Abbreviated Answers to Problems A-13 In (a), V max doubles, but K m is constant. In (b), V max halves, but K m is constant. In (c), V max is constant, but the apparent K m increases. In (d), V max decreases, but K m is constant. In (e), V max decreases, K m decreases, but the ratio K m /V max is constant. 6. a. The slope is given by K m B /V max (K S A /[A] ϩ 1). b. y-intercept ϭ ((K m A /[A]) ϩ 1)1/V max . c. The horizontal and vertical coordinates of the point of intersection are 1/[B] ϭϪK m A /K S A K m B and 1/v ϭ 1/V max (1 Ϫ (K m A /K S A )). 7. Top left: (1) Competitive inhibition (I competes with S for binding to E). (2) I binds to and forms a complex with S. Top right: (1) Pure noncompetitive inhibition. (2) Random, single-displacement bisubstrate reaction, where A doesn’t affect B binding, and vice versa. (Other possibilities include [3] Irre- versible inhibition of E by I; [4] 1/v vs. 1/[S] plot at two differ- ent concentrations of enzyme, E.) Bottom left: (1) Mixed noncompetitive inhibition. (2) Ordered single-displacement bisubstrate mechanism. Bottom right: (1) Uncompetitive inhibition. (2) Double- displacement (ping-pong) bisubstrate mechanism. 8. Clancy must drink 694 mL of wine, or about one 750-mL bottle. 9. a. K S ϭ 5 M; b. K m ϭ 30 M; c. k cat ϭ 5 ϫ 10 3 sec Ϫ1 ; d. k cat /K m ϭ 1.67 ϫ 10 8 M Ϫ1 sec Ϫ1 ; e. Yes, because k cat /K m approaches the limiting value of 10 9 M Ϫ1 sec Ϫ1 ; f. V max ϭ 10 Ϫ5 mol/mL и sec; g. [S] ϭ 90 M; h. V max would equal 2 ϫ 10 Ϫ5 mol/mL и sec, but K m ϭ 30 M, as before. 10. a. V max ϭ 132 mol mL Ϫ1 sec Ϫ1 ; b. k cat ϭ 44,000 sec Ϫ1 ; c. k cat /K m ϭ 24.4 ϫ 10 8 M Ϫ1 sec Ϫ1 ; d. Yes! k cat /K m actually exceeds the theoreti- cal limit of 10 9 M Ϫ1 sec Ϫ1 in this problem; e. The rate at which E encounters S; the ultimate limit is the rate of diffusion of S. 11. a. V max ϭ 1.6 mol mL Ϫ1 sec Ϫ1 ; b. v ϭ 1.45 mol mL Ϫ1 sec Ϫ1 ; c. k cat /K m ϭ 1.6 ϫ 10 8 M Ϫ1 sec Ϫ1 ; d. Yes. (e) v [S] O ϩI (d) v [S] O ϩI (c) v [S] O ϩI 12. a. V max ϭ 6 mol mL Ϫ1 sec Ϫ1 ; b. k cat ϭ 1.2 ϫ 10 6 sec Ϫ1 ; c. k cat /K m ϭ 1 ϫ 10 8 M Ϫ1 sec Ϫ1 ; d. Yes! k cat /K m approaches the theoretical limit of 10 9 M Ϫ1 sec Ϫ1 ; e. The rate at which E en- counters S; the ultimate limit is the rate of diffusion of S. 13. a. V max ϭ 64 mol mL Ϫ1 sec Ϫ1 ; b. k cat ϭ 1.28 ϫ 10 4 sec Ϫ1 ; c. k cat /K m ϭ 1.4 ϫ 10 8 M Ϫ1 sec Ϫ1 ; d. Yes! k cat /K m approaches the theoretical limit of 10 9 M Ϫ1 sec Ϫ1 . 14. a. V max ϭ 120 mmol mL Ϫ1 sec Ϫ1 ; b. v ϭ 104.7 mmol mL Ϫ1 sec Ϫ1 ; c. k cat /K m ϭ 3.64 ϫ 10 8 M Ϫ1 sec Ϫ1 ; d. Yes! k cat /K m approaches the theoretical limit of 10 9 M Ϫ1 sec Ϫ1 . 15. a. Starting from V max f ϭ k 2 [E T ] and V max r ϭ k Ϫ1 [E T ] for the maximal rates of the forward and reverse reactions respec- tively, and the Michaelis constants K m S ϭ(k Ϫ1 ϩ k 2 )/k 1 and K m P ϭ(k Ϫ1 ϩ k 2 )/k Ϫ2 for S and P, respectively, v ϭ (V max f [S]/K m S )ϪV max r [P]/K m p )/(1 ϩ [S]/K m S ϩ [P]/K m P ) b. At equilibrium, v ϭ 0 and K eq ϭ [P]/[S] ϭ V max f K m P /V max r K m S . 16. a. S is the preferred substrate; its K m is smaller than the K m for T, so a lower [S] will give v ϭ V max /2, compared with [T]. b. k cat /K m defines catalytic efficiency. k cat /K m for S ϭ 2 ϫ 10 7 M Ϫ1 sec Ϫ1 ; k cat /K m for T ϭ 4 ϫ 10 7 M Ϫ1 sec Ϫ1 , so the enzyme is a more efficient catalysis with T as substrate. 17. a. Because the enzyme shows maximal activity at or below 40°C, it seems more like a mammalian enzyme than a plant enzyme, which would be expected to have a broader temper- ature optimum because plants experience a broader range of temperatures. b. An enzyme from a thermophilic bacterium growing at 80°C would show an activity versus temperature profile similar to this one but shifted much farther to the right. Chapter 14 1. a. Nucleophilic attack by an imidazole nitrogen of His 57 on the OCH 2 O carbon of the chloromethyl group of TPCK cova- lently inactivates chymotrypsin. (See The Student Solutions Manual, Study Guide and Problems Book for structures.) b. TPCK is specific for chymotrypsin because the phenyl ring of the phenylalanine residue interacts effectively with the binding pocket of the chymotrypsin active site. This positions the chloromethyl group to react with His 57 . c. Replacement of the phenylalanine residue of TPCK with arginine or lysine produces reagents that are specific for trypsin. 2. a. The structures proposed by Craik et al., 1987 (Science 237:905–907) are shown here. (If you look up this reference, note that the letters A and B of the figure legend for Figure 3 of this article actually refer to parts B and A, respectively. Reverse either the letters in the figure or the letters in the figure legend and it will make sense.) O N G N H H H H O O O N H G Ser 195 57 Wild type 56 N Asp 102 Ser 214 His 57 A Ϫ A-14 Abbreviated Answers to Problems b. Asn 102 of the mutant enzyme can serve only as a hydrogen- bond donor to His 57 . It is unable to act as a hydrogen-bond acceptor, as aspartate does in native trypsin. As a result, His 57 is unable to act as a general base in transferring a proton from Ser 195 . This presumably accounts for the diminished ac- tivity of the mutant trypsin. 