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Lehninger Principles of Biochemistry Seventh Edition Solution Manual

Lehninger Principles of Biochemistry Seventh Edition Solution Manual gives you access to textbook solutions that make learning easier and more efficient.

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    The Absolute, Ultimate Guide to Lehninger Principles of Biochemistry Study Guide and Solutions Manual Marcy Osgood University of New Mexico School of Medicine Karen Ocorr Sanford Burnham Prebys Medical Discovery Institute Solutions Manual based on a previous edition by Frederick Wedler Robert Bernlohr Ross Hardison Teh-Hui Kao Ming Tien Pennsylvania State University Seventh Edition

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    Preface vii About the Authors ix 1 The Foundations of Biochemistry 1 Part I Structure and Catalysis 2 Water 11 3 Amino Acids, Peptides, and Proteins 21 4 The Three-Dimensional Structure of Proteins 41 5 Protein Function 54 6 Enzymes 70 7 Carbohydrates and Glycobiology 87 8 Nucleotides and Nucleic Acids 100 9 DNA-Based Information Technologies 113 10 Lipids 123 11 Biological Membranes and Transport 134 12 Biosignaling 151 Part II Bioenergetics and Metabolism 13 Bioenergetics and Biochemical Reaction Types 172 14 Glycolysis, Gluconeogenesis, and the Pentose Phosphate Pathway 183 15 Principles of Metabolic Regulation 195 16 The Citric Acid Cycle 206 17 Fatty Acid Catabolism 218 18 Amino Acid Oxidation and the Production of Urea 228 19 Oxidative Phosphorylation 238 20 Photosynthesis and Carbohydrate Synthesis in Plants 248 21 Lipid Biosynthesis 261 22 Biosynthesis of Amino Acids, Nucleotides, and Related Molecules 274 23 Hormonal Regulation and Integration of Mammalian Metabolism 286 Part III Information Pathways 24 Genes and Chromosomes 299 25 DNA Metabolism 307 26 RNA Metabolism 319 27 Protein Metabolism 331 28 Regulation of Gene Expression 343 v Contents Study Guide

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    1 The Foundations of Biochemistry S-1 Part I Structure and Catalysis 2 Water S-14 3 Amino Acids, Peptides, and Proteins S-29 4 The Three-Dimensional Structure of Proteins S-44 5 Protein Function S-54 6 Enzymes S-63 7 Carbohydrates and Glycobiology S-78 8 Nucleotides and Nucleic Acids S-88 9 DNA-Based Information Technologies S-98 10 Lipids S-112 11 Biological Membranes and Transport S-120 12 Biosignaling S-132 Part II Bioenergetics and Metabolism 13 Bioenergetics and Biochemical Reaction Types S-141 14 Glycolysis, Gluconeogenesis, and the Pentose Phosphate Pathway S-161 15 Principles of Metabolic Regulation S-174 16 The Citric Acid Cycle S-184 17 Fatty Acid Catabolism S-200 18 Amino Acid Oxidation and the Production of Urea S-212 19 Oxidative Phosphorylation S-224 20 Photosynthesis and Carbohydrate Synthesis in Plants S-236 21 Lipid Biosynthesis S-249 22 Biosynthesis of Amino Acids, Nucleotides, and Related Molecules S-260 23 Hormonal Regulation and Integration of Mammalian Metabolism S-268 Part III Information Pathways 24 Genes and Chromosomes S-276 25 DNA Metabolism S-285 26 RNA Metabolism S-297 27 Protein Metabolism S-304 28 Regulation of Gene Expression S-314 vi Contents Solutions Manual

