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Principles of Biochemistry
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Principles of Biochemistry FOURTH EDITION PHOTO TO COME
H. Robert Horton North Carolina State University
Laurence A. Moran University of Toronto
K. Gray Scrimgeour University of Toronto
Marc D. Perry University of Toronto
J. David Rawn Towson State University
Upper Saddle River, NJ 07458
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Library of Congress Cataloging-in-Publication Data Principles of biochemistry / H. Robert Horton . . . [et al.].—4th ed. p. cm. Includes bibliographical references and index. ISBN 0-13-145306-8 I. Biochemistry. I. Horton, H. Robert. QP514.2.P745 2006 612'.015—dc22 550-dc22 2005007745
Executive Editor: Gary Carlson Executive Managing Editor: Kathleen Schiaparelli Marketing Manager: Andrew Gilfillan Media Editor: Patrick Shriner Assistant Managing Editor, Science Media: Nicole Bush Assistant Managing Editor, Science Supplements: Becca Richter Development Editor: John Murdzek Project Editor: Crissy Dudonis Director of Creative Services: Paul Belfanti Creative Director: Carole Anson Art Director: Kenny Beck Cover and Interior Design: Koala Bear Design Cover Illustrator: Jonathan C. Parrish Manufacturing Manager: Alexis Heydt-Long Manufacturing Buyer: Alan Fischer Photo Researcher: Diane Austin Director, Image Resource Center: Melinda Reo Manager, Rights and Permissions: Zina Arabia Interior Image Specialist: Beth Boyd-Brenzel Cover Image Specialist: Karen Sanatar Image Permission Coordinator: Robert Farrell Managing Editor, Audio Visual Assets and Production: Patricia Burns Art Editors: Jay McElroy, Connie Long Art Studio: Artworks/Jonathan Parrish Editorial Assistants: Nancy Bauer, Jennifer Hart Production Supervision/Composition: Marty Sopher/CMyK Associates About the cover: Complex III (ubiquinol:cytochrome c oxidoreductase). This membrane-bound complex plays a key role in membrane-associated electron transport and the generation of the proton gradient that eventually gives rise to new ATP molecules. Complex III catalyzes the Q-cycle reactions—one of the most important pathways in biochemistry. (See page 427.) © 2006, 2002, 1996, 1993 by Pearson Education, Inc. Pearson Prentice Hall Pearson Education, Inc. Upper Saddle River, New Jersey 07458 Pearson Prentice Hall™ is a trademark of Pearson Education, Inc. All rights reserved. No part of this book may be reproduced, in any form or by any means, without permission in writing from the publisher. Printed in the United States of America 10 9 8 7 6 5 4 3 2 1 ISBN 0-13-145306-8 Pearson Education Ltd., London Pearson Education Australia, PTY. Limited, Sydney Pearson Education Singapore, Pte. Ltd Pearson Education North Asia Ltd, Hong Kong
Pearson Education Canada, Ltd., Toronto Pearson Educacion de Mexico, S.A. de C.V. Pearson Education—Japan, Tokyo Pearson Education Malaysia, Pte. Ltd
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Science should be as simple as possible, but not simpler. —Albert Einstein
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PHOTO TO COME
The Authors H. Robert Horton Dr. Horton, who received his Ph.D from the University of Missouri in 1962, is William Neal Reynolds Professor Emeritus and Alumni Distinguished Professor Emeritus in the Department of Biochemistry at North Carolina State University, where he served on the faculty for over 30 years. Most of Professor Horton’s research was in protein and enzyme mechanisms. Laurence A. Moran After earning his Ph.D from Princeton University in 1974, Professor Moran spent four years at the Université dè Geneve in Switzerland. He has been a member of the Department of Biochemistry at the University of Toronto since 1978, specializing in molecular biology and molecular evolution. His research findings on heat-shock genes have been published in many scholarly journals. K. Gray Scrimgeour Professor Scrimgeour received his doctorate from the University of Washington in 1961 and has been a faculty member at the University of Toronto since 1967. He is the author of The Chemistry and Control of Enzymatic Reactions (1977, Academic Press), and his work on enzymatic systems has been published in more than 50 professional journal articles during the past 40 years. From 1984–1992, he was editor of the journal Biochemistry and Cell Biology.
Marc D. Perry After earning his Ph.D. from the University of Toronto in 1988, Dr. Perry trained at the University of Colorado, where he studied sex determination in the nematode C. elegans. In 1994 he returned to the University of Toronto as a faculty member in the department of Molecular and Medical Genetics. His research has focused on developmental genetics, meiosis and bioinformatics. In 2004 he joined the Heart & Stroke / Richard Lewar Centre of Excellence in Cardiovascular Research in the University of Toronto’s Faculty of Medicine. J. David Rawn Professor Rawn, who received his Ph.D from Ohio State University in 1971, has taught and done research in the Department of Chemistry at Towson State University for the past 25 years. He did not write chapters for Principles of Biochemistry, but his textbook Biochemistry (1989, Neil Patterson) served as a source of information and ideas concerning content and organization.
New problems and solutions for the fourth edition were created by Drs. Laurence A. Moran, University of Toronto and Elizabeth S. Roberts-Kirchhoff, University of Detroit Mercy. The remaining problems were created by Drs. Robert N. Lindquist, San Francisco State University, Marc Perry and Diane M. De Abreu of the University of Toronto.
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Student Supplements
THE BIOCHEMISTRY STUDENT COMPANION by Allen J. Scism Central Missouri State University
No student should be without this helpful resource. Contents include the following: • carefully constructed drill problems for each chapter, including short-answer, multiplechoice, and challenge problems • comprehensive, step-by-step solutions and explanations for all problems • a remedial chapter that reviews the general and organic chemistry that students require for biochemistry—topics are ingeniously presented in the context of a metabolic pathway • tables of essential data
Please order through your college bookstore or call Prentice Hall at 1-800-947-7700. The Biochemistry Student Companion ISBN 0-13-147605-X
COMPANION WEBSITE An online student tool that includes 3-D modules to help visualize biochemistry and MediaLabs to investigate important issues related to its particular chapter. Please visit the site at http://www. prenhall.com/horton.
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Brief Contents PART ONE Introduction 1 Introduction to Biochemistry 2 Water PART TWO Structure and Function 3 Amino Acids and the Primary Structures of Proteins 4 Proteins: Three-Dimensional Structure and Function 5 Properties of Enzymes 6 Mechanisms of Enzymes 7 Coenzymes and Vitamins 8 Carbohydrates 9 Lipids and Membranes PART THREE Metabolism and Bioenergetics 10 Introduction to Metabolism 11 Glycolysis 12 Gluconeogenesis, The Pentose Phosphate Pathway, and Glycogen Metabolism 13 The Citric Acid Cycle 14 Electron Transport and ATP Synthesis 15 Photosynthesis 16 Lipid Metabolism 17 Amino Acid Metabolism 18 Nucleotide Metabolism PART FOUR Biological Information Flow 19 Nucleic Acids 20 DNA Replication, Repair, and Recombination 21 Transcription and RNA Processing 22 Protein Synthesis 23 Recombinant DNA Technology
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Contents Preface
xxv
PART ONE Introduction
1
Introduction to Biochemistry
1.1
Biochemistry Is a Modern Science
1.2
The Chemical Elements of Life
1.3
Many Important Macromolecules Are Polymers A. Proteins
2 3 5
6
B. Polysaccharides C. Nucleic Acids
7 9
D. Lipids and Membranes 1.4
1
The Energetics of Life
10
11
A. Reaction Rates and Equilibria B. Thermodynamics
12
13
C. Equilibrium Constants and Standard Gibbs Free Energy Changes 1.5
Biochemistry and Evolution
1.6
The Cell Is the Basic Unit of Life
1.7
Prokaryotic Cells: Structural Features
17
1.8
Eukaryotic Cells: Structural Features
18
A. The Nucleus
15 16
18
B. The Endoplasmic Reticulum and Golgi Apparatus C. Mitochondria and Chloroplasts D. Specialized Vesicles E. The Cytoskeleton
20
21 22
1.9
A Picture of the Living Cell
1.10
Biochemistry Is Multidisciplinary
22 24
Appendix: The Special Terminology of Biochemistry Selected Readings
19
25
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2
Water
2.1
The Water Molecule Is Polar
27
2.2
Hydrogen Bonding in Water
28
2.3
Water Is an Excellent Solvent
26
30
A. Ionic and Polar Substances Dissolve in Water B. Cellular Concentrations and Diffusion C. Osmotic Pressure
31
31
2.4
Nonpolar Substances Are Insoluble in Water
2.5
Noncovalent Interactions B. Hydrogen Bonds
33
34
C. Van der Waals Forces
35
D. Hydrophobic Interactions 2.6
Water Is Nucleophilic
2.7
Ionization of Water
2.8
The pH Scale
36
36 37
39
Box 2.1 The little “p” in pH.
40
2.9
Acid Dissociation Constants of Weak Acids
2.10
Buffered Solutions Resist Changes in pH 49
Problems
49
32
33
A. Charge–charge Interactions
Summary
30
Selected Readings
41 46
51
PART TWO Structure and Function
3
Amino Acids and the Primary Structures of Proteins
3.1
General Structure of Amino Acids
3.2
Structures of the 20 Common Amino Acids
53
Box 3.1 An Alternative Nomenclature A. Aliphatic R Groups
56 57
58
C. Sulfur-Containing R Groups
58
D. Side Chains with Alcohol Groups E. Basic R Groups
55
57
Box 3.2 Common Names of Amino Acids B. Aromatic R Groups
59
59
F. Acidic R Groups and Their Amide Derivatives
60
G. The Hydrophobicity of Amino Acid Side Chains 3.3
52
Other Amino Acids and Amino Acid Derivatives
60
61
3.4
Ionization of Amino Acids
3.5
Peptide Bonds Link Amino Acids in Proteins
62
3.6
Protein Purification Techniques
3.7
Analytical Techniques
3.8
Amino Acid Composition of Proteins
3.9
Determining the Sequence of Amino Acid Residues
3.10
Protein Sequencing Strategies
3.11
Comparisons of the Primary Structures of Proteins Reveal Evolutionary Relationships 78
66
67
69 72 73
75
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Summary
81
Problems
81
Selected Readings
83
4
Proteins: Three-Dimensional Structure and Function 84
4.1
There Are Four Levels of Protein Structure
86
4.2
Methods for Determining Protein Structure
87
4.3
The Conformation of the Peptide Group
4.4
The a Helix
4.5
b Strands and b Sheets
95
4.6
Loops and Turns
4.7
Tertiary Structure of Proteins
97 98
A. Supersecondary Structures B. Domains
99
100
C. Domain Structure and Function 4.8
90
92
Quaternary Structure
104
104
4.9
Protein Denaturation and Renaturation
4.10
Protein Folding and Stability
110
A. The Hydrophobic Effect
110
B. Hydrogen Bonding
107
111
C. Van der Waals Interactions and Charge–Charge Interactions D. Protein Folding Is Assisted by Molecular Chaperones 4.11
Collagen, a Fibrous Protein
4.12
Structures of Myoglobin and Hemoglobin
4.13
Oxygen Binding to Myoglobin and Hemoglobin
115
A. Oxygen Binds Reversibly to Heme
116 118
118
B. Oxygen-Binding Curves of Myoglobin and Hemoglobin C. Hemoglobin Is an Allosteric Protein 4.14
Antibodies Bind Specific Antigens Summary
125
Problems
125
Selected Readings
123
127
Properties of Enzymes
5.1
The Six Classes of Enzymes
5.2
Kinetic Experiments Reveal Enzyme Properties A. Chemical Kinetics
5.3
119
121
5
B. Enzyme Kinetics
112
112
129
130 132
133 134
The Michaelis–Menten Equation
135
A. Derivation of the Michaelis–Menten Equation B. The Catalytic Constant kcat C. The Meanings of Km
137
138
138
5.4
Kinetic Constants Indicate Enzyme Activity and Catalytic Proficiency
5.5
Measurement of Km and Vmax
5.6
Kinetics of Multisubstrate Reactions
140 141
Box 5.1 Hyperbolas versus Straight Lines
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5.7
Reversible Enzyme Inhibition A. Competitive Inhibition
142
143
B. Uncompetitive Inhibition
145
C. Noncompetitive Inhibition
146
D. Uses of Enzyme Inhibition
146
5.8
Irreversible Enzyme Inhibition
147
5.9
Allosteric Enzymes
5.10
Regulation of Enzyme Activity
148 148
A. Phosphofructokinase Is an Allosteric Enzyme B. General Properties of Allosteric Enzymes C. Two Theories of Allosteric Regulation
152
D. Regulation by Covalent Modification 5.11
149
150
153
Multienzyme Complexes and Multifunctional Enzymes Summary
154
Problems
155
Selected Readings
157
6
Mechanisms of Enzymes
6.1
The Terminology of Mechanistic Chemistry A. Nucleophilic Substitutions B. Cleavage Reactions
158 158
159
160
C. Oxidation—Reduction Reactions
160
6.2
Catalysts Stabilize Transition States
6.3
Chemical Modes of Enzymatic Catalysis
160 162
Box 6.1 Site-Directed Mutagenesis Modifies Enzymes A. Polar Amino Acid Residues in Active Sites B. Acid–Base Catalysis C. Covalent Catalysis
164 166
Diffusion-Controlled Reactions
167
A. Triose Phosphate Isomerase
167
B. Superoxide Dismutase 6.5
170
Binding Modes of Enzymatic Catalysis A. The Proximity Effect
171
172
B. Weak Binding of Substrates to Enzymes C. Induced Fit
163
163
165
D. pH Affects Enzymatic Rates 6.4
154
172
174
D. Transition-State Stabilization 6.6
Lysozyme
6.7
Properties of Serine Proteases
175
178
Box 6.2 Proposed Transition State for a Bimolecular Reaction
181
182
A. Zymogens Are Inactive Enzyme Precursors B. Substrate Specificity of Serine Proteases
182 183
C. Serine Proteases Use Both the Chemical and the Binding Modes of Catalysis 184 Summary
188
Problems
188
Selected Readings
191
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Contents
7
Coenzymes and Vitamins
7.1
Many Enzymes Require Inorganic Cations
7.2
Coenzyme Classification
7.3
ATP and Other Nucleotide Cosubstrates
7.4
NADP and NAD
192 193
193
Box 7.1 Vitamin C: A Vitamin but Not a Coenzyme 197
Box 7.2 NAD Binding to Dehydrogenases 7.5
FAD and FMN
7.6
Coenzyme A
7.7
Thiamine Pyrophosphate
7.8
Pyridoxal Phosphate
199
200 201 202
203
7.9
Biotin
7.10
Tetrahydrofolate
7.11
Cobalamin
210
7.12
Lipoamide
211
7.13
Lipid Vitamins
212
A. Vitamin A
213
207 208
B. Vitamin D
213
C. Vitamin E
213
D. Vitamin K
214
7.14
Ubiquinone
7.15
Protein Coenzymes
7.16
Cytochromes
214 215
216
Summary
218
Problems
219
Selected Readings
221
8
Carbohydrates
8.1
Most Monosaccharides Are Chiral Compounds
8.2
Cyclization of Aldoses and Ketoses
226
8.3
Conformations of Monosaccharides
229
8.4
Derivatives of Monosaccharides
222
A. Sugar Phosphates
231
C. Amino Sugars
231
E. Sugar Acids F. Ascorbic Acid
8.7
231
232
233 234
Disaccharides and Other Glycosides A. Structures of Disaccharides
8.6
234
234
B. Reducing and Nonreducing Sugars
236
C. Nucleosides and Other Glycosides
236
Polysaccharides
237
A. Starch and Glycogen
237
B. Cellulose and Chitin
239
Glycoconjugates A. Proteoglycans
223
231
B. Deoxy Sugars D. Sugar Alcohols
8.5
195
196
241 241
Box 8.1 Nodulation Factors Are Lipo-oligosaccharides
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B. Peptidoglycans
243
C. Glycoproteins
244
Box 8.2 ABO Blood Group Summary
249
Problems
250
Selected Readings
248
252
9
Lipids and Membranes
9.1
Structural and Functional Diversity of Lipids
9.2
Fatty Acids
Inner leaflet
253 253
254
Box 9.1 Common Names of Fatty Acids.
255
Box 9.2 Trans Fatty Acids and Margarine
256
9.3
Triacylglycerols
258
9.4
Glycerophospholipids
9.5
Sphingolipids
9.6
Steroids
9.7
Other Biologically Important Lipids
9.8
Biological Membranes Are Composed of Lipid Bilayers and Proteins
259
262
Outer leaflet
264 265
Box 9.3 Special Nonaqueous Techniques Must Be Used to Study Lipids A. Lipid Bilayers
270
9.9
Lipid Bilayers and Membranes Are Dynamic Structures
9.10
Three Classes of Membrane Proteins
9.11
Membrane Transport
275
278
A. Thermodynamics of Membrane Transport B. Pores and Channels C. Passive Transport
279
280 281 281
E. Endocytosis and Exocytosis
283
Box 9.5 The Hot Spice of Chili Peppers Transduction of Extracellular Signals
284 284
A. G Proteins Are Signal Transducers
285
B. The Adenylyl Cyclase Signaling Pathway
287
C. The Inositol–Phospholipid Signaling Pathway Box 9.6 Bacterial Toxins and G Proteins D. Receptor Tyrosine Kinases Summary
292
Problems
292
Selected Readings
271
274
Box 9.4 New Lipid Vesicles, or Liposomes
9.12
268
269
B. Fluid Mosaic Model of Biological Membranes
D. Active Transport
267
288
289
291
294
PART THREE Metabolism and Bioenergetics
10 Introduction to Metabolism
296
10.1
Metabolism Is the Sum of Cellular Reactions
10.2
Metabolic Pathways
296
298
A. Pathways Are Sequences of Reactions
299
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B. Metabolism Proceeds by Discrete Steps C. Metabolic Pathways Are Regulated D. Evolution of Metabolic Pathways
300
301 303
10.3
Major Pathways in Cells
304
10.4
Compartmentation and Interorgan Metabolism
10.5
Actual Gibbs Free Energy Change, Not Standard Free Energy Change, Determines the Spontaneity of Metabolic Reactions 308
10.6
The Free Energy of ATP
10.7
The Metabolic Roles of ATP
306
310 313
A. Phosphoryl-Group Transfer
314
B. Production of ATP by Phosphoryl Group Transfer C. Nucleotidyl Group Transfer
315
316
10.8
Thioesters Have High Free Energies of Hydrolysis
317
10.9
Reduced Coenzymes Conserve Energy from Biological Oxidations A. Gibbs Free Energy Change Is Related to Reduction Potential B. Electron Transfer from NADH Provides Free Energy
318 319
322
Box 10.1 NAD and NADH Differ in Their Ultraviolet Absorption Spectra 10.10 Experimental Methods for Studying Metabolism Summary
324
Problems
324
Selected Readings
Insulin α S S α S S S S β
11 Glycolysis
327
The Enzymatic Reactions of Glycolysis
11.2
The Ten Enzyme-Catalyzed Steps of Glycolysis
328 328
Box 11.1 A Brief History of the Glycolytic Pathway
332
Box 11.2 Formation of 2,3-Bisphosphoglycerate in Red Blood Cells Box 11.3 Arsenate Poisoning 11.3
Tyrosinekinase domains
323
326
11.1
β
The Fate of Pyruvate
338
340
340
A. Metabolism of Pyruvate to Ethanol B. Reduction of Pyruvate to Lactate 11.4
Free Energy Changes in Glycolysis
11.5
Regulation of Glycolysis
341 342
343
344
A. Regulation of Hexose Transporters B. Regulation of Hexokinase
344
346
Box 11.4 Glucose 6-Phosphate Has a Pivotal Metabolic Role in the Liver C. Regulation of Phosphofructokinase-1 D. Regulation of Pyruvate Kinase E. The Pasteur Effect 11.6
347
348
350
Other Sugars Can Enter Glycolysis
350
A. Fructose Is Converted to Glyceraldehyde 3-Phosphate
11.7
B. Galactose Is Converted to Glucose 1-Phosphate
351
C. Mannose Is Converted to Fructose 6-Phosphate
352
The Entner–Doudoroff Pathway in Bacteria Summary
354
Problems
354
Selected Readings
322
355
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350
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12 Gluconeogenesis, The Pentose Phosphate Pathway, and Glycogen Metabolism 12.1
Gluconeogenesis
359
B. Phosphoenolpyruvate Carboxykinase C. Fructose 1,6-bisphosphatase D. Glucose 6-phosphatase
C. Glycerol
Allosteric binding site
362
363
363
D. Propionate and Lactate 12.3
Phosphorylation site
361
362
B. Amino Acids
E. Acetate
360
361
Precursors for Gluconeogenesis A. Lactate
363
364
Regulation of Gluconeogenesis
364
Box 12.1 Glucose Is Sometimes Converted to Sorbitol 12.4
The Pentose Phosphate Pathway A. Oxidative Stage
366
366
368
B. Nonoxidative Stage
368
Box 12.2 Glucose 6-phosphate Dehydrogenase Deficiency in Humans C. Interconversions Catalyzed by Transketolase and Transaldolase 12.5
Glycogen Metabolism
369 370
371
A. Glycogen Synthesis B. Glycogen Degradation 12.6
371 372
Regulation of Glycogen Metabolism
374
A. Hormones Regulate Glycogen Metabolism
375
B. Reciprocal Regulation of Glycogen Phosphorylase and Glycogen Synthase 375 C. Intracellular Regulation of Glycogen Metabolism Involves Interconvertible Enzymes 376 Box 12.3 Glycogen Storage Diseases 12.7
378
Maintenance of Glucose Levels in Mammals Summary
381
Problems
382
Selected Readings
Catalytic site
357
358
A. Pyruvate Carboxylase
12.2
xvii
379
383
13 The Citric Acid Cycle
384
13.1
Conversion of Pyruvate to Acetyl CoA
385
13.2
The Citric Acid Cycle Oxidizes Acetyl CoA
13.3
The Citric Acid Cycle Enzymes
391
393
Box 13.1 Where Do the Electrons Come From?
394
Box 13.2 Three-point Attachment of Prochiral Substrates to Enzymes Box 13.3 Converting One Enzyme into Another
397
402
13.4
Reduced Coenzymes Can Fuel the Production of ATP
13.5
Regulation of the Citric Acid Cycle
13.6
The Citric Acid Cycle Isn’t Always a “Cycle”
13.7
The Glyoxylate Pathway
13.8
Evolution of the Citric Acid Cycle
403
404 406
407 410
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Contents
Summary
354
Problems
354
Selected Readings OUTSIDE
Heme b
Membrane
355
14 Electron Transport and ATP Synthesis 14.1
Overview of Membrane-associated Electron Transport and ATP Synthesis
14.2
The Mitochondrion
14.3
The Chemiosmotic Theory and the Protonmotive Force
418
A. Historical Background: The Chemiosmotic Theory
418
INSIDE
14.4
Electron Transport
420
421
A. Complexes I Through IV
Fe·S clusters
421
B. Cofactors in Electron Transport
FAD
14.5
Complex I
14.6
Complex II
14.7
Complex III
427
14.8
Complex IV
429
14.9
Complex V: ATP Synthase
424
424 425
432
Box 14.1 Proton Leaks and Heat Production
435
14.10 Active Transport of ATP, ADP, and Pi Across the Mitochondrial Membrane 14.11 The P/O Ratio
436
14.12 NADH Shuttle Mechanisms in Eukaryotes Box 14.2 The High Cost of Living
436
439
14.13 Other Terminal Electron Acceptors and Donors 14.14 Superoxide Anions Summary
441
Problems
441 442
15 Photosynthesis
444
15.1
Light-Gathering Pigments
15.2
Bacterial Photosystems A. Photosystem II
445 449
449
B. Photosystem I
452
C. Coupled Photosystems and Cytochrome bf
454
D. Reduction Potentials and Gibbs Free Energy in Photosynthesis E. Photosynthesis Takes Place within Internal Membranes 15.3
Plant Photosynthesis A. Chloroplasts
CF0
B. Plant Photosystems
460 461
463
C. Organization of Chloroplast Photosystems 15.4
Fixation of CO2: The Calvin Cycle A. The Calvin Cycle
463
464
465
B. Rubisco: Ribulose 1,5-bisphosphate Carboxylase–oxygenase ADP Pi
ATP H2O
458
459
460
Box 15.1 Bacteriorhodopsin
CF1
439
440
Selected Readings
H
416
416
B. The Protonmotive Force
QH2
415
C. Oxygenation of Ribulose 1,5-Bisphosphate
469
D. Calvin Cycle: Reduction and Regeneration Stages Box 15.2 Building a Better Rubisco
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470
465
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15.5
Sucrose and Starch Metabolism in Plants
15.6
Additional Carbon-Fixation Pathways
471 473
Box 15.3 Gregor Mendel’s Wrinkled Peas A. The C4 Pathway
473
474
B. Crassulacean Acid Metabolism (CAM) C. Carbon Fixation in Bacteria Summary
477
Problems
477
Selected Readings
478
16 Lipid Metabolism 16.1
474
476
Fatty Acid Synthesis
479
480
A. Synthesis of Malonyl ACP and Acetyl ACP
480
B. The Initiation Reaction of Fatty Acid Synthesis
481
C. The Elongation Reactions of Fatty Acid Synthesis D. Activation of Fatty Acids
482
483
E. Fatty Acid Extension and Desaturation
484
16.2
Synthesis of Triacylglycerols and Glycerophospholipids
16.3
Synthesis of Eicosanoids
16.4
Synthesis of Ether Lipids
16.5
Synthesis of Sphingolipids
Box 16.1 The Search for a Replacement for Aspirin 491 493
Box 16.3 Regulating Cholesterol Levels Synthesis of Cholesterol
490
490
Box 16.2 Lysosomal Storage Diseases 16.6
494
495
A. Stage 1: Acetyl CoA to Isopentenyl Diphosphate B. Stage 2: Isopentenyl Diphosphate to Squalene C. Stage 3: Squalene to Cholesterol Fatty Acid Oxidation
495 496
496
D. Other Products of Isoprenoid Metabolism 16.7
485
488
496
498
A. The Reactions of b-Oxidation
499
B. Fatty Acid Synthesis and b-Oxidation
500
C. Transport of Fatty Acyl CoA into Mitochondria
501
Box 16.4 A Trifunctional Enzyme for b-Oxidation
502
D. ATP Generation from Fatty Acid Oxidation
502
E. b-Oxidation of Odd-Chain and Unsaturated Fatty Acids 16.8
Eukaryotic Lipids Are Made at a Variety of Sites
16.9
Lipid Metabolism Is Regulated by Hormones in Mammals
16.10 Absorption and Mobilization of Fuel Lipids in Mammals A. Absorption of Dietary Lipids B. Lipoproteins
504
506 507 509
509
510
Box 16.5 Lipoprotein Lipase and Coronary Heart Disease C. Serum Albumin
513
513
16.11 Ketone Bodies Are Fuel Molecules
513
A. Ketone Bodies Are Synthesized in the Liver B. Ketone Bodies Are Oxidized in Mitochondria
514 515
Box 16.6 Altered Carbohydrate and Lipid Metabolism in Diabetes Summary
516
517
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Problems
517
Selected Readings
519
17 Amino Acid Metabolism
520
17.1
The Nitrogen Cycle and Nitrogen Fixation
17.2
Assimilation of Ammonia
521
523
A. Ammonia Is Incorporated into Glutamate and Glutamine B. Transamination Reactions 17.3
Synthesis of Amino Acids
524
524
526
A. Aspartate and Asparagine
526
Box 17.1 Childhood Acute Lymphoblastic Leukemia Can Be Treated with Asparaginase 526 B. Lysine, Methionine, and Threonine
527
C. Alanine, Valine, Leucine, and Isoleucine
528
D. Glutamate, Glutamine, Arginine, and Proline E. Serine, Glycine, and Cysteine
529
530
F. Phenylalanine, Tyrosine, and Tryptophan
531
Box 17.2 Genetically Modified Food Box 17.3 Essential and Nonessential Amino Acids in Animals G. Histidine 17.4
535
Amino Acids as Metabolic Precursors
536
A. Products Derived from Glutamate, Glutamine, and Aspartate B. Products Derived from Serine and Glycine C. Synthesis of Nitric Oxide from Arginine 17.5
Protein Turnover
536
538
Box 17.4 Apoptosis—Programmed Cell Death 17.6
Amino Acid Catabolism
538
539
A. Alanine, Asparagine, Aspartate, Glutamate, and Glutamine B. Arginine, Histidine, and Proline C. Glycine and Serine D. Threonine F. Methionine
542 543
545 546
H. Phenylalanine, Tryptophan, and Tyrosine
546
Box 17.5 Phenylketonuria, a Defect in Tyrosine Formation I. Lysine
546
548
Box 17.6 Diseases of Amino Acid Metabolism 17.7
The Urea Cycle Converts Ammonia into Urea A. Synthesis of Carbamoyl Phosphate B. The Reactions of the Urea Cycle
548 549
549 549
C. Ancillary Reactions of the Urea Cycle
550
Box 17.7 The Liver Is Organized for Removing Toxic Ammonia 17.8
541
541
543
E. The Branched-Chain Amino Acids G. Cysteine
536
536
Renal Glutamine Metabolism Produces Bicarbonate Summary
554
Problems
555
Selected Readings
556
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551
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18 Nucleotide Metabolism
557
18.1
Synthesis of Purine Nucleotides
18.2
Other Purine Nucleotides Are Synthesized from IMP
558
18.3
Synthesis of Pyrimidine Nucleotides
Box 18.1 Common Names of the Bases
561
561 563
A. The Pathway for Pyrimidine Synthesis
564
Box 18.2 How Some Enzymes Transfer Ammonia from Glutamine B. Regulation of Pyrimidine Synthesis
566
18.4
CTP Is Synthesized from UMP
18.5
Reduction of Ribonucleotides to Deoxyribonucleotides
18.6
Methylation of dUMP Produces dTMP
568 569
570
Box 18.3 Free Radicals in the Reduction of Ribonucleotides Box 18.4 Cancer Drugs Inhibit dTTP Synthesis 18.7
Salvage of Purines and Pyrimidines
18.8
Purine Catabolism
18.9
The Purine Nucleotide Cycle in Muscle
18.10 Pyrimidine Catabolism 580
Problems
580
570
572
573
574
Box 18.5 Lesch–Nyhan Syndrome and Gout
Summary
565
Selected Readings
574
578
579
581
PART FOUR Biological Information Flow
19 Nucleic Acids 19.1
19.2
583
Nucleotides Are the Building Blocks of Nucleic Acids A. Ribose and Deoxyribose
584
B. Purines and Pyrimidines
584
C. Nucleosides
586
D. Nucleotides
587
DNA Is Double-Stranded
584
590
A. Nucleotides Are Joined by 3¿ – 5¿ Phosphodiester Linkages B. Two Antiparallel Strands Form a Double Helix C. Weak Forces Stabilize the Double Helix DNA Can Be Supercoiled
592
595
D. Conformations of Double-Stranded DNA 19.3
597
597
19.4
Cells Contain Several Kinds of RNA
19.5
DNA Is Packaged in Chromatin in Eukaryotic Cells A. Nucleosomes
599 599
600
Box 19.1 Histones Can Be Acetylated and Deacetylated B. Higher Levels of Chromatin Structure C. Bacterial DNA Packaging 19.6
590
601
603
604
Nucleases and Hydrolysis of Nucleic Acids A. Alkaline Hydrolysis of RNA
605
605
B. Ribonuclease-Catalyzed Hydrolysis of RNA C. Restriction Endonucleases D. EcoRI Binds Tightly to DNA
605
608 610
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19.7
Uses of Restriction Endonucleases Summary
612
Problems
612
Selected Readings
610
613
20 DNA Replication, Repair, and Recombination 20.1
Chromosomal DNA Replication Is Bidirectional
20.2
DNA Polymerase
615
616
618
A. Chain Elongation Is a Nucleotidyl-Group-Transfer Reaction
619
B. DNA Polymerase III Remains Bound to the Replication Fork C. Proofreading Corrects Polymerization Errors 20.3
621
DNA Polymerase Synthesizes Two Strands Simultaneously A. Lagging-Strand Synthesis Is Discontinuous
621
622
623
B. Each Okazaki Fragment Begins with an RNA Primer
623
C. Okazaki Fragments Are Joined by the Action of DNA Polymerase I and DNA Ligase 624 20.4
Model of the Replisome
626
20.5
Initiation and Termination of DNA Replication
20.6
DNA Replication in Eukaryotes
20.7
Repair of Damaged DNA
629
630
Box 20.1 Sequencing DNA Using Dideoxynucleotides
632
634
A. Repair after Photodimerization: An Example of Direct Repair B. Excision Repair 20.8
Homologous Recombination
639
A. The Holliday Model of General Recombination B. Recombination in E. coli
639
640
C. Recombination Can Be a Form of Repair
641
Box 20.2 Molecular Links Between DNA Repair and Breast Cancer Summary
644
Problems
645
Selected Readings
646
21 Transcription and RNA Processing 21.1
Types of RNA
21.2
RNA Polymerase
647
648 649
A. RNA Polymerase Is an Oligomeric Protein B. The Chain Elongation Reaction 21.3
Transcription Initiation
649
650
652
A. Genes Have a 5¿ : 3¿ Orientation
652
B. The Transcription Complex Assembles at a Promoter C. The s Subunit Recognizes the Promoter D. RNA Polymerase Changes Conformation 21.4
Transcription Termination
21.5
Transcription in Eukaryotes
652
655 655
656 659
A. Eukaryotic RNA Polymerases B. Eukaryotic Transcription Factors
659 662
C. The Role of Chromatin in Eukaryotic Transcription 21.6
635
635
Transcription of Genes Is Regulated
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663
643
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21.7
The lac Operon, an Example of Negative and Positive Regulation A. lac Repressor Blocks Transcription B. The Structure of lac Repressor
667
C. cAMP Regulatory Protein Activates Transcription 21.8
Posttranscriptional Modification of RNA A. Transfer RNA Processing
668
670
671
B. Ribosomal RNA Processing 21.9
Eukaryotic mRNA Processing
672 674
A. Eukaryotic mRNA Molecules Have Modified Ends
674
B. Some Eukaryotic mRNA Precursors Are Spliced Summary
680
Problems
680
Selected Readings
The Genetic Code
22.2
Transfer RNA
677
682
22 Protein Synthesis 22.1
683
683
686
A. The Three-Dimensional Structure of tRNA
686
B. tRNA Anticodons Base-Pair with mRNA Codons 22.3
Aminoacyl-tRNA Synthetases
688 M16
688
A. The Aminoacyl-tRNA Synthetase Reaction
689
B. Specificity of Aminoacyl-tRNA Synthetases
RNase III
689
C. Proofreading Activity of Aminoacyl-tRNA Synthetases 22.4
Ribosomes
665
665
691
P
692
A. Ribosomes Are Composed of Both Ribosomal RNA and Protein B. Ribosomes Contain Two Aminoacyl-tRNA Binding Sites 22.5
Initiation of Translation A. Initiator tRNA
695
3′
P P
695
695
B. Initiation Complexes Assemble Only at Initiation Codons C. Initiation Factors Help Form the Initiation Complex D. Translation Initiation in Eukaryotes 22.6
693
5′
695
696
697
Chain Elongation Is a Three-Step Microcycle
697
A. Elongation Factors Dock an Aminoacyl-tRNA in the A Site B. Peptidyl Transferase Catalyzes Peptide Bond Formation C. Translocation Moves the Ribosome by One Codon 22.7
Termination of Translation
22.8
Protein Synthesis Is Energetically Expensive
22.9
Regulation of Protein Synthesis
699
700
701
705
21S particle 705
705
A. Ribosomal Protein Synthesis Is Coupled to Ribosome Assembly in E. coli 706 Box 22.1 Some Antibiotics Inhibit Protein Synthesis 707 B. Globin Synthesis Depends on Heme Availability
707
C. The E. coli trp Operon Is Regulated by Repression and Attenuation 22.10 Posttranslational Processing 712 A. The Signal Hypothesis B. Glycosylation of Proteins Summary
716
Problems
717
Selected Readings
708
712 716
Complete 30S subunit
718
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23 Recombinant DNA Technology 23.1
Making Recombinant DNA
23.2
Cloning Vectors
719
719
721
A. Plasmid Vectors
723
B. Bacteriophage l Vectors C. Shuttle Vectors
723
724
D. Yeast Artificial Chromosomes as Vectors 23.3
724
Identification of Host Cells Containing Recombinant DNA A. Selection Strategies Use Marker Genes B. Selection in Eukaryotes
727
727
727
C. Visual Markers: Insertional Inactivation of the b – Galactosidase Gene 23.4
Genomic Libraries
728
Box 23.1 The Human Genome Project
728
23.5
cDNA Libraries Are Made from Messenger RNA
23.6
Screening a Library
23.7
Chromosome Walking
23.8
Expression of Proteins Using Recombinant DNA Technology
733
A. Prokaryotic Expression Vectors
734
Applications of Recombinant DNA Technology A. Genetic Engineering of Plants
735
737
B. Genetic Engineering in Prokaryotes 23.10 Applications to Human Diseases
734
734
B. Expression of Proteins in Eukaryotes 23.9
729
730
737
739
23.11 The Polymerase Chain Reaction Amplifies Selected DNA Sequences Box 23.2 Medical Uses of PCR
741
23.12 Site-Directed Mutagenesis of Cloned DNA Summary
744
Problems
745
Selected Readings
Solutions
749
Illustration Credits Glossary Index
747
809
811 827
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741
727
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PHOTO TO COME
Preface To the Student Welcome to biochemistry—the study of the of life at the molecular level. As you venture into this exciting and dynamic discipline, you’ll discover many new and wonderful things. You’ll learn how some enzymes can catalyze chemical reactions at speeds close to theoretical limits—reactions that would otherwise occur only at imperceptibly low rates. You’ll learn about the forces that maintain biomolecular structure and how even some of the weakest of those forces make life possible. You’ll also learn how biochemistry has thousands of applications in day-to-day life—in medicine, drug design, nutrition, forensic science, agriculture, and manufacturing. In short, you’ll begin a journey of discovery about how biochemistry makes life both possible and better. Before we begin, we would like to offer a few words of advice: Don’t just memorize facts; instead, understand principles In this book, we have tried to identify the most important principles of biochemistry. Every year, a million or so research papers are published. Half of them describe the results of research in some area of biochemistry. Because the knowledge base of biochemistry is continuously expanding, we must grasp the underlying themes of this science in order to understand it. This textbook is designed to expand on the foundation you have acquired in your chemistry and biology courses and to provide you with a biochemical framework that will allow you to understand new phenomena as you meet them. As you progress in your studies, you will encounter many examples that flesh out the framework we describe in this book. These individual facts are useful in illuminating the basic principles. Be prepared to learn a new vocabulary An understanding of biochemical facts requires that you learn a biochemical vocabulary. This vocabulary includes the chemical structures of a number of key molecules. These molecules are grouped into families based on their structures and functions. You will also learn how to distinguish among members of each family and how small molecules combine to form macromolecules such as proteins and nucleic acids. As with any newly studied discipline, the more familiar you are with the vocabulary the more easily you can learn and appreciate the scientific literature. Test your understanding True mastery of biochemistry lies with learning how to apply your knowledge and how to solve problems. Each chapter concludes with a set of carefully crafted problems that test your understanding of core principles. Many of these problems are mini case studies that present the problem within the context of a real biochemical puzzle.