3. a. The usual explanation for the inhibitory properties of pep- statin is that the central amino acid, statine, mimics the tetra- hedral amide hydrate transition state of a good pepsin substrate with its unique hydroxyl group. b. Pepsin and other aspartic proteases prefer to cleave peptide chains between a pair of hydrophobic residues, whereas HIV-1 protease preferentially cleaves a Tyr-Pro amide bond. Because pepstatin more closely fits the profile of a pepsin substrate, we would surmise that it is a better inhibitor of pepsin than of HIV-1 protease. In fact, pepstatin is a potent inhibitor of pepsin (K I Ͻ 1 nm) but only a moderately good inhibitor of HIV-1 protease, with a K I of about 1 M. 4. The enzyme-catalyzed rate is given by: 5. This problem is solved best by using the equation derived in problem 4. ϭ e (⌬G u ‡ Ϫ⌬G e ‡ )/RT k e ᎏ k u So or Similarly, for the uncatalyzed reaction: A ssuming that Х Then k e k u ϭ e (⌬G ‡ u Ϫ⌬G ‡ e )͞RT k e k u ϭ e Ϫ⌬G ‡ e ͞RT e Ϫ⌬G ‡ u ͞RT k Ј e k Ј u k u ϭ k Ј u e Ϫ⌬G ‡ u ͞RT k e ϭ k Ј e e Ϫ⌬G ‡ e ͞RT k e [ES] ϭ k Ј e e Ϫ⌬G ‡ e ͞RT [ES] [EX ‡ ] ϭ K ‡ e [ES] ϭ e Ϫ⌬G ‡ e ͞RT [ES] K ‡ e ϭ e Ϫ⌬G ‡ e ͞RT ⌬G ‡ e ϭϪRT ln K ‡ e K ‡ e ϭ [EX ‡ ] [ES] v ϭ k e [ES] ϭ k Ј e [EX ‡ ] O N N H H H H O O N N H G Ser 195 57 D102N mutant 56 N Asn 102 Ser 214 His 57 B H H Using this equation, we can show that the difference in activa- tion energies for the uncatalyzed and catalyzed hydrolysis reac- tions (⌬G u Ϫ⌬G c ) is 92 kJ/mol. 6. Trypsin catalyzes the conversion of chymotrypsinogen to -chymotrypsin, and chymotrypsin itself catalyzes the conversion of -chymotrypsin to ␣-chymotrypsin. 7. The mechanism suggested by Lipscomb is a general base pathway in which Glu 270 promotes the attack of water on the carbonyl carbon of the substrate: 8. Using the equation derived in problem 4, it is possible to calcu- late the ratio k e /k u as 1.86 ϫ 10 12 . 9. If the concentration of free enzyme is equal to the concentra- tion of enzyme–ligand complex, the concentration of ligand would be 1 ϫ 10 Ϫ27 M. This corresponds to 1.67 ϫ 10 3 liters per molecule. 10. ⌬G ϭϪ154 kJ/mol. This value is intermediate between nonco- valent forces (H bonds are typically 10–30 kJ/mol) and covalent bonds (300–400 kJ/mol). 11. Assuming that the rate of gluconeogenesis would be equal to the slowest step in the pathway, with k ϭ 2 ϫ 10 Ϫ20 /sec, and as- suming a cellular concentration of fructose-1,6-bisphosphatase of 0.031 (see Table 18.2), the rate of sugar synthesis would be 2 ϫ 10 Ϫ20 /sec ϫ 0.031 mM, or 6.2 ϫ 10 Ϫ22 mM/sec. Assuming a total human cell volume of 40 L, this corresponds to 2.48 ϫ 10 Ϫ23 moles glucose/sec. Assuming 30 ATP per glucose (under cellular conditions) and 50 kJ/mole of ATP hydrolyzed, we find that the rate of energy production is (2.48 ϫ 10 Ϫ23 ) ϫ (30 ATP/glucose) ϫ (30.5 kJ/mole), or 2.269 ϫ 10 Ϫ20 kJ/sec. Converting to kilocalories and years, we find that the time to synthesize the needed 480 kilocalories with an uncatalyzed reac- tion would be 2.8 ϫ 10 15 years, or roughly 200,000 times the life- time of the universe so far. 12. The correct answer is c. The stomach is a very acidic environ- ment, whereas the small intestine is slightly alkaline. The lower part of the figure shows that enzyme X has optimal activity near pH 2, whereas enzyme Y works best at a pH near 8. 13. The correct answer is a. The two enzymes have nonoverlapping pH ranges, so it is highly unlikely that they could operate in the same place at the same time. P P OO O C CO O H N H H NH BH Zu 2ϩ Zu 2ϩ COO Ϫ O OO O CHR Ј H R A OO CO Ϫ Gla 270 O A A A OO O C C Ϫ O O N H H HN ϩ H COO Ϫ OO O CHR Ј H R A A A A O C O OH N H O O CHR Ј A Carboxypeptidase P OOO COHGla 270 O OO C NH 2 ϩ COO Ϫ H R A A P Abbreviated Answers to Problems A-15 14. The correct answer is b. Only enzyme A has a temperature range that encompasses human body temperature (37°C). 15. The correct answer is d. The activities of the two enzymes overlap between 40° and 50°C. 16. The correct answer is c. We have no information on the pH be- havior of enzymes A and B, nor on the behavior of X and Y as a function of temperature. The only answer that is appropriate to the data shown is c. 17. The only possible answer is b, because a “leveling off” implies that all the enzyme is saturated with S. 18. The correct answer is c. In order to bring the substrate into the transition state, an enzyme must enjoy environmental conditions that favor catalysis. There is no activity apparent for enzyme Y below pH 5.5. Chapter 15 1. a. As [P] rises, the rate of P formation shows an apparent decline, as enzyme-catalyzed conversion of P ⎯→ S becomes more likely. b. Availability of substrates and cofactors. c. Changes in [enzyme] due to enzyme synthesis and degradation. d. Covalent modification. e. Allosteric regulation. f. Specialized controls, such as zymogen activation, isozyme variability, and modulator protein influences. 