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    vii Preface Learning a complex subject, such as biochemistry, is very much like learning a foreign language. In the study of a foreign language there are several distinct components that must be mastered: the vo- cabulary, the grammatical rules, and the integration of these words and rules so they can be used to commu- nicate ideas. Similarly, in the study of biochemistry there is a very large (some would say vast) number of new terms and concepts, as well as a complex set of “rules” governing biochemical reactions that must all be memorized. All of this information must be inte- grated into an interrelated whole that describes bio- logical systems. Memorizing vocabulary and grammatical rules will not make you fluent in a foreign language; nei- ther will such memorization make you fluent in biochemistry. Similarly, listening to someone speak a foreign lan- guage will not, by itself, make you more capable of producing those same sounds, words, and sentences. The key to mastery of any new subject area, whether it is a foreign language or biochemistry, is the interac- tion of memorization, practice, and application until the information fits together into a coherent whole. In this workbook we attempt to guide you through the material presented in Lehninger Principles of Biochemistry, Seventh Edition by Nelson and Cox. The Step-by-Step Guide to each chapter includes three parts: • A one- to two-page summary of the Major Concepts helps you to see the “big picture” for each chapter. • What to Review helps you make direct connections between the current material and related informa- tion presented elsewhere in the text. • Topics for Discussion for each section and subsec- tion in the chapter focus your attention on the main points being presented and help you internalize the information by using it. • Discussion Questions for Study Groups are questions that are especially suited to Study Groups, either because they pull together several points made in the chapter or because they are more in- volved questions that would benefit from collabora- tive insight. Each chapter also includes a Self-Test for you to assess your progress in mastering biochemical ter- minology and facts, and learning to integrate and apply that information. • Do You Know the Terms? asks you to complete a crossword puzzle using the new vocabulary intro- duced in the chapter. • Do You Know the Facts? tests how well you have learned the “rules” of biochemistry. • Applying What You Know tests how well you “speak the language” of biochemistry, often in an experimental or metabolically relevant context. Two especially popular features of the Absolute, Ulti- mate Guide are: • The Biochemistry Online problems will expose you to just a few of the analytical resources that are available to scientists on the Internet. The molecu- lar models are fun to play with, and many of the questions provide you with an opportunity to analyze “data” as you might in an actual laboratory setting. • The Cell Map is based on many semesters of use by our students. It is designed to help you place the biochemical pathways that you are learning about into their proper cellular perspective. The Cell Map questions tell you what to include in your Map, but you can be creative! Students have told us it is a great study aid, that really helped them to make the connections between the 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 people who get together on a regular basis (weekly) to “speak biochemistry.” This type of interaction is critical to flu- ency in a foreign language and is no less critical in the successful assimilation of biochemistry. We designed our Step-by-Step Guide to each chapter with study groups in mind. The questions posed for each section can be used as springboards for study group discus- sions. We purposefully have not supplied answers to this 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 of the text, the questions will focus your attention on the more important aspects of the material. Detailed Solutions to all the end-of-chapter textbook problems are included as a separate section in the Absolute, Ultimate Guide. We have taken great care to ensure that the solutions are correct, complete, and informative. The final answer to numerical problems has been rounded off to reflect the number of significant figures in the data. We thank each and every one of our students for their invaluable feedback and input, which have helped to make this study guide its absolute and ultimate best. viii Preface

