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Book Problems Chapter 2 1. Identify the potential hydrogen bond donors and acceptors in the following molecules: (a) (b) (c) 2. Occasionally, a C−H group can form a hydrogen bond. Why would such a group be more likely to be a hydrogen bond donor group when the C is next to N? 8. Draw the structures of the conjugate bases of the following acids: a. b. c. d. 8. Draw the structures of the conjugate bases of the following acids: a. b. c. d. 10. 2 mL solutiion of pure water to whichh has been addded 50 mL off 1 mM HCl. Calculate the pH of a 200 14. w acid HA A if a solutionn containing 0.1 0 M HA andd 0.2 M A− haas a pH of 6.55? What is thhe pK of the weak 15. How manny grams of so odium succinaate (formula weight w 140 g · mol−1) and disodium succcinate (formuula weight 162 g · mool−1) must be added to 1 L of water to produce p a soluution with pH H 6.0 and a tottal solute concentration of 50 mM? 16. t volume off a solution of 5 M NaOH that must be added to adjuust the pH froom 4 to 9 in 1000 mL of a Estimate the 100 mM solution s of ph hosphoric acidd. 17. (a) Wouldd phosphoric acid or succinnic acid be a better b buffer at a pH 5? (b) Would W ammoonia or piperiddine be a better bufffer at pH 9? (c) ( Would HE EPES or Tris be b a better buuffer at pH 7.55? Chaapter 2 Answers: A : 1. (a) D Donors: NH1, NH2 at C2, NH9; N acceptorrs: N3, O at C6, C N7. (b) D Donors: NHl, NH N 2 at C4; accceptors: O att C2, N3. (c) D Donors: group, OH group; g acceptoors: COO− grroup, OH grouup. 2. A protonaated (and therrefore positiveely charged) nitrogen n would promote thhe separation of charge in the t adjacent C−H bondd so that the C would havee a partial neggative charge and the H woould have a paartial positivee charge. Thiss would maake the H morre likely to bee donated to a hydrogen boond acceptor group. g 8. (a) (b) (c) (d) 10. ue to the addiition of HCl iss (50 mL) (l mM)/(250 m mL L) = 0.2 mM = 2 × 10−4 M. Because thee The increaase in [H+] du − M, is relativ − ) or 3.7. [H+] of puure water, 10−7 vely insignificant, the pH of o the solutionn is equal to −log − (2 × 10−4 14. Use the Henderson–Ha H asselbalch equuation (Eq. 2--9) and solve for pK: 15. Let HA = sodium succinate and A− = disodium succinate. HA] = 0.05 M,, so [A−] = 0.005 M − [HA] [A−] + [H From Eq. 2-9 and Tablle 2-4, H − pK = 6.0 − 5.64 = 0.366 log ([A−]/[HA]) = pH [A−]/[H HA] = antilog 0.36 = 2.29 (0.05 M − [HA])/[HA A] = 2.29 [HA] = 0.015 M 0 M − 0.01 15 M = 0.0355 M [A−] = 0.05 grams of o sodium succcinate = (0.015 mol ⋅ L−1)((140 g ⋅ mol−1) × (1 L) = 2.1 2 g − )(162 g ⋅ mo grams of o disodium su uccinate = (0.035 mol ⋅ L−1 ol−1) × (1 L) = 5.7 g 16. At pH 4, essentially e alll the phosphooric acid is in the form, and at pH 9, essentiially all is in the t − form (Fig. 2-18). Thereefore, the conncentration off OH required is equivalennt to the conccentration of the t acid: (0.1 mol ⋅ L−1 phosphoric accid)(0.1 L) = 0.01 mol NaO OH required = (0.01 mol)((1 L/5 mol NaaOH) = 0.0022 L = 2 mL 17. (a) Suuccinic acid; (b) mmonia; am (c) PES. HEP Chaapter 4 1. Identify the t amino accids that difffer from eachh other by a single methyyl or methyllene group. 3. Identify the t hydrogen n bond donoor and accepttor groups inn asparagine. 4. Draw thee dipeptide Asp-His A at pH H 7.0. 6. Determinne the net ch harge of the predominant p t form of Aspp at (a) pH 1.0, 1 (b) pH 3.0, 3 (c) pH 6.0, and (d) pH 11.0. 7. H and (c) Glu. G Calculatee the pI of (aa) Ala, (b) His, 9. Circle thee chiral carb bons in the foollowing com mpounds: 13. Some am mino acids arre synthesizeed by replaciing the keto group (C=O O) of an orgaanic acid knoown as an αketo acidd with an am mino group ( from the following α-keto α acids: ). Identify thee amino acidds that can bee produced this way 14. t amino accid residue from fr which the t followingg compoundds are synthesized: Identify the 15. Draw thee peptide AT TLDAK. (a) Calculate itss approximaate pI. (b) Whhat is its net charge at pH H 7.0? 16. The proteein insulin consists of tw wo polypeptiides termed the t A and B chains. Insuulins from diifferent organism ms have been n isolated andd sequencedd. Human andd duck insulins have thee same aminoo acid sequencee with the ex xception of six amino aciid residues, as a shown below. Is the pI p of human insulin lower thaan or higher than that of duck insulinn? Amino acid d residue A8 A9 A10 B1 B2 B B27 Human Thhr Ser Ile Phe Val T Thr Duck Gluu Asn Pro Ala Ala S Ser Chapter 4 Answers: 1. Ser and Thr; Val, Leu, and Ile; Asn and Gln; Asp and Glu. 3. Hydrogen bond donors: α-amino group, amide nitrogen. Hydrogen bond acceptors: α-carboxylate group, amide carbonyl. 4. 6. (a) +1; (b) 0; (c) −1; 7. (a) pI = (2.35 + 9.87)/2 = 6.11 (b) pI = (6.04 + 9.33)/2 = 7.68 (c) pI = (2.10 + 4.07)/2 = 3.08 13. (a) Glutamate; 14. (a) Serine (N-acetylserine); (c) methionine (N-formylmethionine). (d) −2. 