2. Proteolytic enzymes have the potential to degrade the proteins of the cell in which they are synthesized. Synthesis of these enzymes as zymogens is a way of delaying expression of their ac- tivity to the appropriate time and place. 3. Monod, Wyman, Changeux allosteric system: Hanes–Woolf plot I O A [S] [S]/v Lineweaver–Burk plot 1/v I O A 1/[S] 4. Using the curves for negative cooperativity as shown in Figure 15.8 as a guide: 5. For n ϭ 2.8, Y lungs ϭ 0.98 and Y capillaries ϭ 0.77. For n ϭ 1.0, Y lungs ϭ 0.79 and Y capillaries ϭ 0.61. Thus, with an n of 2.8 and a P 50 of 26 torr, hemoglobin becomes almost fully saturated with O 2 in the lungs and drops to 77% saturation in resting tissue, a change of 21%. If n of 1.0 and a P 50 of 26 torr, hemoglobin would become only 79% saturated with O 2 in the lungs and would drop to 61% in resting tissue, a change of 18%. The difference in hemoglobin O 2 saturation conditions between the values for n ϭ 2.8 and 1.0 (21% Ϫ 18%, or 3% saturation) seems small, but note that the potential for O 2 delivery (98% saturation versus 79% saturation) is large and becomes crucial when pO 2 in actively metabolizing tissue falls below 40 torr. 6. More glycogen phosphorylase will be in the glycogen phos- phorylase a (more active) form, but caffeine promotes the less active T conformation of glycogen phosphorylase. 7. Over time, stored erythrocytes will metabolize 2,3-BPG via the pathway of glycolysis. If [BPG] drops, hemoglobin may bind O 2 with such great affinity that it will not be released to the tissues (see Figure 15.29). The patient receiving a transfusion of [BPG]-depleted blood may actually suffocate. 8. a. By definition, when [P i ] ϭ K 0.5 , v ϭ 0.5 V max . b. In the presence of AMP, at [P i ] ϭ K 0.5 , v ϭ 0.85 V max (from Figure 15.14c). c. In the presence of ATP, at [P i ] ϭ K 0.5 , v ϭ 0.12 V max (from Figure 15.14b). 9. If G ␣ –GTPase activity is inactivated, the interaction between G ␣ and adenylyl cyclase will be persistent, adenylyl cyclase will be active, [cAMP] will rise, and glycogen levels will fall because [S] [S] Negative cooperativity Negative cooperativit y Negative cooperativity Michaelis– Menten Michaelis– Menten Lineweaver–Burk Hanes–Woolf M-M v 1 — v 1 / [S] [S] — v A-16 Abbreviated Answers to Problems glycogen phosphorylase will be predominantly in the active, phosphorylated a form. 10. An excess of a negatively cooperative allosteric inhibitor could never completely shut down the enzyme. Because the enzyme leads to several essential products, inhibition by one product might starve the cell for the others. 11. a. -R(R/K)X(S*/T*)- b. -KRKQIAVRGL- 12. Ligand binding is the basis of allosteric regulation, and allosteric effectors are common metabolites whose concentrations reflect prevailing cellular conditions. Through reversible binding of such ligands, enzymatic activity can be adjusted to the momen- tary needs of the cell. On the other hand, allosteric regulation is inevitably determined by the amounts of allosteric effectors at any moment, which can be disadvantageous. Covalent modifica- tion, like allosteric regulation, is also rapid and reversible, because the converter enzymes act catalytically. Furthermore, covalent modification allows cells to escape allosteric regulation by covalently locking the modified enzyme in an active (or inac- tive) state, regardless of effector concentrations. One disadvan- tage is that covalent modification systems are often elaborate cascades that require many participants. 13. Sickle-cell anemia is the consequence of Hb S polymerization through hydrophobic contacts between the side chain of Val6 and a pocket in the EF corner of -subunits. Potential drugs might target this interaction directly or indirectly. For example, a drug might compete with Val6 side chain for binding in the EF corner. Alternatively, a useful drug might alter the conforma- tion of Hb S such that the EF corner was no longer accessible or accommodating to the Val6 side chain. Another possibility might be to create drug s that deter the polymerization process in other ways through alterations in the surface properties between Hb S molecules. 14. Nitric oxide is covalently attached to Cys93. Thus, the interac- tion is not reversible binding, as in allosteric regulation. On the other hand, the reaction of NOj with this cysteine residue is ap- parently spontaneous, and no converter enzyme is needed to add or remove it. Thus, the regulation of covalent modification that is afforded by converter enzyme involvement is obviated. The HbϺNOj interaction illustrates that nature does not always neatly fit the definitions that we create. 15. Pro: Lactate is a metabolic indicator of the need for oxygen; it binds to Mb at a distinct site (the allosteric site?) and it lowers Mb’s affinity for O 2 . Con: Mb is a monomeric protein. Traditionally, allosteric phe- nomena have been considered the realm of oligomeric proteins. How then is Mb “allosteric”? Since a ligand-induced conforma- tional change in a monomeric proteins can affect binding of “substrate” (i.e., O 2 ), the definition of allostery may need to be broadened. 