Lehninger Principles of Biochemistry Seventh Edition Solution Manual

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The Absolute, Ultimate Guide toLehninger Principles of BiochemistryStudy Guide and Solutions ManualMarcy OsgoodUniversity of New Mexico School of MedicineKaren OcorrSanford Burnham Prebys Medical Discovery InstituteSolutions Manual based ona previous edition byFrederick WedlerRobert BernlohrRoss HardisonTeh-Hui KaoMing TienPennsylvania State UniversitySeventh Edition

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PrefaceviiAbout the Authorsix1The Foundations of Biochemistry1Part IStructure and Catalysis2Water113Amino Acids, Peptides, and Proteins214The Three-Dimensional Structureof Proteins415Protein Function546Enzymes707Carbohydrates and Glycobiology878Nucleotides and Nucleic Acids1009DNA-Based Information Technologies11310Lipids12311Biological Membranes and Transport13412Biosignaling151Part IIBioenergetics andMetabolism13Bioenergetics and Biochemical ReactionTypes17214Glycolysis, Gluconeogenesis, and thePentose Phosphate Pathway18315Principles of Metabolic Regulation19516The Citric Acid Cycle20617Fatty Acid Catabolism21818Amino Acid Oxidation and the Productionof Urea22819Oxidative Phosphorylation23820Photosynthesis and CarbohydrateSynthesis in Plants24821Lipid Biosynthesis26122Biosynthesis of Amino Acids, Nucleotides,and Related Molecules27423Hormonal Regulation and Integration ofMammalian Metabolism286Part IIIInformation Pathways24Genes and Chromosomes29925DNA Metabolism30726RNA Metabolism31927Protein Metabolism33128Regulation of Gene Expression343vContentsStudy Guide

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1The Foundations of BiochemistryS-1Part IStructure and Catalysis2WaterS-143Amino Acids, Peptides, and ProteinsS-294The Three-Dimensional Structure ofProteinsS-445Protein FunctionS-546EnzymesS-637Carbohydrates and GlycobiologyS-788Nucleotides and Nucleic AcidsS-889DNA-Based Information TechnologiesS-9810LipidsS-11211Biological Membranes and TransportS-12012BiosignalingS-132Part IIBioenergetics andMetabolism13Bioenergetics and Biochemical ReactionTypesS-14114Glycolysis, Gluconeogenesis, and thePentose Phosphate PathwayS-16115Principles of Metabolic RegulationS-17416The Citric Acid CycleS-18417Fatty Acid CatabolismS-20018Amino Acid Oxidation and the Productionof UreaS-21219Oxidative PhosphorylationS-22420Photosynthesis and CarbohydrateSynthesis in PlantsS-23621Lipid BiosynthesisS-24922Biosynthesis of Amino Acids, Nucleotides,and Related MoleculesS-26023Hormonal Regulation and Integration ofMammalian MetabolismS-268Part IIIInformation Pathways24Genes and ChromosomesS-27625DNA MetabolismS-28526RNA MetabolismS-29727Protein MetabolismS-30428Regulation of Gene ExpressionS-314viContentsSolutions Manual

