Solution Manual For Essential Biochemistry, 4th Edition

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1SolutionsChapter 11.a.carboxylic acid;b.amine;c.ester;d.alcohol.2.a.ether;b.phosphoric acid ester;c.thiol;d.ketone.3.Hydroxyl groupImino groupEther linkageCH3ONOH[From Li, S.-Y., Wang, X.-B., and Kong, L.-Y.Eur. J. Med. Chem.71,36–45 (2014).]4.HydroxylCoenzyme QEtherNicotinic acid (niacin)NOCOHH3COOCH3H3COOCH2Vitamin CCarboxylCOHCHCHHOOHHOCarbonylCarbonylEsterOCOCH(CH2CHCCH2)10CH35.Amino acids, monosaccharides, nucleotides, and lipids are thefour types of biological small molecules. Amino acids, monosacchar-ides, and nucleotides can form polymers of proteins, polysaccharides,and nucleic acids, respectively.6.a.N-acetylglucosamine is a monosaccharide.b.CMP is a nucle-otide.c.Homocysteine is an amino acid.d.Cholesteryl ester is a lipid.7.a.C and H plus some O;b.C, H, and O;c.C, H, O, and N plussmall amounts of S.8.It is a lipid (it is actually lecithin). It is mostly C and H, withrelatively little O and only one N and one P. It has too little O to bea carbohydrate, too little N to be a protein, and too little P to be anucleic acid.9.You should measure the nitrogen content, since this would indi-cate the presence of protein (neither lipids nor carbohydrates containappreciable amounts of nitrogen).10.You could add the compound that contains the most nitrogen,compound B, which is melamine. [Melamine is a substance that inthe past was added to some pet foods and milk products from Chinaso that they would appear to contain more protein. Melamine is toxicto pets and children.] Compound C is an amino acid, so it wouldalready be present in protein-containing food.11.A diet high in protein results in a high urea concentration, sinceurea is the body’s method of ridding itself of extra nitrogen. Nitrogenis found in proteins but is not found in significant amounts in lipidsor carbohydrates. A low-protein diet provides the patient with justenough protein for tissue repair and growth. In the absence of excessprotein consumption, urea production decreases, and this puts lessstrain on the patient’s weakened kidneys.12.Asn has an amido group and Cys has a sulfhydryl group.13.HHOHHOHHOHCH2OHHOHCOHydroxylCarbonyl14.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 anamino functional group.16.Nucleotides consist of a five-carbon sugar, a nitrogenous ring,and one or more phosphoryl groups linked covalently together.17.As described in the text, palmitate and cholesterol are highlynonpolar and are therefore insoluble in water. Both are highly aliphatic.Alanine is water soluble because its amino group and carboxylategroup are ionized, which render the molecule “saltlike.” Glucose isalso water soluble because its aldehyde group and many hydroxylgroups are able to form hydrogen bonds with water.18.Glucose has several hydroxyl groups and is a polar molecule.As such, it will have difficulty crossing the nonpolar membrane.The 2,4-dinitrophenol molecule consists of a substituted benzenering and has greater nonpolar character. Of the two molecules, the2,4-dinitrophenol will traverse the membrane more easily.19.DNA forms a more regular structure because DNA consists of onlyfour different nucleotides, whereas proteins are made up of as many as20 different amino acids. In addition, the 20 amino acids have muchmore individual variation in their structures than do the four nucleotides.Both of these factors result in a more regular structure for DNA. The cel-lular role of DNA relies on thesequenceof the nucleotides that make upthe DNA, not on the overall shape of the DNA molecule itself. On theother hand, proteins fold into unique shapes, as illustrated by endothelinin Figure 1.4. The ability of proteins to fold into a wide variety of shapesmeans that proteins can also serve a wide variety of biochemical rolesin the cell. According to Table 1.2, the major roles of proteins in the cellare to carry out metabolic reactions and to support cellular structures.20.Polysaccharides serve as fuel-storage molecules and can alsoserve as structural support for the cell.21.Pancreatic amylase is unable to digest the glycosidic bonds thatlink the glucose residues in cellulose. Figure 1.6 shows the structuraldifferences between starch and cellulose. Pancreatic amylase binds tostarch prior to catalyzing the hydrolysis of the glycosidic bond; thusthe enzyme and the starch must have shapes that are complementary.The enzyme would be unable to bind to the cellulose, whose structureis much different from that of starch.22.Cellulose cannot be digested by mammals and therefore the energyyield is 0 kilocalories per gram. Although both starch and glycogen

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2Solutionsare polymers of glucose, the glucose residues are linked differently inthe two molecules, and pancreatic amylase is unable to hydrolyze theglycosidic bonds in cellulose (see Solution 21). Cellulose provides noenergy to the diet but is an important component of the diet as fiber.23.A positive entropy change indicates that the system has becomemore disordered; a negative entropy change indicates that the system hasbecome more ordered.a.negative;b.positive;c.positive;d.positive;e.negative.24.a.decrease;b.increase.25.The polymeric molecule is more ordered and thus has less entropy.A mixture of constituent monomers has a large number of differentarrangements (like the balls scattered on a pool table) and thus hasgreater entropy.26.Entropy increases as the reactants (7 molecules) are converted toproducts (12 molecules).27.The dissolution of ammonium nitrate in water is a highly endo-thermic process, as indicated by the positive value of ΔH. This meansthat when ammonium nitrate dissolves in water, the system absorbsheat from the surroundings and the surroundings become cold. Theplastic bag containing the ammonium nitrate becomes cold and canbe used as a cold pack to treat an injury.28.The dissolution of calcium chloride in water is a highly exo-thermic process, as indicated by the negative value of ΔH. Thismeans that when calcium chloride dissolves in water, the system losesheat to the surroundings and the surroundings become warm. Theplastic bag holding the calcium chloride solution becomes warm andcan be used as a hot pack by the camper at cold temperatures.29.The dissolution of urea in water is an endothermic process andhas a positive ΔHvalue. In order for the process to be spontaneous,the process must also have a positive ΔSvalue in order for the freeenergy change of the process to be negative. Solutions have a higherdegree of entropy than the solvent and solute alone.30.First, calculate ΔHand ΔS, as described in Sample Calculation 1.1:ΔH=HB–HAΔH= 60 kJ · mol–1– 54 kJ · mol–1ΔH= 6 kJ · mol–1ΔS=SB–SAΔS= 43 J · K–1· mol–1– 22 J · K–1· mol–1ΔS= 21 J · K–1· mol–1a.ΔG= (6000 J · mol–1) – (4 + 273 K)(21 J · K–1· mol–1)ΔG= 180 J · mol–1The reaction is not favorable at 4°C.b.ΔG= (6000 J · mol–1) – (37 + 273 K)(21 J · K–1· mol–1)ΔG= –510 J · mol–1The reaction is favorable at 37°C.31.0 > 15,000 J · mol–1– (T)(51 J · K–1· mol–1)–15,000 > –(T)(51 K–1)15,000 < (T)(51 K–1)294 K <TThe reaction is favorable at temperatures of 21°C and higher.32.Processais always spontaneous; processesbandcare likely tobe spontaneous, depending on the temperature, and processdis neverspontaneous.33.0 > –14.3 kJ · mol–1– (273 + 25 K)(ΔS)14.3 kJ · mol–1> – (273 + 25 K)(ΔS)–48 J · K–1· mol–1> ΔSΔScould be any positive value, or it could have a negative valuesmaller than – 48 J · K–1· mol–1.34.–63 kJ ·mol–1= ΔH– (273 + 25 K)(190 J ·K–1·mol–1)ΔH= –63 kJ ·mol–1+ 56.6 kJ ·mol–1ΔH= –6.4 kJ ·mol–1The reaction releases heat to the surroundings.35.a.Entropy decreases when the antibody–protein complex bindsbecause the value of ΔSis negative.b.ΔG= ΔH–TΔSΔG= –87,900 J·mol–1– (298 K)(–118 J ·K–1·mol–1)ΔG= –52.7 kJ ·mol–1The negative value of ΔGindicates that the complex forms spon-taneously.c.The second antibody binds to cytochromecmore readily thanthe first because the change in free energy of binding is a morenegative value. [From Raman, C. S., Allen, M. J., and Nall, B. T.Biochemistry34,5831–5838 (1995).]36.a.The reaction releases heat to the surroundings because thevalue of ΔHis negative.b.ΔG= ΔH–TΔS–17,200 J · mol–1= –9500 J · mol–1– (310 K)(ΔS)ΔS= 25 J ·K–1· mol–1The positive value of ΔSindicates that the reaction proceeds withan increase in entropy.c.The ΔHterm makes a greater contribution to the ΔGvalue.This indicates that the reaction is spontaneous largely because thereaction is exothermic.37.a.The conversion of glucose to glucose-6-phosphate is notfavorable because the ΔGvalue for the reaction is positive, indi-cating an endergonic process.b.If the two reactions are coupled, the overall reaction is the sumof the two individual reactions. The ΔGvalue is the sum of theΔGvalues for the two individual reactions.ATP + glucose → ADP + glucose-6-phosphateΔG=–16.7 kJ · mol–1Coupling the conversion of glucose to glucose-6-phosphate withthe hydrolysis of ATP converts an unfavorable reaction to a favor-able reaction. The ΔGvalue of the coupled reaction is negative,which indicates that the reaction as written is favorable.38.a.The reaction is not favorable because the ΔGvalue for thereaction is positive, indicating an endergonic process.b.GAP + Pi+ NAD+→ 1,3BPG + NADHΔG= +6.7 kJ·mol–11,3BPG + ADP → 3PG + ATPΔG= –18.8 kJ·mol–1GAP + Pi+ NAD++ ADP → 3PG + NADH + ATPΔG= –12.1 kJ·mol–1The coupled reaction is spontaneous because the ΔGvalue isnegative.39.C (most oxidized), A, B (most reduced)40.a.reduction;b.oxidation.41.a.oxidized;b.oxidized;c.oxidized;d.reduced.42.a.oxidizing agent;b.oxidizing agent;c.oxidizing agent;d.reducing agent.43.a.Palmitate’s carbon atoms, which have the formula —CH2—,are more reduced than CO2, so their reoxidation to CO2releasesfree energy.

