 2 2cyt c DG8¿52nFDE8¿ (Fe21) DG8¿52(2)(96,485 J?V21?mol21)(0.360 V) E89 5 0.235 V [cyt c DG8¿5269.5 kJ?mol21 E 5 E8 ¿ 2

The phosphorylation of ADP to ATP requires 30.5 kJ mol 1 of free energy. Assuming that this reaction is 35% effi cient (see Problem 11), 0.8 mol ATP can be synthesized under standard conditions.

(Fe 21 )] 0.026 V ln n [cyt c(Fe31)]

E50.235 V2

0.026 V 2

ln

1 5

E50.235 V10.021 V50.256 V E 5 20.075 V

QH2 S Q 1 2 H1 1 2 e2 2 cyt c (Fe31) 1 2 e2 S 2 cyt c (Fe21) QH 21 2 cyt c (Fe31) S Q 1 2 H1 1 2 cyt c (Fe1)

2

E 5 0.256 V DE 5 0.181 V

(b) Use Equation 15-4 to calculate DG for this reaction: DG52nFDE DG52(2)(96,485 J?V21?mol21)(0.181 V) DG5234.9 kJ?mol21 10. Reverse the half-reaction for the iron–sulfur protein to indicate that it is being oxidized. Add the two half-reactions to obtain the E for the reaction.

E89 5 20.280 V FeS (red) S FeS (ox) 1 e2 21 12S cyt c 1 (Fe) e E89 5 0.215 V 1 cyt c (Fe31) 21 FeS (red ) 1 cyt c (Fe31) S FeS (ox)1 DE89 5 20.065 V cyt c1 (Fe) 1 Use Equation 15-4 to calculate DG 89 for this reaction: DG8¿52nFDE8¿ DG8¿52(1)(96,485 J?V21?mol21)(20.065 V) DG8¿56.27 kJ?mol21

69.5 kJ?mol21 mo1(0.35)50.80 mol l2 30.5 kJ ? (b) Consult Table 15-1 for the relevant half-reactions involving ubiquinol and cytochrome c. Reverse the ubiquinol half-reaction and the sign of its E value to indicate oxidation, multiply the coeffi cients in the cytochrome c equation by 2 so that the number of electrons transferred will be equal, then combine the half-reactions and their E values.

E89 5 20.045 V E89 5 0.235 V

QH2 S Q 1 2 H1 1 2 e2 2 cyt c (Fe31) 1 2 e2 S 2 cyt c (Fe21) QH 21 2 cyt c (Fe31) S Q 1 2

H1 1

2 cyt c (Fe)21DE89 5 0.190 V

Use Equation 15-4 to calculate DG 89 for this reaction: DG8¿52nFDE8¿ DG8¿52(2)(96,485 J?V21?mol21)(0.190 V) DG8¿5236.7 kJ?mol21 The phosphorylation of ADP to ATP requires 30.5 kJ/mol of free energy. Assuming that this reaction is 35% effi cient, 0.42 mol ATP can be synthesized.

36.7 kJ?mol21 mo1(0.35)50.42 mol l2

30.5 kJ ?

(c) Consult Table 15-1 for the relevant half-reactions involving cytochrome c and oxygen. Reverse the cytochrome c half-reaction and the sign of its E value to indicate oxidation, multiply the coeffi cients by 2 so that the number of electrons transferred will be equal, then combine the half-reactions and their E values.

2cyt c (Fe21)12 e2S2 cyt c (Fe31) V ¿ E8520.235 850.815 V ½ O 21 2 H1 1 2 e2 SH2O ¿ E

The positive G indicates that the electron transfer is unfavorable under standard conditions. However, cellular conditions are not necessarily standard conditions, and the G for this reaction is likely to be negative. Also, since this reaction occurs as part of the electron transport chain, the electrons 2 cyt c (Fe21) 1 ½ O1 2 2H1 2

cyt c (Fe31 )1H2O D850.580 E¿V

72

Ch 15 Solutions Use Equation 15-4 to calculate G

for this reaction:

DG8¿52nFDE8¿ DG8¿52(2)(96,485 J?V21?mol21)(0.580 V) DG8¿52112 kJ?mol21 The phosphorylation of ADP to ATP requires 30.5 kJ/mol of free energy. Assuming that this reaction is 35% effi cient, 1.3 mol ATP can be synthesized.

112 kJ?mol21 30.5 kJ?mol2

1 (0.35)51.3 mol

13. The relevant reactions and their E values are obtained from Table 15-1: E8¿520.031 V succinateSfumarate12 H112 e2

cyanide binds to the Fe2 in hemoglobin, where it does not interfere with respiration (although it does interfere with oxygen delivery).

HbO 2 (Fe2 ) HbO 2 (Fe3 ) cytochrome a(Fe2 3

Cu)

cytochrome a(Fe3 3

Cu)

CN

HbO 2 (Fe2 )

22. (a) Pyruvate and malate are oxidized by pyruvate dehydrogenase and malate dehydrogenase, respectively, and both reactions produce E8¿50.045 V Q12 H112 e2SQH2 NADH, which can enter electron transport.

DE8¿50.014 V succinate1QSfumarate1QH2 (b) Fluoxetine inhibits electron transport generally, since the rate of electron transport falls from an average of 163 to 77 when 0.15 mM Use Equation 15-4 to calculate G for this reaction: fl uoxetine is present. The inhibition is primarily at Complex I, since the rate of electron transport in the presence of fl uoxetine is not decreased DG 8¿52nFDE8¿ substantially when succinate (which donates its electrons to Complex II) 2 V) D G ¿8 5 2 (2)(96,485 JV1? 2 ? mol1)(0.014 and rotenone (which inhibits electron transfer in Complex I) are added.

2 DG¿ 8522.7 kJ?mol1 However, fl uoxetine also has the ability to inhibit Complex IV somewhat, since electron transport in the presence of fl uoxetine decreases in 14. This is not enough free energy to drive ATP synthesis under standard the presence of ascorbate (which donates its electrons to cytochrome c) conditions ( G 30.5 kJ ? mol 1). and TMPD (which donates its electrons to Complex IV).

15. (a) Since all of these inhibitors interfere with electron transfer some(c) By decreasing both the rate of electron transport and ATP synthesis, where in the electron transport chain, oxygen consumption will decrease fl uoxetine decreases the rate of ATP production in the brain. The brain when any of the inhibitors are added to a suspension of respiring mito- relies on a constant source of ATP for proper function, so decreased ATP chondria. Adding any of these inhibitors prevents electrons from being production could lead to an impairment of brain function. [From Curti, transferred to the oxygen, the fi nal electron acceptor.

C., Mingatto, F. E., Polizello, A. C. M., Galastri, L. O., Uyemura, S. A., (b) In rotenone- or amytal-blocked mitochondria, NADH and Comand Santos, A. C., Mol. Cell. Biochem. 199, 103–109 (1999).] plex I redox centers are reduced while components from ubiquinone on are oxidized. In antimycin A–blocked mitochondria, NADH, Complex I 23. All these enzymes catalyze reactions in which electrons are transferred redox centers, ubiquinol, and Complex III redox centers are reduced whilefrom reduced substances, such as NADH, to ubiquinone, a compound with cytochrome c and Complex IV redox centers are oxidized. In cyanide- a higher reduction potential. The fl avin group, whose reduction potential is blocked mitochondria, all of the electron transport components are reduced lower than that of ubiquinone (Table 15-1), is ideally suited to shuttle elecand only oxygen remains oxidized. trons between the reduced NADH and the ubiquinone.

16. Adding succinate to rotenone-blocked mitochondria effectively bypasses 24. Side chains with ionizable groups could accept and give up a proton as the block as succinate donates its electrons to ubiquinone and electron trans- part of a proton wire (see Table 4-1): Arg, Asp, Cys, Glu, His, Lys, and Tyr. port resumes. Adding succinate is not an effective bypass for antimycin A– 25. Like the lipids that compose the membrane, coenzyme Q is amphiphilic, with a hydrophilic head and a hydrophobic tail. “Like dissolves like,” or cyanide-blocked mitochondria because succinate donates its electrons and coenzyme Q literally dissolves in the membrane, which facilitates rapid upstream diffusion. of the block.

26. Ubiquinone is a lipid-soluble redox group, so it binds to Complex I 17. The donation of a pair of electrons to Complex IV will result in the from within the membrane rather than from the matrix (in mitochondria) or synthesis of about 1.3 ATP per atom of oxygen (½ O2). Therefore, the P:O ratio of this the cytoplasm (in bacteria). compound is 1.3 (see Solution 12c).

27. Cytochrome c is a water-soluble, peripheral membrane protein and is easily dissociated from the membrane by adding salt solutions that inter18. The donation of a pair of electrons to cytochrome c and then to fere with the ionic interactions that tether it to the inner mitochondrial Complex IV will result in the synthesis of about 1.3 ATP per atom of membrane. Cytochrome c1 is an integral membrane protein and is largely oxygen (½ O2). Therefore, the P:O ratio of this compound is 1.3 (see water insoluble due to the nonpolar amino acids that interact with the Solution 12c). acyl chains of the membrane lipids. Detergents are required to dissociate 19. Adding tetramethyl-p-phenylenediamine to rotenone-blocked and cytochrome c1 from the membrane because amphiphilic detergents can antimycin A–blocked mitochondria effectively bypasses the block as the disrupt the membrane and coat membrane proteins, acting as substitute compound donates its electrons to Complex IV and electron transport lipids in the solubilization process. resumes. Adding tetramethyl-p-phenylenediamine is not an effective by- 28. See Solution 27. Cytochrome c is the most easily released of the mitopass for cyanide-blocked mitochondria because cyanide inhibits electron chondrial electron transport proteins because it is a peripheral membrane transport in Complex IV. Similarly, ascorbate, which donates its elecprotein that dissolves easily in the intermembrane space. A damaged outer trons to cytochrome c and then to Complex IV, can act as an effective mitochondrial membrane allows cytochrome c into the cytosol, where it bypass for antimycin A–blocked mitochondria but not cyanide-blocked triggers apoptosis. mitochondria.

29. The dead algae are a source of food for aerobic microorganisms lower 20. Myxothiazol inhibits electron transfer in Complex III (specifi cally, it in- in the water column. As the growth of these organisms increases, the rates hibits the transfer of electrons from reduced ubiquinol to both cytochrome b of respiration and O2 consumption increase to the point where the conelectron to reoxidize the III). hemoglobin to Fe 2 . centration of O2 in the water becomes too low to sustain larger aerobic cytochrome a3 can donate and the iron–sulfur proteinanredox centers in Complex This oxidizes the iron in cytochrome a 3 to Fe 3 . Cyanide does not bind to organisms.

21. to and the Fe2 in theIV Fe–Cu centerfunction of cytochrome a3 The (see Fe3Cyanide , so it is binds released Complex can again normally.

Problem 15). When the iron in hemoglobin is oxidized from Fe2 to Fe3 ,

Ch 15 Solutions

73

30. The molasses and oil are food for microorganisms. As the food is con(c) Uncouplers such as DNP ferry protons across the inner mitochonsumed, the rates of respiration and oxygen consumption increase. Eventually, drial membrane and dissipate the proton gradient established by electhe depletion of oxygen creates a more reducing environment that favors the tron transport. In the presence of DNP, electron transport still occurs, reduction of Cr(VI) compounds to Cr(III) compounds. but the free energy released by the process is dissipated as heat instead of being harnessed to synthesize ATP. 31. In the absence of myoglobin function, the mice developed several compensatory mechanisms to ensure adequate oxygen delivery to tissues. The 40. (a) The impermeability of the membrane to ions is an important feasymptoms described all involve increasing the amount of available hemoture of chemiosmosis because the free movement of ions other than H globin. In this manner, hemoglobin takes over some of the functions usually would dissipate the electrical component of the proton gradient and performed by myoglobin. thereby compromise the ability of the gradient to supply free energy for 32. Myoglobin content and cytochrome c oxidase activity are roughly correATP synthesis. lated. This provides indirect evidence for myoglobin’s role in delivering (b) If the membrane were permeable to other ions, the proton gradient oxygen to cytochrome c oxidase to be used as the fi nal electron acceptor in would be a simple chemical gradient rather than an electrochemical graelectron transport. [From Wittenberg, J. B., and Wittenberg B. A., J. Exp. Biol. dient. However, it would still be able to serve as a source of free energy 206, 2011–2020 (2003).] for ATP synthesis.

33. Myoglobin functions in muscle cells to facilitate oxygen diffusion throughout the cell and possibly assumes this same role in tumor cells. The 41. (a) K+ resulting increase in oxygen concentration may allow the tumor cells to oxidize a higher percentage of the available glucose aerobically and thus obtain more ATP per molecule of glucose than if glucose were oxidized anaerobicalNiger icin ly. [From Kristiansen, G., et al., J. Biol. Chem., 286, 43417–43428 (2011).]

34. Fructose-2,6-bisphosphate is a potent stimulator of PFK (see Section 13-2).

This large results in greater fl ux through the glycolytic pathway and generation of amounts of pyruvate. The citric acid cycle cannot accommodate the

H+

increased pyruvate.

Consequently, pyruvate is converted to lactate instead of being con-

K+

Valinomycin MATRIX

verted to acetyl-CoA and entering the citric acid cycle.

INTERMEMBRANE SPACE 35. As a result of the exercise program, the number of mitochondria in the CYTOSOL muscle cells of the participants increased, as indicated by an increase in DNA content. The increase in Complex II activity is of similar magnitude, which (b) Potassium ions enter the matrix with the assistance of valinomymight be the result of the increased number of mitochondria. However, the cin. These ions are then exported by nigericin in exchange for total electron transport chain activity is twofold greater after the exercise protons. Importing protons into the mitochondrial matrix dissipates intervention, indicating that mitochondrial function increased as well. These the proton gradient. Since the proton gradient serves as the energy results suggest that even though oxidative damage to mitochondria increases reservoir that drives ATP synthesis, no ATP can be synthesized. with age, exercise can help maintain or increase mitochondrial function.

42. A phosphate ion enters the mitochondrial matrix in symport with a pro[From Menshikova, E. V., Ritov, V. B., Fairfull, L., Ferrell, R. E., Kelley, D. E., and ton. This proton moves in a direction opposite to the protons translocated by Goodpaster, B. H., J. Gerontol. A Biol. Sci. Med. Sci. 61, 534–540 (2006)]. 36. Complexes I, II, and IV. Thus, the difference in pH between the mitochondrial matrix and the intermembrane space becomes smaller.

Because Complex I did not decrease in concentration, it is possible that it becomes damaged in the course of the disease. The decrease in Complex I 43. A total of 32 ATP are obtained from the exergonic oxidation of glucose activity affects the electron transport chain as a whole, which will also show under aerobic conditions: decreased activity and decreased aerobic ATP production in the muscle. This may contribute to the muscle damage that is one of the symptoms of the glycolysis 2 ATP 2 ATP disease. [From Jung, C., Higgins, C. M. J., and Xu, Z., J. Neurochem. 83, 2 NADH 2 3 2.5 5 5 ATP 535–545 (2002).] 37. The free energy change for generating the electrical 2 NADH 2 3 2.5 5 5 ATP 2 pyruvate S 2 acetyl-CoA imbalance is calcuDG5ZFDc lated using Equation 15-6: DG5(1)(96,485 J?V21?mol21)(0.081 V) 2 3 3 NADH 6 3 2.5 5 15 ATP citric acid cycle (2 rounds) DG57.8 kJ?mol21 38. Use the rearrangement of Equation 15-7 as shown in Sample Calculation 15-3.

DG52.303RT(pHin2pHout)1ZF Dc DG52.303(8.314531023 J?K21?mol21)(310 K) (7.627.2)1(1)(96,485 J?V21?mol21)(0.200 V) DG52.4 kJ?mol21119.3 kJ?mol21 DG521.7 kJ?mol21 39. (a) The pH of the intermembrane space is lower than the pH of the mitochondrial matrix because protons are pumped out of the matrix, across the inner membrane, and into the intermembrane space. The increase in concentration of protons in the intermembrane space decreases the pH; the defi cit of protons in the matrix results in an increase in pH. (b) Detergents disrupt membranes. An intact inner mitochondrial membrane is required for oxidative phosphorylation to take place.

Without an intact membrane, an electrochemical gradient, which is the energy reservoir that drives ATP synthesis, cannot be established, and ATP synthesis does not occur.

2 3 1 QH2 2 3 1 GTP Total

2 3 1.5 5 3 ATP 2 3 1 5 ATP 32 ATP

Atotal of 2 ATP per glucose are obtained when glucose is oxidized in the absence of oxygen by conversion to lactate or ethanol (Section 13-1).

44. Since two NADH are generated in the cytosol per mole of glucose, and because the electrons are donated to FAD instead of NAD , the number of ATP generated is reduced by one ATP per NADH “transferred.” Therefore, 30 ATP are generated per glucose instead of 32 (see Solution 43).

45. Aerobic respiration yields 32 ATP (see Problem 43), each of which requires 30.5 kJ ? mol 1 to synthesize: 32 ATP330.5 kJ?mol21 ATP 2850 kJ? mol2 1 Lactate fermentation yields 2 ATP: 2 ATP330.5 kJ?mol21 ATP 196 kJ ? mol2 1

3 100 5 34%

3 100 5 31%

74

Ch 15 Solutions

Alcoholic fermentation yields 2 ATP: 2 ATP330.5 kJ?mol21 ATP

3100526% 235 kJ ? mol12 46. Organisms that can oxidize glucose in the presence of oxygen have an advantage over anaerobic organisms because these organisms can extract more energy per glucose. This may have been important in evolution.

47. (a) Aerobic oxidation of glucose yields 32 ATP per glucose, whereas alcoholic fermentation of glucose by the yeast yields only 2 ATP per glucose. Assuming that the energy needs of the yeast cell remain constant under both aerobic and anaerobic conditions, the catabolism of glucose by the yeast will be 16-fold greater in the absence of oxygen than in the presence of oxygen in order to produce the same amount of ATP. Thus, the rate of consumption of glucose decreases when the cells are exposed to oxygen because fewer glucose molecules must be oxidized to yield the same amount of ATP. (b) Both ratios will initially increase, as the citric acid cycle (which does not operate under anaerobic conditions) produces more NADH equivalents for electron transport. The [ATP]/[ADP] ratio will also increase, since aerobic oxidation of glucose produces more ATP per mole of glucose than anaerobic oxidation [as described in part (a)]. ATP and NADH will “reset” the equilibrium by inhibiting the regulatory enzymes of glycolysis and the citric acid cycle, slowing down these processes. Eventually, the [NADH]/[NAD ] and [ATP]/[ADP] ratios return to their “original” values.

48. Stimulation of the glycolytic enzymes hexokinase and phosphofructokinase increases glycolytic fl ux and allows the yeast cells to obtain ATP more quickly under anaerobic conditions. [From Beitner, R., Cohen, T. J., Nordenberg, J., and Haberman, S., Biochim. Biophys. Acta 586, 266–277 (1979).]

49. (a) The import of ADP (net charge –3) and the export of ATP (net charge –4) represent a loss of negative charge inside the mitochondria.

This decreases the difference in electrical charge across the membrane, since the outside is positive due to the translocation of protons during electron transport. Consequently, the gradient is diminished by the activity of the adenine nucleotide translocase. (b) The activity of the Pi H symport protein diminishes the proton gradient by allowing protons from the intermembrane space to reenter the matrix. (c) Both transport systems are driven by the free energy of the electrochemical proton gradient.

53. DNP uncouples the electron transport chain from oxidative phosphorylation by dissipating the proton gradient. Electron transport still occurs, but the energy released by electron transport is dissipated as heat instead of being harnessed to synthesize ATP. One might think that DNP would be an effective diet aid because the sources of the electrons that go down the electron transport chain are dietary carbohydrates and fatty acids. If the energy of these compounds is dissipated as heat instead of used to synthesize ATP (which would then be used for, among other processes, the synthesis of fatty acids in adipocytes), weight gain from the ingestion of food could theoretically be prevented.

54. FCCP is a lipid-soluble compound that can pass through both the outer and inner mitochondrial membranes and into the matrix. FCCP has an acidic hydrogen on the central nitrogen that dissociates in the basic environment of the matrix. The FCCP passes back through the inner membrane to the intermembrane space, where it is reprotonated and the entire cycle begins again. In this manner, the FCCP shuttles protons into the mitochondrial matrix and dissipates the proton gradient.

55. Organic compounds are oxidized by oxygen through the activity of the electron transport complexes of mitochondria. This activity generates the proton gradient that is dissipated during phosphorylation of ADP. When there is no more ADP, ATP synthase is inactive and unable to dissipate the proton gradient. As a result of the tight coupling between oxidative phosphorylation and electron transport, electron transport and oxygen consumption also come to a halt.

56. (a) The P:O ratio is decreased. For some reason, electron transport and oxidative phosphorylation have been uncoupled. NADH is oxidized in the electron transport chain, oxygen is reduced to water, but ATP may not be synthesized if ADP is not available. This decreases the P:O ratio because oxygen consumption occurs to a greater extent than ADP phosphorylation. (b) If the energy released in electron transport is not used to synthesize ATP, it is released as heat, which accounts for the elevated body temperature. If suffi cient ATP is not synthesized to meet energy needs, the rate of electron transport increases, which leads to an increased consumption of O2 and an increase in the concentration of reduced coenzymes that enter electron transport, thus increasing metabolic rate. (c) The patient will not be able to carry out strenuous exercise under aerobic conditions because her muscle cells are incapable of generating enough ATP to power the muscle.

57. F0 acts as a proton channel as the c ring rotates, feeding protons through the a subunit (see Fig. 15-23). The addition of F1 blocks proton movement because the shaft rotates along with the c ring. In this system, the subunit 50. If the translocase is unable to function, ATP will not be able to exit the and the c ring can rotate only when the binding change mechanism is in mitochondrial matrix and ADP will not be transported inside. Without ADP, operation, that is, when the subunits are binding and releasing nucleotides. the substrate for ATP synthase, ATP synthesis will not occur. Since electron ATP or ADP Pi must be added to the system in order for the subunit transport and oxidative phosphorylation are coupled, a decrease in the rate of to move. oxidative phosphorylation will decrease the rate of electron transport.

58. For yeast, the ratio is 10 H per 3 ATP, or 3.33. For the chloroplast, the 51. Inactivation of one c subunit by DCCD blocks all proton translocaratio is 14 H per 3 ATP, or 4.67. tion by F0 since the movement of protons across the membrane requires 59. Since it has 10 c subunits, the bacterial enzyme can theoretically produce continuous rotation of the c ring. Without this rotation, the subunit of F1 3 ATP for every 10 protons translocated. In the chloroplast, 3 ATP are syncannot move, and therefore the subunits cannot undergo the conformathesized for every 14 protons. Thus, the bacterium is more effi cient in its use tional changes necessary to synthesize or hydrolyze ATP by the binding of the proton gradient established during electron transport and has a higher change mechanism. ratio of ATP produced per oxygen consumed.