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viiPrefaceLearning a complex subject, such as biochemistry, isvery much like learning a foreign language.In the study of a foreign language there are severaldistinct components that must be mastered: the vo-cabulary, the grammatical rules, and the integration ofthese words and rules so they can be used to commu-nicate ideas. Similarly, in the study of biochemistrythere is a very large (some would say vast) number ofnew terms and concepts, as well as a complex set of“rules” governing biochemical reactions that must allbe memorized. All of this information must be inte-grated into an interrelated whole that describes bio-logical systems.Memorizingvocabularyandgrammaticalruleswill not make you fluent in a foreign language; nei-ther will such memorization make you fluent inbiochemistry.Similarly, listening to someone speak a foreign lan-guage will not, by itself, make you more capable ofproducing those same sounds, words, and sentences.The key to mastery of any new subject area, whetherit is a foreign language or biochemistry, is the interac-tion of memorization, practice, and application untilthe information fits together into a coherent whole.In this workbook we attempt to guide you throughthematerialpresentedinLehninger Principles ofBiochemistry, Seventh Editionby Nelson and Cox.TheStep-by-Step Guideto each chapter includesthree parts:• A one- to two-page summary of theMajor Conceptshelps you to see the “big picture” for each chapter.•What to Reviewhelps you make direct connectionsbetween the current material and related informa-tion presented elsewhere in the text.•Topics for Discussionfor each section and subsec-tion in the chapter focus your attention on the mainpoints being presented and help you internalize theinformation byusingit.•DiscussionQuestionsforStudyGroupsarequestions that are especially suited to Study Groups,eitherbecausetheypulltogetherseveralpointsmade in the chapter or because they are more in-volved questions that would benefit from collabora-tive insight.Each chapter also includes aSelf-Testfor you toassess your progress in mastering biochemical ter-minology and facts, and learning to integrate andapply that information.•Do You Know the Terms?asks you to complete acrossword puzzle using the new vocabulary intro-duced in the chapter.•Do You Know the Facts?tests how well you havelearned the “rules” of biochemistry.•Applying What You Knowtests how well you“speak the language” of biochemistry, often in anexperimental or metabolically relevant context.Two especially popular features of theAbsolute, Ulti-mate Guideare:• TheBiochemistry Onlineproblems will expose youto just a few of the analytical resources that areavailable to scientists on the Internet. The molecu-lar models are fun to play with, and many of thequestionsprovideyouwithanopportunitytoanalyze “data” as you might in an actual laboratorysetting.• TheCell Mapis based on many semesters of use byour students. It is designed to help you place thebiochemical pathways that you are learning aboutinto their proper cellular perspective. The Cell Mapquestions tell you what to include in your Map, butyou can be creative!Students have told us it is a great study aid, thatreally helped them to make the connections betweenthe various pathways.

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In our experience, this material can best be assimi-lated when it is discussed in a study group.A study group is nothing more than two to four peoplewho get together on a regular basis (weekly) to “speakbiochemistry.” This type of interaction is critical to flu-ency in a foreign language and is no less critical in thesuccessful assimilation of biochemistry. We designedour Step-by-Step Guide to each chapter with studygroups in mind. The questions posed for each sectioncan be used as springboards for study group discus-sions. We purposefully havenotsupplied answers tothis section to force you to wrestle with the concepts.It is the struggle that will make you learn the material.In addition, because most of the answers can be read-ily worked out by a careful reading of the section ofthe text, the questions will focus your attention on themore important aspects of the material.DetailedSolutionsto all the end-of-chapter textbookproblems are included as a separate section in theAbsolute, Ultimate Guide.We have taken great care to ensure that the solutionsarecorrect,complete,andinformative.Thefinalanswer to numerical problems has been roundedoff to reflect the number of significant figures inthe data.We thank each and every one of our students for theirinvaluable feedback and input, which have helped tomake this study guide its absolute and ultimate best.viiiPreface