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For more practice, including solving more traditional review exercises in biochemistry, we are pleased to refer you to The Biochemistry Student Companion by Allen Scism (see page viii). This book presents a variety of supplementary questions that you may find helpful. You will also find additional problems on the Principles of Biochemistry Companion Website (http://www.prenhall.com/horton). Learn to visualize in 3-D Biochemicals are three-dimensional objects. Understanding what happens in a biochemical reaction at the molecular level requires that you be able to “see” what happens in three dimensions. We present the structures of simple molecules in several different ways in order to illustrate their three-dimensional conformation. The structures of proteins and protein complexes are also drawn as three-dimensional objects. Throughout this book you will find many excellent drawings by our experienced team of artists. In addition, on the website you will find many animations and interactive molecular models that you can manipulate in real-time on a computer. We strongly suggest you look at these movies and do the exercises that accompany them as well as participate in the molecular visualization tutorials. Feedback Finally, please let us know of any errors or omissions you encounter as you use this text. Tell us what you would like to see in the next edition. Many of the changes in this fourth edition began with suggestions, and criticisms, from students like you. With your help, we will continue to evolve this work into an even more useful tool. Our e-mail addresses are at the end of this Preface. Good luck, and enjoy!
To the Instructor We welcome all our loyal users and those instructors who are, for the first time, teaching courses in biochemistry. This textbook is intended for students in onesemester courses but it is becoming more and more popular in two-semester courses. We hope you will find this the ideal book for your course. If you have any questions or comments for us, please get in touch with us by e-mail. A book for your students We have organized this book as we would organize our own courses. We have included all the topics that are usually covered in introductory biochemistry courses although we are well aware of the fact that no single course covers every topic and every chapter. At some point, students cease to be intimidated by the need to skip around in a book. In our experience, that transition lies somewhere after they have begun a course like yours. We have tried to make the text as modular as possible. Each large topic resides in its own section. Reaction mechanisms are often separated from the main thread of the text and can be passed over by those who prefer not to cover this level of detail. The text is extensively cross-referenced to make it easier for you to reorganize the chapters and for students to see the interrelationships among various topics and to drill down to deeper levels of understanding. We built the book explicitly for the beginning student taking a first course in this burgeoning subject. Parts One and Two lay a solid foundation of chemical knowledge that will help students understand, rather than merely memorize, the dynamics of metabolic and genetic processes. These sections assume that students have taken prerequisite courses in general and organic chemistry and have acquired a rudimentary knowledge of the organic chemistry of carboxylic acids, amines,
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alcohols, and aldehydes. Even so, key functional groups and chemical properties of each type of biomolecule are carefully explained as their structures and functions are presented. A focus on principles There are, in essence, two kinds of biochemistry textbooks: those for reference and those for teaching. It is difficult for one book to be both as it is those same thickets of detail sought by the professional that ensnare the struggling novice on his or her first trip through the bush. This text is unapologetically a text for teaching. It has been designed to foster student understanding and is not an encyclopedia of biochemistry. This book focuses unwaveringly on teaching basic principles, each principle supported by carefully chosen examples. A focus on chemistry with rigorous biology When we first wrote this text, we decided to take the time to explain in chemical terms the principles we underscore. To that end, we offer detailed chemical explanations of most of the chemistry herein, including mechanisms (which tell students how and why things happen). While we emphasize chemistry, we also stress the bio in biochemistry. We point out that biochemical systems evolve and that the reactions that occur in some species are variations on a larger theme. In this edition, we increase our emphasis on the similarities of prokaryotic and eukaryotic systems, while we continue to avoid making generalizations about all organisms based on reactions that occur in a few. We are proud of the fact that this is the most scientifically accurate biochemistry textbook. We have gone to great lengths to ensure that our facts are correct and our explanations of basic concepts reflect the modern consensus among active researchers. Our success is due, in large part, to the dedication of our many reviewers and editors. A focus on the visual Biochemistry is a three-dimensional science. Our inclusion of the latest computergenerated images is intended to clarify the shape and function of molecules and to leave students with an appreciation for the relationship between the structure and function. Most of the protein images in this edition are new; they have been skillfully prepared by Jonathan Parrish of the University of Alberta. For those students with access to a computer, we offer a number of other opportunities. We have included Protein Data Bank (PDB) reference numbers for the coordinates from which all protein images were derived. This allows students to explore the structures further on their own. In addition, we have a gallery of prepared PDB files that students can view using Chime or any other molecular viewer; these are posted on the text’s Companion Website (http://www.prenhall. com/horton) as are animations of key dynamic processes as well as visualization tutorials using Chime. Finally, we offer instructors a CD that contains the gallery of animations, all the PDB files, as well as most of the art from the textbook in .jpg format. This makes the preparation of in-class electronic presentations (e.g., using PowerPoint) much easier. Please see the listing of supplements on page xxxii. The organization We strive to present the story of biochemistry in a clear, coherent, well-integrated manner, building at each stage the background needed for the next stage. This book is arranged in four parts, each building on those that come before. Part One forms an introduction to all that follows. In Chapter 1 we have added sections on thermodynamics and reaction kinetics in response to suggestions from your colleagues. These short descriptions will help students who need to review
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basic chemical principles. Chapter 1 also includes a brief survey of cellular structure and the four classes of macromolecules—proteins, carbohydrates, lipids, and nucleic acids. In this chapter, and throughout the book, we sustain a consistent pattern of presentation, from basic chemistry to biochemical function. We first show the chemistry of monomeric units and then explore the properties and functions of the biopolymers and aggregates formed from those monomers—amino acids to proteins, monosaccharides to polysaccharides and glycoconjugates, lipids to membranes, and nucleotides to nucleic acids. Part Two includes three chapters on enzyme properties, enzyme mechanisms, and coenzymes (Chapters 5, 6, and 7). We encourage a firm grasp of these critical subjects since they are required for later appreciation of the role of enzymes and coenzymes in metabolic pathways. Chapter 8 covers carbohydrates and their derivatives. We have added new information on the ABO blood groups in the section on glycoproteins. Lipids and membranes are described in Chapter 9 and this is where we introduce students to the exciting field of signal transduction. Our description of membrane transport and membrane potentials has been improved in order to better prepare students for chapters on electron transport and proton gradients. Part Three features nine chapters that focus on metabolic pathways. In Chapter 10, we introduce the intricate molecular symphonies of metabolism by considering how pathways are energized, interrelated, and regulated. In this edition we have added new material on how metabolic pathways evolve—a theme that we return to in subsequent chapters. Chapter 11 describes glycolysis in detail. At this stage, we establish a format that recurs in subsequent metabolic chapters—first describing the basic pathway in chemical and enzymatic terms, then demonstrating the bioenergetic requirements and sources of energy, and concluding with an account of regulatory mechanisms. Note that the early introduction of signal transduction (Chapter 9) allows integration of regulation into each chapter of metabolism and allows discussion of flux through reciprocal pathways. Chapter 12 has been completely reorganized. We now place a greater emphasis on gluconeogenesis and we explain how the pathways of glycolysis and gluconeogenesis are related. Most biochemistry teachers are spending more time on gluconeogenesis since it is the fundamental pathway in most species. In the following chapters on lipid metabolism (Chapter 16), amino acid metabolism (Chapter 17), and nucleotide metabolism (Chapter 18), we describe the biosynthesis pathways before showing the degradation pathways. Again, this is consistent with our focus on the basic principles of biochemistry that apply to most species and it reflects a new trend in teaching biochemistry. Whenever possible we explain where human biochemical pathways differ from the ones seen in other species. We hope you enjoy the new look of Chapter 13 (Citric Acid Cycle) and Chapter 14 (Electron Transport and ATP Synthesis). We have put far more emphasis on the evolution of these pathways and the relationship between prokaryotic and eukaryotic versions. We have also incorporated new structures of membrane complexes as part of a strategy to relate structure and function in a more meaningful way than was possible with previous editions. Part Three also includes a completely revised chapter on photosynthesis (Chapter 15). We have taken advantage of the tremendous amount of new information on photosynthesis to describe how simple bacterial systems evolved into the more complex plant systems. We have also paid more attention to integrating the principles of photosynthesis and the material on chemiosmotic theory introduced in the preceding chapter on electron transport. Part Four concludes the book with five chapters on the flow of biological information. Our main emphasis throughout this series of chapters is on the basic processes that govern pathways of information flow. We take pains to incorporate within our story of genes and gene expression the principles of biochemistry taught in preceding chapters. The final chapter (Chapter 23) is a substantial and contemporary presentation of techniques used in recombinant DNA technology.
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New features in this edition We are grateful for all the input we received on the first three editions of this text. In addition to the changes noted above, you’ll notice the following improvements in this fourth edition: • expanded descriptions of oxidation-reduction reactions in several chapters, including more explanation of where electrons come from. • additional problems at the end of every chapter, with detailed solutions to all problems in an appendix at the end of the book. • more sample solutions to problems. • more extensive marginal cross-referencing to enhance the connection of concepts and the integration of material, to permit instructors to cover the material in a different order, and to improve the efficiency of students’ studying. • the addition of special topic, clinical application, and deeper-look boxes to most chapters. • extra boxes on the origins of biochemical names and terms in order to make the vocabulary more friendly and interesting. • expansion of the complete glossary of biochemical terms at the end of the book. • update of all Selected Readings at the end of each chapter. Since publication of the third edition, major advances have been made in a number of areas. For this edition, we made changes to each section of the text to reflect these advances. For example, we added many new protein structures, each with Protein Data Bank cross-references; expanded the discussion of protein structure (with more examples); expanded the discussion of known multifunctional enzymes; increased the emphasis on molecular evolution; enhanced the discussion of the mechanism of energy conservation as ATP; incorporated the results of the various genome initiatives; and added new information on transcription and translation, including new information on the structures of the ribosome and RNA polymerase. We have created a student learning package that includes the following: • The Biochemistry Student Companion • Companion Website The instructor’s packages consist of • Instructor Resource Center on CD/DVD • test item file, • Transparency Set of 150 full-color acetates and masters, • OneKey solution to course management.
Acknowledgments We are grateful to our many talented and thoughtful reviewers who have helped shape this book. Reviewers who helped in the Fourth Edition: Accuracy Reviewers David Watt, University of Kentucky Neil Haave, University of Alberta
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Content Reviewers Consuelo Alvarez, Longwood University Marilee Benore Parsons, University of Michigan Albert M Bobst, University of Cincinnati Gary J. Blomquist, University of Nevada, Reno Kelly Drew, University of Alaska, Fairbanks Andrew Feig, Indiana University Giovanni Gadda, Georgia State University Donna L. Gosnell, Valdosta State University Charles Hardin, North Carolina State University Jane E. Hobson, Kwantlen University College Ramji L. Khandelwal, University of Saskatchewan Scott Lefler, Arizona State Kathleen Nolta, University of Michigan Jeffrey Schineller, Humboldt State University Richard Shingles, Johns Hopkins University Michael A. Sypes, Pennsylvania State University Martin T. Tuck, Ohio University Julio F. Turrens, University of South Alabama David Watt, University of Kentucky James Zimmerman, Clemson University Reviewers who have helped with previous editions: Lawrence Aaronson, Utica College of Syracuse; Stephen L. Bearne, University of North Carolina at Chapel Hill; Robert Bergen, University of Central Arkansas; Gary J. Blomquist, University of Nevada–Reno; Robert I. Bolla, Saint Louis University; Lori Bolyard, University of Evansville; John Brosnan, Memorial University of Newfoundland; Ronald Callahan, New York University; Martin F. Chaplin, South Bank University; William Coleman, University of Hartford; Harold Cook, Dalhousie University; Gary W Daughdrill, University of Idaho; Ruthellen M. Dawley, University of Evansville; John Durham, West Virginia School of Medicine; Yves Engelborghs, University of Leuven; Edward Funkhouser, Texas A&M University; Milton Gordon, University of Washington; Kenneth, E. Guyer, Marshall University; Robert Harris, Indiana University–School of Medicine; Jan Hoek, Thomas Jefferson University; J. Kenneth Hoober, Arizona State University; Cristi Hunnes, Rocky Mountain College; Mahendra K. Jain, University of Delaware; Thomas D. Kim, Youngstown State University; David Koetje, SUNY–Fredonia; James A. Knopp, North Carolina State University; Susan Lees-Miller, Roger A. Lewis, University of Nevada at Reno; University of Calgary; Robert N. Lindquist, San Francisco State University; Richard Lomneth, University of Nebraska at Omaha; Ray Lutgring, University of Evansville; George Marzluf, Ohio State University; Barbara Olson, University of Calgary; Thomas Prasthofer, Albany College of Pharmacy; Thomas Reilly, California State University–Dominguez Hills; Carl Rhodes, University of Illinois–Urbana-Champagne; Gale Rhodes, University of Southern Maine; Duane L. Rohlfing, University of South Carolina; Douglas Russell, Dalhousie University; Aziz Sancar, University of North Carolina-School of Medicine; Larry Scheve, California State University—Hayward; Allen Scism, Central Missouri State University; Steven Seifried, University of Hawaii at Manoa; Thomas Sherman, University of Pittsburgh; Timothy A. Sherwood, Arkansas Tech University; Dean Sherry, University of Texas at Dallas; David Skalnik, Indiana University School of Medicine; Anthony F. Sky, Lawrence Technological University; Gary D. Small, University of South Dakota; Ronald L. Somerville, Purdue University; Ralph Stephani, St. John’s University; Laurence Tate, University of South Alabama; William Thompson, University of Toronto; Richard W. Topham, University of Richmond; Arrel Towes, University of South Carolina; Julio F. Turrens, University of South Alabama; Jack Y. Vanderhoek, Charles Waechter,
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University of Kentucky; Jubran M. Wakim, Middle Tennessee State University; George Washington University; William Wolodko, University of Alberta; David Yang, Georgetown University We would also like to thank our colleagues who have previously contributed material for particular chapters and whose careful work still inhabits this book: Roy Baker, University of Toronto Roger W. Brownsey, University of British Columbia Willy Kalt, Agriculture Canada Robert K. Murray, University of Toronto Frances Sharom, University of Guelph Malcolm Watford, Rutgers, The State University of New Jersey Putting this book together was a collaborative effort, and we would like to thank various members of the team who have helped give this project life: Jonathan Parrish, Jay McElroy, Lisa Shoemaker, and the artists of Prentice Hall; Jennifer Hart, editorial assistant, Crissy Dudonis, Project Editor in charge of supplements, Patrick Shriner, Media Editor, Andrew Gilfillan, Marketing Manager; and a special thanks to Marty Sopher, our Production Editor, whose organizational skills made this book happen. We would also like to thank Gary Carlson, our Executive Editor at Prentice Hall. Finally, we close with an invitation for feedback. Despite our best efforts (and a terrific track record in the previous editions), there are bound to be mistakes in a work this size. We are committed to making this the best biochemistry text available; please know that all comments are welcome. Laurence A. Moran
[email protected]
K. Gray Scrimgeour
[email protected]
Marc D. Perry
[email protected]
CMYK ASSOCIATES, INC. FIRST PROOF
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Guide to the Features of the Text Most students who purchase this book aim for mastery of the material. They also want a good grade in the course. Here’s some advice to help you attain these goals. • Attend lectures and take notes. • Keep up with assignments. • Read this textbook as assigned by your instructor. • Practice your skills by solving problems—many are like the ones on your exams.
348
CHAPTER 11
2
Glycolysis
■
6
⌷3POCH2 5
OH
O
2
HO
H 4
3
CH2OH 1
H
OH
b-D-Fructose 6-phosphate Pi
ATP
Fructose 2,6bisphosphatase
PFK-2
H 2O 2
ADP 6
⌷3POCH2 5
H
H 4
OPO3 2
O
2
HO 3
CH2OH 1
OH H b-D-Fructose 2,6-bisphosphate ▼
Figure 11.16 Interconversion of b -D-fructose 6-phosphate and b -D-fructose 2,6-bisphosphate.
The adenylyl cyclase signaling pathway is described in Section 9.12B.
Section 10.6, significant changes in the concentrations of ADP and AMP do occur because these molecules are present in cells in much lower concentrations than ATP, and small changes in the level of ATP cause proportionally larger changes in the levels of ADP and AMP. The steady-state concentrations of these compounds are therefore able to control flux through PFK-1. Citrate, an intermediate of the citric acid cycle, is another physiologically important inhibitor of PFK-1. An elevated concentration of citrate indicates that the citric acid cycle is blocked and further production of pyruvate would be pointless. The regulatory effect of citrate on PFK-1 is an example of feedback inhibition that regulates the supply of pyruvate to the citric acid cycle. The activity of PFK-1 is also regulated by the intracellular pH. For example, during heavy exercise, anaerobic glycolysis produces lactic acid in muscle cells. If the lactic acid is not removed fast enough by the blood, the pH decreases and PFK-1 is inhibited by the excess hydrogen ions. Fructose 2,6-bisphosphate (Figure 11.16) is a potent activator of PFK-1, effective in the micromolar range. This compound is present in mammals, fungi, and plants, but not prokaryotes. Fructose 2,6-bisphosphate is formed from fructose 6-phosphate by the action of the enzyme phosphofructokinase-2 (PFK-2, or fructose 6-phosphate, 2-kinase). PFK-2 is stimulated by inorganic phosphate and inhibited by citrate. Surprisingly, in mammalian liver, a different active site on the same protein catalyzes the hydrolytic dephosphorylation of fructose 2,6-bisphosphate, re-forming fructose 6-phosphate. This activity of the enzyme is called fructose 2,6-bisphosphatase. The dual activities of PFK-2 control the steady-state concentration of fructose 2,6-bis phosphate. In the liver, the activity of PFK-2 is linked to the action of glucagon, a hormone produced by the pancreas in response to low blood glucose levels. An elevation of the glucagon concentration in the blood triggers the adenylyl cyclase signaling pathway in liver cells, culminating in the phosphorylation of a serine residue in PFK-2 (Figure 11.17). Phosphorylation inactivates the kinase activity of the bifunctional enzyme and activates its phosphatase activity. Under these conditions, the level of fructose 2,6-bisphosphate falls, PFK-1 becomes less active, and glycolysis is depressed. Under conditions in which glucose is rapidly metabolized, the glucagon level falls, the concentration of fructose 6-phosphate increases, and more fructose 2,6-bisphosphate is formed since fructose 6-phosphate is both a substrate of PFK-2 and a potent inhibitor of fructose 2,6-bisphosphatase. A phosphoprotein phosphatase catalyzes the dephosphorylation of PFK-2. Thus, in liver cells, control of glycolysis by glucagon and glucose is accomplished through control of the bifunctional enzyme whose activity establishes the steady-state concentration of fructose 2,6-bisphosphate. D. Regulation of Pyruvate Kinase The third site of allosteric regulation of glycolysis is the reaction catalyzed by pyruvate kinase. Four different isozymes of pyruvate kinase are present in mammalian tissues. The isozymes found in liver, kidney, and red blood cells yield a sigmoidal curve when initial velocity is plotted against phosphoenolpyruvate concentration (Figure 11.18a). These enzymes are allosterically activated by fructose 1,6-bisphosphate (Figure 11.19, on page 350) and inhibited by ATP. In the absence of fructose 1,6-bisphosphate, physiological concentrations of ATP almost completely inhibit the isolated enzyme. The presence of fructose 1,6-bisphosphate—probably the most important modulator in vivo—shifts the curve to the left. With sufficient fructose 1,6-bisphosphate, the curve becomes hyperbolic. Figure 11.18a shows that enzyme activity is greater in the presence of the allosteric activator for a range of substrate concentrations. Recall that fructose 1,6-bisphosphate is the product of the reaction catalyzed by PFK-1. Its concentration increases when the activity of PFK-1 increases. Since fructose 1,6-bisphosphate activates pyruvate
▲ Complete Explanations of the Chemistry There are thousands of metabolic reactions in humans. You might try to memorize them all, but eventually you’ll run out of memory. What’s more, memorization won’t help you if you encounter something you haven’t seen before. In this book, we show you some of the basic mechanisms of enzyme-catalyzed reactions—an extension of what you learned in organic chemistry. If you understand the mechanism, you’ll understand the chemistry. You’ll have less to memorize, and you’ll retain the information more effectively. xxxii
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▲
Interest Boxes
The Hot Spice of Chili Peppers
BOX 9.5
Biochemistry is at the root of a number of applied sciences, including medicine, forensic science, biotechnology, and bioengineering. There are many interesting stories to tell, you’ll find them boxed off from the main text so they don’t distract you from the main topic. They will give you insight to the commercial, practical and medical applications of biochemistry.
Biochemists now know the mechanism by which spice from “hot” peppers exerts its action, causing a burning pain. The active factor in capsicum peppers is a lipophilic vanilloid compound called capsaicin.
pain. Control of the capsaicin receptor might have value in the relief of chronic pain in conditions such as arthritis.
O H 3 CO
N H
HO Capsaicin
▲
A nerve-cell protein receptor that responds to capsaicin has been identified and characterized. It is an ion channel, and its amino acid sequence suggests that it has six transmembrane domains. Activation of the receptor by capsaicin causes the channel to open so that calcium and sodium ions can flow into the nerve cell and send an impulse to the brain. The receptor is activated not only by vanilloid spices but also by rapid increases in temperature. In fact, the probable in vivo role of the receptor is detection of heat. Whereas the action of opioids suppresses pain, the action of vanilloids produces
360
CHAPTER 12
■
Gluconeogenesis, The Pentose Phosphate Pathway, and Glycogen Metabolism
pyruvate carboxylase catalyzes the conversion of pyruvate to oxaloacetate. The reaction is coupled to the hydrolysis of one molecule of ATP. COO C
O
COO +
HCO 3 Bicarbonate
CH 3
Pyruvate carboxylase
ATP
ADP + Pi
Pyruvate
C
+
O
H 2O
(12.2)
CH 2 COO Oxaloacetate
The reaction mechanism for pyruvate carboxylase was described in Section 7.9.
Pyruvate carboxylase is a large, complex enzyme composed of four identical subunits. Each subunit has a biotin prosthetic group covalently linked to a lysine residue. The biotin is required for the addition of bicarbonate to pyruvate. Pyruvate carboxylase catalyzes a metabolically irreversible reaction and can be allosterically activated by acetyl CoA. This is the only regulatory mechanism known for the enzyme. Accumulation of acetyl CoA indicates that it is not being efficiently metabolized by the citric acid cycle. Pyruvate carboxylase is stimulated in order to direct pyruvate to oxaloacetate instead of acetyl CoA. Oxaloacetate can enter the citric acid cycle or serve as a precursor for glucose biosynthesis. Bicarbonate is one of the substrates in Reaction 12.2. Bicarbonate is formed when carbon dioxide dissolves in water and the reaction is sometimes written with CO2 as a substrate. The pyruvate carboxylase reaction plays an important role in fixing carbon dioxide in bacteria and some eukaryotes. This role may not be so obvious when we examine gluconeogenesis since the carbon dioxide is released in the very next reaction. However, much of the oxaloacetate that is made is not used for gluconeogenesis. Instead it replenishes the pool of citric acid cycle intermediates that serve as precursors to the biosynthesis of amino acids and lipids (Section 13.7).
Margin Notes There is a great deal of detail in biochemistry, but we want you to see both the forest and the trees. When we need to cross-reference something discussed earlier in the book, or something that we will come back to later, we put it in the margin. Backward references offer a review of concepts you may have forgotten. Forward references will help you see the big picture.
B. Phosphoenolpyruvate Carboxykinase Phosphoenolpyruvate carboxykinase (PEPCK) catalyzes the conversion of oxaloacetate to phosphoenolpyruvate.
COO C
O
CH 2 COO Oxaloacetate
(ATP) GTP
(ADP) GDP
Phosphoenolpyruvate carboxykinase (PEPCK)
CO 2
COO C
OPO 3
2
CH 2
(12.3)
Phosphoenolpyruvate (PEP)
There are two different versions of PEPCK. The enzyme found in bacteria, protists, fungi, and plants uses ATP as the phosphoryl group donor in the decarboxylation reaction. The animal version uses GTP. In most species, the enzyme displays no allosteric kinetic properties and has no known physiological modulators. Its activity is most often affected by controls at the level of transcription of its gene. The level of PEPCK activity in cells influences the rate of gluconeogenesis. This is especially true in mammals in which gluconeogenesis occurs primarily in cells of the liver, kidneys, and small intestine. During fasting in mammals, prolonged release of glucagon from the pancreas leads to an elevation of intracellular cAMP, which triggers increased transcription of the PEPCK gene in the liver and increased synthesis of PEPCK. This triggering of gene expression by a hormone is called hormonal induction. After several hours, the amount of PEPCK rises, and the rate of gluconeogenesis increases. Insulin, abundant in the fed state, acts in opposition to glucagon at the level of the gene, reducing the rate of synthesis of PEPCK.
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▲
End-of-Chapter Problems
Problems
Beginning with chapter 2, a number of biochemical puzzles are included that can be solved using the information contained in the chapter. Answers are found at the back of the book. If you want some additional practice, we recommend you get a copy of The Biochemistry Student Companion by Allen Scism. This book features hundreds of additional problems, including drill questions and chapter review exercises.
