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2. 2. Denise R. Ferrier, PhD Professor of Biochemistry Department of Biochemistry and Molecular Biology Drexel University College of Medicine Philadelphia, Pennsylvania Lippincott Illustrated Reviews Flash Cards BIOCHEMISTRY Bradford A. Jameson, PhD Professor of Biochemistry Department of Biochemistry and Molecular Biology Drexel University College of Medicine Philadelphia, Pennsylvania Ferrier_FM.indd iFerrier_FM.indd i 5/3/14 4:48 AM5/3/14 4:48 AM
3. 3. Acquisitions Editor: Tari Broderick Product Development Editor: Stephanie Roulias Production Project Manager: David Orzechowski Design Coordinator: Holly McLaughlin Illustration Coordinator: Doug Smock Manufacturing Coordinator: Margie Orzech Prepress Vendor: Absolute Service, Inc. Copyright 2015 Wolters Kluwer Health All rights reserved. This book is protected by copyright. No part of this book may be reproduced or transmitted in any form or by any means, including as photocopies or scanned-in or other electronic copies, or utilized by any information storage and retrieval system without written permission from the copyright owner, except for brief quotations embodied in critical articles and reviews. Materials appearing in this book prepared by individuals as part of their ofcial duties as U.S. government employees are not covered by the above-mentioned copyright. To request permission, please contact Wolters Kluwer Health at Two Commerce Square, 2001 Market Street, Philadelphia, PA 19103, via email at permissions@lww.com, or via our website at lww.com (products and services). 9 8 7 6 5 4 3 2 1 Printed in China 978-1-4511-9111-0 1-4511-9111-1 Library of Congress Cataloging-in-Publication Data is available upon request Care has been taken to conrm the accuracy of the information presented and to describe generally accepted practices. However, the author(s), editors, and publisher are not responsible for errors or omissions or for any consequences from application of the information in this book and make no warranty, expressed or implied, with respect to the currency, completeness, or accuracy of the contents of the publication. Application of this information in a particular situation remains the professional responsibility of the practitioner; the clinical treatments described and recommended may not be considered absolute and universal recommendations. The author(s), editors, and publisher have exerted every effort to ensure that drug selection and dosage set forth in this text are in accordance with the current recommendations and practice at the time of publication. However, in view of ongoing research, changes in government regulations, and the constant ow of information relating to drug therapy and drug reactions, the reader is urged to check the package insert for each drug for any change in indications and dosage and for added warnings and precautions. This is particularly important when the recommended agent is a new or infrequently employed drug. Some drugs and medical devices presented in this publication have Food and Drug Administration (FDA) clearance for limited use in restricted research settings. It is the responsibility of the health care provider to ascertain the FDA status of each drug or device planned for use in his or her clinical practice. Ferrier_FM.indd iiFerrier_FM.indd ii 5/23/14 1:09 AM5/23/14 1:09 AM
4. 4. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer Features: Three-Step Review SPOT FLASH Test your grasp of key concepts or equations on a lecture-by-lecture basis! COURSE REVIEW Ensure a thorough understanding of course material through in-depth questions. High-yield facts for course- and Board-exam review! CLINICAL CORRELATIONS Explain how the basic science helps predict outcomes in a clinical setting! Featuring the same visionary artwork found in Lippincott Illustrated Reviews: Biochemistry With Lippincott Illustrated Reviews, Seeing is Understanding. Ferrier_FM.indd iiiFerrier_FM.indd iii 5/3/14 4:48 AM5/3/14 4:48 AM
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6. 6. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer Preface Lippincott Illustrated Reviews Flash Cards: Biochemistry is a portable study tool designed for self-assessment and review of medical biochemistry. The ash cards were developed primarily for use by medical students studying biochemistry and preparing for United States licensing exams, but information is presented with a clarity and level of detail that makes them ideal supplements for any of the allied health sciences. The deck contains three card types: Question (Q) cards, Case cards, and Summary cards. Q CARDS The majority of cards are Q cards that prompt the reader with questions (on the front) to assess level of understanding, depth of knowledge, and ability to apply biochemical concepts. The answers (on the back) are more inclusive than those found on typical ash cards. Most Q cards contain three questions or sets of questions on a common topic: The rst tests for retention of basic facts, whereas the next two test understanding and/or application of related concepts and clinical correlations. Each question type is denoted by icons. SPOT FLASH: Illustration-based questions test your grasp of key facts and are intended for use on a lecture-by-lecture assessment and review basis. COURSE REVIEW: In-depth questions promote a thorough understanding of related concepts. The answers focus on high-yield facts to help consolidate and apply material during course- and licensing-exam review. CLINICAL CORRELATIONS: Clinical questions highlight the basic science foundations of medicine. They help students apply biochemi- cal concepts to clinical problems and are particularly useful when studying for licensing exams. Continued, over Ferrier_FM.indd vFerrier_FM.indd v 5/3/14 4:48 AM5/3/14 4:48 AM