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About the AuthorsixKaren Ocorrreceived her Ph.D. from Wesleyan University, where she studied the physiologyand neurochemistry of the lobster cardiac ganglion. As an NIH postdoctoral research fellowat the University of Texas, she examined the roles of enzymes and second messengersin neuronal plasticity inAplysia californica.She continued these investigations at theCalifornia Institute of Technology and Stanford University, examining the role of intracellularsignaling underlying long-term potentiation in the vertebrate hippocampus. She taughtIntroductory Biochemistry for 10 years at the University of Michigan, Ann Arbor, where shealso taught Animal Physiology, Cell Biology, and Introductory Biology for Non-Majors. Shealso taught Introductory Biochemistry for four years as a visiting Professor at the HunanNormal University in Changsha, China. Currently she is an Assistant Professor at SBPMedical Discovery Institute in La Jolla, California, where she is researching the roles of ionchannels in cardiomyopathies. She currently teaches graduate and medical students at boththe SBP Medical Discovery Institute and the University of California–San Diego.Marcy Osgoodreceived her Ph.D. from Rensselaer Polytechnic Institute. After twopostdoctoral positions, she became an Assistant Professor in the Department of PhysicalSciences at Albany College of Pharmacy, Union University. She was then a Lecturer in theBiology Department at the University of Michigan, Ann Arbor for nine years, where herresearch interests began to focus on educational issues. Her current position is as AssociateProfessor in the Department of Biochemistry and Molecular Biology and Assistant Dean ofUndergraduate Medical Education at the University of New Mexico School of Medicine. Hercurrent research efforts deal with assessment of problem-solving strategies.Osgood and Ocorr have collaborated for 10 years to develop effective techniques for teachingbiochemistry, all of which are based on educational research. This study guide is the result ofthese efforts and embodies much of what they have found to be effective over years ofinstruction at many levels and in many areas of the biological sciences. It includes thousandsof discussion, quiz, and exam questions from their biochemistry courses. Osgood and Ocorrhave been responsible for teaching biochemistry to more than 10,000 undergraduatestudents.

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The Foundationsof Biochemistrychapter1S-11.The Size of Cells and Their Components(a)If you were to magnify a cell 10,000-fold (typical of the magnification achieved using an electronmicroscope), how big would it appear? Assume you are viewing a “typical” eukaryotic cell with acellular diameter of 50mm.(b)If this cell were a muscle cell (myocyte), how many molecules of actin could it hold? Assumethe cell is spherical and no other cellular components are present; actin molecules are spherical,with a diameter of 3.6 nm. (The volume of a sphere is 4/3pr3.)(c)If this were a liver cell (hepatocyte) of the same dimensions, how many mitochondria could ithold? Assume the cell is spherical; no other cellular components are present; and themitochondria are spherical, with a diameter of 1.5mm.(d)Glucose is the major energy-yielding nutrient for most cells. Assuming a cellular concentration of1 mM(i.e., 1 millimole/L), calculate how many molecules of glucose would be present in ourhypothetical (and spherical) eukaryotic cell. (Avogadro’s number, the number of molecules in 1mol of a nonionized substance, is 6.021023.)(e)Hexokinase is an important enzyme in the metabolism of glucose. If the concentration of hexokinasein our eukaryotic cell is 20mM, how many glucose molecules are present per hexokinase molecule?Answer(a)The magnified cell would have a diameter of 50104mm500103mm500 mm,or 20 inches—about the diameter of a large pizza.(b)The radius of a globular actin molecule is 3.6 nm/21.8 nm; the volume of themolecule, in cubic meters, is (4/3)(3.14)(1.8109m)32.41026m3.*The number of actin molecules that could fit inside the cell is found by dividing the cellvolume (radius25mm) by the actin molecule volume. Cell volume(4/3)(3.14)(25106m)36.51014m3. Thus, the number of actin molecules in the hypotheticalmuscle cell is(6.51014m3)/(2.41026m3)2.71012moleculesor 2.7 trillion actin molecules.*Significant figures:In multiplication and division, the answer can be expressed with nomore significant figures than the least precise value in the calculation. Because some of thedata in these problems are derived from measured values, we must round off the calculatedanswer to reflect this. In this first example, the radius of the actin (1.8 nm) has two significantfigures, so the answer (volume of actin2.41026m3) can be expressed with no morethan two significant figures. It will be standard practice in these expanded answers to roundoff answers to the proper number of significant figures.