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    About the Authors ix Karen Ocorr received her Ph.D. from Wesleyan University, where she studied the physiology and neurochemistry of the lobster cardiac ganglion. As an NIH postdoctoral research fellow at the University of Texas, she examined the roles of enzymes and second messengers in neuronal plasticity in Aplysia californica. She continued these investigations at the California Institute of Technology and Stanford University, examining the role of intracellular signaling underlying long-term potentiation in the vertebrate hippocampus. She taught Introductory Biochemistry for 10 years at the University of Michigan, Ann Arbor, where she also taught Animal Physiology, Cell Biology, and Introductory Biology for Non-Majors. She also taught Introductory Biochemistry for four years as a visiting Professor at the Hunan Normal University in Changsha, China. Currently she is an Assistant Professor at SBP Medical Discovery Institute in La Jolla, California, where she is researching the roles of ion channels in cardiomyopathies. She currently teaches graduate and medical students at both the SBP Medical Discovery Institute and the University of California–San Diego. Marcy Osgood received her Ph.D. from Rensselaer Polytechnic Institute. After two postdoctoral positions, she became an Assistant Professor in the Department of Physical Sciences at Albany College of Pharmacy, Union University. She was then a Lecturer in the Biology Department at the University of Michigan, Ann Arbor for nine years, where her research interests began to focus on educational issues. Her current position is as Associate Professor in the Department of Biochemistry and Molecular Biology and Assistant Dean of Undergraduate Medical Education at the University of New Mexico School of Medicine. Her current research efforts deal with assessment of problem-solving strategies. Osgood and Ocorr have collaborated for 10 years to develop effective techniques for teaching biochemistry, all of which are based on educational research. This study guide is the result of these efforts and embodies much of what they have found to be effective over years of instruction at many levels and in many areas of the biological sciences. It includes thousands of discussion, quiz, and exam questions from their biochemistry courses. Osgood and Ocorr have been responsible for teaching biochemistry to more than 10,000 undergraduate students.

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    The Foundations of Biochemistry chapter 1 S-1 1. 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 electron microscope), how big would it appear? Assume you are viewing a “typical” eukaryotic cell with a cellular diameter of 50 m m. (b) If this cell were a muscle cell (myocyte), how many molecules of actin could it hold? Assume the 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 4/3 p r 3 .) (c) If this were a liver cell (hepatocyte) of the same dimensions, how many mitochondria could it hold? Assume the cell is spherical; no other cellular components are present; and the mitochondria are spherical, with a diameter of 1.5 m m. (d) Glucose is the major energy-yielding nutrient for most cells. Assuming a cellular concentration of 1 m M (i.e., 1 millimole/L), calculate how many molecules of glucose would be present in our hypothetical (and spherical) eukaryotic cell. (Avogadro’s number, the number of molecules in 1 mol of a nonionized substance, is 6.02  10 23 .) (e) Hexokinase is an important enzyme in the metabolism of glucose. If the concentration of hexokinase in our eukaryotic cell is 20 m M , how many glucose molecules are present per hexokinase molecule? Answer (a) The magnified cell would have a diameter of 50  10 4 m m  500  10 3 m m  500 mm, or 20 inches—about the diameter of a large pizza. (b) The radius of a globular actin molecule is 3.6 nm/2  1.8 nm; the volume of the molecule, in cubic meters, is (4/3)(3.14)(1.8  10  9 m) 3  2.4  10  26 m 3 . * The number of actin molecules that could fit inside the cell is found by dividing the cell volume (radius  25 m m) by the actin molecule volume. Cell volume  (4/3)(3.14)(25  10  6 m) 3  6.5  10  14 m 3 . Thus, the number of actin molecules in the hypothetical muscle cell is (6.5  10  14 m 3 )/(2.4  10  26 m 3 )  2.7  10 12 molecules or 2.7 trillion actin molecules. * Significant figures: In multiplication and division, the answer can be expressed with no more significant figures than the least precise value in the calculation. Because some of the data in these problems are derived from measured values, we must round off the calculated answer to reflect this. In this first example, the radius of the actin (1.8 nm) has two significant figures, so the answer (volume of actin  2.4  10  26 m 3 ) can be expressed with no more than two significant figures. It will be standard practice in these expanded answers to round off answers to the proper number of significant figures.