52. (a) ATP synthesis decreases dramatically in the presence of oligomycin, 60. One complete revolution of the c ring generates 3 ATP. since proton transfer is required to stimulate rotation of the subunit of ATP synthase, which causes the sequential conformational change of the 3 ATP 1 min 6000 revolutions subunits that catalyze the phosphorylation of ADP to ATP. 5300 ATP?s21 3 3 (b) Since oxidative phosphorylation and electron transport are coupled, revolution 60 s min a decrease in the rate of oxidative phosphorylation will also affect the 61. -Ketoglutarate is an intermediate of the citric acid cycle, so increasrate of electron transport. If ATP synthesis is not occurring, the proton ing its concentration can increase fl ux through the citric acid cycle, congradient is not “discharged” and the rate of electron transport decreases. tributing to ATP synthesis by substrate-level phosphorylation. However, the (c) A decrease in the rate of electron transport will also decrease the rate -ketoglutarate dehydrogenase reaction generates NADH, which cannot be of oxygen consumption. effi ciently reoxidized when electron transport and oxidative phosphorylation (d) If dinitrophenol is added, the proton gradient is dissipated, or are not operating normally. The added aspartate helps to eliminate the excess “discharged,” but not in a way that leads to ATP synthesis. Therefore, NADH: Aspartate is transaminated to oxaloacetate (Section 14-3), which is ATP synthesis still does not occur, but electron transport and oxygen then converted to malate by malate dehydrogenase in a reaction that conconsumption resume, with the free energy of the process released as sumes the NADH. [From Sgarbi, G., Casalena, G. A., Baracca, A., Lenaz, G., heat.

DiMauro, S., and Solaini, G., Arch. Neurol. 66, 951–957 (2009).]

Ch 16 Solutions 75 62. An increase in the amount of glucose led to an increase in glycolytic fl ux M., Klein, M., Riesmeier, J. W., Müller-Röber, B., Fleury, C., Bouillaud, F., and an increase in the concentration of NADH, which was subsequently re- and Ricquier, D., Nature 389, 135–136 (1997); Knutson, R. M., Science oxidized during electron transport. Because this did not occur with the pdc– 186, 746–747 (1974); and Seymour, R. S., and Schultze-Motel, P., Nature mutants, pyruvate decarboxylase must play a role in reoxidizing the NADH 383, 305 (1996).] produced during glycolysis. In the absence of the enzyme, glycolytic fl ux 67. (a) Glutamate enters the mitochondrion through a specifi c transporter could not increase due to the defi ciency of NAD . The NAD defi ciency also and is then oxidized to -ketoglutarate by glutamate dehydrogenase in prevented the decarboxylation of pyruvate by pyruvate dehydrogenase. the mitochondrial matrix. Concomitantly, NAD is reduced to NADH, 63. During anaerobic fermentation, only 2 ATP molecules can be obtained which can then enter electron transport. from each glucose molecule. Thus, in order to get enough ATP to satisfy (b) Ceramide could inhibit the glutamate transporter, the glutamate the energy needs of the cell, glucose consumption by the glycolytic pathway dehydrogenase enzyme, any of the three complexes involved in the elecpredominates, and only a small amount is left over for the pentose phosphate tron transport chain, ATP synthase, or the ATP translocase. pathway. But during aerobic respiration, 32 ATP are available per glucose (c) Ceramide doesn’t interfere with ATP synthase. Ceramide must block molecule. Thus, less glucose is needed to synthesize the same amount of ATP, respiration at an earlier step because the rate of respiration did not increase in the presence of the uncoupler. so a larger fraction of glucose can enter the pentose phosphate pathway. (d) Ceramide does not inhibit the glutamate transporter or glutamate 64. (a) When UCP1 is stimulated in normal mice, oxidative phosphoryladehydrogenase. Ceramide must inhibit one of the three complexes intion is uncoupled from electron transport. This means that the ATP volved in electron transport. (In fact, ceramide inhibits the activity of yield per substrate molecule oxidized is decreased, since ATP must then Complex III.) [From Gudz, T. I., Tserng, K.-Y., and Hoppel, C. L., J. be produced via substrate-level phosphorylation. The cell’s energy needs Biol. Chem. 272, 24154–24158 (1997).] are not met, which increases the rate of glycolysis and the citric acid cycle, and eventually electron transport, in a vain attempt to synthe- 68. (a) Substrate-level phosphorylation is catalyzed by phosphoglycerate size more ATP. Since oxygen is the fi nal electron acceptor in electron kinase and pyruvate kinase in glycolysis, and by succinyl-CoA synthetase transport, oxygen consumption increases in order to keep up with the in the citric acid cycle. increased rate of electron transport. In knockout mice, since there is no (b) C6H12O6 6 O2 S 6 CO2 6 H2O UCP1, oxidative phosphorylation is not uncoupled. Therefore, ATP can be synthesized via oxidative phosphorylation. Since the energy needs of the cell are being met, the rate of electron transport and oxygen conChapter 16 sumption do not increase.

1. proton translocation C, M (b) If UCP1 is not present, then uncoupling of oxidative phosphorylaC C photophosphorylation tion does not occur. Thus, ATP can be synthesized via oxidative phosC, M photooxidation phorylation, so less lipid needs to be broken down in order to meet the M C quinones organism’s energy needs.

C, M oxygen reduction (c) In the normal mice, the cold temperature stimulated the uncouM C water oxidation pling protein. As a result, the energy of electron transport was dissiM C electron transport pated as heat rather than used to synthesize ATP. This helped the mice C, M oxidative phosphorylation maintain normal body temperature. But the UCP1-knockout mice C, M carbon fi xation lacked the uncoupling protein and were unable to uncouple oxidation C, M NADH oxidation from phosphorylation. Thus, they could not generate “extra” heat, and C Mn cofactor their body temperatures decreased as a result. heme groups (d) When overfeeding occurs, adipose tissue content should increase in binding change mechanism the absence of UCP1, especially if energy expenditure is less than the iron–sulfur clusters caloric intake. Since the mice did not gain weight, it is likely that overNADP1 reduction feeding induced the expression of UCP2. Uncoupling electron transport and ATP synthesis would allow some of the energy in the excess food 2. Both cellular organelles are membrane-bound structures with their own to be converted to heat rather than stored as fat. [From Enerbäck, S., DNA that codes for proteins essential for organelle function (respiration for et al., Nature 387, 90–94 (1997); Fleury, C., et al., Nature Genetics 15, the mitochondrion, photosynthesis for the chloroplast). Both organelles have 269–272 (1997); Lowell, B. B., et al., Nature 366, 740–742 (1993); and an inner and an outer membrane that enclose an intermembrane space. The Hirsch, J., Nature 387, 27–28 (1997).] outer membrane is fairly porous in both organelles, whereas the inner membrane is fairly impermeable. The inner compartment of the mitochondrion is 65. (a) In the root system of the skunk cabbage, starch is broken down to referred to as the mitochondrial matrix, while the inner compartment of the yield glucose, which is catabolized via glycolysis, the citric acid cycle, chloroplast is called the stroma. Contained within the stroma is a third memand the electron transport chain. The oxidation of glucose provides the branous structure called the thylakoid, which is analogous to mitochondrial reduced cofactors (NADH and QH2) required to keep electron transinner-membrane cristae. The thylakoid membrane and the cristae differ in port going so that thermogenesis can occur. shape; the thylakoid consists of stacks of fl attened vesicles, whereas the cristae (b) In the skunk cabbage, thermogenesis increases as the temperature are tubular or highly folded. decreases. Thus, the rate of aerobic oxidation of glucose increases to increase the fl ux of NADH and QH2 through the electron transport 3.

CHOH 2 chain. Since oxygen is the fi nal electron acceptor, an increase in electron O transport will also increase oxygen consumption. When the ambient air O O CH2 HO H H H temperature is warmer, the need for heat production is less, so there is OH HC O C R1 less fl ux through the electron transport chain. Thus, oxygen consumpH tion decreases during the day.

O 66. The synthesis of the uncoupling protein increases with decreasing tem-

H

OH

CH2

O

C

R2

perature, presumably by an increase in transcription of the mRNA that codes for the uncoupling protein (although a decrease in the rate of mRNA deg- 4. Lipids with highly unsaturated fatty acids have low melting points, since radation would produce the same results). The increased amount of mRNA the cis double bonds create bends that make it diffi cult for the lipids to pack together. Because there are few intermolecular forces binding these lipids likely results in an increase in concentration of the uncoupling protein so that the mitochondrial proton gradient would be discharged. Thus, at low together, they generate a highly fl uid membrane, which must be essential for temperatures, the potato could generate heat rather than ATP. [From Laloi, photosynthesis in some way.

76

Ch 16 Solutions

5. (a) Use Planck’s law multiplied by Avogadro’s number (N ) to calculate the energy of the photons: E5 E5

hc

3 N (6.626310234 J?s)(2.9983108 m?s21)

13. The buildup of the proton gradient indicates a high level of activity of the photosystems. A steep gradient could therefore trigger photoprotective activity to prevent further photooxidation when the proton-translocating machinery is operating at maximal capacity.

14. Photooxidation would not be a good protective mechanism since it

431027 m 3(6.02231023 photons?mol21)

E5300 kJ?mol21 hc

(b) E5

l

E5

3N

might interfere with the normal redox balance among the electron-carrying groups in the thylakoid membrane. Releasing the energy by exciton transfer or fl uorescence (emitting light of a longer wavelength) could potentially funnel light energy back to the overactive photosystems. Dissipation of excess energy as heat would be the safest mechanism, since the photosystems do not have any way to harvest thermal energy to drive chemical reactions.

(6.626310234 J?s)(2.9983108 m?s21) 731027 m 3(6.02231023 photons?mol21)

E5170 kJ?mol21 6. (a) If the synthesis of each ATP requires 30.5 kJ · mol 1, then 9.8 mol ATP (300/30.5) could be synthesized. (b) About 5.6 mol ATP (170/30.5) could be synthesized.

7. The difference in reduction potential between P680* and P680 is –0.8 V 1.15 V 1.95 V. DG 8¿52nFDE8¿ 52(1)(96,485 J?V21?mol21)(21.95 V) 5188,000 J?mol215188 kJ?mol21 hc l 3 N

l5 5 5

hcN E (6.6263

102 34 J ?s)(2.9983108ms2

? 1)(6.022 3 10 188,000 J?mol21

23

photosn?ml o21 )

636 3 102 9 m 5 636 nm

9. Because the algae appear red, red light is transmitted rather than absorbed. Therefore, the photosynthetic pigments in the red algae do not absorb red light but absorb light of other wavelengths.

10. Photosynthetic bacteria can absorb light in the infrared and ultraviolet regions of the electromagnetic spectrum.

11.

A

D

N

N

N

N

15. The order of action is water–plastoquinone oxidoreductase (Photosystem II), then plastoquinone–plastocyanin oxidoreductase (cytochrome b6 f ), then plastocyanin–ferredoxin oxidoreductase (Photosystem I).

16. If the photosystems were in close proximity, then Photosystem II might not undergo photooxidation and instead act as a light-harvesting complex for Photosystem I. Because the reaction center of Photosystem II absorbs light with a peak wavelength of 680 nm, it could pass energy to the reaction center of Photosystem I, which absorbs lower-energy (longer-wavelength) light at its P700 group.

17. If electrons cannot be transferred to Photosystem I, then Photosystem II remains reduced and cannot be reoxidized. The photosynthetic production of oxygen ceases. No proton gradient is generated, so ATP synthesis does not occur in the presence of DCMU. 18. Myxothiazol inhibits electron transfer in the cytochrome b6f complex.

8. Since the energy of one mole of photons is E5

(b) Low-resolution models that included the heme groups would be more useful, since the positions of the heme groups within the proteins might indicate whether they function similarly in accepting and donat- ing electrons to their redox partners. It is possible that the heme groups could have similar orientations even if the apoproteins did not resemble each other in overall tertiary structure.

B

C

• The central metal ion in chlorophyll a is Mg2 , whereas in heme a the central metal ion is Fe2 .

• In chlorophyll a, there is a cyclopentanone ring fused to ring C. • Ring B in chlorophyll a has an ethyl side chain; the chain in heme b is unsaturated.

• The propionyl side chain in ring D of chlorophyll a is esterifi ed to a long, branched-chain alcohol.

12. (a) The sequences would provide only limited information, since even highly homologous proteins such as myoglobin and hemoglobin (which have very similar structures) have limited sequence identity (see Section 5-1). The three-dimensional structures of the proteins would be more likely to indicate whether they function similarly.

19. Like the lipids that compose the membrane, plastoquinone is amphiphilic, with a hydrophilic head and a hydrophobic tail. “Like dissolves like,” so plastoquinone literally dissolves in the membrane, which facilitates rapid diffusion.

20. The alanine residue at position 251 is essential for photosynthesis and photoautotrophic growth and may be part of the binding site for herbicides.

When the alanine is mutated to a cysteine, which resembles alanine except that a sulfhydryl group has replaced a hydrogen, the mutated protein is similar to the wild-type. When the alanine is changed to an amino acid that is similar in size but more polar (such as Gly, Pro, or Ser), photosynthesis is impaired. Replacement with a larger, nonpolar amino acid results in a mutant that is further impaired, and replacement with a larger polar amino acid results in an organism that is not photosynthetically competent.

Therefore, the residue at position 251 must be relatively small and nonpolar.

Residues of increasing size and polarity result in photosynthetically impaired organ- isms. [From Lardans, A., Förster, B., Prásil, O., Falkowski, P. G., Sobolev, V., Edelman, M., Osmond, C. B., Gillman, N. W., and Boynton, J. E., J. Biol. Chem. 273, 11082–11091 (1998).]

21. Use Equation 15-7, as applied in Sample Calculation 15-3.

The matrix and the stroma are both in.

DG52.303 RT (pHin2pHout)1ZFDc DG52.303(8.3145 J?K21?mol21)(298 K)(3.5) 1(1)(96,485 J?V21?mol21)(20.05 V) DG520,000 J?mol2124800 J?mol21 DG515.2 kJ?mol21 22.

DG52.303 RT (pHin2pHout)1ZFDc DG52.303(8.3145 J?K21?mol21)(298 K)(0.75) 1(1)(96,485 J?V21?mol21)(0.20 V) DG54300 J?mol21119,300 J?mol21 DG523.6 kJ?mol21

Ch 16 Solutions 77 For both organelles, the translocation is endergonic. In the chloroplast, most The cytosolic ratio of ATP/ADP decreases because less ATP is exported to of this energy is due to the concentration (pH) difference on the two sides of the cytosol from the matrix when ATP synthesis is inhibited. Because the membrane; in the mitochondrion, most of the energy is due to the oligomembrane potential ( c). mycin does not inhibit CF0, ATP synthesis is not inhibited, so the 23. Consult Table 15-1 for the reduction potentials of the relevant halfATP/ADP ratio is not affected. In some cases the decreased cytosolic ATP/ADP ratio reactions, reversing the sign for the water oxidation half-reaction. might stimulate ATP synthesis in the chloroplast, resulting in an increased E89 5 20.815 V chloroplastic ATP/ADP ratio. [From Gardeström, P., and Lernmark, U., H2O S ½ O1 22 H1 1 2 e2 J. Bioenerg. Biomembr. 27, 415–421 (1995).]

E89 5 20.320 V NADP1 1 H1 1 2 e2 S NADPH 31. This statement is false. The “dark” reactions do not require darkDE89 5 21.135 V H2O 1 NADP1 S ½ O 1 NAD PH 1 H 1 2 ness in order to proceed. Sometimes the “dark” reactions are called “lightindependent” reactions in order to specify that these reactions do not Use Equation 15-4 to calculate G : directly DG 8¿52nFDE8¿ require light energy. This term is also misleading because the “dark” reactions 2 2 5 J?V1?mol1)(21.135 V) of the Calvin cycle do require the products of the light reactions—ATP and DG852(2)(96,48 ¿ NADPH—in order to proceed. Thus, for a majority of plants, the “dark” re2 219,000 J?mol1 D G¿ 8 5 actions actually occur during the day when the light reactions are operational the needed ATP and NADPH. H2SS(CH2O)1S21 H2O Divide by Avogadro’s number to obtain the free energy change per molecule: and can produceCO212 32.

The net equation would be 33. 3-Phosphoglycerate is the fi rst stable radioactive intermediate that 219,000 J?mol21 forms when algal cells are exposed to 14CO2. The radioactive label is found 53.6310219 J?molecule21 on the carboxyl group of the compound.

6.02231023 molecules?mol21 24. Use Equation 16-1 and multiply by 2 to calculate the energy of the photons: E5

3 2

hc l

(6.626310234 J?s)(2.9983108 m?s21)(2) E5 631027 m E56.6310219 J

34. When rubisco reacts with oxygen in a process called photorespiration, the products are 3-phosphoglycerate and 2-phosphoglycolate. The latter product is further metabolized by pathways that consume ATP and NADPH and thus waste some of the energy captured by photosynthesis. If rubisco could bind only CO2, the enzyme would produce two molecules of 3-phosphoglycerate, which is a precursor to many biological macromolecules. In a plant with a rubisco that binds only CO2, photorespiration would not occur, and more of the energy obtained in photosynthesis would be available for biosynthetic reactions. 35. The increase in mass comes from carbon dioxide.

CO2 is the carbon source for cellulose, a major structural component of the tree. Water also contributes to the increase in mass. Soil nutrients contribute a very small percentage of the mass of the full-grown oak tree.

25. The fi nal electron acceptor in photosynthesis is NADP . The fi nal elec- 36.

DG5DG 8¿1RT ln Q tron acceptor in mitochondrial electron transport is oxygen.

241.0 kJ?mol215235.1 kJ?mol211(8.3145 26. Some of the label appears in molecular oxygen released by the water31023 kJ?K21?mol21)(298 K)ln Q splitting reaction of Photosystem II. Some of the label appears in other com25.9 kJ?mol2152.48 kJ?mol21 ln Q pounds, including intermediates of the Calvin cycle, since H2O participates in the rubisco reaction (see Fig. 16-24).

22.385In Q 27. (a) An uncoupler dissipates the transmembrane proton gradient by proe22.385Q viding a route for translocation other than ATP synthase. Therefore, The energy of two photons is nearly twofold greater than the energy change for the oxidation of one molecule of water by NADP , so, in theory, the two photons supply suffi cient energy to drive the oxidation.

chloroplast ATP production would decrease.

Q50.093 (b) The uncoupler would not affect NADP reduction since lightdriven electron-transfer reactions would continue regardless of the state The ratio of products to reactants is 0.093 to 1. of the proton gradient.

37. Normally, plants must synthesize large quantities of rubisco, a protein whose constituent amino acids all contain nitrogen. If rubisco had greater 28. Complex III of the mitochondrial electron transport chain is analogous catalytic activity, the plant might produce less of the enzyme, thereby deto the cytochrome b6f complex. In the presence of antimycin A, electron creasing its need for nitrogen. fl ow in the cytochrome b6f complex would be inhibited. Electrons would not reach Photosystem I, and NADPH would not be produced. Proton transloca- 38. (a) The compound resembles the transition state of the rubisco carboxylase reaction (see Fig. 16-24) and therefore inhibits the enzyme by tion from the stroma to the thylakoid lumen also would not occur, so CF1 binding in the active site. would not be stimulated and ATP synthesis would not occur. (b) At night, the compound inhibits rubisco in order to prevent the 29. (a) More c subunits means that more protons are required to rotate the Calvin cycle from consuming NADPH and ATP. During the day, when ATP synthase through one ATP-synthesizing step. Therefore, more phothe light reactions are supplying NADPH and ATP, the inhibitor is brotons must be absorbed to drive the translocation of more protons, so the ken down to reactivate rubisco so that it can fi x carbon. quantum yield decreases.

(b) Cyclic electron fl ow contributes to the proton gradient and there39. The unprotonated Lys side chain serves as a nucleophile when reacting fore leads to ATP synthesis. However, carbon fixation by the Calvin with CO2. At high pH, a higher percentage of ´-amino groups are in the cycle requires NADPH also, so the additional photons that drive cyclic unprotonated form. fl ow do not lead to more carbon fi xed. Consequently, the quantum yield 40. ATP and NADPH are produced by the light reactions during the day. decreases.

When photosynthesis is occurring, synthesis of starch (an ATP-requiring 30. Because oligomycin inhibits the F0 subunit of mitochondrial ATP synprocess) occurs, and glycolysis, an opposing process, is inhibited. At night, thase, protons from the intermembrane space cannot translocate to the mi- when the light reactions do not occur and ATP and NADPH concentrations tochondrial matrix. F1 is not stimulated, and ATP synthesis does not occur. are low, the inhibition of phosphofructokinase is relieved. Starch is broken

78

Ch 17 Solutions

down, and the resulting glucose-1-phosphate enters glycolysis to produce 45. (a) PEPC catalyzes the formation of oxaloacetate, one of the two reacATP to meet the energy needs of the plant. [From Cséke, C., Nishizawa, tants for the fi rst reaction of the citric acid cycle. Anaplerotic reactions A. N., and Buchanan, B. B., Plant Physiol. 70, 658–661 (1982).] are important because they replenish citric acid cycle intermediates (see Fig. 14-18). If oxaloacetate is unavailable, the citric acid cycle cannot 41. Grasses turn brown because they undergo photorespiration in hot, continue. (b) Acetyl-CoA is an allosteric activator of PEPC. When the dry conditions. Rubisco reacts with oxygen to form 2-phosphoglycolate, concen- tration of acetyl-CoA rises, additional oxaloacetate will be which subsequently consumes large amounts of ATP and NADPH. CO2 required to react with it in the fi rst reaction of the citric acid cycle.

Activa- tion of PEPC by acetyl-CoA will lead to increased production of concentrations are low because the plants close their stomata in order to the required oxaloacetate. avoid loss of water when the weather is hot and dry (see Box 16-A). Without CO2, photosynthesis does not occur and the grass turns brown. But C4 plants such as crab grass can generate CO2 from oxaloacetate, which 46. PEPC supplies oxaloacetate for the citric acid cycle, which can generate can enter the Calvin cycle. Carbon fi xation occurs, and the crab grass ATP thrives even in hot, dry weather.

The for the biosynthetic processes occurring in the germinating seedling.

PEPC enzyme also plays a role in providing oxaloacetate for the glyoxylate 42. cycle (see Box 14-B), which provides a route for the synthesis of glucose from Carbon fatty acids, a pathway that is lacking in animals. 47. The primers should have Light fixatio n

P700* red

Reductasered

Ferredoxinox

Thioredoxinox

SHSH

P700

ox

Reductaseox

Ferredoxinred

S

S

Enzymered SHSH

S

Thioredoxinred

Enzyme ox

SHSH

S

S

S

the following sequences: GTAGTGGGATTGTGCGTC and GCTCCTACAAATGCCATC (see Fig. 3-17).

48. The Bt toxin is not toxic to humans because the stomach is acidic (see Table 2-3). The toxin forms a pore only under high pH conditions.

49. Glyphosate herbicides are effective at killing weeds because glyphosate inhibits the plant EPSPS enzyme required for the synthesis of aromatic amino acids. The transgenic crops are protected from this inhibitor because these crops contain the bacterial enzyme that is not subject to inhibition by glyphosate. Using this strategy allows for weed eradication while preserving the desired crop.

50. Glyphosate is not toxic to humans because humans do not contain the

The activity of Photosystem I generates reduced ferredoxin, which is a enzyme EPSPS. This pathway was lost during evolution, and the result is substrate for the reductase. The product of the reductase reaction, reduced that phenylalanine is an essential amino acid that must be obtained in the thioredoxin, can then reduce the disulfi des of the Calvin cycle enzymes. diet (see Table 12-1).

Conformational changes in the enzymes upon exposure of free sulfhydryl groups could increase their activity.

43.

NADP+ NADPH Glyceraldehyde 1,3-Bisphospho3-phosphate glycerate ADP

ATP 3-Phosphoglycerate

Dihydroxyacetone phosphate

Chapter 17 1. The lipoproteins increase in density as the percentage of protein content increases and the percentages of lipid content decreases. Thus, chylomicrons have the lowest density and HDL have the highest density.