7. 7. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer Preface Q cards include several features to facilitate learning and retaining the material: Illustrations: Richly detailed illustrations from the popular companion text, Lippincott Illustrated Reviews: Biochemistry, appear on both sides of the cards. Many of the illustrations include narrative boxes that guide readers through complex concepts. Notes: Answers may be supplemented with information that goes beyond the need-to-know basics to provide context or to enrich and help anchor a concept. Emphasis: Key terms, disease names, and pathologic ndings are bolded for rapid review and assimilation. CASE CARDS AND SUMMARY CARDS Case cards use common clinical presentations to highlight biochemical concepts. Summary cards (for the vitamins and the fed/fasted states) highlight key features of these information-rich areas of medical biochemistry. The card deck is designed to be comprehensive, covering all signicant biochemical concepts. Ferrier_FM.indd viFerrier_FM.indd vi 5/3/14 4:48 AM5/3/14 4:48 AM
8. 8. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer Acknowledgments The authors wish to thank John Swaney, PhD, our colleague at Drexel University College of Medicine, for his careful reading of the manuscript and constructive comments. Any errors are ours alone. We thank the publishing team assembled by Wolters Kluwer. Stephanie Roulias, product development editor, and Kelly Horvath, freelance development editor, along with Doug Smock, Teresa Exley, and David Orzechowski, gave invaluable assistance in the development and production of the nished product. We also thank Robin R. Preston, PhD, for his design of the ash card format. Dedication The authors dedicate this work to the medical, biomedical graduate, and professional studies students of Drexel University. You have challenged and inspired us, and have made us better teachers. Ferrier_FM.indd viiFerrier_FM.indd vii 5/3/14 4:48 AM5/3/14 4:48 AM
9. 9. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer Figure Credits Card 3.6 Question and Answer: Modied photo courtesy of Photodyne Incorporated, Hartland, WI. Card 4.2 Answer: Kronauer and Buhler, Images in Clinical Medicine, The New England Journal of Medicine, June 15, 1995, Vol. 332, No. 24, p. 1611. Card 4.5 Question and Answer: 1. Modied photo from Web site Derma.de. 2. Modied from Jorde LB, Carey JC, Bamshad MJ, et al. Medical Genetics. 2nd ed. St. Louis, MO: Mosby; 2000. http://medgen.genetics. utah.edu/index.htm Card 13.6 Answer: From the Crookston Collection, University of Toronto. Card 21.2 Answer: Modied from Rich MW. Porphyria cutanea tarda. Postgrad Med. 1999;105:208214. Card 21.4 Question and Answer: From Custom Medical School Stock Photo, Inc. Card 22 Case Card Question: Modied from WebMD Inc. http://www.samed.com/sam/ forms/index.htm. Card 23.6 Question and Answer: Modied from Cryer PE, Fisher JN, Shamoon H. Hypoglycemia. Diabetes Care. 1994;17: 734753. Ferrier_FM.indd viiiFerrier_FM.indd viii 5/3/14 4:48 AM5/3/14 4:48 AM
10. 10. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer Contents UNIT 1 Protein Structure and Function 1.1 UNIT 2 Bioenergetics and Carbohydrate Metabolism 6.1 UNIT 3 Lipid Metabolism 15.1 UNIT 4 Nitrogen Metabolism 19.1 UNIT 5 Metabolism Integration 23.1 UNIT 6 Genetic Information Storage and Expression 29.1 CHAPTER 34 Blood Clotting 34.1 APPENDIX Abbreviations A-1 Ferrier_FM.indd ixFerrier_FM.indd ix 5/3/14 4:48 AM5/3/14 4:48 AM
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12. 12. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer 1.1 QuestionAmino Acid Structure What effect will raising pH from an acidic value to the physiologic value of 7.4 have on the structural features shown in red at right? At physiologic pH, what will be the charge on the side chain (R group) of free Asp? Of Lys? Which amino acid(s) contains a side-chain hydroxyl group that can be glycosylated? A secondary amino group? Is Val ionized when incorporated into a protein? C+H3N COOH HC+H3N COOH H These are common to all `-amino acids. Free amino acid RAmino group Carboxyl group `C H` RRAmino group R Side chain is distinctive for each amino acid. `-Carbon is linked to the carboxyl, amino, and R groups. Ferrier_Unit01.indd 1Ferrier_Unit01.indd 1 5/2/14 7:08 PM5/2/14 7:08 PM