16. The wedge-shaped protein monomers (red) assemble into trimers, but the alternative conformation for the monomer (square, green) forms tetramers. The substrate or allosteric reg- ulator (yellow) binds only to the square conformation, and its binding prevents the square from adopting the wedge confor- mation. Thus, if S or the allosteric regulator is present, equilib- rium favors a greater population of square tetramers among the morpheein ensemble at the expense of round trimers. 17. 18. Negative cooperativity in NAD ϩ binding to glyceraldehyde-3-P dehydrogenase: 0.2 0.4 0.6 V max v 0.8 1.0 30 69 Glutamine ATP or CTP K 0.5 [S] S ⌬ S ⌬ Where affinity for NADH is ϽϽ NAD ϩ NAD ϩ NAD ϩ NAD ϩ NAD ϩ Induced conformation change Abbreviated Answers to Problems A-17 19. Hyperventilation results in decreased blood pCO 2 , which in turn leads to an increase in pH. Thus, the affinity of Hb for O 2 in- creases (less O 2 will be released to tissues). Hypoventilation has exactly the opposite effects on pCO 2 and pH; thus, hypoventila- tion leads to a diminished affinity of Hb for O 2 (more O 2 will be released to the tissues). 20. When hormone disappears, the hormoneϺreceptor complex dis- sociates, the GTPase activity of the G ␣ subunit cleaves bound GTP to GDP ϩ P i , and the affinity of G ␣ –GDP for adenylyl cyclase is low, so it dissociates and adenylyl cyclase is no longer activated. Residual cAMP will be converted to 5Ј-AMP by phos- phodiesterase, and the catalytic subunits of protein kinase A will be bound again by the regulatory subunits, whose cAMP ligands have dissociated. When protein kinase A becomes inactive, phos- phorylase kinase will revert to the unphosphorylated, inactive form through loss of phosphoryl groups. Phophoprotein phos- phatase 1 will act on glycogen phosphorylase a, removing the phosphoryl group from Ser 14 and thereby converting glycogen phosphorylase to the less active, allosterically regulated b form. Chapter 16 1. The pronghorn antelope is truly a remarkable animal, with nu- merous specially evolved anatomical and molecular features. These include a large windpipe (to draw in more oxygen and exhale more carbon dioxide), lungs that are three times the size of those of comparable animals (such as goats), and lung alveoli with five times the surface area so that oxygen can diffuse more rapidly into the capillaries. The blood contains larger numbers of red blood cells and thus more hemoglobin. The skeletal and heart muscles are likewise adapted for speed and endurance. The heart is three times the size of that of comparable animals and pumps a proportionally larger volume of blood per contrac- tion. Significantly, the muscles contain much larger numbers of energy-producing mitochondria, and the muscle fibers them- selves are shorter and thus designed for faster contractions. All these characteristics enable the pronghorn antelope to run at a speed nearly twice the top speed of a thoroughbred racehorse and to sustain such speed for up to 1 hour. 2. Refer to Figure 16.9. The step in which the myosin head confor- mation change occurs, is the step that should be blocked by ,␥-methylene-ATP, because hydrolysis of ATP should occur in this step and ,␥-methylene-ATP is nonhydrolyzable. 3. Phosphocreatine is synthesized from creatine (via creatine kinase) primarily in muscle mitochondria (where ATP is readily generated) and then transported to the sarcoplasm, where it can act as an ATP buffer. The creatine kinase reaction in the sar- coplasm yields the product creatine, which is transported back into the mitochondria to complete the cycle. Like many mito- chondrial proteins, the expression of mitochondrial creatine kinase is directed by mitochondrial DNA, whereas sarcoplasmic creatine kinase is derived from information encoded in nuclear DNA. 4. Note in step 3 of Figure 16.9 that it is ATP that stimulates disso- ciation of myosin heads from the actin filaments—the dissocia- tion of the cross-bridge complex. When ATP levels decline (as happens rapidly after death), large numbers of myosin heads are unable to dissociate from actin and the muscle becomes stiff and unable to relax. 5. The skeletal muscles of the average adult male have a cross- sectional area of approximately 35,000 cm 2 . The gluteus maximus muscles represent approximately 300 cm 2 of this total. Assuming 4 kg of maximal tension per square centimeter of cross-sectional area, one calculates a total tension for the gluteus maximus of 1200 kg (as stated in the problem). The same calcu- lation shows that the total tension that could be developed by all the muscles in the body is 140,000 kg (or 154 tons)! 6. Taking 55,000 g/mol divided by 6.02 ϫ 10 23 /mol and by 1.3 g/mL, one obtains a volume of 7.03 ϫ 10 Ϫ20 mL. Assume a sphere and use the volume of a sphere (V ϭ (4/3)r 3 ) to obtain a radius of 25.6 Å. The diameter of the tubulin dimer (see Figure 16.12) is 8 nm, or 80 Å, and two times the radius we calculated here is approximately 51 Å, a reasonable value by comparison. 