9. (b) aspartate (b) lysine (5-hydroxylysine); 15. (a) The pK’s of the ionizable side chains (Table 4-1) are 3.90 (Asp) and 10.54 (Lys); assume that the terminal Lys carboxyl group has a pK of 3.5 and the terminal Ala amino group has a pK of 8.0 (Section 4-1D). Thhe pI is apprroximately midway m betw ween the pK’’s of the twoo ionizations involving thhe neutral species (tthe pK of Assp and the N-terminal N pK K): (b) 16. T net charg The ge at pH 7.0 is 0 (as draw wn above). At positioon A8, duck k insulin has a Glu residuue, whereas human h insullin has a Thrr residue. Sinnce Glu is negativelly charged att physiological pH and Thr T is neutraal, human inssulin has a higher h pI thann duck insulin. (The ( other am mino acids thhat differ beetween the prroteins do noot affect the pI because they t are unchargeed.) Chaapter 5 1. Which peeptide has grreater absorbbance at 2800 nm? A. G Gln–Leu–Glu u–Phe–Thr–L Leu–Asp–G Gly–Tyr B. Ser–Val–Trp– –Asp–Phe–G Gly–Tyr–Trpp–Ala 2. (a) In whhat order wou uld the aminno acids Arg, His, and Leeu be eluted from a carbboxymethyl column c at pH 6? (b)) In what ord der would Glu, G Lys, andd Val be eluted from a diiethylaminoeethyl columnn at pH 8? 4. Determinne the subun nit compositiion of a proteein from the following innformation: Molecuular mass by y gel filtration: 200 kD Molecuular mass by y SDS-PAGE E: 100 kD Molecuular mass by y SDS-PAGE E with 2-merrcaptoethanool: 40 kD and a 60 kD 6. What fractionation prrocedure couuld be used to t purify prootein 1 from a mixture off three proteeins whose f amino accid composittions are as follows? 1. 25% Ala, 20% % Gly, 20% Ser, 10% Ilee, 10% Val, 5% Asn, 5% % Gln, 5% Prro 2. 30% Gln, 25% % Glu, 20% Lys, 15% Ser, S 10% Cyss 3. % Gly, 20% Asp, 20% Ser, S 10% Lys, 5% Tyr 25% Asn, 20% All three proteins aree similar in size s and pI, and a there is no n antibody available foor protein 1. 9. You must cleave c the fo ollowing pepptide into sm maller fragmeents. Which of the proteaases listed inn Table 5-3 would be liikely to yield the most fragments? fr T fewest? The NMTQGRCKPVNTFVHEPLVDVQNVCFKE 11. Separate cleavage reactions of a polypeptide by CNBr and chymotrypsin yield fragments with the following amino acid sequences. What is the sequence of the intact polypeptide? CNBr treatment 1. Arg–Ala–Tyr–Gly–Asn 2. Leu–Phe–Met 3. Asp–Met 6. Gly–Asn Chymotrypsin 4. 13. Met–Arg–Ala–Tyr 5. Asp–Met–Leu–Phe Treatment of a polypeptide with 2-mercaptoethanol yields two polypeptides: 1. Ala-Val-Cys-Arg-Thr-Gly-Cys-Lys-Asn-Phe-Leu 2. Tyr-Lys-Cys-Phe-Arg-His-Thr-Lys-Cys-Ser Treatment of the intact polypeptide with trypsin yields fragments with the following amino acid compositions: 3. (Ala, Arg, Cys2, Ser, Val) 4. (Arg, Cys2, Gly, Lys, Thr, Phe) 5. (Asn, Leu, Phe) 6. (His, Lys, Thr) 7. (Lys, Tyr) Indicate the positions of the disulfide bonds in the intact polypeptide Chapter 5 Answers 1. Peptide B, because it contains more Trp and other aromatic residues. 2. (a) Leu, His, Arg. (b) Lys, Val, Glu. 4. The protein contains two 60-kD polypeptides and two 40-kD polypeptides. Each 40-kD chain is disulfide bonded to a 60-kD chain. The 100-kD units associate noncovalently to form a protein with a molecular mass of 200 kD. 6. Because protein 1 has a greater proportion of hydrophobic residues (Ala, Ile, Pro, Val) than do proteins 2 and 3, hydrophobic interaction chromatography could be used to isolate it. 9. Thermolysin would yield the most fragments (9) and endopeptidase V8 would yield the fewest (2). 11. Asp–Met–Leu–Phe–Met–Arg–Ala–Tyr–Gly–Asn 13. Chapter 6 5. Globular proteins are typically constructed from several layers of secondary structure, with a hydrophobic core and a hydrophilic surface. Is this true for a fibrous protein such as α keratin? 9. Is it possible for a native protein to be entirely irregular, that is, without α helices, β sheets, or other repetitive secondary structure? 12. You are performing site-directed mutagenesis to test predictions about which residues are essential for a protein’s function. Which of each pair of amino acid substitutions listed below would you expect to disrupt protein structure the most? Explain. (a) Val replaced by Ala or Phe. (b) Lys replaced by Asp or Arg. (c) Gln replaced by Glu or Asn. (d) Pro replaced by His or Gly. 15. Describe the intra- and intermolecular bonds/interactions that are broken or retained when collagen is heated to produce gelatin. 16. Under physiological conditions, polylysine assumes a random coil conformation. Under what conditions might it form an α helix? 18. Which of the following polypeptides is most likely to form an α helix? Which is least likely to form a β strand? (a) CRAGNRKIVLETY (c) QKASVEMAVRNSG (b) SEDNFGAPKSILW Chapter 6 Answers 5. A fibrous protein such as α keratin does not have a discrete globular core. Most of the residues in its coiled coil structure are exposed to the solvent. The exception is the strip of nonpolar side chains at the interface of the two coils. 9. Yes, although such irregularity should not be construed as random. 12. (a) Phe. Ala and Phe are both hydrophobic, but Phe is much larger and might not fit as well in Val’s place. (b) Asp. Replacing a positively charged Lys residue with an oppositely charged Asp residue would be more disruptive. (c) Glu. The amide-containing Asn would be a better substitute for Gln than the acidic Glu. (d) His. Pro’s constrained geometry is best approximated by Gly, which lacks a side chain, rather than a residue with a bulkier side chain such as His. 15. Hydrophobic effects, van der Waals interactions, and hydrogen bonds are destroyed during denaturation. Covalent cross-links are retained. 