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S-2Chapter 1The Foundations of Biochemistry(c)The radius of the spherical mitochondrion is 1.5mm/20.75mm, therefore the volumeis (4/3)(3.14)(0.75106m)31.81018m3. The number of mitochondria in thehypothetical liver cell is(6.51014m3)/(1.81018m3)36103mitochondria(d)The volume of the eukaryotic cell is 6.51014m3, which is 6.5108cm3or 6.5108mL. One liter of a 1 mMsolution of glucose has (0.001 mol/1000 mL)(6.021023molecules/mol)6.021017molecules/mL. The number of glucose molecules in thecell is the product of the cell volume and glucose concentration:(6.5108mL)(6.021017molecules/mL)3.91010moleculesor 39 billion glucose molecules.(e)The concentration ratio of glucose/hexokinase is 0.001M/0.00002M, or 50/1, meaning thateach enzyme molecule would have about 50 molecules of glucose available as substrate.2.Components ofE. coliE. colicells are rod-shaped, about 2mm long and 0.8mm in diameter. Thevolume of a cylinder ispr2h,wherehis the height of the cylinder.(a)If the average density ofE. coli(mostly water) is 1.1103g/L, what is the mass of a single cell?(b)E. colihas a protective cell envelope 10 nm thick. What percentage of the total volume of thebacterium does the cell envelope occupy?(c)E. coliis capable of growing and multiplying rapidly because it contains some 15,000 sphericalribosomes (diameter 18 nm), which carry out protein synthesis. What percentage of the cellvolume do the ribosomes occupy?Answer(a)The volume of a singleE. colicell can be calculated frompr2h(radius0.4mm):3.14(4105cm)2(2104cm)1.01012cm311015m311015LDensity (g/L) multiplied by volume (L) gives the mass of a single cell:(1.1103g/L)(11015L)11012gor a mass of 1 pg.(b)First, calculate the proportion of cell volume that doesnotinclude the cell envelope,that is, the cell volumewithoutthe envelope—withr0.4mm0.01mm; andh2mm2(0.01mm)—divided by the total volume.Volume without envelopep(0.39mm)2(1.98mm)Volume with envelopep(0.4mm)2(2mm)So the percentage of cell that doesnotinclude the envelope is90%(Note that we had to calculate to one significant figure, rounding down the 94% to 90%,which here makes a large difference to the answer.) The cell envelope must account for10% of the total volume of this bacterium.(c)The volume of all the ribosomes (each ribosome of radius 9 nm)15,000(4/3)p(9103mm)3The volume of the cellp(0.4mm)2(2mm)So the percentage of cell volume occupied by the ribosomes is5%15,000(4/3)p(9103mm)3100p(0.4mm)2(2mm)p(0.39mm)2(1.98mm)100p(0.4mm)2(2mm)

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3.Genetic Information inE. ColiDNAThe genetic information contained in DNA consists of alinear sequence of coding units, known as codons. Each codon is a specific sequence of three deoxyri-bonucleotides (three deoxyribonucleotide pairs in double-stranded DNA), and each codon codes for asingle amino acid unit in a protein. The molecular weight of anE. coliDNA molecule is about3.1109g/mol. The average molecular weight of a nucleotide pair is 660 g/mol, and each nucleotidepair contributes 0.34 nm to the length of DNA.(a)Calculate the length of anE. coliDNA molecule. Compare the length of the DNA molecule withthe cell dimensions (see Problem 2). How does the DNA molecule fit into the cell?(b)Assume that the average protein inE. coliconsists of a chain of 400 amino acids. What is themaximum number of proteins that can be coded by anE. coliDNA molecule?Answer(a)The number of nucleotide pairs in the DNA molecule is calculated by dividing the molec-ular weight of DNA by that of a single pair:(3.1109g/mol)/(0.66103g/mol)4.7106pairsMultiplying the number of pairs by the length per pair gives(4.7106pairs)(0.34 nm/pair)1.6106nm1.6 mmThe length of the cell is 2mm (from Problem 2), or 0.002 mm, which means the DNA is(1.6 mm)/(0.002 mm)800 times longer than the cell. The DNA must be tightly coiledto fit into the cell.(b)Because the DNA molecule has 4.7106nucleotide pairs, as calculated in (a), it musthave one-third this number of triplet codons:(4.7106)/31.6106codonsIf each protein has an average of 400 amino acids, each requiring one codon, the numberof proteins that can be coded byE. coliDNA is(1.6106codons)(1 amino acid/codon)/(400 amino acids/protein)4,000 proteins4.The High Rate of Bacterial MetabolismBacterial cells have a much higher rate of metabolismthan animal cells. Under ideal conditions some bacteria double in size and divide every 20 min,whereas most animal cells under rapid growth conditions require 24 hours. The high rate of bacterialmetabolism requires a high ratio of surface area to cell volume.(a)Why does surface-to-volume ratio affect the maximum rate of metabolism?(b)Calculate the surface-to-volume ratio for the spherical bacteriumNeisseria gonorrhoeae(diameter0.5mm), responsible for the disease gonorrhea. Compare it with the surface-to-volume ratio for aglobular amoeba, a large eukaryotic cell (diameter 150mm). The surface area of a sphere is 4pr2.Answer(a)Metabolic rate is limited by diffusion of fuels into the cell and waste products out of thecell. This diffusion in turn is limited by the surface area of the cell. As the ratio ofsurface area to volume decreases, the rate of diffusion cannot keep up with the rate ofmetabolism within the cell.(b)For a sphere, surface area4pr2and volume4/3pr3. The ratio of the two is thesurface-to-volume ratio,S/V,which is 3/ror 6/D, whereDdiameter. Thus, rather thancalculatingSandVseparately for each cell, we can rapidly calculate and compareS/Vratios for cells of different diameters.S/VforN. gonorrhoeae6/(0.5mm)12mm1S/Vfor amoeba6/(150mm)0.04mm1300Thus, the surface-to-volume ratio is 300 times greater for the bacterium.12mm10.04mm1S/Vfor bacteriumS/Vfor amoebaChapter 1The Foundations of BiochemistryS-3