1. Identify each of the following: (a) Two aldoses whose configuration at carbons 3, 4, and 5 matches that of D-fructose. (b) The enantiomer of D-galactose. (c) An epimer of D-galactose that is also an epimer of D-mannose. (d) A ketose that has no chiral centers. (e) A ketose that has only one chiral center. (f) Monosaccharide residues of cellulose, amylose, and glycogen. (g) Monosaccharide residues of chitin.
Answer the following questions about the structure of sialic acid. (a) Is it an a or a b anomeric form? (b) Will sialic acid mutorotate between a and b anomeric forms? (c) Is this a “deoxy” sugar? (d) Will the open chain form of sialic acid be an aldehyde or a ketone? (e) How many chiral carbons are there in the sugar ring? 9
CH 3 O
2. Draw Fischer projections for (a) L-mannose, (b) L-fucose (6-deoxy-L-galactose), (c) D-xylitol, and (d) D-iduronate.
C HN 5
3. Describe the general structural features of glycosaminoglycans.
CH 2 OH H
5. Sialic acid (N-Acetyl-a-D-neuraminic acid) is often found in N-linked oligosaccharides that are involved in cell-cell interactions. Cancer cells synthesize much greater amounts of sialic acid than normal cells and derivatives of sialic acid have been proposed as anticancer agents to block cellsurface interactions between normal and cancerous cells.
CHOH
6
O 7 CHOH H H
H
1
COOH 2
OH 4
OH
4. Honey is an emulsion of microcrystalline D-fructose and D-glucose. Although D-fructose in polysaccharides exists mainly in the furanose form, solution or crystalline Dfructose (as in honey) is a mixture of several forms with b -D-fructopyranose (67%) and b -D-fructofuranose (25%) predominating. Draw the Fischer projection for D-fructose and show how it can cyclize to form both of the cyclized forms above.
8
3
H
Sialic acid
6. How many stereoisomers are possible for glucopyranose and for fructofuranose? How many are D sugars in each case, and how many are L sugars? 7. Draw the structure of each of the following molecules and label each chiral carbon with an asterisk. (a) a-D-Glucose 1-phosphate. (b) 2-Deoxy- b -D-ribose 5-phosphate. (c) D-Glyceraldehyde 3-phosphate. (d) L-Glucuronate. 8. In aqueous solution, virtually all D-glucose molecules 1799%2 are in the pyranose form. Other aldoses have a
OUTSIDE (Periplasm)
Bacterial membrane (a)
INSIDE (Cytoplasm) OUTSIDE (Periplasm)
Bacterial membrane
INSIDE (Cytoplasm)
Pitch (advance 0.54 nm per turn) 0.15 nm Rise (advance per amino acid residue)
(b)
Cytochrome c heme
e hν
Special pair (P870)
e
Bacteriochlorophyll a
Right-handed α-helix
e
Axis
α-carbon Carbonyl carbon Hydrogen Nitrogen Oxygen Side chain
▲ Figures e Bacteriopheophytin
e Quinone
Q QH2 e 2Hin
xxxiv
e
Biochemistry is a three-dimensional science and we have placed great emphasis on helping you visualize abstract concepts and molecules too small to see. We have taken advantage of new rendering technologies to make illustrative figures both informative and beautiful. For additional views of molecules in 3-D go to the web site at www.prenhall.com/horton and view the many PDB files.
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Companion Website The new edition of the textbook offers a Companion Website at www.prenhall.com/horton. This website is focused on the visualization of Biochemistry and includes hundreds of 3-D models of biomolecules, including nearly all of those that appear in your textbook. Viewing these is a great way to ensure that you understand the key features of macromolecular architecture. Also included on the Companion Website are MediaLabs that help you use the internet to practice and apply what you have learned. With each MediaLab you are able to investigate the background of an important issue related to the chapter you have just read and then answer questions that test your comprehension of the topic. MediaLab topics have been carefully chosen to focus on information that is timely and relevant in your daily life. These investigations are designed to become a dynamic discovery activity in which you may participate and report, individually or in groups.
Instructor Resources Test Item File (0-13-147601-7), by William Coleman, University of Hartford and Donna Gosnell, Valdosta State University. Provides a selection of more than 1500 questions referenced to the text. TestGen-EQ Computerized Testing Software (0-13-147602-5) In addition to the printed test bank, the test questions are available as part of the TestGen-EQ testing software, a text specific testing program that can reside on a network for administering tests. It allows instructors to view and edit questions, export the questions as tests, and print them out in a variety of formats. Transparency Pack (0-13-147616-5) Over 100 full-color acetates and masters available free to qualified adopters. Instructor Resource Center on CD (0-13-133055-1) Includes most of the art from the textbook in a format for easy presentation on-screen. Also includes pre-built PowerPoint slides. Course Management Instructors can access Prentice Hall OneKey content by downloading it to a school’s server in cartridge or epack format by visiting www.prenhall.com/onekey, clicking “OneKey For Instructors”, clicking “Getting Started”, and selecting the desired platform.
Student Resources Biochemistry Student Companion (0-13-147605-X), by Allen Scism. Contains hundreds of additional, carefully constructed, short answer, multiple choice, and challenge problems for each chapter, comprehensive, step-by-step solutions to all problems, lists of abbreviations and tables of essential data. Companion Website An online student tool that includes 3-D modules to help visualize biochemistry and MediaLabs to investigate important issues related to a particular chapter. Please visit the site at http://www.prenhall.com/horton.
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Principles of Biochemistry
xxxvii
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Solutions Chapter 2 Water 1. Hydrogen bonds involve strongly electronegative atoms such as nitrogen, oxygen, or sulfur. O
(a) H CH2
(b)
O
H
H
(c) H
H
O
H
H
H
O
O H
O CH2
O H
C
.. N
H CH2
N H
H
H
N
H
O
H
H
H O
O H
H
2. (a) Glycerol is polar; it is not amphipathic; and it readily dissolves in water. (b) Hexadecanoyl phosphate is polar; it is amphipathic, and it does not readily dissolve in water but forms micelles. (c) Laurate is polar; it is amphipathic; and it does not readily dissolve in water but forms micelles. (d) Glycine is polar; it is not amphipathic; and it readily dissolves in water. 3. There is a larger osmotic pressure inside the cells than outside because the molar concentration of solutes is much greater inside cells than outside. This results in a diffusion of water into cells, causing them to swell and burst. 4. If the pH of a solution is below the pKa of any given ionizable group, the predominant species will be the one with the dissociable proton on that group. If the pH of a solution is above the pKa of any given ionizable group, the predominant species will be the one with the dissociable proton off of that group. (a) pH = 11 where ¬ COO form predominates. (b) pH = 2 where the H form predominates. (c) pH = 2 where the H form predominates. (d) pH = 11 where the R ¬ O form predominates. 5. (a) Tomato juice. For pH = 4.2, if pH = -log [H ], then [H ] = 10-pH[H ] = 10-4.2 = 6.3 * 10-5 M. The ion-product constant of water 1Kw2 relates the concentrations of OH and H (Equation 2.6). [OH ] = Kw2/[H ] = 1.0 * 10-14 M2/6.3 x-5 M = 1.6 * 10-10 M. (b) Human blood plasma. If the pH = 7.4, then [H ] = 10-7.4 = 4.0 * 10-8 M. [OH ] = Kw/[H ] = 1.0 * 10-14 M2/4.0 x-8 M = 2.5 * 10-7 M. (c) 1M Ammonia. If the pH = 11.7, then [H ] = 10-11.7 = 2.0 * 10-12 M. [OH ] = Kw/[H ] = 1.0 * 10-14 M2/2.0 x-12 M = 5 * 10-3 M. 6. H R R′ O O H H N
O CH2
C
CMYK ASSOCIATES, INC. FIRST PROOF
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Solutions
7. The total buffer species = [weak acid 1HA2] + [conjugate base 1A 2] Total buffer concentration = 0.25 M + 0.15 M = 0.4 M.
The pH can be calculated from the pKa and the concentrations given using the Henderson-Hasselbalch equation. pH = pKa + log
10.15 M2 [A ] = 3.90 + log = 3.90 - 0.22 = 3.68 [HA] 10.25 M2
8. The pKa for the ionization of H2PO4 is 7.2. The Henderson-Hasselbalch equation (Equation 2.18) indicates that when the concentrations of the acidic form 1H2PO4 2 22 are equivalent, the pH is equal to the pKa, because the and its conjugate base 1HPO4~ log term is zero 1log 1 = 02. Therefore, mixing 50 milliliters of solution A with 50 milliliters of solution B gives a buffer of pH 7.2. Since the concentration of each solution is 0.02 M, mixing equal volumes gives a buffer whose phosphate concentration is also 0.02 M. The reason why this is an effective buffer is that the final pH is at the pKa value. This means that the buffer will resist changes in pH over a considerable range. 9. (a) The effective range of a buffer is from approximately one pH unit below to one pH unit above the pKa. The buffering range for MOPS is therefore 6.2–8.2, and the buffering range for SHS is 4.5–6.5. Use the Henderson-Hasselbalch equation to calculate the ratios of basic to acidic species. [R3N] For MOPS: pH = pKa + log For SHS: [R3NH ] [RCOO ] 6.5 = 5.5 + log [RCOOH] 6.5 = 7.2 + log [R3N] [R3NH ]
=
[RCOO ] 10 = [RCOOH] 1
[R3N] [R3
NH ]
1 5
(b) An SHS buffer solution at pH 6.5 contains a much greater proportion of conjugate base relative to acid (10:1) than MOPS does (1:5). Therefore, an SHS buffer would more effectively maintain the pH upon addition of acid: H + RCOO IJ RCOOH. Conversely, a MOPS buffer at pH 6.5 contains a greater proportion of acid than SHS does; therefore, MOPS would more effectively maintain the pH upon addition of base: R3NH + OH IJ R3N + H2O. 10. (a)
O HO
O
P
O
5
CH2
OH
H
O
H
O
H
H
OH
pKa 1.2
O
P
O
CH2
OH
H
O
H
H
H
OH OH Fully protonated
OH Partially ionized (monoanion)
OH OH
12
(b)
10 pK2 = 6.6 8 pH
750
5/10/05
Second endpoint
First endpoint
6 4
pK1 = 1.2
2 0 0
0.5
1.0
1.5
2.0
Equivalents of OH
CMYK ASSOCIATES, INC. FIRST PROOF
O pKa 6.6
P
O
CH2
O
H
H
O H
H OH Fully ionized (dianion)
OH OH
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11. Excess gaseous CO2 rapidly equilibrates with aqueous CO2 (Equation 2.22), leading to formation of carbonic acid (Equation 2.21). Carbonic acid ionizes to H and HCO3 (Equation 2.20). The excess acid, in the form of H , can accumulate in bodily fluids, producing acidosis. 12. Although the metabolism of lactate and other organic acids in the diet can lead to production of CO2 as shown, CO2 is efficiently expired from the lungs (except during respiratory acidosis). Thus, the net product of the metabolic process is bicarbonate 1HCO3 2, a base. Excess H present during metabolic acidosis can be removed when it combines with HCO3 to form H2CO3 (Equation 2.20), which then forms aqueous CO2 and H2O (Equation 2.21). 13. The acidic and conjugate base species of aspirin can be represented as RCOOH and RCOO . Use the Henderson-Hasselbalch equation to calculate the ratio of the two species at pH 2.0 and pH 5.0. Then calculate the fraction of the total species that is unionized and available for absorption. In the stomach at pH 2.0, pH = pKa + log 2.0 = 3.5 + log
[RCOO ] [RCOOH]
[RCOO ] [RCOOH]
[RCOO ] 0.03 = [RCOOH] 1
The percentage of the uncharged species (RCOOH) is equal to the amount of RCOOH divided by the total of RCOOH and RCOO , times 100%. [RCOOH] [RCOOH] + [RCOO ]
* 100% =
1 * 100% = 97% 1 + 0.03
Therefore, nearly all aspirin in the stomach is in a form available for absorption. In the upper intestine at pH 5.0, however, only a small percentage of aspirin is available for absorption. 5.0 = 3.5 + log
[RCOO ] [RCOOH]
[RCOO ] 32 = [RCOOH] 1 [RCOOH]
* 100% =
[RCOOH] + [RCOO ]
1 * 100% = 3% 1 + 32
Note that aspirin must be in solution in order to be absorbed. For this reason, coated or slow-release forms of aspirin may alter the availability of aspirin in the stomach and intestine.
(a)
COO H3N
C
H
COO H
CH2
D-Phe
COO H3N
C
H
CH2SH L-Cys
NH3
CH2
L-Phe
(b)
C
COO H
C
NH3
CH2SH D-Cys
CMYK ASSOCIATES, INC. FIRST PROOF
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14. Use the Henderson-Hasselbach equation to calculate the ratio of the two species at each pH At pH = 7.5 [H2NCH2CONH2]
pH = pKa + log
7.5 = 8.2 + log
log
[+H3NCH2CONH2] [H2NCH2CONH2]
[+H3NCH2CONH2]
[H2NCH2CONH2]
= 7.5 - 8.2 = -0.7
[+H3NCH2CONH2]
[H2NCH2CONH2]
1 5
= [+H3NCH2CONH2]
The ratio of [H2NCH2CONH2] to [+H3NCH2CONH2] is 1 to 5. To determine the percent in the conjugate base form: 1/11 + 52*100 = 17%. Therefore 17% is unprotonated at pH 7.5. At pH = 8.2 [H2NCH2CONH2]
pH = pKa + log
[H2NCH2CONH2]
8.2 = 8.2 + log
log
[+H3NCH2CONH2]
[+H3NCH2CONH2]
[H2NCH2CONH2] [+H3NCH2CONH2]
[H2NCH2CONH2] = [+H3NCH2CONH2]
= 8.2 - 8.2 = 0 1 1
The ratio of [H2NCH2CONH2] to [+H3NCH2CONH2] is 1.0 to 1.0. To determine the percent in the conjugate base form: 1/11 + 12*100 = 50%. Therefore 50% is unprotonated at pH 8.2. At pH 9.0: pH = pKa + log
9.0 = 8.2 + log
log
[H2NCH2CONH2] [+H3NCH2CONH2]
[H2NCH2CONH2] [+H3NCH2CONH2]
[H2NCH2CONH2] [+H3NCH2CONH2]
[H2NCH2CONH2] = [+H3NCH2CONH2]
= 9.0 - 8.2 = 0.8 6.3 1
The ratio of [H2NCH2CONH2] to [+H3NCH2CONH2] is 6.3 to 1. To determine the percent of the conjugate base: 6.3/16.3 + 12*100 = 86%. That is, 86% is unprotonated at pH 9.0. 15. This titration curve represents a compound with two pKa values, shown by the two plateaus (near pH 2 and pH 10). Glycine has two pKa values at 2.4 and at 9.8. 16. Only (a) vitamin C would be soluble in water. Vitamin C contains several hydroxyl groups, each of which can hydrogen bond with water.
Chapter 3 Amino Acids and the Primary Structures of Proteins 1.
3 COO 1 H3N
C
H 4
CH2SH 2
CMYK ASSOCIATES, INC. FIRST PROOF
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2. The stereochemistry of each chiral carbon must be examined to determine whether it has the R or S configuration. 2 COO 1 H3N
C
2 CH(NH3)COO
H 4
4 H
OH 1
C
CH(OH)CH3 3
CH3 3
C-2, S-configuration
C-3, R-configuration
3. CH2CH2 Cl
HN
NH3Cl
NH
4. Methionine. 5. (a) Serine; phosphorylation of the hydroxyl group. (b) Glutamate; carboxylation of the g-carbon. (c) Lysine; acetylation of the e-amino group. 6. By convention, peptides are designated from the N-terminus : C-terminus, therefore Glu is the N-terminus and Gly is the C-terminus. SH COO H3NCH
CH2
O
CH2CH2C γ-Glu
NH
CH Cys
O C
NH
CH2 Gly
COO
7. The 6 residues at the C-terminus of melittin are highly hydrophilic (Table 3.1). Of the remaining 20 amino acid residues, nearly all are hydrophobic, including 9 with highly hydrophobic side chains (leucine, isoleucine, valine). The hydrophilic portion of melittin is more soluble in aqueous solution, while the hydrophobic portion is more soluble in the membrane lipids. 8. Use Table 3.2 to determine the net charge at each pKa value. The pH at which the net charge is 0 lies midway between the two pKa values at which the average charges are +0.5 and -0.5. (a) At pH 2.2, the net charge of serine is +0.5, and at pH 9.2 the net charge is -0.5. Therefore, pISer = 12.2 + 9.22/2 = 5.7. (b) At pH 9.0, the net charge of arginine is +0.5, and at pH 12.5, the net charge is -0.5. Therefore, pIArg = 19.0 + 12.52 , 2 = 10.8. (c) At pH 2.1, the net charge of glutamate is +0.5, and at pH 4.1, the net charge is -0.5. Therefore, pIGlu = 12.1 + 4.12 , 2 = 3.1. 9. The ionizable groups are the free amino group of the N-terminal cysteine residue 1pKa = 10.72, the glutamate side chain 1pKa = 4.12, and the histidine side chain 1pKa = 6.02. (a) At pH 2.0, the N-terminus and the histidine side chain have positive charges, and the glutamate side chain is uncharged. The net charge is +2. (b) At pH 8.5, the N-terminus has a positive charge, the histidine side chain is uncharged, and the glutamate side chain has a negative charge. The net charge is 0. (c) At pH 10.7, the charge of the N-terminus is +0.5, the histidine side chain is uncharged, and the glutamate side chain has a negative charge. The net charge is -0.5. 10. (a) COOH
COO
COO
H H3N
C
H3N
H
CH2
pKa = 1.8
C
H
CH2
NH HN A
COO
H
H H3N
pKa = 6.0
C CH2
NH HN B
H2N
H pKa = 9.3
C CH2
N HN C
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(b) (c) (d) (e) (f)
A, 1; B, 3; C, 5; D, 7 1, 4, 5, 7 4 5 Histidine would be a good buffer within one pH unit of any of its three pKa values: 0.8–2.8, 5.0–7.0, and 8.3–10.3.
11.
CH3
(a)
CH3 CH CH2 O
S H N
(b)
H N
C
C
C
CH3 H N
CH
COO
S C
N
H
C
H
N C
CH2OH
O
(c)
S C
N
N C O
C H
12. (a) Gly–Ala–Trp–Arg, Asp–Ala–Lys, Glu–Phe–Gly–Gln (b) Gly–Ala–Trp, Arg–Asp–Ala–Lys–Glu–Phe, Gly–Gln (c) Gly–Ala–Trp–Arg–Asp, Ala–Lys–Glu, Phe–Gly–Gln 13. (a) Because there are two N-terminal groups, there must be two peptide chains, each having an N-terminal aspartate residue. (b) 2-Mercaptoethanol reduces disulfide bonds, and trypsin catalyzes cleavage on the carboxyl side of arginine residues. Since aspartate is found at both N-termini of FP, the sequence of the dipeptide is Asp–Arg, and the sequence of the pentapeptide is Asp–(Cys, Gly, Met, Phe). The tripeptide has the sequence Cys–(Ala, Phe) and is derived from trypsin–catalyzed cleavage of a pentapeptide whose sequence is Asp–Arg–Cys–(Ala, Phe). (c) The C-terminal residue of each peptide chain is phenylalanine. Now that the terminal residues are known, one peptide must have the sequence Asp–(Cys, Gly, Met)–Phe, and the other must have the sequence Asp–Arg–Cys–Ala–Phe. (d) BrCN cleaves on the carbonyl side of methionine residues to produce C-terminal homoserine lactone residues. The peptides are therefore Asp–Met and (Cys, Gly)–Phe. Glycine is the N-terminal residue of the tripeptide, so that pentapeptide sequence is Asp–Met–Gly–Cys–Phe. The complete FP structure is Asp
Arg
Cys
Ala
Phe
S S Asp
Met
Gly
Cys
Phe
14. (a) The substitution of aspartate (D) for glutamate (E) at position 50 is an example of a conservative change. The amino acids aspartate and glutamate both contain acidic side chains that are negatively charged at physiological pH.
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(b) The substitution of tyrosine (Y) for histidine (H) is an example of a nonconservative substitution since tyrosine contains an aromatic side chain and histidine contains a hydrophilic side chain consisting of an imidazole group. 15. The neurotransmitter serotonin is derived from the amino acid tryptophan. In the conversion, the carboxyl group from tryptophan is removed and a hydroxyl group is added to the aromatic ring. COO H 3N
C
H H3N
H
C
H
CH2
CH2 HO
N H Tryptophan
N H
Serotonin
16. (a) There are two peptide bonds present in TRH. They are marked with the dashed lines. (b) TRH is derived from the tripeptide Glu-His-Pro. The proline carboxyl group has been modified to an amide (marked with an *). The side chain carboxyl group of the amino terminal Glu forms an amide with the residue’s a-amino group (marked with a **). (c) The amino- and carboxyl-terminal groups have been modified to amide groups and thus are uncharged. 17. (a) L-Dopa is in the S configuration. (b) They are both derived from the amino acid tyrosine.
Chapter 4 Proteins: Three-Dimensional Structure and Function 1. (a)
O H3N
C α1 R1
H
H
R2 C α2
C
O
H N
C
N
C
C α3
H
O
R3 H
O
(b) The R groups represent the side chains of the amino acid residues. (c) The partial double-bond character of the C ¬ N amide bonds prevents free rotation. (d) Both peptide groups in this tripeptide are in the trans conformation, since the a-carbon atoms are on opposite sides of the peptide bonds. (e) The peptide groups may rotate around the N ¬ Ca and Ca ¬ C bonds. 2. (a) (1) In an a helix, intrachain hydrogen bonds form between carbonyl oxygens of certain residues and amide hydrogens of other residues. The hydrogen bonds are approximately parallel to the helix axis (Figure 4.10). (2) In a collagen triple helix, interchain hydrogen bonds form between amide hydrogens of the glycines in one chain and carbonyl oxygen atoms of residues (which are often proline) in an adjacent chain (Figure 4.36). There are no intrachain hydrogen bonds in a collagen helix. (b) The side chains of an a helix point outward from the cylinder of the helix (Figure 4.11). In collagen, three chains coil around each other so that every third residue of a given chain makes contact with the other two chains along the central axis of the triple helix (Figure 4.34). Only the small side chain of glycine can fit at these positions. The other side chains point outward from the triple helical coil. 3. (1) The presence of glycine in an a helix destabilizes the helix due to the greater freedom of movement allowed by the small side chain. For this reason, many a helices begin or end with glycine. (2) Proline tends to disrupt a helices because its rigid, cyclic side chain stereochemically interferes with the space that would normally be occupied by a neighboring residue in the a helix. In addition, proline lacks a hydrogen on its amide nitrogen and cannot participate in normal intrahelical hydrogen bonding.
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4. (a) Due to the flexibility resulting from a small side chain 1 ¬ H2, glycine is often found in “hairpin loops” that connect sequential antiparallel b strands. The glycine residues (G) in positions 8 and 14 provide two hairpin-loop regions to connect the three b strands in Betanova. N G G COO
(b) b-sheet structures are stabilized by hydrogen bonds that form between a carbonyl oxygen of one strand and an amide nitrogen of an adjacent strand (Figure 4.15). H3N
C
N
O
H
H
O
N
C
G N
C
H
O
O
H
C
N
G COO
5. Helix-loop-helix (HLH) motif (Figure 4.19). 6. (a) a/b. Regions of a helix and b strand alternate in the polypeptide chain. (b) a/b barrel. Parallel b strands are surrounded by a layer of a helices in a cylindrical shape. (c) Yeast FMN oxidoreductase and E. coli enzyme required for tryptophan biosynthesis (Figure 4.24 (i) and (j) respectively). 7. Protein disulfide isomerase contains two reduced cysteine residues at the active site, and these participate in a reduction and disulfide exchange which allows the misfolded protein to refold into the lower energy native conformation. A
B
A
B
S
S
S
S
D
C Native
SH
SH S S
PDI SH
D
S S
PDI SH
C Non-native (1)
S PDI SH
(3)
A
B
S SH
S S
D
C
S (2)
PDI SH
A
B
S SH
SH S
D
C
8. The highly hydrophobic side chains of methionine, leucine, phenylalanine, and isoleucine are most likely to be on the side of the helix that faces the interior of the protein. Most of the other side chains are polar or charged and can interact with the aqueous solvent. Since the a helix is a repeating structure with approximately 3.6 residues per turn, the hydrophobic groups must be found every three or four residues along the sequence, so that one side of the helix is hydrophobic. 9. Covalent cross-linking contributes significantly to the strength and rigidity of collagen fibers. In one type of cross-link, allysine residues in a collagen molecule condense with lysine residues in an adjacent molecule, forming Schiff bases (Figure 4.38a). When an allysine residue reacts with homocysteine, it is unable to participate in the normal
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cross-linking of collagen molecules. High levels of homocysteine in blood probably lead to defective collagen structure and skeletal deformities. O
C
COO
O
CH
CH 2
CH 2
CH 2
+
C
H 2N
CH
H
HN
CH2
CH2
SH
Homocysteine
Allysine residue in collagen H2O
O
COO
C CH
CH 2
CH 2
CH 2
CH
N
CH
CH 2
CH 2
SH
HN
10. The sequence –Gly–Pro–X–Y– occurs frequently in collagen, which is found throughout the body, including in the skin. Because the larval enzyme can catalyze cleavage of collagen chains, the parasite is able to enter the host. 11. The reaction of carbon dioxide with water explains why there is a concomitant lowering of pH when the concentration of CO2 increases. Carbon dioxide produced by rapidly metabolizing tissue reacts with water to produce bicarbonate ions and H . (a) CO2 + H2O · H2CO3 · HCO3 + H The H generated in this reaction decreases the pH of the blood and thus stabilizes the deoxy form (T conformation) of hemoglobin. The net effect is an increase in the P50, that is, a lower affinity of hemoglobin for oxygen, so that more oxygen is released to the tissue (Figure 4.50). Carbon dioxide also lowers the affinity of hemoglobin for oxygen by forming carbamate adducts with the N-termini of the four chains (Figure 4.51). These adducts contribute to the stability of the deoxy (T) conformation, thereby further increasing the P50 and promoting the release of oxygen to the tissue. (b) Shock victims suffer a critical deficit of oxygen supply to their tissues. Bicarbonate administered intravenously provides a source of carbon dioxide to the tissues. By lowering the affinity of hemoglobin for oxygen, carbon dioxide facilitates a release of oxygen from oxyhemoglobin to the tissues. 12. (a) 2,3BPG binds to positively charged side chains in the central cavity of deoxyhemoglobin (Figure 4.49). Since Hb F lacks two positively charged groups (His-143 of each b chain), 2,3BPG binds less tightly to Hb F than to Hb A. (b) 2,3BPG stabilizes the deoxy form of hemoglobin, increasing the fraction of molecules in the deoxy form. Since Hb F binds 2,3BPG less tightly than does Hb A, Hb F is less affected by 2,3BPG in the blood and has a greater fraction of molecules in the oxy form. Hb F therefore has a greater affinity than Hb A for oxygen at any oxygen pressure. (c) At the oxygen pressure of tissues, 20 –40 torr, Hb F has a greater affinity for oxygen than does Hb A. The difference in affinity allows efficient transfer of oxygen from maternal blood to the fetus. 13. The low P50 value of HbYakima indicates a greater than normal affinity for oxygen even at the oxygen pressures found in working muscle. The increased affinity means that HbYakima gives up less oxygen to the working muscle. 14. (a) The most effective binding of selenoprotein P to heparin is seen at a pH below 6. The binding of selenoprotein P to heparin decreases as the pH is increased to 7. There is very little binding of selenoprotein P to heparin at pH values greater than 7. (b) Heparin is negatively charged. If selenoprotein P is positively charged, it can bind to heparin. Histidine residues are abundant in selenoprotein P. Histidine has an imidazole side chain that has a pKa value of 6.0. That is, at a pH of 6.0, 50% of the histidine residues would be protonated and positively charged and 50% would be unprotonated and uncharged. Below a pH of 6.0, there would be a net positive charge on the histidine residues, resulting in effective electrostatic interactions with
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the heparin. At pH values above 7, almost all of the histidine residues would be unprotonated and uncharged and will not effectively interact with the negativelycharged heparin molecule. 15. (a) Hydrophilic (italicized) and hydrophobic (underlined) residues are identified: ECGKFMWKCKNSNDCCKDLVCSSRWKWCVLASPF (b) In the three dimensional structure of proteins, amino acids that are far from each other in the primary sequence can interact in the globular structure of the protein. Thus the hydrophobic amino acids can be very close to each other in the threedimensional structure and provide a “hydrophobic” face for interaction with the membrane. 16. The replacement of lysine by methionine results in one less positive charge on each beta subunit in the central cavity (see Figure 4.49). 2,3BPG binds less tightly to HbH. This causes more of the mutant protein to be in the R state (oxy hemoglobin is stabilized). The curve is shifted towards the left (more like myoglobin). Since more is in the R state, the affinity for oxygen has increased.
Chapter 5 Properties of Enzymes 1. (a) When [S] W Km, Km + [S] L [S]. Substrate concentration has no effect on velocity, and v0 = Vmax, as shown in the upper part of the curve in Figure 5.3a. v0 =
Vmax[S]
Vmax[S] Km + [S]
L
[S]
= Vmax
(b) When [S] V Km, Km + [S] L Km, and the Michaelis-Menten equation simplifies to v0 =
Vmax[S]
Vmax[S] Km + [S]
L
Km
Velocity is related to [S] by a constant value, and the reaction is first order with respect to S, as shown in the lower part of the curve in Figure 5.3a. (c) When v0 = Vmax/2, Km = [S]. v0 =
Vmax 2
Vmax[S] =
Km + [S]
Km + [S] = 2[S] Km = [S]
2. The initial velocities are approaching a constant value at the higher substrate concentrations, so we can estimate the Vmax as 70 mM/min. Since Km equals the concentration of substrate [S] required to reach half the maximum velocity, we can estimate the Km to be 0.1 mM from the 35 mM/min initial-velocity value. 3. (a) The ratio kcat/Km, or specificity constant, is a measure of the preference of an enzyme for different substrates. When two substrates at the same concentration compete for the active site of an enzyme, the ratio of their rates of conversion to product is equal to the ratio of the kcat/Km values, since v0 = 1kcat/Km2[E][S] for each substrate and [E] and [S] are the same. v01S12 v01S22
=
1kcat>Km21>[E][S]
1kcat>Km22>[E][S]
(b) The upper limit of kcat/Km approaches 108 to 109 s-1, the fastest rate at which two uncharged molecules can approach each other by diffusion at physiological temperatures. (c) The catalytic efficiency of an enzyme cannot exceed the rate for the formation of ES from E and S. The most efficient enzymes have kcat/Km values approaching the rate at which they encounter a substrate, and at this limiting velocity they have become as efficient catalysts as possible.