16. At physiological pH, the positively charged Lys side chains repel each other. Increasing the pH above the pK (>10.5) would neutralize the side chains and allow an α helix to form. 18. Peptide c is most likely to form an α helix with its three charged residues (Lys, Glu, and Arg) aligned on one face of the helix. Peptide a has adjacent basic residues (Arg and Lys), which would destabilize a helix. Peptide b contains Gly and Pro, both of which are helix-breaking (Table 6-1). The presence of Gly and Pro would also inhibit the formation of β strands, so peptide b is least likely to form a β strand. Chapter 9 4. Draw the structure of a glycerophospholipid that has a saturated C16 fatty acyl group at position 1, a monounsaturated C18 fatty acyl group at position 2, and an ethanolamine head group. 13. When bacteria growing at 20°C are warmed to 30°C, are they more likely to synthesize membrane lipids with (a) saturated or unsaturated fatty acids, and (b) short-chain or long-chain fatty acids? Explain. 14. (a) How many turns of an α helix are required to span a lipid bilayer (∼30 Å across)? (b) What is the minimum number of residues required? (c) Why do most transmembrane helices contain more than the minimum number of residues? Chapter 9 Answers 4. 13. (a) Saturated; (b) long-chain. By increasing the proportion of saturated and long-chain fatty acids, which have higher melting points, the bacteria can maintain constant membrane fluidity at the higher temperature. 14. (a) (1 turn/5.4 Å)(30 Å) = 5.6 turns (b) (3.6 residues/turn)(5.6 turns) = 20 residues (c) The additional residues form a helix, which partially satisfies backbone hydrogen bonding requirements, where the lipid head groups do not offer hydrogen bonding partners. Chapter 10 2. Rank the rate of transmembrane diffusion of the following compounds: 3. Calculate the free energy change for glucose entry into cells when the extracellular concentration is 5 mM and the intracellular concentration is 3 mM. 4. (a) Calculate the chemical potential difference when intracellular [Na+] = 10 mM and extracellular [Na+] = 150 mM at 37°C. (b) What would the electrochemical potential be if the membrane potential were −60 mV (inside negative)? 8. The rate of movement (flux) of a substance X into cells was measured at different concentrations of X to construct the graph below. (a) Does this information suggest that the movement of X into the cells is mediated by a protein transporter? Explain. (b) What additional experiment could you perform to verify that a transport protein is or is not involved? 10. Endothelial cells and pericytes in the retina of the eye have different mechanisms for glucose uptake. The figure shows the rate of glucose uptake for each type of cell in the presence of increasing amounts of sodium. What do these results reveal about the glucose transporter in each cell type? 11. T compoun The nd shown below is the anntiparasitic drug d miltefosine. (a) Iss this compo ound a glycerrophospholippid? (b) H does miiltefosine likkely cross thee parasite cell membranee? How (c) Inn what part of o the cell would the druug tend to acccumulate? Explain. E (d) Miltefosine M binds b to a prootein that alsso binds som me sphingolippids and som me glyceroophospholipiids. What feature common to all thesse compounds is recognnized by the protein? p Thee proteinn does not bin nd triacylglyycerols. 15. Proteins known as multidrug m resiistance (MD DR) transportters use the free f energy of o ATP hydrrolysis to pump a variety v of hy ydrophobic suubstances ouut of cells. (aa) Write the reaction for ATP hydrollysis and explain, in i structural terms, how the transporrter might takke advantage of this proocess to drivee transport. There is no n phosphorrylated proteein intermediate, as in thhe (Na+–K+)––ATPase. (b) Why would overexprression of an n MDR transpporter in a cancer c cell make m the canccer more diffficult to treaat? Chaapter 10 Answerss 2. 3. The less polar a subsstance, the faaster it can diffuse d througgh the lipid bilayer. b From m slowest too fastest: C, A, B. 4. (a) (b) 8. T data do not The n indicate the t involvem ment of a trannsport proteiin, since the rate of transsport does (a) not approoach a maxim mum as [X] increases. (b) T verify thatt a transport protein is innvolved, incrrease [X] to demonstratee saturation of To o the transportter at high [X X], or add a structural s annalog of X too compete with X for binnding to the transporter, t resulting in a lower flux f of X. 10. The hypeerbolic curvee for glucosee transport innto pericytess indicates a protein-meddiated sodium m-dependennt + process. The transporrt protein haas binding sittes for sodiuum ions. At low l [Na ], glucose transpport is + + + directly proportional p to [Na ]. Hoowever, at high h [Na ], alll Na bindinng sites on thhe transport protein are occupiedd, and thus gllucose transpport reaches a maximum m velocity. Glucose G transsport into endothelial + cells is not sodium-d dependent annd occurs at a high rate whether w or noot Na is preesent. There is not enough innformation in i the figure to determinne whether gllucose transpport into enddothelial cellls is protein-mediatedd. 