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5.Fast Axonal TransportNeurons have long thin processes called axons, structures specialized forconducting signals throughout the organism’s nervous system. Some axonal processes can be as longas 2 m—for example, the axons that originate in your spinal cord and terminate in the muscles of yourtoes. Small membrane-enclosed vesicles carrying materials essential to axonal function move along mi-crotubules of the cytoskeleton, from the cell body to the tips of the axons. If the average velocity of avesicle is 1mm/s, how long does it take a vesicle to move from a cell body in the spinal cord to theaxonal tip in the toes?AnswerTransport time equals distance traveled/velocity, or(2106mm)/(1mm/s)2106sor about 23 days!6.Is Synthetic Vitamin C as Good as the Natural Vitamin?A claim put forth by some purveyors ofhealth foods is that vitamins obtained from natural sources are more healthful than those obtained bychemical synthesis. For example, pureL-ascorbic acid (vitamin C) extracted from rose hips is betterthan pureL-ascorbic acid manufactured in a chemical plant. Are the vitamins from the two sources dif-ferent? Can the body distinguish a vitamin’s source?AnswerThe properties of the vitamin—like any other compound—are determined by itschemical structure. Because vitamin molecules from the two sources are structurally identical,their properties are identical, and no organism can distinguish between them. If different vitaminpreparations contain different impurities, the biological effects of themixturesmay vary withthe source. The ascorbic acid in such preparations, however, is identical.7.Identification of Functional GroupsFigures 1–16 and 1–17 show some common functional groupsof biomolecules. Because the properties and biological activities of biomolecules are largely deter-mined by their functional groups, it is important to be able to identify them. In each of the compoundsbelow, circle and identify by name each functional group.S-4Chapter 1The Foundations of BiochemistryHHEthanolamine(a)CHHCOHH3NGlycerol(b)HHCOHHCOHHCOHHThreonine, anamino acid(d)HCH3COHCHPantothenate,a vitamin(e)H3CCH2OHCCH3HCOHCONHPhosphoenolpyruvate,an intermediate inglucose metabolism(c)CCOHOPOHHCH2CH2COD-Glucosamine(f )HCH2OHCOHHCOHHOCHHCNH3COHH3N