7. A liver cell is 20,000 nm long, which would correspond to 5000 tubulin monomers if using the value of Figure 16.12, and about 7800 tubulin monomers using the value of 25.6 Å calculated in problem 6. 8. 4 inches (length of the giant axon) ϭ 10.16 cm. Movement at 2 to 5 /sec would correspond to a time of 5.6 to 14 hours to traverse this distance. 9. 14 nm is 140 Å, which would be approximately 93 residues of a coiled coil. 10. Using Equation 9.1, one can calculate a ⌬G of 17,100 J/mol for this calcium gradient. 11. Using Equation 3.13, one can calculate a cellular ⌬G of Ϫ48,600 J/mol for ATP hydrolysis. 12. 17,100 J are required to transport 1 mole of calcium ions. Two moles of calcium would require 34,200 J, and three would cost 51,300 J. Thus, the gradient of problem 10 would provide enough energy to drive the transport of two calcium ions per ATP hydrolyzed. 13. Energy (or work) ϭ force ϫ distance. If 1 ATP is hydrolyzed per step and the cellular value of ⌬G°Ј for ATP hydrolysis is Ϫ50 kJ/mol, then the calculation of force exerted by a motor is force ϭ 50 kJ/mol/(step size). The SI unit of force is the newton, and 1 newton и meter ϭ 1 J. For the kinesin-1 motor, 50 kJ/mol/(8 ϫ 10 Ϫ9 m) ϭ 6.25 ϫ 10 12 newtons. For the myosin-V motor, 50 kJ/mol/(36 ϫ 10 Ϫ9 m) ϭ 1.39 ϫ 10 12 newtons. For the dynein motor, 50 kJ/mol/(28 ϫ 10 Ϫ9 m) ϭ 1.78 ϫ 10 12 newtons. 14. Perhaps the simplest way to view this problem is to consider the potential energy change for lifting a 10-kg weight 0.4 m. E ϭ mgh ϭ (10 kg)и(9.8 m/sec 2 )и(0.4 m) ϭ 39.2 J. Now, if 1 ATP is expended per myosin step along an actin filament, 39.2 J/(50 kJ/mol) ϭ 7.8 ϫ 10 Ϫ4 mol. Then, (7.8 ϫ 10 Ϫ4 mol)и(6.02 ϫ 10 23 ) ϭ 4.72 ϫ 10 20 molecules of ATP expended, one per step. So there must be 4.72 ϫ 10 20 myosin steps along actin. However, the stepping process is almost certainly not 100% efficient. As shown in the box on page 489, a step size of 11 nm against a force of 4 pN would cor- respond to an energy expended per myosin per step of 4.4 ϫ 10 Ϫ20 J. With this assumed energy expended per step, we can calculate 39 J/(4.4 ϫ 10 Ϫ20 J) ϭ 8.86 ϫ 10 20 steps total. If we take the step size value for skeletal myosin from the box on page 489 as 11 nm, one myosin head would have to take 0.4 m/11 ϫ 10 Ϫ9 m or 3.64 ϫ 10 7 steps to raise the weight a dis- tance of 0.4 m. Taking (8.86 ϫ 10 20 steps total)/(3.64 ϫ 10 7 steps per myosin) ϭ 2.43 ϫ 10 13 myosin heads involved per 11-nm step. A-18 Abbreviated Answers to Problems 15. All these are smooth muscle except for the diaphragm. (See http://chanteur.net/contribu/index.htm#http://chanteur.net/contribu/ cJMdiaph.htm for an explanation of the common misconception that the diaphragm is smooth muscle.) 16. 1, a; 2, e; 3, d; 4, c; 5, b. 17. This exercise is left to the student and will presumably be differ- ent for every student. 18. The 80 Å step of the kinesin motor is a 0.008 m step. To cover 10.16 cm would require 12.7 million steps. 19. The correct answer is d, although each answer is reasonable. ATP is needed for several processes involved in muscle relax- ation. Salt imbalances can also prevent normal muscle function, and interrupted blood flow could prevent efficient delivery of oxygen needed for ATP production during cellular respiration. 20. The correct answer is a. Understanding the inheritance pattern of sex-linked traits is essential. Males always inherit sex-linked traits (on the X chromosome) from their mothers. In addition, sex-linked traits are much more commonly expressed in males because they have only one X chromosome. Chapter 17 1. 6.5 ϫ 10 12 (6.5 trillion) people. 2. Consult Table 17.2. 3. O 2 , H 2 O, and CO 2 . 4. See Section 17.2. 5. Consult Figure 17.8. 6. Consult Corresponding Pathways of Catabolism and Anabolism Differ in Important Ways, p. 520. See also Figure 17.8. 7. See The ATP Cycle, p. 521; NAD ϩ Collects Electrons Released in Ca- tabolism, p. 522; and NADPH Provides the Reducing Power for Ana- bolic Processes, p. 523. 8. In terms of quickness of response, the order is allosteric regula- tion Ͼ covalent modification Ͼ enzyme synthesis and degrada- tion. See The Student Solutions Manual, Study Guide and Problems Book for further discussions. 9. See Metabolic Pathways Are Compartmentalized within Cells, p. 527, and discussions in The Student Solutions Manual, Study Guide and Problems Book. 10. Many answers are possible here. Some examples: Large numbers of metabolites in tissues and fluids; great diversity of structure of biomolecules; need for analytical methods that can detect and distinguish many different metabolites, often at very low concen- trations; need to understand relationships between certain metabolites; time and cost required to analyze and quantitate many metabolites in a tissue or fluid. 11. Mass spectrometry provides great sensitivity for detection of metabolites. NMR offers a variety of methods for resolving and discriminating metabolites in complex mixtures. Other compar- isons and contrasts are discussed in the references at the end of the chapter. 12. See Figure 13.23; The Coenzymes of the Pyruvate Dehydroge- nase Complex, page 568, and the Activation of Vitamin B 12 , page 712. 