11. (a) N there is no No; n glycerol backbone. b (b) M Miltefosine iss amphipathiic and thereffore cannot cross c the parrasite cell meembrane by diffusion. Since it is i not a norm mal cell compponent, it proobably does not have a dedicated d acctive transporrter. It most likely entters the cell via a passivee transport protein. p (c) T amphipaathic molecuule most likely accumulaates in membbranes, with its hydrophoobic tail This buried inn the bilayer and its polarr head groupp exposed to the solvent. (d) T protein reecognizes thhe phosphochholine head group, which also occurrs in some spphingolipids The and somee glyceropho ospholipids. Since the prrotein does not n bind all phospholipid p ds or triacylgglycerols, it does not recognize th he hydrocarbbon tail. 15. (a) ATP + H2O → ADP + Pi The transporter must include a cytosolic nucleotide binding site that changes its conformation when its bound ATP is hydrolyzed to ADP. This conformational change must be communicated to the membrane-spanning portion of the protein, where the transported substrate binds. (b) Overexpression of an MDR transporter would increase the ability of the cancer cell to excrete anticancer drugs. Higher concentrations of the drugs or different drugs would then be required to kill the drug-resistant cells. Chapter 11 7. Studies at different pH’s show that an enzyme has two catalytically important residues whose pK’s are ∼4 and ∼10. Chemical modification experiments indicate that a Glu and a Lys residue are essential for activity. Match the residues to their pK’s and explain whether they are likely to act as acid or base catalysts. 18. Predict the effect of mutating Asp 102 of trypsin to Asn (a) on substrate binding and (b) on catalysis. 20. Why is the broad substrate specificity of chymotrypsin advantageous in vivo? Why would this be a disadvantage for some other proteases? Chapter 11 Answers 7. Glu has a pK of ∼4 and, in its ionized form, acts as a base catalyst. Lys has a pK of ∼10 and, in its protonated form, acts as an acid catalyst. 18. (a) Little or no effect; (b) catalysis would be much slower because the mutation disrupts the function of the catalytic triad. 20. As a digestive enzyme, chymotrypsin’s function is to indiscriminately degrade a wide variety of ingested proteins, so that their component amino acids can be recovered. Broad substrate specificity would be dangerous for a protease that functions outside of the digestive system, since it might degrade proteins other than its intended target. Chapter 12 6. Explain why it is usually easier to calculate an enzyme’s reaction velocity from the rate of appearance of product rather than the rate of disappearance of a substrate. 7. At what concentration of S (expressed as a multiple of KM) will vo = 0.95 Vmax? 10. Calculate KM and Vmax from the following data: [S] (μM) vo (mM · s−1) 0.1 0.34 0.2 0.53 0.4 0.74 0.8 0.91 1.6 1.04 13. You are constructing a velocity versus [substrate] curve for an enzyme whose KM is believed to be about 2 μM. The enzyme concentration is 200 nM and the substrate concentrations range from 0.1 μM to 10 μM. What is wrong with this experimental setup and how could you fix it? 15. Is it necessary to know [E]T in order to determine (a) KM, (b) Vmax, or (c) kcat? 16. The KM for the reaction of chymotrypsin with N-acetylvaline ethyl ester is 8.8 × 10−2 M, and the KM for the reaction of chymotrypsin with N-acetyltyrosine ethyl ester is 6.6 × 10−4 M. (a) Which substrate has the higher apparent affinity for the enzyme? (b) Which substrate is likely to give a higher value for Vmax? 19. Determine the type of inhibition of an enzymatic reaction from the following data collected in the presence and absence of the inhibitor. [S] (mM) vo (mM · min−1) vo with I present (mM · min−1) 1 1.3 0.8 2 2.0 1.2 4 2.8 1.7 8 3.6 2.2 12 4.0 2.4 25. Sphingosine-1-phosphate (SPP) is important for cell survival. The synthesis of SPP from sphingosine and ATP is catalyzed by the enzyme sphingosine kinase. An understanding of the kinetics of the sphingosine kinase reaction may be important in the development of drugs to treat cancer. The velocity of the sphingosine kinase reaction was measured in the presence and absence of threo-sphingosine, a stereoisomer of sphingosine that inhibits the enzyme. The results are shown below. [Sphingosine] (μM) vo (mg · min−1) (no inhibitor) vo (mg · min−1) (with threo-sphingosine) 2.5 32.3 8.5 3.5 40 11.5 5 50.8 14.6 10 72 25.4 20 87.7 43.9 50 115.4 70.8 Construct a Lineweaver–Burk plot to answer the following questions: (a) What are the apparent KM and Vmax values in the presence and absence of the inhibitor? (b) What kind of an inhibitor is threo-sphingosine? Explain. Chapter 12 Answers 6. Enzyme activity is measured as an initial reaction velocity, the velocity before much substrate has been depleted and before much product has been generated. It is easier to measure the appearance of a small amount of product from a baseline of zero product than to measure the disappearance of a small amount of substrate against a background of a high concentration of substrate. 