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    S-2 Chapter 1 The Foundations of Biochemistry (c) The radius of the spherical mitochondrion is 1.5 m m/2  0.75 m m, therefore the volume is (4/3)(3.14)(0.75  10  6 m) 3  1.8  10  18 m 3 . The number of mitochondria in the hypothetical liver cell is (6.5  10  14 m 3 )/(1.8  10  18 m 3 )  36  10 3 mitochondria (d) The volume of the eukaryotic cell is 6.5  10  14 m 3 , which is 6.5  10  8 cm 3 or 6.5  10  8 mL. One liter of a 1 m M solution of glucose has (0.001 mol/1000 mL)(6.02  10 23 molecules/mol)  6.02  10 17 molecules/mL. The number of glucose molecules in the cell is the product of the cell volume and glucose concentration: (6.5  10  8 mL)(6.02  10 17 molecules/mL)  3.9  10 10 molecules or 39 billion glucose molecules. (e) The concentration ratio of glucose/hexokinase is 0.001 M /0.00002 M , or 50/1, meaning that each enzyme molecule would have about 50 molecules of glucose available as substrate. 2. Components of E. coli E. coli cells are rod-shaped, about 2 m m long and 0.8 m m in diameter. The volume of a cylinder is p r 2 h, where h is the height of the cylinder. (a) If the average density of E. coli (mostly water) is 1.1  10 3 g/L, what is the mass of a single cell? (b) E. coli has a protective cell envelope 10 nm thick. What percentage of the total volume of the bacterium does the cell envelope occupy? (c) E. coli is capable of growing and multiplying rapidly because it contains some 15,000 spherical ribosomes (diameter 18 nm), which carry out protein synthesis. What percentage of the cell volume do the ribosomes occupy? Answer (a) The volume of a single E. coli cell can be calculated from p r 2 h (radius  0.4 m m): 3.14(4  10  5 cm) 2 (2  10  4 cm)  1.0  10  12 cm 3  1  10  15 m 3  1  10  15 L Density (g/L) multiplied by volume (L) gives the mass of a single cell: (1.1  10 3 g/L)(1  10  15 L)  1  10  12 g or a mass of 1 pg. (b) First, calculate the proportion of cell volume that does not include the cell envelope, that is, the cell volume without the envelope—with r  0.4 m m  0.01 m m; and h  2 m m  2(0.01 m m)—divided by the total volume. Volume without envelope  p (0.39 m m) 2 (1.98 m m) Volume with envelope  p (0.4 m m) 2 (2 m m) So the percentage of cell that does not include the envelope is  90% (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 for 10% 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 (9  10  3 m m) 3 The volume of the cell  p (0.4 m m) 2 (2 m m) So the percentage of cell volume occupied by the ribosomes is  5% 15,000  (4/3) p (9  10  3 m m) 3  100  p (0.4 m m) 2 (2 m m) p (0.39 m m) 2 (1.98 m m)  100  p (0.4 m m) 2 (2 m m)