13. The genome is the entire hereditary information in an organism, as encoded in its DNA or (for some viruses) RNA. The transcrip- tome is the set of all messenger RNA molecules (transcripts) pro- duced in one cell or a population of cells under a defined set of conditions. The proteome is the entire complement of proteins produced by a genome, cell, tissue, or organism under a defined set of conditions. The metabolome is the complete set of low molecular weight metabolites present in an organism or excreted by it under a given set of physiological conditions. 14. A mechanism for liver alcohol dehydrogenase: 15. TCA cycle: in mitochondria; converts acetate units to CO 2 plus NADH and FADH 2 . Glycolysis: in cytosol; converts glucose to pyruvate. Oxidative phosphorylation: in mitochondria; uses electrons (from NADH and FADH 2 ) to produce ATP. Fatty acid synthesis: in cytosol; uses acetate units to synthesize fatty acids. 16. Most discussions of ocean sequestration of CO 2 are devoted to capture of CO 2 by phytoplankton or injection of CO 2 deep in the ocean. However, the box on page 512 offers another possibility—carbon sequestration in the shells of corals, mol- lusks, and crustaceans. This approach may be feasible. However, there are indications that these classes of sea organisms are already suffering from global warming and from pollution of ocean water. 17. 32 P and 35 S both decay via beta particle emission. A beta particle is merely an electron emitted by a neutron in the nucleus. Thus, beta decay does not affect the atomic mass, but it does convert a neutron to a proton. Thus, beta emission changes 32 P to 32 S and converts 35 S to 35 Cl: 32 P ⎯⎯→  Ϫ ϩ 32 S, t 1/2 ϭ 14.3 days. 35 S ⎯⎯→  Ϫ ϩ 35 Cl, t 1/2 ϭ 87.1 days. The decay equation is a first-order decay equation: A/A 0 ϭ e Ϫ0.693/t y 2 Thus, after 100 days of decay, the fraction of 32 P remaining would be 0.786% and the fraction of 35 S remaining would be 45%. 18. The correct answer is b. Obligate anaerobes can survive only in the absence of oxygen. 19. The correct answer is a. Fiber provides little or no nutrition. The foods containing fiber, however, have other nutritional sub- stances that we are able to digest and absorb. Chapter 18 1. a. Phosphoglucoisomerase, fructose bisphosphate aldolase, triose phosphate isomerase, glyceraldehyde-3-P dehydroge- nase, phosphoglycerate mutase, and enolase. b. Hexokinase/glucokinase, phosphofructokinase, phospho- glycerate kinase, pyruvate kinase, and lactate dehydrogenase. c. Hexokinase and phosphofructokinase. d. Phosphoglycerate kinase and pyruvate kinase. P P C O OOO C H H OH H O H 3 C O CH 3 NH 2 H A A A A A A A B N P C O HH NH 2 A A A A A N Abbreviated Answers to Problems A-19 e. According to Equation 3.13, reactions in which the number of reactant molecules differs from the number of product molecules exhibit a strong dependence on concentration. That is, such reactions are extremely sensitive to changes in concentration. Using this criterion, we can predict from Table 19.1 that the free energy changes of the fructose bis- phosphate aldolase and glyceraldehyde-3-P dehydrogenase reactions will be strongly influenced by changes in concentration. f. Reactions that occur with ⌬G near zero operate at or near equilibrium. See Table 18.1 and Figure 18.22. 2. The carboxyl carbon of pyruvate derives from carbons 3 and 4 of glucose. The keto carbon of pyruvate derives from carbons 2 and 5 of glucose. The methyl carbon of pyruvate is obtained from carbons 1 and 6 of glucose. 3. Increased [ATP], [citrate], or [glucose-6-phosphate] inhibits glycolysis. Increased [AMP], [fructose-1,6-bisphosphate], or [fructose-2,6-bisphosphate] stimulates glycolysis. 4. See The Student Solutions Manual, Study Guide and Problems Book for discussion. 5. The mechanisms for fructose bisphosphate aldolase and glyceraldehyde-3-P dehydrogenase are shown in Figures 18.12 and 18.13, respectively. 6. The relevant reactions of galactose metabolism are shown in Figure 18.24. See The Student Solutions Manual, Study Guide and Problems Book for mechanisms. 7. Iodoacetic acid would be expected to alkylate the reactive active- site cysteine that is vital to the glyceraldehyde-3-P dehydrogenase reaction. This alkylation would irreversibly inactivate the enzyme. 8. Ignoring the possibility that 32 P might be incorporated into ATP, the only reaction of glycolysis that utilizes P i is glyceraldehyde- 3-P dehydrogenase, which converts glyceraldehyde-3-P to 1,3-bisphosphoglycerate. 32 P i would label the phosphate at carbon 1 of 1,3-bisphosphoglycerate. The label will be lost in the next reaction, and no other glycolytic intermediates will be directly labeled. (Once the label is incorporated into ATP, it will also show up in glucose-6-P, fructose-6-P, and fructose-1, 6-bisphosphate.) 