13. 13. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer 1.1 Answer Amino Acid Structure Raising the pH from an acidic value to the physiologic value of 7.4 will result in deprotonation (ionization) of the -carboxyl group (pK2) to COO . The -amino group (pK9) will remain protonated. At physiologic pH, the charge on the side chain (R group) of free Asp is negative. Lys is positive. Ser and Thr each contain a hydroxyl group that can be O-glycosylated. [Note: The hydroxyl group can also be phosphorylated.] Pro contains a secondary amino group. Its -amino N and R group form a rigid ring. Val is not ionized when incorporated into a protein because (1) the -amino and -carboxyl groups are involved in peptide bonds and, consequently, are unavailable for ionization, and (2) the side chain is nonpolar. C+H3N COO- HC+H3N CCOO- H These are common to all `-amino acids. Free amino acid RAmino group Carboxyl group `C H` RRAmino group R Side chain is distinctive for each amino acid. `-Carbon is linked to the carboxyl, amino, and R groups. COOH H Proline C CH2 +H2N H2C CH2 Ferrier_Unit01.indd 2Ferrier_Unit01.indd 2 5/2/14 7:08 PM5/2/14 7:08 PM
14. 14. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer 1.2 QuestionAmino Acid Structure Based on the gure, where would Leu likely be located in a protein that spans the membrane? In a soluble protein? What term refers to the tendency of nonpolar molecules (or regions of molecules such as amino acid side chains) to cluster together in a polar environment such as an aqueous solution? In sickle cell anemia (SCA), why does the replacement of a Glu by a Val on the surface of the deoxyHb molecule result in the association of these molecules? Cell membrane Polar amino acids ( ) cluster on the surface of soluble proteins. CellCC ll Nonpolar amino acids ( ) cluster on the surface of membrane proteins. Nonpolar amino acids ( ) cluster in the interior of soluble proteins. Soluble protein Membrane protein Ferrier_Unit01.indd 3Ferrier_Unit01.indd 3 5/2/14 7:08 PM5/2/14 7:08 PM
15. 15. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer Amino Acid Structure1.2 Answer Leu, a nonpolar amino acid, would likely be located within the hydrophobic membrane-spanning domain of the protein. It would likely be located in the interior of a soluble protein. The term hydrophobic effect refers to the tendency of nonpolar molecules (or regions of molecules such as amino acid side chains) to cluster together in a polar environment such as an aqueous solution. The replacement of polar Glu by nonpolar Val creates a hydrophobic region on the surface of the deoxyHb molecule that will interact with a hydrophobic region on other deoxyHb molecules. This interaction creates rigid polymers of deoxyHb that deform RBCs. Thus, it is the hydrophobic effect that drives the association of deoxyHb molecules in SCA. Cell membrane Cell Leu Polar amino acids ( ) cluster on the surface of soluble proteins. Nonpolar amino acids ( ) cluster on the surface of membrane proteins. Nonpolar amino acids ( ) cluster in the interior of soluble proteins. Soluble protein Membrane protein Ferrier_Unit01.indd 4Ferrier_Unit01.indd 4 5/2/14 7:08 PM5/2/14 7:08 PM
16. 16. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer 1.3 QuestionAmino Acid Structure Which structure shown (A or B) represents L-Ala? Which amino acid does not possess a chiral (asymmetric) carbon? Which peptide is less soluble in an aqueous (polar) environment, Ala-Gly-Asn-Ser-Tyr or Gly-Met-Phe-Leu-Ala? H 3C HOOC B H C NH 3 + CH3 COOH A H C +H3 N Ferrier_Unit01.indd 5Ferrier_Unit01.indd 5 5/2/14 7:08 PM5/2/14 7:08 PM
17. 17. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer 1.3 Answer Amino Acid Structure Structure A represents L-Ala. The L isomer of an amino acid has the -amino group on the left. The D isomer has the -amino group on the right. D and L isomers are mirror images of each other (enantiomers). Gly, with its two H substituents, does not possess a chiral (asymmetric) carbon. Because the Gly-Met-Phe-Leu-Ala peptide contains no charged or polar uncharged amino acids, it is less soluble than Ala-Gly-Asn-Ser-Tyr in an aqueous (polar) environment. H 3C HOOC D-Alanine H C NH 3 + CH3 COOH L-Alanine H C +H3 N Ferrier_Unit01.indd 6Ferrier_Unit01.indd 6 5/2/14 7:08 PM5/2/14 7:08 PM
18. 18. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer 1.4 QuestionAcidic and Basic Properties of Amino Acids What relationship is described by the HendersonHasselbalch equation shown? Is an acid with a large pKa stronger or weaker than one with a small pKa? The pKa of acetic acid (CH3COOH) is 4.8. What is the pH of a solution containing acetic acid and its conjugate base (CH3COO ) in a ratio of 10 to 1? Physiologic buffers are important in resisting blood pH changes. Maximal buffering occurs when the pH is equal to the , while effective buffering can occur within . pH pKa log [A] [HA] + Ferrier_Unit01.indd 7Ferrier_Unit01.indd 7 5/2/14 7:08 PM5/2/14 7:08 PM