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Solutions3b.Because the —CH2— groups of palmitate are more reducedthan those of glucose (—HCOH—), their conversion to the fullyoxidized CO2would be even more thermodynamically favorable(have a larger negative value of ΔG) than the conversion of glu-cose carbons to CO2. Therefore, palmitate carbons provide morefree energy than glucose carbons.44.The complete oxidation of stearate to CO2yields more energybecause 17 of the 18 carbons of stearate are fully reduced. The con-version of these carbons to CO2provides more free energy than someof the carbons of α-linolenate, which participate in double bonds andare therefore already partially oxidized.45.Morphological differences, which are useful for classifying largeorganisms, are not useful for bacteria, which often look alike. Further-more, microscopic organisms do not leave an easily interpreted imprintin the fossil record, as vertebrates do. Thus, molecular information isoften the only means for tracing the evolutionary history of bacteria.46.It is difficult to envision how a single engulfment event could havegiven rise to a stable and heritable association of the eukaryotic hostand the bacterial dependent within a single generation. It is much morelikely that natural selection gradually promoted the interdependenceof the cells. Over many generations, genetic information supportingthe association would have become widespread.47.ABC48.a.H15 and H7 are closely related, as are H4 and H14.b.H4 and H14 are most closely related to H3.Chapter 21.The water molecule is not perfectly tetrahedral because the elec-trons in the nonbonding orbitals repel the electrons in the bondingorbitals more than the bonding electrons repel each other. The anglebetween 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 isnot polar.OCO3.Water has the higher boiling point because, although each mol-ecule has the same geometry and can form hydrogen bonds with itsneighbors, the hydrogen bonds formed between water molecules arestronger than those formed between H2S molecules. The electroneg-ativity difference between H and O is greater than that between H andS and results in greater differences in the partial charges on the atomsin the water molecule.4.Water has the highest melting point because each water moleculeforms hydrogen bonds with four neighboring water molecules, and hy-drogen bonds are among the strongest intermolecular forces. Ammoniais also capable of forming hydrogen bonds, but they are not as strong(due to the smaller electronegativity difference between hydrogenand nitrogen). Methane cannot form hydrogen bonds; the moleculesare attracted to their neighbors only via weak London dispersion forces.5.The arrows point toward hydrogen acceptors and away fromhydrogen donors:OCNNNNNHHHHHOOSOOOOOCH3NH2NCHCHCOOCH2NH2CH2CH3AspartameUric acidSulfanilamide6.Arrows point toward hydrogen acceptors and away from hydrogendonors. [From Kubiny, H., in3D QSAR in Drug Design: Volume 1:Theory Methods and Application,Springer Science & BusinessMedia (1993).]MTXDHFNNNNNHHNRHOHHNHHHNHHNNNNNRH3C7.Identical hydrogen bonding patterns in the two molecules areshown as open arrows in Solution 6.8.[From Puschner, P., Poppenga, R. H., Lowenstine, L. J., Filigenzi,M. S., and Pesavento, P. A.J. Vet. Diagn. Invest.19,616–624 (2007).]NNNNNNHOHHHHHMelamine cyanurateHONHNNHO9. a.H < C < S < N < O < Fb.The greater an atom’s electronegativity, the more polar its bondwith H and the greater its ability to act as a hydrogen bond acceptor.Thus, N, O, and F, which have relatively high electronegativities, canact as hydrogen bond acceptors, whereas C and S, whose electroneg-ativities are only slightly greater than hydrogen’s, cannot.10.Compound A does not form hydrogen bonds (the molecule has ahydrogen bond acceptor but no hydrogen bond donor). CompoundsB and C form hydrogen bonds as shown because each molecule con-tains at least one hydrogen bond donor and a hydrogen bond acceptor.The molecules in D do not form hydrogen bonds with each other be-cause ethyl chloride lacks both a hydrogen bond donor and a hydro-gen bond acceptor. The molecules in E do because ammonia has ahydrogen bond donor and diethyl ether has a hydrogen bond acceptor:

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4Solutions18.Methanol, which has the highest dielectric constant, would bethe best solvent for the cationic NH4+. The polarity of the alcohols,which all contain a primary OH group, varies with the size of thehydrocarbon portion. 1-Butanol, with the largest hydrophobic group,is the least polar and therefore has the lowest dielectric constant.19.Structure A depicts a polar compound, while structure B depictsan ionic compound similar to a salt like sodium chloride. This is moreconsistent with glycine’s physical properties as a white crystallinesolid with a high melting point. While structure A could be water sol-uble because of its ability to form hydrogen bonds, the high solubilityof glycine in water is more consistent with an ionic compound whosepositively and negatively charged groups are hydrated in aqueoussolution by water molecules.20.a.Surface tension is defined as the force that must be appliedto surface molecules in a liquid so that they may experience thesame forces as the molecules in the interior of the liquid. Water’ssurface tension is greater than ethanol’s because the strength andnumber of water’s intermolecular forces (hydrogen bonds) areboth greater. Ethanol’s OH group also forms hydrogen bonds,but the hydrocarbon portion of the molecule cannot interactfavorably with water, and weaker London dispersion forces forminstead.b.The kinetic energy of the water molecules increases when tem-perature increases. Intermolecular forces are weaker in strength asa consequence of the increased molecular motion. Because sur-face tension increases when the strength of intermolecular forcesincreases as described in part (a), surface tension decreases whentemperature increases.21.The waxed car is a hydrophobic surface. To minimize its inter-action with the hydrophobic molecules (wax), each water drop mini-mizes its surface area by becoming a sphere (the geometrical shapewith the lowest possible ratio of surface to volume). Water does not beadon glass, because the glass presents a hydrophilic surface with which thewater molecules can interact. This allows the water to spread out.22.The paper clip, although composed of a metal with a greaterdensity than water, fl oats due to the strong hydrogen bonding thatoccurs among water molecules on the surface of the liquid. Soap dis-rupts these strong intermolecular forces and as a result the paper clipsinks to the bottom of the container.23.Polar and nonpolar regions of the detergents are indicated.HexadecyltrimethylammoniumHOOHOOnonpolarpolarOHCholateCH3CH3CH3(H2C)15polarnonpolarH3CCH3CH3CH3N24.Polar and nonpolar regions of the detergents are indicated.Dimethyldecylphosphine oxideOOHHHHHOOHOHCH2OH(CH2)7CH3n-OctylglucosideNonpolar(CH2)9H3CPOCH3CH3NonpolarPolarPolar25.Compounds A and D are amphiphilic, compound B is nonpolar,and compounds C and E are polar.26.Compound A has a polar head and a nonpolar tail as indicatedand can form a micelle (see Fig. 2.9). Compound D has a polar headBCENNCH2OH3CHNHHHCH2OH3CHNNHHCH3OCH2CH3H2C11. a.van der Waals forces (dipole–dipole interactions);b.hydrogenbonding;c.van der Waals forces (London dispersion forces);d.ionicinteractions.12. a.hydrogen bonding;b.ionic interactions;c.van der Waalsforces (London dispersion forces).13.Solubility in water decreases as the number of carbons in thealcohol increases. The hydroxyl group of the alcohol is able to formhydrogen bonds with water, but water cannot interact favorably withthe hydrocarbon chain. Increasing the length of the chain increasesthe number of potentially unfavorable interactions of the alcohol withwater and solubility decreases as a result.14.Solubility in water and dielectric constants are correlated; lowmolecular weight alcohols are miscible with water and have a highdielectric constant; high molecular weight alcohols are not very sol-uble in water and have a low dielectric constant.15.Aquatic organisms that live in the pond are able to survive thewinter. Since the water at the bottom of the pond remains in the liquidform instead of freezing, the organisms are able to move around. Theice on top of the pond also serves as an insulating layer from the coldwinter air.16.Water is unique in that its liquid form is more dense than its solidform. The weight of the skater puts pressure on the thin blade of theice skate. The ice melts under the blade because of this increasedpressure. A higher pressure favors the liquid form of water over thesolid form because the liquid form is more dense and takes up lessvolume.17.The positively charged ammonium ion is surrounded by a shell ofwater molecules that are oriented so that the partially negativelycharged oxygen atoms interact with the positive charge on the am-monium ion. Similarly, the negatively charged sulfate ion is hydratedwith water molecules oriented so that the partially positively chargedhydrogen atoms interact with the negative charge on the sulfate anion.(Not shown in the diagram is the fact that the ammonium ions out-number the sulfate ions by a 2:1 ratio. Also note that the exact numberof water molecules shown is unimportant.)OOOONH4SO42HHHOHHOHHHOHHOHHHHHHOHHOHH