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4. The catalytic constant 1kcat2 is the first-order rate constant for the conversion of ES to E + P under saturating substrate concentrations (Equation 5.26), and CA has a much higher catalytic activity in converting substrate to product than does OMPD. However, the efficiency of an enzyme can also be measured by the rate acceleration provided by the enzyme over the corresponding uncatalyzed reaction (kcat/kn Table 5.2). The reaction of the substrate for OMPD in the absence of enzyme is very slow 1kn = 3 * 10-16 s-12 compared to the reaction for the CA substrate in the absence of enzyme 1kn = 1 * 10-1 s-12. Therefore, while the OMPD reaction is much slower than the CA reaction in terms of kcat, OMPD is one of the most proficient enzymes known and provides a much higher rate acceleration than does CA when the reactions of each enzyme are compared to the corresponding uncatalyzed reactions. 5. When [S] = 100 mM, [S] W Km, so v0 = Vmax = 0.1 mM min-1. (a) For any substrate concentration greater than 100 mM, v0 = Vmax = 0.1 mM min-1. (b) When [S] = Km, v0 = Vmax/2, or 0.05 mM min-1. (c) Since Km and Vmax are known, the Michaelis-Menten equation can be used to calculate v0 at any substrate concentration. For [S] = 2 mM, v0 =
Vmax[S] Km + [S]
=
10.1 mM min-1212 mM2 11 mM + 2 mM2
=
0.2 mM min-1 = 0.067 mM min-1 3
6. (a) Determine [E]total in moles per liter, then calculate Vmax. [E]total = 0.2 g l-1 a
1 mol b = 9.3 * 10-6 M 21 500 g
Vmax = kcat[E]total = 1000 s-119.3 * 10-6 M2 = 9.3 * 10-3 M s-1
(b) Since Vmax is unchanged in the presence of the inhibitor, competitive inhibition is occurring. Because the inhibitor closely resembles the heptapeptide substrate, competitive inhibition by binding to the enzyme active site is expected (i.e., classical competitive inhibition). 7. Curve A represents the reaction in the absence of inhibitors. In the presence of a competitive inhibitor (curve B), Km increases and Vmax is unchanged. In the presence of a noncompetitive inhibitor (curve C), Vmax decreases and Km is unchanged. A B v0 C
[S]
8. Since the inhibitor sulfonamides structurally resemble the PABA substrate we would predict that sulfonamides bind to the enzyme active site in place of PABA and act as competitive inhibitors (Figure 5.9). + Sulfonamide No sulfonamide 1 v0
1 [PABA]
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9. (a) To plot the kinetic data for fumarase first calculate the reciprocals of substrate concentrations and initial rates of product formation. (Note the importance of including correct units in calculating and plotting the data.) Fumarate [S] (mM)
1 1mM-12 [S]
Rate of product formation v0 1mmol l1 min12
1 1mmol1 l min2 v0
0.50 0.30 0.20 0.10
2.5 3.1 3.6 4.2
0.40 0.32 0.28 0.24
02.0 03.3 05.0 10.0
1 −1 v0 (mmol l min)
0.5 0.4 0.3 0.2 − 1 Km
1 Vmax
0.1
− 0.5 − 0.4 − 0.3 − 0.2 − 0.1
0
0.1
0.2 0.3 0.4 1 (mM−1) [S]
0.5
Vmax is obtained by taking the reciprocal of 1/Vmax from the y intercept (Figure 5.6). 1/Vmax = 0.20 mmol-1 l min, so Vmax = 5.0 mmol l-1 min-1
Km is obtained by taking the reciprocal of -1/Km from the x intercept. -1/Km = -0.5 mM-1, so Km = 2.0 mM or 2 * 10-3 M
(b) The value of kcat represents the number of reactions per second that one enzyme active site can catalyze. Although the concentration of enzyme is 1 * 10-8 M, fumarase is a tetramer with four active sites per molecule so the total concentration of enzyme active sites [Etotal] is 4 * 10-8 M. Using Equation 5.26: kcat =
Vmax [Etotal]
5.0 mmol l-1 min-1 =
* 4 * 10-5 mmol l-1
1 min = 2 * 103 s-1 60 s
10. Both APHS and aspirin transfer an acetyl group to a serine residue on COX enzymes. Since APHS is an irreversible inhibitor, it does not exhibit competitive inhibition kinetics even though it acts at the active site of COX enzymes. O (COX-2)
CH2OH Active site serine
O
HO O
CH 3 (COX-2)
S CH2C APHS
C(CH2)3CH3
CH2O
C
CH3
+
S CH2C
C(CH2)3CH3
Irreversibly inhibited enzyme
11. Like pyruvate dehydrogenase (PDH) (Figure 5.25), glycogen phosphorylase (GP) activity is regulated by alternate phosphorylation by a kinase and dephosphorylation by a phosphatase. However, unlike PDH, the active form of GP has two phosphorylated serine residues; in the inactive GP form, two serine residues are not phosphorylated.
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(2) ATP
(2) ADP GP Kinase
OH
O P
Glycogen phosphorylase
Glycogen
Glycogen phosphorylase O P
OH (less active)
G1P
(more active) GP Phosphatase
(2)
(2) H2O
Pi
12. Inhibition of the first committed step of a multistep pathway allows the pathway to proceed only when the end product is needed. Since the first committed step is regulated, no intermediates accumulate. This type of regulation conserves raw material and energy. 13. When [aspartate] = 5 mM, v0 = Vmax/2. Therefore, in the absence of allosteric modulators, Km = [S] = 5 mM. ATP increases v0, and CTP decreases v0.
+ATP Native ATCase v0
0
+ CTP
5
10 15 [Aspartate] (mM)
20
14. Collagen is protein consisting of three polypeptide chains that are wound together in a triple helix. The protease bromelin is an enzyme that cleaves some of the peptide bonds in the polypeptide chains. The polypeptide chains are necessary to trap the water molecules in a semi solid state when gelatin cools and if these are cleaved, the gelatin will not set properly. The cleavage of the polypeptide chains in collagen by bromelin destroys the ability of the gelatin to harden. If the pineapple is first cooked, the heat will denature the protein and thus the enzyme activity will be destroyed. Therefore cooked pineapple can be added to slightly thickened gelatin, and the gelatin will proceed to the semi solid state as desired. 15. (a) To plot the kinetic data for P450 3A4, first calculate the reciprocals of substrate concentrations and initial rates of product formation. The data are plotted in the double reciprocal plot and are shown with the dashed line. Midazolam [S] 1mM2
1/[S] 1mM-12
Rate of product formation v0 1pmol l-1 min-12
1
10.00
100
0.0100
2
0.500
156
0.0064
4
0.250
222
0.0045
8
0.125
323
0.0031
1/v0 1pmol-1 l min2
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0.08 0.06
0.04 0.02 0.4 0.2 0.02
0.2
0.4
0.6
0.8
1.0
1.2
1/[S](?M1)
Vmax is obtained by taking the reciprocal of 1/Vmax from the y intercept (Figure 5.6). 1/Vmax - 0.0025 pmol-1 l min, so Vmax = 400 pmol l-1 min-1
Km is obtained by taking the reciprocal of -1/Km from the x intercept -1/Km = -0.3 mM-1, so Km = 3.3 mM
(b) The reciprocals of the substrate concentration and activity in the presence of ketoconazole are given in the table.
Midazolam [S] 1mM2
1/[S] 1mM-12
Rate of product formation in the presence of 0.1 mM ketoconazole v0 1pmol l-1 min-12
1 2 4 8
1.000 0.500 0.250 0.125
11 18 27 40
1/v0 1pmol-1 l min2 0.091 0.056 0.037 0.025
The plot of the data (solid line) is given in the double reciprocal plot shown in (a). There is an increase in the y intercept and no apparent change in the x intercept. From the double reciprocal plot, it appears that ketoconazole is a noncompetitive inhibitor (see Figure 5.12). These inhibitors are characterized by an apparent decrease in Vmax (increase in 1/Vmax) with no change in Km. 16. (a) Bergamottin appears to inhibit the activity of P450 3A4 since the P450 activity measured in the presence of 0.1 and 5 mM bergamottin is less than that of the P450 activity in the absence of bergamottin. (b) It might be dangerous for a patient to take their medication with grapefruit juice since there appears to be an inhibition of P450 activity in the presence of bergamottin. If the bergamottin decreases the P450 activity, and the P450 enzyme is known to metabolize the drug to an inactive form, the time it takes to convert the drug to its inactive form maybe increased. This may prolong the effects of the drug which may lead to adverse consequences for the patient.
Chapter 6 Mechanisms of Enzymes 1. (a) The major binding forces in ES complexes include charge-charge interactions, hydrogen bonds, hydrophobic interactions, and van der Waals forces. (About 20% of enzymes bind a substrate molecule or part of it covalently.) (b) Tight binding of a substrate would produce an ES complex that lies in a thermodynamic pit, effectively increasing the activation energy and thereby slowing down the reaction. Tight binding of the transition state, however, lowers the energy of the ES‡ complex, thereby decreasing the activation energy and increasing the rate of the reaction.
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2. The activation barrier for the reaction is lowered by (1) raising the ground state energy level (ES) and (2) lowering the transition state energy level 1ES‡2, resulting in a reaction rate increase.
Free energy
S++ (2)
Uncatalyzed reaction barrier
ES++ ES
E+S
Catalyzed reaction barrier EP
(1)
E+P
Course of the reaction
3. The rate-determining step of a multistep reaction is the slowest step, which is the step with the highest activation energy. For Reaction 1, Step 2 is the rate-determining step. For Reaction 2, Step 1 is the rate-determining step. 4. The reactive groups in Reaction 2 ( ¬ OH and ¬ COOH) are held at close proximity. They are oriented in a manner suitable for catalysis by steric crowding of the bulky methyl groups of the ring. The reactive —COOH group cannot rotate away as freely as it can in Reaction 1. Model systems such as these are relevant because they indicate potential rate increases that might be obtained by enzymes that bring substrates and the enzyme’s catalytic groups into positions that are optimal for reaction. 5. (a) Proximity effect. Lysozyme binds the substrate so that the glycosidic bond to be cleaved is very close to both of the enzyme catalytic groups (Glu-35 and Asp-52). (b) Substrate distortion. The energy of the ground-state sugar ring is raised because it is distorted into a half-chair conformation. (c) Acid-base catalysis. Glu-35 first donates a proton to an oxygen of the leaving sugar (general acid catalysis), and then accepts a proton from the attacking water molecule (general base catalysis). (d) Transition-state stabilization. Asp-52 stabilizes the developing positive charge on the oxocarbocation intermediate, and subsite D favors the half-chair sugar conformation of this intermediate. The structure proposed for the transition state includes both this charge and sugar conformation in addition to hydrogen bonding to several active-site residues. 6. Serine 195 is the only serine residue in the enzyme that participates in the catalytic triad at the active site of a-chymotrypsin. The resulting increase in the nucleophilic character of Ser-195 oxygen allows it to react rapidly with DFP. 7. (a) The catalytic triad is composed of an aspartate, a histidine, and a serine residue. Histidine acts as a general acid-base catalyst, removing a proton from serine to make serine a more powerful nucleophile in the initial step. Aspartate forms a low-barrier hydrogen bond with histidine, stabilizing the transition state. An acid catalyst, histidine donates a proton to generate the leaving amine group. (b) The oxyanion hole contains backbone ¬ NH ¬ groups that form hydrogen bonds with the negatively charged oxygen of the tetrahedral intermediate. The oxyanion hole mediates transition-state stabilization since it binds the transition state more tightly than it binds the substrate. (c) During trypsin catalysis, aspartate forms a low-barrier hydrogen bond with the imidazolium form of histidine. Because asparagine lacks a carboxylate group to form the stabilizing hydrogen bond with histidine, enzyme activity is dramatically decreased.
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8.
(a) Human cytomegalovirus protease: His, His, Ser His
His N
..
HN
Ser
HN
HO
N
CH2
(b) β-Lactamase: Glu, Lys, Ser Glu
Ser
..
COO
Lys HN H
HO
CH2
(c) Asparaginase: Asp, Lys, Thr Asp
Thr
..
COO
Lys HN H
HO
CH
CH3
(d) Hepatitis A protease: Asp H
O
HN
Cys
..
COO
His
H
HS
N
CH2
9. In the first step, Asp-25 acts as a base and Asp-25¿ acts as an acid. In the second step, the roles of the two residues are reversed, leading to formation of the acid and amine products without a covalent enzyme-substrate intermediate. O C
O
H
H
R2
O
C
O O
HO
NHR1
Asp-25
C
Asp-25′
R2
HO O
O
C OH
NHR1 OH
O
Asp-25
C Asp-25′
R2
HO
O C
C
O
Asp-25
C O
O HO
C Asp-25′
+ H2NR1
10. Asp-25 1pKa = 3.32 must be unprotonated 1RCOO 2 to act as a base, while Asp-25¿ 1pKa = 5.32 must be protonated (RCOOH) to donate a proton to a peptide carbonyl group. The aspartate residues exist predominantly in the correct forms in the pH region between the two pKa values.
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pKa = 3.3 pKa = 5.3 2
3
4
5
6
7
pH
11. Invirase resembles the tetrahedral intermediate in the hydrolysis of peptides by HIV protease. However, invirase possesses a nonhydrolyzable C ¬ C bond in place of the susceptible C ¬ N peptide bond of a substrate molecule. Invirase thus acts as a transition state analog, and binds much more tightly to the active site than does a corresponding substrate. 12. The derivative is a transition-state analog for lysozyme. The conformation of the lactone ring of the inhibitor resembles the conformation of the sugar residue in the D site of lysozyme. In addition, the resonance form of the derivative is a positively charged oxonium ion, which is analogous to the oxocarbocation transition state (below) that is stabilized by the negatively charged aspartate at the active site. CH 2 OH
H (GlcNAc)3
H
O
HO
HO
O
NH H O
C CH 3
13. When tyrosine was mutated to phenylalanine, the activity of the mutant enzyme was less than 1% of the wild type enzyme. Thus, the tyrosine residue is involved in the catalytic activity of DDP-IV. Tyrosine contains an -OH group on the aromatic ring of the side chain. As previously stated, this tyrosine is found in the oxyanion hole of the active site. Hydrogen bonds in the oxyanion hole of serine proteases are known to stabilize the tetrahedral intermediate. Tyrosine with an -OH group on the side chain can hydrogen bond and stabilize the tetrahedral intermediate. Phenylalanine does not have a side chain that can hydrogen bond. Therefore the tetrahedral intermediate will not be stabilized resulting in a loss of enzyme activity. 14. (a) Acetylcholinesterase catalytic triad: Glu-His-Ser His
Glu
Ser N
..
HN
COO
(b)
HO
CH2
Ser H F
O
O P
i-PrO
OCH3
Ser
CH2 H
O
O
CH2
F P i-PrO OCH3
Ser O
O
CH2
P i-PrO
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OCH3
+ F
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15. Transition state analogs bound to carrier proteins are used as antigens to induce the formation of antibodies with catalytic activity. The tetrahedral phosphonate ester molecule is an analog of the tetrahedral intermediate structure in the transition state for hydrolysis of the benzyl ester moiety of cocaine. An antibody raised against the phosphonate structure that was able to stabilize the transition state of the cocaine benzyl ester hydrolysis could effectively catalyze this reaction. O H3C
N
OCH3 O
O OH
Transition state
16. (a) Wild-type a1-proteinase inhibitor is given as treatment to individuals who produce an a1-proteinase inhibitor with substitutions in the amino acid sequence. These changes result in a protein that does not effectively inhibit the protease elastase. Uncontrolled elastase activity leads to increased breakdown of elastin leading to destructive lung disease. Therefore, these patients are given a functional elastase inhibitor. (b) The treatment for a1-proteinase inhibitor deficiency is to administer the wild-type protein intravenously. If the protein is given orally, the enzymes present in the digestive tract will cleave the peptide bonds in the a1-proteinase inhibitor. By administering the drug directly into the bloodstream, the protein can circulate to the lungs to act at the site of the neutrophil elastase.
Chapter 7 Coenzymes and Vitamins 1. (a) NAD , NADP , FAD, FMN, lipoamide, ubiquinone. Protein coenzymes such as thioredoxin and the cytochromes. (b) Coenzyme A, lipoamide. (c) Tetrahydrofolate, S-adenosylmethionine, methylcobalamin (d) Pyridoxal phosphate (e) Biotin, thiamine pyrophosphate, vitamin K 2. (a) Oxidation; NAD , FAD, or FMN. (The coenzyme for the reaction shown is NAD .) (b) Decarboxylation of an a-keto acid; thiamine pyrophosphate. (c) Carboxylation reaction requiring bicarbonate and ATP; biotin. (d) Molecular rearrangement; adenosylcobalamin. (e) Transfer of a hydroxyethyl group from TPP to CoA as an acyl group; lipoic acid. 3. No. NAD acquires two electrons but only one proton. The second proton is released into solution, and is reutilized by other proton-requiring reactions. 4.
H H3C H3C
O
N
NH
5
1
N
N
R
H
O
5. NAD , FAD, and coenzyme A all contain an ADP group (or ADP with 3¿-phosphate for coenzyme A).
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6.
N Isoniazid addition H
C
O CONH2
4 5
NAD
6
3
..
2
1
N
R
7. Vitamin B6 is converted to pyridoxal phosphate, which is the coenzyme for a large number of reactions involving amino acids, including the decarboxylation reactions in the pathways that produce serotonin and norepinephrine from tryptophan and tyrosine, respectively. Insufficient vitamin B6 can lead to decreased levels of PLP and a decrease in the synthesis of the neurotransmitters. 8. Vitamin K, which is required for the modification of some of the proteins involved in blood coagulation. 9. The synthesis of thymidylate (dTMP) requires a tetrahydrofolate (folic acid) derivative. Deficiency of folic acid decreases the amount of dTMP available for the synthesis of DNA. Decreased DNA synthesis in red blood cell precursors results in slower cell division, producing macrocytic red blood cells. The loss of cells by rupturing causes anemia. 10. (a) Cobalamin. (b) The cobalamin derivative adenosylcobalamin is a coenzyme for the intramolecular rearrangement of methylmalonyl CoA to succinyl CoA (Figure 7.25). A deficiency of adenosylcobalamin results in increased levels of methylmalonyl CoA and its hydrolysis product, methylmalonic acid. Another cobalamin derivative, methylcobalamin, is a coenzyme for the synthesis of methionine from homocysteine (Reaction 7.5), and a deficiency of cobalamin results in an excess of homocysteine and a deficiency of methionine. (c) Plants do not synthesize cobalamin and are therefore not a source of this vitamin. 11. (a) In one proposed mechanism, a water molecule bound to the zinc ion of alcohol dehydrogenase forms OH-, in the same manner as the water bound to carbonic anhydrase (Figure 7.2). The basic hydroxide ion abstracts the proton from the hydroxyl group of ethanol to form H2O. (Another mechanism proposes that the zinc also binds to the alcoholic oxygen of the ethanol, polarizing it.)
O R N
NH 2 H
NAD (oxidized coenzyme)
H
2
O
O
H
Zn
H
C CH3 Ethanol
H2O
H2O
O R N
..
NH 2
2
H NADH (oxidized coenzyme)
H
Zn H
O C CH3
Acetaldehyde
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(b) No, a residue such as arginine is not required. Ethanol, unlike lactate, lacks a carboxylate group that can bind electrostatically to the arginine side chain. 12. A carboxyl group is transferred from methylmalonyl CoA to biotin to form carboxybiotin and propionyl CoA. O C O
O N
O NH
+
HN
CH2
C
S-CoA
Enz
S Carboxybiotin
13. (a)
CH3
N CH2
H COO C
H
N C
2
H O
O3PO H2C N H
CH 3
(b) Racemization would not occur. Although a Schiff base forms during decarboxylation as well as racemization, the reactive groups in the histidine decarboxylase active site specifically catalyze decarboxylation, not racemization, of histidine. 14. (a)
HETPP TPP
CH3
Acetyl-TPP TPP
CH
OH
CH3
+
C
O
+
+ R
S
TPP
HS
S
Lipoamide
R
R H3C
SH
C
S
SH
O Acetyl-dihydrolipoamide
Dihydrolipoamide
(b) TPP HOCH2
..C
HOCH2 HC
..
C
OH
H
2
CH2OPO3
O +
HOCH OH OH
OH OH
CH2OPO3
2
CH2OPO3 F-6-P
E-4-P
Chapter 8 Carbohydrates D-Glucose
C
HOC
O OH OH
1. (a) (b) (c) (d) (e) (f) (g)
CH2OH
TPP
OH
and D-mannose
L-Galactose D-Glucose
or D-talose Dihydroxyacetone Erythrulose (either D or L) D-Glucose N-Acetylglucosamine
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2
TPP
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2. (a)
(b)
O
H
H
C
(c)
O
H
C
OH
HO
C
H
H
C
OH
H
C
OH
HO
C
H
H
C
OH
HO
C
H
HO
C
H
(d)
CH2OH
C
H
C
OH
HO
C
H
H
C
OH
H
O C
CH2OH
HO
C
H
H
C
OH
HO
C
H
H
C
OH
CH3
CH2OH
COO
3. Glycosaminoglycans are unbranched heteroglycans of repeating disaccharide units. One component of the disaccharide is an amino sugar and the other component is usually an alduronic acid. Specific hydroxyl and amino groups of many glycosaminoglycans are sulfated 4.
CH2OH HOCH2
O
H
(a)
HO
H
CH2OH OH
C
O
HO
C
H (a) (b)
H
C
OH
H
OH H
(b)
O H H
OH
HO CH2OH
HO
H
H
C
OH
OH
H
CH2OH β-D-Fructofuranose
5. (a) (b) (c) (d) (e)
β-D-Fructopyranose
D-Fructose
a-Anomer Yes, it will mutorotate Yes, it is a deoxy sugar A ketone Four chiral carbons
6. Glucopyranose has five chiral carbons and 25, or 32, possible stereoisomers; 16 are D sugars and 16 are L sugars. Fructofuranose has four chiral carbons and 24, or 16, possible stereoisomers; 8 are D sugars and 8 are L sugars. 7. (a) H * HO
CH2OH * O H OH H * H
* OH
(b)
2
O3 PO CH 2
H * O
O P
* H H * OH
O
O
(c) O
H
OH H H
* H
(d)
O
COO
C H
*C
HO
*C
H
H
*C
OH
HO
*C
H
HO
*C
H
OH
H2 COPO3
2
CH2OH
8. Only the open-chain forms of aldoses have free aldehyde groups that can form Schiff bases with amino groups of proteins. Because relatively few molecules of D-glucose are found in the open-chain form, D-glucose is less likely than other aldoses to react with proteins. 9. A pyranose is most stable when the bulkiest ring substituents are equatorial, minimizing steric repulsion. In the most stable conformer of b- D-glucopyranose, all the hydroxyl groups and the ¬ CH2OH group are equatorial; in the most stable conformer of a- Dglucopyranose, the C-1 hydroxyl group is axial. 10.
O O
P
4
O
O H
5
OCH 2 H
(B) 1
H
2
OH H
3
HO Envelope conformation
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11. The a and b anomers of glucose are in rapid equilibrium. As b- D-glucose is depleted by the glucose oxidase reaction, more b anomer is formed from the a anomer until all the glucose has been converted to gluconolactone. 12. Sucralose is a derivative of the disaccharide sucrose (see figure 8.20). The two hydroxyl groups on C-1 and C-6 of the fructose molecule have been replaced with chlorine. The hydroxyl group on C-4 of the glucose molecule was removed and then chlorine added. In the chemical synthesis of sucralose from sugar, the configuration of the C-4 substituent of the glucose moiety is reversed. 13. (a)
CH2OH H
O H OH
CH2OH
H H
H
HO
O H
N
(b)
HO H
O
O H H
H H
H OH
O
HO
H OH
H H
H
CH2
CH2OH CH
O
O
HO
HO
NH3
C
CH2
H
HO
CH2 H OH
O H OH
(c)
H
CH2
NH
OH H
H OH
H
HO
CH
H
H
HO
O
CH2 CH
C O
HO
14. H2C
OSO3 O
H OH
CH2OH
H H
O
O
HO
OH
H
H OH
H
H
OH
OH
H
15. (a) (b) (c) (d)
a, b, and c; these oligosaccharides contain GlcNAc ¬ Asn bonds. b and c; these oligosaccharides contain b-galactosidic bonds. b; this oligosaccharide contains sialic acid. None, since none of the oligosaccharides shown contains fucose.
16.
9
CH3 O
C HN 5
CH2OH H 8 CHOH 1 O COOH 7 CHOH H 2 H
6
H 4
3
H
OH
HO
CH2OH H
CH2OH O H O
H
H H
α (2→3) linkage
(X) H
O
H
H
O H OH OH
OH β (1→4) linkage
Chapter 9 Lipids and Membranes 1. (a) CH3(CH 2)7
H
H
C
C
(b) (CH2)13COO
CH3(CH 2)5
H
H
C
C
(c) (CH2)9COO
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H CH3CH 2(C
H CCH2)5(CH 2)2COO
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2.
(a) CH3CH 2CH 2
O
(b)
(CH2)9COO
CH3
CH 2
(c)
CH3
CH3(CH 2)5CH
CH3(CHCH 2CH 2CH 2)3CHCH 2COO
3. (a) v-3; (b) v-6; (c) v-6; (d) neither 1v-92; (e) v-6. 4.
O O
CH3
P
O
CH2CH2N
O H 2C O
CH
CH3
CH3
CH 2
O C
O
CH3
5. (a)
H3C HOOC
(b) Docosahexaenoic acid is classified as an v-3 fatty acid 6. O O
P O
1
H 2C
O
2
CH
O
O
C
C
O
NH3 O
O
CH2CH COO
CH 2
O
H 2C
CH
O
OH
(R 2 ) PS
CH2CH COO
CH 2 O C
C
Phospholipase A2
(R 1)
O
O
3
O
P
NH3
O
+
(R 2)
(R 1) A lysolecithin
Fatty acid
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CH(CH 2)9COO
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7. (a)
CH3
(b)
NH3
N
H3C
CH2
P
1
O
2
O
O
O
C
C
H
H
O
H
H
C
C (CH2)12
(H2C)12
OH
H
CH3
3
CH
CH
O
H
C
H
C
NH
C (CH2)12
(H2C)14
C
C
NH
OH
O
H
O
CH
CH3
2
H2C
CH3 C
β
3
CH
O
1
(H2C)16 (CH2)7
H OH
2
H2C
HO
O
CH2
1
O
H
P
3
CH
H
O
O
O H2C
CH3
CH2
O
OH CH 2 OH
CH2
CH2
O
(c)
CH3
CH3
(CH2)7 CH3
8. PE contains docosahexaenoic acid at position C-2 on the glycerol-3-phosphate backbone at both temperatures. At lower temperatures, the percent of the monounsaturated fatty acyl groups at position C-1 increased from 14% at 30°C to 39% at 10°C. The membrane fluidity must be maintained for the organism, and this is accomplished by changing the composition of the membrane lipids. The increase in the unsaturated lipids at the lower temperature will allow for the proper membrane fluidity. 9. CH2 OH
(a)
(b)
O O
HO
C
O OH
17
16
12 11
HO HO 19 HO CH2 1
13 1 10
O HO
OH
3
O
17
14 15 8
5 4
OH
16
13
11
10
CH3
6
18
9
2
7
5 4
15
8
3
O
14
9
2
12
OH 7
6
HO
10. Farnesyl transferase adds a farnesyl or “prenyl” group to a cysteine side chain of the ras protein (Figure 9.29b). The ras protein is subsequently anchored to the plasma and endoplasmic reticulum membranes and is active in cell signaling processes. Farnesyl transferase is a chemotherapy target because inhibition of this enzyme in tumor cells would disrupt the signaling activity of the mutated ras protein. In fact, farnesyl transferase (FT) inhibitors are potent suppressors of tumor growth in mice. 11. Line A represents diffusion of glucose through a channel or pore, and line B represents passive transport. Diffusion through a channel or pore is generally not saturable, with the rate increasing linearly with the concentration of the solute. Transport via a transport protein is saturable at high solute concentrations, much like an enzyme is saturated at high substrate concentrations (Section 9.11 B). 12. Theobromine is structurally related to caffeine and theophylline (Figure 9.42). The methylated purines, including the obromine, inhibit cAMP phophodiesterase, a soluble enzyme that catalyzes the hydrolysis of cAMP to AMP (Figure 9.40). These methylated purines inhibit the breakdown of the intracellular messenger cAMP to AMP. Therefore,
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the effects of the cAMP are prolonged. For dogs, this is combined with the fact that they have slower clearance of the ingested theobromine from their system. Both of these result in the toxicity associated with ingesting the chocolate. 13.
Cl
K
K
H ADP + Pi
ATP
Gastric mucosal cells
Stomach Cl
K
K
H
HCl
14. The two second messengers IP3 and DAG are complementary in that they both promote the activation of cellular kinases, which then activate intracellular target proteins by causing their phosphorylation. Diacylglycerol activates protein kinase C directly, 2+ 2+ whereas IP3 elevates Ca~ levels by opening a Ca~ channel in the membrane of the en2+ doplasmic reticulum releasing stored Ca~ into the cytosol (Figure 9.46). The increased 2+ levels activate other kinases leading to a phosphorylation and activation of certain Ca~ target proteins. 15. Insulin can still bind normally to the a subunits of the insulin receptor, but due to the mutation, the b subunits lack tyrosine-kinase activity and cannot catalyze autophosphorylation or other phosphorylation reactions. Therefore, insulin does not elicit an intracellular response. The presence of more insulin will have no effect. 16. G proteins are molecular switches with two interconvertible forms, an active GTPbound form and an inactive GDP-bound form (Figure 9.39). In normal G proteins, GTPase activity converts the active G protein to the inactive form. Because the ras protein lacks GTPase activity, it cannot be inactivated. The result is continuous activation of adenylyl cyclase and prolonged responses to certain extracellular signals.
Chapter 10 Introduction to Metabolism 1. (a) − A −
B
C
D
+
−
F
G
−
H
I
E J
(b) Inhibition of the first step in the common pathway by either G or J prevents the needless accumulation of intermediates in the pathway. When there is ample G or J, fewer molecules of A enter the pathway. By regulating an enzyme after the branch point, G or J inhibits its own production without inhibiting production of the other. 2. Compartmentalizing metabolic processes allows optimal concentrations of substrates and products for each pathway to exist independently in each compartment. In addition, separation of pathway enzymes also permits independent regulation of each pathway without interference by regulators from the other pathway. 3. In a multistep enzymatic pathway, the product from one enzyme will be the substrate for the next enzyme in the pathway. For independent soluble enzymes, the product of each enzyme must find the next enzyme by random diffusion in solution. By having sequential enzymes located in close proximity to each other, either in a multienzyme complex or on a membrane, the product of each enzyme can be passed directly on to the next enzyme without losing the substrate by diffusion into solution. 4. (a) ¢G°¿ = RT ln Keq ln Keq = -
¢G°¿ -9000 J mol-1 = 3.63 = RT 18.315 J K-1 mol-121298 K2
Keq = 38
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(b) ¢G°¿ = -RT ln Keq Keq =
[Glucose][Pi] [Glucose 6-P][H2O]
=
10.1 M210.1 M2
13.5 * 10-5 M2112
= 286
¢G°¿ = -18.315 JK-1 mol-121298 K2 ln 286 ¢G°¿ = -14 000 J mol-1 = -14 KJ mol-1
5. (a) ¢G = ¢G°¿ + RT ln
[Arginine][Pi] [Phosphoarginine][H2O]
¢G = -32 000 J mol-1 + 18.315 J K-1 mol-121298 K2ln
12.6 * 10-3215 * 10-32 16.8 * 10-32112
¢G = -48 kJ mol-1
(b) ¢G°¿ is defined under standard conditions of 1 M concentrations of reactants and products. (The concentration of water is assigned a value of 1.) ¢G depends on the actual concentrations of the reactants and products. (c) Molecules with high free energies of hydrolysis, such as phosphoarginine and acetyl CoA, are thermodynamically unstable but may be kinetically stable. These molecules are hydrolyzed very slowly in the absence of an appropriate catalyst. ¢G°¿1kJ mol-12 0 -33.5 ¢G°¿ -33.5
6. Glucose 1-phosphate + UTP ¡ UDP-glucose + PPi PPi + H2O ¡ 2 Pi
7. (a) Although ATP is rapidly utilized for energy purposes such as muscle contraction and membrane transport, it is also rapidly resynthesized from ADP and Pi through intermediary metabolic routes. Energy for this process is supplied from the degradation of carbohydrates, fats, and amino acids or from energy storage molecules such as muscle creatine phosphate 1CP + ADP : ATP + C2. With this rapid recycling, 50 grams total of ATP and ADP is sufficient for the chemical energy needs of the body. (b) The role of ATP is that of a free energy transmitter rather than an energy storage molecule. As indicated in part (a), ATP is not stored, but is rapidly utilized in energy-requiring reactions. 8. ¢G°¿ for the reaction of ATP and creatine is calculated as Creatine + Pi · Phosphocreatine + H2O ATP + H2O · ADP + Pi Creatine + ATP · Phosphocreatine + ADP
¢G°¿1kJ mol-12 +43 -30 +13
The ratio of ATP to ADP needed to maintain a 20:1 ratio of phosphocreatine to creatine is calculated from Equation 10.13. At equilibrium, ¢G = 0, so ¢G°¿ = -RT ln ln
1202[ADP] 112[ATP]
1202[ADP] 112[ATP]
[Phosphocreatine][ADP] [Creatine][ATP]
= -
113 000 J mol-12 ¢G°¿ = -5.24 = RT 18.315 J K-1 mol-121298 K2
= 5.3 * 10-3
[ATP] = 3800 : 1 [ADP]
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9. O R
H C
C
(Acyl adenylate)
O O
+
P
P
O
P
Adenosine
R PPi
NH3 2 Pi
Pyrophosphatase
H C
P
O
C
O
Adenosine
NH3 tRNA
H2O
AMP O R
H C
C
O
NH3
10. ¢G°¿ = -RT ln Keq Keq =
[fructose-6-phosphate] [glucose-6-phosphate]
=
2 1
¢G°¿ = -18.315 J K-1 mol-121298 K2 ln 2 ¢G°¿ = -1.7 kJ mol-1
11. (a) ln Keq = -
125 000 J mol-12 ¢G°¿ = = -10.1 RT 18.315 J K-1 mol-121298 K2
Keq = 4.1 * 10-5
(b) ¢G°¿ for the coupled reaction is calculated as A ¡ B ATP + H2O · ADP + Pi A + ATP + H2O · B + ADP + Pi ln Keq = -
¢G°¿1kJ mol-12 +25 -30 -5
¢G°¿ = 2.0 RT
Keq = 7.5 Keq for the coupled reaction is about 180000 times larger than Keq in part (a).