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    3. Genetic Information in E. Coli DNA The genetic information contained in DNA consists of a linear 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 a single amino acid unit in a protein. The molecular weight of an E. coli DNA molecule is about 3.1  10 9 g/mol. The average molecular weight of a nucleotide pair is 660 g/mol, and each nucleotide pair contributes 0.34 nm to the length of DNA. (a) Calculate the length of an E. coli DNA molecule. Compare the length of the DNA molecule with the cell dimensions (see Problem 2). How does the DNA molecule fit into the cell? (b) Assume that the average protein in E. coli consists of a chain of 400 amino acids. What is the maximum number of proteins that can be coded by an E. coli DNA 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: (3.1  10 9 g/mol)/(0.66  10 3 g/mol)  4.7  10 6 pairs Multiplying the number of pairs by the length per pair gives (4.7  10 6 pairs)(0.34 nm/pair)  1.6  10 6 nm  1.6 mm The length of the cell is 2 m m (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 coiled to fit into the cell. (b) Because the DNA molecule has 4.7  10 6 nucleotide pairs, as calculated in (a), it must have one-third this number of triplet codons: (4.7  10 6 )/3  1.6  10 6 codons If each protein has an average of 400 amino acids, each requiring one codon, the number of proteins that can be coded by E. coli DNA is (1.6  10 6 codons)(1 amino acid/codon)/(400 amino acids/protein)  4,000 proteins 4. The High Rate of Bacterial Metabolism Bacterial cells have a much higher rate of metabolism than 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 bacterial metabolism 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 bacterium Neisseria gonorrhoeae (diameter 0.5 m m), responsible for the disease gonorrhea. Compare it with the surface-to-volume ratio for a globular amoeba, a large eukaryotic cell (diameter 150 m m). The surface area of a sphere is 4 p r 2 . Answer (a) Metabolic rate is limited by diffusion of fuels into the cell and waste products out of the cell. This diffusion in turn is limited by the surface area of the cell. As the ratio of surface area to volume decreases, the rate of diffusion cannot keep up with the rate of metabolism within the cell. (b) For a sphere, surface area  4 p r 2 and volume  4/3 p r 3 . The ratio of the two is the surface-to-volume ratio, S / V, which is 3/ r or 6/ D , where D  diameter. Thus, rather than calculating S and V separately for each cell, we can rapidly calculate and compare S / V ratios for cells of different diameters. S / V for N. gonorrhoeae  6/(0.5 m m)  12 m m  1 S / V for amoeba  6/(150 m m)  0.04 m m  1   300 Thus, the surface-to-volume ratio is 300 times greater for the bacterium. 12 m m  1  0.04 m m  1 S / V for bacterium  S / V for amoeba Chapter 1 The Foundations of Biochemistry S-3

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    5. Fast Axonal Transport Neurons have long thin processes called axons, structures specialized for conducting signals throughout the organism’s nervous system. Some axonal processes can be as long as 2 m—for example, the axons that originate in your spinal cord and terminate in the muscles of your toes. 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 a vesicle is 1 m m/s, how long does it take a vesicle to move from a cell body in the spinal cord to the axonal tip in the toes? Answer Transport time equals distance traveled/velocity, or (2  10 6 m m)/(1 m m/s)  2  10 6 s or about 23 days! 6. Is Synthetic Vitamin C as Good as the Natural Vitamin? A claim put forth by some purveyors of health foods is that vitamins obtained from natural sources are more healthful than those obtained by chemical synthesis. For example, pure L -ascorbic acid (vitamin C) extracted from rose hips is better than pure L -ascorbic acid manufactured in a chemical plant. Are the vitamins from the two sources dif- ferent? Can the body distinguish a vitamin’s source? Answer The properties of the vitamin—like any other compound—are determined by its chemical structure. Because vitamin molecules from the two sources are structurally identical, their properties are identical, and no organism can distinguish between them. If different vitamin preparations contain different impurities, the biological effects of the mixtures may vary with the source. The ascorbic acid in such preparations, however, is identical. 7. Identification of Functional Groups Figures 1–16 and 1–17 show some common functional groups of 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 compounds below, circle and identify by name each functional group. S-4 Chapter 1 The Foundations of Biochemistry H H Ethanolamine (a) C H H C OH H 3 N Glycerol (b) H H C OH H C OH H C OH H Threonine, an amino acid (d) H CH 3 C OH C H Pantothenate, a vitamin (e) H 3 C CH 2 OH C CH 3 H C OH C O NH Phosphoenolpyruvate, an intermediate in glucose metabolism (c) C C O HO P O H H CH 2 CH 2 C O D -Glucosamine (f ) H CH 2 OH C OH H C OH HO C H H C NH 3 C O H H 3 N