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Solutions5c.The hydrophobic grease moves into the hydrophobic core of thewater-soluble soap micelle. The “dissolved” grease can then bewashed away with the micelle.29. a.The nonpolar core of the lipid bilayer helps prevent the passageof water since the polar water molecules cannot easily penetrate thehydrophobic core of the bilayer.b.Most human cells are surroundedby a fluid containing about 150 mM Na+and slightly less Cl–(seeFig. 2.12). A solution containing 150 mM NaCl mimics the extracel-lular fluid and therefore helps maintain the isolated cells in near-normal conditions. If the cells were placed in pure water, waterwould tend to enter the cells by osmosis; this might cause the cells toburst.30.In reverse osmosis, water moves from an area of low concen-tration (high solute concentration) to an area of high concentration(low solute concentration). This movement is opposite that describedfor osmosis in Problem 29. This is a non-spontaneous process thatrequires an input of energy in order to proceed, unlike osmosis, whichoccurs spontaneously without input of energy.31. a.CO2is nonpolar and would be able to cross a bilayer.b.Glu-cose is polar and would not be able to pass through a bilayer becausethe presence of the hydroxyl groups means glucose is highly hydratedand would not be able to pass through the nonpolar tails of the mol-ecules forming the bilayer.c.DNP is nonpolar and would be ableto cross a bilayer.d.Calcium ions are charged and are, like glucose,highly hydrated and would not be able to cross a lipid bilayer.32.Vesicles consist of a lipid bilayer that closes up to enclose anaqueous compartment. The polar drug readily dissolves in thisaqueous compartment. Delivery to the cell is accomplished when thevesicle membrane fuses with the cell membrane, releasing the druginto the cytosol.33.Substances present at high concentration move to an area of lowconcentration spontaneously, or “down” a concentration gradient ina process that increases their entropy. The export of Na+ions fromthe cell requires that the sodium ions be transported from an areaof low concentration to an area of high concentration. The same istrue for potassium transport. Thus, these processes are not sponta-neous, and an input of cellular energy is required to accomplish thetransport.34.The amount of Na+(atomic weight 23 g · mol–1) lost in 15 minutes,assuming a fluid loss rate of 2 L per hour and a sweat Na+concentrationof 50 mM (Box 2.B), is0.25h × 2Lh× 0.05molL× 23gmol × 1000mg Na+gNa+×loz chips200mg Na+= 2.9 oz chipsIt would take 2.9 ounces of potato chips (about a handful) to replacethe lost sodium ions.35. a.In a high-solute medium, the cytoplasm loses water and there-fore its volume decreases.b.In a low-solute medium, the cytoplasmgains water and therefore its volume increases.36.E. coliaccumulates water when grown in a low-salt medium.However, regulation of water content only would cause a large in-crease in cytoplasmic volume. To avoid this large increase in volume,E. colialso exports K+ions. The opposite occurs whenE. coliisgrown in a high-salt medium: The cytoplasmic water content is de-creased, but cytoplasmic osmolarity increases asE. coliimports K+ions. [From Record, M.T., et al.,Trends Biochem. Sci.23, 143–148(1998).]37.Since the molecular mass of H2O is 18.0 g · mol–1, a given volume(for example, 1 L or 1000 g) has a molar concentration of 1000 g · L–1÷18.0 g · mol–1= 55.5 M. By definition, a liter of water at pH 7.0 hasand two nonpolar tails as indicated and can form a bilayer (see Fig.2.10). Compounds B, C, and E form neither micelles nor bilayers.ADH3C(CH2)11(CH2)11CH3CH3CH3(CH2)11CH3CH2CH2NCH2COOHCCCOOOHOOPolar headPolar headNonpolar tailNonpolartails27.a.In the nonpolar solvent, AOT’s polar head group faces the interiorof the micelle, and its nonpolar tails face the solvent.NonpolartailsPolarheadH3C(CH2)3CHCH2CH3CH2H3C(CH2)3CHCH2CH3CH2CH2CHOOOOOOOCCSAOTb.The protein, which contains numerous polar groups, interactswith the polar AOT groups in the micelle interior.WaterProteinIsooctane28.a.b.a)b)Nonpolar tailH3C(CH2)11ONaPolar headOOOSNaOSOOO(H2C)11CH3SOOO(CH2)11CH3NaONaOSOOO(CH2)11CH3SOOH3C(CH2)11OONaSOOCH3(CH2)11OONaSOOCH3(CH2)11OONaSOOH3CONa(CH2)11OSOOH3CONa(CH2)11O

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6Solutionsitself, 1.0 × 10−7M, can be ignored because it is much smallerthan the hydroxide ion concentration contributed by the KOH.)Kw= 1.0 × 10−14= [H+] [OH−][H+] = 1.0 × 10−14[OH−][H+] = 1.0 × 10−14(0.029 M)[H+] = 3.4 × 10−13MpH = −log[H+]pH = −log(3.4 × 10−13)pH = 12.548. a.The final concentration of HCl is (0.0015 L)(3 mol/L) ÷ 1 L= 0.0045 M. Since HCl is a strong acid and dissociates com-pletely, the added [H+] is equal to [HCl]. (The existing hydrogenion concentration in the water itself, 1.0 × 10–7M, can be ignoredbecause it is much smaller than the hydrogen ion concentrationcontributed by the hydrochloric acid.)pH = −log[H+]pH = −log(0.0045)pH = 2.3b.The final concentration of NaOH is (0.0015 L)(3 mol/L) ÷ 1L = 0.0045 M. Since NaOH dissociates completely, the added[OH–] is equal to the [NaOH]. (The existing hydroxide ion con-centration in the water itself, 1.0 × 10–7M, can be ignored becauseit is much smaller than the hydroxide ion concentration contrib-uted by the NaOH.)Kw= 1.0 × 10−14= [H+] [OH−][H+] = 1.0 × 10−14[OH−][H+] = 1.0 × 10−14(0.0045 M)[H+] = 2.2 × 10−12MpH = −log[H+]pH = −log(2.2 × 10−12)pH = 11.649.HOCH2CH2COOHCOOHCOOHCCitric acidHHNPiperidineOOHHOCOCOxalic acidCH2CH2OOOOSNHOHCHCH2CH2CH2CH2OCHHH2NH2NLysineNHNHOOOBarbituric acid4-Morphine ethanesulfonic acid (MES)a hydrogen ion concentration of 1.0 × 10–7M. Therefore the ratio of[H2O] to [H+] is 55.5 M/(1.0 × 10–7M) = 5.55 × 108.38.[H+] = [OH−] = 1.47 × 10−14[H+] =√1.47 × 10−14[H+] = 1.21 × 10−7MpH = −log(1.21 × 10−7M)pH = 6.9239.The HCl is a strong acid and dissociates completely. This meansthat the concentration of hydrogen ions contributed by the HCl is1.0 × 10–9M. But the concentration of the hydrogen ions contrib-uted by the dissociation of water is 100-fold greater than this: 1.0 ×10–7M. The concentration of the hydrogen ions contributed by theHCl is negligible in comparison. Therefore, the pH of the solutionis equal to 7.0.40.The pH of this solution is 7.0 (see Solution 39).The concen-tration of hydroxide ions contributed by the dissociation of wateris 100-fold greater than that contributed by the dissociation of thedilute NaOH.41.−42.Acid, base,or neutral?pH[H+] (M)[OH–] (M)Bloodbase7.423.8 × 10–82.6 × 10–7Salivaneutral7.001.0 × 10–71.0 × 10–7Urineacid6.801.6 × 10–76.3 × 10–8Gastric juiceacid2.107.9 × 10–31.3 × 10–1243.The stomach contents have a low pH due to the contribution ofgastric juice (pH 1.5–3.0). When the partially digested material entersthe small intestine, the addition of pancreatic juice (pH 7.8–8.0) neu-tralizes the acid and increases the pH.44.The carbonate ions accept protons from water and form hydrox-ide ions (as shown in the equation below), resulting in basic urine.CO2−3(aq) + H2O(l) → HCO−3(aq) + OH−(aq)45. a.C2O42−b.SO32−c.HPO42−d.CO32−e.AsO43−f.PO43−g.O22−46.a.H2C2O4b.H2SO3c.H3PO4d.H2CO3e.H2AsO4–f.H2PO4–g.H2O247. a.The final concentration of HNO3is (0.020L)(1.0M)0.520L=0.038 M. Since HNO3is a strong acid and dissociates completely,the added [H+] is equal to [HNO3]. (The existing hydrogen ionconcentration in the water itself, 1.0 × 10−7M, can be ignoredbecause it is much smaller than the hydrogen ion concentrationcontributed by the nitric acid.)pH = −log[H+]pH = −log(0.038)pH = 1.4b.The final concentration of KOH is (0.015 L)(1.0 M)0.515 L= 0.029 MSince KOH dissociates completely, the added [OH−] is equal tothe [KOH]. (The existing hydroxide ion concentration in the water

