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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
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3. Acquisitions Editor: Tari Broderick Product Development
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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
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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
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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
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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.
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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
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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
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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
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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
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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
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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
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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
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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
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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] +
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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
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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
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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.]
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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. 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
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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
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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
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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.
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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
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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
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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
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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
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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
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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