(c) Keq = 7.5 =
[B]112 [B][ADP][Pi] [B][ADP] = = [A][ATP][H2O] [A][ATP] [A]14002
[B] = 3000 : 1 [A]
Coupling the reaction to ATP hydrolysis increases the ratio of [B] to [A] by a factor of about 73 million 13000 , 14.1 * 10-52 = 7.3 * 1072. 12. Electrons flow from the molecule with a more negative standard reduction potential to the molecule with a more positive standard reduction potential. 2+ 3+ (a) Cytochrome b51Fe~ 2 + Cytochrome f1Fe~ 2: 3+ 2+ Cytochrome b51Fe~ 2 + Cytochrome f1Fe~ 2
(b) Succinate + Q : Fumarate + QH2 (c) Isocitrate + NAD : a-Ketoglutarate + NADH 13. The expected results are as shown in the graph on the right. As NADH is formed in the reaction mixture, the absorbance at 340 nm will increase (see Box 10.2). 14. The standard reduction potentials in Table 10.4 refer to half-reactions that are written as Sox + n e : Sred. Two half-reactions can be added to obtain the coupled oxidationreduction reaction by reversing the direction of the half-reaction involving the reduced species and reversing the sign of its reduction potential. 3+ 2+ 2 + 2e ¡ 2 Cyt c1Fe~ 2 (a) 2 Cyt c1Fe~
QH2 ¡ Q + 2 H + 2e
3+ 2+ 2 Cyt c1Fe~ 2 + QH2 ¡ 2 Cyt c1Fe~ 2 + Q + 2 H
E°¿1V2 +0.23 -0.04 ¢E°¿ = 0.19 V
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¢G°¿ = -nF¢E°¿ = -122196.48 kJ V-1 mol-1210.19 V2 ¢G°¿ = -37 kJ mol-1
(b) 12 O2 + 2 H + 2 e ¡ H2O H
Succinate ¡ Fumarate 2 + 2e 1 2 O2 + Succinate ¡ H2O + Fumarate ¢G°¿ = -122196.48 kJ V-1 mol-1210.79 V2
E°1V2 +0.82 -0.03 ¢E°¿ = 0.79 V
¢G°¿ = -150 kJ mol-1
15. The oxidation of NADH and resulting electron transfer to O2 via the electron-transport chain provides free energy that is coupled to ATP synthesis from ADP and Pi (Equations 10.32–10.36). A mutant cytochrome oxidase with defective electron transport could decrease ATP production, resulting in impaired energy metabolism in the brain. 16. Q + 2 H + 2 e ¡ QH2 H
FADH2 ¡ FAD + 2 + 2 Q + FADH2 ¡ QH2 + FAD
E°¿1V2 +0.04 +0.22 ¢E°¿ = 0.26 V
e
¢E = ¢E°¿ -
RT [QH2][FAD] ln nF [Q][FADH2]
¢E = 0.26 V -
-5 -4 0.026 V 15 * 10 212 * 10 2 ln -4 2 11 * 10 215 * 10-32
¢E = 0.26 V - 0.0131-3.92 = 0.31 V ¢G = -nF¢E = -122196.48 kJ V-1 mol-1210.31 V2 ¢G = -60 kJ mol-1
Theoretically, the oxidation of FADH2 by ubiquinone liberates more than enough free energy to drive ATP synthesis from ADP and Pi.
Chapter 11 Glycolysis 1. Glycolysis occurs in the cytosol. All of the intermediates of glycolysis are phosphorylated and thus cannot easily cross the plasma membrane. 2. (a) 2 (see Figure 11.2 and Reaction 11.12) (b) 2 (1 ATP is consumed by the fructokinase reaction, 1 ATP is consumed by the triose kinase reaction, and 4 ATP are generated by the triose stage of glycolysis) (c) 2 (2 ATP are consumed in the hexose stage, and 4 ATP are generated by the triose stage) (d) 5 (2 ATP are obtained from fructose, as in part (b), and 3 ATP—rather than 2—are obtained from the glucose moiety since glucose 1-phosphate, not glucose, is formed when sucrose is cleaved) 3. The individuals who are lactose intolerant are deficient in the enzyme lactase. Lactase is the intestinal enzyme that catalyzes the hydrolysis of lactose to glucose and galactose. In lactase-deficient individuals, lactose is metabolized by bacteria in the large intestine with the production of gases such as CO2.
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4. (a)
H
O C 1
H
H
C 2
OH
HO
C 3
H
H
4
H
5
6
C
OH
C
OH
1
CH3
2
C
3
COO
OH
+
Glycolysis
HO
CH 2 OH
Glucose
4
COO
5
C
6
CH3
H
2 Lactate 14
(b) Glucose labeled at either C-3 or C-4 yields pyruvate. H
O 1
H HO H H
CO2 from the decarboxylation of
2
3
C C
OH
C
H
C 4
OH
C
OH
5
6
H
O
C (3,4) H
CH 2 OH
C
(2,5)
COO
(3,4)
OH
CH2OPO3
C
CoA
C
O
CH3
CH3
(2) Pyruvate
(2) Glyceraldehyde 3-phosphate
S (2,5)
(1,6)
(1,6)
(1,6)
Glucose
O
(2,5) 2
2(3,4)CO2
(2) Acetyl CoA
5. Inorganic phosphate 132Pi2 will be incorporated into 1,3-bisphosphoglycerate (1,3 BPG) at the C-1 carbon in the glyceraldehyde 3-phosphate dehydrogenase (GADPH) reaction: glyceraldehyde 3-phosphate + NAD + Pi : 1,3 BPG, and then transferred to the g-position of ATP in the next step: 1,3 BPG + ADP : ATP + 3-phosphoglycerate. 6. Since the brain relies almost solely on glucose for energy, it is dependent on glycolysis as the major pathway for glucose catabolism. Since the Huntington protein binds tightly to GAPDH, this suggests that it might inhibit this crucial glycolytic enzyme, and thereby impair the production of ATP. Decreased ATP levels would be detrimental to neuronal cells in the brain. 7. (a)
1
HO
2
3
CH2OH C
H
CH2OH
Glycerol
ATP
CH2OH
ADP
NAD
1
HO
C
2
CH2OPO3
L-Glycerol 3-phosphate
CH2OH
1
C
H
3
NADH, H
2 2
O
CH2OPO3
2
3
Dihydroxyacetone phosphate
(b) C-2 and C-3 of glycerol 3-phosphate must be labeled. Once dihydroxyacetone phosphate is converted to glyceraldehyde 3-phosphate, C-1 is oxidized to an aldehyde and subsequently lost as CO2 (Problem 2). 8. Glucose freely enters the liver via the GLUT2 transporter proteins and then must be converted to glucose 6-phosphate by the action of glucokinase, the hexokinase isozyme IV that predominates in the liver. Since glucokinase has a high Km for glucose, it is most active only at elevated glucose levels and can therefore respond to increases in blood glucose after a meal. Liver glucokinase participates in the regulation of blood glucose levels by converting glucose to glucose 6-phosphate, and decreased levels of glucokinase due to insufficient insulin would impair utilization of glucose for glycolysis or glycogen synthesis in the liver. 9. As discussed in chapter 10, ADP and ATP are usually present at the active site of enzymes as complexes with magnesium ions. This would include the glycolytic enzymes hexokinase, phosphofructokinase, and pyruvate kinase. In addition, enolase is another 2+ enzyme that requires Mg~ for activity. When an individual is deficient in magnesium, several reactions in glycolysis will be affected.
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10. Cells that metabolize glucose to lactate by anaerobic glycolysis produce far less ATP per glucose than do cells that metabolize glucose aerobically to CO2 via glycolysis and the citric acid cycle (Figure 11.1). More glucose must be utilized via anaerobic glycolysis to produce a sufficient amount of ATP for cellular needs, and the rate of conversion of glucose to lactate is much higher than under aerobic conditions. Cancer cells in an anaerobic environment take up far more glucose, and may overproduce some glycolytic enzymes to compensate for the increase in the activity of this pathway of carbohydrate metabolism. 11. Under vigorous exercise, glycolysis rapidly produces ATP to power muscle contraction and consumes NAD . To regenerate NAD , pyruvate is reduced to lactate and NADH is oxidized to NAD . The regenerated NAD is required for glycolysis to continue. Under vigorous exercise, a deficiency in the enzyme lactate dehydrogenase would result in a rise in the concentration of pyruvate and less regeneration of NAD . This would result in muscle fatigue. Under intense exercise, this deficiency could lead to actual damage of the muscle tissue resulting in myoglobinuria. This is rust-colored urine resulting from the loss of myoglobin from the damaged muscle tissue. 12. No. The conversion of pyruvate to lactate, catalyzed by lactate dehydrogenase, oxidizes NADH to NAD , which is required for the glyceraldehyde 3-phosphate dehydrogenase reaction of glycolysis. 13. In the reactions catalyzed by these enzymes, the bond between the g-phosphorus atom and the oxygen of the b-phosphoryl group is cleaved when the g-phosphoryl group of ATP is transferred (Figure 11.3). The analog cannot be cleaved in this way and therefore inhibits the enzymes by competing with ATP for the active site. 14. The free-energy change for the aldolase reaction under standard conditions 1¢G°¿2 is +22.8 kJ mol1. The concentrations of fructose 1,6-bisphosphate, dihydroxyacetone phosphate, and glyceraldehyde 3-phosphate in heart muscle, however, are much different than the 1 M concentrations assumed under standard conditions. The actual freeenergy change under cellular concentrations 1¢G°¿ = -5.9 kJ mol-12 is much different than ¢G°¿, and the aldolase reaction readily proceeds in the direction necessary for glycolysis: Fructose 1, 6-bisphosphate : glyceraldehyde 3-phosphate + dihydroxyacetone phosphate. 15. ATP is both a substrate and an allosteric inhibitor for PFK-1. Higher concentrations of ATP result in a decrease in the activity of PFK-1 due to an increase in the Km. AMP is an allosteric activator that acts by relieving the inhibition caused by ATP, thus raising the curve when AMP is present with ATP. 16. Increased [cAMP] activates protein kinase A, which catalyzes the phosphorylation and inactivation of pyruvate kinase. cAMP
Pyruvate kinase (more active)
+ OH
Protein kinase A ATP
Pyruvate kinase (less active)
P
ADP
17. (a) A decrease in glycolysis in the liver makes more glucose available for export to other tissues. (b) Decreased activity of the glucagon transducer system decreases the amount of cAMP formed. As existing cAMP is hydrolyzed by the activity of a phosphodiesterase, cAMP-dependent protein kinase A becomes less active. Under these conditions, PFK-2 activity increases and fructose 2,6-bisphosphatase activity decreases (Figure 11.18). The resulting increase in fructose 2,6-bisphosphate activates PFK-1, increasing the overall rate of glycolysis. A decrease in cAMP also leads to the activation of pyruvate kinase (Problem 12).
Chapter 12 Gluconeogenesis, The Pentose Phosphate Pathway, and Glycogen Metabolism 1. (a) After glycogen is depleted during fasting, lactate, glycerol, and alanine are initially used as precursors for glucose synthesis. As the fasting continues, muscle proteins are degraded to amino acids that are used for gluconeogenic precursors.
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(b) Intense exercise requires high rates of anaerobic glycolysis in muscles, which results in large amounts of blood lactate. Lactate is taken up by the liver and converted to pyruvate, which is used for glucose synthesis (Cori Cycle, Figure 13.12). 2. Reducing power in the form of NADH (2), and ATP (4) and GTP (2) are required for the synthesis of glucose from pyruvate (Equation 13.5). The NADH and GTP are direct products of the citric acid cycle, and ATP can be generated from NADH and QH21FADH22 during the oxidative phosphorylation process. 3. Epinephrine interacts with the liver b-adrenergic receptors and activates the adenylyl cyclase signaling pathway leading to cAMP production and activation of protein kinase A (Figure 13.6). Protein kinase A activates phosphorylase kinase, which in turn activates glycogen phosphorylase (GP) leading to glycogen degradation (Figure 13.7). Glucose can then be transported out of the liver and into the bloodstream, where it is taken up by muscles for needed energy production. Liver[Glycogen
1GP2 "
GIP ¡ G6P ¡ Glucose] ¡ Bloodstream ¡ Muscles
4. (a) Protein phosphatase-1 activated by insulin catalyzes the hydrolysis of the phosphate ester bonds on glycogen synthase (activating it) and on glycogen phosphorylase and phosphorylase kinase (inactivating them), as shown in Figure 13.8. Therefore, insulin stimulates glycogen synthesis and inhibits glycogen degradation in muscle cells. (b) Only liver cells are rich in glucagon receptors, so glucagon selectively exerts its effects on liver enzymes. (c) The binding of glucose to the glycogen phosphorylase–protein phosphatase-1 complex in liver cells relieves the inhibition of protein phosphatase-1 and makes glycogen phosphorylase more susceptible to dephosphorylation (inactivation) by protein phosphatase-1 (Figure 13.9). Protein phosphatase-1 also catalyzes the dephosphorylation of glycogen synthase, making it more active. Therefore, glucose stimulates glycogen synthesis and inhibits glycogen degradation in the liver. 5. Increased concentrations of fructose 2,6-bisphosphate (F2,6BP) lead to a decreased rate of glycolysis and an increased rate of gluconeogenesis. F2,6BP is an activator of the glycolytic enzyme phosphofructokinase-1 (PFK-1), and lower F2,6BP levels will result in decreased rates of glycolysis. In addition, F2,6BP is an inhibitor of the gluconeogenic enzyme fructose 1,6-bisphosphatase, and therefore decreased levels of F2,6BP will decrease the inhibition and increase the rate of gluconeogenesis (Figure 13.15). 6. When glucagon binds to its receptor, it activates adenylyl cyclase. Adenylyl cyclase catalyzes the synthesis of cAMP from ATP. The cAMP activates protein kinase A. Protein kinase A catalyzes the phosphorylation of PFK-2 which inactivates the kinase activity and activates the phosphatase activity. Fructose 2,6-bisphosphatase catalyzes the hydrolytic dephosphorylation of fructose 2,6-bisphosphate to form fructose 6-phosphate. The resulting decrease in the concentration of fructose 2,6-bisphosphate relieves the inhibition of fructose 1,6-bisphosphatase, thereby activating gluconeogenesis. Thus, the kinase activity of PFK-2 is decreased. 7. (a) Yes. The synthesis of glycogen from glucose 6-phosphate requires the energy of one phosphoanhydride bond (in the hydrolysis of PPi; Figure 13.4). However, when glycogen is degraded to glucose 6-phosphate, inorganic phosphate 1Pi2 is used in the phosphorolysis reaction. No high-energy phosphate bond is used. (b) One fewer ATP molecule is available for use in the muscle when liver glycogen is the source of the glucose utilized. Liver glycogen is degraded to glucose phosphates and then to glucose without consuming ATP. After transport to muscle cells, the glucose is converted to glucose 6-phosphate by the action of hexokinase in a reaction that consumes one molecule of ATP. Muscle glycogen, however, is converted directly to glucose 1-phosphate by the action of glycogen phosphorylase, which does not consume ATP. Glucose 1-phosphate is isomerized to glucose 6-phosphate by the action of phosphoglucomutase. 8. A deficiency of glycogen phosphorylase in the muscle prevents the mobilization of glycogen to glucose. Insufficient glucose prevents the production of ATP by glycolysis. Existing ATP used for muscle contraction is not replenished, thus increasing the levels of ADP and Pi. Since no glucose is available from glycogen in the muscle, no lactate is produced.
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9. Converting glucose 1-phosphate to two molecules of lactate yields 3 ATP equivalents (1 ATP expended in the phosphofructokinase-1 reaction, 2 ATP produced in the phosphoglycerate kinase reaction, and 2 ATP produced in the pyruvate kinase reaction). Converting two molecules of lactate to one molecule of glucose 1-phosphate requires 6 ATP equivalents (2 ATP in the pyruvate carboxylase reaction, 2 GTP in the PEP carboxykinase reaction, and 2 ATP in the phosphoglycerate kinase reaction). 10. (a) Muscle pyruvate from glycolysis or amino acid catabolism is converted to alanine by transamination. Alanine travels to the liver, where it is reconverted to pyruvate by transamination with a-ketoglutarate. Gluconeogenesis converts pyruvate to glucose, which can be returned to muscles. (b) NADH is required to reduce pyruvate to lactate in the Cori cycle, but it is not required to convert pyruvate to alanine in the glucose-alanine cycle. Thus, the glucose-alanine cycle makes more NADH available in muscles for the production of ATP by oxidative phosphorylation. 11. (a) Inadequate glucose 6-phosphatase activity 1G6P : glucose + Pi2 leads to accumulation of intracellular G6P, which inhibits glycogen phosphorylase and activates glycogen synthase. This prevents liver glycogen from being mobilized. This results in increased glycogen storage (and enlargement of the liver), and low blood glucose levels (hypoglycemia). (b) Yes. A defective branching enzyme leads to accumulation of glycogen molecules with defective, short outer branches. These molecules cannot be degraded, so there will be much less efficient glycogen degradation for glucose formation. Low blood glucose levels result due to the impaired glycogen degradation. 12. Glucose 6-phosphate, glyceraldehyde 3-phosphate, and fructose 6-phosphate. 13. The repair of tissue injury requires cell proliferation and synthesis of scar tissue. NADPH is needed for the synthesis of cholesterol and fatty acids (components of cellular membranes), and ribose 5-phosphate is needed for the synthesis of DNA and RNA. Since the pentose phosphate pathway is the primary source of NADPH and ribose 5phosphate, injured tissue responds to the increased demands for these products by increasing the level of synthesis of the enzymes in the pentose phosphate pathway. 14. (a) CH2OH CH2OH
HO H
C
O
C
H
C
O C H +
OH 2
CH2 OPO3 Xylulose 5-phosphate
H
C
O
HO
C
H
H
C
OH
H
C
OH
H
C C
O OH
C
Transketolase
H
OH 2
CH2 OPO3 Erythrose 4-phosphate
H
C
+
OH 2
CH2 OPO3 Glyceraldehyde 3-phosphate
2
CH2 OPO3 Fructose 6-phosphate
(b) C-2 of glucose 6-phosphate becomes C-1 of xylulose 5-phosphate. After C-1 and C-2 of xylulose 5-phosphate are transferred to erythrose 4-phosphate, the label appears at C-1 of fructose 6-phosphate, as shown in part (a).
Chapter 13 The Citric Acid Cycle 1. (a) No net synthesis is possible since two carbons from acetyl CoA enter the cycle in the citrate synthase reaction and two carbons leave as CO2 in the isocitrate dehydrogenase and a-ketoglutarate dehydrogenase reactions. (b) Oxaloacetate can be replenished by the pyruvate carboxylase reaction, which carries out a net synthesis of OAA, Pyruvate + CO2 + ATP + H2O ¡ Oxaloacetate + ADP + Pi
This is the major anaplerotic reaction in some mammalian tissues. Many plants and some bacteria supply oxaloacetate via the phosphoenolpyruvate carboxykinase reaction, Phosphoenolpyruvate + HCO3 - ¡ Oxaloacetate + Pi
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2. Aconitase would be inhibited by fluorocitrate formed from fluoroacetate, leading to increased levels of citric acid and decreased levels of all subsequent citric acid cycle intermediates from isocitrate to oxaloacetate. Since fluorocitrate is a competitive inhibitor, very high levels of citrate would at least partially overcome the inhibition of aconitase by fluorocitrate and permit the cycle to continue at some level. 3. (a) 12.5; 10.0 from the cycle and 2.5 from the pyruvate dehydrogenase reaction. (b) 10.0; 7.5 from oxidation of 3 NADH, 1.5 from oxidation of 1 QH2, and 1.0 from the substrate-level phosphorylation catalyzed by CoA synthetase. 4. 87.5% (28 of 32) of the ATP is produced by oxidative phosphorylation, and 12.5% (4 of 32) is produced by substrate-level phosphorylation. 5. Thiamine is the precursor of the coenzyme thiamine pyrophosphate (TPP), which is found in two enzyme complexes associated with the citric acid cycle: the pyruvate dehydrogenase complex and the a-ketoglutarate dehydrogenase complex. A deficiency of TPP decreases the activities of these enzyme complexes. Decreasing the conversion of pyruvate to acetyl CoA and of a-ketoglutarate to succinyl CoA causes accumulation of pyruvate and a-ketoglutarate. 6. Since C-1 of pyruvate is converted to CO2 in the reaction catalyzed by the pyruvate dehydrogenase complex, 1-[14C]-pyruvate is the first to yield 14CO2. Neither of the two acetyl carbon atoms of acetyl CoA is converted to CO2 during the first turn of the citric acid cycle (Figure 12.5). However, the carboxylate carbon atoms of oxaloacetate, which arise from C-2 of pyruvate, become the two carboxylates of citrate that are removed as CO2 during a second turn of the cycle. Therefore, 2-[14C]-pyruvate is the second labeled molecule to yield 14CO2. 3-[14C]-Pyruvate is the last to yield 14CO2, in the third turn of the cycle. First turn 2 COO 1 COO 2C
O
3 CH 3
1 CO 2
S-CoA 2C
3 CH 2
O
HO
3 CH 3
C
CO2 CO2
3 CH 2
COO
3 CH 2
CH2 COO
Pyruvate
Acetyl CoA
2 COO
2 COO Succinate
Citrate
Second turn COO 2 COO 3C
O
CH2 HO
3C
2 COO
3 CH 2
3 CH 2
2 COO
2 COO
Oxaloacetate
Citrate
2 CO 2 2 CO 2
3 COO 3 CH 2 3 CH 2 3 COO
Succinate
7. (a) The NADH produced by the oxidative reactions of the citric acid cycle must be recycled back to NAD , which is required for the pyruvate dehydrogenase reaction. When O2 levels are low, fewer NADH molecules are reoxidized by O2 (via the process of oxidative phosphorylation), so the activity of the pyruvate dehydrogenase complex decreases. (b) Pyruvate dehydrogenase kinase catalyzes phosphorylation of the pyruvate dehydrogenase complex, thereby inactivating it (Figure 12.13). Inhibiting the kinase shifts the pyruvate dehydrogenase complex to its more active form. 8. A deficiency in the citric acid cycle enzyme fumarase would result in abnormally high concentrations of fumarate and prior cycle intermediates including succinate and a-ketoglutarate, which could lead to excretion of these molecules. 9. The different actions of acetyl CoA on two components of the pyruvate dehydrogenase (PDH) complex both lead to an inhibition of the pyruvate to acetyl CoA reaction. Acetyl
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CoA inhibits the E2 component of the PDH complex directly (Figure 12.12). Acetyl CoA causes inhibition of the E1 component indirectly by activating the pyruvate kinase (PK) component of the PDH complex, and PK phosphorylates the E1 component of the PDH complex thus inactivating it (Figure 12.13). 10. The pyruvate dehydrogenase complex catalyzes the oxidation of pyruvate to form acetyl CoA and CO2. If there is reduced activity of this complex, then the pyruvate concentration will increase. Pyruvate will be converted to lactate through the action of lactate dehydrogenase. Lactate builds up since glycolytic metabolism is increased to synthesize ATP since oxidation of pyruvate to acetyl CoA is impaired. In addition, pyruvate is converted to alanine as shown in equation 12.6. 11. Calcium activates both isocitrate dehydrogenase and a-ketoglutarate dehydrogenase in the citric acid cycle (Figure 12.16) thereby increasing this catabolic process and pro2+ ducing more ATP. In addition, Ca~ activates the pyruvate dehydrogenase phosphatase enzyme of the PDH complex, which activates the E1 component (Figure 12.13). Activation of the PDH complex converts more pyruvate into acetyl CoA for entry into the citric acid cycle resulting in an increased production of ATP. 12. (a) Alanine degradation replenishes citric acid cycle intermediates, since pyruvate can be converted to oxaloacetate via the pyruvate carboxylase reaction, the major anaplerotic reaction in mammals (Reaction 12.18). Leucine degradation cannot replenish intermediates of the citric acid cycle, since for every molecule of acetyl CoA that enters the cycle, two molecules of CO2 are lost. (b) By activating pyruvate carboxylase, acetyl CoA increases the amount of oxaloacetate produced directly from pyruvate. The oxaloacetate can react with the acetyl CoA produced by the degradation of fatty acids. As a result, flux through the citric acid cycle increases to recover the energy stored in the fatty acids. 13. (a)
CH2 COO 14
(b)
(c)
Ala
14
CH 2
COO CH 2
O C COO α-Ketoglutarate
O CH3 O
(Pyruvate)
14
C COO Oxaloacetate
COO
C
CO2 CH3 O
14
C
(Acetyl SCoA) SCoA
14
CH2 COO HO
C CH 2
COO
Citrate
COO
14. (a) Two molecules of acetyl CoA yield 20 ATP molecules via the citric acid cycle (Equation 12.16) or 6.5 ATP molecules via the glyoxylate cycle (from the oxidation of two molecules of NADH and one molecule of QH2; Reaction 12.21). (b) The primary function of the citric acid cycle is to oxidize acetyl CoA to provide the reduced coenzymes necessary for the generation of energy-rich molecules such as ATP. The primary function of the glyoxylate cycle is not to produce ATP, but to convert acetyl groups to four-carbon molecules that can be used to produce glucose. 15. The protein that controls the activity of isocitrate dehydrogenase in E. coli is a bifunctional enzyme with kinase and phosphatase activities in the same protein molecule. The kinase activity phosphorylates isocitrate dehydrogenase to inhibit the activity of isocitrate dehydrogenase, and the phosphatase activity dephosphorylates isocitrate dehydrogenase to activate isocitrate dehydrogenase. When concentrations of glycolytic and citric acid cycle intermediates are high, isocitrate dehydrogenase is not phosphorylated and is active. When phosphorylation decreases the activity of isocitrate dehydrogenase, isocitrate is diverted to the glyoxylate cycle.
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Chapter 14 Electron Transport and Oxidative Phosphorylation 1. The reduction potential of an iron atom in a heme group depends on the surrounding protein environment, which differs for each cytochrome. The differences in reduction potentials allow electrons to pass through a series of cytochromes. 2. Refer to Figure 14.10. (a) Complex III. The absence of cytochrome c prevents further electron flow. (b) No reaction occurs since Complex I, which accepts electrons from NADH, is missing. (c) O2 (d) Cytochrome c. The absence of Complex IV prevents further electron flow. 3. UCP-2 leaks protons back into the mitochondria thereby decreasing the protonmotive force. The metabolism of foodstuffs provides the energy for electron transport, which in turn creates the protonmotive gradient used to produce ATP. An increase in UCP-2 levels would make the tissue less metabolically efficient (i.e., less ATP would be produced per gram of foodstuff metabolized). As a result, more carbohydrates, fats, and proteins would have to be metabolized in order to satisfy the basic metabolic needs, and this could “burn off” more calories and potentially cause weight loss. 4. (a) Demerol interacts with Complex I and prevents electron transfer from NADH to Q. The concentration of NADH increases since it cannot be reoxidized to NAD . The concentration of Q increases since electrons from QH2 are transferred to O2 but Q is not reduced back to QH2. (b) Myxothiazole inhibits electron transfer from QH2 to cytochrome c1 and from QH2 (via # Q- ) to cytochrome b566 in Complex III (Figure 14.14). The oxidized forms of 3+ both cytochromes predominate since Fe~ cannot be reduced by electrons from QH2.
3+ 5. (a) Oxygen 1O22 must bind to the Fe~ of cytochrome a3 in order to accept electrons (Figure 14.16), and it is prevented from doing so by the binding of CN- to the iron atom. 3+ (b) The methemoglobin 1Fe~ 2 generated from nitrite treatment competes with cytochrome a3 for the CN ions. This competition effectively lowers the concentration of cyanide available to inhibit cytochrome a3 in complex IV, and decreases the inhibition of the electron-transport chains in the presence of CN .
6. A substrate is usually oxidized by a compound with a more positive reduction potential. Since E°¿ for the fatty acid is close to E°¿ for FAD in Complex II (0.0 V, as shown in Table 14.2), electron transfer from the fatty acid to FAD is energetically favorable. ¢E°¿ = 0.0 V - 1-0.05 V2 = +0.05 V ¢G°¿ = -nF¢E°¿ ¢G°¿ = -122196.48 kJ V-1210.05 V2 = -9.6 kJ mol-1
Since E°¿ for NADH in Complex I is -0.32 V, the transfer of electrons from the fatty acid to NADH is unfavorable. ¢E°¿ = -0.32 V - 1-0.05 V2 = -0.27 V
¢G°¿ = -122196.48 kJ V-1 mol-121-0.27 V2 = 52 kJ mol-1
7. (a) 10 protons; 2.5 ATP; P : O = 2.5. (b) 6 protons; 1.5 ATP; P : O = 1.5. (c) 4 protons; 1.0 ATP; P : O = 1.0. 8. (a) The inner mitochondrial membrane has a net positive charge on the cytosolic side 43(outside). The exchange of one ATP~ transferred out for one ADP ~ transferred in yields a net movement of one negative charge from the inner matrix side to the positive cytosolic side. The membrane potential thereby assures that outward transport of a negatively charged ATP is favored by the outside positive charge. (b) Yes. The electrochemical potential with a net positive charge outside the membrane is a result of proton pumping, which is driven by the electron-transport chain. This in turn requires oxidation of metabolites to generate NADH and QH2 as electron donors.