19. 19. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer The HendersonHasselbalch equation describes the relationship between the pH of a solution and the concentration of a weak acid [HA] and its conjugate base [A ]. An acid with a large pKa is weaker than one with a small pKa because the large pKa reects less ioniza- tion (fewer H released). This is because pKa log Ka. Because pH pKa log [A ]/[HA], when pKa is 4.8 and the ratio of the acid to its conjugate base is 10 to 1, the pH is equal to 4.8 log of 0.1. Therefore, pH 4.8 (1) 3.8. Physiologic buffers are important in resisting blood pH changes. Maximal buffering occurs when the pH is equal to the pKa, while effective buffering can occur within 1 pH unit of the pKa. 1.4 Answer Acidic and Basic Properties of Amino Acids 0 3 4 5 6 7 0 0.5 1.0 pH EquivalentsOH added Buffer region CH3COOH CH3COO H2O FORM I (acetic acid, HA) FORM II (acetate, A ) pKa = 4.8[I] = [II] OH H+ [I] > [II] [II] > [I] pH pKa log [A] [HA] + Ferrier_Unit01.indd 8Ferrier_Unit01.indd 8 5/2/14 7:08 PM5/2/14 7:08 PM
20. 20. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer Which FORM (I, II, or III) shown represents the isoelectric form of Ala? Calculate the pI for Arg, which has three pKs: pK1 2.2, pK2 9.2, and pK3 12.5. What will happen to the charge on His residues in a protein that moves from the cytoplasm (pH 7.4) to a lysosome (pH 5.0)? 1.5 QuestionAcidic and Basic Properties of Amino Acids COOH FORM I of Ala FORM II of Ala FORM III of Ala CH3 C+H3N H COO CH3 C+ H3N H COO CH3 CH2N H H2OOH H+ H2OOH H+ pK1 = 2.3 pK2 = 9.1 Ferrier_Unit01.indd 9Ferrier_Unit01.indd 9 5/2/14 7:08 PM5/2/14 7:08 PM
21. 21. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer 1.5 Answer Acidic and Basic Properties of Amino Acids COOH FORM I of Ala FORM II of Ala FORM III of Ala CH3 C+H3N H COO CH3 C+ H3N H COO CH3 CH2N H H2OOH H+ H2OOH H+ pK1 = 2.3 pK2 = 9.1 The isoelectric form has no net charge. It is the zwitterionic (two ion) form. Therefore, FORM II is the isoelectric form of Ala. The pI corresponds to the pH at which an amino acid is electrically neutral, that is, the average of the pKs on either side of the isoelectric form. For Arg, a dibasic amino acid with pK1 (most acidic group) 2.2, pK2 9.2, and pK3 (least acidic group) 12.5, the pI is 10.8 (the average of 9.2 and 12.5). In a protein, the imidazole R group of His can be charged or uncharged depending on the local environment. It will be uncharged (deprotonated) at pH 7.4 and charged (protonated) at pH 5.0. [Note: In free His the pK of the R group is 6.0.] Ferrier_Unit01.indd 10Ferrier_Unit01.indd 10 5/2/14 7:08 PM5/2/14 7:08 PM
22. 22. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer 1.6 QuestionAcidic and Basic Properties of Amino Acids Based on the bicarbonate buffer system shown, what will happen to the availability of HCO3 when H is lost, such as with emesis (vomiting)? Use the HendersonHasselbalch equation to determine what will happen to pH when HCO3 is lost (e.g., with diarrhea) and when CO2 is increased (e.g., with pulmonary obstruction). Aspirin (pKa 3.5) is largely protonated and uncharged in the stomach (pH 1.5). What percentage of the aspirin will be in this lipid-soluble form at pH 1.5? H2CO3 HCO3 -H+H2OCO2 + + Ferrier_Unit01.indd 11Ferrier_Unit01.indd 11 5/2/14 7:08 PM5/2/14 7:08 PM
23. 23. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer 1.6 Answer Acidic and Basic Properties of Amino Acids With emesis (vomiting), the loss of H (rise in pH) results in increased availability of HCO3 as the result of a compensatory rightward shift in the bicarbonate buffer system. The HendersonHasselbalch equation is used to calculate how the pH of a system changes in response to changes in the concentration of an acid or its conjugate base. For the bicarbonate buffer system, pH pK log [HCO3 ]/[CO2]. Therefore, both the loss of HCO3 (base) with diarrhea and the increase in CO2 (acid) because of decreased elimination with pulmonary obstruction result in decreased pH. pH pK log [Drug ]/[Drug-H]. Therefore, for aspirin in the stomach, 1.5 3.5 (2). Because the antilog of 2 is 0.01, the ratio of [Drug ]/[Drug-H] is 1/100. This means that 1 out of 100 (1%) of the aspirin molecules will be the Drug form and 99 out of 100 (99%) will be the uncharged, lipid-soluble, Drug-H form. H2CO3 HCO3 -H+H2OCO2 + + DRUG ABSORPTION At the pH of the stomach (1.5), a drug like aspirin (weak acid, pK = 3.5) will be largely protonated (COOH) and, thus, uncharged. Uncharged drugs generally cross membranes more rapidly than do charged molecules. pH = pK + log [Drug-H] [Drug ] A HA - Lipid membrane LUMEN OF STOMACH STOMACH BLOOD H+ H+ H+ A HA - H+ Remove B Ferrier_Unit01.indd 12Ferrier_Unit01.indd 12 5/2/14 7:08 PM5/2/14 7:08 PM
24. 24. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer 2.1 QuestionProtein Structure Which level of protein structure depicted can be correctly described as the three-dimensional shape of a folded polypeptide chain? Mutations that insert, delete, or replace amino acids change this level of protein structure. How many different isoforms of the tetrameric enzyme PK can be made from M and/or L subunits? How many different tetrapeptides could be generated from three different amino acids? CN C H H CN C H CH3O H N H C O C O CN C N H H CO C C N H O C C O O H N C C N H N H R CR C R C R 3 2 1 H 4 Ferrier_Unit01.indd 13Ferrier_Unit01.indd 13 5/2/14 7:08 PM5/2/14 7:08 PM