2. Hydrophobic amino acid side chains, such as Ile, Val, Leu, and Phe, would be expected to contact the nonpolar lipids.

3. The products are one palmitoyl methyl ester, two oleoyl methyl esters, and one glycerol:

STROMA

O H3C

CYTOSOL

Dihydroxyacetone phosphate

3-Phosphoglycerate ATP

Glyceraldehyde 3-phosphate NAD+

44.

ADP 1,3-Bisphosphoglycerate NADH

DG5DG 8¿1RT ln Q 229.7 kJ?mol215214.2 kJ?mol211(8.3145 31023 kJ?K21?mol21)(298 K)ln Q 215.5 kJ?mol2152.48 kJ?mol21 ln Q 26.255In Q e26.255Q Q50.0019

The enzyme is likely to be regulated because it catalyzes an irreversible step of the Calvin cycle (as evidenced by the large negative value of G ).

(CH2)14

C

O

CH

CH

H3C

(CH2)7

H2C

CH

CH2

HO

OH

OH

CH3 O (CH)2 7

C

O

CH3

4. Water would participate in hydrolysis reactions, yielding glycerol plus three fatty acids.

5. The hydrolysis of pyrophosphate releases 19.2 kJ ? mol 1. (a) fatty acid1CoA1ATPΔacyl-CoA1AMP1PPi DG 8¿50 kJ?mol21 D G 8 ¿ 5 219.2 kJ?mol2 PPi1H2OS2 Pi fatty acid1CoA1ATP1H2OSacyl-C oA1AMP12 Pi (b)

DG8¿5219.2 kJ?mol21

K eq 5 e 2 D G 8 ¿ y R T K eq 5e2(219.2 kJ?mol21)y(8.314531023 kJ?K21?mol21)(310 K) K eq5e7.4 3 K eq51.7310

1

Ch 17 Solutions 79 6.

12. Oxidation of the methylene groups is accomplished by the action of a dehydrogenase, which removes electrons. The oxygen atom that becomes part of the carbonyl group is derived from a water molecule added by enoyl- CoA hydratase.

13. (a) Benzoate was produced when the dogs were fed phenylpropionate. (b) Phenylacetate was produced when the dogs were fed phenylbutyrate.

O R

C O

Fatty Acid

O OP

O O

O

P

O O

P

O

CH2

Adenosine

Phenylpropionate

O

O

ATP inorganic pyrophosphatase

PPi

O

R C

C

S

CH2

COO

CoA

Acetyl-CoA Benzoate

O O

P

O

Adenosine H

O

CH2

SCoA

O C

CH2

Phenylbutyrate

Acyladenylate mixed anhydride

R

H3C

COO

2 Pi

H2O O

COO

CH2

O SCoA

O

Acyl-CoA

P

O

Adenosine

O

O H3C

COO

CH2

C

S

CoA

Acetyl-CoA Phenylacetate

AMP

14. The products are three molecules of propionyl-CoA, three molecules of 7. If carnitine is defi cient, fatty acid transport from the cytosol to the miacetyl-CoA, and one molecule of 2-methylpropionyl-CoA. tochondrial matrix (the site of oxidation) is impaired. Fatty acid oxidation generates a great deal of ATP to power the muscle, so in the absence of fatty acid oxidation the muscle must rely on stored glycogen or uptake of circuSCoA C lating glucose to obtain the necessary ATP. Muscle cramping is exacerbated Pristanate by fasting because the concentration of circulating glucose is decreased and O glycogen stores are depleted. Exercise also increases muscle cramping because O the demand for ATP by the muscle is greater. oxidation H3C CH 2 C S CoA 8. If medium-chain fatty acids are metabolized normally in the absence of carnitine, then fatty acids of this size must be able to cross the inner mitochondrial membrane and enter the mitochondrial matrix without the aid of carnitine.

Carnitine is required for transport of fatty acids longer than 10 carbons.

9. Medium-chain acyl-CoA (4–12 carbons) would accumulate in individuals with MCAD defi ciency, since the conversion of fatty acyl-CoA to enoyl-CoA is blocked. Acyl-carnitine esters would also accumulate.

10. The patient should avoid fasting and sustained exercise, since both of these conditions result in the release of free fatty acids into the bloodstream for uptake by the tissues. The patient should also consume a diet that is low in fat and high in carbohydrates. 11. The conversion of fatty acyl-CoA to enoyl-CoA is similar to the conversion of succinate to fumarate because both reactions involve oxidation of the substrate and concomitant reduction of FAD to FADH2 (see Section 14-2). The conversion of enoyl-CoA to hydroxyacyl-CoA is similar to the conversion of fumarate to malate because both reactions involve the addition of water across a trans double bond. The conversion of hydroxyacyl-CoA to ketoacyl-CoA is similar to the conversion of malate to oxaloacetate because both reactions involve the oxidation of an alcohol to a ketone with concomitant reduction of NAD to NADH. Oxidation Citric acid cycle

Fatty acyl-CoA

Enoyl-CoA Hydroxyacyl-CoA

Succinate

Fumarate

Propionyl-CoA

C O O oxidation

CH 3

SCoA

SCoA O oxidation

Propionyl-CoA

oxidation

Acetyl-CoA

oxidation

Propionyl-CoA

oxidation

Acetyl-CoA

CH3

Malate Oxaloacetate

C

Acetyl-CoA

SCoA

H3C

Ketoacyl-CoA

SCoA

O

80

Ch 17 Solutions

15. (a) Palmitate goes through seven cycles of oxidation. The fi rst six cycles produce 1 QH2, 1 NADH, and 1 acetyl-CoA. The seventh cycle produces 1 QH2, 1 NADH, and 2 acetyl-CoA. Each QH2 generates 1.5 ATP in the electron transport chain, each NADH generates 2.5 ATP in the electron transport chain, and each acetyl-CoA generates a total of 10 ATP (1 QH2 1.5 1.5 ATP; 3 NADH 2.5 7.5 ATP; 1 GTP 1 ATP for a total of 10 ATP per acetyl-CoA). The total is 108 ATP. Two ATP must be subtracted from this total to account for the ATP spent in activating palmitate to palmitoyl-CoA. This gives a total of 106 ATP. (b) The same logic is used for stearate, except that stearate goes through eight cycles of oxidation. The total is 120 ATP.

O H3C

(CH)2 4

CH

CH

CH2

C

C

H

H

H3C

(CH)2 4

CH

CH

CH2

CH2

H3C

CH

(CH)2 7

CH

(CH2)7

C

SCoA

Oleoyl-CoA

3 (1 QH 2, 1 NADH, 1 acetyl-CoA) 3(1.5 2.5 10) 42 ATP O

3 cycles of oxidation

H3C

(CH)2 7

C H

C

C

CH2

1 cycle of oxidation (acyl-CoA dehydrogenase reaction is bypassed)

H3C

(CH)2 8

C

H

O

C

C

CH

SCoA

C

C

C

CoA

1 NADH, 1 acetyl-CoA 2.5 10 12.5 ATP

CH

CH2

CH2

C

S

S

CoA

CoA

acyl-CoA dehydrogenase 1 QH2 1.5 ATP

H3C

(CH)2 4

CH

CH

O C

H C

C

H NADPH NADP

Costs 2.5 ATP O

H3C

(CH)2 4

CH2 CH

CH2

CH

C

S

S

CoA

CoA

enoyl-CoA isomerase

H3C

(CH)2 4

CH2

CH2

C

H

O

C

C

H

3 cycles of oxidation

H 1 cycle of oxidation (acyl-CoA dehydrogenase reaction is bypassed)

(CH)2 4

1 cycle of oxidation (acyl CoA dehydrogenase reaction is bypassed)

H

O

O H3C

SCoA

enoyl-CoA isomerase

H

CoA

H

2,4-dienoyl-CoA reductase

O

C

enoyl-CoA isomerase

16. (a) Oleate is an 18-carbon fatty acid. If it were completely saturated, 120 ATP would be generated via oxidation (see Problem 15b). The double bond at the 9,10 position must be converted from the cis to the trans isomer. oxidation continues after this isomerization step, but with the loss of 1 QH2. Thus 1.5 ATP should be subtracted from the total to yield 118.5 ATP. The pathway below shows a total of 120.5 ATP, but 2 ATP are subtracted to account for the ATP spent in fatty acid activation.

CH2

1 NADH, 1 acetyl-CoA (2.5 10) 12.5 ATP 3 (1 QH2, 1 NADH, 1 acetyl-CoA) 3(1.5 2.5 10) 42 ATP

Acetyl-CoA (10 ATP)

1 NADH, 1 acetyl-CoA 2.5 10 12.5 ATP O

17. A C17 fatty acid goes through seven cycles of oxidation. The fi rst six cycles produce 1 QH2, 1 NADH, and 1 acetyl-CoA. The seventh cycle produces 1 QH2, 1 NADH, 1 acetyl-CoA, and 1 propionyl-CoA. Each QH2 generates 1.5 ATP in the electron transport chain, each NADH generates 4 cycles of oxidation 4 (1 QH2, 1 NADH, 1 acetyl-CoA) 2.5 ATP, and each acetyl-CoA generates a total of 10 ATP (1 QH2 1.5 4(1.5 2.5 10) 56 ATP 1.5 ATP; 3 NADH 2.5 7.5 ATP; 1 GTP 1 ATP for a total of 10 Acetyl-CoA (10 ATP) ATP per acetyl-CoA). The total is 98 ATP. Propionyl-CoA is metabolized to succinyl-CoA (at a cost of 1 ATP; see Fig. 17-7) and enters the citric acid cycle. Conversion of succinyl-CoA to succinate yields 1 GTP (which (b) Linoleate is also an 18-carbon fatty acid and would yield 120 ATP offsets the cost of the propionyl-CoA S succinyl-CoA conversion), and via oxidation if it were saturated. The double bond at the 9,10 posiconversion of succinate to fumarate yields 1 QH2 (equivalent to 1.5 ATP). tion must be converted from the cis to the trans isomer, resulting in the Fumarate is converted to malate, then malate is converted to pyruvate, loss of 1 QH2 (1.5 ATP). In addition, an NADPH-dependent reduction yielding 1 NADPH (equivalent to 2.5 ATP). The pyruvate dehydrogenase of the 12,13 double bond costs 2.5 ATP. This reduces the ATP total by reaction converts pyruvate to acetyl-CoA (which is subsequently oxidized 4, so oxidation of linoleate yields 116 ATP. (Subtract 2 ATP for activaby the citric acid cycle to yield 10 ATP) and 1 NADH, which yields 2.5 tion from the total shown in the fi gure.)

ATP. Therefore, oxidation of propionyl-CoA yields an additional 16.5 ATP. The total is 98 ATP 16.5 ATP 114.5 ATP. Two ATP must be subO tracted from this total to account for the ATP spent in activating the C17 fatty acid to fatty acyl-CoA. This gives a fi nal total of 112.5 ATP. Note that H3C (CH)2 4 CH CH CH2 CH CH (CH2)7 C SCoA a C17 fatty acid yields more ATP than palmitate (106 ATP) and less than Linoleyl-CoA oleate (118.5 ATP).

18. Adding up the ATP in the fi gure gives a total of 155 ATP. Subtracting 3 (1 QH, 1 NADH, 1 acetyl-CoA)

H3C

3 cycles of

(CH)2 8

oxidation

SCoA

C

2

3(1.5 2.5 10) 42 ATP

the 2 ATP required for activation brings the total to 153 ATP.

Ch 17 Solutions 81 ATP requirement

O H3C

CH2

(CH)220

acyl-CoA oxidase

H3C

C

(CH)220

enoyl-CoA

CH2

C

FAD FADH2 H

O

C

C

SCoA H2O2 O2

2 H2O O2

SCoA

H

6 cycles of oxidation (acyl-CoA dehydrogenase reaction is bypassed)

6 NADH, 6 acetyl-CoA 6(2.5) 6(10) ATP 75 ATP CH2 O

H3C

(CH2)8

5 cycles of oxidation

One ATP (two highConsumes one ATP energy phosphoper two carbons anhydride bonds) incorporated into required to activate the growing fatty the fatty acid acyl chain.

Two-carbon acetyl units C3 intermediate (acetyl-CoA) (malonyl-CoA)

CH2

C

SCoA PEROXISOME MITOCHONDRION

5 QH 2, 5 NADH, 5 acetyl-CoA 5(1.5) 5(2.5) 5(10) ATP 70 ATP

acetyl-CoA (10 ATP)

Unit product/unit donor Confi guration of hydroxyacyl intermediate Shortening/growth occurs at which end of the fatty acyl chain?

L

D

Thioester end

Thioester end

24. Acetyl-CoA carboxylase generates malonyl-CoA, the substrate for fatty acid synthase. Mice without acetyl-CoA carboxylase are unable to synthesize fatty acids and hence store less fat in their bodies. Malonyl-CoA also inhibits carnitine acyltransferase. In the absence of acetyl-CoA carboxylase to produce this inhibitor, transport of fatty acids into mitochondria cannot be regulated, and mitochondrial oxidation proceeds continuously.

25.

Phase I

19. The patient could be treated with injections of vitamin B12 directly into the bloodstream. Alternatively, the patient could be treated with high doses of oral vitamin B12. In the presence of high concentrations of the vitamin, suffi cient amounts may be absorbed even in the absence of intrinsic factor.

20. A patient with pernicious anemia would have an excess of methylmalo-

O P

Adenosine

O

HO P C O O

Cellular location

O

Acyl-group carrier Electron carrier(s)

Acyl-carrier protein NADPH donates elec-

NAD1 accept electrons to become ubiquinol and NADH

trons and becomes oxidized to NADP1

C OH

O

O

Pi

O

biotinyl-enzyme

O

C

NH

HN

O

O (CH2)4

S

C

NH

(CH2)4

E

NH

(CH2)4

Biotinyl-enzyme

O O

O C

NH

N

O C

(CH2)4

S

E

Carboxybiotinyl-enzyme

Phase II CoAS

O C

Acetyl-

CH 2

C CH

CoA

O O

O N

O

CoAS

H C

Coenzyme A Ubiquinone and

O ADP

O O

Carboxyphosphate

22. (a) The free energy cost of synthesizing ATP from ADP Pi is 30.5 kJ ?

Fatty acid degradation Fatty acid synthesis Mitochondrial matrix Cytosol

O

O

ment to coenzyme A in an ATP-requiring step. The fi rst phase of glycolysis also requires the investment of free energy in the form of ATP. Consequently, neither oxidation nor glycolysis can produce any ATP unless some ATP is already available to initiate these catabolic pathways.

23.

O P

O

ATP

nate in the blood or urine because the methylmalonyl-CoA mutase reaction (Figure 17-7), which requires the cobalamin cofactor, would be unable to take place, leading to accumulation of reactants.

21. A fatty acid cannot be oxidized until it has been activated by its attach-

mol 1. Glucose oxidation could theoretically yield 2850/30.5, or 93.4 ATP. The ATP yield per carbon atom is 93.4/6 15.6. Palmitate oxidation could theoretically yield 9781/30.5, or 320.7 ATP. The ATP yield per carbon atom is 320.7/16 20.4 ATP. (b) In vivo, the catabolism of glucose leads to the production of 32 ATP (see Solution 15-43), which is 32/6, or 5.3 ATP per carbon atom. The catabolism of palmitate leads to the production of 106 ATP (see Solution 15a), which is 106/16, or 6.6 ATP per carbon atom. (c) In theory as well as in vivo, fatty acid oxidation yields more ATP per carbon atom than glucose oxidation (this is primarily because the carbons of carbohydrates are already partially oxidized, whereas fatty acid carbons are usually fully reduced). Both pathways are equally effi cient from a thermodynamic point of view, recovering slightly more than 30% of the free energy available (for glucose 5.3/15.6, and for palmitate, 6.6/20.4).

O P

O

O C O

NH

O HN

2

H

O N

NH

Carboxybiotinyl-enzyme Acetyl-CoA enolate

CoAS

O C O

O

CoAS

O

C

C

CH2

CH2 C O

O

Malonyl-CoA

Biotinyl-enzyme

NH

82

Ch 17 Solutions Alkyl side chain (R)

Breast cancer ID50 Normal cells ID50

¬C13H27

2.7

¬C11H23

6.0

¬C9H19

2.5

27. All three enzymes involve the ATP-dependent addition of a carboxylate

D

¬C8H17

4.3

group (donated by bicarbonate) to a substrate. The bicarbonate is activated via its attachment to biotin, a cofactor required by all three enzymes. The carboxybiotin, a prosthetic group of all three enzymes, transfers the carboxylate group to the substrate, increasing the number of carbons in the substrate by one.

28. Epinephrine signaling via an adrenergic receptor activates a G protein,

E

¬C7H15

4.5

F

¬C6H13

1.5

26. As shown in the mechanism for malonyl-CoA formation (Problem 25), ATP is required to generate a carboxyphosphate intermediate. The result is Compound that malonyl-CoA contains some of the free energy of ATP. This free energy A is released when the acetyl group condenses with malonyl-ACP and CO2 is B released. The release of CO2 is entropically favored and drives the reaction to completion.

C

which activates adenylate cyclase to produce cyclic AMP to activate protein kinase A (PKA). PKA phosphorylates its substrates, including acetyl-CoA carboxylase, thereby inactivating the enzyme. As a result of lower acetylCoA carboxylase activity, the rate of fatty acid synthesis drops. Epinephrine signaling also leads to phosphorylation and activation of glycogen phosphorylase, which mobilizes glucose from glycogen. These responses are consistent: When the cell needs to mobilize metabolic fuel (for example, by glycogenolysis), storage of fuel (for example, by fatty acid synthesis) is inhibited.

29. The K I is lower when the enzyme is phosphorylated. Phosphorylation de-

(c) Inhibitor D at a concentration of 2 g/mL signifi cantly inhibits incorporation of radiolabeled acetate into triacylglycerols. At a concentration of 5 g/mL, 80% of total acylglyceride synthesis is inhibited. Phospholipid synthesis is not signifi cantly affected, supporting the hypothesis that the drug’s target is fatty acid synthase. [From Kuhajda, F. P., Pizer, E. S., Li, J. N., Mani, N. S., Frehywot, G. L., and Townsend, C., Proc. Natl.

Acad. Sci. 97, 3450–3454 (2000).]

35. The shorthand form for fatty acids is described in Problem 8-1. Fatty acids that cannot be synthesized from palmitate using cellular elongases and desaturases are essential fatty acids and must be obtained from the diet. Mammals do not have a desaturase enzyme that can introduce double bonds beyond C9. Oleate and palmitoleate, with a double bond at the 9,10 position, are not essential fatty acids. Linoleate has a second double bond at the 12,13 position. Because a doucreases the activity of the enzyme (see Solution 28). Palmitoyl-CoA (a pathway ble bond could not be introduced at the 12,13 position, linoleate is an essential product) allosterically inhibits the enzyme; a lower concentration is required fatty acid. -Linolenate has double bonds at positions 9,10, 12,13, and 15,16 for inhibition if the enzyme is already phosphorylated. A low K I value indi- and is also essential. cates more effi cient inhibition (see Section 7-3); therefore, palmitoyl-CoA O binds more tightly to the phosphorylated enzyme. The allosteric inhibition of acetyl-CoA carboxylase by palmitoyl-CoA is a method of “fi ne-tuning” the C O regulation of the enzyme by phosphorylation.

(CH2)7

30. Acetyl-CoA carboxylase is stimulated by citrate (which signals abundant

O

acetyl-CoA, the enzyme’s substrate). The phosphorylated form of acetyl-CoA carboxylase is less active than the dephosphorylated form (see Solution 28); therefore, a greater concentration of citrate would be required to increase the activity of the enzyme.

31. The synthesis of palmitate from acetyl-CoA costs 42 ATP. Seven rounds of the synthase reaction are required. ATP is required to convert each of 7 acetyl-CoA to malonyl-CoA for a total of 7 ATP. Two NADPH are required for seven rounds of synthesis, which is equivalent to 2 7 2.5 35 ATP. 32. The label does not appear in palmitate because 14CO2 is released in Reaction 3 of fatty acid synthesis (Fig. 17-13).

33. Mammalian fatty acid synthase is structurally different from bacterial fatty acid synthase; thus, triclosan can act as an inhibitor of the bacterial enzyme but not the mammalian enzyme. The mammalian fatty acid synthase is a multifunctional enzyme made up of two identical polypeptides. In bacteria, the enzymes of the fatty acid synthetic pathway are separate proteins.

Triclosan actually inhibits the bacterial enoyl-ACP reductase. The enzymes of the mammalian multifunctional enzyme must be arranged in such a way as to preclude the binding of triclosan to the active site of the enoyl-ACP reductase.

34. (a) A portion of the inhibitor molecule mimics the structure of a fatty acid. The compound may act as a competitive inhibitor, binding to the enzyme active site to preclude binding of the substrate. (b) The lower the ID50 value, the lower the concentration of inhibitor required to kill the cells. It is desirable for the inhibitor to inhibit fatty acid synthase in cancer cells but not normal cells. Thus, an effective inhibitor would have a low ID50 value in cancer cells but a high ID50 value in normal cells. The ratios of ID50 values for normal and cancer cells were calculated to determine which of the inhibitors tested was most effective. As shown in the table below, compounds B, D, and E were the most effective inhibitors. These compounds have side chains of 11, 7, and 8 carbons, respectively. Inhibitor D has the shortest alkyl chain and is the most soluble of the three.

C

CH O

CH

(CH2)7

CH2

O

CH

CH

C

O

CH

CH

C

(CH2)7

CH2

CH2

(CH2)7

CH

CH

CH

CH

CH

CH

CH

CH

(CH2)4

CH2

(CH2)5

CH3

CH3

CH3

CH3

Oleate

Linoleate

-Linolenate

Palmitoleate

(CH2)7

O O

36. DHA is an n-3 fatty acid and is essential for growth. A well-nourished mother will produce breast milk containing DHA, but a bottle-fed infant whose sole source of food is infant formula will not obtain this essential fatty acid unless it is added to the formula.

37. In gluconeogenesis, the input of free energy is required to undo the exergonic pyruvate kinase reaction of glycolysis. Pyruvate is carboxylated to produce oxaloacetate, and then oxaloacetate is decarboxylated to produce phosphoenolpyruvate. Each of these reactions requires the cleavage of one phosphoanhydride bond (in ATP and GTP, respectively). In fatty acid synthesis, ATP is consumed in the acetyl-CoA carboxylase reaction, which produces malonyl-CoA. The decarboxylation reaction is accompanied by cleavage of a thioester bond, which has a similar change in free energy to cleaving a phosphoanhydride bond.

38. (a) Liver fatty acid synthase activity increases with consumption of a high-carbohydrate diet. Glucose in excess of what is required to meet immediate energy needs is oxidized to pyruvate by glycolysis, then converted to acetyl-CoA by pyruvate dehydrogenase. Excess acetyl-CoA is used to synthesize fatty acids, which are ultimately used to synthesize triacylglycerols for storage in adipose tissue.

Ch 17 Solutions 83 (b) Liver fatty acid synthase activity is decreased with consumption of HMG-CoA reductase on the essential serine residue. The phosphorylated a high-fat diet. Endogenous fatty acid synthesis is not required if fatty form of the enzyme is less active. In the mutant cells, the alanine cannot be acids are being obtained from the diet. (c) Mammary gland fatty acid phosphorylated, so the enzyme activity was unaffected even in the presence synthase activity increases in mid to late pregnancy to provide fatty of increasing cellular cholesterol. acids for triacylglycerols in breast milk for the neonate.