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Answer(a)ONH3amino;OOHhydroxyl(b)OOHhydroxyl (three)(c)OP(OH)O2phosphoryl (in its ionized form);OCOOcarboxyl(d)OCOOcarboxyl;ONH3amino;OOHhydroxyl;OCH3methyl (two)(e)OCOOcarboxyl;OCOONHOamide;OOHhydroxyl (two);OCH3methyl(two)(f)OCHOaldehyde;ONH3amino;OOHhydroxyl (four)8.Drug Activity and StereochemistryThe quantitative differences in biological activity between thetwo enantiomers of a compound are sometimes quite large. For example, theDisomer of the drug iso-proterenol, used to treat mild asthma, is 50 to 80 times more effective as a bronchodilator than theLisomer. Identify the chiral center in isoproterenol. Why do the two enantiomers have such radicallydifferent bioactivity?AnswerA chiral center, or chiral carbon, is a carbon atom that is bonded to four differentgroups. A molecule with a single chiral center has two enantiomers, designatedDandL(or in theRS system,SandR). In isoproterenol, only one carbon (asterisk) has four different groupsaround it; this is the chiral center:The bioactivity of a drug is the result of interaction with a biological “receptor,” a proteinmolecule with a binding site that is also chiral and stereospecific. The interaction of theDisomerof a drug with a chiral receptor site will differ from the interaction of theLisomer with that site.9.Separating BiomoleculesIn studying a particular biomolecule (a protein, nucleic acid, carbohy-drate, or lipid) in the laboratory, the biochemist first needs to separate it from other biomolecules inthe sample—that is, topurifyit. Specific purification techniques are described later in the book. How-ever, by looking at the monomeric subunits of a biomolecule, you should have some ideas about thecharacteristics of the molecule that would allow you to separate it from other molecules. For example,how would you separate(a)amino acids from fatty acids and(b)nucleotides from glucose?Answer(a)Amino acids and fatty acids have carboxyl groups, whereas only the amino acids have aminogroups. Thus, you could use a technique that separates molecules on the basis of the prop-erties (charge or binding affinity) of amino groups. Fatty acids have long hydrocarbonchains and therefore are less soluble in water than amino acids. And finally, the sizes andshapes of these two types of molecules are quite different. Any one or more of these prop-erties may provide ways to separate the two types of compounds.(b)A nucleotide molecule has three components: a nitrogenous organic base, a five-carbonsugar, and phosphate. Glucose is a six-carbon sugar; it is smaller than a nucleotide. The sizedifference could be used to separate the molecules. Alternatively, you could use the nitroge-nous bases and/or the phosphate groups characteristic of the nucleotides to separate them(based on differences in solubility, charge) from glucose.Chapter 1The Foundations of BiochemistryS-5OHHIsoproterenolCHOCH2HNCH3CCH3HHOHOHOC*HOHCCH2CH3CH3NHH