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9. Oxidative phosphorylation is normally tightly coupled to electron transport. Unless ADP can continue to be translocated into the mitochondrial matrix for the FOF1 ATP synthase reaction 1ADP + Pi : ATP2, oxidative phosphorylation is not possible and respiration will cease (i.e., electron transport will stop and no oxygen will be reduced to water). 10. (a) ¢pH = pHin - pHout = 7.5 - 6.7 = 0.08 ¢p = ¢c - 10.059 V2¢pH
¢p = -0.18 V - 10.059 V210.82 = -0.23 V
(b) ¢pchem =
10.059 V210.82
* 100% = 21% -0.23 V -0.18 V * 100% = 78% ¢pelec = -0.23 V (c) ¢G = nF¢p ¢G = 112196.48 kJ V-1 mol-121-0.23 V2 = -22 kJ mol-1
11. (a) In the malate-aspartate shuttle, the reduction of oxaloacetate in the cytosol consumes a proton that is released in the matrix by the oxidation of malate (Figure 14.22). Therefore, one fewer proton is contributed to the proton concentration gradient for every cytosolic NADH oxidized (9 versus 10 for mitochondrial NADH). The ATP yield from two molecules of cytoplasmic NADH is about 4.5 rather than 5.0. (b) Cytoplasmic reactions Glucose ¡ 2 Pyruvate 2.0 ATP 2 NADH ¡ 4.5 ATP Mitochondrial reactions 2 Pyruvate ¡ 2 Acetyl CoA + 2 CO2 2 NADH ¡ 5.0 ATP 2 Acetyl CoA ¡ 4 CO2 2.0 GTP 6 NADH ¡ 15.0 ATP 2 QH2 ¡ 3.0 ATP Total 31.5 ATP
Chapter 15 Photosynthesis 1. Plant chlorophylls absorb energy in the red region of the spectrum (Figure 15.3). The dragonfish chlorophyll derivatives absorb the red light energy (667 nm), and pass the signals on to the visual pigments in much the same manner that plant antenna chlorophylls and related molecules capture light energy and transfer it to a reaction center where electrons are promoted into excited states for transfer to acceptors of the electron-transport chain. 2. (a) Rubisco is the world’s most abundant protein and the principal catalyst for photosynthesis, the basic means by which living organisms acquire the carbon necessary for life. Its importance in the process of providing food for all living things can be well justified. (b) Photorespiration is a process that wastes ribulose 1,5-bisphosphate, consumes the NADPH and ATP generated by the light reactions, and can greatly reduce crop yields. As much as 20 to 30% of the carbon fixed in photosynthesis can be lost to photorespiration. This process results from the lack of specificity of Rubisco, which can use O2 instead of CO2 (Figure 15.8) to produce phosphoglycolate and 3-phosphoglycerate (Figure 15.18) instead of two triose phosphate molecules. In addition, Rubisco has low catalytic activity 1Kcat L 3 s-12. This lack of specificity and low activity earns Rubisco the title of a relatively incompetent, inefficient enzyme. 3. O2 is produced exclusively during the light reactions of photosynthesis, and both oxygen atoms are derived from H2O (Figure 15.7). Labeled H2 18O or C18O2 can be used to trace the fates of particular oxygen atoms. CO2 + 2 H2O*
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Light
" 1CH O2 + O* + H O 2 2 2
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4. (a) CO2 + 2 H2 S CO2 + 2 CH3CH2OH Ethanol
" 1CH O2 + H O + 2 S 2 2 " 1CH O2 + H O + 2 CH CHO 2 2 3 Acetaldehyde
Light Light
(b) When H2O is the proton donor, O2 is the product, but when other proton donors such as H2S and ethanol are used, oxygen cannot be produced. Most photosynthetic bacteria do not produce O2 and are obligate anaerobes that are poisoned by O2. Light " 1CH O2 + H O + 2A (c) CO + 2 H A 2
2
2
2
5. Because the synthesis of glucose from CO2 and H2O is driven by the products of the light reactions (Equation 15.2), isolated chloroplasts cannot synthesize glucose in the dark. When the products of the light reactions (NADPH and ATP) are added, chloroplasts can synthesize glucose in the absence of light. 6. (a) Two H2O molecules provide the oxygens for one O2 during the photosynthetic process. A total of 4 electrons must be removed from 2 H2O and passed through an electron-transport system to 2 NADPH. One quantum of light is required to transfer one electron through PSI and one quantum for PSII. Therefore a total of 8 hn will be required to move 4 electrons through both reaction centers (4 hn for PHI and 4 hn for PHII). (b) Six NADPH are required for the RPP synthesis of one triose phosphate (Equation 15.5). Therefore 12 electrons must be transferred through the two reaction centers of the electron-transport system and this will require the absorption of 24 hn. 7. (a) Yes. (Refer to the Z-scheme, Figure 15.8). When DCMU blocks electron flow, PSII in the P680* state will not be reoxidized to the P680 state, which is required as an acceptor of electrons from H2O. If H2O is not oxidized by P680 then no O2 will be produced. In the absence of electron flow through the cytochrome bf complex, no protons will be translocated across the membrane. Without a pH gradient no photophosphorylation (ATP synthesis) will be possible. (b) External electron acceptors for PSII will permit P680 to be reoxidized to P680 and will restore O2 evolution. No electrons will flow through the cytochrome bf complex, however, so that no photophosphorylation will occur. 8. (a) When the external pH rises to 8.0, the stromal pH also rises quickly, but the luminal pH remains low initially because the thylakoid membrane is relatively impermeable to protons. The pH gradient across the thylakoid membrane drives the production of ATP via proton translocation through chloroplast ATP synthase (Figure 15.12). (b) Protons are transferred from the lumen to the stroma by ATP synthase, driving ATP synthesis. The pH gradient across the membrane decreases until it is insufficient to drive the phosphorylation of ADP, and ATP synthesis stops. 9. During cyclic electron transport, reduced ferredoxin donates its electrons back to P700 via the cytochrome bf complex (Equation 15.4). As these electrons cycle again through photosystem I, the proton concentration gradient generated by the cytochrome bf complex drives ATP synthesis. However, no NADPH is produced because there is no net flow of electrons from H2O to ferredoxin. No O2 is produced because photosystem II, the site of O2 production, is not involved in cyclic electron transport. 10. The light absorbing complexes, electron-transport chain, and chloroplast ATP synthase all reside in the thylakoid membranes, and the structure and interactions of any of these photosynthetic components could be affected by a change in the physical nature of the membrane lipids. 11. The compound is acting as an uncoupler. The electron transfer is occurring without the synthesis of ATP. The compound destroys the proton gradient that is produced through electron transfer. 12. (a) The synthesis of one triose phosphate from CO2 requires 9 molecules of ATP and 6 molecules of NADPH (Equation 15.5). Since two molecules of triose phosphate can be converted to glucose, glucose synthesis requires 18 molecules of ATP and 12 molecules of NADPH. (b) Incorporating glucose 1-phosphate into starch requires one ATP equivalent during the conversion of glucose 1-phosphate to ADP-glucose (Figure 15.22), bringing the total requirement to 19 molecules of ATP and 12 molecules of NADPH.
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13. Refer to Figure 15.17. (a) C-1. (b) C-3 and C-4. (c) C-1 and C-2. C-1 and C-2 of fructose 6-phosphate are transferred to glyceraldehyde 3-phosphate to form xylulose 5-phosphate and C-3 and C-4 of fructose 6-phosphate become C-1 and C-2 of erythrose 4-phosphate. 14. (a) In the C4 pathway (Figure 15.19), the pyruvate-phosphate dikinase reaction consumes two ATP equivalents for each CO2 fixed (since PPi is hydroyzed to 2 Pi). Therefore, C4 plants require 12 more molecules of ATP per molecule of glucose synthesized than C3 plants require. (b) Because C4 plants minimize photorespiration, they are more efficient than C3 plants in using light energy to fix CO2 into carbohydrates, even though the chemical reactions for fixing CO2 in C4 plants require more ATP. 15. (a) An increase in stromal pH increases the rate of the RPP cycle in two ways. (1) An increase in stromal pH increases the activity of ribulose 1,5-bisphosphate carboxylase-oxygenase (RuBisCO), the central regulatory enzyme of the RPP cycle, and the activities of fructose 1,6-bisphosphatase and sedoheptulose 1,7bisphosphatase. It also increases the activity of phosphoribulokinase. Phosphoribulokinase is inhibited by 3-phosphoglycerate (3PG) in the 3PG2- ionization state but not in the 3PG3- ionization state, which predominates at higher pH. (2) An increase in stromal pH also increases the proton gradient that drives the synthesis of ATP in chloroplasts. Since the reactions of the RPP cycle are driven by ATP, an increase in ATP production increases the rate of the RPP cycle. 2+ (b) A decrease in the stromal concentration of Mg~ decreases the rate of the RPP cycle by decreasing the activity of RuBisCO, fructose 1,6-bisphosphatase, and sedoheptulose 1,7-bisphosphatase.
Chapter 16 Lipid Metabolism 1. (a) LDLs are rich in cholesterol and cholesterol esters, and transport these lipids to peripheral tissues. Delivery of cholesterol to tissues is moderated by LDL receptors on the cell membranes. When LDL receptors are defective, receptor-mediated uptake of cholesterol does not occur (Figure 16.6). Because cholesterol is not cleared from the blood it accumulates and contributes to the formation of atherosclerotic plaques. (b) Increased cholesterol levels normally repress transcription of HMG-CoA reductase and stimulate the proteolysis of this enzyme as well. With defective LDL, however, cholesterol synthesis continues in spite of high blood cholesterol levels because the extracellular cholesterol cannot enter the cells to regulate intracellular synthesis. (c) HDLs remove cholesterol from plasma and cells of nonhepatic tissues and transport it to the liver where it can be converted into bile salts for disposal. In Tangier patients, defective cholesterol-poor HDLs cannot absorb cholesterol, and the normal transport process to the liver is disrupted. 2. (a) Carnitine is required to transport fatty acyl CoA into the mitochondrial matrix for b-oxidation (Figure 16.8). The inhibition of fatty acid transport caused by a deficiency in carnitine diminishes energy production from fats for muscular work. Excess fatty acyl CoA can be converted to triacylglycerols in the muscle cells. (b) Since carnitine is not required to transport pyruvate, a product of glycolysis, into mitochondria for oxidation, muscle glycogen metabolism is not affected in individuals with a carnitine deficiency. 3. (a) Activation of the C12 fatty acid to a fatty acyl CoA consumes 2 ATP. Five rounds of b-oxidation generate 6 acetyl CoA, 5 QH2 (which yield 7.5 ATP via oxidative phosphorylation), and 5 NADH (which yield 12.5 ATP). Oxidation of the 6 acetyl CoA by the citric acid cycle yields 60 ATP. Therefore, the net yield is 78 ATP equivalents. (b) Activation of the C16 monounsaturated fatty acid to a fatty acyl CoA consumes 2 ATP. Seven rounds of b-oxidation generate 8 acetyl CoA, 6 QH2 (which yield 9 ATP via oxidative phosphorylation), and 7 NADH (which yield 17.5 ATP). The fatty acid contains a cis-b,g double bond that is converted to a trans-a,b double bond, so the acyl-CoA dehydrogenase-catalyzed reaction, which generates QH2, is bypassed in the fifth round. Oxidation of the 8 acetyl CoA by the citric acid cycle yields 80 ATP. Therefore, the net yield is 104.5 ATP equivalents.
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4. When triacylglycerols are ingested in our diets, the hydrolysis of the dietary lipids occurs mainly in the small intestine. Pancreatic lipase catalyzes the hydrolysis at the C-1 and C-3 positions of triacylglycerol, producing free fatty acids and 2-monoacylglycerol. These molecules are transported in bile-salt micelles to the intestine, where they are absorbed by intestinal cells. Within these cells, the fatty acids are converted to fatty acyl CoA molecules which eventually form a triacylglycerol that is incorporated into chylomicrons for transport to other tissues. If the pancreatic lipase is inhibited, the ingested dietary triglyceride cannot be absorbed. The triglyceride will move through the digestive tract and will be excreted without absorption. 5. (a) Oleate has a cis-¢ 9 double bond, so oxidation requires enoyl-CoA isomerase (as in Step 2 of Figure 16.11). (b) Arachidonate has cis double bonds at both odd 1¢ 5, ¢ 112 and even 1¢ 8, ¢ 142 carbons, so oxidation requires both enoyl-CoA isomerase and 2,4-dienoyl-CoA reductase (as in Step 5 of Figure 16.11). (c) This C17 fatty acid contains a cis double bond at an even-numbered carbon 1¢ 62, so oxidation requires 2,4-dienoyl-CoA reductase. In addition, three enzymes are required to convert the propionyl CoA product into succinyl CoA: propionyl-CoA carboxylase, methylmalonyl-CoA racemase, and methylmalonyl-CoA mutase (Figure 16.10). 6. Even-chain fatty acids are degraded to acetyl CoA, which is not a gluconeogenic precursor. Acetyl CoA cannot be converted directly to pyruvate because for every two carbons of acetyl CoA, which enter the citric acid cycle, two carbons in the form of two CO2 molecules leave as products. The last three carbons of odd-chain fatty acids, on the other hand, yield a molecule of propionyl CoA upon degradation in the fatty acid oxidation cycle. Propionyl CoA can be carboxylated and converted to succinyl CoA in three steps (Figure 16.10). Succinyl CoA can be converted to oxaloacetate by citric acid cycle enzymes, and oxaloacetate can be a gluconeogenic precursor for glucose synthesis. 7. (a) The labeled carbon remains in H14CO3 ; none is incorporated into palmitate. Although H14CO3 is incorporated into malonyl CoA, the same carbon is lost as CO2 during the ketoacyl-ACP synthase reaction in each turn of the cycle (Figure 16.17). (b) All the even-numbered carbons are labeled. Except for the acetyl CoA that becomes C-15 and C-16 of palmitate, the acetyl CoA is converted to malonyl CoA and then to malonyl-ACP before being incorporated into a growing fatty acid chain with the loss of CO2. 8. (a) Enoyl ACP reductase catalyzes the second reductive step in the fatty acid biosynthesis pathway, converting a trans-2,3 enoyl moiety into a saturated acyl chain, and uses NADPH as cofactor.
R
C
H
O
C
C
S
ACP
H enoyl-ACP reductase
NADPH H NADP O
R
CH2
CH2
C
S
ACP
(b) Fatty acids are essential for membranes in bacteria. If fatty acid synthesis is inhibited there will be no new membranes and no growth of the bacteria. (c) The fatty acid synthesis systems are different in animals and bacteria. Animals contain a type I fatty acid synthesis system (FAS I) where the various enzymatic activities are localized to individual domains in a large multifunctional enzyme. In bacteria, each reaction in fatty acid synthesis is catalyzed by a separate monofunctional enzyme. Understanding some of the differences in these two systems, would allow for the design of specific inhibitors of the bacterial FAS II. 9. Eating stimulates the production of acetyl CoA from the metabolism of carbohydrates (glycolysis and pyruvate dehydrogenase) and fats (FA oxidation). Normally, increased acetyl CoA results in the elevation of malonyl CoA levels (acetyl CoA carboxylase reaction, Figure 16.16), which may act to inhibit appetite. By blocking fatty acid synthase
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enzyme, C75 prevents the removal of malonyl CoA for the synthesis of fatty acids (Figure 16.17), thereby elevating the levels of malonyl-CoA and further suppressing appetite. 10. (a)
MITOCHONDRION
Carbohydrates
Citrate
Citrate
Glucose
Acetyl CoA
Acetyl CoA Fatty acid synthesis
Glycolysis
Pyruvate
Pyruvate
Fatty acids
(b) The NADH generated by glycolysis can be transformed into NADPH by the combined action of two cytosolic enzymes of the citrate transport system (Figure 16.15). 11. (a) Plentiful citrate and ATP levels promote fatty acid synthesis. High citrate levels activate ACC by preferential binding and stabilization of the active dephosphorylated filamentous form. On the other hand, high levels of fatty acyl CoAs indicate that there is no further need for more fatty acid synthesis. Palmitoyl CoA inactivates ACC by preferential binding to the inactive protomeric dephosphorylated form. (b) Glucagon and epinephrine inhibit fatty acid synthesis by inhibiting the activity of acetyl CoA carboxylase. Both hormones bind to cell receptors and activate cAMP synthesis which in turn activates protein kinases. Phosphorylation of ACC by protein kinases converts it to the inactive form, thus inhibiting fatty acid synthesis. On the other hand, the active protein kinases catalyze phosphorylation and activation of triacylglycerol lipases that catalyze hydrolysis of triacylglycerols, releasing fatty acids for b-oxidation. 12. (a) An inhibitor of acetyl-CoA acetylase will affect a key regulatory reaction for fatty acid synthesis. The concentration of malonyl CoA, the product of the acetyl-CoA carboxylase-catalyzed reaction, will be decreased in the presence of the inhibitor. The decrease in the concentration of malonyl CoA will relieve the inhibition of carnitine acyltransferase I, which is a key regulatory site for the oxidation of fatty acids. Thus, with an active carrier system, fatty acids will be translocated to the mitochondrial matrix where the reactions of b-oxidation occur. In the presence of an inhibitor of acetyl-CoA carboxylase, fatty acid synthesis will decrease and b-oxidation will increase. (b) CABI is a structural analog of biotin. Acetyl-CoA carboxylase is a biotin-dependent enzyme. A biotin analog may bind in place of biotin and inhibit the activity of acetyl-CoA carboxylase. 13. The overall reaction for the synthesis of palmitate from acetyl CoA is the sum of two processes: (1) the formation of seven malonyl CoA by the action of acetyl-CoA carboxylase and (2) seven cycles of the fatty acid biosynthetic pathway (Equation 17.4). 7 Acetyl CoA + 7 CO2 + 7 ATP ¡ 7 Malonyl CoA + 7 ADP + 7 Pi Acetyl CoA + 7 Malonyl CoA + 14 NADPH + 14 H ¡ Palmitate + 7 CO2 + 14 NADP + 8 HS - CoA + 6 H2O 8 Acetyl CoA + 7 ATP + 14 NADPH + 14 H ¡ Palmitate + 7 ADP + 7 Pi + 14 NADP + 8 HS - CoA + 6 H2O 14. (a) Arachidonic acid is a precursor for synthesis of eicosanoids including “local regulators” such as prostaglandins, thromboxanes, and leukotrienes (Figure 16.20). These regulators are involved in mediation of pain, inflammation, and swelling responses resulting from injured tissues. (b) Both prostaglandins and leukotrienes are derived from arachidonate, which is released from membrane phospholipids by the action of phospholipases. By inhibiting a phospholipase, steroidal drugs block the biosynthesis of both prostaglandins and leukotrienes. Aspirin-like drugs block the conversion of arachidonate to prostaglandin precursors by inhibiting cyclooxygenase but do not affect leukotriene synthesis.
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15. (a)
O O R2
C
CH 2 O
O
CH
(b)
CH 2
O
CR1
R2
C
O
O
CH
H C
H C
R1
O
O CH 2 O
CH 2
P O
O
O
CH2
P
CH2CH2NH3
O
CHOH CH2OH
OH
(c)
R
C
H C
CH
O
C H
(CH2)12
CH3
CH
NH
CH 2
O
HOCH2 O HO
OH
H OH
16. Palmitate is converted to eight molecules of acetyl CoA labeled at C-1. Three acetyl CoA molecules are used to synthesize one molecule of mevalonate (Figure 16.29). O H3C
(CH2CH2)7 Palmitate
COO
8 H3C
C
S-CoA
Acetyl CoA
O 3 H3C
C
OH S-CoA
OOC
CH2
C
CH2
CH2
OH
CH3 Acetyl CoA
Mevalonate
Chapter 17 Amino Acid Metabolism 1. PSII contains the oxygen evolving complex and oxygen is produced during photosynthesis. Since oxygen inhibits nitrogenase the synthesis of O2 in hetocysts must be avoided. PSI is retained because it can still generate a light-induced proton gradient by cyclic electron transport and it is not involved in the production of O2. 2. (a) Glutamate dehydrogenase + glutamine synthetase NH4 + a-Ketoglutarate + NAD1P2H + H ¡ Glutamate + NAD1P2 + H2O NH3 + Glutamate + ATP ¡ Glutamine + ADP + Pi 2 NH4 + a-Ketoglutarate + NAD1P2H + ATP ¡ Glutamine + NAD1P2 + ADP + Pi + H2O (b) Glutamine synthetase + glutamate synthase 2 NH3 + 2 Glutamate + 2 ATP ¡ 2 Glutamine + 2 ADP + 2 Pi Glutamine + a-Ketoglutarate + NAD1P2H + H ¡ 2 Glutamate + NAD1P2 2 NH3 + a-Ketoglutarate + NAD1P2H + 2 ATP + H ¡ Glutamine + NAP1P2 + 2 ADP + 2 Pi The coupled reactions in (b) consume one more ATP molecule than the coupled reactions in (a). Because the Km of glutamine synthetase for NH3 is much lower than the Km of glutamate dehydrogenase for NH4 , the coupled reactions in (b) predominate when
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NH4 levels are low. Thus, more energy is spent to assimilate ammonia when its concentration is low. 3. The 15N-labeled amino group is transferred from aspartate to a-ketoglutarate, producing glutamate in a reaction catalyzed by aspartate transaminase (Reaction 17.10). Since transaminases catalyze near-equilibrium reactions and many transaminases use glutamate as the a-amino group donor, the labeled nitrogen is quickly distributed among the other amino acids that are substrates of glutamate-dependent transaminases. 4. (a) a-Ketoglutarate + Amino acid IJ Glutamate + a-Keto acid Oxaloacetate + Amino acid IJ Aspartate + a-Keto acid Pyruvate + Amino acid IJ Alanine + a-Keto acid (b) NAD(P)H, H
NAD(P) α-Ketoglutarate + NH4
Glutamate + H2 O
Glutamate dehydrogenase
5.
(Sulfide)S 2 O-acetylserine
Serine
Cysteine-SH (Fig 17.12)
(Plants) Homoserine
Methionine-S-CH3 (Fig 17.16)
Homocysteine-SH
Methionine-S-CH3
Homocysteine-SH (Fig 17.34)
(Animals) Serine
Cysteine-SH (Fig 17.13)
Cystathionine (S)
6. (a) C-3 of serine is transferred to tetrahydrofolate during the synthesis of glycine, and C-2 is transferred to tetrahydrofolate when glycine is cleaved to produce ammonia
H3 N
1
COO
2
CH
3
CH2 OH
+
Tetrahydrofolate
H3 N
Serine
1
COO
2
CH2
+
5,10-Methylenetetrahydrofolate
+
H2 O
Glycine
COO H3 N
CH2
+ Tetrahydrofolate + NAD
+ H2 O
5,10-Methylenetetrahydrofolate
+
NADH
+
HCO3
+
NH4
+
H
Glycine
and bicarbonate. (b) Serine is synthesized from 3-phosphoglycerate (Figure 17.9), an intermediate of glycolysis. C-3 of both 3-phosphoglycerate and serine is derived from either C-1 or C-6 of glucose, and C-2 of both 3-phosphoglycerate and serine is derived from either C-2 or C-5 of glucose. 7. (a)
COO H3 N
(b)
NH3 CH2
CH
CH
CH H3 C
(c)
COO CH H2N H2 C
N H
OH
CH2
(d)
NH3 CH2
CH2
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COO
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8. (a) Glutamic acid. PPI inhibits glutamine synthetase. (b) Histidine biosynthesis pathway (Figure 17.22). 9. Aspartame is a dipeptide consisting of an asparate and a phenylalanine residue joined by a peptide bond. This bond is eventually hydrolyzed inside the cell producing aspartate and phenylalanine. Phenylketonuria patients must avoid any excess phenylalanine 10. (a)
Leucine
Valine
O
H3C CH H3C
CH2
C
O
H3C COO
CH H3C
Isoleucine
C
H3C
O
H2C
COO
CH H3C
(b) Lysine degradation pathway. a-Aminoadipate d-semialdehyde synthase is deficient (Figure 17.35) (c) Urea cycle. Argininosuccinate synthetase is deficient (Figure 17.39). 11. (a) (b) (c) (d)
Alanine Aspartate Glycine Cysteine
12. The urea cycle does not operate in muscle, so ammonia from the deamination of amino acids cannot be converted to urea. Because high concentrations of ammonia are toxic, ammonia is converted to other products for disposal. In the first pathway, ammonia is incorporated into glutamine by the action of glutamine synthetase (Figure 17.4). Glutamine can then be transported to the liver or kidneys. The second pathway is the glucosealanine cycle (Figure 17.41). Pyruvate accepts the amino group of amino acids by transamination, and the alanine produced is transported to the liver where it can be deaminated back to pyruvate. The amino group is used for urea synthesis, and the pyruvate can be converted to glucose. 13. Inhibition of nitric oxide synthase (NOS) can prevent excess amounts of nitric oxide from being produced in cells lining the blood vessels. Nitric oxide causes relaxation of the vessels and in excess amounts can cause reduced blood pressure leading to shock. Thiocitrulline and S-methylthocitrulline inhibit NOS because they are unreactive analogs of the NOS reaction product citrulline (Figure 17.24). 14. The 2004 Nobel Prize went to Aaron Ciechanover, Avram Hershko, and Irwin Rose for Afor the discovery of ubiquitin-mediated protein degradation.” Ubiquination is described in Section 17.4. It leads to protein degradation and protein degradation is a source of amino acids that can be broken down to produce energy. 15. There are two reasons. Firstly, many of the amino acid biosynthesis pathways aren’t found in humans so there won’t be any metabolic diseases of non-existent Aessential amino acid pathways. Secondly, the remaining pathways are probably crucial pathways during development so that any defects in these pathways are likely to be lethal. This is the same reasoning that we used to explain the lack of metabolic diseases in the sphingolipid biosynthesis pathways (Box 16.2). 16. The 21st and 22nd amino acids are selenocysteine and pyrrolysine. Selenocysteine is synthesized during translation on aminoaclyated tRNA and not by the standard metabolic pathways covered in this chapter. Pyrrolysine may also be synthesized on aminoacylated tRNA. The precursors are serine and lysine. 17. The precursor in the serine biosynthesis pathway is 3-phosphoglycerate. This precursor can be derived from glyceraldehyde-3-phosphate (G3P) in the glycolytic pathway where the conversion is associated with the gain of 1 ATP + 1 NADH. This gain must be subtracted from the total cost of G3P synthesis. Therefore, the cost of making 3phosphoglycerate is 24 - 3.5 = 20.5 ATP equivalents, assuming that each NADH is equivalent to 2.5 ATPs. (The same cost can be derived from the Calvin cycle pathway in Figure 15.19.) The serine biosynthesis pathway produces one NADH when 3-phosphoglycerate is oxidized to 3-phosphohydroxypyruvate so the next cost of making serine is 20.5 - 2.5 = 18 ATP equivalents. This value is identical to the value given in Box 17.3. (Note that the transamination reaction in the serine biosynthesis pathway is cost-free.)
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COO
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Alanine is made from pyruvate in a simple, cost-free, transamination reaction. The cost of making pyruvate can be estimated from the conversion of 3-phosphoglycerate to pyruvate in the glycolytic pathway. This conversion is associated with a gain of 1 ATP so the cost of pyruvate is 20.5 - 1 = 19.5 ATP equivalents. Thus, the cost of synthesizing alanine is 19.5 ATP equivalents, or 20 ATP equivalents when rounded to two significant figures. This value is the same as that given in Box 17.3.
Chapter 18 Nucleotide Metabolism 1. (a)
* NH2
H2 N *
Ribose 5-phosphate
Ribose 5-phosphate
N
HN
N
N
N
N
O
(c)
* C N
N
N
*N
NH2
(b)
N *
N* Ribose 5-phosphate
See Figure 18.10 for the reactions in the pathway of UMP synthesis. (d) Labeled C-2 from aspartate, which is incorporated into carbamoyl aspartate, appears at C-6 of the uracil of UMP. (e) The labeled carbon from HCO3 , which is incorporated into carbamoyl phosphate, appears at C-2 of the pyrimidine ring of UMP. O (b) from HCO3
HN
C
C2
O3POCH2
6
(a) from C-2 of aspartate
O
H
CH
N
O 2−
CH
H
H
H OH OH UMP
2. Seven ATP equivalents are required. One ATP is cleaved to AMP when PRPP is synthesized (Figure 18.3). The pyrophosphoryl group of PRPP is released in Step 1 of the IMP biosynthetic pathway and subsequently hydrolyzed to 2 Pi (Figure 18.5), accounting for the second ATP equivalent. Five ATP molecules are consumed in Steps 2, 4, 5, 6, and 7. 3. Purines: Reaction 3: GAR transformylase 10-formyl-THF, C-8 position. Reaction 9: AICAR transformylase, 10-formyl-THF, C-2 position. Pyrimidines: Thymidylate synthase, 5,10-methylene-THF, 5-CH3 of thymidylate. 4. (a)
COO H3 N
C
H
COO H3 N
C
CH O N
CH 2 CH2
CH2 C
Cl Acivicin
H
H2 N
C
O Glutamine
(b) Acivicin inhibits glutamine-PRPP amidotransferase, the first enzyme in the purine biosynthetic pathway, so PRPP accumulates. (c) Acivicin inhibits the carbamoyl phosphate synthetase II activity of dihydroorotate synthase, which catalyzes the first step in the pyrimidine biosynthetic pathway
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5. (a) When b-alanine is used instead of aspartate, no decarboxylation reaction (Step 6 of the E. coli pathway) would be required. O
(b) HN
C* * CH *CH
C O
N Ribose 5-phosphate
6. (a) dUMP + NH4 (b) Synthesis of DNA requires certain ratios of A, T, G and C. If dTTP levels are higher than necessary, dTTP will act to decrease its own synthesis pathway by inhibiting the conversion of dCMP to dUMP by dCMP deaminase. dUMP is the precursor to dTMP (thymidylate synthase, Figure 18.15), and the subsequent conversion to dTDP and dTTP (needed for DNA synthesis). On the other hand, if dCTP levels are high, activation of dCMP deaminase will lead to an increased conversion of dCMP to dUMP and this diverts any dCMP that might have been converted to more dCTP by phosphorylation (Figure 18.19). 7. Four ATP equivalents are required. One ATP equivalent is required for the synthesis of PRPP from ribose 5-phosphate (Figure 18.3). Carbamoyl phosphate synthesis requires 2 ATP (Figure 18.10, Step 1). One ATP equivalent is consumed in Step 5, when PPi is hydrolyzed to 2 Pi. 8. In the absence of adenosine deaminase, adenosine and deoxyadenosine are not degraded via inosine and hypoxanthine to uric acid (Figure 18.18 and 18.20). This leads to an increase in the concentration of deoxyadenosine, which can be converted to dATP. High concentrations of dATP inhibit ribonucleotide reductase (Table 18.1). The inhibition of ribonucleotide reductase results in decreased production of all deoxynucleotides and therefore inhibits DNA synthesis. 9. The purine nucleotide cycle in muscle (Figure 18.22) supports some of the energy requirements in exercising muscle. A deficiency of myoadenylate deaminase limits the amount of fumarate provided to the citric acid cycle and therefore decreases the amounts of ATP available for muscle contraction. 10. Glutamine-PRPP amidotransferase is the first enzyme and the principal site of regulation in the de novo pathway to IMP (Figure 18.5). In humans, PRPP is both a substrate and a positive effector of this enzyme. An increase in the cellular levels of PRPP due to increased PRPP synthetase activity will therefore enhance the activity of the amidotransferase. This will result in an increased synthesis of IMP and other purine nucleosides and nucleotides. Overproduction of purine nucleotides and subsequent degradation can lead to elevated uric acid levels characteristic of gout. 11. (a) ATP (b) ATP (c) ATP (d) GTP (e) UTP (f) GTP (g) C TP (h) UTP ( i) ATP (j) IMP (k) IMP 12. Purines and pyrimidines are not significant sources of energy. The carbon atoms of fatty acids and carbohydrates can be oxidized to yield ATP, but there are no comparable energy-yielding pathways for nitrogen-containing purines and pyrimidines. However, the NADH produced when hypoxanthine is converted to uric acid may indirectly generate ATP via oxidative phosphorylation. The degradation of uracil and thymine yields acetyl CoA and succinyl CoA, respectively, which can be metabolized via the citric acid cycle to generate ATP. 13. The sugar D-ribose exists as an equilibrium mixture of a-D-ribopyranose, a-D-ribofuranose. b-D-ribopyranose, and b-D-ribofuranose. These forms freely interconvert with each through the open-chain form (Section 8.2). 14. Xanthine is 2,6-dioxopurine; hypoxanthine is 6-oxopurine; orotate is 2,4-dioxo-6carboxyl-pyrimidine. 15. SAICAR synthetase + adenylosuccinate lyase in the IMP biosynthesis pathway (Figure 8.5) and argininosuccinate synthetase + argininosuccinate lyase in the arginine biosynthesis pathway (urea cycle: Figure 17.39).