7. vo = Vmax[S]/(KM + [S]) vo/Vmax = [S]/(KM + [S]) 0.95 = [S]/(KM + [S]) [S] = 0.95KM + 0.95[S] 0.05[S] = 0.95 KM [S] = (0.95/0.05)KM = 19KM 10. Construct a Lineweaver–Burk plot. KM = −1/x-intercept = −1/(−4 μM−1) = 0.25 μM Vmax = 1/y-intercept = 1/(0.8 mM−1 ⋅ s) = 1.25 mM ⋅ s−1 13. The enzyme concentration is comparable to the lowest substrate concentration and therefore does not meet the requirement that [E] ≪ [S]. You could fix this problem by decreasing the amount of enzyme used for each measurement. 15. (a, b) It is not necessary to know [E]T. The only variables required to determine KM and Vmax (for example, by constructing a Lineweaver–Burk plot) are [S] and vo. (c) The value of [E]T is required to calculate kcat since kcat = Vmax/[E]T. 16. (a) N-Acetyltyrosine ethyl ester, with the lower value of KM, has greater apparent affinity for chymotrypsin. (b) The value of Vmax is not related to the value of KM, so no conclusion can be drawn. 19. The lines of the double-reciprocal plots intersect to the left of the 1/vo axis (on the 1/[S] axis). Hence, inhibition is mixed (with α = α′). [S] 1/[S] 1/vo 1/vo with I 1 1.00 0.7692 1.2500 2 0.50 0.5000 0.8333 4 0.25 0.3571 0.5882 8 0.125 0.2778 0.4545 12 0.083 0.2500 0.4167 25. (a) KM is determiined from thhe x-interceptt (= −1/KM). In the absennce of inhibiitor, KM = 1//0.14 μM−1 = 7 μM. In the presence of inhibitoor, . Vmax is i determinedd from the yy−1 intercept (= 1/Vmax). In I the absennce of inhibittor, Vmax = 1//0.008 mg ⋅ min = 1255 mg ⋅ min−1. In the presence of inhibitor, . (b) T lines in th The he double-reeciprocal ploots intersect very v close too the 1/vo axxis. Hence, thhreosphingossine is most likely l a com mpetitive inhiibitor. Comppetitive inhibbition is likelly also becauuse of the structural similarity between b the inhibitor and the substraate, which alllows them to t compete for f binding to the enzzyme active site. Chaapter 13 2. A certainn metabolic reaction r takees the form A → B. Its sttandard free energy channge is 7.5 kJ · mol−1. (a) Calculatee the equilibrium constannt for the reaaction at 25°°C. (b) Calcuulate ∆G at 37°C 3 when the t concentraation of A iss 0.5 mM and the concenntration of B is 0.1 mM. Is the reactiion spontaneeous under these connditions? (c) How might the reactionn proceed in the cell? 3. Choose the t best defin nition for a near-equilibr n rium reactionn: 7. a. allways operattes with a faavorable freee energy channge. b. has a free eneergy change near zero. c. iss usually a co ontrol point in a metabolic pathway.. d. opperates very y slowly in viivo. Predict whether w creattine kinase will w operate in i the directiion of ATP synthesis s or phosphocreatine synthesiss at 25°C wh hen [ATP] = 4 mM, [AD DP] = 0.15 mM, m [phosphoocreatine] = 2.5 mM, annd [creatine] = 1 mM. 8. If intraceellular [ATP]] = 5 mM, [A ADP] = 0.5 mM, m and [Pi] = 1.0 mM,, calculate thhe concentraation of AMP P at pH 7 and a 25°C und der the conddition that thee adenylate kinase k reactiion is at equuilibrium. 13. A hypothhetical three--step metaboolic pathwayy consists of intermediatees W, X, Y, and Z and enzymes e A, B, and C. Deduce thee order of the enzymatic steps in the pathway froom the follow wing inform mation: 14. 1. C Compound Q, Q a metaboliic inhibitor of o enzyme B, causes Z too build up. 2. A mutant in enzyme e C requires Y forr growth. 3. A inhibitor of An o enzyme A causes W, Y, and Z to accumulate.. 4. C Compound P, a metabolicc inhibitor of enzyme C,, causes W and a Z to build up. A certainn metabolic pathway p cann be diagram med as where A,, B, C, and D are the inteermediates, and a X, Y, annd Z are the enzymes thaat catalyze thhe reactions.. The physsiological freee energy chhanges for thhe reactions are a (a) Which reaction iss likely to bee a major reggulatory poinnt for the patthway? (b) Iff your answeer in Part a was in faact the case, in the presennce of an inhhibitor that blocks b the acctivity of enzzyme Z, wouuld the concentraations of A, B, C, and D increase, deecrease, or not n be affecteed? Chaapter 13 Answerss 2. (a) Since ∆G°′ = −RT ln K, (b) The reaaction is not spontaneous since ∆G > 0. (c) The T reaction can proceedd in the cell if the producct B is the substrate for a second reacction such that thee second reacction continuually draws off B, causinng the first reaction r to coontinually prroduce moree B from m A. 3. b 7. Calculatiing ∆G for th he reaction ATP A + creatiine ⇌ phosphhocreatine + ADP, usingg Eq. 13-1: Since ∆G G > 0, the reaaction will proceed p in thhe opposite direction d as written w above, that is, in the directionn of ATP synthesis. s 8. Using thee data in Tab ble 13-2, we calculate ∆G G°′ for the adenylate a kinnase reactionn. Since ∆G G for a reactiion at equilibbrium is zeroo, Eq. 13-1 becomes b ∆G G°′ = −RT ln Keq so that Keq = −∆G°′/RT e . 13. 14. s catalyzed by enzymee Y is likely to be the maajor flux-conntrol point, since s this steep operates (a) The step farthest from f equilibrrium (it is ann irreversiblee step). (b) Inhibition off enzyme Z would w cause the concentraation of D, th he reaction’s product, too decrease, and a it would cause C, thee reaction’s substrate, s to accumulaate. The concentrations of o A and B would w not chhange becausse the steps catalyzed byy enzymes X and Y woould not be affected. a Thee accumulateed C would not be transfformed backk to B since the t step catalyzedd by enzymee Y is irreverrsible. Chaapter 14 1. Which off the 10 reacctions of glyccolysis are (aa) phosphoryylations, (b) isomerizatioons, (c) oxiddation– reductionns, (d) dehyd drations, andd (e) carbon––carbon bondd cleavages?? 