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10.Silicon-Based Life?Silicon is in the same group of the periodic table as carbon and, like carbon, canform up to four single bonds. Many science fiction stories have been based on the premise of silicon-based life. Is this realistic? What characteristics of silicon make itlesswell adapted than carbon as thecentral organizing element for life? To answer this question, consider what you have learned about car-bon’s bonding versatility, and refer to a beginning inorganic chemistry textbook for silicon’s bondingproperties.AnswerIt is improbable that silicon could serve as the central organizing element for life undersuch conditions as those found on Earth for several reasons. Long chains of silicon atoms are notreadily synthesized, and thus the polymeric macromolecules necessary for more complex func-tions would not readily form. Also, oxygen disrupts bonds between two silicon atoms, so silicon-based life-forms would be unstable in an oxygen-containing atmosphere. Once formed, the bondsbetween silicon and oxygen are extremely stable and difficult to break, which would prevent thebreaking and making (degradation and synthesis) of biomolecules that is essential to theprocesses of living organisms.11.Drug Action and Shape of MoleculesSome years ago two drug companies marketed a drug underthe trade names Dexedrine and Benzedrine. The structure of the drug is shown below.The physical properties (C, H, and N analysis, melting point, solubility, etc.) of Dexedrine andBenzedrine were identical. The recommended oral dosage of Dexedrine (which is still available)was 5 mg/day, but the recommended dosage of Benzedrine (no longer available) was twice that.Apparently, it required considerably more Benzedrine than Dexedrine to yield the same physiologi-cal response. Explain this apparent contradiction.AnswerOnly one of the two enantiomers of the drug molecule (which has a chiral center) isphysiologically active, for reasons described in the answer to Problem 3 (interaction with astereospecific receptor site). Dexedrine, as manufactured, consists of the single enantiomer(D-amphetamine) recognized by the receptor site. Benzedrine was a racemic mixture (equalamounts ofDandLisomers), so a much larger dose was required to obtain the same effect.12.Components of Complex BiomoleculesFigure 1–11 shows the major components of complex bio-molecules. For each of the three important biomolecules below (shown in their ionized forms at physi-ological pH), identify the constituents.(a)Guanosine triphosphate (GTP), an energy-rich nucleotide that serves as a precursor to RNA:S-6Chapter 1The Foundations of BiochemistryCCH2CH3NH2HNCOOOOPOPONNNHNH2OPCH2OOOOHHHHOHOHO

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(b)Methionine enkephalin, the brain’s own opiate:(c)Phosphatidylcholine, a component of many membranes:Answer(a)Three phosphoric acid groups (linked by two anhydride bonds), esterified to ana-D-ribose (at the 5position), which is attached at C-1 to guanine.(b)Tyrosine, two glycine, phenylalanine, and methionine residues, all linked by peptide bonds.(c)Choline esterified to a phosphoric acid group, which is esterified to glycerol, which isesterified to two fatty acids, oleic acid and palmitic acid.13.Determination of the Structure of a BiomoleculeAn unknown substance, X, was isolated fromrabbit muscle. Its structure was determined from the following observations and experiments. Qualita-tive analysis showed that X was composed entirely of C, H, and O. A weighed sample of X was com-pletely oxidized, and the H2O and CO2produced were measured; this quantitative analysis revealedthat X contained 40.00% C, 6.71% H, and 53.29% O by weight. The molecular mass of X, determinedby mass spectrometry, was 90.00 u (atomic mass units; see Box 1–1). Infrared spectroscopy showedthat X contained one double bond. X dissolved readily in water to give an acidic solution; the solutiondemonstrated optical activity when tested in a polarimeter.(a)Determine the empirical and molecular formula of X.(b)Draw the possible structures of X that fit the molecular formula and contain one double bond.Consideronlylinear or branched structures and disregard cyclic structures. Note that oxygenmakes very poor bonds to itself.(c)What is the structural significance of the observed optical activity? Which structures in(b)areconsistent with the observation?(d)What is the structural significance of the observation that a solution of X was acidic? Whichstructures in(b)are consistent with the observation?(e)What is the structure of X? Is more than one structure consistent with all the data?Chapter 1The Foundations of BiochemistryS-7CH3CH3CH3CH2CH2HCCH2CH3C(CH2)14OOCCC(CH2)7(CH2)7CH3CH2OOPNOOHHOOCH2HOCCCNCNCCNCCNCCOOHHONH2HHHOHHHHHHOCH2CH2CH2CH3SO