9. The sucrose phosphorylase reaction leads to glucose-6-P without the need for the hexokinase reaction. Direct production offers the obvious advantage of saving a molecule of ATP. 10. All of the kinases involved in glycolysis, as well as enolase, are activated by Mg 2ϩ ion. A Mg 2ϩ deficiency could lead to reduced activity for some or all of these enzymes. However, other sys- temic effects of a Mg 2ϩ deficiency might cause even more serious problems. 11. a. 7.5. b. 0.0266 mM. 12. ϩ1.08 kJ/mol. 13. a. Ϫ13.8 kJ/mol. b. K eq ϭ 194,850. c. [Pyr]/[PEP] ϭ 24,356. 14. a. Ϫ13.8 kJ/mol. b. K eq ϭ 211. c. [FBP]/[F-6-P] ϭ 528. 15. a. Ϫ19.1 kJ/mol. b. K eq ϭ 1651. c. [ATP]/[ADP] ϭ 13.8. 16. ⌬G°ЈϭϪ33.9 kJ/mol. 17. An 8% increase in [ATP] changes the [AMP] concentration to 20 M (from 5 M). 18. Formation of Schiff base intermediate Borohydride reduction of Schiff base intermediate Stable (trapped) E–S derivative Degradation of enzyme (acid hydrolysis) N 6 -dihydroxypropyl-L-lysine PO 3 2 – PO 3 2 – – PO 3 2 – CH 2 CH 2 CO O B Lys Lys OH CH 2 CH 2 CN O H HH H H OH + + Lys CH 2 CH 2 CNH O H OH Lys CH 2 CH 2 CNH H OH OH H 2 N A-20 Abbreviated Answers to Problems 20. Pyruvate kinase deficiency primarily affects red blood cells, which lack mitochondria and can produce ATP only via glycoly- sis. Absence of pyruvate kinase reduces the production of ATP by glycolysis in red cells, which in turn reduces the activity of the Na,K-ATPase (see Chapter 9). This results in alterations of elec- trolyte (Na ϩ and K ϩ ) concentrations, which leads to cellular dis- tortion, rigidity, and dehydration. Compromised red blood cells are destroyed by the spleen and liver. 21. The best answer is d. PFK is more active at low ATP than at high ATP, and F-2,6-bisP activates the enzyme. 22. Hexokinase is inhibited by high concentrations of glucose-6-P, so glycolysis would probably stop at such a high level of glucose- 6-P. Also, the increased concentration of glucose-6-P would change the cellular ⌬G of the hexokinase reaction to approxi- mately Ϫ22,000 J/mol. Chapter 19 1. Glutamate enters the TCA cycle via a transamination to form ␣-ketoglutarate. The ␥-carbon of glutamate entering the cycle is equivalent to the methyl carbon of an entering acetate. Thus, no radioactivity from such a label would be lost in the first or second cycle, but in each subsequent cycle, 50% of the total label would be lost (see Figure 19.15). 2. NAD ϩ activates pyruvate dehydrogenase and isocitrate dehydro- genase and thus would increase TCA cycle activity. If NAD ϩ increases at the expense of NADH, the resulting decrease in NADH would likewise activate the cycle by stimulating citrate synthase and ␣-ketoglutarate dehydrogenase. ATP inhibits pyru- vate dehydrogenase, citrate synthase, and isocitrate dehydroge- nase; reducing the ATP concentration would thus activate the cycle. Isocitrate is not a regulator of the cycle, but increasing its concentration would mimic an increase in acetate flux through the cycle and increase overall cycle activity. 3. For most enzymes that are regulated by phosphorylation, the co- valent binding of a phosphate group at a distant site induces a conformation change at the active site that either activates or in- hibits the enzyme activity. On the other hand, X-ray crystallo- graphic studies reveal that the phosphorylated and unphosphorylated forms of isocitrate dehydrogenase share iden- tical structures with only small (and probably insignificant) con- formation changes at Ser 113 , the locus of phosphorylation. What phosphorylation does do is block isocitrate binding (with no effect on the binding affinity of NADP ϩ ). As shown in Figure 2 of the paper cited in the problem (Barford, D., 1991. Molecular mechanisms for the control of enzymic activity by protein phos- phorylation. Biochimica et Biophysica Acta 1133:55–62), the ␥-carboxyl group of bound isocitrate forms a hydrogen bond with the hydroxyl group of Ser 113 . Phosphorylation apparently prevents isocitrate binding by a combination of a loss of the crucial H bond between substrate and enzyme and by repulsive electrostatic and steric effects. 4. A mechanism for the first step of the ␣-ketoglutarate dehydroge- nase reaction: 5. Aconitase is inhibited by fluorocitrate, the product of citrate syn- thase action on fluoroacetate. In a tissue where inhibition has occurred, all TCA cycle metabolites should be reduced in con- centration. Fluorocitrate would replace citrate, and the concen- trations of isocitrate and all subsequent metabolites would be reduced because of aconitase inhibition. 6. FADH 2 is colorless, but FAD is yellow, with a maximal ab- sorbance at 450 nm. Succinate dehydrogenase could be conve- niently assayed by measuring the decrease in absorbance in a solution of the flavoenzyme and succinate. 7. The central (C-3) carbon of citrate is reduced, and an adjacent carbon is oxidized in the aconitase reaction. The carbon bearing the hydroxyl group is obviously oxidized by isocitrate dehydrogenase. In the ␣-ketoglutarate dehydrogenase reaction, the departing carbon atom and the carbon adjacent to it are both oxidized. Both of the OCH 2 O carbons of succinate are oxidized in the succinate dehydrogenase reaction. Of these four molecules, all but citrate undergo a net oxidation in the TCA cycle. 