47. (a) Fumonisin inhibits ceramide synthase. The concentration of the fi nal product, ceramide, decreased, while other lipid synthetic pathways were 39. (a) Acetoacetate is a ketone body. It is converted to acetyl-CoA, not affected. The fi rst enzyme in the pathway, serine palmitoyl transferwhich can be oxidized by the citric acid cycle to supply free energy ase, was not the target of fumonisin B1 because there was no signifi cant to the cell. decrease in the production of its product, 3-ketosphinganine. The sec(b) Intermediates of the citric acid cycle are also substrates for other ond enzyme in the pathway, 3-ketosphinganine reductase, was also not metabolic pathways. Unless they are replenished, the catalytic activity a target because if it were, the substrate of this reaction would have acof the cycle is diminished. Ketone bodies are metabolic fuels, but they cumulated in the presence of fumonisin. Accumulation of sphinganine cannot be converted to citric acid cycle intermediates. A three-carbon indicates that ceramide synthase is inhibited. When ceramide synthase glucose-derived compound such as pyruvate can be converted to oxalois inhibited, sphinganine cannot be converted to dihydroceramide and acetate to increase the pool of citric acid cycle intermediates and keep instead accumulates. the cycle operating at a high rate. (b) Fumonisin likely acts as a competitive inhibitor. It is structurally similar to sphingosine and its derivatives and thus can bind to the ac40. Fatty acid degradation produces acetyl-CoA, but in the absence of glutive site and prevent substrate from binding. The fumonisin may form cose, acetyl-CoA cannot enter the citric acid cycle due to insuffi cient oxaa covalent bond with the enzyme, or it may bind to the enzyme’s acloacetate, a substrate for the fi rst step. When carbohydrate concentrations tive site noncovalently with high affi nity. Alternatively, fumonisin may drop, any available oxaloacetate is diverted to gluconeogenesis to produce the act as a substrate and be converted to a product that cannot be subseglucose necessary for red blood cell and brain function. If acetyl-CoA cannot quently converted to ceramide. [From Wang, E., Norred, W. P., Bacon, enter the citric acid cycle (and acetyl-CoA cannot be converted to pyruvate C. W., Riley, R. T., and Merrill, A. H., J. Biol. Chem. 266, 14486 –14490 in mammals), it is converted to ketone bodies.

(1991).]

41. The synthesis of the ketone body acetoacetate does not require the input of free energy (the thioester bonds of 2 acetyl-CoA are cleaved; see Fig. 17-16). 48. Three phosphoanhydride bonds must be cleaved in order to synthesize Conversion of acetoacetate to 3-hydroxybutyrate consumes NADH (which phosphatidylcholine from choline and diacylglycerol (see Fig. 17-19). Chocould otherwise generate 2.5 ATP by oxidative phosphorylation). However, line is activated by conversion to phosphocholine, with a phosphate group the conversion of 3-hydroxybutyrate back to 2 acetyl-CoA regenerates the donated by ATP. The phosphocholine is subsequently converted to CDP– NADH (see Fig. 17-17). This pathway also requires a CoA group donated by choline, a reaction in which a second phosphoanhydride bond is cleaved. succinyl-CoA. The conversion of succinyl-CoA to succinate by the citric acid Pyrophosphate is the leaving group in this reaction, and the third phosphocycle enzyme succinyl-CoA synthetase generates GTP from GDP Pi , so the anhydride bond is cleaved when the pyrophosphate is hydrolyzed to orthoconversion of ketone bodies to acetyl-CoA has a free energy cost equivalent phosphate. to 49. A hydrolysis reaction removes a fatty acyl group from a triacylglycerol, one phosphoanhydride bond. leaving a diacylglyerol. The intent is to reduce the total fatty acid content 42. Oxidation of fatty acids yields acetyl-CoA, which enters the citric acid of the oil, thereby reducing its caloric value, without drastically altering the cycle by reacting with oxaloacetate. Oxaloacetate is produced by the action fl uidity of the product. of pyruvate carboxylase on pyruvate produced by glycolysis. Without glucose 50. Choline kinase catalyzes the ATP-dependent phosphorylation of chometabolism to generate pyruvate, the citric acid cycle cannot proceed. Thus, the statement “fats burn in the fl ame of carbohydrates” refers to the fact that line to produce phosphocholine, which is subsequently activated by CTP as part of the phosphatidylcholine synthetic pathway (Fig. 17-19). normal carbohydrate metabolism (leading to the production of oxaloacetate) is absolutely required for normal fatty acid metabolism, which provides 51. (a) The enzyme transfers a methyl group from a donor molecule three acetyl-CoA for the citric acid cycle. times in order to convert phosphoethanolamine to phosphocholine.

43. In the absence of pyruvate carboxylase, pyruvate cannot be converted 2 O3P to oxaloacetate. Without suffi cient oxaloacetate to react with acetyl-CoA in 2 O3P O CH 2 CH 2 NH 3 O CH 2 CH 2 N(CH 3) 3 the fi rst reaction of the citric acid cycle, acetyl-CoA accumulates. The excess Phosphoethanolamine Phosphocholine acetyl-CoA is converted to ketone bodies.

44. A patient with a pyruvate carboxylase defi ciency has elevated levels of (b) The methyltransferase would be a suitable drug target only if the enzyme were unique to the parasite and was not expressed in humans.

[In fact, this is the case.] py-

ruvate (see Solution 14-50), which also results in an increase in lactate as the 52. Ergosterol has two additional double bonds (one in a ring and one in the cells rely on anaerobic metabolism in the absence of proper citric acid cycle function. NADH is required to transform pyruvate to lactate (see Section extended chain) and one additional methyl group.

53. (a) Because cholesterol is water insoluble, it is commonly found associ13-1), resulting in a decrease in the cellular [NADH]/[NAD ] ratio. NADH ated with other lipids in cell membranes. Only an integral membrane is required to convert acetoacetate to -hydroxybutyrate (see Reaction 4 in protein would be able to recognize cholesterol, which has a small OH Figure 17-16); therefore, the equilibrium of this reaction favors acetoacetate head group and is mostly buried within the lipid bilayer. when cellular NADH concentrations are diminished. [From Saudubray, (b) Proteolysis releases a soluble fragment of the SREBP that can travel J. M., Marsac, C., Charpentier, C., Cathelineau, L., Besson Leaud, M., and from the cholesterol-sensing site to other areas of the cell, such as the Leroux. J. P. Acta Pædiat. Scand. 65, 717–724 (1976).] nucleus.

45. All cells can obtain glycerol-3-phosphate from glucose because glucose (c) The DNA-binding portion of the protein might bind to a DNA enters glycolysis to form dihydroxyacetone phosphate, which is subsequently sequence near the start of certain genes to mark them for transcription. transformed to glycerol-3-phosphate via the glycerol-3-phosphate dehydrogeIn this way, the absence of cholesterol could stimulate the expression of nase reaction.

Cells capable of carrying out gluconeogenesis, such as liver proteins required to synthesize or take up cholesterol. cells, can transform pyruvate to dihydroxyacetone phosphate. (Interestingly, 54. (a) HDL remove excess cholesterol from tissues and transport it back adipoto the liver. This prevents the accumulation of cholesterol in vessel walls cytes able do not carry out gluconeogenesis but do express PEPCK and are thus that leads to atherosclerosis. to convert pyruvate to glycerol-3-phosphate.)

46. The experimental results indicate that HMG-CoA reductase is regulated by phosphorylation. In normal cells, delivery of LDL particles results in an increase in cellular cholesterol, which stimulates a kinase that phosphorylates

Ch 18 Solutions

84

(b) HDL level alone does not indicate the risk of developing atheroscle- 8. (a) rosis, since the level of LDL, the activity of the LDL receptor, and other Glutamine synthetase factors such as smoking or vessel wall injuries resulting from infection (more active) can all infl uence the likelihood of developing the disease.

CH2 55. (a) Insufficient apolipoprotein B-100 affects the production of ATP LDL. Without enough of the protein portion of the lipoprotein, these individuals are less able to export triacylglycerols and other lipids from the liver in the form of LDL, and the lipids remain in the liver.

OH (b) LDL are a major vehicle for cholesterol transport from the liver to other tissues, so low levels of LDL would cause hypocholesterolemia.

Glutamine synthetase (less active) CH2

O

56. Chylomicrons are produced by intestinal cells to package dietary lipids for delivery to the rest of the body. Fat-soluble vitamins such as vitamin A are also transported via chylomicrons. Without chylomicrons, the body lacks an effi cient way to distribute vitamin A obtained from the diet.

(a)

O

H3N

6. (a) The low K M of glutamine synthetase for ammonium ions ensures

results in the effective uptake of glutamine from the circulation. Another strategy used by cancer cells is upregulation of the glutamine synthetase enzyme, which converts glutamate to glutamine. Glutamine analogs (compounds that are structurally similar to glutamine) could be used to either block the transport proteins or to bind to the enzyme and interfere with substrate binding).

C

CH2

Glu O

O

H

H

OH

OH

(b) HN

O C

CH

Glu O -Ketoglutarate

CH2

CH CH2

(d) CH CH2

10. (a) CH CH2

C

CH CH3

(c) H3N Glu

CH CH2 CH2 COO

NH 2

NH2 O C

Glu O

COO C

-Ketoglutarate

O

CH2 OH

O C

Glu O

O C

COO O

C -Ketoglutarate

Glu

CH2

COO C

O -Ketoglutarate

COO

(b) H 3N

O

CH2

NH NH 2

OH

Ala

C

CH2

(c)

Asp

COO

CH2

NH2

H 3N

H

CH2 C

H 3N

O

CH2 NH

H 3N

COO C

-Ketoglutarate

Glu 1 ATP 1 NH14 S Gln 1 Pi 1 ADP Gln1a-ketoglutarate1NADPHS2 Glu1NADP1 Glu1a-keto acidΔa-ketoglutarate1amino acid

that the fi rst method (a) will be used when the concentration of available ammonia is low. The second method (b) will be used when the concentration of ammonia is high. (b) Incorporation of one mole of ammonia into an amino acid using the fi rst method (a) costs one mole of ATP, whereas ATP is not required when the prokaryotic cell uses the second method (b). The prokaryotic cell is at a disadvantage when the ammonia concentration is low because energy must be expended in order to synthesize amino acids under these conditions.

7. Cancer cells upregulate glutamine membrane transport proteins, which

CH2

Adenine H

(b) -Ketoglutarate would inhibit adenylylation, preserving the enzyme in its unmodifi ed, active form. Transamination reactions involving ketoglu- tarate produce glutamate, a reactant for the glutamine synthetase reaction.

3. Leghemoglobin, like myoglobin, is an O2-binding protein. Its presence decreases the concentration of free O2 that would otherwise inactivate the bacterial nitrogenase.

4. Bacterial nitrogenase is inactivated by oxygen, so nitrogen fi xation takes place in cells that lack Photosystem II, the complex where water is oxidized to form molecular oxygen (see Section 16-2).

NH141ATP1NADPH1a-keto acidS ADP1P1NADP11amino acid (b) a-ketoglutarate 1 NH14 1 NAD(P)H Δ Glu1HO1NAD(P)1 2 Glu1a-keto acidΔa-ketoglutarate1amino acid a-keto acid1NAD(P)H1NH14Δ NAD(P)11amino acid1H2O

O

H

Chapter 18 1. The ATP-induced conformational change must decrease the E, from 0.29 V to about 0.40 V. The decrease in reduction potential allows the 9. protein to donate electrons to N2, since electrons fl ow spontaneously from a substance with a lower reduction potential to a substance with a higher reduction potential. Without the conformational change, nitrogenase could not reduce N2.

2. Alfalfa is a leguminous plant (see Fig. 18-1) that harbors nitrogen-fi xing bacteria in its root nodules. Crop growing depletes the soil of fi xed nitrogen; planting alfalfa every few years allows the associated nitrogen-fi xing bacteria to replenish the nitrogen supply in the soil and decreases the need for nitrogenbased fertilizers.

5. (a)

O P O

PPi

O

CH2 COO

O C

Glu -Ketoglutarate

O C

COO C

O

Asp Oxaloacetate

O Pyruvate

CH3 COO C

O

Oxaloacetate

O

CH2 CH2 COO

-Ketoglutarate

Ch 18 Solutions The products are all intermediates of the citric acid cycle (oxaloacetate and 17.

-ketoglutarate) or closely associated with the citric acid cycle (pyruvate).

11. (a) Leucine

(b) Methionine (c) Tyrosine (d) Valine

12.

E

(CH)2 4

NH2 H 3C

H O

3-Phosphoglycerate

C C

O2 3PO

N

transamination

Ala Gly

Ser

Cys

H

C H H

Pyruvate

85

Oxaloacetate

COO

-Ketoglutarate

H O

transamination

Glu

Gln synthetase

transamination

Gln

Asp

Asn synthetase

Asn

N H O C

E H

(CH2)4

H

N

NH 3

H

H3C

C C

O2 3PO

COO H O

Glu

Arg

Pro

18. Evolution has shaped the biosynthetic pathway to perform under the conditions where its activity is most needed—when the cognate amino acids are in short supply. If the enzymes contained large numbers of Cys and Met residues, then a scarcity of these amino acids would hinder the production of the enzymatic machinery to synthesize more of them.

19. Cysteine is the source of taurine. The cysteine sulfhydryl group is oxidized to a sulfonic acid, and the amino acid is decarboxylated.

20. (a) The sulfonamide is a structural analog of the p-aminobenzoate group of folate.

N H

O H2N

C

O

13. In the glutamine synthetase reaction, ATP donates a phosphoryl group to p-Aminobenzoate glutamate, which is then displaced by an ammonium ion, pro- ducing glutamine and phosphate. The ammonium ion is the nitrogen source. The (b) Mammals obtain folate from their diet and lack the enzymes necesasparagine synthetase reaction also requires ATP as an energy source, but the sary to synthesize folate. They are therefore not affected by the drug. nitrogen donor is glutamine, not an ammonium ion. As- partate is converted 21. If an essential amino acid is absent from the diet, then the rate of proto asparagine, and the glutamine becomes glutamate after donating an amino tein synthesis drops signifi cantly, since most proteins contain an assortment of group. ATP is hydrolyzed to AMP and pyro- phosphate instead of ADP and amino acids, including the defi cient one. The other amino acids that would norphosphate as in the glutamine synthetase reaction. mally be used for protein synthesis are therefore broken down and their nitrogen excreted. The decrease in protein synthesis, coupled with the normal turnover of body proteins, leads to the excretion of nitrogen in excess of the intake.

22. (a) Collagen contains a high proportion of Gly and Pro residues and 14. (a) relatively few of the other amino acids (Section 5-2). Therefore, it does O O NH 4 not provide a good mix of amino acids.

H3N CH C O H3N CH C O (b) Because collagen is a protein, it supplies amino acids that can be ATP AMP PPi used as metabolic fuels or converted to carbohydrates, lipids, or nucleoCH2 CH2 tides. Sucrose (glucose and fructose) is a metabolic fuel, but its carbons C O C O cannot be used to synthesize amino acids or nucleotides unless a source of amino groups is also provided.

O NH 2

23. (a) The sigmoidal shape of the velocity versus substrate concentration plot indicates that threonine deaminase binds its substrate in a positively cooperative manner. As the threonine concentration increases, threo(b) The above reaction shows NH 4 rather than glutamate as the nitronine binds to the enzyme with increasing affi nity. gen donor. ATP is hydrolyzed to yield ATP and PPi. This is similar to the glutamine synthetase reaction with the exception that ATP is hydro- (b) Isoleucine is an allosteric inhibitor of threonine deaminase and binds to the T form of the enzyme. Velocity decreases by about 15%, but the lyzed to ADP and Pi (see Solution 13). nearly 10-fold increase in K M is more dramatic. The decrease in velocity and increase in K M indicate that isoleucine, the end product of the path15. Phosphoglycerate dehydrogenase converts 3-phosphoglycerate to way, acts as a negative allosteric inhibitor of the enzyme that catalyzes an 3-phosphohydroxypyruvate, which then undergoes transamination with early, committed step of its own synthesis. The velocity versus substrate glutamate to yield 3-phosphoserine and -ketoglutarate, a citric acid cycle intermediate. The additional -ketoglutarate allows greater fl ux through the concentration curve obtained for threonine deaminase in the presence cycle to meet the cell’s energy needs. [From Possemato, R. et al., Nature 476, of isoleucine has greater sigmoidal character, which means that binding of threonine is even more cooperative in the presence of the inhibitor.

346–350 (2011).] (c) Valine stimulates threonine deaminase by binding to the R form.

16. (a)H The maximal velocity is somewhat increased, but the K M is decreased, CH2 COO 3C NH2 indicating that the threonine substrate has a higher affi nity for the (b) The conversion of sarcosine to glycine requires the methyl-group enzyme in the presence of valine. The cooperative binding of threodonor tetrahydrofolate, which contains folate and will be scarce when nine to threonine deaminase is abolished in the presence of valine, however, as indicated by the hyperbolic shape of the curve. folate is in short supply.

Aspartate

Asparagine

86

Ch 18 Solutions

24. (a) Lysine might regulate its own synthesis by acting as an allosteric inhibitor of dihydropicolinate synthase. This enzyme catalyzes the fi rst unique reaction in a series of eight reactions that lead to lysine. (b) Using the same reasoning, methionine might acts as a feedback inhibitor for acyltransferase. Methionine might also regulate the activity of homoserine dehydrogenase, another major branch point of the pathway, even though this enzyme is also involved in threonine biosynthesis. (c) Similarly, threonine might act as a feedback inhibitor for homoserine kinase and might also inhibit homoserine dehydrogenase.

34. 3-Methylhistidine cannot be reused to synthesize new proteins because it is a modifi ed amino acid for which there is no corresponding codon. The concentration of 3-methylhistidine offers only an approximation of the rate of muscle degradation because, although actin is a major component of muscle cells, it is found in nonmuscle cells as well (see Section 5-2).

35. Pyruvate can be transaminated to alanine, carboxylated to oxaloacetate, or oxidized to acetyl-CoA to enter the citric acid cycle. -Ketoglutarate, succinyl-CoA, fumarate, and oxaloacetate are all citric acid cycle intermediates; they can also all enter gluconeogenesis. Acetyl-CoA can 25. In an autoimmune disease, the body’s own white blood cells become activated to mount an immune response that leads to pain, infl ammation, and enter the citric acid cycle, be converted to acetoacetate, or be used for tissue damage. The activity of the white blood cells, which proliferate rapidly, fatty acid synthesis. Acetoacetate is a ketone body and can be converted to can be diminished by methotrexate, as occurs in rapidly dividing cancer cells. acetyl-CoA for the citric acid cycle or fatty acid synthesis.

36. The ketogenic amino acids are broken down to acetyl-CoA or acetoacetate, which can be converted to acetyl-CoA. For glucogenic amino acids, dTMP and dihydrofolate. DHFR can convert DHF back to THF to participate the carbon skeletons are broken down to either pyruvate (which is converted in additional rounds of dTMP synthesis. Because DHFR can catalyze a reaction to acetyl-CoA by the pyruvate dehydrogenase complex) or a citric acid cycle each time TS catalyzes a reaction, it makes sense for the enzymes to be closely intermediate (which can be converted to phosphoenolpyruvate by the reaclinked: The enzymes can be present in equal amounts and the DHF product of tions of gluconeogenesis; phosphoenolpyruvate is the precursor of pyruvate the TS reaction can be effi ciently channeled from one active site to the other. and therefore of acetyl-CoA).

27. (a) ADP and GDP both serve as allosteric inhibitors of ribose phosphate 37. Threonine catabolism yields glycine and acetyl-CoA. The acetyl-CoA pyrophosphokinase. is a substrate for the citric acid cycle, which ultimately provides ATP for (b) PRPP, the substrate of the amidophosphoribosyltransferase, stimulates the rapidly dividing cell. Glycine is a source of one-carbon groups, which the enzyme by feed-forward activation. AMP, ADP, ATP, GMP, GDP, and become incorporated into methylene-tetrahydrofolate via the glycine cleavGTP are all products and inhibit the enzyme by feedback inhibition. age system. THF delivers one-carbon groups for the synthesis of purine 28. (a) A 10-fold increase in the KM for APRT results in decreased affi nity nucleotides and for the methylation of dUMP to produce dTMP; nucleoof the enzyme for one of its substrates. The result is the accumulation of tides are needed in large amounts in rapidly dividing cells. [From Wang, adenine, which is oxidized to dihydroxyadenine and forms kidney stones.

J., Alexander, P., Wu, L., Hammer, R., Cleaver, O., and McKnight, S. L., (b) One way to treat this condition is to administer a compound that Science 325, 435–439 (2009).] would bind to xanthine dehydrogenase and prevent adenine from binding. This would prevent the conversion of adenine to dihydroxyadenine.

38. (a) 29. Inhibiting HGPRT would block production of IMP, which is a precurNH 3 26. TS uses methylene-tetrahydrofolate to methylate dUMP to produce

sor of AMP and GMP. In order to be an effective drug target, HGPRT must be essential for parasite growth; that is, the parasite cannot synthesize its own purine nucleotides de novo but instead relies on salvage reactions using the host cell’s hypoxanthine.

30. Phosphoribosyl pyrophosphate (PRPP) is a reactant in the salvage reactions involving IMP and GMP, so if these reactions cannot occur, the PRPP that would normally be used in the salvage reactions would accumulate.

PRPP stimulates amidophosphoribosyltransferase by feed-forward activation (see Problem 27), which accelerates the synthesis of purine nucleotides and increases the concentration of their degradation product, uric acid.

31. Lymphocytes that have not fused with a myeloma cell are unable to use the de novo synthetic pathway because it is blocked. These cells are still able to use the HGPRT salvage pathway, but the cells will not survive beyond 7–10 days. Myeloma cells cannot survive in HAT medium because the aminopterin blocks the de novo pathway and these cells lack HGPRT and cannot use the salvage pathway. Only hybridomas that result from the fusion of a lymphocyte (which can carry out the salvage pathway) and a myeloma cell (which can divide in culture indefi nitely) will survive in HAT medium.

32. 5-Fluorouracil structurally resembles uracil and competitively inhibits thymidylate synthase. When the enzyme is inhibited, dUMP is not converted to dTMP and there is no subsequent phosphorylation to dTTP, so the concentration of dTTP in the cells declines. In the presence of the inhibitor, dUMP accumulates and is converted to dUTP. The concentration of dUTP in cells is normally low because thymidylate synthase acts quickly on dUMP to convert it to dTMP. Cancer cells die in the presence of 5-fl uorouracil because the cells lack the dTTP required for DNA synthesis. Normal cells are not affected because DNA synthesis is not as rapid as in cancer cells.

33. (a) Arginine residues are converted to citrulline residues by a process of deamination (water is a reactant and ammonia is a product). Note that free citrulline produced by the urea cycle or in the generation of nitric oxide is not incorporated into polypeptides by ribosomes since there is no codon for this nonstandard amino acid. (b) The nonstandard amino acid citrulline is not normally incorporated into polypeptides, so its presence appears foreign to the immune system, increasing the risk of triggering an autoimmune response.

CH3

CH2

Isoleucine

COO

CH

CH

CH3

1 transaminase O

CH3

CH

CH2

C

COO

CH3

2

branched-chain -keto-acid dehydrogenase

O CH3

CH2

CH

C

SCoA 3

CH3

acyl-CoA dehydrogenase

O CH3

CH

C

C

SCoA enoyl-CoA hydratase

CH34 H

O

C

CH3

CH

C

5 hydroxyacyl-CoA dehydrogenase

OH CH3 O

O

C

CH3

SCoA

CH CH3

C

SCoA 6 thiolase O

O CH3

C

SCoA

Acetyl-CoA

CH3

CH2

C

Propionyl-CoA

SCoA

Ch 18 Solutions (b) Reaction 2 is analogous to the pyruvate dehydrogenase reaction. The 44. (a) dehydrogenase catalyzes the release of carbon dioxide and the formation of a high-energy bond with coenzyme A. (c) Reaction 3 is analogous to the acyl-CoA dehydrogenase enzyme of fatty acid biosynthesis.