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Solutions754.pH = pK+ log[A−][ HA ]log [A−][ HA ] = pH − pK= 5.0 − 4.76 = 0.24[A−][ HA ] = 1.74 or [A−] = 1.74 [HA ][A−] = 1.74[HA ] = 1.74(0.05 M − [A−] )[A−] = 0.087 − 1.74[A−]2.74[A−] = 0.087 M[A−] = 0.032 M or 32 mM55.First, determine the ratio of [A−] to [HA]:pH = pK+ log[A−][ HA ]log [A−][ HA ] = pH − pK[A−][ HA ] = 10(pH−pK)Substitute the values for the desired pH (5.0) and the pK(4.76):[A−][ HA ] = 10(5.0−4.76)= 100.24= 1.74Calculate the number of moles of acetate (A−) already present:(0.50 L)(0.20 mol · L−1) = 0.10 moles acetateCalculate the moles of acetic acid needed, based on the calculated ratio:[A−][ HA ] = 1.74[ HA ] = 0.10 moles1.74[HA ] = 0.057 molesFinally, calculate the volume of glacial acetic acid needed:0.057 moles17.4 mol · L−1= 0.0033 L, or 3.3 mLThe addition of 3.3 mL to a 500-mL solution dilutes the solution byless than 1%, which doesn’t introduce significant error.56.Adding NaOH to the acetic acid will convert some of the aceticacid (HA) to acetate (A–):NaOH + CH3COOH →Na++ CH3COO−+ H2OFor every mole of NaOH added, one mole of CH3COOH will beconsumed, and one mole of CH3COO–will be generated. Ifxis thenumber of moles of NaOH added, thenxwill also be the number ofmoles of A–generated.Calculate the initial amount of acetic acid:The initial amount of acetic acid is 0.10 mol (see Solution 55), so thefinal amount of acetic acid will be 0.10 mol –x.[A−][ HA ] = 1.74 =x0.10 mol −xx= 1.74(0.10 mol −x) = 0.174 mol − 1.74x2.74x= 0.174 molx= 0.174 mol/2.74 = 0.0635 molCalculate the mass of NaOH to add:0.0635 mol × 40.0 g · mol−1= 2.54 g50.Convert all the data to eitherKaor pKvalues to evaluate (pK=–logKa). The greater theKavalue, the stronger the acid—that is, thegreater the tendency for the proton to be donated. (The lower the pKvalue, 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 acid is a stronger acid than citrate, and succinicacid is a stronger acid than succinate.AcidKapKAcitrate1.74 × 10−54.76Bsuccinic acid6.17 × 10–54.21Csuccinate2.29 × 10–65.64Dformic acid1.78 × 10–43.75Ecitric acid7.41 × 10−43.1351.Calculate the final concentrations of the weak acid (H2PO−4) andconjugate base (HPO42−). Note that K+is a spectator ion.[H2PO−4] = (0.025 L)(2.0 M)0.200 L= 0.25 M[HPO2−4] = (0.050 L)(2.0 M)0.200 L= 0.50 MNext, substitute these values into the Henderson–Hasselbalch equa-tion using the pKvalues in Table 2.4:pH = pK+ log [A−][ HA ]pH = 6.82 + log(0.50 M)/(0.25 M)pH = 6.82 + 0.30pH = 7.1252.Use the pKvalue in Table 2.4 and the Henderson–Hasselbalchequation to calculate the ratio of the concentrations of imidazole (A–)and the imidazolium ion (HA):pH = pK+ log[A−][ HA ]log[A−][ HA ] = pH − pK[A−][ HA ] = 10(pH−pK)[A−][ HA ] = 10(7.4−7.0)[A−][ HA ] = 2.5153.The fi nal volume is 500 mL + 10 mL + 20 mL = 0.53 L[boric acid] = [HA ] = (0.01 L)(0.05 M)0.53 L= 9.4 × 10−4M[borate] = [A−] = (0.02 L)(0.02 M)0.53 L= 7.5 × 10−4MpH = pK+ log[A−][ HA ]= 9.24 + log 7.5 × 10−49.4 × 10−4= 9.24 − 0.10 = 9.14

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8Solutions0.51.01.52.02.53.0H+ions dissociatedpH14121086420[H3PO4][H2PO4][HPO42][PO43−]pK1Midpoint one[H3PO4][H2PO4]pK2Midpoint two[H2PO4][HPO42]pK3Midpoint three[HPO42−][PO43−]b.H3PO4→H++ H2PO−4H2PO−4→H++ HPO2−4HPO2−4→H++ PO3−4c.The dissociation of the second proton has a pKof 6.82, whichis closest to the pH of blood. Therefore the weak acid present inblood is H2PO4–and the weak base is HPO42–.d.The dissociation of the third proton has a pKof 12.38. There-fore, a buffer solution at pH 11 would consist of the weak acidHPO42−and its conjugate base, PO43−(supplied as the sodium saltsNa2HPO4and Na3PO4).64.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 lipidbilayer. At the pH of the small intestine, the carboxylate group ismostly in the ionized form and will be negatively charged. Chargedspecies are more polar than uncharged species (and are likely to behydrated) and will have difficulty traversing a lipid bilayer.65. a.HO(H2C)2HNN(CH2)2SO3Weak acid (HA)Conjugate base (A)HO(H2C)2(CH2)2SO3HNNb.The pKfor HEPES is 7.55; therefore, its effective bufferingrange is 6.55–8.55.c.1.0 L × 0.10 moleL× 260.3 gmol= 26 gWeigh 26 g of the HEPES salt and add to a beaker. Dissolve inslightly less than 1.0 liter of water (leave “room” for the HCl solu-tion that will be added in the next step).d.At the final pH,[A−][ HA ] = 10(pH−pK)= 10(8.0−7.55)= 100.45= 2.82For each mole of HCl added,x, one mole of HEPES salt (A–) willbe converted to a mole of HEPES acid (HA). The starting amountof A–is (1.0 L)(0.10 mol · L–1) = 0.10 moles. After the HCl isadded, the amount of A−will be 0.10 moles –x, and the amountof HA will bex. Consequently,57. a.H2CO3→H++ HCO−3HCO−3→H++ CO2−3b.The pKof the first dissociation is closer to the pH; thereforethe weak acid present in blood is H2CO3and the conjugate baseis HCO3–.c.pH = pK+ log[HCO−3][H2CO3]7.40 = 6.35 + log24 × 10−3M[H2CO3]1.05 = log24 × 10−3M[H2CO3]11.2 = 24 × 10−3M[H2CO3][H2CO3] = 2.1 × 10−3M = 2.1 mM58.The pKof the fluorinated compound would be lower (it is 9.0);that is, the compound becomes less basic and more acidic. Thisoccurs because the F atom, which is highly electronegative, pulls onthe nitrogen’s electrons, loosening its hold on the proton.59. a.CH3COCOOPyruvateb.Pyruvate predominates in the cell at pH 7.4. The pKvalues forcarboxylic acid groups are typically in the 2–3 range; therefore,the carboxylate group will be unprotonated at physiological pH.60.pH 2+H3NCH2COOHpH 7+H3NCH2COO−pH 10H2NCH2COO−The carboxylic acid group has a pKof 2.35, and the amino group hasa pKof 9.78. The Henderson–Hasselbalch equation can be used tocalculate the exact percentage of protonated/unprotonated forms ofeach functional group, but that really isn’t necessary. Instead, the pKvalues for each group should be compared to the pH. At pH = 2, thepH is below both pKvalues, so both functional groups are mostly pro-tonated. At pH = 7, the pH is well above the pKfor the carboxylic acidgroup but below the pKfor the amino group. Therefore the carboxylicacid group is unprotonated and the amino group is protonated. AtpH = 10, the pH is above the pKvalues of both functional groups.Thus, both groups are mostly unprotonated.61. a.10 mM glycinamide buffer, because its pKis closer to thedesired pH.b.20 mM Tris buffer, because the higher the concen-tration of the buffering species, the more acid or base it can neutral-ize.c.Neither. Both a weak acid and a conjugate base are requiredbuffer constituents. Neither the weak acid alone (boric acid) northe conjugate base alone (sodium borate) can serve as an effectivebuffer.62. a.10 mM acetic acid buffer, because its pKis closer to thedesired pH.b.20 mM acetic acid buffer, because the higher concen-tration of buffer species will allow it to neutralize a greater amountof acid or base.c.Neither. Both a weak acid and a conjugate baseare required buffer constituents. Neither the weak acid alone (aceticacid) nor the conjugate base alone (sodium acetate) can serve as aneffective buffer.63. a.The three ionizable protons of phosphoric acid have pKvaluesof 2.15, 6.82, and 12.38 (Table 2.4). The pKvalues are the mid-points of the titration curve:

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Solutions9pH = pK+ log[A−][ HA ]pH = 8.3 + log48.5 mmol ÷ 1001.5 mL51.5 mmol ÷ 1001.5 mLpH = 8.3 + (−0.026)pH = 8.27The buffer is effective: The pH increases only 0.07 unit (from 8.2to 8.27) with the addition of the strong base. In comparison, theaddition of the same amount of base to water, which is not buf-fered, results in a pH change from approximately 7.0 to 11.6 (seeProblem 48b).67.pH = pK+ log [A−][ HA ]log[A−][ HA ] = pH − pK[A−][ HA ] = 10(pH−pK)[A−][ HA ] = 10(6.5−7.0)= 10−0.5= 0.316Since the starting solution contains (0.5 L)(0.01 mol · L−1) = 0.005mole of imidazole (A–), the amount of imidazolium chloride (HA)needed is 0.005 mol/0.316 = 0.016 moles. The stock imidazoliumchloride is 1 M, so the volume of imidazolium chloride to be added is0.016 mol1.0 mol · L−1= 0.016 L or16 mL68. a.First, calculate the ratio of [A−] to [HA]. Rearranging theHenderson–Hasselbalch equation gives[A−][ HA ] = 10(pH−pK)= 10(2.0−8.3)= 10−6.3= 5 × 10−7Virtually all of the Tris is in the weak acid form. Therefore, theconcentration of the weak acid, HA, is 0.10 M and the concentra-tion of the conjugate base, A−, is 5.0 × 10−8M.b.The added HCl dissociates completely, so the amount of H+added is (0.0015 L)(3.0 mol · L−1) = 0.0045 mol. In an effectivebuffer, the acid would convert some of the conjugate Tris base toweak acid. But the concentration of conjugate base is already neg-ligible. Therefore, the moles of additional H+should be added tothe concentration of hydrogen ions already present (1.0 × 10−2M),for a total concentration of 0.0145 M.pH = −log[H+] = log(0.0145 M) = 1.84The buffer has not functioned effectively. There was not enoughconjugate base to react with the additional hydrogen ions added.The result is a decrease in pH from 2.0 to 1.84.c.When NaOH is added, an equivalent amount of Tris acid (HA)is converted to Tris base (A−). Letx= moles of OH−added =(0.0015 L)(3.0 mol · L−1) = 0.0045 moles = 4.5 mmol.The final amount of A−is 5.0 × 10−8mol + 4.5 mmol = 4.5 mmol.The final amount of HA is 100 mmol – 4.5 mmol = 95.5 mmol.The new pH is determined by substituting the new concentrationsof H−and HA into the Henderson–Hasselbalch equation:pH = pK+ log[A−][ HA ]pH = 8.3 + log(4.5 mmol ÷ 1001.5 mL)(95.5 mmol ÷ 1001.5 mL)pH = 8.3 + (−1.3) = 7.0[A−][ HA ] = 2.82 = 0.10 mole −xx2.82x= 0.10 mol −x3.82x= 0.10 molx= 0.10 mol/3.82 = 0.0262 molCalculate how much 6.0 M HCl to add:0.0262 mol6.0 mol · L−1= 0.0044 L, or 4.4 mLTo make the buffer, dissolve 26 g of HEPES salt [see part c] in lessthan 1.0 L. Add 4.4 mL of 6.0 M HCl, then add water to bring thefinal volume to 1.0 L.66. a.Weak acid (HA)CH2OHCH2OHHOH2CCNH3Conjugate base (A)HNH2CH2OHCH2OHHOH2CCb.The pKof Tris is 8.30; therefore, its effective buffering rangeis 7.30–9.30.c.Rearranging the Henderson–Hasselbalch equation gives[A−][ HA ] = 10(pH−pK)= 10(8.2−8.3)= 10−0.1= 0.79Since [A−] + [HA] = 0.10 M, [A−] = 0.10 M − [HA ],and (0.10 M − [HA ] )[ HA ]= 0.790.79 [HA ] = 0.10 M − [HA ]1.79[HA ] = 0.10 M[ HA ] = 0.10 M1.79= 0.056 M = 56 mM[A−] + [HA ] = 0.10 M = 100 mM, so [A−] = 44 mMd.When HCl is added, an equivalent amount of Tris base (A–) isconverted to Tris acid (HA).Letx= moles of H+added = (0.0015 L)(3.0 mol · L–1) = 0.0045moles = 4.5 mmol.The final amount of A–is 44 mmol – 4.5 mmol = 39.5 mmol.The final amount of HA is 56 mmol + 4.5 mmol = 60.5 mmol.Use the Henderson–Hasselbalch equation to calculate the new pH:pH = pK+ log[A−][ HA ]pH = 8.3 + log39.5 mmol ÷ 1001.5 mL60.5 mmol ÷ 1001.5 mLpH = 8.3 + (−0.2)pH = 8.1The buffer is effective: The pH decreases about 0.1 unit (from8.2 to 8.1) with the addition of the strong acid. In comparison,the addition of the same amount of acid to water, which is notbuffered, results in a pH change from approximately 7.0 to 2.3(see Problem 48a).e.When NaOH is added, an equivalent amount of Tris acid (HA)is converted to Tris base (A–). Letx= moles of OH–added =(0.0015 L)(3.0 mol · L–1) = 0.0045 moles = 4.5 mmol.The final amount of A–is 44 mmol + 4.5 mmol = 48.5 mmol.The final amount of HA is 56 mmol – 4.5 mmol = 51.5 mmol.Use the Henderson–Hasselbalch equation to calculate the new pH:

Solution Manual For Essential Biochemistry, 4th Edition - Page 11 preview image

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10Solutions101.05= [HCO−3][H2CO3][HCO−3][H2CO3] = 11.21In 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 ratio of conjugate base to weak acid does not liewithin an effective buffering range. The bicarbonate concentration(conjugate base) is too high relative to the carbonic acid (weak acid)concentration; thus the relative amount of weak acid is insufficient.[Krause, D. S., Wolf, B. A., and Shaw, L. M.,Ther. Drug Monit.14,441– 451 (1992).]73.Ammonia and ammonium ions are in equilibrium, as representedby the following equation:NH+4⇌H++ NH3Carbonic acid and bicarbonate ions are in equilibrium, as representedby the following equation:H2CO3⇌H++ HCO−3Phosphate ions are in equilibrium, according to the following equation:H2PO−4⇌H++ HPO2−4In metabolic acidosis, the concentration of protons increases, so theequilibrium shifts to form H2PO4–, carbonic acid, and ammoniumions. In order to bring the pH back to normal, the kidney excretesH2PO4–and ammonium ions, and bicarbonate ions are reabsorbed. Theresult is a decrease in the concentration of protons and an increasein blood pH.74.The relevant equations are shown in Solution 73. In metabolicalkalosis there is an excess of hydroxide ions, which react with pro-tons to form water. This causes the equilibria to shift to form HPO42–,NH3, and HCO3–. In order to bring the pH back to normal, the kidneyreabsorbs NH4+and H2PO4–and excretes HCO3–.75.The concentrations of both Na+and Cl–are greater outside thecell than inside (see Fig. 2.12). Therefore the movement of these ionsinto the cell is thermodynamically favorable. Na+movement into thecell drives the exit of H+via an exchange protein in the plasma mem-brane (the favorable movement of Na+into the cell “pays for” theunfavorable movement of H+out of the cell). Similarly, the move-ment of Cl–into the cell drives the movement of HCO3–out of the cellthrough another exchange protein.76.Acetoacetate and 3-hydroxybutyrate are acids (they are ionizedat physiological pH). The accumulation of the ketone bodies there-fore causes metabolic acidosis. The body attempts to compensate byincreasing the breathing rate in order to eliminate more CO2.77.The cell-surface carbonic anhydrase can catalyze the conversionof H++ HCO3–to CO2, which can then diffuse into the cell (the ionicH+and HCO3–cannot cross the hydrophobic lipid bilayer on theirown). Inside the cell, carbonic anhydrase converts the CO2back toH++ HCO3–.78.The lungs can compensate for metabolic acidosis through anincrease in the breathing rate in order to eliminate more CO2. Thekidneys can compensate for respiratory acidosis by increasing thebreakdown of glutamine to produce NH3(excreted in urine as NH4+,see Solution 73); however, this mechanism requires the synthesis ofenzymes, which takes several hours at least.Tris is not an effective buffer at pH 2.0, a pH more than 6 unitslower than its pKvalue. Virtually all of the Tris is in the weak acidform at this pH. If acid is added, there is not enough base to absorbthe excess added hydrogen ions, and the pH decreases. If base isadded, some of the weak acid is converted to the conjugate baseand the pH approaches the value of the pK.69.H+(aq) + HCO−3(aq)⇌H2CO3(aq)⇌H2O(l) + CO2(aq)Failure to eliminate CO2in the lungs would cause a buildup of CO2(aq).This would shift the equilibrium of the above equations to the left.The increase in CO2(aq) would lead to the increased production ofcarbonic acid, which would in turn dissociate to form additional hy-drogen ions, causing acidosis.70. a.Mechanical hyperventilation removes CO2from the patient’slungs. Carbonic acid in the blood would produce more water andCO2to make up for the loss of CO2. This in turn would causeadditional hydrogen ions and bicarbonate ions to form morecarbonic acid. The loss of hydrogen ions would result in an in-creased pH, bringing the patient’s pH back to normal.b.The additional bicarbonate would combine with hydrogen ionsto form carbonic acid. The additional carbonic acid would disso-ciate to form water and carbon dioxide. This helps alleviate theacidosis because the bicarbonate combines with excess hydrogenions, thus decreasing the hydrogen ion concentration and increas-ing the pH. However, it is not acceptable for use in patients withALI because of the increased production of aqueous CO2in theblood. The CO2produced would need to be exhaled in the lungs,which would be difficult in patients with ALI.c.Tris becomes protonated to form its conjugate acid. This re-moves H+from circulation and brings the pH back to normal. Theprotonated form of Tris is excreted in the urine. This method ofacidosis treatment does not involve exhalation of CO2and is there-fore an acceptable treatment for patients with ALI.CH2OHCH2OHHOH2CCNH3NH2CH2OHCH2OHHOH2CCH[From Kallet, R. H., et al.,Am. J. Respir. Crit. Care Med.161,1149–1153 (2000).]71.During hyperventilation, too much CO2(which is equivalent toH+in the form of carbonic acid) is given off, resulting in respiratoryalkalosis. By repeatedly inhaling the expired air, the individual canrecover some of this CO2and restore acid–base balance.72.The ratio of bicarbonate to carbonic acid in the patient’s bloodcan be determined using the Henderson–Hasselbalch equation:pH = pK+ log[A−][ HA ]7.55 = 6.35 + log[HCO−3][H2CO3]101.2= [HCO−3][H2CO3][HCO−3][H2CO3] = 15.81Similarly, the ratio of bicarbonate to carbonic acid in a normal person’sblood can be determined:pH = pK+ log[A−][ HA ]7.4 = 6.35 + log[HCO−3][H2CO3]

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Solutions118-ChloroadenosineNH2NNNNClOHHHHOHOHCH2OH12.The compound is a thymidine analog. [From Maity, J. and Stromberg,R.,Molecules18,12740–12750 (2013).]5-Bromo-2′-deoxyuridineONOHNBrOHHHHOHHCH2OH13. a.A diphosphate bridge links the ribose groups in each dinuc-leotide. This linkage is a variation of the monophosphate bridge(phosphodiester linkage) in DNA and RNA.b.The adenosine groupin CoA bears a phosphoryl group on C3ʹ.14.OOHHHHOOHCH2OOOPNH2NNONH2OHHHHOHOHCH2OOOPNNNNPhosphodiesterbondIf the dinucleotide were DNA, it would lack OH groups at each riboseC2ʹ position.15.OOHOHOOOONH2NHNH2NHHHHHHHHNNGNNNNACH2OOOPOOOP35CH252Chapter 31.The heat treatment destroys the polysaccharide capsule of the wild-typePneumococcus, but the DNA survives the heat treatment. TheDNA then “invades” the mutantPneumococcusand supplies the genesencoding the enzymes needed for capsule synthesis that the mutantlacks. The mutant is now able to synthesize a capsule and has the ca-pacity to cause disease, which results in the death of the mice and theappearance of encapsulatedPneumococcusin the mouse tissue.2.Theseexperimentsshowedthatthetransformingfactorwasneither a protein nor RNA.3.Some of the labeled “parent” DNA appears in the progeny, butnone of the labeled protein appears in the progeny. This indicatesthat the bacteriophage DNA is involved in the production of progenybacteriophages, but bacteriophage protein is not required.4.The triple-helical model is not consistent with the hydrophobiceffect, which suggests that the nonpolar nitrogenous bases would residein the center of the DNA structure and the hydrophilic phosphateswould reside on the surface. The triple-helical model also assumesthat the phosphate groups are protonated and form stabilizing hydro-gen bonds in the DNA interior. But the pKvalue for phosphate iswell below 7, so the phosphate groups would not be protonated atphysiological pH. In the absence of hydrogen bonds, there are no ad-ditional forces that would hold the strands of the triple helix together.5.Thymine (5-methyl uracil) contains a methyl group attached to C5of the pyrimidine ring of uracil.6.POOHHHOOHHOOH2CAdenine7.NHNH3CO5-MethylcytosineNH28. a.NNNHNHCH3NN6-Methyladenineb.TheN6-DNA methyltransferase might be a good drug target. Ifmethylation of certain adenine residues is required for virulence,then it is possible that inhibition of the bacterial transferase enzymemight prevent adenine methylation and thus prevent disease causedby pathogenic bacteria.9.The base 5-chlorouracil is a substitute for thymine (5-methyluracil).10.A chlorine is substituted for a hydrogen in 5-chorouracil, whichclosely resembles thymine (see Solution 9). Therefore, the culturecontaining 5-chlorouracil will incorporate this base in place of thymineas the DNA replicates. The 5-chlorouracil has a greater mass thanthymine, so DNA isolated from this culture will have a greater massthan DNA isolated from the control culture.11.[From Jordheim, L. P., Durantel, D., Zoulim, F., and Dumontet,C.Nat. Rev. Drug Discov.12,447–464 (2013).]

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12Solutions25.The sugar–phosphate backbone is on the outside of the molecule.The polar sugar groups can form hydrogen bonds with the surround-ing water molecules. The negatively charged phosphate groups inter-act favorably with positively charged ions. The nonpolar nitrogenousbases are found on the inside of the molecule and interact favorablyvia stacking interactions. In this way, contact with the aqueous solu-tion is minimized, as described by the hydrophobic effect.26. a.Proteins are more likely to bind to the major groove, whichcan easily accommodate proteins. The larger surface area of the ma-jor groove allows multiple favorable interactions between the DNAand protein.b.The positively charged side chains of the Lys and Argresidues form ion pairs with the negatively charged phosphate groupson the DNA backbone. These are strong interactions, so the histoneshave a high affinity for DNA.27. a.TheTmis approximately 72°C.b.The melting curves areshown below.1.01.41.31.21.1Temperature (°C)5030Relative absorbance at 260 nm70TmTmD. discoideumS. albus9028. a.The DNA contains 50% G + C, so its melting point would beapproximately 90°C.b.The DNA would need to be cooled graduallyto 65–70°C (20–25°C below its melting temperature).GC content (%)70010203040506070808090100110Tm(°C)29.The DNA from the organisms that thrive in hot environmentswould contain more G and C than DNA from species living in a moretemperate environment. The higher GC content increases the stabilityof DNA at high temperatures.30.The positively charged sodium ions can form ion pairs with thenegatively charged phosphate groups on the DNA backbone and“shield” the negative charges from one another. This increases theoverall stability of DNA and makes it more difficult to melt.31. a.You should increase the temperature to melt out imperfectmatches between the probe and the DNA.b.You should decrease thetemperature to increase the chances that the two strands will align,despite the mismatch.32.Heating denatures the target DNA (separates its two strands) sothat a single-stranded probe can more easily form sequence-specifichydrogen bonds with it.16.OOOHOHOHHHHN1NHOHOHNNHHHHN9OCH2OPOOPOOH2C17.The organism must also contain 19% A (since [A] = [T] accord-ing 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 × 107bases. Therefore,[A] = [T] = (0.19)(6 ×107bases) = 1.14 × 107bases[C] = [G] = (0.31)(6×107bases) = 1.86×107bases18. a.Using Chargaff’s rules (see Solution 17), the number of Cresidues must also be 24,182. Subtracting (2 × 24,182) from 97,004yields 48,640 (A + T) residues. Dividing this number by 2 yields24,320 residues each of A and T.b.GenBank reports only the sequenceof a single strand of DNA, since the sequence of the complementarystrand can easily be deduced using Watson–Crick base-pairing.19.The total amount of purines (A + G) in DNA must equal thetotal amount of pyrimidines (C + T) because each base pair in thedouble-stranded DNA molecule consists of a purine and a pyrimidine.This is not true for RNA, which is single-stranded.20.The genome contains 28.7% T, 21.5% G, 29.3% A, and 20.6% C.Chargaff ’s rules do not apply because the viral genome is composedof single-stranded DNA.21.It is a G:C base pair.22.NHNNNHNHONNHHOCytosineHypoxanthineNNNHNNNHNHONNHHAdenineHypoxanthineNHNOUracilNNHNHONHOHypoxanthine23.The statement is false because the greater stability of GC-richDNA is due to the stronger stacking interactions involving G:C basepairs and does not depend on the number of hydrogen bonds in thebase pairs.24.It is certainly the case that hydrogen bonds hold A:T and G:Cbase pairs together and that these interactions are very favorable. Butupon denaturation of the DNA, each nitrogenous base has the oppor-tunity to form equally favorable hydrogen bonds with water. There-fore, forces other than hydrogen bonds must contribute to the overallstability of the DNA molecule.