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    Answer (a) O NH 3   amino; O OH  hydroxyl (b) O OH  hydroxyl (three) (c) O P(OH)O 2   phosphoryl (in its ionized form); O COO   carboxyl (d) O COO   carboxyl; O NH 3   amino; O OH  hydroxyl; O CH 3  methyl (two) (e) O COO   carboxyl; O CO O NH O  amide; O OH  hydroxyl (two); O CH 3  methyl (two) (f) O CHO  aldehyde; O NH 3   amino; O OH  hydroxyl (four) 8. Drug Activity and Stereochemistry The quantitative differences in biological activity between the two enantiomers of a compound are sometimes quite large. For example, the D isomer of the drug iso- proterenol, used to treat mild asthma, is 50 to 80 times more effective as a bronchodilator than the L isomer. Identify the chiral center in isoproterenol. Why do the two enantiomers have such radically different bioactivity? Answer A chiral center, or chiral carbon, is a carbon atom that is bonded to four different groups. A molecule with a single chiral center has two enantiomers, designated D and L (or in the RS system, S and R ). In isoproterenol, only one carbon (asterisk) has four different groups around it; this is the chiral center: The bioactivity of a drug is the result of interaction with a biological “receptor,” a protein molecule with a binding site that is also chiral and stereospecific. The interaction of the D isomer of a drug with a chiral receptor site will differ from the interaction of the L isomer with that site. 9. Separating Biomolecules In 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 in the sample—that is, to purify it. 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 the characteristics 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 amino groups. 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 hydrocarbon chains and therefore are less soluble in water than amino acids. And finally, the sizes and shapes 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-carbon sugar, and phosphate. Glucose is a six-carbon sugar; it is smaller than a nucleotide. The size difference 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 1 The Foundations of Biochemistry S-5 OH H Isoproterenol C HO CH 2 H N CH 3 C CH 3 H HO HO HO C * H OH C CH 2 CH 3 CH 3 N H H

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    10. Silicon-Based Life? Silicon is in the same group of the periodic table as carbon and, like carbon, can form 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 it less well adapted than carbon as the central 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 bonding properties. Answer It is improbable that silicon could serve as the central organizing element for life under such conditions as those found on Earth for several reasons. Long chains of silicon atoms are not readily 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 bonds between silicon and oxygen are extremely stable and difficult to break, which would prevent the breaking and making (degradation and synthesis) of biomolecules that is essential to the processes of living organisms. 11. Drug Action and Shape of Molecules Some years ago two drug companies marketed a drug under the 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 and Benzedrine 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. Answer Only one of the two enantiomers of the drug molecule (which has a chiral center) is physiologically active, for reasons described in the answer to Problem 3 (interaction with a stereospecific receptor site). Dexedrine, as manufactured, consists of the single enantiomer ( D -amphetamine) recognized by the receptor site. Benzedrine was a racemic mixture (equal amounts of D and L isomers), so a much larger dose was required to obtain the same effect. 12. Components of Complex Biomolecules Figure 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-6 Chapter 1 The Foundations of Biochemistry C CH 2 CH 3 NH 2 H N C  O  O O O P O P O N N NH NH 2 O P CH 2  O O  O O H H H H OH OH O

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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 an a - D -ribose (at the 5  position), 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 is esterified to two fatty acids, oleic acid and palmitic acid. 13. Determination of the Structure of a Biomolecule An unknown substance, X, was isolated from rabbit 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 H 2 O and CO 2 produced were measured; this quantitative analysis revealed that X contained 40.00% C, 6.71% H, and 53.29% O by weight. The molecular mass of X, determined by mass spectrometry, was 90.00 u (atomic mass units; see Box 1–1). Infrared spectroscopy showed that X contained one double bond. X dissolved readily in water to give an acidic solution; the solution demonstrated 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. Consider only linear or branched structures and disregard cyclic structures. Note that oxygen makes very poor bonds to itself. (c) What is the structural significance of the observed optical activity? Which structures in (b) are consistent with the observation? (d) What is the structural significance of the observation that a solution of X was acidic? Which structures 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 1 The Foundations of Biochemistry S-7 CH 3 CH 3 CH 3 CH 2 CH 2 HC CH 2 CH 3 C (CH 2 ) 14 O O C C C (CH 2 ) 7 (CH 2 ) 7 CH 3 CH 2 O O P  N O  O H H O O CH 2 HO C C C N C N C C N C C N C COO  H H O NH 2 H H H O H H H H H H O CH 2 CH 2 CH 2 CH 3 S O