Hydrogens are removed from the and car- bons of the substrate.

39. The fate of propionyl-CoA produced upon degradation of isoleucine is identical to that of propionyl-CoA produced in the oxidation of odd-chain fatty acids (see Fig. 17-7). Propionyl-CoA is converted to (S )-methylmalonylCoA by propionyl-CoA carboxylase. A racemase converts the (S )-methylmalonyl-CoA to the (R ) form. A mutase enzyme converts the (R )-methylmalonyl-CoA to succinyl-CoA, which enters the citric acid cycle.

40.

Leucine

H3C

OC

NH 3 C H

CH

CH2

CH3

1 transaminase

COO

OOC

CH2

O

CH

CH2

CH3

2

C

SCoA

6 lyase

COO

branched chain -keto acid dehydrogenase

C

Acetyl-CoA

Acetoacetate

O H3C

H3C

CH3

OH OOC

CH2

C

87

O CH2

C

COO

Phenylpyruvate

(b) A tetrahydrobiopterin defi ciency prevents the conversion of phenylalanine to tyrosine in the phenylalanine catabolic pathway.

Consequently, phenylalanine accumulates and undergoes transamination to phenylpyru- vate, which is excreted. (c) The growing child still requires some phenylalanine for growth. The low-phenylalanine diet should provide enough phenylalanine for growth but should not exceed amounts needed for growth since the excess cannot be metabolized due to the lack of phenylalanine hydroxylase. (d) The artifi cial sweetener aspartame consists of a methylated Asp–Phe dipeptide. (The C-terminal carboxyl group is methylated.) Aspartame is broken down to Asp, Phe, and methanol. Since aspartame is a source of phenylalanine, patients with PKU should not use this product, and physicians should advise their patients to check labels carefully and to avoid use of this product. (e) Phenylalanine is the precursor of tyrosine. If the phenylalanine intake is low, supplemental tyrosine may be needed.

O CH2

C

SCoA

45. Persons with NKH lack a functioning glycine cleavage system.

This is the major route for the disposal of glycine, and in its absence, glycine accumulates in body fl uids. The presence of excessive glycine, a H2O CH3 3 acyl-CoA dehydrogenase neurotransmitter, in the cerebrospinal fl uid explains the effects on the 5 hydratase nervous system.

O O 46. Proteins, a polymeric form of amino acids, could be considered as a stor4 H3C C CH C SCoA OOC CH2 C CH C SCoA carboxylase age depot for amino acids, since the proteins can be degraded to release amino acids for use as metabolic fuels. However, proteins have functions CH3 CH3 ATP other than fuel storage, which is not the case for glycogen and triacylglycerols (although triacylglycerols also function as thermal insulation 41. (a) Acetyl-CoA can enter the citric acid cycle if suffi cient oxaloacetate is available; if not, excess acetyl-CoA is converted to ketone in some species). bodies. Leucine differs from isoleucine in that leucine is 47. The glutamate dehydrogenase reaction converts -ketoglutarate to exclusively ketogenic, generating a ketone body and a ketone glutamate. In the presence of excess ammonia, -ketoglutarate in the brain body precursor upon degradation. Isoleucine produces propionyl- could be depleted, diminishing fl ux through the citric acid cycle.

CoA along with acetyl-CoA; the former can be converted to 48. (a) GTP (representing a high level of cellular ATP) inhibits glutamate dehydrogenase to favor glutamate formation. succinyl-CoA (see Solu- tion 39) and then to glucose. Thus, (b) ADP (a signal for low cellular energy) activates glutamate dehydroisoleucine is glucogenic as well as ketogenic. (b) Persons defi cient genase to favor synthesis of -ketoglutarate so that citric acid cycle fl ux in HMG-CoA lyase are unable to degrade leucine and must can increase. restrict this amino acid in their diets. A low-fat diet is also rec(c) NADH inhibits glutamate dehydrogenase to favor glutamate formaommended because this same enzyme is involved in the tion, since a high level of NADH means that the cell does not need the production of ketone bodies (see Reaction 3 in Fig. 17-16). A diet citric acid cycle to produce additional reduced cofactors. high in fat would generate a high concentration of acetyl-CoA, O

H3C

CH

CH2

C

SCoA

CH3

which would not be able to be converted to ketone bodies in the 49. Glutamine is degraded to glutamate with the release of ammonia, which absence of this enzyme. can bind protons to form NH 4. This helps counteract the acidosis that oc42. The missing enzyme is the branched chain -keto-acid dehydrogenase curs when the concentration of acidic ketone bodies in the blood increases (see Fig. 18-11). This enzyme is common to the catabolic pathways of valine, during starvation. 50. One nitrogen atom is derived from ammonia that is isoleucine (see Problem 38), and leucine (see Problem 40) and explains why incorporated into the corresponding -keto acids of these amino acids accumulate. Because the carbamoyl phosphate for entrance into the urea cycle. The other nitrogen patients are unable to break down branched-chain amino acids, they should atom comes from aspartate, which serves as a substrate in the argininosucciconsume a diet that contains only enough of these amino acids to meet the nase reaction. Ultimately, both nitrogen atoms that appear in urea originated needs of growth and development. from excess dietary protein. 51. The reaction of serine and homocysteine to 43. (a) Insulin inhibits the enzyme, whereas glucagon stimulates the enzyme. produce cysteine and (b) In the presence of Phe, the activity of the enzyme increases dra- -ketobutyrate, the catabolism of the pyrimidine breakdown products matically, more so in the presence of glucagon. Phe acts as an -ureidopropionate and -ureidoisobutyrate, the reaction catalyzed by asallosteric activator of phenylalanine hydroxylase and plays a role in paraginase, the conversion of serine to pyruvate, the conversion of cysteine converting the enzyme from the inactive dimeric form to the active to pyruvate, the glycine cleavage system, and the glutamate dehydrogenase tetrameric form. (c) The incorporation of phosphate into the active reaction all generate free ammonia. form of phenylalanine hydroxylase indicates that the enzyme is 52. Glutamate dehydrogenase, glutamine synthetase, and carbamoyl phosregulated by phosphorylation as well as allosteric control. Glucagon signaling must lead to phosphoryla- tion of the enzyme. (d) phate synthetase can potentially “mop up” NH14.

Phenylalanine hydroxylase is most active when the glucagon con- 53. (a) A urea cycle enzyme defi ciency decreases the rate at which nitrogen centration is high, corresponding to the fasting state. Under these circumstances, phenylalanine can be degraded to produce acetoacetate (a ketone body) and fumarate (which can be converted to glucose); both compounds provide necessary resources in the fasting state.

can be eliminated as urea. Since the sources of nitrogen for urea synthesis include free ammonia, low urea cycle activity may lead to high levels of ammonia in the body. (b) A low-protein diet might reduce the amount of nitrogen to be excreted.

88

Ch 19 Solutions

54. Excretion of the nitrogen-containing amino acids glycine and glutamine (an important source of amino groups) helps reduce the amount of amino groups in the body. With Ucephan treatment, nitrogen is excreted as phenylacetylglutamine or hippuric acid.

O

O C

O

C Glycine

Benzoate

CH2

Hippuric acid

NH

CH2

COO

Glycine residue

cellular nitrogen level. When nitrogen levels are adequate, glutamine synthetase activity is inhibited. (c) The fi rst batch of glutamine synthetase was adenylylated and was in its less active form and more susceptible to inhibition by allosteric mod- ulators. The enzyme was in the adenylylated form because the growth medium contained glutamate, and under these conditions, glutamine synthetase activity is inhibited. The second batch of enzyme was not adenylylated and was in its fully active form, which is what would be expected when NH14 is the sole nitrogen source. Because the second batch of enzyme was not adenylylated, it was not susceptible to inhibition by the allosteric modulators. [From Stadtman, E. R., J. Biol. Chem. 276, 44357–44363 (2001).]

62. (a) H. pylori urease converts urea to NH3 and CO2. The ammonia has a

COO Glutamate

Phenylacetate

O CH2

C

Phenylacetylglutamine

CH COO

NH

pK of 9.25, so it combines with protons to produce NH14. The resulting decrease in hydrogen ion concentration helps the bacteria maintain an intracellular pH higher than the environmental pH. (b) Urease on the cell surface increases the pH of the fl uid surrounding the cell, creating a more hospitable microenvironment for bacterial growth.

(CH2)2 COO

Chapter 19 1. The two main metabolites at the “crossroads” are pyruvate and acetyl55. Adding arginine, the product of the argininosuccinase reaction, would CoA. Pyruvate is the main product of glycolysis. It can be convert- ed to acetyl-CoA by pyruvate dehydrogenase. Pyruvate is produced from a increase fl ux through the urea cycle. transamination reaction involving alanine. Pyruvate can be carboxylated to 56. An individual consuming a high-protein diet uses amino acids as metabolic fuels. As the amino acid skeletons are converted to glucogenic or oxaloacetate for gluconeogenesis. Acetyl-CoA is a product of fatty acid ketogenic compounds, the amino groups are disposed of as urea, leading to degradation and one of the reactants in the citric acid cycle. Acetyl-CoA is a increased fl ux through the urea cycle. During starvation, proteins (primarily product of the degradation of ketogenic amino acids. Acetyl-CoA can be from muscle) are degraded to provide precursors for gluconeogenesis. Ni- used to synthesize fatty acids and ketone bodies.

Glutamate residue

trogen from these protein-derived amino acids must be eliminated, which demands a high level of urea cycle activity.

57. (a) The condensation of ammonia and -ketoglutarate produces the amino acid glutamate. (b) The reverse reaction, in which glutamate is deaminated to produce -ketoglutarate, replenishes the citric acid cycle intermediate.

58. Glutaminase:

glutamine S glutamate 1 NH14 Glutamate dehydrogenase: glutamate1NAD(P)1S a-ketoglutarate1NH141NAD(P)H net: glutamine1NAD(P)1S a-ketoglutarate12 NH141NAD(P)H 59. The amino acids released by protein degradation are used as metabolic fuels. The amino groups are removed and the nitrogens are eventually excreted in the form of urea. The carbon skeletons can be completely catabolized to CO2 by muscle cells to produce ATP via the citric acid cycle and oxidative phosphorylation, or they can be partially broken down and transported to the liver to be used in gluconeogenesis, which indirectly supplies the muscle cells with glucose.

60.

COO

H3C

NH2

C

H

CH2 COO NMDA

61. (a) UTase is stimulated by -ketoglutarate and ATP, both substrates (either directly or indirectly) of glutamine synthetase. UTase is inhibited by glutamine and inorganic phosphate, both products of the reaction.

UR is stimulated by glutamine, a product of the glutamine synthetase reaction. High concentrations of glutamine decrease the activity of glutamine synthetase. (b) Histidine, tryptophan, carbamoyl phosphate, glucosamine-6phosphate, AMP, CTP, and NAD are all end products of glutamine metabolic pathways. Alanine, serine, and glycine refl ect the overall

2. Glucose-6-phosphate is the product of the hexokinase reaction and is produced when glucose is transported into cells. G6P can then continue along the pathway of glycolysis to produce ATP and biosynthetic intermediates. G6P can be reversibly converted to glucose-1-phosphate for incorporation into glycogen. Glycogen degradation produces glucose-1-phosphate, which is isomerized to glucose-6-phosphate. G6P is also the product of gluconeogenesis. Glucose-6-phosphatase catalyzes the removal of the phosphate group to produce glucose, which can leave the cell. Glucose-6-phosphate can also enter the pentose phosphate pathway to produce NADPH and ribose.

3. The Na,K-ATPase pump requires ATP to expel Na ions while im- porting K ions, both against their concentration gradients. Inhibition by oubain reveals that the brain devotes half of its ATP production solely to power this pump. Because the brain does not store much glycogen, it must obtain glucose from the circulation. Glucose is oxidized aerobically in order to maximize ATP production.

4. Glucose enters the red blood cell via specifi c transport proteins and undergoes glycolysis to produce pyruvate. The next step is the conversion of pyruvate to lactate with concomitant production of NAD . The NAD is used in the glyceraldehyde-3-phosphate dehydrogenase reaction, and its production is essential for the continuation of the glycolytic pathway.

Despite the presence of oxygen in the red blood cell, glucose is oxidized anaerobically due to the lack of mitochondria. Two ATP are produced per glucose molecule.

5. (a) Phosphorolytic cleavage glycogen(n residues) Pi S glycogen(n21 residues) glucose-1-phosphate Hydrolytic cleavage glycogen(n residues) H2O S glycogen(n21 residues) glucose (b) Phosphorolytic cleavage yields glucose-1-phosphate, which is negatively charged due to its phosphate group and cannot exit the cell via the glucose transporter. In addition, glucose-1-phosphate can be isomerized to glucose-6-phosphate (and can enter glycolysis) without the expenditure of ATP. Hydrolytic cleavage yields neutral glucose, which can leave the cell via the glucose transporter. Converting free glucose to glucose-6phosphate so that it can enter glycolysis requires the expenditure of ATP in the hexokinase reaction.

Ch 19 Solutions 89 14. Increased levels of citrulline indicate that the cytosolic argininosuccinate 6. (a) The reaction is probably a near-equilibrium reaction because the reactants and products have the same total number of phosphoanhydride synthetase reaction in the urea cycle (see Section 18-5) is not occurring norbonds. (b) In highly active muscle, ATP is rapidly converted to ADP. mally. This reaction requires aspartate as a reactant in addition to citrulline.

Adenyl- ate kinase catalyzes the conversion of two ADP to ATP and The accumulation of citrulline may occur because there is a shortage of asparAMP as a way to generate additional ATP to power the actin–myosin tate. A defi ciency of pyruvate carboxylase results in a decreased concentration of oxaloacetate that could otherwise be transaminated to aspartate. Hyperamcontractile mechanism. monemia is the result of urea cycle impairment, since ammonia is not being converted to urea for excretion by the kidneys. [From Coude, F. X., Ogier, H., 7. (a) The fumarate is used to boost the level of citric acid cycle intermediMarsac, C., Munnich, A., Charpentier, C., and Saudubray, J. M., Pediatrics ates to increase fl ux through this pathway and increase ATP synthesis by 68, 914 (1981).] oxidative phosphorylation.

15. (a) The antibiotics affect the composition of the intestinal microbiome, (b) Transamination of aspartate would produce oxaloacetate, but anfavoring species that allow the animal to gain more weight from a given other -keto acid, such as pyruvate or -ketoglutarate, would become amount of food. an amino acid in the process. Consequently, there would be no net (b) The presence of antibiotics selects for the growth of microbial spechange in the level of citric acid cycle intermediates. cies that can resist the antibiotic. The result is a large number of bacteria 8. When AMP concentrations are high, the energy status of the cell is low, with antibiotic-resistance genes that can be transferred to other species, and the glycolytic pathway is needed to provide ATP for the cell. AMP de- including those that cause disease in humans. aminase acts on the increased concentrations of AMP to produce ammonium ions. These, along with AMP, stimulate the activities of phosphofructokinase 16. Fatty acid oxidation must be a major source of metabolic free energy and pyruvate kinase and increase the activity of the glycolytic pathway as a during metamorphosis. 17. The increase in glycolytic activity and the whole. Factors that infl uence the activity of adenosine deaminase will thus decrease in citric acid cycle also activity indicate that aerobic metabolism is decreased. Instead, metabolic inindirectly affect the rate of glycolysis. [From Yoshino, M., and Murakami, K., termediates are funneled into pathways that will ultimately produce the gluJ. Biol. Chem. 260, 4729–4732 (1985).] 9. Glycolysis produces two moles of cose and amino acids required for milk production. The main source of these intermediates is the muscle. Amino acids released from the muscle travel via ATP per mole of glucose. Synthesis of the bloodstream to the liver, where they are converted to glucose via one mole of glucose via gluconeogenesis costs six moles of ATP. Therefore, the cost of running one round of the Cori cycle is four ATP. The extra ATP gluconeogenesis. Rather than returning to the muscle (which would occur during the is generated from the oxidation of fatty acids in the liver.

Cori cycle), the glucose and amino acids are diverted to the mammary gland 10. The lactate dehydrogenase reaction, which reduces pyruvate to lactate, for milk production. [From Khula, B., Nuernberg, G., Albrecht, D., Goers, occurs with concomitant oxidation of NADH to NAD . NAD serves as a S., Hammon, H. M., and Metges, C. C., J. Proteome Res., 10, 4252–4262 reactant in the glyceraldehyde-3-phosphate reaction in glycolysis. If pyruvate (2011).] 18. (a) were the end product of glycolysis, all of the cellular NAD would become reduced to NADH and the glycolytic pathway would grind to a halt for lack CH 2OPO32 CH2OH of NAD .

ADP O O 11. During starvation, muscle proteins are broken down to produce glucoATP H H H H H HO H O H neogenic precursors. The amino groups of the amino acids are transferred to HO H H O H pyruvate via transamination reactions. The resulting alanine travels to the hexokinase OH OH H liver, H glucosewhich can dispose of the nitrogen via the urea cycle and produce H H from the alanine skeleton (pyruvate) and other amino acid skeletons. This glucose circulates not just to the muscles but to all tissues that need it, so the 2-Deoxy-D-glucose 2-Deoxy-D-glucose-6-phosphare metabolic pathway is not truly a cycle involving just the liver and muscles.

12. Plasma alanine would be elevated as well. The increased plasma con(b) Cancer cells utilize anaerobic metabolism even in the presence of centration of pyruvate would accelerate the operation of the glucose–alanine oxygen (the Warburg effect; Box 19-A). The cells obtain ATP mainly cycle. Pyruvate is taken up by the muscle and transaminated to alanine, from glycolysis, which explains why a glycolytic inhibitor interferes which is then released to the circulation to be taken up by the liver. Since with ATP production, whereas an inhibitor of electron transport plasma pyruvate levels are elevated, alanine levels would also be elevated. does not.

19. Because glucokinase is not saturated at physiological glucose concentra13. (a)oxaloacetate, Since pyruvate carboxylase thewould carboxylation pyruvatepyruto tions, it can respond to changes in glucose availability with an increase or dea defi ciency of catalyzes the enzyme result in of increased vate levels and decreased oxaloacetate levels. Some of the excess pyruvate crease in reaction velocity. Consequently, the entry of glucose into glycolysis would also be converted to alanine, so alanine levels would be elevated. and subsequent metabolic pathways depends on the glucose concentration. (b) Some of the excess pyruvate is converted to lactate, which explains Because hexokinase is saturated at physiological glucose concentrations, its rate does not change with changes in glucose concentration. why the patient suffers from lactic acidosis. Decreased oxaloacetate levels decrease the activity of the citrate synthase reaction, the fi rst step of20. Indirect evidence indicates that glucokinase functions as a glucose the citric acid cycle. This causes the accumulation of acetyl-CoA, which sensor, so increasing its activity might be benefi cial in treating diabetes. forms ketone bodies that accumulate in the blood to cause ketosis.

Allosteric activators could be designed to bind to the high-affi nity form of (c) A pyruvate carboxylase defi ciency results in decreased oxaloacetate glucokinase, increasing its catalytic activity even at low glucose concentralevels. Aspartate is formed by transamination of oxaloacetate, so astions. The result is an increased uptake of glucose, an increased rate of phospartate levels are decreased as well. Low levels of oxaloacetate decrease phorylation of glucose to glucose-6-phosphate, an enhanced capacity of the the levels of all the citric acid cycle intermediates due to the decreased liver to store glucose as glycogen, and a decrease in the circulating blood activity of the citrate synthase reaction. This results in lower levels of glucose concentration.

-ketoglutarate, which can be transaminated to glutamate. Therefore, glutamate levels are low, as are GABA levels, since GABA is produced 21. Insulin binding to its receptor stimulates the tyrosine kinase activity from glutamate (see Section 18-2). of the receptor. Proteins whose tyrosine residues are phosphorylated by (d) Acetyl-CoA stimulates pyruvate carboxylase activity. Adding acetyl- the receptor tyrosine kinase can then interact with additional compoCoA would allow the investigators to determine whether there was a slightnents of the signaling pathway. These interactions could not occur if a amount of pyruvate carboxylase activity that could be detected by adding tyrosine phosphatase removed the phosphoryl groups attached to the Tyr this activator. [From Stanbury, J. B., Wyngaarden, J. B., Fredrickson, D. S., residues.

Goldstein, J. L., and Brown, M. S., The Metabolic Basis of Inherited Disease, 22. Insulin stimulates uptake of glucose via GLUT4 receptors in adipopp. 196–198, McGraw-Hill Book Company, New York (1983).] cytes. Glucose is converted to glycerol-3-phosphate via glycolysis and the

90

Ch 19 Solutions

glycerol-3-phosphate dehydrogenase reaction. The glycerol-3-phosphate is used as the backbone of the triacylglycerol molecule.

23. Phosphorylation of glycogen synthase by GSK3 inactivates the enzyme so that glycogen synthesis does not occur. But when insulin activates protein kinase B, the kinase phosphorylates GSK3. Phosphorylated GSK3 is inactive and unable to phosphorylate glycogen synthase. In the dephosphorylated state glycogen synthase is active and glycogen synthesis can occur.

24. Phosphorylation and inactivation of GSK3 indirectly results in the activation of glycogen synthase as described in Solution 23. Promoting the storage of glucose as glycogen in muscle and liver reverses the symptoms of insulin resistance.

25. If the phosphorylation of GSK3 is blocked, GSK3 remains active and can phosphorylate glycogen synthase, rendering it inactive (see Solution 23).

This decreases the ability of the cell to take up glucose and convert it to a storage form and exacerbates insulin resistance (see Problem 24).

26. Ingesting large amounts of glucose stimulates the cells of the pancreas

(e) If the patient ingests glucose or fructose, blood sugar concentrations can be maintained at a high level. The muscles will thus be able to take up glucose or fructose and oxidize these sugars via glycolysis to yield the ATP required of active muscles. In this way, the muscles do not have to rely on stored glycogen as a fuel source.

29. Phosphorylase kinase is regulated by phosphorylation, which causes a conformational change that activates the enzyme. The enzyme is phosphorylated by protein kinase A (see Figure 19-11). The activity of the enzyme therefore depends somewhat on the G protein–coupled receptor pathway activated by either glucagon or epinephrine. But phosphorylase kinase is not fully active until calmodulin (a calcium-binding protein that is part of its structure) binds calcium and undergoes its own conformational change. The intracellular concentration of calcium increases when the phosphoinositide signaling system is activated (see Section 10-2), so the activation of phosphorylase kinase depends on this signaling pathway as well.

30. The increased demand for glucose is met by an increase in the rate of

to release insulin, which causes liver and muscle cells to use the glucose to gluconeogenesis in the liver, which is stimulated by glucagon in the fasted synthesize glycogen and causes adipose tissue to synthesize fatty acids. Insulin state. Other body tissues contribute by providing gluconeogenic substrates; also inhibits the breakdown of metabolic fuels. The body is in a state of rest- for example, protein breakdown in muscle provides amino acids. Catecholing and digestion and is not prepared for running. amines stimulate glycogenolysis in muscle, leading to increased production of lactate, which is a gluconeogenic precursor after being transported to the 27. (a) Stimulation of the rate of phosphate removal from glucose-6liver. Glucagon stimulates triacylglycerol breakdown in adipose tissue; the phosphate explains why the glucose concentration increased and the glycerol can serve as a gluconeogenic precursor, whereas the mobilized fatty glucose-6-phosphate concentration decreased. The phosphoenolpyruacids are used as substrates for oxidation by the muscle. [From Mitrou, P., vate concentration increased because fl ux from dihydroxyacetone phosRaptis, S. A., and Dimitriadis, G., Endocr. Rev., 31, 663–679 (2010).] phate to phosphoenolpyruvate increased (the glycolytic reactions are all 31. (a) AMPK increases the expression of GLUT4, since this will increase near equilibrium).