25. 25. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer 2.1 Answer Protein Structure The three-dimensional shape of a folded polypeptide chain describes a proteins tertiary structure (No. 3 shown). At a minimum, the primary structure (amino acid sequence) will change with mutations that insert, delete, or replace amino acids. [Note: Changes in the primary structure can also affect the higher levels of protein structure (No. 2 to 4 shown). Such changes frequently result in protein misfolding and can lead to loss of function, aggregation, or degradation.] Five different forms of tetrameric PK can be made from M and/or L subunits: M4, M3L, M2L2, ML3, and L4. Because PK is composed of more than one subunit, it has a quaternary structure. There are 34 or 81 (where 3 the number of amino acids and 4 the chain length) different tetrapeptides that could be generated from three different amino acids. CN C H H CN C H CH3O H N H C O C O CN C N H H CO C C N H O C C O O H N C C N H N H R CR C R C R Quaternary structure4 Tertiary structure3 2 Secondary structure Primary structure1 H Ferrier_Unit01.indd 14Ferrier_Unit01.indd 14 5/2/14 7:08 PM5/2/14 7:08 PM
26. 26. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer 2.2 QuestionPrimary Structure of Proteins What is the name given to the bond outlined by the black box shown? What are the characteristics of this bond? With fever, why might proteins begin to unfold but not be hydrolyzed to peptides and free amino acids? C COO H Valine Valylalanine C+ H3N COO H CH3 Alanine C C H CN COO H CH3O H Free carboxyl end of peptide CHH3C CH3 H2O Free amino end of peptide + H3N CHH3C CH3 +H3N Ferrier_Unit01.indd 15Ferrier_Unit01.indd 15 5/2/14 7:08 PM5/2/14 7:08 PM
27. 27. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer 2.2 Answer Primary Structure of Proteins A peptide bond, a type of amide bond, is outlined by the black box. Peptide bonds link the amino acid residues in a peptide or protein by joining the -amino group of one amino acid to the -carboxyl group of the next as water is released. The peptide bond has partial double-bond character, is rigid and planar, uncharged but polar, and almost always in the trans conguration that reduces steric interference by the R groups. Peptide bonds are resistant to conditions (such as the heat from a fever) that can denature proteins and cause them to unfold. However, they are susceptible to cleavage by enzymes known as proteases or peptidases. [Note: Strong acids or bases at high temperatures can nonenzymatically cleave peptide bonds.] C COO H Valine Valylalanine C+ H3N COO H CH3 Alanine C C H CN COO H CH3O H Free carboxyl end of peptide CHH3C CH3 H2O Free amino end of peptide Peptide bond + H3N CHH3C CH3 +H3NTrans peptide bond C N H O C C C N HO CC Cis peptide bond R RR R Ferrier_Unit01.indd 16Ferrier_Unit01.indd 16 5/2/14 7:08 PM5/2/14 7:08 PM
28. 28. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer 2.3 QuestionPrimary Structure of Proteins Sequencing large polypeptides involves cleavage reactions, as shown. Which sites in a peptide are susceptible to cleavage by the endopeptidase trypsin? By cyanogen bromide? What is the Edman degradation method? What is the amino acid sequence of a nonapeptide if trypsin digestion yields three products (Asn, Met-Gln-Lys, and Ala-Gly-Met-Leu-Arg) and cyanogen bromide cleavage yields three products (Leu-Arg-Met, Gln-Lys-Asn, and Ala-Gly-Met)? 1. Cleave with trypsin Peptide of unknown sequence 2. Determine sequence of peptides using the Edman method What is the correct order? Peptide BPeptide A Peptide X Peptide Y Peptide C 1. Cleave with cyanogen bromide 2. Determine sequence of peptides using the Edman method 1 2 Original sequence of peptide A B C ? A C B ? B A C ? B C A ? C A B ? C B A ? Peptide of unknown sequence Ferrier_Unit01.indd 17Ferrier_Unit01.indd 17 5/2/14 7:08 PM5/2/14 7:08 PM
29. 29. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer 2.3 Answer Primary Structure of Proteins Trypsin, an endopeptidase, cleaves at the carboxyl side of Lys and Arg residues within a peptide. [Note: Exopeptidases remove the terminal amino acid.] Cyanogen bromide cleaves at the carboxyl side of Met residues. The Edman degradation method chemically determines the sequence of amino acids through the sequential removal and identication of the N-terminal amino acids in the small peptides generated from a polypeptide by cleavage reactions. Based on the overlapping amino acids in the products of the trypsin (Asn, Met-Gln-Lys, and Ala-Gly-Met-Leu-Arg) and the cyanogen bromide (Leu-Arg-Met, Gln-Lys-Asn, and Ala-Gly-Met) cleav- age reactions, the amino acid sequence of the nonapeptide is Ala-Gly-Met-Leu-Arg-Met-Gln-Lys-Asn. [Note: The sequence of amino acids in a protein is always written from the N-terminal to the C-terminal amino acid.] 1. Cleave with trypsin at lysine and arginine Peptide of unknown sequence 2. Determine sequence of peptides using the Edman method What is the correct order? Peptide BPeptide A Peptide X Peptide Y Peptide C 1. Cleave with cyanogen bromide at methionine 2. Determine sequence of peptides using the Edman method 1 2 Original sequence of peptide A B C ? A C B ? B A C ? B C A ? C A B ? C B A ? Peptide of unknown sequence Ferrier_Unit01.indd 18Ferrier_Unit01.indd 18 5/2/14 7:08 PM5/2/14 7:08 PM