Solution Manual For Essential Biochemistry, 4th Edition - Page 14 preview image

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Solutions1335. a.3′-TGTGGTACCACGTAGACTGA-5′b.5′-ACACCAUGGUGCAUCUGACU-3′36. a.The top strand is the coding strand and the bottom strand isthe noncoding strand.b.Only coding strands are published becausethe mRNA sequence is identical to the sequence of the coding strand,with the exception that U replaces T in the mRNA.37. a.A poly-Phe polypeptide was produced.b.Poly A producespoly-Lys; poly C yields poly-Pro; and poly G yields poly-Gly.38. a.The cell-free system produces polypeptides consisting of al-ternating Val and Cys residues. Some of the polypeptides begin withVal; others begin with Cys, depending on which reading frame isused.b.The production of this peptide does not allow one to un-ambiguously assign the GUG and UGU codons. Additional experi-mental data are required to make this assignment.39.The number of possible sequences of four different nucleotidestakennat a time is 4n; herenis the number of nucleotides in thesequence.a.41= 4b.42= 16c.43= 64d.44= 256. At least threenucleotides are necessary to code for 20 amino acids.40.There are 216 codons possible: 63= 216. [From Malyshev, D. A.,Dhami, K., Lavergne, T., Chen, T., Dai, N., Foster, J. M., Corrêa Jr., I.R., and Romesberg, F. E.,Nature509,385–388 (2014).]41. a.First reading frame:AGG TCT TCA GGG AAT GCC TGG CGA GAG GGG AGCArg - Ser - Ser - Gly - Asn - Ala - Trp - Arg - Glu - Gly - Ser-AGC TGG TAT CGC TGG GCC CAA AGG CSer - Trp - Tyr - Arg - Trp - Ala - Gln - ArgSecond reading frame:A GGT CTT CAG GGA ATG CCT GGC GAG AGG GGA GCA- Gly - Leu - Gln - Gly - Met - Pro - Gly - Glu - Arg - Gly - Ala-GCT GGT ATC GCT GGG CCC AAA GGCAla - Gly - Ile - Ala - Gly - Pro - Lys - GlyThird reading frame:AG GTC TTC AGG GAA TGC CTG GCG AGA GGG GAG CAG--Val - Phe - Arg - Glu - Cys - Leu - Ala - Arg - Gly - Glu - Gln-CTG GTA TCG CTG GGC CCA AAG GCLeu - Val - Ser - Leu - Gly - Pro - Lys-b.The second reading frame, which produces a protein in which everythird amino acid is Gly, is the correct reading frame.42. a.Asparagine has two codons, AAU and AAC (see Table 3.3).An A→G mutation at the second position could generate a codonfor serine (AGU or AGC).b.The CGA codon codes for the aminoacid arginine; the mutation (C→T in the DNA) converts the codonto a Stop codon. When the mRNA for the gene is translated, pro-tein synthesis terminates prematurely and the truncated protein isnonfunctional.43.The genetic code (shown in Table 3.3) is redundant. Since thereare 64 different possibilities for 3-base codons and only 20 aminoacids, most amino acids have more than one codon. If a mutationhappens to occur in the third position (3ʹ end), the mutation might notalter the protein sequence. For example, GUU, GUC, GUA, and GUGall code for valine. A mutation in the third position of a valine codonwould still result in the selection of valine and would have no effecton the amino acid sequence of the protein.44.The same segment of DNA can encode two different proteins ifeach strand is a coding strand.45.First, identify the translation start site, the Met residue whosecodon is AUG in the mRNA (see Table 3.3) or ATG in the DNA.Translation stops at the DNA sequence TAA, which corresponds to33. a.An inherited characteristic could be determined by more thanone gene.b.Some sequences of DNA encode RNA molecules that arenot translated into protein (for example, rRNA and tRNA).c.Somegenes are not transcribed during a cell’s lifetime. This can occur if thegene is expressed only under certain environmental conditions or in cer-tain specialized cells in a multicellular organism.34. a.The DNA isolated after one generation is a homogeneoussample of DNA with a density intermediate between DNA containingall14N and all15N.b.The DNA isolated after the second generation isheterogeneous. Half of the DNA has the same density as the first gener-ation; half of the DNA consists of all14N DNA and has a lower density.15N14NFirst-generationdaughter moleculesSecond-generationdaughter moleculesc.A hypothetical scheme for conservative DNA replication is shownbelow. The DNA isolated after both the first and the second genera-tions is heterogeneous. Half of the DNA has the same high density asthe original DNA (all15N); half of the DNA consists of all14N DNAand has a lower density.15N14NFirst-generationdaughter moleculesSecond-generationdaughter molecules