8. Several TCA metabolite analogs are known, including malonate, an analog of succinate, and 3-nitro-2-S-hydroxypropionate, the anion of which is a transition-state analog for the fumarase reaction. OH A COO Ϫ A CH 2 COO Ϫ O C H H ϩ A C H E [ ( ð Δ E H Ϫ OOC H Ϫ B OH A C H N ϩ C H E ( ð E H Ϫ OOC O Ϫ B O H O Ϫ N ϩ Malonate (succinate analog) 3-Nitro-2-S- hydroxypropionate Transition-state analog for fumarase S Nϩ S S N S A O P CH 2 COO Ϫ COH TPP A O A A OCH 2 ϩ A OP CH 2 COO Ϫ CO A OOCH 2 COO Ϫ Ϫ A CO A O CH 2 COO Ϫ COH OOCH 2 Nϩ Nϩ H ϩ B A A O CH 2 COO Ϫ COH OOCH 2 H H O Ϫ B H AA H ␣ -KG Covalent TPP intermediate þ 19. UDPGlc k 1 k –1 E E E • UDPGlc k – 2 k 2 E-UMP • Glc-1-P Glc-1-P k 3 k –3 E-UMP Gal-1-P k 4 k –4 E-UMP • Gal-1-P k –5 k 5 E-UDPGal UDPGal k 6 k – 6 Abbreviated Answers to Problems A-21 9. A mechanism for pyruvate decarboxylase: 12. 14 C incorporated in a reversed TCA cycle would label the two carboxyl carbons of oxaloacetate in the first pass through the cycle. One of these would be eliminated in its second pass through the cycle. The other would persist for more than two cycles and would be eliminated slowly as methyl carbons in acetyl-CoA in the reversed citrate synthase reaction. 13. The labeling pattern would be the same as if methyl-labeled acetyl-CoA was fed to the conventional TCA cycle (see Figure 19.15). C CH 3 COO – O S N R" R R' B + S N R" R R' + BH S N R" R R' + C CH 3 CO 2 C OH OO S N R" R R' + C CH 3 OH S N R" R R' C CH 3 OH Resonance-stabilized carbanion on substrate H + S N R" R R' + C CH 3 O H H Hydroxyethyl-TPP S N R" R R' + – + CH 3 CH O E E H Acetaldehyde Pyruvate – – – 10. [isocitrate]/[citrate] ϭ 0.1. When [isocitrate] ϭ 0.03 mM, [citrate] ϭ 0.3 mM. 11. 14 CO 2 incorporation into TCA via the pyruvate carboxylase reac- tion would label the OCH 2 OCOOH carboxyl carbon in oxalo- acetate. When this entered the TCA cycle, the labeled carbon would survive only to the ␣-ketoglutarate dehydrogenase reac- tion, where it would be eliminated as 14 CO 2 . 14. Acetyl-CoA CH 2 C H SCoA O E B HC O COO – Glyoxalate OH Malate C H 2 C SCoA O HC O COO – E B + H HC COO – CH 2 C SCoA CoASH O OH HC COO – H 2 C H 2 O COO – – A-22 Abbreviated Answers to Problems 15. + H + Pyruvate Cytosol Malate + H + Oxaloacetate Acetyl-CoA + Citrate ATP–Citrate lyase Mitochondrial membrane Malate Oxaloacetate Citrate + H + Acetyl-CoA + Citrate synthase Malate dehydrogenase Malate dehydrogenase Pyruvate Mitochondrion ATP NADH NADPH NADH NADP + NAD + NAD + CoA ADP P i CoA CO 2 CO 2 CO 2 H 2 O 16. For the malate dehydrogenase reaction, Malate ϩ NAD ϩ 34 oxaloacetate ϩ NADH ϩ H ϩ with the value of ⌬G°Ј being ϩ30 kJ/mol. Then ⌬G°ЈϭϪRT ln K eq ϭϪ(8.314 J/molиK)(298)ln () ϭ ln (x/4.4 ϫ 10 Ϫ3 ) Ϫ12.1 ϭ ln (x/4.4 ϫ 10 Ϫ3 ) x ϭϪRT ln K eq x ϭ [oxaloacetate] ϭ 0.024 M Using the dimensions given in the problem, one can calculate a mitochondrial volume of 1.57 ϫ 10 Ϫ15 L. (0.024 ϫ 10 Ϫ6 M)и (1.57 ϫ 10 Ϫ15 L)и(6.02 ϫ 10 23 molecules/mole) ϭ 14.4, or about 14 molecules of OAA in a mitochondrion (“pOAA” ϭ 7.62). 17. This exercise is left to the student. Review of the calculation of oxidation numbers should be a prerequisite for answering this problem. 18. This exercise is left to the student. 19. 2R, 3R-fluorocitrate is converted by aconitase to 2-fluoro-cis- aconitate. This intermediate then rotates 180 degrees in the active site. Addition of hydroxide at the C4 position is followed by double bond migration from C3-C4 to C2-C3, to form 4-hydroxy-trans-aconitate. This product remains tightly bound at the aconitase active site, inactivating the enzyme. Studies by Trauble et al. have shown that the inhibitory product can be dis- placed by a 10 6 -fold excess of isocitrate. Ϫ30,000 J/mol ᎏᎏ 2478 J/mol [1]x ᎏᎏ [20][2.2 ϫ 10 Ϫ4 ] 20. Only eight ATPs would be generated in the succinyl-CoA synthe- tase (and nucleoside diphosphate kinase) reaction of the TCA cycle itself. 21. d is false. Succinyl-CoA is an inhibitor of citrate synthase. Chapter 20 1. The cytochrome couple is the acceptor, and the donor is the (bound) FAD/FADH 2 couple, because the cytochrome couple has a higher (more positive) reduction potential. ⌬Ᏹ o Јϭ0.254 VϪ0.02 V* *This is a typical value for enzyme-bound [FAD]. ⌬G ϭ ϪnᏲ⌬Ᏹ o Ј ϭϪ43.4 kJ/mol. 2. ⌬Ᏹ o Ј ϭϪ0.03 V; ⌬G ϭ ϩ5790 J/mol. 3. This situation is analogous to that described in Section 3.3. The net result of the reduction of NAD ϩ is that a proton is con- sumed. The effect on the calculation of free energy change is similar to that described in Equation 3.26. Adding an appropri- ate term to Equation 20.12 yields: Ᏹ ϭ Ᏹ o Ј ϪRT ln [H ϩ ] ϩ (RT/nᏲ ln ([ox]/[red]) 4. Cyanide acts primarily via binding to cytochrome a 3 , and the amount of cytochrome a 3 in the body is much lower than the amount of hemoglobin. Nitrite anion is an effective antidote for cyanide poisoning because of its unique ability to oxidize ferro- hemoglobin to ferrihemoglobin, a form of hemoglobin that competes very effectively with cytochrome a 3 for cyanide. The amount of ferrohemoglobin needed to neutralize an otherwise