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Answer(a)From the C, H, and O analysis, and knowing the mass of X is 90.00 u, we can calculatethe relative atomic proportions by dividing the weight percents by the atomic weights:S-8Chapter 1The Foundations of BiochemistryHHHOHCCCOHHO1HOHHOHCCCOHH5HHHCCCHOH6HOHHHCCCOHHO2HOHHOHCCCOHH3HOOHHOHCCCHH4OOHHHHOCCCHH7OOHHHHOCCCHOH8OHHHHCCCOHOH9OHHHOHOCCCHH10OHHHHHOOCCCHOH1112HHHHOCCCOHOHAtomRelative atomic proportionNo. of atoms relative to OC(90.00 u)(40.00/100)/(12 u)33/31H(90.00 u)(6.71/100)/(1.008 u)66/32O(90.00 u)(53.29/100)/(16.0 u)33/31Thus, the empirical formula is CH2O, with a formula weight of 1221630. Themolecular formula, based on X having a mass of 90.00 u, must be C3H6O3.(b)Twelve possible structures are shown below. Structures1through5can be eliminatedbecause they are unstable enol isomers of the corresponding carbonyl derivatives.Structures9,10, and12can also be eliminated on the basis of their instability: they arehydrated carbonyl derivatives (vicinal diols).(c)Optical activity indicates the presence of a chiral center (a carbon atom surrounded byfour different groups). Only structures6and8have chiral centers.(d)Of structures6and8, only6contains an acidic group: a carboxyl group.(e)Structure6is substance X. This compound exists in two enantiomeric forms that cannotbe distinguished, even by measuring specific rotation. One could determine absolutestereochemistry by x-ray crystallography.14.Naming Stereoisomers with One Chiral Carbon Using the RS SystemPropranolol is a chiralcom-pound. (R)-Propranolol is used as a contraceptive; (S)-propranolol is used to treat hypertension. Identifythe chiral carbon in the structure below. Is this the (R) or the (S) isomer? Draw the other isomer.

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AnswerDetermining the chirality of an asymmetric carbon requires ranking the four sub-stituents in order of decreasing atomic number or decreasing molecular weight with immediateattachments. The asymmetric carbon is propranolol in the question because it is shown with ablack triangular wedge bond to the hydroxyl. This means that the bond is in front of the planeof the page.The four substituents of the alcohol carbon would be ranked (1) the hydroxyl, (2) C—O,(3) C—N, and (4) a hydrogen (not shown), which is behind the plane of the page. With thislowest-priority “group” pointing away, the other groups in decreasing priority (1 to 3) are in thesequence down, left, right—clockwise—so the configuration is (R).15.Naming Stereoisomers with Two Chiral Carbons Using the RS SystemThe (R,R) isomer ofmethylphenidate (Ritalin) is used to treat attention deficit hyperactivity disorder (ADHD). The (S,S)isomer is an antidepressant. Identify the two chiral carbons in the structure below. Is this the (R,R) orthe (S,S) isomer? Draw the other isomer.Chapter 1The Foundations of BiochemistryS-9OOHNHOOHHNOOHNHAnswerThe asymmetric carbons can be identified by the presence of the wedge-shaped bondsindicating the spatial relationship of the bound groups. Wedge bonds always have the narrowend at the chiral carbon and the wide end at the attached atom or group. Solid wedge bondsproject toward the reader; dashed wedge bonds project away from the reader, behind the planeof the paper. In this molecule, one wedge bond is solid black; the benzene ring at the wide endis coming out of the plane of the paper toward the reader. The hydrogen at the wide end of thedashed wedge bond projects behind the paper.The chiral center on the ring has the lowest priority group, the hydrogen atom, projectingaway, so we can evaluate it as it stands. The remaining attached atoms, in priority order, are(1) nitrogen, (2) the carbon with two carbons attached to it, and (3) the ring carbon with onecarbon attached. In decreasing priority (1 to 3) the sequence is left, down, right—counterclock-wise—so the configuration is (S). The second chiral center also has the lowest priority group,the hydrogen atom, projecting away from the reader. The priority order of the other sub-stituents is (1) the carboxyl group (with two oxygens attached), (2) the nitrogen ring, and (3)the benzene ring. In decreasing priority (1 to 3) the sequence is right, left, down—counter-clockwise—so the configuration around the second chiral center is also (S). This is the (S,S)configuration of methylphenidate.To draw the (R,R) configuration, given the (S,S) configuration, make a complete mirrorimage.To maintain the original orientation of the molecule, it is often possible to begin by making thedashed wedge bonds solid and the solid wedge bonds dashed. However, it is essential to verifythe configuration and make adjustments as needed, because this method is not foolproof.

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