4. Arsenatee ( ), a structural analog of phhosphate, caan act as a suubstrate for any a reaction in which phosphatte is a substrrate. Arsenatte esters, unlike phosphaate esters, aree kinetically as well as thermodyynamically unstable u and hydrolyze almost a instanntaneously. Write W a balannced overalll equation for the coonversion off glucose to pyruvate p in the t presencee of ATP, AD DP, NAD+, and a either (aa) phosphate or (b) arssenate. (c) Why W is arsenaate a poison?? 7. ∆G°′ for the aldolasee reaction is 22.8 kJ · mool−1. In the ceell at 37°C, [DHAP]/[GA [ AP] = 5.5. Calculate C the −4 equilibriuum ratio of [FBP]/[GAP [ P] when [GA AP] = 10 M. M 8. The half--reactions in nvolved in thhe lactate dehhydrogenasee (LDH) reacction and theeir standard reduction r potentials are Calculatee ∆G at pH 7.0 7 for the LDH-catalyzeed reductionn of pyruvatee under the following fo connditions: (a) [llactate]/[pyru uvate] = 1 annd [NAD+]/[[NADH] = 1 (b) [llactate]/[pyru uvate] = 1600 and [NAD+]/[NADH] = 160 (c) [llactate]/[pyru uvate] = 10000 and [NAD D+]/[NADH]] = 1000 (d) D Discuss the effect of the concentratio c n ratios in parts a–c on the t directionn of the reacttion. 9. Althoughh it is not thee primary fluux-control pooint for glyccolysis, pyruvvate kinase is i subject to allosteric regulation. (a) What is the metabbolic importaance of regullating flux thhrough the pyruvate p kinaase reaction?? (b) Whatt is the advan ntage of activvating pyruvvate kinase with w fructosee-1,6-bisphoosphate? 10. Comparee the ATP yield of three glucose mollecules that enter e glycolyysis and are converted too pyruvate with thatt of three glu ucose molecuules that proceed throughh the pentose phosphate pathway such that their carbon skkeletons (as two F6P andd one GAP) re-enter glycolysis and are a metaboliized to pyruvvate. Chaapter 14 Answerss 1. (a) Reacttions 1, 3, 7, and 10; (b) Reactions 2, 2 5, and 8; (cc) Reaction 6; 6 (d) Reactiion 9; (e) Reeaction 4. 4. (a) G Glucose + 2 NAD N + + 2 ADP A + 2 Pi → 2 pyruvate + 2 NADH + 2 ATP + 2 H2O (b) (c) Arsenate A is a poison becaause it uncouuples ATP geeneration froom glycolysiis. Consequeently, glycolyytic energy generation g caannot occur. 7. − M. Accord When [G GAP] = 10−4 M, [DHAP] = 5.5 × 10−4 ding to Eq. 1-17, 1 K = e−∆∆G°′/RT 8. For the coupled c reacttion Pyruvatee + NADH + H + → lacttate + NAD D + ∆ℰ°′ = (− −0.185 V) − (−0.315 V) = 0.130 V. Accordinng to Eq. 13--8 and ∆G = −nℱ∆ℰ (Eq. 13-7). Sinnce two electtrons are trannsferred in the t above reaaction, n = 2. 2 (a) (b) (c) (d) At A the concen ntration ratioos of Part a, ∆G ∆ is negatiive and the reaction r procceeds as writtten. As [lactatee]/[pyruvate]] increases, the t reaction ∆G increasees even thouggh [NAD+]/[[NADH] also increases so that in Part b thee reaction is at equilibriuum (∆G = 0) and in Part c ∆G is posiitive and thee reaction proceedds spontaneo ously in the opposite direection. 9. (a) Pyruvate kinaase regulatioon is importaant for controolling the fluux of metaboolites, such as a fructose (in liver), which enteer glycolysis after the PF FK step. (b) F is the pro FBP oduct of the third reactioon of glycolyysis, so it acts as a feed-forward actiivator of the enzyme that t catalyzees Step 10. This T regulatoory mechanissm helps enssure that oncce metabolitees pass the PFK stepp of glycolyssis, they willl continue thhrough the paathway. 10. The three glucose molecules that proceed through glycolysis yield 6 ATP. The bypass through the pentose phosphate pathway results in a yield of 5 ATP. Chapter 15 1. Indicate the energy yield or cost, in ATP equivalents, for the following processes: (a) glycogen (3 residues) → 6 pyruvate (b) 3 glucose → 6 pyruvate (c) 6 pyruvate → 3 glucose 2. Write the balanced equation for (a) the sequential conversion of glucose to pyruvate and of pyruvate to glucose and (b) the catabolism of six molecules of G6P by the pentose phosphate pathway followed by conversion of ribulose-5-phosphate back to G6P by gluconeogenesis. 10. Individuals with McArdle’s disease often experience a “second wind” resulting from cardiovascular adjustments that allow glucose mobilized from liver glycogen to fuel muscle contraction. Explain why the amount of ATP derived in the muscle from circulating glucose is less than the amount of ATP that would be obtained by mobilizing the same amount of glucose from muscle glycogen. Chapter 15 Answers 1. (a) +9 ATP, (b) +6 ATP, 2. (a) The equation for glycolysis is (c) −18 ATP. Glucose +2 NAD + + 2 ADP + 2 Pi → 2 pyruvate + 2 NADH + 4 H + + 2 ATP + 2 H2 O The equation for gluconeogenesis is 2 Pyruvate + 2 NADH + 4 H + + 4 ATP + 2 GTP + 6 H2 O → glucose + 2 NAD + + 4 ADP + 2 GDP + 6 Pi For the two processes operating sequentially, 2 ATP + 2 GTP + 4 H2 O → 2 ADP + 2 GDP + 4 Pi (b) The equation for catabolism of 6 G6P by the pentose phosphate pathway is 6 G 6 P + 12 NADP + + 