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Chapter 19 Nucleic Acids 1. In the a helix, hydrogen bonds form between the carbonyl oxygen of one residue and the amine hydrogen four residues, or one turn, away. These hydrogen bonds between atoms in the backbone are roughly parallel to the axis of the helix. The amino acid side chains, which point away from the backbone, do not participate in intrahelical hydrogen bonding. In double-stranded DNA, the sugar-phosphate backbone is not involved in hydrogen bonding. Instead, two or three hydrogen bonds, which are roughly perpendicular to the helix axis, form between complementary bases in opposite strands. In the a helix, the individual hydrogen bonds are weak, but the cumulative forces of these bonds stabilize the helical structure, especially within the hydrophobic interior of a protein where water does not compete for hydrogen bonding. In DNA, the principal role of hydrogen bonding is to allow each strand to act as a template for the other. Although the hydrogen bonds between complementary bases help stabilize the helix, stacking interactions between base pairs in the hydrophobic interior make a greater contribution to helix stability. 2. If 58% of the residues are 1G + C2, 42% of the residues must be 1A + T2. Since every A pairs with a T on the opposite strand, the number of adenine residues equals the number of thymine residues. Therefore, 21%, or 420, of the residues are thymine 12000 * 0.21 = 4202. 3. (a) The base compositions of complementary strands of DNA are usually quite different. For example, if one strand is poly dA (100% A), the other strand must be poly dT (100% T). However, since the two strands are complementary, the amount of 1A + T2 must be the same for each strand, and the amount of 1G + C2 must be the same for each strand. (b) 1A + G2 = 1T + C2. Complementarity dictates that for every purine (A or G) on one strand, there must be a pyrimidine (T or C) on the complementary strand. 4. Since the DNA strands are anti-parallel, the complementary strand runs in the opposite direction. The sequence of the double-stranded DNA is ATCGCGTAACATGGATTCGG TAGCGCATTGTACCTAAGCC
By convention, DNA sequences are written in the 5¿ : 3¿ direction. Therefore the sequence of the complementary strand is CCGAATCCATGTTACGCGAT
5. The stability of the single-stranded helix is largely due to stacking interactions between adjacent purines. Hydrophobic effects also contribute, since the stacked bases form an environment that is partially shielded from water molecules. 6.
H
H N
N
6 1N
N
H
N 4
H
N3
N Adenine (imino form)
N O Cytosine
7. There will be two discrete melting points separated by a plateau. When the extra strand of poly dT is released the absorbance of the solution at 260 nm will increase as the stacked bases leave the largely hydrophobic interior of the triple helix. A second increase in the absorbance occurs when the remaining two DNA strands denature. 8. The sequence is 5¿ ACGCACGUAUAUGUACUUAUACGUGGCU 3¿
The underlined sequences are palindromic. 9. The main products will be a mixture of mononucleotides and pieces of single-stranded DNA approximately 500 bp in length. A piece of DNA with an enzyme molecule bound
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at each end will be degraded until the two strands can no longer base-pair, at that point the single strands cease to be a substrate for the enzyme. 10. In the 30 nm fiber, DNA is packaged in nucleosomes, each containing about 200 bp of DNA; therefore, the DNA in a nucleosome has a molecular weight of 130 000 1200 * 650 = 130 0002. Assuming there is one molecule of histone H1 per nucleosome, the molecular weight of the protein component of the nucleosome would be 129 800. Histone H1
Histone H2A 1*22
Histone H2 B 1*22 Histone H3 1 *22 Histone H4 1 *22 Total
021000 028000 027600 030600 022600 129800
Thus, the ratio by weight of protein to DNA is 129800:130000, or approximately 1:1. 11. Nucleosomes are composed of histones plus 200 base pairs of DNA. Since you inherited half your chromosomes from your mother, the oocyte contained 13.2 * 109 bp2 *
1 nucleosome = 8 * 106 nucleosomes 200 bp
(You inherited no nucleosomes from your father, since nucleosomes are replaced by small, positively charged polypeptides during spermatogenesis.) 12. (a) pdApdGpdT + pdC (b) pdAp + dGpdTpdC (c) pdA + pdGpdTpdC 13. Since the supercoiled plasmid DNA is in equilibrium with relaxed DNA containing short unwound regions the Aspergillus enzyme will slowly convert the DNA into nicked circles. Eventually the enzyme will convert the relaxed circles into unit-length linear fragments of double-stranded DNA. 14. Yes. The sugar-phosphate backbone in both RNA and DNA contains phosphodiester bonds that link the sugar residues. 15. pppApCpUpCpApUpApGp + CpUpApUpGp + ApGp + U 16. Bacteriophages have evolved several mechanisms to protect their DNA from restriction endonucleases. In general, bacteriophage DNA contains few restriction sites. Restriction endonuclease recognition sites are strongly selected against and any mutations that alter these sites will ne favored. In addition, restriction sites are often methylated, as in the bacterial chromosome. This is presumably due to a fortuitous event in the distant past when the phage DNA became methylated before it could be cleaved. Some bacteriophages incorporate modified nucleotides into their DNA. The modified nucleotides (e.g., 5-hydroxymethylcytosine in bacteriophage T4) are not recognized by restriction endonucleases. Phage genomes may also encode an enzyme that inactivates restriction endonucleases, or they may encode proteins that bind to restriction sites to prevent cleavage. 17. (a) The probability can be estimated from the probability of each nucleotide in the Hind III restriction site. 1G = C = 0.18 and A = T = 0.322 For the sequence AAGCTT there will be, on average, one Hind III site every 1/10.32210.32210.18210.18210.32210.322 = 2943 bp Thus, in a 100 Mb genome there will be, on average, 100 000/2943 = 33 070 sites (b) 24414
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Chapter 20 DNA Replication, Repair, and Recombination 1. (a) Two replication forks form at the origin of replication and move in opposite directions until they meet at a point opposite the origin. Therefore, each replisome replicates half the genome (2.6 * 106 base pairs). The time required to replicate the entire chromosome is 2.6 * 106 base pairs 1 000 base pairs s-1
= 2600 s = 43 min and 20 s
(b) Although there is only one origin (O), replication can be reinitiated before the previous replication forks have reached the termination site. Thus, the chromosome can contain more than two replication forks. Replication of a single chromosome still requires approximately 43 minutes, but completed copies of each chromosome can appear at shorter intervals, depending on the rate of initiation. Replication forks initiated before completion of first round of DNA replication O First replication fork O O
O
First replication fork
2. T4 DNA polymerase should be an early gene product because it is required for replication of the viral genome. 3. (a) The single-stranded DNA template used for DNA synthesis in vitro can form secondary structures such as hairpins. SSB prevents the formation of double-stranded structure by binding to the single-stranded template. SSB thus renders the DNA a better substrate for DNA polymerase. (b) The yield of DNA in vitro is improved at higher temperatures because formation of secondary structure in the template is less likely. A temperature of 65°C is high enough to prevent formation of secondary structure but not high enough to denature the newly synthesized DNA. DNA polymerases from bacteria that grow at high temperatures are used because they are active at 65°C, a temperature at which DNA polymerases from other bacteria would be inactive. 4. Extremely accurate DNA replication requires a proofreading mechanism to remove errors introduced during the polymerization reaction. Synthesis of an RNA primer by a primase, which does not have proofreading activity, is more error prone than DNA synthesis. However, because the primer is RNA, it can be removed by the 5¿ : 3¿ exonuclease activity of DNA polymerase I and replaced with accurately synthesized DNA when Okazaki fragments are joined. If the primer were composed of DNA made by a primase without proofreading activity, it would not be removed by DNA polymerase I, and the error rate of DNA replication would be higher at sites of primer synthesis. 5. (a) In the hypothetical nucleotidyl-group-transfer reaction, the nucleophilic 3¿-hydroxyl group of the incoming nucleotide would attack the triphosphate group of the growing chain. Pyrophosphate would be released when a new phosphodiester linkage was formed.
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O O
P
O O
O
P
O O
P
O
5′
CH2
O
O
B
O
H
H
H
H
Incoming nucleoside triphosphate
3′
H
..
O
H O
O
O 5′
O
P
O
O
P
O
P
O
CH2
O
O
B
O
H
H
H
H
Growing chain
3′
O
H
3′ DNA
PPi O
O
O 5′
O
P O
O
P O
O
P
O
CH2
O
B
O
H
H
H
H 3′
O
H
O O
P
O
O 5′
CH2
B
O
H
H
H
H 3′
O
H
3′ DNA
(b) If the hypothetical enzyme had 5¿ : 3¿ proofreading activity, removal of a mismatched nucleotide would leave a 5¿-monophosphate group at the end of the growing chain. Further DNA synthesis, which would require a terminal triphosphate group, could not occur. 6. Topisomerase II or gyrase relieves supercoiling ahead of and behind the replication fork. If this enzyme is inhibited, the unwinding of the parental DNA cannot occur. Therefore, the DNA of the E. coli cannot be replicated. 7. (a) Assume that the genome is one large linear molecule of DNA and that the origin of replication is at the midpoint of this chromosome. Since the replication forks move in opposite directions, 60 base pairs can be replicated per second. The time required to replicate the entire genome would be 1.65 * 108 base pairs 60 base pairs s-1
= 2.75 * 106 s = 764 h = 32 days
(b) Assuming that the 2000 bidirectional origins are equally spaced along the DNA molecule and that initiation occurs simultaneously at all origins, the rate would be 2000 * 2 * 30 base pairs per second, or 1.2 * 105 base pairs per second. The time required to replicate the entire genome would be
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1.65 * 108 base pairs 1.2 * 105 base pairs s-1
= 1375 s = 23 min
(c) Assume that the origins are equally spaced and that initiation at all origins is simultaneous. The required rate of replication is 1.65 * 108 base pairs 300 s
= 5.5 * 105 base pairs s-1
Bidirectional replication from each fork proceeds at an overall rate of 60 base pairs per second. The minimum number of origins would be 5.5 * 105 base pairs s-1 60 base pairs s-1 origin-1
= 9170 origins
8. The modified G can no longer form a productive Watson-Crick base pair with C but can now base-pair with T. Therefore, one of the daughter strands of DNA will contain a T across from the modified base. After further rounds of replication, the T will base-pair with A and what was originally a G/C base pair will have mutated into an A/T base pair. 9. Ultraviolet light can damage DNA by causing dimerization of thymidylate residues. One mechanism for repairing thymine dimers is enzymatic photoreactivation, catalyzed by DNA photolyase. This enzyme uses energy from visible light to cleave the dimer and repair the DNA. Thus, cells that are exposed to visible light following ultraviolet irradiation are better able to repair DNA than cells kept in the dark. 10. (a) DNA from a dut- strain will appear normal because the Ung enzyme will remove any uracil that gets incorporated. (b) DNA from a dut-, ung- strain will contain dU residues in the place of some dT residues. 11. The DNA repair enzyme uracil N-glycosylase removes uracil formed by the hydrolytic deamination of cytosine. Because the enzyme does not recognize thymine or the other three bases normally found in DNA, it cannot repair the damage when 5-methylcytosine is deaminated to thymine. 12. High mutation rates occur at methylcytosine-containing regions because the product of deamination of 5-methylcytosine is thymine, which cannot be recognized as abnormal. When the mismatched T/G base pair that results from deamination of methylcytosine is repaired, the repair enzymes may delete either the incorrect thymine or the correct guanine. When the guanine is replaced by adenine, the resulting A/T base pair is a mutation. Normal sequence
m5C
G
Deamination
T
G
C
G
T
A
Repair
Mispaired bases Parental strand Daughter strand
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13. Proofreading during replication results in excision of 99% of misincorporated nucleotides, thus reducing the overall error rate to 10-7. Of those errors that escape the proofreading step, a further 99% are corrected by repair enzymes. The overall mutation rate is therefore 10-9. 14. Yes. The E. coli enzyme DNA ligase is required to seal the nicks left in the DNA strands following DNA repair. This enzyme has a strict requirement for NAD { . 15. The dimers can be removed by excision repair. UvrABC endonuclease removes a 12-13 residue segment containing the pyrimidine dimer. The DNA oligonucleotide is removed with the help of a helicase. The gap is filled by the action of DNA polymerase I, and the nick sealed by the action of DNA ligase. The dimers can also be repaired through direct repair. DNA photolyase binds to the distorted double helix at the site of the dimer. As the DNA-enzyme complex absorbs light, the dimerization reaction is reversed. 16. The repair enzymes need an undamaged template in order to repair mutations in DNA. If both strands of the DNA molecule have been damaged, there is not a template to use for repair. 17. The proteins that catalyze strand exchange recognize regions of high sequence similarity and promote formation of a triple-stranded intermediate in which the invading strand base pairs with a complementary strand. This pairing would not be possible if the sequences of the two DNA molecules were different. 18. DNA polymerase III is a component of the replisome that synthesizes the leading strand and the lagging strand during replication of the E. coli chromosome. DNA polymerase I is required to remove the short RNA primers on the lagging strand.
Chapter 21 Transcription and RNA Processing 1. (a) Since the rate of transcription is 70 nucleotides per second and each transcription complex covers 70 base pairs of DNA, an RNA polymerase completes a transcript and leaves the DNA template each second (assuming that the complexes are densely packed). Therefore, when the gene is loaded with transcription complexes, 60 molecules of RNA are produced per minute. (b) Since each transcription complex covers 70 base pairs, the maximum number of complexes is 6000 base pairs 70 base pairs transcription complex
= 86 transcription complexes
2. (a) Since the average E. coli gene is 1 kb (1000 bp) long, 4000 genes account for 4000 kb of DNA. The percentage of DNA that is not transcribed is 500 kb * 100% = 10.9% 4600 kb
Most of the nontranscribed DNA consists of promoters and regions that regulate transcription initiation. (b) Since the gene products in mammals and bacteria are similar in size, the amount of DNA in the exons of a typical mammalian gene must also be 1000 bp. The total amount of DNA in exons is 5 * 104 genes * 1.0 kb gene-1 = 5 * 104 kb
This DNA represents about 1.7% of the mammalian genome. 5 * 104 kb 3 * 106 kb
* 100% = 1.7%
The remaining 97.5% of DNA consists of introns and other sequences. 3. No. It is extremely unlikely that the eukaryotic gene’s promoter will contain the correct sequences in the correct location to permit accurate initiation by the prokaryotic RNA
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polymerases. Likewise, it is extremely unlikely that the prokaryotic gene’s promoter will contain the correct sequence in the correct location to permit accurate initiation by RNA polymerase II. 4. No. A typical eukaryotic triose phosphate isomerase gene contains introns. The prokaryotic cell contains no spliceosomes and therefore will not be able to correctly process the primary transcript. Therefore translation of the RNA will yield an aberrant protein fragment. 5. (a) In the presence of both lactose and glucose, the lac operon is transcribed at a low level because lac repressor forms a complex with allolactose (an isomer of lactose). Because the allolactose-repressor complex cannot bind to the promoter region of the lac operon, the repressor does not prevent initiation of transcription. (b) In the absence of lactose, no allolactose is formed. Thus, lac repressor binds near the lac operon promoter and prevents transcription. (c) When lactose is the sole carbon source, the lac operon is transcribed at the maximum rate. In the presence of allolactose, transcription is allowed since lac repressor does not bind to the promoter region of the lac operon. Also, in the absence of glucose, the transcription rate increases because cAMP production increases, making more CRP-cAMP available to bind to the promoter region of the lac operon. The absence of the repressor and the enhancement of transcription initiation by CRPcAMP allow the cell to synthesize the quantities of enzymes required to support growth when lactose is the only carbon source. 6. Since the wild-type lac promoter is relatively weak, maximal transcription requires the activator CRP. The UV5 mutations alters the -10 region such that it now resembles the consensus -10 sequence making it a much stronger promoter. In the absence of the lac repressor, the promoter is independent of CRP. 7.
32 P appears only at the 5¿ end of mRNA molecules that have ATP as the first residue. It does not appear in any other residues because pyrophosphate, which includes the b-phosphoryl group, is released when nucleoside triphosphates are added to the 3¿ end of a growing RNA chain (Figure 21.3).
When the 5¿ end of mRNA is capped, only the g-phosphoryl group of the initial residue is removed when the cap forms. The b-phosphoryl group, which contains the label, is retained and receives the GMP group from GTP (Figure 21.26). 8. The lack of proofreading activity in RNA polymerase makes the error rate of transcription greater than the error rate of DNA replication. However, the defective RNA molecules produced are not likely to affect cell viability because most copies of RNA synthesized from a given gene are normal. In the case of defective mRNA, the number of defective proteins is only a small percentage of the total number of proteins synthesized. Also, mistakes made during transcription are quickly eliminated since most mRNA molecules have a short half-life. 9. During maturation, eukaryotic mRNA precursors are modified at their 3¿ ends by the addition of a poly A tail. When a mixture of components from a cell extract is passed over the column, the poly A tail will hybridize with oligo dT on the column. The other components in the cell extract will pass through the column. The bound mature mRNA with the poly A tail is removed from the column by changing the pH or the ionic strength of the buffer. This will disrupt the hydrogen bonds between the A and T nucleotides. 10. (a) A much lower concentration of rifampicin stopped the growth of the wild type E. coli 165 mg/mL2 as compared to the concentration of rifampicin that stopped the growth of the mutant 1 750 mg/mL2. (b) RNA polymerase consists of a core enzyme with a stoichiometry of a2 bb¿v that participates in many of the transcription reactions. The large b and b¿ subunits make up the active site of the enzyme. (c) The rifampicin-resistant bacteria could arise from mutations that occur in the gene for the b subunit of RNA polymerase.
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11. Since either strand can serve as a template, two mRNA molecules can be transcribed from this DNA segment. When the bottom strand is the template, the mRNA sequence is complementary to the bottom strand.
5′
C
3′
G
mRNA 5′
C
G GC TAAGAT C TGAC TA
G
C
3′
G
5′
C CGGCUAAGAUCUGACUAGC mRNA
3′
C
G
C
C G
CGA T T C T AGAC T GA
T
C
GC UAAGAUCUGA OH 3′
Direction of transcription
5′
When the top strand is the template, the mRNA sequence is complementary to the top strand.
3′
HO
5′
C
3′
G
5′
C
Direction of transcription
A U U C U A G A C U G AU
G GC TAAGAT C TGAC TA
C G
G
5′ mRNA
C
3′
G
5′
GCUAGUC AGAUCUUAGC C GG mRNA
3′
G
C
CGA T T C T AGAC T GA
T
C
12. A gene was defined as a DNA sequence that is transcribed. By this definition, the entire ribosomal RNA operon is a gene. However, it is sometimes more convenient to restrict the term gene to the segment of RNA that encodes a functional product, for example, one of the enzymes encoded by the lac operon. The operon in Figure 21.25 therefore contains tRNA and 16S, 23S, and 5S rRNA genes. The DNA sequences between these genes, although transcribed, are not considered part of any gene. 13. The genomic DNA sequence provides an accurate rendition of the primary RNA sequence as expected. However, sequencing a purified tRNA reveals that many of the nucleotides have been specifically modified post-transcriptionally. The same is true for eukaryotes. 14. The gene for triose phosphate isomerase in maize contains about 3400 base pairs. If the spliceosome assembles at the first intron, then 2900 base pairs remain to be transcribed. The time required to transcribe 2900 base pairs is 97 seconds (2900 nucleotides , 30 nucleotides per second). If the spliceosome assembles immediately after transcription of the first intron, and if splicing cannot begin until transcription of the entire gene is complete, the spliceosome must be stable for at least 97 seconds. 15. The CRP-cAMP binding site probably overlaps the promoter of the gene. When CRPcAMP binds, the promoter is blocked and transcription cannot occur. 16. When the sequence of the 5¿ or 3¿ splice site or the branch point is altered by mutation, proper splicing cannot occur and no functional mRNA can be produced. 17. Yes. Once the U2 snRNP binds to the branch site it will occlude the U5 snRNP from binding to the 3¿ splice acceptor and interfere with splicing. Furthermore, the deletion will have removed a large part of the pyrimidine stretch required for binding to the 3¿ splice site. Both of these will prevent proper mRNA processing and the aberrant RNA will not be properly translated.
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Chapter 22 Protein Synthesis 1. One strand of DNA has three different overlapping reading frames, therefore a doublestranded DNA has six reading frames. This can be seen by examining the DNA sequence beginning at the 5¿ end of each strand and marking off the triplet codons. This identifies one reading frame on each strand. Now start at the second nucleotide in from the 5¿ ends and mark off triplet codons; that is reading frame 2. The third reading frame on each strand begins at the third nucleotide in from the 5¿ ends. The “fourth” reading frame is identical to the first—test this for yourself. Using similar logic it follows that if the genetic code were read in codons four nucleotides in length, then one strand of DNA could be read in four different reading frames and therefore a double-stranded piece of DNA would contain eight reading frames (four on each strand). 2. Each mRNA sequence could be translated in three different reading frames. For the first mRNA sequence, the possible codons and polypeptide sequences are Reading Frame 1
5′
CCGGCUAAGAUCUGACUAGC Ala Lys Ile STOP Pro
3′
Reading Frame 2
5′
CCGGCUAAGAUCUGACUAGC Ala Leu Arg Ser Asp STOP
3′
Reading Frame 3
5′
CCGGCUAAGAUCUGACUAGC Gly STOP
3′
For the second mRNA sequence, the possible codons and polypeptide sequences are Reading Frame 1
5′
GCUAGUCAGAUCUUAGCCGG Gln Ile Leu Ala Gly Ala Ser
3′
Reading Frame 2
5′
GCUAGUCAGAUCUUAGCCGG Leu Val Arg Ser STOP
3′
Reading Frame 3
5′
GCUAGUCAGAUCUUAGCCGG
3′
STOP
Since only a reading frame without a stop codon can encode a polypeptide, the second mRNA sequence corresponds to the actual transcript. The sequence of the encoded polypeptide is –Ala–Ser–Gln–Ile–Leu–Ala–Gly–. 3. Two phosphoanhydride bonds are hydrolyzed for each amino acid activated by an aminoacyl-tRNA synthetase. Amino acid + tRNA + ATP ¡ Aminoacyl - tRNA + AMP + PPi PPi + H2O ¡ 2 Pi
The rest of the energy needed to synthesize the protein is provided by hydrolysis of GTP: one high-energy bond is hydrolyzed in the formation of the 70S initiation complex, another during the insertion of each aminoacyl-tRNA into the A site of the ribosome, and another at each translocation step. Since the initial methionyl-tRNA is inserted into the P site, 599 new insertions and 599 translocations occur during the synthesis of a 600-residue protein. Finally, one phosphoanhydride bond is hydrolyzed during release of the completed polypeptide chain from the ribosome. The total number of phosphoanhydride bonds hydrolyzed during synthesis of the protein is Activation 1600 * 22 Initiation
1200 1
Insertion
599
Translocation
599
Termination Total
1 2400
4. The answer depends on your frame of reference. For example, relative to the ribosome, the mRNA and both tRNAs get translocated by one triplet codon. Relative to the mRNA it is the ribosome that is shifted by three nucleotides.
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5. The region of the mRNA molecule upstream of the true initiation codon contains the purine-rich Shine-Dalgarno sequence, which is complementary to a pyrimidine-rich sequence at the 3¿ end of the 16S rRNA component of the 30S ribosomal subunit (Figure 22.17). By correctly positioning the 30S subunit on the mRNA transcript, the ShineDalgarno sequence allows fMct-tRNAMct to bind to the initiation codon. Once protein f synthesis begins, all subsequent methionine codons are recognized by Met-tRNAMet. 6. No, because proper translation initiation in an E. coli cell requires a Shine-Dalgarno sequence located in the 5¿ untranslated region of the mRNA. Since eukaryotic ribosomes do not have this requirement, it is extremely unlikely that an mRNA from a plant would fortuitously contain a Shine-Dalgarno sequence in the proper location. If, however, the part of the gene encoding the plant mRNA were fused to a bacterial Shine-Dalgarno sequence, then the open reading frame for the plant protein would be properly translated in the bacterial cell. 7. The transcript of each rRNA gene is an rRNA molecule that is directly incorporated into a ribosome. Thus, multiple copies of rRNA genes are needed to assemble the large number of ribosomes that the cell requires. In contrast, the transcript of each ribosomal protein gene is an mRNA that can be translated many times. Because of this amplification of RNA to protein, fewer genes are needed for each ribosomal protein than for rRNA. 8. Possible suppressor tRNA species include all those that recognize codons differing from UAG by a single nucleotide, that is, tRNAs whose anticodons differ by a single nucleotide from the sequence CUA, which is complementary to the stop codon UAG. tRNAGln, tRNALys, and tRNAGln all recognize codons that differ only at the first position (codons CAG, AAG, and GAG, respectively). tRNALeu, tRnaSer, and tRNATrp recognize codons that differ only at the second position (UUG, UCG, and UGG, respectively). tRNATyr recognizes codons that differ only at the third position (UAU or UAC). A cell that contains a suppressor tRNA can survive despite the loss of a normal tRNA because the cell also contains isoacceptor tRNA molecules that carry the same amino acid. Although the suppressor tRNA may occasionally insert an amino acid at a normally occurring stop codon, the resulting protein, which is larger than the normal gene product, is usually not lethal to the cell. In fact, strains of E. coli that contain suppressor tRNAs do survive but are often not as healthy as wild-type strains. 9. (a) Aminoacyl-tRNA synthetases—the enzymes that bind to tRNAs and catalyze aminoacylation. (b) IF-2 in bacteria and eIF-2 in eukaryotes, a protein that binds to aminoacylated initiator-tRNA and loads it into the ribosome’s P site during translation initiation. (c) EF-Tu in bacteria and EF-1a in eukaryotes—a protein that binds to charged tRNAs and loads them into the ribosome’s A site during polypeptide elongation. (d) Ribosomes. These large complexes of RNA and protein contain two sites that can bind specifically to tRNAs, the A site and the P site. (e) mRNA—tRNAs bind to mRNA through codon-anticodon hydrogen bonds. The enzymes that modify specific residues on individual tRNAs during the maturation process must also be able to bind to tRNAs. 10. Under normal circumstances, when the translation machinery encounters UGA in RF-2 mRNA, RF-2 recognizes the stop codon and terminates protein synthesis. When the cellular concentration of RF-2 is low, however, the ribosome pauses at the termination codon, shifts frame, and continues translating the RF-2 mRNA to produce the fulllength functional protein. Thus, the presence of the stop codon encourages translational frameshifting in the absence of RF-2, and allows RF-2 to regulate its own production. 11. (a) If the entire leader region were deleted, attenuation would not be possible, and transcription would be controlled exclusively by trp repressor. The overall rate of transcription of the trp operon would increase. (b) If the region encoding the leader peptide were deleted, transcription would be controlled exclusively by trp repressor. Deletion of the sequence encoding the leader peptide would remove Sequence 1, thus allowing the stable 2–3 hairpin to form. Since neither the pause site (1–2 hairpin) nor the terminator (3–4 hairpin) could form, initiated transcripts would always continue into the trp operon.
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(c) If the leader region did not contain an AUG codon, the operon would be rarely transcribed. Because of the absence of the initiation codon, the leader peptide would not be synthesized, and 1–2 hairpins and 3–4 hairpins would almost always form, leading to termination of transcription. 12. No, this is difficult to imagine. One of the important features of the attenuation model is that one or more codons in the leader peptide usually encode the amino acid that is synthesized by that operon. It is the relative shortage or abundance of particular aminoacylated tRNAs that modulates the attenuation. The products of the lac operon are not directly involved in amino acid biosynthesis so we would not expect cellular levels of one class of aminoacylated tRNAs to vary with the activity of the operon. 13. The presence of codons specifying valine and leucine in the leader regions of isoleucine operons suggests that a scarcity of these amino acids would promote transcription of the genes for isoleucine biosynthesis. Many of the enzymes required to synthesize isoleucine are also required in the pathways to valine and leucine (Section 18.5A). Thus, even when the isoleucine concentration is high, a low concentration of valine or leucine ensures that transcription of the isoleucine operon does not terminate prematurely. 14. As the newly synthesized protein is extruded from the ribosome, the N-terminal signal peptide is recognized and bound by a signal-recognition particle (SRP). Further translation is inhibited until the SRP binds to its receptor on the cytosolic face of the endoplasmic reticulum. Ribophorins anchor the ribosome to the endoplasmic reticulum. When translation resumes, the polypeptide chain passes through a pore into the lumen. If the polypeptide does not pass completely through the membrane, the result is an integral membrane protein with its N-terminus in the lumen of the endoplasmic reticulum and its C-terminus in the cytosol. Glycosylation of specific residues takes places in the lumen of the endoplasmic reticulum and in the Golgi apparatus. The protein, still embedded in the membrane, is transported between the endoplasmic reticulum and the Golgi apparatus in transfer vesicles that bud off the endoplasmic reticulum. Secretory vesicles transport the fully glycosylated protein from the Golgi apparatus to the plasma membrane. When the vesicles fuse with the plasma membrane, the Nterminal portion of the protein, which was in the lumen, is now exposed to the extracellular space, and the C-terminal portion remains in the cytosol. 15. Yes. A hydrophobic secretion signal sequence located at the N-terminus of a protein is necessary and sufficient for entry into the cell’s secretory pathway. 16. The initiator tRNA anticodon pairs with GUG by forming a G/U base pair between the 5¿ nucleotide of the codon and the 3¿ position of the anticodon. 3′ Initiator tRNA anticodon mRNA codon 5′
5′ UAC GUG
3′
This interaction is unrelated to wobble, since the 5¿ position of the anticodon is the wobble position.
Chapter 23 Recombinant DNA Technology 1. Only cells that have taken up intact plasmids form colonies, because untransformed cells are sensitive to tetracycline. There are two classes of plasmids: intact pBR322 formed by ligation of the ends cut by PstI, and a recombinant molecule formed by insertion of the PstI restriction fragment into the vector. Cells containing the intact vector are resistant to ampicillin, whereas those containing the recombinant are sensitive to ampicillin, since the PstI site is within the ampR gene (Figure 23.3). These two types of cells can be distinguished by transferring part of the colonies to another petri plate with medium containing ampicillin. Cells from colonies that fail to grow in the presence of ampicillin contain the recombinant plasmid.