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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 calculate the relative atomic proportions by dividing the weight percents by the atomic weights: S-8 Chapter 1 The Foundations of Biochemistry H H H OH C C C OH HO 1 HO H H OH C C C OH H 5 H H H C C C H OH 6 H OH H H C C C OH HO 2 H OH H OH C C C OH H 3 HO OH H OH C C C H H 4 O OH H H HO C C C H H 7 O OH H H HO C C C H OH 8 O H H H H C C C OH OH 9 O H H HO HO C C C H H 10 O H H H H HO O C C C H OH 11 12 H H H H O C C C OH OH Atom Relative atomic proportion No. of atoms relative to O C (90.00 u)(40.00/100)/(12 u)  3 3/3  1 H (90.00 u)(6.71/100)/(1.008 u)  6 6/3  2 O (90.00 u)(53.29/100)/(16.0 u)  3 3/3  1 Thus, the empirical formula is CH 2 O, with a formula weight of 12  2  16  30. The molecular formula, based on X having a mass of 90.00 u, must be C 3 H 6 O 3 . (b) Twelve possible structures are shown below. Structures 1 through 5 can be eliminated because they are unstable enol isomers of the corresponding carbonyl derivatives. Structures 9 , 10 , and 12 can also be eliminated on the basis of their instability: they are hydrated carbonyl derivatives (vicinal diols). (c) Optical activity indicates the presence of a chiral center (a carbon atom surrounded by four different groups). Only structures 6 and 8 have chiral centers. (d) Of structures 6 and 8 , only 6 contains an acidic group: a carboxyl group. (e) Structure 6 is substance X. This compound exists in two enantiomeric forms that cannot be distinguished, even by measuring specific rotation. One could determine absolute stereochemistry by x-ray crystallography. 14. Naming Stereoisomers with One Chiral Carbon Using the RS System Propranolol is a chiral com- pound. ( R )-Propranolol is used as a contraceptive; ( S )-propranolol is used to treat hypertension. Identify the chiral carbon in the structure below. Is this the ( R ) or the ( S ) isomer? Draw the other isomer.

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    Answer Determining the chirality of an asymmetric carbon requires ranking the four sub- stituents in order of decreasing atomic number or decreasing molecular weight with immediate attachments. The asymmetric carbon is propranolol in the question because it is shown with a black triangular wedge bond to the hydroxyl. This means that the bond is in front of the plane of 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 this lowest-priority “group” pointing away, the other groups in decreasing priority (1 to 3) are in the sequence down, left, right—clockwise—so the configuration is ( R ). 15. Naming Stereoisomers with Two Chiral Carbons Using the RS System The ( R , R ) isomer of methylphenidate (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 ) or the ( S,S ) isomer? Draw the other isomer. Chapter 1 The Foundations of Biochemistry S-9 O OH N H O O H HN O O H NH Answer The asymmetric carbons can be identified by the presence of the wedge-shaped bonds indicating the spatial relationship of the bound groups. Wedge bonds always have the narrow end at the chiral carbon and the wide end at the attached atom or group. Solid wedge bonds project toward the reader; dashed wedge bonds project away from the reader, behind the plane of the paper. In this molecule, one wedge bond is solid black; the benzene ring at the wide end is coming out of the plane of the paper toward the reader. The hydrogen at the wide end of the dashed wedge bond projects behind the paper. The chiral center on the ring has the lowest priority group, the hydrogen atom, projecting away, 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 one carbon 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 mirror image. To maintain the original orientation of the molecule, it is often possible to begin by making the dashed wedge bonds solid and the solid wedge bonds dashed. However, it is essential to verify the configuration and make adjustments as needed, because this method is not foolproof.

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