30. 30. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer 2.4 QuestionSecondary Structure of Proteins Which type of secondary structure is illustrated at right? How does the orientation of the hydrogen bonds differ between the -helix and the -sheet structures? In proteins (e.g., the GPCRs for glucagon and the catecholamines) that contain several -helical membrane-spanning domains, why would Pro not be one of the amino acids found in these domains? Side chains of amino acids extend outward N H C O C O CN C N H H CO C C N H O C C O O H N C C N H N H R C C C R R R Ferrier_Unit01.indd 19Ferrier_Unit01.indd 19 5/2/14 7:08 PM5/2/14 7:08 PM
31. 31. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer 2.4 Answer Secondary Structure of Proteins The gure illustrates an -helix, a right-handed, helical, secondary structural element commonly encountered in both brous and globular proteins. The hydrogen bonds in a coiled -helix are intrachain bonds that are parallel to the polypeptide back- bone, whereas those in a -sheet (an extended structure) can be intra- or interchain bonds (depending on whether they form between sections of one polypeptide or between two polypeptides) that are perpendicular to the backbone. [Note: -Helices and -sheets may be components of supersecondary structures (motifs), such as a -barrel.] Pro contains a secondary amino group that is not compatible with the right-handed spiral of the -helix because (1) it cannot participate in the hydrogen bonding and (2) it causes a kink in the protein. Consequently, Pro is not found in the membrane-spanning domains of proteins such as GPCRs. [Note: Amino acids with bulky or charged R groups can also disrupt formation of an -helix.] Side chains of amino acids extend outward Intrachain hydrogen bond N H C O C O CN C N H H CO C C N H O C C O O H N C C N H N H R C C C R R R COOH H Proline C CH2 +H2N H2C CH2 Ferrier_Unit01.indd 20Ferrier_Unit01.indd 20 5/2/14 7:08 PM5/2/14 7:08 PM
32. 32. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer 2.5 QuestionTertiary Structure of Proteins What type of molecular interaction involved in stabilizing the tertiary structure of a protein is shown? What type of interaction would likely occur between Asp and Lys? The tertiary structures of proteins (such as albumin) that function in the extracellular environment are stabilized by the formation of covalent links between the oxidized side chains of which sulfur-containing amino acid(s)? CH2 C CH3 CH3CH3 CH2 CH H3C CH3 H C C H N OH H CN H C O Polypeptide backbone Isoleucine Leucine Ferrier_Unit01.indd 21Ferrier_Unit01.indd 21 5/2/14 7:08 PM5/2/14 7:08 PM
33. 33. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer 2.5 Answer Tertiary Structure of Proteins Shown are hydrophobic interactions between Ile and Leu, two amino acids with nonpolar R groups. Ionic interactions (salt bridges) would likely occur between Asp (acidic R group) and Lys (basic R group). Two sulfur-containing Cys residues, brought into close proximity by the folding of the peptide(s), are covalently linked through oxidation of their thiol side chains. The disulde bonds formed stabilize the tertiary structure of the folded peptide, preventing it from becoming denatured in the oxidizing extracellular environment. [Note: Cys-containing albumin transports hydrophobic molecules (e.g., fatty acids and bilirubin) in the blood. Its levels are used as an indicator of nutritional status.] CH2 C CH3 CH3CH3 CH2 CH H3C CH3 H C C H N OH H CN H C O Polypeptide backbone Isoleucine Leucine CCCCCCCCCCC 333333333333333333333333HHHHHHHHCHCHCHCHCHCHCHCHCHCHCHCCCCCCHCHCHCHCHCHCHCHCHCHCHCHCHCHCHCHCHCHCHCHCHCHCHCHCHCHCHCHCHCHCHCHCH333333333333333333333333333333333333333333333333333333333333333333 CH2 CH HHH333CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC CHCCCCCCCCCCCCCCCCCC 3 H CN H C O peptide kbone Leucine Hydrophobic interactions Cystine residue H CN CH2H S C C H C O CH2 N O H Two cysteine residues H CN CH2H SH SH C C H C O CH2 N O H S Polypeptide backbone Cystine residue H CN HHCHCHCHCHCHCHCHCHCHCHCHCHCHCHCHCHCHCC 2HHHHHH S C O Disulfide bond Oxidant (for example, O2) Ferrier_Unit01.indd 22Ferrier_Unit01.indd 22 5/2/14 7:08 PM5/2/14 7:08 PM
34. 34. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer 2.6 QuestionProtein Misfolding As illustrated, what secondary structural feature is enriched in the infectious form of a prion protein (PrP) as compared to the noninfectious form? Why do most large denatured proteins not revert to their native conformations even under favor- able environmental conditions? What misfolded peptide formed by abnormal proteolytic cleavage is the dominant component of the plaque that accumulates in the brains of individuals with Alzheimer disease? Infectious PrPSc Infectious PrPSc Infeectious PrP cSc Interaction of the infectious PrP molecule with a normal PrP causes the normal form to fold into the infectious form. Noninfectious PrPC Ferrier_Unit01.indd 23Ferrier_Unit01.indd 23 5/2/14 7:08 PM5/2/14 7:08 PM