6 H2 O → 6 Ru 5 P + 12 NADPH + 12 H + + 6 CO2 Ru5P can be converted to G6P by transaldolase, transketolase, and gluconeogenesis: 6 Ru 5 P + H2 O → 5 G 6 P + Pi The net equation is therefore G 6 P + 12 NAD DP + + 7 H2 O → 12 NAD DPH + 12 H H + + 6 CO2 + Pi 10. The convversion of cirrculating gluucose to lacttate in the muuscle generaates 2 ATP. If I muscle glyycogen could be mobilized, the t energy yield y would be b 3 ATP, since phosphoorolysis of glycogen g byppasses the hexokinaase-catalyzed d step that coonsumes AT TP in the firsst stage of glyycolysis. Chaapter 16 12. Given thee following information, calculate thhe physiologgical ∆G of the t isocitratee dehydrogennase reactionn + at 25°C and a pH 7.0: [NAD ]/[NA ADH] = 8, [α α-ketoglutarate] = 0.1 mM; m and [isoccitrate] = 0.002 mM. Assume standard s con nditions for CO C 2 (∆G°′ iss given in Taable 16-2). Is I this reactioon a likely siite for metabolic control? 13. Althoughh animals cannot synthessize glucose from acetyl-CoA, if a raat is fed 14C--labeled acettate, some off the label appears in glycogen g exttracted from its muscles.. Explain. Chaapter 16 Answerss 12. For the reeaction isociitrate + NAD D+ ⇌ α-ketogglutarate + NADH N + CO O2 + H+, we assume a [H+] and [CO2] = 1. Accoording to Eq q. 13-1, With succh a large neg gative free energy e of reaaction under physiologiccal conditionns, isocitrate dehydroggenase is likeely to be a metabolic m conntrol point. 13. Animals cannot carry y out the nett synthesis off glucose froom acetyl-CooA (to whichh acetate is converted). c 14 Howeverr, C-labeled d acetyl-CoA A enters the citric acid cycle and is converted c to oxaloacetatte. Some of this oxalooacetate may y exchange with w the celllular pool off oxaloacetate to be convverted to gluccose throughh gluconeoogenesis and d subsequentlly taken up by b muscle annd incorporaated into glycogen. Chaapter 17 2. How manny ATPs aree synthesizedd for every cytoplasmic c N NADH that participates in the glyceerophosphatee shuttle inn insect fligh ht muscle? How H does this compare too the ATP yiield when NADH N reduciing equivalennts are transfferred into thhe matrix viaa the malate–aspartate shhuttle? 3. Calculatee ∆G°′ for th he oxidation of free FAD DH2 by O2. What W is the maximum m nuumber of AT TPs that can be syntheesized, assum ming standarrd conditions and 100% conservationn of energy?? 5. Why is itt possible for electrons too flow from a redox cennter with a more m positive ℰ°′ to one with w a more negative ℰ°′ within an a electron-trransfer compplex? 7. The diffeerence in pH H between thee internal annd external suurfaces of thhe inner mitoochondrial membrane m is 1.4 pH unnits (externaal side acidicc). If the mem mbrane poteential is 0.06 V (inside neegative) whaat is the free energy reeleased on trransporting 1 mol of prottons back accross the mem mbrane? Hoow many prootons must be transpported to provide enoughh free energyy for the syntthesis of 1 mol m of ATP (assuming ( sttandard biochemiical conditio ons)? 11. Explain why w compou unds such ass DNP increaase metaboliic rates. Chaapter 17 Answerss 2. When NA ADH particiipates in the glycerophossphate shuttlle, the electroons of NAD DH flow to FA AD and thenn to CoQ, bypassing b Complex C I. Thhus, about 2 ATP are syynthesized peer NADH. About A three ATP A are producedd when NAD DH participattes in the maalate–aspartaate shuttle. 3. The relevvant half-reaactions (Tablle 13-3) are Since thee O2/H2O hallf-reaction has h the more positive ∆ℰ ℰ°′, the FAD half-reaction is reversedd and the overall reeaction is Since ∆G G°′ = −nℱ∆ℰ ℰ°′, ∆ G °′ = − (2)(96,4 485 J ⋅ V − 1 ⋅ mol − 1)(1.0034 V ) = − 200 kJ ⋅ mol − 1 The maximum numb ber of ATP thhat could be synthesizedd under standdard conditioons is therefo fore 200 kJ ⋅ mol−1/30.5 kJ ⋅ mol−11 = 6.6 mol ATP/mol A FA ADH2 oxidizzed by O2. 5. ℰ may diiffer from ℰ°°′, dependingg on the redoox center’s microenviron m nment and thhe concentraations of reactantss and productts. In additioon, the tight coupling bettween succeessive electroon transfers within w a complex may “pull” electrons soo that the oveerall process is spontaneous. 7. For the trransport of a proton from m outside to inside (Eq. 17-1), 1 ∆ G = 2..3 RT[ pH ( in i ) − pH ( out o )] + Zℱ ∆ Ψ The diffeerence in pH H is −1.4. Sinnce an ion is transported from the positive to the negative sidde of the membranne, ∆Ψ is neg gative. Since ∆G G°′ for ATP synthesis is 30.5 kJ ⋅ mool−1 and 30.55/13.8 = 2.2, between tw wo and three moles m of protons must m be transsported to prrovide the frree energy too synthesize one mole off ATP under standard biochemiical conditio ons. 11. DNP andd related com mpounds disssipate the prroton gradiennt required for fo ATP syntthesis. The dissipation d off this gradiient decreasees the rate of synthesis of o ATP, decrreasing the ATP A mass acction ratio. Decreasing D this ratio relieves thee inhibition of o the electroon transport chain, causinng an increaase in metaboolic rate. Chaapter 19 1. Explain why w individu uals with a hereditary h deeficiency of carnitine c pallmitoyl transsferase II havve muscle weaknesss. Why are th hese symptooms more sevvere during fasting? 