ATP production via glucose catabolism. (b) Removing the phosphate from glucose-6-phosphate is the fi nal step (b) AMPK decreases glucose-6-phosphatase expression, since this is of gluconeogenesis (and glycogenolysis) and allows glucose to leave the a gluconeogenic enzyme that would contribute to cellular ATP conliver. Increasing the rate of this step enhances the overall rate of glucosumption. neogenesis. At the same time, glycolysis is inhibited because the very fi rst step of glycolysis is the phosphorylation of glucose to glucose-632. (a) inhibit (b) inhibit (c) stimulate (d) stimulate 33. AMPK stimulates phosphate. If phosphate groups are continually detached, glucose-6pathways involved in ATP production, so the rates of phosphate cannot enter glycolysis and thus glycolysis is effectively inhibglycolysis and fatty acid oxidation increase. Cancer cells normally rely on anited. [From Ichai, C., Guignot, L., El-Mir, M. Y., Nogueira, V., Guigas, aerobic metabolism to meet their energy needs, so the stimulation of AMPK B., Chauvin, C., Fontaine, E., Mithieux, G., and Leverve, X. M., J. Biol. has Chem. 276, 28126–28133 (2001).] the effect of increasing aerobic metabolism, which increases the production of 28. (a) Normally, glucagon binds to cell-surface receptors on the liver, stim- reactive oxygen species. 34. Insulin activation of cAMP phosphodiesterase ulating adenylate cyclase to produce cAMP and activate protein kinase destroys cAMP so that it A, which subsequently activates glycogen phosphorylase via phosphorycan no longer activate protein kinase A, the enzyme that activates lation. Glycogen phosphorylase catalyzes the degradation of glycogen to phosphoryglucose, which is released into the bloodstream. Blood glucose concen- lase kinase, as a step toward activating glycogen phosphorylase and inactivattrations should rise shortly after an intravenous injection of glucagon. ing glycogen synthase. The result is that insulin promotes glycogen synthesis.

Glycogen degradation in the patient’s liver thus appears to be normal. (b) Glycogen metabolism in the liver appears to be normal, since glyco- 35. Overexpression of IRS-1 would increase the rate of translocation of gen content is normal and the patient’s response to the glucagon test is GLUT4 transporters to the cell surface. IRS-1 triggers the downstream actinormal. However, muscle glycogen is elevated, which indicates a defect vation of proteins such as the phosphatase enzyme that removes phosphate in muscle glycogen metabolism. Most likely glycogen synthesis is nor- groups from glycogen phosphorylase (which inactivates the enzyme) and mal, since the biochemical structure of the glycogen is normal. A defi - glycogen synthase (which activates the enzyme). The result is that glycogen ciency in the muscle glycogen phosphorylase enzyme is the most likely synthesis in the cultured muscle cells would increase. explanation.

36. The liver is a “glucose buffer” because it can absorb glucose when (c) Blood alanine concentrations normally increase as part of the glucose– glucose is plentiful and release glucose when glucose is scarce. The liver alanine cycle. Glucose-6-phosphate, the product of glycogenolysis, enters responds to hormones and the changing supply of glucose in order to glycolysis and produces pyruvate. Pyruvate undergoes a transamination maintain the blood glucose concentration at a relatively constant level of reaction to form alanine, which is released from the muscle. Alanine ,5 mM. If the blood glucose concentration becomes greater than 5 mM, then enters the liver, where the transamination reaction takes place in the liver takes up glucose and converts it to glucose-6-phosphate. This reacreverse to re-form pyruvate. Pyruvate can then enter gluconeogenesis, tion is catalyzed by glucokinase, whose K M for glucose is higher than that the resulting glucose product is released, and the cycle begins again. But the patient’s tissues cannot perform the glucose–alanine cycle because his of hexokinase and is in the range of the elevated blood glucose concentramuscles are unable to produce glucose-6-phosphate. Instead, the muscles tion. If the glucose concentration drops below 5 mM, the liver responds by take up alanine and use it as a fuel. Thus, plasma alanine levels in the degrading glycogen and releasing glucose to bring the glucose concentration back up to ,5 mM. patient decrease rather than increase. (d) Blood glucose concentrations are regulated by pancreatic hormones acting on the liver to stimulate glycogen synthesis or degradation, whatever is appropriate. Since the patient’s liver enzymes appear to function normally, his blood glucose concentration is properly regulated and he is neither hypo- or hyperglycemic.

37. These two enzymes are part of the gluconeogenic pathway. Their concentrations increase when dietary fuels are not available so that the liver can supply other tissues with newly synthesized glucose.

38. Several days into a fast, muscle and liver glycogen have been depleted.

In the absence of dietary glucose, the main source of endogenous glucose is gluconeogenesis. Citric acid cycle intermediates are used to synthesize oxaloacetate and then pyruvate, which enters the gluconeogenic pathway.

Catalytic amounts of citric acid cycle intermediates are required for proper

Ch 19 Solutions 91 functioning of the cycle, so when these intermediates are diverted to gluconeogenesis, the citric acid cycle cannot function properly.

39. If 3,000 g of fat are utilized at a rate of 75/g day, the fast can last for 40 days before death occurs.

40. During a prolonged fast, body protein is conserved as fats are mobilized to provide energy in the absence of food intake. Triacylglyercols are degraded to glycerol and fatty acids. The glycerol can serve as a gluconeogenic precursor, that and the fatty acids are degraded to acetyl-CoA to form ketone bodies are used by the heart and brain as a substitute for glucose. In this manner, body protein is spared and the chance for survival is increased.

41. After a few days of a diet low in carbohydrate, glycogen stores are depleted and the liver converts fatty acids to ketone bodies to be used as fuel for muscle and other tissues. Acetone is produced from the nonenzymatic decarboxylation of the ketone body acetoacetate. The relatively nonpolar acetone passes from blood capillaries into the lung alveoli, and its smell can be detected in exhaled breath.

42. Regular exercise maintains muscle mass so that muscle proteins are less likely to be broken down to provide amino acids as fuel, and the body uses its stored fat instead. Inactivity while dieting may promote loss of muscle mass instead of fat.

43. (a) stimulates is glucose uptake by skeletal muscle.inhibition of gly(b)Leptin Glycogenolysis inhibited, probably by direct

heart for oxidation, since the muscle and heart cannot synthesize their own fatty acids. This accounts for the increased blood levels of triacylglycerols. (d) Insulin stimulates the activity of ACC2 in the muscle cells of nor- mal mice, which promotes fatty acid synthesis and inhibits fatty acid oxidation (due to increased malonyl-CoA levels). The muscle cells in the knockout mice lack ACC2 and are not subject to insulin-mediated control. Fatty acid synthesis does not occur, malonyl-CoA levels do not rise, and fatty acid oxidation proceeds normally, even in the presence of insulin. (e) Knockout mice are leaner because their heart and muscle tissue cannot synthesize fatty acids, so triacylglycerols are mobilized to provide fatty acids for these tissues, as described in part (d). Knockout mice have a higher rate of fatty acid oxidation and a lower rate of synthesis, as described in part (c), which also accounts for their lower weight gain despite the increased caloric intake. (f) Molecular modeling techniques could be used to design a drug that inhibits the enzyme activity of ACC2 but not ACC1.

The drug would have to be targeted in such a way that it would be delivered to the mitochondrial matrix, where ACC2 is located.

[From Abu-Elheiga, L., Matzuk, M. M., Abo-Hashema, K. A. H., and Wakhil, S. J., Science 291, 2613–2616 (2001).]

48. (a) Acetate is converted to acetyl-CoA, which serves as a substrate for acetyl-CoA carboxylase and is converted to malonyl-CoA. Fatty acid synthase normally acts on malonyl-CoA for incorporation into fatty acids, but in the presence of the inhibitor C75, fatty acid synthesis does not occur. Therefore, radioactive malonate accumulates, and virtually no label appears in fatty acids. (b) C75 causes hypothalamic levels of NPY to decrease. This decreases appetite, which leads to weight loss. In this manner, C75 fools the body into thinking that it is in the fed state, when in fact it is being 44. (a) The cytochrome c content is high because of the large number starved. of mitochondria that allow brown adipose tissue to oxidize meta(c) Acetyl-CoA carboxylase (ACC) catalyzes the reaction that converts bolic fuels aerobically, funneling the reduced coenzymes through acetyl-CoA to malonyl-CoA. In the presence of the ACC inhibitor, the electron transport chain. Uncoupling oxidative phosphorylation malonyl-CoA cannot be produced and hepatic malonyl-CoA levels permits the energy of electron transport to be dissipated as heat (see would decrease. Subsequent administration of C75 would have no efBox 15-B). fect on the ACC inhibitor–treated animals because C75, a fatty acid (b) When subjects were exposed to cold, uptake of the labeled glucose synthase inhibitor, acts downstream of ACC. If malonyl-CoA inhibits into brown adipose tissue increased to provide reduced coenzymes for feeding, the ACC inhibitor–treated and C75-treated mice would have electron transport and thermogenesis. lower levels of malonyl-CoA, and feeding would not be inhibited, that 45. Both of these observations indicate a decrease in the ability of the mitois, the mice would eat normally. These experiments were carried out and chondrion to oxidize fatty acids. Carnitine acyltransferase transports fatty acids this is in fact what occurred. from the cytosol to the mitochondrial matrix, the site of oxidation. Products (d) When malonyl-CoA accumulates, long-chain fatty acyl-CoA molecules also accumulate in the cytosol. Malonyl-CoA inhibits carnitine of oxidation, the reduced coenzymes NADH and QH2, are reoxidized by the electron transport chain, with concomitant production of ATP. If fatty acid acyltransferase, which prevents translocation of fatty acyl-CoAs into the oxidation cannot occur, fatty acids will be used to synthesize triacylglycerols for mitochondrial matrix for oxidation. It is possible that long-chain fatty storage in adipose tissue instead. acyl-CoAs serve as signaling molecules to begin the pathway that suppresses appetite. [From Loftus, T. M., Jaworksy, D. E., Grehywot, G. 46. Endogenous fatty acid synthesis is required when dietary fatty acid L., Townsend, C. A., Ronnett, G. V., Lane, M. D., and Kuhajda, F. P., intake is insuffi cient. Stimulation of acetyl-CoA carboxylase ensures that Science 288, 2379–2381 (2000).] the body will have enough fatty acids in the absence of dietary lipids (although essential fatty acids will still be lacking under these circumstances). During starvation (and untreated diabetes, which is similar to 49. The drugs can activate the intracellular tyrosine kinase domains of the the starved state), body tissues do not have the resources to synthesize insulin receptor, bypassing the need for insulin to bind to the receptor. fatty acids, so acetyl-CoA carboxylase is inhibited. In the starved state, 50. (a) In the absence of the phosphatase, the intracellular domain of the infatty acids are mobilized to provide fuel to body tissues. sulin receptor remains phosphorylated longer. The IRS-1 substrate may 47. (a) Malonyl-CoA, the product of the acetyl-CoA carboxylase reaction also retain its phosphate group, enabling it to stimulate the intracellular (the committed reaction of fatty acid synthesis), inhibits simultaneous processes that allow glucose to be brought into cells via the GLUT4 fatty acid oxidation by inhibiting carnitine acyltransferase, an enzyme receptor. In the absence of phosphatase activity to turn off the signaling required for shuttling fatty acids into the mitochondrial matrix for pathway, the action of insulin is potentiated, and a normal physiological oxidation. This mechanism prevents the oxidation of newly synthesized effect is observed with a lower concentration of the hormone. fatty acids because they cannot enter the mitochondrion. (b) Injecting the mice with insulin would produce greater phosphoryla(b) In the absence of ACC2, heart and muscle are unable to synthesize tion of the insulin receptor and possibly IRS-1, since there is no phosfatty acids. This increases the demand on the liver to provide fatty acids phatase enzyme to remove the phosphate groups. for heart and muscle. In the knockout mice, liver glycogen is degraded (c) Compounds that inhibit the activity of the PTP-1B phosphatase to glucose, then oxidized to pyruvate and acetyl-CoA in order to provide might be effective in treating diabetes. A concern would be the specifi cacetyl-CoA for fatty acid synthesis. ity of these compounds, since unpredictable side effects would result if (c) Fatty acid levels decrease because of the lack of ACC2. Triacylthe drug inhibited other cellular phosphatases or if PTP-1B has other glycerols are released by adipose tissue and travel to the muscle and cellular targets unrelated to insulin signaling. cogen phosphorylase, which catalyzes the committed step in glycogenolysis. (c) Leptin increases the activity of cAMP phosphodiesterase; the result is that cAMP cellular concentration decreases. In this way, leptin acts as a glucagon antagonist in the same manner that insulin does; glucagon’s signal transduction pathway leads to an increase in cAMP concentration.

92 51. (a)

Ch 20 Solutions

53. AMPK phosphorylates and activates phosphofructokinase-2, the en- zyme that catalyzes the synthesis of fructose-2,6-bisphosphate. This metabo- lite is Glucagon G a potent activator of the glycolytic enzyme phosphofructokinase and an Receptor inhibitor of fructose-1,6-bisphosphatase, which catalyzes the opposing reaction for gluconeogenesis. Stimulation of AMPK would increase the concentration of fructose-2,6-bisphosphate and would therefore stimulate glyadenylate colysis and inhibit gluconeogenesis. The increase in glucose utilization and Glucose cyclase G protein decrease in glucose production would lower the level of glucose in the blood in the diabetic patient. [From Hardie, D. G., Hawley, S. A., and Scott, J. W., + glycogenolysis J. Physiol. 574, 7–15 (2006).]

ATP Protein 54. Metformin activates AMPK and, in so doing, promotes the phosphorycAMP kinase A − lation of acetyl-CoA carboxylase, the enzyme that catalyzes the fi rst glycogen commitLiver cell synthesis ted step of lipid synthesis (see Section 17-2). Phosphorylation inactivates acetyl-CoA carboxylase, which decreases the concentration of malonyl-CoA (b) These amino acids are positively charged. Since a negatively charged and inhibits lipid synthesis. The decrease in malonyl-CoA relieves the inaspartate residue in the glucagon receptor has been shown to be essen- hibition of fatty acid transport and allows oxidation to occur. The result tial for binding, it’s possible that an ion pair forms between a positively is a decrease in the concentration of the plasma triacylglycerols that predischarged amino acid side chain (His, Lys, or Arg) and the essential aspar- pose the patient to atherosclerosis. Phosphorylation of protein kinase B by tate. This hypothesis can be tested by modifying His1, Lys12, and Arg18 AMPK would increase translocation of GLUT4 transporters to the plasma to neutral or negatively charged side chains and assessing the resulting membrane. In this manner, AMPK is acting as insulin does, promoting the analogs’ binding and signal-transducing capabilities. (c) Eliminating the uptake of glucose into muscle and adipose tissue. Stimulation of AMPK by Asp at position 9 results in an analog with de- creased affi nity for the metformin benefi ts the patient with metabolic syndrome, who is resistant receptor and with little biological activity, indicating that the Asp plays to insulin, because AMPK triggers the same responses as insulin signaling. a role in both binding and signal trans- duction. Substituting the Asp [From Hardie, D. G., FEBS Lett. 582, 81–89 (2008).] with a positively charged Lys decreases the binding affi nity by about half but completely eliminates the bio- logical activity. The Asp evidently plays an important role in binding, but conservation of the negative charge does not seem to be critical, since a positive charge does Chapter 20 not abolish binding. Hence some other aspect of the Asp side chain 1. Topoisomerase I reactions are driven by the free energy change of DNA structure is important for binding. The Asp at position 9 does seem to shifting from a supercoiled conformation to a relaxed conformation, so no be important for biological activity, since deletion or substitution of the external source of free energy is needed. The enzyme merely accelerates a Asp greatly decreases biological activity. reaction that is already favorable. Topoisomerase II reactions involve more extensive mechanical intervention because both strands of the DNA are cleaved and held apart while another segment of DNA passes through the Abolishing the positive charge at position 12 greatly decreases binding affi nity. But once the analogs are bound, they are still capable of elic- break. This process requires the free energy of ATP, since it is not thermodyiting a biological response. The addition of a negative charge at position namically favorable on its own.

12 virtually abolishes binding, so it’s possible that the positively charged 2. E. coli topoisomerase IV is a type II topoisomerase. Both strands of group at position 12 forms an ion pair with a negatively charged amino the DNA helix need to be cut in order to separate the two circular DNA acid on the glucagon receptor. molecules.

Leu18 binds more effectively to receptors than does Ala18, supporting the hypothesis that hydrophobic interactions between the hormone 3. Novobiocin and ciprofl oxacin are useful as antibiotics because they and the receptor are important, since leucine has a more hydrophobic inhibit prokaryotic DNA gyrase but not eukaryotic topoisomerases. They side chain than alanine. Substitution with a Glu residue also decreases can kill disease-causing prokaryotes without harming host eukaryotic cells. binding, but not as much as at position 12. The positive charge is im- Doxorubicin and etoposide inhibit eukaryotic topoisomerases and can be portant, since replacement with the negatively charged Glu abolishes used as anticancer drugs. Although these drugs inhibit topoisomerases from both cancer cells and normal cells, cancer cells have a higher rate of DNA more than 90% of the binding ability of the analog. replication and are more susceptible to the effects of the inhibitors than are The des-His1-glucagon has decreased binding affi nity and has a normal cells. greater decrease in biological activity, indicating that the histidine at 4. (a) In all the variants, a neutral or negatively charged amino acid has position 1 is important for binding but plays a greater role in signal transduction. This is also supported by the other des-His1 analogs. The been replaced with a positively charged amino acid (Lys or Arg). The 1 des-His1-des-Asp9 analog does not bind well (only 7% of the control) abbreviation means that one additional positive charge was introduced, and has no biological activity. Interestingly, the des-His1-Lys9 derivative a 2 indicates the introduction of two additional positive charges, and binds well (70%) but has no biological activity. This indicates that the so on. DNA is negatively charged because of its phosphodiester backsubstitution of aspartate for lysine at position 9 preserves characteristics bone. It’s reasonable to hypothesize that an enzyme with an increased that are important for binding. However, once bound, the analog does number of positive charges might bind more effectively to the negatively not trigger signal transduction. charged DNA. (d) The des-His1-Lys9 is the best antagonist because it binds to the (b) Intact, double-stranded DNA has a lower absorbance at 260 nm receptor with 70% of the affi nity of the native hormone but has no than does single-stranded DNA. An increase in absorbance at 260 nm biological activity. In this derivative, the two amino acids important in over time is a useful measurement of the catalytic activity of DNase, signal transduction have been modifi ed, while the positively charged since the products of the reaction are short, single-stranded oligonucleresidues at positions 12 and 18, which are critical for binding, have been otides. retained. [From Unson, C. G., et al., J. Biol. Chem. 266, 2763–2766 (c) All of the variants have lower K M values than the wild-type DNase, (1991); and Unson, C. G., et al., J. Biol. Chem. 273, 10308–10312 indicating that the DNase variants bind more tightly to the DNA sub(1998).] strate than does the wild-type enzyme. The tighter binding is no doubt due to the additional positive charges on the variants, which allow the 52. The bypass surgery eliminates a portion of the stomach and the small formation of ion pairs between the variant enzymes and the DNA. intestine, which are normally sources of hormones that regulate fuel metabo- There seems to be a rough correlation between the number of posilism and insulin sensitivity. The surgery apparently alters hormonal signaling tively charged residues and the K M value: An increase in the number of enough to restore insulin sensitivity.

Ch 20 Solutions

93

positively charged residues results in a lower K M value, which indicates of the replication fork, generating positive supercoils that would hinder tighter binding. However, the Vmax values must also be assessed. All of DNA unwinding. the variants have a higher Vmax value than the wild-type enzyme, indi- 7. Negatively supercoiled DNA is more easily unwound, and thus the sepacating that the tighter binding leads to greater catalytic activity. The ration of the DNA strands occurs more easily, which facilitates the replicareplacement of three amino acids yields a variant enzyme in which the tion process.

K M and Vmax values are optimized. When four or fi ve amino acids are 8. The origin is more likely to be richer in A:T base pairs, since these expereplaced, the velocity is lower than when only three amino acids are rerience fewer stacking interactions and are more easily separated, which would placed. Since the K M value for the 4 and 5 variants is also generally allow easier access for the replication proteins. smaller, it is possible that the enzyme and substrate bind to each other 9. (a) Yes. By moving along a single DNA strand, the helicase can act as a so tightly that catalytic effi ciency is compromised. wedge to push apart the double-stranded DNA ahead of it. (d) The plasmid DNA normally exists in a supercoiled circle, as shown in (b) The free energy of dTTP hydrolysis is similar to the free energy of the control lane. The wild-type DNase can nick the DNA on one strand ATP hydrolysis. Each hydrolysis reaction drives the helicase along two to convert the plasmid to the relaxed circular DNA. The 3 mutant is to three bases of DNA. the best of the three in its ability to cleave supercoiled DNA. All three (c) The T7 helicase is probably a processive enzyme. Its hexameric ring mutants are able to produce linear DNA, whereas the wild-type DNase structure is reminiscent of the clamp structure that promotes the prodoes not. This indicates that the variants can cut both strands, whereas cessivity of DNA polymerase (see Fig. 20–12). [From Kim, D.-E., the wild-type enzyme cuts only one strand.

Narayan, M., and Patel, S. S., J. Mol. Biol. 321, 807–819 (2002).] (e) All of the mutants have a greater nicking activity than the wild-type for low- and high-molecular-weight DNA at low concentrations. The 10. Helicase would unwind the parental DNA, exposing extensive segments of single-stranded DNA. Without a polymerase to convert these template 2 mutant is especially active under these conditions and would be a strands into double-stranded DNA, the single strands would be susceptible good choice to use for a lupus patient. The 1 and 2 mutants can degrade low-molecular-weight DNA at a high DNA concentration to endonucleases, especially if the supply of replication protein A was limited. better than the wild-type enzyme, but the 3 and 4 mutants are not as active.

Not all of the data are given, so it is diffi cult to draw conclusions about the ability of the mutants to degrade high-molecular-weight DNA, but the 2 mutant is clearly more effective than the wild-type DNase and would be a good choice to use for a CF patient. [From Pan, C. Q., and Lazarus,15N-labeled R. A., J. Biol.

Chem.

273, 11701–11708 (1998).]

5. Parental DNA strands are shown in black, and newly synthesized 14N DNA strands are shown in gray. The original 15N-labeled parental DNA strands persist throughout succeeding generations, but their proportion of the total DNA decreases as new DNA is synthesized.