35. 35. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer 2.6 Answer Protein Misfolding The -sheet secondary structure is enriched in the infectious PrPSc form of a PrP, which causes the transmissible spongiform encephalopathies, as compared to the noninfectious PrPC form that is -helical rich. The folding of most large proteins is a facilitated process that requires the assistance of proteins known as chaperones and ATP hydrolysis. A is the misfolded peptide produced by abnormal proteolytic cleavage of amyloid precursor protein by secretases. A forms an extended -sheet and spontaneously aggregates to form brils that are the dominant component of the amyloid plaque that accumulates in the brains of individuals with Alzheimer disease. [Note: The -sheets in A have exposed hydrophobic amino acid residues. The hydrophobic effect drives the aggregation and precipitation of A.] Interaction of the infectious PrP molecule with a normal PrP causes the normal form to fold into the infectious form. Infectious PrPSc (contains a-sheets) Infectious PrPSc (contains a-sheets) Noninfectious PrPC (contains `-helix) Aa Cell membrane Amyloid Spontaneous aggregation to form insoluble fibrils of a-pleated sheets Ferrier_Unit01.indd 24Ferrier_Unit01.indd 24 5/2/14 7:08 PM5/2/14 7:08 PM
36. 36. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer 3.1 QuestionMyoglobin Structure and Function Which His residue (A or B), as shown, is the proximal His? What is its function? What is special about the location of this amino acid? What type of secondary structure is most abundant in Mb? Does Mb have a quaternary structure? Rhabdomyolysis (muscle destruction) caused by trauma, for example, is characterized by muscle pain, muscle weakness, and dark-colored urine. The dark color of the urine is the result of excretion of , a condition known as . Oxygen molecule (O2) Heme F Helix E Helix A B Fe Ferrier_Unit01.indd 25Ferrier_Unit01.indd 25 5/2/14 7:08 PM5/2/14 7:08 PM
37. 37. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer Choice A is the proximal His. It forms a coordination bond with the Fe2 in the heme prosthetic group. Polar His is located in the nonpolar crevice where heme binds. Mb is rich in -helices. Because it is a monomeric protein, Mb does not have a quaternary structure. Rhabdomyolysis (muscle destruction) caused by trauma, for example, is characterized by muscle pain, muscle weakness, and dark-colored urine (shown). The dark color of the urine is the result of excretion of Mb, a condition known as myoglobinuria. Oxygen molecule (O2) Heme F Helix E Helix Fe Proximal histidine (F8) Distal histidine (E7) 3.1 Answer Myoglobin Structure and Function Ferrier_Unit01.indd 26Ferrier_Unit01.indd 26 5/2/14 7:08 PM5/2/14 7:08 PM
38. 38. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer 3.2 QuestionHemoglobin Structure and Function Which form of Hb (deoxygenated or oxygenated) is referred to as the R form? What determines the equilibrium concentrations of deoxyHb and oxyHb? How does the structure of Hb change as O2 binds to the heme Fe2 ? What condition, characterized by a chocolate cyanosis, results from the oxidation of Fe2 to Fe3 in Hb? Why might replacement of the distal His cause this condition? 4 O2 O2 O2 O2O2 4 O2 dimer 2 dimer 1 dimer 1 dimer 2 dim O2 4 O2 4 O2 mememerr 2 mer 1m Weak ionic and hydrogen bonds occur between dimer pairs in the deoxygenated state. Some ionic and hydrogen bonds between dimers are broken in the oxygenated state. Strong interactions, primarily hydrophobic, between and chains form stable dimers. Ferrier_Unit01.indd 27Ferrier_Unit01.indd 27 5/2/14 7:08 PM5/2/14 7:08 PM
39. 39. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer 3.2 Answer Hemoglobin Structure and Function The oxygenated, high-O2-afnity form of Hb is referred to as the R form. The availability of O2 determines the equilibrium concentrations. The binding of O2 to the heme Fe2 pulls the Fe2 into the plane of the heme. This causes salt bridges between the two dimers to rupture, thereby allowing movement that converts the T to the R form. Methemoglobinemia, characterized by a chocolate cyanosis (dark-colored blood, bluish colored skin), results from the oxidation of Fe2 to Fe3 in Hb. Because the distal His stabilizes the binding of O2 to the heme Fe2 , its replacement with another amino acid will favor oxidation of Fe2 to Fe3 and decreased binding of O2. 4 O2 O2 O2 O2O2 4 O2 "R," or relaxed, structure of oxyhemoglobin"T," or taut, structure of deoxyhemoglobin dimer 2 dimer 1 dimer 1 dimer 2 dim O2 4 O2 4 O2 mememerr 2 mer 1m Weak ionic and hydrogen bonds occur between dimer pairs in the deoxygenated state. Some ionic and hydrogen bonds between dimers are broken in the oxygenated state. Strong interactions, primarily hydrophobic, between and chains form stable dimers. Ferrier_Unit01.indd 28Ferrier_Unit01.indd 28 5/2/14 7:08 PM5/2/14 7:08 PM
40. 40. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer 3.3 QuestionO2 Binding to Myoglobin and Hemoglobin Use the gure to determine the approximate amount of O2 that would be delivered by Mb and Hb when the pO2 in the capillary bed is 26 mm Hg. Why is the O2-dissociation curve for Hb sigmoidal and that for Mb hyperbolic? How might RBC production be altered to compensate for changes to Hb that result in an abnormally high afnity for O2? %SaturationwithO2(Y) 0 0 40 80 120 100 Hemoglobin Myoglobin pO2 in tissues pO2 in lungs 50 Partial pressure of oxygen (pO2) (mm Hg) Ferrier_Unit01.indd 29Ferrier_Unit01.indd 29 5/2/14 7:08 PM5/2/14 7:08 PM