3. The first three steps of o β oxidatioon (Fig. 19-112) chemicallly resemble three successive steps of o the citric acid cyclle. Which steeps are thesee? 5. Why are unsaturated fats preferabble to saturaated fats for an a individuaal whose calooric intake must m be limited? 10. 1 CO used to On what carbon atom ms does the 14 t synthesizee malonyl-CooA from aceetyl-CoA apppear in 2 palmitatee? 11. Explain why w adipocy ytes need gluucose as well as fatty aciids in order to t synthesizee triacylglyccerols. 12. Is the fattty acid show wn below likkely to be synnthesized in animals? Exxplain. 13. Comparee the energy cost, in ATP P equivalents, of synthessizing stearate from mitoochondrial accetyl-CoA to the eneergy recoverred by degraading stearatee (a) to acetyyl-CoA and (b) to CO2. Chaapter 19 Answerss 1. A defect in carnitine palmitoyl trransferase II prevents noormal transpoort of activatted fatty acidds into the xidation. Tisssues such as muscle thaat use fatty acids a as metaabolic fuels therefore t mitochonndria for β ox cannot geenerate ATP P as needed. The problem m is more sevvere during a fast because other fuells, such as dietary glucose, are not n readily avvailable. 3. The first three steps of o β oxidatioon resemble the reactions that convert succinate to oxaloacettate (Sectionss 16-3F–16-3 3H). 5. There are not as many usable nutritional calories per gram in unsaturated fatty acids as there are in saturated fatty acids. This is because oxidation of fatty acids containing double bonds yields fewer reduced coenzymes whose oxidation drives the synthesis of ATP. In the oxidation of fatty acids with a double bond at an odd-numbered carbon, the enoyl-CoA isomerase reaction bypasses the acyl-CoA dehydrogenase reaction and therefore does not generate FADH2 (equivalent to 2 ATP). A double bond at an even-numbered carbon must be reduced by NADPH (equivalent to the loss of 3 ATP). 10. The label does not appear in palmitate because 14CO2 is released in Reaction 2b of fatty acid synthesis (Fig. 19-26). 11. The breakdown of glucose by glycolysis generates the dihydroxyacetone phosphate that becomes the glycerol backbone of triacylglycerols (Fig. 19-30). 12. This fatty acid (linolenate) cannot be synthesized by animals because it contains a double bond closer than 6 carbons from its noncarboxylate end. 13. The synthesis of stearate (18:0) from mitochondrial acetyl-CoA requires 9 ATP to transport 9 acetylCoA from the mitochondria to the cytosol. Seven rounds of fatty acid synthesis consume 7 ATP (in the acetyl-CoA carboxylase reaction) and 14 NADPH (equivalent to 42 ATP). Elongation of palmitate to stearate requires 1 NADH and 1 NADPH (equivalent to 6 ATP). The energy cost is therefore 9 + 7 + 42 + 6 = 64 ATP. (a) The degradation of stearate to 9 acetyl-CoA consumes 2 ATP (in the acyl-CoA synthetase reaction) but generates, in eight rounds of β oxidation, 8 FADH2 (equivalent to 16 ATP) and 8 NADH (equivalent to 24 ATP). Thus, the energy yield is 16 + 24 − 2 = 38 ATP. This represents only about half of the energy consumed in synthesizing stearate (38 ATP versus 64 ATP). (b) The complete oxidation of the 9 acetyl-CoA to CO2 by the citric acid cycle yields an additional 9 GTP (equivalent to 9 ATP), 27 NADH (equivalent to 81 ATP), and 9 FADH2 (equivalent to 18 ATP) for a total of 38 + 9 + 81 + 18 = 146 ATP. Thus, more than twice the energy investment of synthesizing stearate is recovered (146 ATP versus 64 ATP). Chapter 20 2. Explain why the symptoms of a partial deficiency in a urea cycle enzyme can be attenuated by a lowprotein diet. 3. Production of the enzymes that catalyze the reactions of the urea cycle can increase or decrease according to the metabolic needs of the organism. High levels of these enzymes are associated with high-protein diets as well as starvation. Explain this apparent paradox. 8. Which of the 20 “standard” amino acids are (a) purely glucogenic, (b) purely ketogenic, and (c) both glucogenic and ketogenic? 9. Alanine, cysteine, glycine, serine, and threonine are amino acids whose breakdown yields pyruvate. Which, if any, of the remaining 15 amino acids also do so? Chapter 20 Answers 2. The urea cycle transforms excess nitrogen from protein breakdown to an excretable form, urea. In a deficiency of a urea cycle enzyme, the preceding urea cycle intermediates may build up to a toxic level. A low-protein diet minimizes the amount of nitrogen that enters the urea cycle and therefore reduces the concentrations of the toxic intermediates. 3. An individual consuming a high-protein diet uses amino acids as metabolic fuels. As the amino acid skeletons are converted to glucogenic or ketogenic compounds, the amino groups are disposed of as urea, leading to increased flux through the urea cycle. During starvation, proteins (primarily from muscle) are degraded to provide precursors for gluconeogenesis. Nitrogen from these protein-derived amino acids must be eliminated, which demands a high level of urea cycle activity. 8. (a) Ala, Arg, Asn, Asp, Cys, Gin, Glu, Gly, His, Met, Pro, Ser, and Val (b) Leu and Lys (c) Ile, Phe, Thr, Trp, and Tyr 9. Tryptophan can be considered a member of this group since one of its degradation products is alanine, which is converted to pyruvate by deamination.