15

First generation

Second generation

+

3 4

Third generation

Fourth generation

7 8

N DNA (heavy)

+

Hybrid DNA

+

+

+

+

14

N DNA (light)

11. (a) DNA replication (and hence bacterial growth) halts immediately at the nonpermissive temperature because the DNA cannot be unwound ahead of the replication fork in the absence of the helicase. (b) Bacterial growth slows and then stops because the role of DnaA is to locate the replication origin (see Problem 7). When the temperature shifts to the nonpermissive temperature, DNA replication already under way is not affected (those cells can complete cell division), but another round of replication cannot begin in the absence of functioning DnaA. 12. (a) SSB, which coats single-stranded DNA exposed at the replication fork, must have a relatively low affi nity for DNA because it is displaced as the polymerase proceeds. The polymerase, along with accessory proteins such as the clamp, has a relatively high affi nity for the DNA. (b) The cell most likely contains large amounts of SSB in order to coat all the single-stranded DNA and prevent it from forming secondary structures. Whereas the cell requires multiple copies of SSB for each replication fork, in theory, only two DNA polymerase enzymes are required (one for the leading strand and one for the lagging strand).

13. The ddNTP lacks the 3 hydroxyl group that serves as the attacking nucleophile for the incoming dNTP. 14. The drug would inhibit DNA synthesis because the polymerization reaction is accompanied by the release and hydrolysis of inorganic pyrophosphate (PPi ; see Fig. 20-9). Failure to hydrolyze the PPi would remove the thermodynamic driving force for the overall process.

15. DNA polymerase ´ would need to have greater processivity because it synthesizes the leading strand continuously. Polymerase can be less processive because it synthesizes only a short DNA segment at the start of each Okazaki fragment.

16. The following proteins are required: an initiator protein, topoisomerase, helicase, SSB, primase, DNA polymerases capable of replicating both the leading and lagging strands, a clamp and clamp loader, and RNAse H. [From Li, J. J., and Kelly, T. J., Proc. Natl. Acad. Sci., 81, 6973-6977 (1984).]

17. First, the cell contains roughly equal concentrations of the four deoxynu-

1 4

1 8

cleotide substrates for DNA synthesis; this minimizes the chance for an overabundant dNTP to take the place of another or for the wrong dNTP to take the place of a scarce dNTP. Second, DNA polymerase requires accurate pairing between the template base and the incoming base. Third, the 3 S 5 exonuclease proofreads the newly formed base pair. Fourth, the removal of the RNA primer and some of the adjacent DNA helps minimize errors introduced by primase and by the DNA polymerase at the 5 end of a new DNA segment.

Finally, DNA repair mechanisms can excise mispaired or damaged nucleotides.

18. (a) The positively charged Mg2 ion could decrease the affi nity of the 6. DNA gyrase is a type II topoisomerase in E. coli. It can introduce nega- DNA’s 3 O atom for H, thereby increasing the nucleophilicity of the tive supercoils into the DNA ahead of the replication fork. In the absence of 3 O atom (making it behave more like an O ion). The Mg2 could DNA gyrase, strand separation would cause overwinding of the DNA ahead also help neutralize the negative charge on the incoming nucleotide.

Ch 20 Solutions

94

(b) The Mg2 could stabilize the negative charge that develops on the pentacovalent transition state.

(DNA) C3 O P

O

O P

O

O

O

O P

O

O

O

C5 (NTP)

O

24. Eukaryotic DNA ligases use ATP, and bacterial ligases use NAD , as a cofactor in the reaction. A drug that inhibited only NAD -dependent ligases would be an effective drug to treat bacterial diseases without harming the host. Bacteria unable to carry out the ligase reaction cannot complete DNA replication and cannot survive.

25. (a) DNA polymerase, (b) reverse transcriptase or telomerase, (c) primase or RNA polymerase.

26. By removing the triphosphate groups from deoxynucleotides, the

19. PPi is the product of the polymerization reaction catalyzed by DNA enzyme depletes the pool of dNTP substrates for the viral reverse tranpolymerase. This reaction also requires a template DNA strand and a primer scriptase. As a result, viral RNA cannot be converted to DNA and viral replication is blocked. with a free 3 end. (a) There is no primer strand, so no PPi is produced.

27. (a) (b) There is no primer strand, so no PPi is produced.

H (c) PPi is produced.

H N N (d) No PPi is produced because there is no 3 end that can be extended.

N H (e) PPi is produced.

N N N H N (f) PPi is produced.

H 20. The enzyme most likely detects the geometry of the polynucleotide N O O chain as it shifts from the wide and shallow A-form characteristic of an RNA N helix to the narrower and steeper B-form of DNA. 21. The DNA molecule H H N H (chromosome) is much longer than an Okazaki H N H H fragment and must be condensed and packaged in some way to fi t inside N the nucleus (in a eukaryote) or cell (in a prokaryote). If the cell waited until O O N the entire DNA molecule had been replicated, the newly synthesized lagging H N strand, in the form of many Okazaki fragments, might already be packaged N N H and inaccessible to the endonuclease, polymerase, and ligase necessary to N N H produce a continuous lagging strand. 22. The DNase can create a nick (a N N H break in one strand). Then the exonuclease H activity of DNA polymerase I can remove nucleotides on the 5 side of the nick. At the same time, the polymerase active site can add radioactive deoxynucleotides to the 3 side of the nick. The removal and replacement of (b) T nucleoT tides translates the nick in the 5 S 3 direction. DNA ligase can then seal the A T gap between the original DNA and the newly synthesized radioactive T A segment.

G G G G 23.

O E

NH2 N

Lys

R

O

G

O

P O O

P O

R A

O

A

NMN

T

Lys

NH2

P

O

28. Because there is a strong link between telomerase activity and aging,

OH

2 O

O O

E

O

E

O O

O

Lys

R A 3

NH2

AMP

O O

P O

O O O P O Lys

NH2

it’s possible that cancer cells might overexpress the telomerase enzyme.

Increased telomerase activity could contribute to the growth of cancer cells. If the G quartet acts as a negative regulator of telomerase activity, inducing its formation could inhibit the telomerase and inhibit growth of cancer cells.

29. The resulting telomeres will have a sequence complementary to the mutated sequence of the telomerase-associated RNA template. This experiment was important because it established the mechanism of the enzyme and verifi ed the role of the RNA template in extending chromosome length. 30. The proteins collectively known as shelterin do not recognize

P

P O

T

R A

O

OH

G G A

T

O E

G

G T

G

NAD

1

G

G

O

the mutated RNA sequence and do not bind to it. This leaves the ends of the DNA unprotected. This has been shown to lead to apoptosis (programmed cell death). [From Blackburn, E. H., Mol. Cancer Res, 3, 477– 482 (2005).] 31. Without functional DNA repair enzymes, additional mutations may arise in genes that are involved in regulating cell growth. In the absence of proper growth controls, cells may begin to proliferate at an accelerated rate.

Ch 20 Solutions 32. The Pt cross-links distort the DNA, much like a thymine dimer does. 40. (a) Unless the nucleotide excision repair pathway restores the normal DNA structure, the DNA containing the cross-link cannot be replicated, and the cell undergoes apoptosis.

33. The triphosphatase removes nucleotides containing the modifi ed base before they can be incorporated into DNA during replication.

O N

H

O N

O

N

N N

H

N

H

N

(b) Since cytosine also pairs with guanine, the deamination of guanine to xanthine does not induce a mutation.

N

41. All of these deaminations produce bases that are foreign to DNA; therefore, they can be quickly spotted and repaired before DNA has replicated and the damage is passed on to the next generation.

H

H

Adenine

42. (a)

8-Oxoguanine

35. The O 6-methylguanine produced by the methylation of guanine produces a residue that can base pair with either cytosine or thymine. If the O 6-methylguanine residue base pairs with thymine, the G:C base pair will eventually be changed to an A:T base pair.

36. The structures of the intercalating agents resemble A:T and G:C base pairs, which explains why they are able to slip in between the stacked base pairs of DNA. This has the effect of creating what appears to the replication machinery as an “extra” base pair. An extra base incorporated into the newly synthesized DNA may eventually lead to a frameshift mutation (in which the additional nucleotide causes the translation apparatus to read a different set of successive three-nucleotide codons).

37. Bromouracil causes an A:T to G:C transition.

O

O

Br

Br

NH N

H

N

H

N

N N

O

N

NH N

H

O

H

5-Methylcytosine

Thymine

(b) Thymine (c) C:G to T:A transition occurs. (d) The cell cannot repair the deaminated 5-methylcytosine since the resulting base is indistinguishable from thymine that occurs normally.

H N

N N N

H Guanine

H

(b)

N

N

O

N

H

Cytosine

2-Aminopurine

H N

H

N N

O

H3C

N

N H

38. The ultraviolet light is an additional precaution against contamination.

Ultraviolet light causes the formation of thymine dimers in bacterial DNA. High levels of exposure to ultraviolet light would overwhelm the cells’ ability to repair the dimers, which would result in the eventual death of the bacteria. Thus, the ultraviolet light helps keep the hood space free from bacteria that could contaminate the cultured cells.

39. (a)

O

N

N

5-Bromouracil (enol tautomer)

5-Bromouracil (keto tautomer)

H3C

N O

H

H

NH2

43. (a)

N

O

H N

O

O

N Xanthine

N N

NH

N

NH 2

34.

95

H

O

H

NH

N

N N

N

N

H

N

Hypoxanthine

H H

(b)

N

H

H N N

O N N

N N

Hypoxanthine

H

N

N

O Cytosine

(c) An A:T base pair is converted to a C:G base pair.

N

O

H

N

N H

N

N

N

H [From Sowers, L. C., Boulard, Y., and Fazakerley, G. V., Biochemistry 29, 7613–7620 (2000).]

Ch 20 Solutions

96 44. (a)

O N N

CH3

H N

N N

CH3

O

H

N

N

O

H

O 6-Methylguanine

Thymine

The methylation of guanine causes a G:C to A:T transition.

H

(b)

histone proteins from the negatively charged DNA by disrupting the ionic bonds that hold these two chromatin components together. The charged sodium and chloride ions can form substitute ionic interactions with the histone protein and DNA. 53.

CH2

(b)

CH2

O

N N

51. The side chains of lysine and arginine residues have high pK values and are positively charged at physiological pH. The positively charged groups can form ion pairs with the negatively charged phosphate groups on the backbone of the DNA molecule.

52. A solution of 0.5 M NaCl effectively dissociates the positively charged

(a)

H N H3C

(d) Overexpression of an error-prone DNA polymerase similar to polymerase would increase the mutation rate. [From Matsuda, T., Bebenek, K., Masutani, C,. Hanaoka, F., and Kunkel, T. A., Nature 404, 1011–1013 (2000).]

H

N O

CH2

N

CH2

N N

O

(c)

CH 2 O P

CH 2 N

O

N O P O

O

NH

N

O6-Methylguanine

H H

C Protonated cytosine

O

CH3 (d)

CH2

(e) CH2

No, since G normally base pairs with C, a mutation does not result.

CH2

CH2

A mutation occurs only when O 6-methylguanine pairs with T. [From Leonard, G. A., Thomson, J., Watson, W. P, and Brown, T., Proc. Natl.

Acad. Sci. 87, 9573–9576 (1990).]

CH2

CH2

CH2

NH

45. Most likely, the thymine–thymine dimer, since this lesion forms upon

NH

C

exposure of the DNA to ultraviolet light.

46. The thymine dimers form when DNA is exposed to ultraviolet light, so

CH3

NH 2

the enzymes that repair the damage are activated by the same stimulus that produced the damage. 47. The mutant bacteria are unable to repair deaminated cytosine (uracil).

In these cells, the rate of change of G:C base pairs to A:T base pairs is much greater than normal. 48. The amino acid code is degenerate in that several

O

NH

CH3

Both acetylation and methylation of the lysine produce a neutral side chain and remove the lysine residue’s positive charge; phosphorylation of the serine and histidine side chains produces a side chain with two negative charges; methylation of the arginine residue increases the size of the side chain somewhat but does not alter the charge of the side chain.

codons may specify 54. The high degree of sequence conservation from cows to peas indicates the same amino acid. For example, the codons GCA, GCC, GCG, and GCU all code for alanine. So if a mutation occurred at the third position that the sequence of the histone H4 protein is so vital to its function that (the 3 end) of the codon, alanine would still be incorporated into the amino acid substitutions, especially those that are nonconservative in nature, protein. 49. DNA polymerase III replicates DNA until a thymine dimer is would disrupt the function of the protein and thus cannot be tolerated.

55. (a) The nonstandard amino acid is homocysteine, which can accept a enmethyl group donated by methyl-tetrahydrofolate to regenerate methiocountered. Polymerase III is accurate but cannot quickly bypass the damnine (see Section 18-2). age. Polymerase V, which can more quickly proceed through the damaged site, does so, but at the cost of misincorporating G rather than A opposite T. Thus, replication can continue at a high rate. The tendency for DNA polymerase V to continue to introduce errors is minimized by its low processivity: Soon after passing the thymine dimer, it dissociates, and the more accurate polymerase III can continue replicating the DNA with high fi delity. 50. (a) The polymerases that lack 3 S 5 exonuclease activity have higher error rates than polymerases that contain the exonuclease activity. (b) , once every 6250 bases; , once every 1493 bases; , once every 100,000 bases; ´, once every 100,000 bases; , once every 29 bases. (c) Polymerase and HIV RT incorporate the correct base with approximately the same effi ciency (for example, 420, 760, and 800 M ? min 1 103). However, polymerase incorporates mispaired bases much more effi ciently than HIV RT does (22, 1.6, and 8.7 vs. 0.07 M ? min 1 103). These results indicate that the high error rate of polymerase results from its ability to incorporate the wrong base rather than its inability to incorporate the correct base.

COO H3N

C

H

CH2 CH2 SH

(b) The other product is methanol, CH3OH. 56. DNA methylation is important in the process of imprinting, in which DNA expression depends on parental origin. In the absence of DNA methylation, imprinting cannot take place. DNA methylation is important in gene expression during embryonic development, and if the DNA cannot be methylated, the embryos do not develop properly and die before birth.

57. No, histones modifi ed with ubiquitin are not marked for proteolytic destruction by the proteasome because the amino acid side chains of the histone

Ch 21 Solutions proteins have only one ubiquitin attached. In order to enter the proteasome for degradation, a chain of at least four ubiquitin molecules is required (see Section 12-1).

58. Histone acetylation has been shown to be associated with transcriptionally active chromatin, so it makes sense that the enzymes that catalyze this particular modifi cation would be associated with transcription factors.

59. (a) Because the protamine–DNA complexes occupy less volume than nucleosomes, DNA can be more easily packed into the small volume of the sperm nucleus. (b) DNA in nucleosomes is less compact and therefore more accessible for transcription, so one would expect nucleosomal DNA to contain genes that would need to be expressed immediately following fertilization, that is, genes essential for early embryogenesis.

60. The C4 amino group of cytosine is involved in base pairing with guanine (Section 3-1), so deamination is more likely to occur when the DNA is single-stranded. However, packaging the DNA in a nucleosome decreases the likelihood of DNA melting, so nucleosomal DNA is protected from cytosine deamination. [From Chen, X., Chen, Z., Chen, H., Su, Z., Yang, J., Lin, F., Shi, S., and He, X., Science 335, 1235–1238 (2012).

Chapter 21

97

cell requires them in equal amounts. Therefore, rRNA gene transcription does not need to be selective, and one promoter suffi ces. In contrast, the protein-coding genes transcribed by RNA polymerase II must be expressed at different times and at different levels. The regulation of transcription of these genes is more elaborate, requiring different promoter sequences that can potentially interact with different sets of transcription factors.

9. A:T base pairs have weaker stacking interactions (see Section 3-1) and are easier to melt apart than G:C base pairs, which have stronger stacking interactions. This facilitates DNA unwinding to expose the template for transcription.

11 10.

GAGCATATAAGGTGAGGTAGGATCAGTTGCTCCTCACATTT TATA box Inr 11.

O NH

CH

C

O NH

CH

CH2

CH2

CH2

CH2

CH2

CH2

CH2

NH

C

1. Messenger RNA is translated into protein, and a single mRNA can be C N CH NH 2 3 used to translate many proteins. In this way, the mRNA is “amplifi ed.”

Ribosomal RNA performs a structural role and is not amplifi ed, so many H CH3 NH2 more rRNA genes are needed to express suffi cient rRNA to meet the needs 12.

The other product is methylamine. of the cell.

2. Organizing genes with related functions in an operon means that O O the genes can be turned on and off together. Furthermore, because the C C H H N CH N CH genes are transcribed as a unit, the products can be generated in roughly equivalent concentrations and in the same region of the cell (where the CH2 CH2 H2N CH3 ribosome is located). Consequently, only one signal is needed to activate Methylamine CH2 CH2 or deactivate the metabolic function carried out by the products of the H2O operon’s genes. A eukaryotic cell can coordinate the transcription of physiCH2 CH2 cally separated but functionally related genes if they are share a set of NH NH enhancers, silencers, activators, or repressors.

Arginine Citrulline 3. Sequence-specifi c interactions require contact with the bases of DNA, C C N CH 3 O which can participate in hydrogen bonding and van der Waals interactions with protein groups. Electrostatic interactions involve the ionic phosphate H NH2 NH2 groups of the DNA backbone and are therefore sequence independent.

4. The protein must identify a gene to be expressed and must acti13. Upregulation of the methylase will increase the extent of methylavate the transcription machinery so that the DNA can be transcribed tion of Lys 4 (K4) and Lys 27 (K27) in histone 3. Since H3K4me3-associto RNA. The gene is identifi ed by the ability of the protein to bind speated genes normally tend to be in transcriptionally active chromatin (see cifi c DNA sequences associated with the gene. The resulting distortion Table 21-2), these genes are hyperactivated in cancer cells. The opposite in the DNA can then be recognized in a sequence-independent manner occurs with H3K27me3-associated genes, which are normally transcripby transcription factors that assemble at the site in order to initiate RNA tionally inactive and will be hypersilenced in cancer cells. [From Stark, G. synthesis.

R., Wang, Y., and Lu, T. Cell Res., 21, 375–380. (2011).]

5. The promoter region is shaded.

14. Both TAFs and histone proteins bind to DNA in a non-sequence1 specifi c manner. Proteins with a high affi nity for DNA likely share structural AAAATAAATGCTTGACTCTGTAGCGGGAAGGCGTATTATCCAACACCC similarities. It is known that histones contain abundant amounts of positively 6. Mutations that result in a promoter sequence that more closely resembles charged Lys and Arg residues that form ionic bonds with negatively charged the consensus sequence will increase the rate of transcription, whereas mu- DNA; perhaps TAFs do as well. tations that result in a promoter sequence that deviates from the consensus 15. (a) The polymerase binds most tightly to the DNA segment with the sequence will have the opposite effect. largest bulge. This DNA mimics the transcription bubble, in which the DNA strands have already been separated.

7. Affi nity chromatography takes advantage of the ability of the desired (b) Since Kd is a dissociation constant, the apparent equilibrium constant protein to bind to a specifi c ligand (see Section 4-5). To purify Sp1, the GGGCGG oligonucleotide is covalently attached to tiny beads, which con- for the binding reaction is 1/Kd. Equation 12-2 gives the relationship between G and K: stitute the stationary phase of a chromatography column. A cellular extract containing the Sp1 protein is loaded onto the column and a buffer (the DG52RT ln K or DG52RT ln(1yKd) mobile phase) is washed through the column to elute proteins that do not bind to the oligonucleotide ligand. Next, a high-salt buffer is applied to the (c) For double-stranded DNA: column to disrupt the strong interactions between the Sp1 and the GC box, DG52(8.3145 J?K21?mol21)(298 K)ln(1y31531029) and the Sp1 protein is eluted. [From Kadonaga, J. T., et al., Trends Biochem.

5237 kJ?mol21 Sci. 11, 20–23 (1986) and Kadonaga, J. T., and Tjian, R., Proc. Natl. Acad.

Sci. 83, 5889–5893 (1986).]

For the eight-base bulge, 8. RNA polymerase I transcribes rRNA genes. Although there are multiple DG52(8.3145 J?K21?mol21)(298 K)ln(1y1.3310212) copies of these genes, there are only a few types of rRNA molecules and the 5268 kJ?mol21

98

Ch 21 Solutions

Polymerase binding to the melted DNA is more favorable than binding the trp operon are not expressed because tryptophan biosynthetic enzymes to double-stranded DNA. (d) Melting open a DNA helix is are not needed when tryptophan is plentiful. But when the concentration thermodynamically unfavorable. Some of the favorable free energy of of tryptophan drops, tryptophan dissociates from the repressor, causing binding the polymerase to the DNA is spent in forming the a conformational change that results in the dissociation of the repressor transcription bubble. When the transcription bubble is preformed (for from the promoter. RNA polymerase can now bind to the promoter, and example, in the DNA with an eight-base bulge), this energy is not spent the genes encoding the enzymes for the tryptophan synthetic pathway can and is refl ected in the apparent energy of polymerase binding. The now be expressed. difference in G values for polymerase binding to double-stranded DNA 23. The accurate transmission of genetic information from one generaand to the eight base bulge is 68 ( 37) 31 kJ ? mol 1. This value estimates the free energy cost ( 31 kJ ? mol 1) of melting open eight base tion to the next requires a high degree of fi delity in DNA replication. A pairs of DNA. (e) The AT-rich sequence is easier to melt open than the higher rate of error in RNA transcription is permitted because the cell’s GC-rich se- quence because GC-rich DNA experiences stronger stacking survival usually does not depend on accurately synthesized RNA. If transinterac- tions. [From Bandwar, R. P., and Patel, S. S., J. Mol. Biol. 324, 63– lated, an RNA transcript containing an error may lead to a defective protein, which is likely to be destroyed by the cell before it can do much 72 (2002).] damage. The gene can be transcribed again and again to generate accurate transcripts.

24. The 5 end of any prokaryotic RNAs ending in A will be labeled.

16. The presence of the factor decreases the affi nity of RNA polymerase The 5 ends of RNA transcripts have 5 triphosphate groups containing for DNA. This allows the polymerase– factor complex to quickly scan long the labeled -phosphate. The phosphodiester bonds in the RNA will not segments of DNA for promoter sequences. Once transcription has begun, be labeled because these phosphate groups come from the -phosphates the factor is no longer needed and dissociates from the enzyme. Now the of the nucleoside triphosphates (the -phosphates are released as pyroRNA polymerase has a high affi nity for DNA, which helps keep it associated phosphate). with the template during transcription. *pppA 1 pppNΔ*pppApN 1 PPi 17. The lactose permease allows lactose to enter the cell, which increases the 25. Cordycepin, which resembles adenosine, can be phosphorylated and intracellular lactose concentration. Allolactose can then bind to the repressor used as a substrate by RNA polymerase. However, it blocks further RNA protein to remove it from the operator. The presence of additional lactose polymerization because it lacks a 3 OH group. assists in the full expression of the operon.

26. The incorporation of cordycepin into a growing RNA chain pro18. (a) When glucose is present, the cAMP concentration is low. CAP canvides evidence that transcription occurs in the 5 S 3 direction. Once not bind to the operon in the absence of cAMP, so the operon is off. incorporated, transcription halts because of the lack of a 3 OH group. If Glucose is the preferred substrate, so lactose is not used as long as glu- transcription occurred in the 3 S 5 direction, cordycepin would not be cose is present. incorporated into the growing RNA chain and would not be able to halt (b) The operon is off, both because glucose is present and CAP is not RNA synthesis. bound [see part (a)] and because lactose is absent and the repressor is 27. Rifampicin inhibits the transition from RNA chain initiation to bound to the operator, preventing RNA polymerase from binding. (c) CAP binds cAMP and can bind to the operon, but the operon remains off because the repressor is also bound in the absence of lactose. (d) The operon is on. In the absence of glucose, cAMP levels rise. cAMP binds to CAP, which can bind to the operon to stimulate transcription. In the presence of lactose, allolactose is formed and binds to the repressor, removing the repressor from the operator. The genes of the operon are expressed and the bacterial cell can use lactose as a food source.

elongation. Normally, RNA polymerase initiates RNA synthesis repeatedly, releasing many short transcripts before committing to elongation.