41. 41. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer At a pO2 of 26 mm Hg, Hb would have delivered 50% of its O2, while Mb would have delivered 10%. Hb has a lower O2 afnity at all pO2 values and a higher P50 than does Mb, as shown. [Note: P50 is that pO2 required to achieve 50% saturation of the O2-binding sites.] Hb is a tetramer. The O2-dissociation curve for Hb is sigmoidal because the four subunits cooperate in binding O2. The rst O2 binds to Hb with low afnity. As subsequent subunits become occupied with O2, the afnity increases such that the last O2 binds with relative ease. Because Mb is a monomeric protein, it does not show cooperativity. Conse- quently, its O2-dissociation curve is hyperbolic, not sigmoidal. RBC production typically is increased (a process known as erythrocytosis) to compensate for changes to Hb that result in an abnormally high afnity for O2: more RBCs more Hb more O2 carried. 3.3 Answer O2 Binding to Myoglobin and Hemoglobin %SaturationwithO2(Y) 0 0 40 80 120 P50 = 1 P50 = 26 100 Hemoglobin Myoglobin pO2 in tissues pO2 in lungs 50 Partial pressure of oxygen (pO2) (mm Hg) tionwithO2(Y) 100 Hemoglo nobin Myoglobin pO2 in tissues pO2 in lungs 50 The oxygen-dissociation curve for Hb is steepest at the oxygen concentrations that occur in the tissues. This permits oxygen delivery to respond to small changes in pO2. Ferrier_Unit01.indd 30Ferrier_Unit01.indd 30 5/2/14 7:08 PM5/2/14 7:08 PM
42. 42. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer 3.4 QuestionAllosteric Effects Which curve (A or B), as shown, represents the lower pH? List two other allosteric effectors that, when increased, result in a rightward shift of the Hb O2-dissociation curve. What does this shift reect? Do these allosteric effectors stabilize the R or the T form of Hb? How does the binding of CO2 to Hb stabilize Hbs deoxygenated form? What is the Bohr effect? %SaturationwithO2(Y) Partial pressure of oxygen (pO2) (mm Hg) 0 0 40 80 120 100 B A 50 Ferrier_Unit01.indd 31Ferrier_Unit01.indd 31 5/2/14 7:08 PM5/2/14 7:08 PM
43. 43. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer Curve B represents the lower pH (higher H concentration). Increased amounts of CO2 and 2,3-BPG also result in a rightward shift of the Hb O2-dissociation curve. The shift reects increased off-loading (delivery) of O2 to the tissues. These allosteric effectors stabilize the T (deoxygenated) form of Hb, enabling O2 delivery. When CO2 binds to the amino termini of the four Hb subunits, forming carbaminohemoglobin, the negative charge is used to form a salt bridge that helps to stabilize Hbs deoxygenated (T) form. Hb NH2 CO2 Hb NH COO H The Bohr effect refers to the increase in O2 delivery when CO2 or H increases. In actively metabolizing tissue, Hb binds CO2 and H and releases O2. The process is reversed in the lungs. 3.4 Answer Allosteric Effects Fe2+ Fe2+ Fe2+ Fe2+ O2 O2 O2 O2 Oxyhemoglobin Fe2+ Fe2+ Fe2+ Fe2+ NHCOO NHCOO Carbaminohemoglobin CO2 O2 O2CO2 O2CO2 C OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO2222222222222222222222222222OCCCCCCCCCCCOOOOOOOOOOOOOOOOOOCOOOOCOCCCCOCCCCCCCCCOCOOCOC 22222222222222222222 CO2 binds to hemoglobin. O2 is released from hemoglobin. O2 binds to hemoglobin. CO2 is released from hemoglobin. TISSUES LUNGS %SaturationwithO2(Y) Partial pressure of oxygen (pO2) (mm Hg) 0 0 40 80 120 100 pH = 7.2 pH = 7.6 50 O2( pH = 7.2 (Y) 100 pH = 7.6 Decrease in pH results in decreased oxygen affinity of hemoglobin and, therefore, a shift to the right in the oxygen-dissociation curve. At lower pH, a greater pO2 is required to achieve any given oxygen saturation. Ferrier_Unit01.indd 32Ferrier_Unit01.indd 32 5/2/14 7:08 PM5/2/14 7:08 PM
44. 44. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer 3.5 QuestionMinor Hemoglobins How does the subunit composition of HbF, as illustrated, inuence the O2 afnity of HbF? What form of Hb replaces HbF, and when does this occur? What form of Hb is measured to assess glycemic control in individuals with diabetes? HbA 22 Form Chain composition HbA1c 22-glucose 22HbF HbA2 22 Ferrier_Unit01.indd 33Ferrier_Unit01.indd 33 5/2/14 7:08 PM5/2/14 7:08 PM
45. 45. Lippincott Illustrated Reviews Flash Cards: Biochemistry Copyright 2015 Wolters Kluwer HbF contains 2 and 2 subunits. Relative to the subunits, the subunits have a reduced afnity for 2,3-BPG. This results in HbF having an increased afnity for O2. [Note: HbF is needed to obtain O2 from maternal HbA, and its increased afnity for O2 enables this process.] HbF is the major Hb found in the fetus and the newborn but represents 2% of the Hb in most adults because it is replaced by HbA (2 and 2 subunits) by about 6 months after birth. Nonenzymatically glycosylated (glycated) Hb, HbA1c, is measured because its concentration in the blood is a reection of the average blood glucose concentration over the previous 3 months. [Note: The goal value for HbA1c in adults with diabetes is 6.5%.] 3.5 Answer Minor Hemoglobins Months before and after birth Percentageoftotalglobinchains 9 6 3 3 6 9 0 25 50 0 25 50 ` a c d f y `-Globin- like chains a-Globin- like chains Time of birth 0 HbA 22 Form Chain composition Fraction of total hemoglobin HbA1c 22-glucose 90% 3%9% 22HbF