Solution Manual For Essential Biochemistry, 4th Edition - Page 15 preview image

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14Solutions•How many copies of each gene are in the organism’s genome?•What is the amount of noncoding DNA in the genome? Is the non-coding DNA really “junk DNA,” or does it have some biological role?•How many transposable elements are in the noncoding DNA?53. a.The fi rst reading frame is the longest ORF.First reading frame:TAT GGG ATG GCT GAG TAC AGC ACG TTG AAT GAGTyr - Gly - Met - Ala - Glu - Tyr - Ser - Thr - Leu - Asn - Glu-GCG ATG GCC GCT GGT GAT GAla - Met - Ala - Ala - Gly - Asp-Second reading frame:T ATG GGA TGG CTG AGT ACA GCA CGT TGA ATG AGG-Met - Gly - Trp - Leu - Ser - Thr - Ala - Arg -Stop - Met - Arg-CGA TGG CCG CTG GTG ATGArg - Trp - Pro - Leu - Val - MetThird reading frame:TA TGG GAT GGC TGA GTA CAG CAC GTT GAA TGA GGC---Trp - Asp - Gly - Stop - Val - Gln - His - Val - Glu - Stop - Gly-GAT GGC CGC TGG TGA TGAsp - Gly - Arg - Trp - Stopb.Assuming the reading frame has been correctly identified, the mostlikely start site is the first Met residue in the first ORF.54.Like the symbiont described in Problem 49, the bacteriophageis parasitic and makes use of the host cell’s tRNAs, protein-synthe-sizing machinery, and replication proteins.55.If a SNP occurs every 300 nucleotides or so, and if there areabout 3 million kb (3 × 109nucleotides) in the human genome (seeTable 3.4), then there are (3 × 109bp ÷ 300 bp/SNP) = 1 × 107(10 million) total SNPs in the human genome. [Source: ghr.nlm.nih.gov/handbook/genomicresearch/snp]56.Because ~1.4% of the human genome consists of protein-codinggenes, about 1.4% of the 10 million SNPs, or 140,000, are likely toaffect these genes. Since the total number of genes is only ~21,000, intheory, every gene would differ.57. a.The strongest associations are located between positions67,370,000 and 67,470,000.b.Gene B contains SNPs associatedwith the disease whereas Genes A and C do not. [From Duerr, R. H.,et al.,Science314,1461–1463 (2006).]58.Chromosomes 10, 11, and 13 have genes carrying SNPs that arecorrelated with the colon disease. [From Garcia-Barcelo, M., et al.,Proc. Natl. Acad. Sci. USA106,2694–2699 (2009).]59.MspI, AsuI, EcoRI, PstI, SauI, and NotI generate sticky ends.AluI and EcoRV generate blunt ends.60.The restriction enzyme with the longer recognition sequencewould be a rare cutter, because it is likely to encounter this sequenceless often and therefore will cleave the DNA less frequently than arestriction enzyme with a shorter recognition sequence.61.AluI cleaves the sequence at two locations and EcoRI and NotIcleave the sequence at one location each. The other enzymes listed inTable 3.5 do not cleave this segment of the plasmid DNA.CGCGGATCCGAATTCGAGCTCCGTCGACAAGCTTGCGGCCGCACTCGAGEcoRIAluIAluINotI62.The enzyme MspI generates sticky ends. The single-stranded re-gions are then removed by the action of the exonuclease, releasing thefree nucleotides C and G:the stop codon UAA in the mRNA. Use Table 3.3 to decode the inter-vening codons, substituting U for T.CTCAGAGTTCACC ATG GGC TCC ATC GGT GCA GCA AGCMetGlySerIleGlyAlaAlaSerATG GAA ··· 1104 bp ·· TTC TTT GGC AGA TGT GTT TCCMetGlu ···················· PhePheGlyArgCysValSerCCT TAA AAAGAAPro*46. a.The siRNA is an “antisense” mRNA, so its sequence should becomplementary to that of the mRNA. The solution below shows ansiRNA that corresponds to the 5ʹ end of the mRNA:mRNA 5ʹ→3ʹ AUG GGC UCC AUC GGU GCA GCA AGC AUG GAAsiRNA 3ʹ→5ʹUAC CCG AGG UAG CCA CGU CGU UCG UAC CUUb.When the antisense RNA binds to the mRNA, the mRNA can nolonger serve as a template for protein synthesis. (In the cell, the bind-ing of antisense RNA to mRNA involves an enzyme that degradesthe mRNA into smaller fragments.)c.The siRNA molecules must beable to cross the cell membrane to enter the cell and find the targetmRNA.47. a.The normal protein sequence is ····Glu-Asn-Ile-Ile-Phe-Gly-Val-Ser-Tyr····. The mutant protein sequence is the same except thePhe at position 508 is deleted. Note that although the deletion of Pheaffects codons 507 and 508, the redundancy of the genetic code meansthat the Ile at position 507 is not affected, and the amino acids down-stream of the mutation are also unaffected.b.The sequence of thenormal protein in this region of the gene is ····Asn-Ile-Asp-Thr····. Theamino acid sequence of the mutated CF gene (in which two bases, Aand T, are missing), is ····Asn-Arg-Tyr····. This is a frameshift muta-tion and all of the amino acids past the deletion will differ from theamino acids in the normal protein.48.In order for a single nucleotide change to switch an argininecodon to a glutamine codon, either a CGA or a CGG Arg codon musthave been replaced with a CAA or CAG glutamine codon. In eithercase, the middle G has been changed to an A. The correct readingframe with the amino acid sequences for the normal and mutant pro-teins are shown below. [From Ludwig, E. H. and McCarthy, B. J.Am.J. Hum. Genet.47,712–720 (1990).]Normal gene· · · CT GGC CGG CTC AAT GGA GAG TCC · · ·GlyArgLeuAsnGlyGluSerMutated gene· · · CT GGC CAG CTC AAT GGA GAG TCC · · ·GlyGlnLeuAsnGlyGluSer49.C. ruddii, with such a small genome and only 182 genes, must besome sort of parasite rather than a free-living bacterium. (In fact,C.ruddiiis an insect symbiont.)50.Prokaryotes tend to have smaller genomes than eukaryotes, soevolution has shaped the prokaryote genome to pack in genes moreefficiently. Eukaryotes, with larger genomes and more noncodingDNA, have more space to arrange genes.51.The 35 million differences out of 3.0 billion total nucleotidesrepresent approximately 1%, or a bit less than the original claim.(This number reflects single-base differences and does not accountfor insertions and deletions of multiple bases.)52.Questions that need to be addressed in order to solve the C-valueparadox:•How does one define organismal complexity? Perhaps humans arenot the most complex organisms.•How much of the organism’s genome codes for RNA? For protein?What are the biological roles of these gene products?

Solution Manual For Essential Biochemistry, 4th Edition - Page 16 preview image

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Solutions15In practice, however, you’d want to embed the recognition sequencerather than having the recognition sequence on the ends; the reason forthis will be explained in Chapter 20.69.Primers with high GC content have highTmvalues. If the annealingtemperature is much lower than the melting temperature, improperbase pairing may occur (see Figure 3.8) and the desired gene frag-ment may not be amplified.70.You can use a DNA polymerase that is not heat-stable. Youwould have to cool the reaction mixture to a temperature at which thepolymerase works best and you would have to add the enzyme at eachreaction cycle because it would be destroyed every time the temper-ature was raised to melt the double-stranded DNA.71.To amplify the protein-coding DNA sequence, the primersshould correspond to the first three and last three residues of the pro-tein (each amino acid represents three nucleotides, so the primerswould each be nine bases long). Use Table 3.3 to find the codons thatcorrespond to the first three residues:MetGlySerAUGGGUUCUGGCUCCGGAUCAGGGUCGAGUAGCUsing just the topmost set of codons, a possible DNA primer wouldtherefore have the sequence 5ʹ-ATGGGTTCT-3ʹ. This primer couldbase pair with the gene’s noncoding strand, and its extension fromits 3ʹ end would yield a copy of the coding strand of the gene (seeFig. 3.18). The other primer must correspond to the last three aminoacids of the protein:ValSerProGUUUCUCCUGUCUCCCCCGUAUCACCAGUGUCGCCGAGUAGCAgain, considering just the topmost set of codons, a probable DNAcoding sequence would be 5ʹ-GTTTCTCCT-3ʹ. This sequence cannotbe used as a primer. However, a suitable primer would be the comple-mentary sequence 5ʹ-AGGAGAAAC-3ʹ, which can then be extendedfrom its 3ʹ end to yield a copy of the noncoding strand of the gene.The number of possible primer pairs is quite large, because all butone of the amino acids has more than one codon. For the first primer,there are 1 × 4 × 6 = 24 possibilities; for the second, 4 × 6 × 4 = 96possibilities. There are 24 × 96 = 2304 different pairs of primers thatcould be used to amplify the gene by PCR.72.To design an oligonucleotide probe for a gene, the researcher mustapply the genetic code in reverse, that is, select codons that correspondto the amino acids in the protein. Most amino acids can be encodedby more than one codon, so the researcher would have to choose oneof them and hope that it matched the DNA well enough for the probeto successfully hybridize with the DNA (see Solution 71). Met andTrp, however, are encoded by only one codon each, so by using thesecodons as part of the probe, the researcher can be assured of a perfectmatch with the DNA, at least for these three nucleotides.73.ATTGTTCCCACAGACCGCGGCGAAGCATTGTTCCACCGTGTTTCCGACCGTTGTTCCCACAGACCGTGTG C T TA G CA C G A ATC G G CC G G A A C G AC T TGC TExonucleasecleavage63. a.A HaeII digest of the linear 2743-bp pGEM-3Z plasmid (seeFigure 3.16) would produce four bands, as shown below (1–323 pro-duces a 323-bp band, 324–693 produces a 370-bp band, 694–2564produces an 1871-bp band, and 2565–2743 produces a 179-bp band).b.A digest of the circular plasmid produces the 370-bp band and the1871-bp band but also produces a 502-bp band (2565–323) in placeof the 323-bp band and the 179-bp band. [From Promega.]LinearCircular3000 bp1500 bp1000 bp500 bp100 bp64.2 kb7 kbPst I8 kb3 kbEcoRIPst I65.Polymerization occurs in the 5ʹ→3ʹ direction and a 3ʹ OH groupmust be available, so the primer must be complementary to the sequenceas shown.5ʹ-AGTCGATCCCTGATCGTACGCTACGGTAACGT-3ʹ3ʹ-TGCCATTGCA-5ʹ66.A restriction endonuclease is often used to prepare fragments ofDNA for insertion into a cloning vector. Since the cloned DNA con-tains the recognition site (whose sequence is known), this sequence canbe used as a starting point to sequence the unknown DNA segment.67.The primers are shown in red below.5ʹ-ATGATTCGCCTCGGGGCTCCCCAGTCGCTGGTGCTGCT GACGCTGCTCGTCG-3ʹ← 3ʹ-ACGACT GCGACGAGCAGC-5ʹ5ʹ-ATGATTCGCCTCGGGGCT-3ʹ →3ʹ-TACTAAGCGGAGCCCCGAGGGGTCAGCGACCACGACGACTGCGACGAGCAGC-5ʹ68. a.The primers are shown in red below.5ʹ-ATGGGCTCCATCGGTGCAGCAAGCATGGAA · · ·TTCTTTGGCAGATGTGT T TCCCCTTAAAAAGAA-3ʹ3ʹ-CAAAGGGGAATT T T TCT T-5ʹ5ʹ-AT GGGCTCCAT CGGT GCA-3ʹ3ʹ-TACCCGAGGTAGCCACGTCGTTCGTACCTT · · ·AAGAAACCGTCTACACAAAGGGGAATTTTTCTT-5ʹb.AddthesequencerecognizedbytheEcoRIendonuclease(GAATTC; see Table 3.5) to the 5ʹ end of each primer. So the forwardprimer (on the left) would have a sequence of 5ʹ-GAATTCATGG-GCTCCATCGGTGCA-3ʹ and the reverse primer (on the right) wouldhaveasequenceof5ʹ-GAATTCTTCTTTTTAAGGGGAAAC-3ʹ.

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