In the presence of rifampicin, the RNA polymerase cannot convert from initiation mode to elongation mode and remains bound to its promoter.

Synthesis of longer RNA transcripts is not possible in the presence of the drug.

28. The prokaryotic RNA polymerase is a six-subunit enzyme with a subunit composition of 2 v and is the only RNA polymerase in prokaryotic cells. In contrast, eukaryotic cells have three different RNA polymerase 19. If the repressor cannot bind to the operator, the genes of the lac operon enzymes that differ in structure from the prokaryotic polymerase. Thus, riare constitutively expressed; that is, the genes are expressed irrespective of fampicin can be effectively used as an antibiotic to treat bacterial diseases whether lactose is present or absent in the growth medium. Adding lactose in eukaryotic organisms because the drug is able to block transcription in has no effect on gene expression. the disease-causing prokaryotic cells without inhibiting transcription in the 20. If the repressor can bind to the operator, the genes of the operon are not eukaryotic organism. expressed. When lactose is added to the growth medium, it cannot bind to 29. G U C C G A U C G A A U G C A U G the repressor. The repressor remains bound to the DNA and the genes of the operon are not expressed. This is an example of a noninducible mutation, 30. (a) mRNA(n residues) 1 Pi S NDP 1 mRNA(n 2 1 residues) because adding lactose does not result in gene expression. 21. Wild-type cells (b) The reverse of the phosphorolysis reaction is an RNA polymerizacannot grow in the presence of phenyl-Gal. The wildtion reaction. PNPase uses an NDP substrate to extend the RNA by type cells produce a small amount of -galactosidase in the absence of lac one nucleotide residue and releases Pi. RNA polymerase uses an NTP operon expression, but not in suffi cient amounts to be able to cleave phenylsubstrate and releases PPi. Gal to phenol and galactose. The lacI mutants, however, will thrive in this (c) High processivity would allow the exonuclease to rapidly degrade growth medium. The mutation in the lacI gene results in the expression of mRNA molecules. This would be important in cases where the gene a nonfunctional repressor (or perhaps no repressor); in any case, the lac opproduct was no longer needed. An mRNA that was degraded more eron is constitutively expressed and -galactosidase is produced in suffi cient slowly could potentially continue to be translated. amounts to act on phenyl-Gal to release galactose. The use of this growth medium permits selection of repressor mutants, since the mutants survive 31. If -amanitin were added to cells in culture, the synthesis of mRNA would be inhibited, but the synthesis of all other types of RNA would be while the wild type cells do not. relatively unaffected. RNA polymerase II is responsible for mRNA synthesis 22. When tryptophan is plentiful, there is “extra” tryptophan available and is the most sensitive to inhibition by -amanitin. Experiments with this to bind to the repressor. With tryptophan bound, the repressor binds to toxin permitted investigators to determine the types of RNA synthesized by the promoter and prevents RNA polymerase from binding. The genes of each polymerase.

Ch 21 Solutions

99

32. Differential binding to the ion exchange column indicates that the RNA hybrid. The ,-imido nucleoside triphosphate cannot be hydrolyzed to its polymerases had different affi nities for the DEAE, indicating that the amino corresponding diphosphate and inorganic phosphate, so Rho-dependent acid sequences of the polymerases were different. The different sensitivities termination cannot occur. to the Mg2 ions and to the mushroom toxin also indicate differences in the enzymatic mechanism for the three enzymes, even though all three enzymes catalyze RNA polymerization.

39. (a) Product absent 33. The C-terminal domain (CTD) is phosphorylated on multiple serine residues when RNA polymerase transitions to elongation mode. The presence of many negatively charged phosphate groups causes charge–charge repulsions that cause this domain to be positioned away from the globular domain of the RNA polymerase as well as away from the negatively charged DNA. 34. The C-terminal domain (CTD) is phosphorylated when RNA poly-

Repressor

RNA polymerase

merase transitions from initiation mode to elongation mode. If the CTD is missing, it cannot be phosphorylated, and elongation cannot occur. Transcripts that do manage to clear the promoter in the absence of a phosphorylated domain will not be properly processed, since the phosphorylated CTD serves as a docking site for enzymes involved in co-transcriptional modifi cation.

transcription

DNA

Operator Operon

35. (a) 59? ? ? NNAAGCGCCGNNNNCCGGCGCUUUUUUNNN ? ? ? 39 (b)

N N

5

N

Product present

N

C G C C G C G G C C G G C A U A U NN UUUU

RNA polymerase no transcription

Product

Repressor

3

36. (a) 5 9 ? ?NNAAICICCINNNNCCIICICUUUUUUNNN ? ? ? 3?9 (b)

N

5

N

N

N

C

I C C I C I A A NN

C I I C I C U U UUUU

(b)

Product absent mRNA translation

DNA

Operon

3

The RNA terminator hairpin is much less stable when I is substituted for G because I:C base pairs have only two hydrogen bonds whereas G:C base pairs have three. The stability of the RNA hairpin is important in termination. When ITP is substituted for GTP in culture, the hairpin might not form; as a consequence, RNA transcripts will not terminate properly.

Product present Product RNA-binding protein

mRNA

37. As transcription proceeds, the nascent RNA forms a variety of secondary structures as portions of the transcript form complementary base pairs. The formation of these secondary structures may cause transcription to pause but not necessarily terminate.

38. The , -imido nucleoside triphosphate can be used as a substrate by RNA polymerase, since it is the phosphoanhydride bond closest to the ribose that is cleaved when the nucleotide is incorporated into the growing RNA transcript. Elongation is therefore not affected by the presence of the modifi ed nucleotide. Rho-dependent termination is affected because Rho is a helicase that uses the energy of ATP hydrolysis to pry apart the DNA–RNA

terminate transcription or block translation DNA

(c) If no protein were involved, the operon’s product would have to interact directly with the mRNA.

Ch 21 Solutions

100 (d)

U A U A U G G C A U C G U A GAU A G C

44.

G C A A C G G C A A C G G C A A C G CAG C A G

exonucleases.

Polymerase DNA polymerase

Type of template DNA

Substrates dATP, dCTP,

Reaction product DNA

dGTP, dTTP dATP, dGTP, Human RNA telomere DNA dTTP telomerase (e) Researchers could probe the conformation of an RNA molecule DNA RNA ATP, CTP, in the presence and absence of FAD by monitoring a conformation- RNA polymerase GTP, UTP dependent property of the RNA molecule, such as its susceptibility none poly(A) tail of to an endonuclease. (f) The FMN component of FAD is the most Poly(A) ATP effective, since it has the lowest dissociation constant. The polymerase mRNA none phosphate group is important for RNA binding, since ribofl avin, Bacterial CCAATP, CTP 3 CCA on tRNA which lacks a phosphate group, binds much less tightly. [From adding enzyme Winkler, W. C., Cohen-Chalamish, S., and Breaker, R. B., Proc. Nat. Acad. Sci. 99, 15908–15913 (2002).]

45. Messenger RNAs are transcribed only by RNA polymerase II. The phos40. (a) CAG codes for Gln. The resulting protein would contain a series of phorylated tail of RNA polymerase II recruits the enzymes needed for capping and polyadenylation. Other types of RNAs are synthesized by different extra Gln residues. These polar residues would most likely be located on RNA polymerases that do not have phosphorylated tails and cannot recruit the protein surface but could interfere with protein folding, stability, enzymes involved in post-transcriptional modifi cation. Thus, only mRNAs interactions with other proteins, and catalytic activity. are capped and polyadenylated. (b) The longer transcripts could be due to transcription initiating upstream of the normal site or failing to stop at the usual termi46. The capped mRNA has a 5 –5 triphosphate linkage, which is not nation point. Longer mRNA molecules could also result from the recognized by exonucleases (which normally cleave 5 –3 phosphodiester addition of an abnormally long poly(A) tail or the failure to undergo bonds). Capping must occur as soon as the 5 end of the mRNA emerges splicing. from the RNA polymerase so that the message will not be degraded by (c)

47. The active site of poly(A) polymerase is narrower because it does not need to accommodate a template strand.

48. Poly(A) polymerase binds only ATP, whereas other RNA polymerases bind ATP, CTP, GTP, and UTP. 49. The PABP binds to the poly(A) tails and protects the mRNA from degradation by the nucleases. Increasing the concentration of PABP extends the half-lives of the mRNAs bound to this protein.

[From Fabre, E., Dujon, B., and Richard, G.-F., Nuc. Acids Res. 30, 3540–3547 (2002).]

41.

RNA DNA

50. Histone synthesis is required for only a very short period of time during the cell cycle when DNA is being replicated; thus, stabilization of the mRNA coding for these proteins is not required.

51. (a) The phosphate groups of the phosphodiester backbone of RNA will be labeled wherever -[32P]-ATP is used as a substrate by RNA polymerase. (b) 32P will appear only at the 5 end of RNA molecules that have A as the fi rst residue (this residue retains its and phosphates). In all other cases where -[32P]-ATP is used as a substrate for RNA synthesis, the - and -phosphates are released as PPi (see Fig. 20-9). (c) No 32P will appear in the RNA chain. During polymerization, the - and -phosphates are released as PPi. The terminal ( ) phosphate of an A residue at the 5 end of an RNA molecule is removed during the capping process.

52. Bacterial genes contain no introns, so the cells, which lack splicing machinery, cannot express eukaryotic genes containing introns and exons.

Mature eukaryotic mRNA, which has already been spliced, contains only exons, which, after they have been converted back to DNA, can be transcribed and translated by the bacteria.

42. Head-on collisions between DNA polymerase and RNA polymerase 53. The splicing reactions are mediated by the spliceosome, a large RNA– would halt both replication and transcription since advancement of the protein complex. The intron must be large enough to include spliceosome replication fork would be blocked by the RNA polymerase moving toward binding site(s). In addition, the formation of a lariat-shaped intermediate (see it. To resolve the stand-off, both enzymes would need to dissociate from Fig. 21-22) requires a segment of RNA long enough to curl back on itself the DNA, and replication would need to be reinitiated, which would be without strain. time-consuming. Cells in which transcription and replication occur in the 54. The mRNA from a gene may be alternatively spliced to yield several difsame direction can undertake both processes more effi ciently. ferent types of proteins. This increases the diversity of the proteins produced by the cell without a correspondingly large number of genes.

43. Bacterial mRNAs have a 5 triphosphate group. The pyrophosphohydrolase removes two of the phosphoryl groups as pyrophosphate (PPi), leaving a 5 monophosphate (this apparently makes the mRNA a better substrate for the endonuclease).

55. Introns are removed co-transcriptionally rather than post-transcriptionally because the former strategy makes it less likely that exons will be skipped during the intron-removal process.

Ch 22 Solutions

101

(b) The possible codons are UUU, UUC, UCC, UCU, CCC, CCU, CUU, and CUC. UUU and UUC code for Phe, UCC and UCU code for Ser, CCC and CCU for Pro, and CUU and CUC for Leu. Therefore, the polypeptide would contain Phe, Ser, Pro, and Leu. (c) 3 splice site The possible codons are UUU, UUG, UGG, UGU, GGG, GGU, GUU, T and GUG. UUU codes for Phe, UUG for Leu, UGG for Trp, UGU for ? ? ? TTTCCCACCCTTAGGCTGCT ? ? ?

Cys, GGG and GGU for Gly, and GUU and GUG for Val. Therefore, the polypeptide would contain Phe, Leu, Trp, Cys, Gly, and Val. 57. (a) U69 is more important for catalysis, since deleting this residue or (d) These results show that the genetic code is redundant because all substituting it with another residue generally resulted in a dramatic dethree templates result in the incorporation of Leu. crease in the catalytic constant (k) but had a modest effect on Kd. (b) Increasing the size of the bulge dramatically decreased substrate 5. (a) A polypeptide consisting of a repeating Tyr–Leu–Ser–Ile tetrapeptide will be produced. binding affi nity, suggesting that although the presence of the bulge is (b) Depending on the reading frame, the polypeptide may begin with necessary for catalysis, the geometry of the bulge is important for subTyr, Ile, or Ser. strate binding. [From Kaye, N. M., Zahler, N. H., Christian, E. L., and Harris, M. E., J. Mol. Biol. 324, 429–442 (2002).]

6. The result is an Ile–Asp–Arg tripeptide. Polypeptides such as the ones described in Solution 5 are not produced because UAG is a stop codon.

NH 2 58.

S 7. Because the tRNAs that match the common codons are most abundant N HN in the yeast cell, protein synthesis is normally effi cient. If a mutation alters a codon so that it is not one of the 25 commonly used codons, it is likely S O N N that the isoacceptor tRNA for that codon is relatively scarce. Consequently, waiting for the appropriate tRNA to deliver the amino acid to the ribosome Ribose Ribose would result in a lower rate of protein synthesis, even though the sequence 4-Thiouridine 2-Thiocytidine of the protein is unchanged.

59.

5 · · · GGAGUACCCUGAUGAGAUC · · · 3 8. (a) For maximum effi ciency in translation, one would expect that the 3 · · · CCUCAUGGGACUACUCUAG · · · 5 highly expressed genes would contain the preferred codons, so that the cell would have high levels of the appropriate tRNAs to deliver the amino [From Takei, Y., Kadomatsu, K., Yuzawa, Y., et al., Cancer Res. 64, 3365– acids to make those proteins. For genes that are expressed more rarely, 3370 (2004).] the delay that would result from using rare codons or rare tRNAs would 60. The conformation of DNA is constrained by the requirement of formprobably not have a large impact on the cell. ing a double-helical structure. Single-stranded RNA, in contrast, can fold (b) Because different organisms have different codon-usage patterns, the into complex three-dimensional shapes that are better suited for catalytic gene’s sequence may reveal whether it contains codons preferred by bacteria activities such as binding substrate molecules, providing chemically reactive or codons preferred by mammals. groups, and stabilizing transition states. In addition, the 2 OH group of 9. The 5 nucleotide is at the wobble position, which can participate in RNA (absent in DNA) participates in certain transesterifi cation reactions, non–Watson–Crick base pairings with the 3 nucleotide of an mRNA codon. such as splicing (see Fig. 21-22). 61. The primary structure of a protein refers Because the fi rst two codon positions are more important for specifying an to its sequence of amino acids; amino acid (see Table 22-1), wobble at the third position may not affect translation. in the RNase P RNA this corresponds to the sequence of 417 nucleotides.

Secondary structure in proteins refers to regular, repeating structural motifs 10. Molecular sequence data indicate that Bacteria split from a group that such as helices and sheets. In the ribozyme, secondary structure refers to later diverged to form the Archaea and Eukarya. One would therefore expect the base-paired stem and loop structures. Tertiary structure in proteins Archaea and Eukarya to be more similar in some ways than Archaea and Bacteria or Eukarya and Bacteria. refers to the overall three-dimensional shape of the macromolecule; similarly for 11.

H the ribozyme, the tertiary structure refers to the three-dimensional shape of the molecule.

O H N 56.

5 splice site T ? ? ? CCCTGGGCAGGTTGGTA ? ? ?

62. The mutation could alter the promoter sequence to decrease the fre-

N

N

quency of transcription; the mutation could increase the binding affi nity N N N N H for a repressor protein (or decrease the binding affi nity for an activator); the N N mutation could introduce secondary structure into the mRNA that would hinder RNA polymerase or promote premature transcription termination; I:A the mutation could slow the rate of mRNA splicing; the mutation could 12. (a) CUG, CUA, and CUU. render the mRNA a better target for RNA interference. (b) The sequence of the polypeptide would not be affected, because all the codons recognized by the tRNA specify leucine. [From Sørensen, M. Chapter 22 A., et al., J. Mol. Biol., 354, 16–24 (2005).]

1. A hypothetical quadruplet code would have 44, or 256, possible com13. If the adenosine of an mRNA codon were changed to inosine, then the binations. codon could pair with a tRNA anticodon containing guanosine rather than 2. Because the genetic code is degenerate, a mutation that alters a codon uridine. This could result in the incorporation of a different amino acid at may not alter the amino acid encoded by that codon. This is particularly true that codon position. A given gene could give rise to several different polypepfor mutations at the third codon position. Changes at the fi rst or second po- tide products, depending on how many A residues were edited to I residues sition almost always change the encoded amino acid, but the new amino acid and how many of these changes resulted in amino acid substitutions. may be chemically similar to the old one (for example, Val S Ala).

14. (a) The RNA-binding site of the synthetase is necessary for binding 3. (a) Poly(Phe), (b) poly(Pro), and (c) poly(Lys). tRNA for aminoacylation and for binding RNA during splicing. (b) Because of their long evolutionary history, aminoacyl–tRNA syn4. (a) The possible codons are UUU, UUA, UAA, UAU, AAA, AAU, thetases have had many opportunities to diversify in structure and in AUU, and AUA. UUU codes for Phe, UUA for Leu, UAA is a stop function in order to assume additional roles in the cell. codon, UAU codes for Tyr, AAA for Lys, AAU for Asn, and both AUU and AUA for Ile. Therefore, the amino acids in the polypeptide would 15. Gly is the smallest amino acid, so the aminoacylation site in GlyRS can be Phe, Leu, Tyr, Lys, Asn, and Ile. be small enough to prevent the entry of any other amino acid.

102

Ch 22 Solutions

16. Gly and Ala; Val and Leu; Ser and Thr; Asn and Gln; and Asp and Glu. 27. Acetylation removes the positive charge on the N-terminus, as shown 17. (a) AlaRS can generate Ala–tRNAAla, Gly–tRNAAla, and Ser–RNAAla. below. If some copies of the protein are acetylated whereas others are not, (b) An editing active site that admits amino acids smaller than alanine two spots will appear on the gel, which separates proteins based on charge would destroy Gly–tRNAAla, but it would not get rid of Ser–RNAAla, as well as size (the difference in size is not signifi cant). [From Kaltschmidt, E. and Wittmann, H. G., Proc. Natl. Acad. Sci. 67, 1276–1282 (1970).] since it cannot fi t into the editing active site (serine is larger than alanine). (c) AlaXp functions as a free-standing editor for AlaRS by elimiO O nating Ser–tRNAAla after it has been synthesized by AlaRS. C

H

C

18. (a) The enzyme itself must act as a template to direct the addition of two H3N CH H3C C N CH C residues followed by one A residue.

O R R (b) The enzyme must recognize only ATP and CTP as substrates, excluding GTP, UTP, and all dNTPs. 28. All three types of proteins have an RNA recognition motif because all of (c) In the two-domain enzyme, one polymerase adds the two C residues, them bind to RNA. The ribosomal proteins bind to rRNA to form the and the other domain then adds the terminal A residue. ribosome. Rho factor is a transcriptional terminator in bacteria that acts as a 19. The two Lys codons are AAA and AAG. Substitution with C would helicase to pry the nascent RNA away from its DNA template (see Section yield CAA and CAG, which code for Gln; substitution with G would yield 21-2). The eukaryotic poly(A) binding protein binds to the poly(A) tail at the GAA and GAG, which code for Glu; and substitution with U would yield 3 end of mRNA (see Figure 21-19).

AUU and UAG, which are stop codons. Replacing a Lys codon with a stop 29. The start codon is likely near the 5 end of the mRNA, and the stop codon would terminate protein synthesis prematurely, most likely producing codon near the 3 end (neither is at the very end of the mRNA molecule). a nonfunctional protein. Replacing Lys with Glu or Gln could disrupt the protein’s structure and therefore its function if the Lys residue was involved I

in III a structurally essential interaction such as an ion pair in the protein interior.

IV VI If the Lys residue was on the surface of the protein, replacing it with Glu or Gln, both of which are hydrophilic, might not have much impact on the 3 5 6 protein’s structure or function.

2 4 1 VII 20. In order for a cell to incorporate a nonstandard amino acid into a poly5 peptide, the amino acid must fi rst be attached to a tRNA corresponding to II one of the 20 standard amino acids. The aminoacyl–tRNA can then bind to the ribosome, and its amino acid can be incorporated into the growing polyStar t 7 DNA peptide at positions corresponding to the codon for the standard amino acid.

The failure of cells to synthesize norleucine-containing peptides most likely poly A V refl ects the inability of LeuRS to effi ciently attach norleucine to tRNALeu. mRNA A mutant LeuRS, which presumably lacks the proofreading activity of the 3 wild-type LeuRS, was able to produce norleucine–tRNALeu, and the cells’ Stop ribosomes used this aminoacylated tRNA to translate Leu codons.

21. Like other nucleic acid–binding proteins we have studied (histones are an example), proteins containing the positively charged Lys and Arg residues 30. interact favorably with the polyanionic RNA. The most important interac- Process Replication Transcription Translation tions between the protein and the nucleic acid are likely to be ion pairs.

Substrates dATP, dCTP, ATP, CTP, GTP, 20 different amino 22. The assembly of functional ribosomes requires equal amounts of the dGTP, dTTP UTP acids, each linked rRNA molecules. Therefore, it is advantageous for the cell to synthesize the to a tRNA rRNAs all at once.

23. In protein-coding genes, a high degree of conservation indicates a seg- Product two identical a single-stranded a polypeptide chain ment of protein that is intolerant of substitution. Mutations that produce double helices RNA molecule amino acid substitutions in these regions are likely to produce nonfunctional Template or both parent one strand of mRNA sequence proteins. It is the same for ribosomal RNA. Mutations in the rRNA that disguide DNA strands DNA is used as specifi es the order rupt its function are likely to be lethal and will not be passed on to progeny. are used as a template of amino acids Consequently, the sequences of the rRNAs are intolerant of substitution and templates are highly conserved.

RNA primer no primer neededMet attached to an 24. Ribosomal inactivating proteins catalyze the removal of adenine residues Primer initiator tRNA from ribosomal RNA. This is analogous to the removal of a side chain from an amino acid residue in a protein. Like proteins, ribosomal RNAs have Enzyme DNA polyRNA polymerase ribosome (rRNA) specifi c residues that are essential to their function; removing these residues merase causes loss of activity.

Cellular nucleus nucleus cytoplasm 25. The small and large ribosomal subunits are held together by RNA–RNA location contacts that are stabilized by Mg2 ions. The tRNAs that bind to the A and P sites in the ribosome contact the rRNA; these interactions have the same structural features. Adding EDTA chelates the magnesium ions and causes 31. The polypeptide would contain all Lys residue with an Asn at the loss of ribosomal stability; this may be accompanied by the unfolding of the C-terminus. If the mRNA were read from 3 S 5 , the polypeptide would ribosomal RNA and the dissociation of the ribosomal proteins, so translation consist of an N-terminal Gln followed by a series of Lys residues. Transcription cannot occur. and translation both take place in the 5 S 3 direction; this allows bacterial 26. The peptidyl transferase activity lies entirely within the 23S rRNA; cells to begin translating nascent mRNA before transcription is complete. If that is, 23rRNA is a ribozyme. The proteins might be necessary to assist the translation took place in the 3 S 5 direction, the ribosome would have to 23S rRNA in forming the necessary three-dimensional structure required wait for mRNA synthesis to be completed before translation could begin. for catalytic activity, just as the proper conformation is required for protein 32. AUG codes for methionine, GUG for valine, and UUG for leucine. enzymes. Extremely strong intermolecular interactions between the proteins and the rRNA confi rm the importance of the proteins and explain why the 33. The sequence on the 16S rRNA that aligns with the Shine–Dalgarno sequence is shown. The initiation codon is highlighted in gray. extraction process failed to remove them.