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2. The transformed cells that grow in the absence of leucine must contain a fragment of yeast DNA that encodes the yeast version of b-isopropylate dehydrogenase. The E. coli cells are able to produce a functional version of this yeast enzyme, suggesting that the yeast and E. coli genes are evolutionarily conserved. Furthermore, the yeast gene must not contain any introns since intron-containing mRNA precursors cannot be processed in bacteria. (The yeast LEU2 gene, described in Section 23.3B, was first cloned in this manner. The technique is called cloning by complementation.) 3. Yes. In fact this was the method used to identify ARS elements in the first place. 4. (a)
SmaI XmaI SacI
EcoRI
KpnI
BamHI
XbaI
AccI HincII SalI
PstI
SphI
HindIII
5′ CAGCTATGACCATGATTACGAATTCGAGCTCGGTACCCGGGGATCCTCTAGAGTCGACCTGCAGGCATGCAAGCTTGGCACTG Met Thr Met Ile Thr Asn Ser Ser Ser Val Pro Gly Asp Pro Leu Glu Ser Thr Cys Arg His Ala Ser Leu Ala Leu
(b) SphI
HindIII
PstI
AccI HincII SalI
SmaI XmaI XbaI
BamHI
KpnI
SacI
EcoRI
5′ CAGCTATGACCATGATTACGCCAAGCTTGCATGCCTGCAGGTCGACTCTAGAGGATCCCCGGGTACCGAGCTCGAATTCACTG Met Thr Met Ile Thr Pro Ser Leu His Ala Cys Arg Ser Thr Leu Glu Asp Pro Arg Val Pro Ser Ser Asn Ser Leu
5. Using Equation 23.1, P = 0.99 and n =
300 kb 3.2 * 106 kb
L 10-4
Thus, N =
ln11 - 0.992
ln11 - 10-42
-4.6 = -10-4
L 46 000
6. The synthetic DNA is inserted into the SmaI site in both orientations. In either orientation, the reading frame of the b-galactosidase gene is maintained. In one orientation, the insert is translated and functional b-galactosidase is produced. Colonies arising from this ligation are blue. Pro 5′
Leu
His
Thr
Thr
Gln
Val
Functional β-galactosidase
Gly
CCCTTACATACTACCCAAGTAGGG
3′
In the opposite orientation, however, a stop codon (TAA) halts translation of the b-galactosidase gene; colonies arising from this ligation are white. Pro 5′
Tyr
Leu
Gly
Ser
Met STOP
CCCTACTTGGGTAGTATGTAAGGG
No β-galactosidase 3′
7. The pBR322 rop gene maintains a relatively low copy number (about 20 plasmids per cell). By inactivating the Rop protein the copy number of the new plasmid would be much higher (several hundred plasmids per cell). In fact, many of the common cloning vectors in use today do not contain the rop gene for just this reason. 8. No. Whenever you transform a plasmid carrying the strong regulated PR promoter from bacteriophage l into an E. coli host that already contains a l lysogen integrated into the E. coli chromosome, the incoming DNA will titrate the cI repressor protein (recall that the l repressor protein effectively keeps the l prophage in a dormant state). As cI repressor protein binds to the sites on the multicopy plasmid DNA the prophage will be induced and enter the lytic pathway. The lytic pathway leads to prophage excision, replication, DNA packaging, and eventual lysis of the E. coli host. Therefore all of the cells that take up the plasmid will die before they can form colonies. 9. An expression system must be chosen that can produce the protein and add the carbohydrate moieties to the expressed protein. Since bacteria lack the enzymes to catalyze
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some of the post translational modifications, the yeast expression system would be a better choice. 10. Although the recognition sites for BglII and BamH1 differ, the enzymes produce fragments with identical sticky ends. These fragments can be ligated as easily as fragments produced by a single enzyme. BglII
A TCTAG
GATCT A
BamHI
G CCTAG
GATCC G
11. Mechanical shearing of DNA produces a random assortment of fragments because the breaks do not occur at the same position in each DNA molecule. In contrast, a restriction endonuclease produces a population of fragments with lengths equal to the distances between restriction sites. Depending on the location of the restriction sites, these fragments may be too large or too small to be cloned, and portions of the genome may not be represented in the library. Shearing DNA overcomes this difficulty. Shearing also ensures that the library contains overlapping fragments of DNA, making it possible to carry out a chromosome walk. 12. E. coli cells that contain a plasmid carrying a full-length copy of the TaqI gene will express the TaqI enzyme. The enzyme will cleave at all of its recognition sites in the host chromosome and kill the cell. Therefore it will be very difficult to find any colonies in this experiment that hybridize to the labeled subfragment from the gene. Since the T. aquaticus genomic DNA was sheared, there may be rare clones in the library that contain incomplete fragments from the gene but these will be of limited use in your experiments. One way to obtain a full-length clone would be to transform E. coli cells that have already been engineered to express the Taq DNA methylase. 13. Restriction enzymes present in normal host cells might cleave newly introduced recombinant molecules, making it impossible to clone certain fragments of DNA. Using a host strain that does not make restriction endonucleases avoids this problem. A mutation in RecA reduces recombination, thereby preventing the rearrangement of recombinant DNA molecules during propagation in the host cells. Rearrangement is often a problem, particularly when the cloned fragment of DNA contains repetitive sequences that can serve as sites for homologous recombination. 14. The DNA fragment was used to probe a Drosophila genomic library in order to isolate overlapping fragments. This process was repeated (a chromosome walk) until the bithorax genes were reached. This was the first example of a chromosome walk. You might want to consult the original paper to see some of the “tricks” used to overcome several technical problems (Bender, W., Spierer, P., and Hogness, D. S. (1983). Chromosomal walking and jumping to isolate DNA from the ace and rosy loci and the bithorax complex in Drosophila melanogaster. J. Mol. Biol. 168:17–33). 15. When a gene responsible for a genetic disease is linked to a particular chromosome, the gene can be cloned by chromosome walking from a nearby site. The ESTs can be used as the start points for chromosome walks. In addition, since the ESTs represent expressed genes, they may actually represent cloned cDNAs from the desired gene. 16. No. There will be no double-stranded DNA amplified if the DNA primers are pointing out instead of pointing inwards toward each other. In each cycle of the reaction, the primers will bind to their target sites and the polymerase will make one copy of the DNA flanking each side of the desired region. There will be no increase in binding sites for the primers and no exponential amplification. 17. Since Parent 1 is homozygous and he received one copy of the locus from each of his parents, we can infer that each grandparent contained at least one identical copy of the locus. In other words, the two grandparents were at least heterozygous at this polymorphic locus. A similar argument can be constructed to demonstrate that the other two grandparents were at least heterozygous for the other copy of this polymorphic locus. 18. PCR could be used to detect the gene for Bt toxin in the corn. The DNA is extracted from a small sample of the food. The PCR detection method uses primer pairs specific
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for the Bt toxin gene. The PCR products are then analyzed by electrophoresis to look for the amplified Bt toxin product. 19. In this experiment, the fidelity of the DNA replication by the wild type Taq polymerase and a mutant Taq polymerase were compared. Both enzymes were used to copy a target gene. The products from the PCR reaction were then sequenced to determine if the PCR products contained the original sequence of the target gene or if there were misincorporated bases that would result in mutations in the DNA sequence of the PCR products. The results show that the wild type had no mutations in 20 clones, 5 of the clones each contained 1 mutation, and 2 clones had 2 mutations within the gene product. When the products from the mutant polymerase were sequenced, more mutations were found in the clones. Only 6 clones had no mutations while 5 clones had 6 mutations. Therefore, the mutant polymerase had a lower fidelity of DNA replication as compared to the wild type since there were more mutations in the PCR products from the mutant Taq polymerase as compared to the wild type enzyme. The mutant Taq polymerase should not be used in further studies.
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chapter one
Introduction to Biochemistry
B
iochemistry is the study of the molecules and chemical reactions of life. It is the discipline that uses the principles and language of chemistry to explain biology at the molecular level. Biochemists have discovered that the same chemical compounds and the same central metabolic processes are found in organisms as distantly related as bacteria, plants, and humans. It is now known that the basic principles of biochemistry are common to all living organisms. Although scientists usually concentrate their research efforts on particular organisms, their results can be applied to many other species. This book is called Principles of Biochemistry because we will focus on the most important and fundamental concepts of biochemistry: those that are common to most species, including bacteria, plants, and mammals such as ourselves. Where appropriate, we will point out features that distinguish particular groups of organisms. Many students and researchers are primarily interested in the biochemistry of humans. The causes of disease and the importance of proper nutrition, for example, are fascinating topics in biochemistry. We share these interests, and that’s why we include many references to human biochemistry in this textbook. However, we will also try to interest you in the biochemistry of other species. As it turns out, it is often easier to understand basic principles of biochemistry by studying many different species in order to recognize common themes and patterns. A knowledge and appreciation of other species will do more than help you learn biochemistry. It will also help you recognize the fundamental nature of life at the molecular level and the ways in which species are related through evolution from a common ancestor. Perhaps future editions of this book will include chapters on the biochemistry of life on other planets. Until then, we will have to be satisfied with learning about the diverse life on our own planet. We begin this introductory chapter with a few highlights of the history of biochemistry, followed by short descriptions of the chemical groups and molecules
Top: Adenovirus. Viruses consist of a nucleic acid molecule surrounded by a protein coat.
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Introduction to Biochemistry
that you will encounter throughout this book. The second half of the chapter is an overview of cell structure in preparation for your study of biochemistry.
1.1 Biochemistry Is a Modern Science Biochemistry has emerged as a dynamic science only within the past 100 years. However, the groundwork for the emergence of biochemistry as a modern science was prepared in earlier centuries. The period before 1900 saw rapid advances in the understanding of basic chemical principles such as reaction kinetics and the atomic composition of molecules. Many chemicals produced in living organisms had been identified by the end of the 19th century. Since then, biochemistry has become an organized discipline, and biochemists have elucidated many of the chemical processes of life. The growth of biochemistry and its influence on other disciplines will continue in the 21st century. In 1828 Friedrich Wöhler synthesized the organic compound urea by heating the inorganic compound ammonium cyanate. O ‘ Heat " H2N ¬ C ¬ NH2 NH4(OCN) ▼
Friedrich Wöhler (1800–1882). By synthesizing urea, Wöhler showed that compounds found in living organisms could be made in the laboratory from inorganic substances.
▼
Eduard Buchner (1860–1917). Buchner discovered cell-free fermentation.
(1.1)
This experiment showed for the first time that compounds found exclusively in living organisms could be synthesized from common inorganic substances. We now understand that the synthesis and degradation of biological substances obey the same chemical and physical laws as those that predominate outside of biology. No special or “vitalistic” processes are required to explain life at the molecular level. Many scientists date the beginnings of biochemistry to Wöhler’s synthesis of urea, although it would be another 75 years before the first biochemistry departments were established at universities. Two major breakthroughs in the history of biochemistry are especially notable—the discovery of the roles of enzymes as catalysts and the role of nucleic acids as information-carrying molecules. The very large size of proteins and nucleic acids made their initial characterization difficult using the techniques available in the early part of the 20th century. With the development of modern technology we now know a great deal about how the structures of proteins and nucleic acids are related to their biological functions. The first breakthrough—identification of enzymes as the catalysts of biological reactions—resulted in part from the research of Eduard Buchner. In 1897 Buchner showed that extracts of yeast cells could catalyze the fermentation of the sugar glucose to alcohol and carbon dioxide. Previously, scientists believed that only living cells could catalyze such complex biological reactions. The nature of biological catalysts was explored by Buchner’s contemporary, Emil Fischer. Fischer studied the catalytic effect of yeast enzymes on a simple reaction, the hydrolysis (breakdown by water) of sucrose (table sugar). Fischer proposed that during catalysis an enzyme and its reactant, or substrate, combine to form an intermediate compound. He also proposed that only a molecule with a suitable structure can serve as a substrate for a given enzyme. Fischer described enzymes as rigid templates, or locks, and substrates as matching keys. Researchers soon realized that almost all the reactions of life are catalyzed by enzymes, and the modified lockand-key theory of enzyme action remains a central tenet of modern biochemistry. As we will see in Chapter 5, enzymatic catalysis produces very high yields with few, if any, by-products. In contrast, many catalyzed reactions in organic chemistry are considered acceptable with yields of 50% to 60%. Biochemical reactions must be efficient since by-products can be toxic to cells and their formation would waste precious energy. Of course, the other key property of enzyme catalysis is that biological reactions occur much faster than they would without a catalyst.
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1.1
DNA ¡ RNA ¡ Protein
Biochemistry Is a Modern Science
▼
The last half of the 20th century saw tremendous advances in the area of structural biology, especially the structure of proteins. The first protein structures were solved in the 1950s and 1960s by scientists at Cambridge University (United Kingdom) led by John C. Kendrew and Max Perutz. Since then, the three-dimensional structures of more than 1000 different proteins have been determined and our understanding of the complex biochemistry of proteins has increased enormously. These rapid advances were made possible by the availability of larger and faster computers and new software that could carry out the many calculations that used to be done by hand using simple calculators. Much of modern biochemistry relies on computers, and this has created a new subdiscipline called bioinformatics. The second major breakthrough in the history of biochemistry—identification of nucleic acids as information molecules—came a half-century after Buchner’s and Fischer’s experiments. In 1944 Oswald Avery, Colin MacLeod, and Maclyn McCarty extracted deoxyribonucleic acid (DNA) from a toxic strain of the bacterium Streptococcus pneumoniae and mixed the DNA with a nontoxic strain of the same organism. The nontoxic strain was thereby permanently transformed into a toxic strain. This experiment provided the first conclusive evidence that DNA is the genetic material. In 1953 James D. Watson and Francis H. C. Crick deduced the three-dimensional structure of DNA. The structure of DNA immediately suggested to Watson and Crick a method whereby DNA could reproduce itself, or replicate, and thus transmit biological information to succeeding generations. Subsequent research showed that information encoded in DNA is transcribed to ribonucleic acid (RNA) and then translated into protein.
■
Emil Fischer (1852–1919). Fischer, who made many contributions to our understanding of the structures and functions of biological molecules, proposed the lock-and-key theory of enzyme action. He received the Nobel Prize in Chemistry in 1902 for his work on sugar and purine metabolism.
(1.2)
The study of genetics at the level of nucleic acid molecules is part of the discipline of molecular biology, and molecular biology is part of the discipline of biochemistry. In order to understand how nucleic acids store and transmit genetic information, you must understand the structure of nucleic acids and their role in encoding the enzyme proteins that catalyze the synthesis and degradation of biomolecules, including the nucleic acids themselves. You will find that much of your study of biochemistry is devoted to considering how enzymes and nucleic acids are central to the chemistry of life. As Crick predicted in 1958, the normal flow of information from nucleic acid to protein is not reversible. He referred to this unidirectional information flow as the Central Dogma of molecular biology. The term “Central Dogma” is often misunderstood. Strictly speaking, it does not refer to the overall flow of information shown above. Instead, it refers to the fact that once information in nucleic acids is transferred to protein it cannot flow backwards from protein to nucleic acids.
1.2 The Chemical Elements of Life Six nonmetallic elements—oxygen, carbon, hydrogen, nitrogen, phosphorus, and sulfur—account for more than 97% of the weight of most organisms. All these elements can form stable covalent bonds. The relative amounts of these six elements vary among organisms. Water is a major component of cells and accounts for the high percentage (by weight) of oxygen. Carbon is much more abundant in living organisms than in the rest of the universe. On the other hand, some elements, such as silicon, aluminum, and iron are very common in the Earth’s crust but are present only in trace amounts in cells. Altogether, a total of 29 different elements are commonly found in living organisms (Figure 1.1, on page 4). These include five ions 2+ 2, potassium 1K 2, sodium 1Na 2, that are essential in all species: calcium 1Ca~ 2+ ~ magnesium 1Mg 2, and chloride 1Cl 2 Note that the additional 29 elements account for only 3% of the weight of living organisms.
3
The synthesis of RNA (transcription) and protein (translation) are described in Chapters 21 and 22, respectively.
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Introduction to Biochemistry
IA 1 H
0 2 He IIA 4 Be
IIIA 5 B
IVA 6 C
VA 7 N
VIA 8 O
VIIA 9 F
4.003
3 Li 6.941
9.012
10.81
12.01
14.01
16.00
19.00
20.18
11 Na
12 Mg
13 Al
14 Si
15 P
16 S
17 Cl
18 Ar
22.99
24.31
IVB 22 Ti
VB 23 V
VIB 24 Cr
VIIB 25 Mn
IIB 30 Zn
32.07
35.45
39.95
28 Ni
IB 29 Cu
30.97
26 Fe
VIIIB 27 Co
28.09
20 Ca
IIIB 21 Sc
26.98
19 K
31 Ga
32 Ge
33 As
34 Se
35 Br
36 Kr
39.10
40.08
44.96
47.87
50.94
52.00
54.94
55.85
58.93
58.69
63.55
65.39
69.72
72.61
74.92
78.96
79.90
83.80
37 Rb
38 Sr
39 Y
40 Zr
41 Nb
42 Mo
43 Tc
44 Ru
45 Rh
46 Pd
47 Ag
48 Cd
49 In
50 Sn
51 Sb
52 Te
53 I
54 Xe
85.47
87.62
88.91
91.22
92.91
95.94
(98)
101.1
102.9
106.4
107.9
112.4
114.8
118.7
121.8
127.6
126.9
131.3
55 Cs
56 Ba
72 Hf
73 Ta
74 W
75 Re
76 Os
77 Ir
78 Pt
79 Au
80 Hg
81 Tl
82 Pb
83 Bi
84 Po
85 At
86 Rn
132.9
137.3
178.5
180.9
183.8
186.2
190.2
192.2
195.1
197.0
200.6
204.4
207.2
209.0
(209)
(210)
(222)
87 Fr
88 Ra
89 ** 104 Ac Rf
105 Db
106 Sg
107 Bh
108 Hs
109 Mt
110
111
112
113
114
115
116
117
118
(223)
(226)
(262)
(263)
(264)
(265)
(268)
(269)
(272)
(277)
58 * Ce
59 Pr
60 Nd
61 Pm
62 Sm
63 Eu
64 Gd
65 Tb
66 Dy
67 Ho
68 Er
69 Tm
70 Yb
71 Lu
140.1
140.9
144.2
(145)
150.4
152.0
157.3
158.9
162.5
164.9
167.3
168.9
173.0
175.0
92 U
93 Np
94 Pu
95 Am
96 Cm
97 Bk
98 Cf
99 Es
100 Fm
101 Md
102 No
103 Lr
238.0
(237)
(244)
(243)
(247)
(247)
(251)
(252)
(257)
(258)
(259)
(262)
1.008
57 * La 138.9
(227)
(261)
90** 91 Th Pa 232.0
231
(285)
(289)
10 Ne
(293)
▼
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Figure 1.1 Periodic table of the elements. The important elements found in living cells are shown in color. The brown elements (CHNOPS) are the six abundant elements. The five essential ions are purple. The trace elements are shown in dark blue (more common) and light blue (less common).
Most of the solid material of cells consists of carbon-containing compounds. The study of such compounds falls into the domain of organic chemistry. There is considerable overlap between the disciplines of organic chemistry and biochemistry, and a course in organic chemistry is helpful in understanding biochemistry. Organic chemists are more interested in reactions that take place in the laboratory, whereas biochemists would like to understand how reactions occur in living cells. The types of organic compounds commonly encountered in biochemistry are shown in Figure 1.2a. Biochemical reactions involve specific chemical bonds or parts of molecules called functional groups. The most important of these sites of reactivity are shown in Figure 1.2b. Figure 1.2c illustrates some types of linkages present in derivatives of the compounds shown in Figure 1.2a. Note that all of these linkages consist of several different atoms and individual bonds between atoms. We will encounter these compounds, functional groups, and linkages throughout this book. Ester and ether linkages are common in fatty acids and lipids. Amide linkages are found in proteins. Phosphate ester and phosphoanhydride linkages occur in nucleotides. An important theme of biochemistry is that the chemical reactions that occur inside cells are the same kinds of reactions that take place in a chemistry laboratory. The most important difference is that almost all reactions that occur in living cells are catalyzed by enzymes and thus proceed at very high rates. One of the main goals of this textbook is to explain how enzymes speed up reactions without violating the fundamental reaction mechanisms of organic chemistry. The catalytic efficiency of enzymes can be observed even when the enzymes and reactants are isolated in a test tube. Researchers often find it useful to distinguish between biochemical reactions that take place in an organism (in vivo) and those that occur under laboratory conditions (in vitro).
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The Chemical Elements of Life
▼
rganic compounds (a) Organic O R OH Alcohol
O
R C H Aldehyde
R
O
C R1 K Ketone
R C OH Carboxylic acid 1
R1 R SH Thiol (Sulfhydryl)
R
NH 2
R
Figure 1.2 General formulas of (a) organic compounds, (b) functional groups, and (c) linkages common in biochemistry. R represents an alkyl group 1CH3 ¬ 1CH22n ¬ 2.
R1
NH
R
Primary
N
R2
T Tertiary
Amines 2
r roups (b) Functional groups O OH Hydroxyl
O
C R Acyl
O
C Carbonyl
C O Carboxylate
O SH Sulfhydryl (Thiol)
NH 2 or NH 3 Amino
O
P
O O
P
O Phosphate
O
O Phosphoryl
(c) Linkages in biochemical compounds
O C
O
O
C
C
Ester
O
C
Ether
O
P
O Phosphate ester
O O
C
Amide
O C
N
O
P
1
Under most biological conditions, carboxylic acids exist as carboxylate anions: O
O O
P
R
O
O O Phosphoanhydride
1.3 Many Important Macromolecules Are Polymers Chemical structures are the vocabulary of biochemistry. We present some of these structures to prepare you for the examples you will encounter in the next few chapters. Much of biochemistry deals with very large molecules that we refer to as macromolecules. Biological macromolecules are usually a form of polymer created by joining many smaller organic molecules, or monomers, via condensation (removal of the elements of water). Each monomer incorporated into a macromolecular chain is termed a residue. In some cases, such as certain carbohydrates, a single residue is repeated many times; in other cases, such as proteins and nucleic acids, a
C
O
2
ammonium ions: R
NH 3 , R
R1
R1
NH 2 , and R
NH
R2
5
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variety of residues are connected in a particular order. Each residue of a given polymer is added by repeating the same enzyme-catalyzed reaction. Thus, all of the residues in a macromolecule are aligned in the same direction, and the ends of the macromolecule are chemically distinct. Macromolecules have properties that are very different from those of their constituent monomers. For example, starch is not soluble in water and does not taste sweet, although it is a polymer of the sugar glucose. Observations such as this have led to the general principle of the hierarchical organization of life. Each new level of organization results in properties that cannot be predicted solely from those of the previous level. The levels of complexity, in increasing order, are atoms, molecules, macromolecules, organelles, cells, tissues, organs, and whole organisms. (Note that many species lack one or more of these levels of complexity. Singlecelled organisms, for example, do not have tissues and organs.) The following sections briefly describe the principal types of macromolecules and how their sequences of residues or three-dimensional shapes grant them unique properties. In discussing molecules and macromolecules, we will often refer to the molecular weight of a compound. A more precise term for molecular weight is relative molecular mass (abbreviated Mr). It is the mass of a molecule relative to 1/12 the mass of an atom of the carbon isotope 12C. (The atomic weight of this isotope has been defined as exactly 12 atomic mass units. Note that the atomic weight of carbon shown in the Periodic Table represents the average of several different isotopes, including 13C and 14C.) Because Mr is a relative quantity, it is dimensionless and has no units associated with its value. The relative molecular mass of a typical protein, for example, is 38 000 1Mr = 38 0002. The absolute molecular mass of a compound has the same magnitude as the molecular weight except that it is expressed in units called daltons (1 dalton = 1 atomic mass unit). The molecular mass is also called the molar mass because it represents the mass (measured in grams) of 1 mole, or 6.023 * 1023 molecules. The molecular mass of a typical protein is 38 000 daltons, which means that 1 mole weighs 38 kilograms. The main source of confusion is that the term “molecular weight” has become common jargon in biochemistry although it refers to relative molecular mass and not to weight. It is a common error to give a molecular weight in daltons when it should be dimensionless. In most cases this isn’t a very important mistake, but you should know the correct terminology. A. Proteins (a)
COO H3 N
C
H
R
O
(b)
H3 N
CH R
C
N
CH
H
R
COO
▼
Figure 1.3 Structure of an amino acid and a dipeptide. (a) All amino acids contain an amino group (blue) and a carboxylate group (red). Different amino acids contain different side chains (designated -R). (b) A dipeptide is produced when the amino group of one amino acid reacts with the carboxylate group of another to form a peptide bond (red).
Twenty common amino acids are incorporated into proteins in all cells. Each amino acid contains an amino group and a carboxylate group, as well as a side chain (R group) that is unique to each amino acid (Figure 1.3a). The amino group of one amino acid and the carboxylate group of another are condensed during protein synthesis to form an amide linkage as shown in Figure 1.3b. The bond between the carbon atom of one amino acid residue and the nitrogen atom of the next residue is called a peptide bond. The end-to-end joining of many amino acids forms a linear polypeptide that may contain hundreds of amino acid residues. A functional protein can be a single polypeptide, or it can consist of several different polypeptide chains that are tightly bound to form a more complex structure. Many proteins function as enzymes. Others are structural components of cells and organisms. The three-dimensional shape of a protein is determined largely by the sequence of its amino acid residues. This sequence information is encoded in the gene for the protein. The function of a protein depends on its three-dimensional structure, or conformation. The structures of many proteins have been determined, and several principles governing the relationship between structure and function have become clear. For example, many enzymes contain a cleft, or groove, that binds the substrates of a reaction. This cavity contains the active site of the enzyme—the region where the chemical reaction takes place. Figure 1.4a shows the structure of the enzyme lysozyme, which catalyzes the hydrolysis of specific
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(b)
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Figure 1.4 Chicken (Gallus gallus) egg white lysozyme. (a) Free lysozyme. Note the characteristic cleft that includes the active site of the enzyme. (b) Lysozyme with bound substrate. [PDB 1LZC].
carbohydrate polymers. Note the prominent cleft in the molecule. This cleft is the site of substrate binding and enzymatic activity. Figure 1.4b shows the structure of the enzyme with the substrate bound in the cleft. We will discuss the relationship between protein structure and function further in Chapters 4 and 6. There are many ways of representing the three-dimensional structures of biopolymers such as proteins. The protein molecule in Figure 1.4 is shown as a cartoon where the conformation of the polypeptide chain is represented as a combination of wires, helical ribbons, and broad arrows. Other kinds of representations in the following chapters include images that show the position of every atom. Computer programs that create these images are freely available on the Internet, and the structural data for proteins can be retrieved from a number of database sites. With a little practice, any student can view these molecules on a computer monitor. B. Polysaccharides Carbohydrates, or saccharides, are composed primarily of carbon, oxygen, and hydrogen. This group of compounds includes simple sugars (monosaccharides) as well as their polymers (polysaccharides). All monosaccharides and all residues of polysaccharides contain several hydroxyl groups and are therefore polyalcohols. The most common monosaccharides contain either five or six carbon atoms. Sugar structures can be represented in several ways. For example, ribose (the most common five-carbon sugar) can be shown as a linear molecule containing four hydroxyl groups and one aldehyde group (Figure 1.5a, on page 8). This linear representation is called a Fischer projection. In its usual biochemical form, however, the structure of ribose is a ring with a covalent bond between the carbon of the aldehyde group (C-1) and the oxygen of the C-4 hydroxyl group as shown in Figure 1.5b. The ring form is most commonly shown as a Haworth projection (Figure 1.5c). This representation is a more accurate way of depicting the actual structure of ribose. The Haworth projection is rotated 90° with respect to the Fischer projection and portrays the carbohydrate ring as a plane with one edge projecting out of the page (represented by the thick lines). However, the ring is not actually planar. It can adopt numerous conformations in which certain ring atoms are out-of-plane. In Figure 1.5d, for example, the C-2 atom of ribose lies above the plane formed by the rest of the ring atoms. Some conformations are more stable than others, so the majority of ribose molecules can be represented by one or two of the many possible conformations. Nevertheless, it’s important to note that most biochemical molecules exist as a
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O
H
(b)
(c)
H
HO
5
HOCH 2
H
C 2
OH
H
C 2
OH
H
C 3
OH
H
3
C
OH
4
C
OH
4
C
5
CH 2 OH
5
CH 2 OH
H
(d)
C 1
C 1
Fischer projection (open-chain form)
H
4
O
H
H
H
3
2
OH
Fischer projection (ring form)
5
OH
O
O H
HOCH 2 4
1
H
H
OH 1
H
2
OH H
3
HO
OH
Haworth projection
Envelope conformation
▼
Conformations of monosaccharides are described in more detail in Section 8.3.
Figure 1.5 Representations of the structure of ribose. (a) In the Fischer projection, ribose is drawn as a linear molecule. (b) In its usual biochemical form, the ribose molecule is in a ring, shown here as a Fischer projection. (c) In a Haworth projection, the ring is depicted as lying perpendicular to the page (as indicated by the thick lines, which represent the bonds closest to the viewer). (d) The ring of ribose is not actually planar but can adopt 20 possible conformations in which certain ring atoms are out-of-plane. In the conformation shown, C-2 lies above the plane formed by the rest of the ring atoms.
Figure 1.6 Glucose and cellulose. (a) Haworth projection of glucose. (b) Cellulose, a linear polymer of glucose residues. Each residue is joined to the next by a glycosidic bond (red).
(a)
6
CH 2 OH 5
H 4
H OH
O
OH 1
H
H
HO 3
H
(b)
2
OH
6
6
CH 2 OH 5
H 4
O
H OH
H 3
O H
O 1
4
H 3
H
2
OH
CH 2 OH
OH
5
2
OH H
H
H O
▼
collection of structures with different conformations. By definition, the change from one conformation to another does not require the breaking of any covalent bonds. In contrast, the two basic forms of carbohydrate structures, linear and ring forms, do require the breaking and forming of covalent bonds. Glucose is the most abundant six-carbon sugar (Figure 1.6a). It is the monomeric unit of cellulose, a structural polysaccharide, and of glycogen and starch, which are storage polysaccharides . In these polysaccharides, each glucose residue is joined covalently to the next by a covalent bond between C-1 of one glucose molecule and one of the hydroxyl groups of another. This bond is called a glycosidic bond. In cellulose, C-1 of each glucose residue is joined to the C-4 hydroxyl group of the next residue (Figure 1.6b). The hydroxyl groups on adjacent chains of cellulose interact noncovalently, creating strong, insoluble fibers. Cellulose is probably the most abundant biopolymer on Earth because it is a major component of flowering plant stems, including tree trunks. We will discuss carbohydrates further in Chapter 8.
H
H
1
4
H OH
CH 2 OH 6
H
O
O 1
H 3
5
O
H
2
OH
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9
C. Nucleic Acids
O O
P O
O γ
O
P O
N
O β
O
P
α
O
CH 2
O
H
N
O H
H
H OH
OH
N N
4
H
H 3
OH
OH
O H 2
1
H
H
Figure 1.7 Deoxyribose, the sugar found in deoxyribonucleotides. Deoxyribose lacks a hydroxyl group at C-2.
▼
NH 2
5
HOCH 2
▼
Nucleic acids are large macromolecules composed of monomers called nucleotides. The term polynucleotide is a more accurate description of a single molecule of nucleic acid, just as polypeptide is a more accurate term than protein for single molecules composed of amino acid residues. The term nucleic acid refers to the fact that these polynucleotides were first detected as acidic molecules in the nucleus of eukaryotic cells. We now know that nucleic acids are not confined to the eukaryotic nucleus but are abundant in the cytoplasm and in prokaryotes that don’t have a nucleus. Nucleotides consist of a five-carbon sugar, a heterocyclic nitrogenous base, and at least one phosphate group. In ribonucleotides, the sugar is ribose; in deoxyribonucleotides, it is the derivative deoxyribose (Figure 1.7). The nitrogenous bases of nucleotides belong to two families known as purines and pyrimidines. The major purines are adenine (A) and guanine (G); the major pyrimidines are cytosine (C), thymine (T), and uracil (U). In a nucleotide, the base is joined to C-1 of the sugar, and the phosphate group is attached to one of the other sugar carbons (usually C-5). The structure of the nucleotide adenosine triphosphate (ATP) is shown in Figure 1.8. ATP consists of an adenine moiety linked to ribose by a glycosidic bond. There are three phosphoryl groups (designated a, b, and g) esterified to the C-5 hydroxyl group of the ribose. The linkage between ribose and the a-phosphoryl group is a phosphoester linkage because it includes a carbon and a phosphorus atom, whereas the b and g phosphoryl groups in ATP are connected by phosphoanhydride linkages that don’t involve carbon atoms. All phosphoanhydrides possess considerable chemical potential energy, and ATP is no exception. It is the central carrier of energy in living cells. The potential energy associated with the bonds between the phosphoryl groups of ATP can be used directly in biochemical reactions or coupled to a reaction in a less obvious way. In condensation reactions, for example, transfer of one of the phosphoryl groups of ATP results in the formation of an activated intermediate which then participates in the synthetic reaction. We will encounter many condensation reactions that involve ATP. In polynucleotides, the phosphate group of one nucleotide is covalently linked to the C-3 oxygen atom of the sugar of another nucleotide, creating a second phosphoester linkage. The entire linkage between the carbons of adjacent nucleotides is called a phosphodiester linkage because it contains two phosphoester linkages (Figure 1.9, on page 10). Nucleic acids contain many nucleotide residues and are characterized by a backbone consisting of alternating sugars and phosphates. In DNA, the bases of two different polynucleotide strands interact to form a helical structure. There are several ways of depicting nucleic acid structures depending on which features are being described. The ball-and-stick model shown in Figure 1.10 on the next page is ideal for showing the individual atoms and the ring structure of the sugars and the bases. In this case the two helices can be traced by following the sugar– phosphate backbone emphasized by the presence of the purple phosphorus atoms surrounded by four red oxygen atoms. The individual base pairs are viewed edge-on in the interior of the molecule. We will see several other DNA models in Chapter 19.
Figure 1.8 Structure of adenosine triphosphate (ATP). The nitrogenous base adenine (blue) is attached to ribose (black). Three phosphoryl groups (red) are also bound to the ribose.
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Figure 1.9 Structure of a dinucleotide. One deoxyribonucleotide residue contains the pyrimidine thymine (top), and the other contains the purine adenine (bottom). The residues are joined by a phosphodiester linkage between the two deoxyribose moieties. (The carbon atoms of deoxyribose are numbered with primes to distinguish them from the atoms of the bases thymine and adenine.)
O O
5′
P
O
O
H 3C
O 4 5
CH 2
4′
H
NH Thymine (T)
2
N
O H
H
O
1′
H 3′
2′
H
O Phosphodiester linkage
3
6 1
O
O
P
NH 2
O 5′
N7
CH 2
4′
H
8
9
N
O H
H
N
6 1 4 3 2
5
A
