medical biochemistry

SIU School o f Medic ine BIOCHEMISTRY

pH and Structural Biology

MEDICAL BIOCHEMISTRY

Problem Unit One1999/2000

pH and Structural BiologyModule 1: Acid/Base Properties of Biomolecules

Module 2: Amino Acids, Peptides, and Proteins

Module 3: Structural Biology and Disease

Faculty: J.W. Shriver Problem Unit 1 -



Faculty: Dr. John W. ShriverDepartment of Biochemistry and Molecular BiologyOffice: 289 Neckers Bldg.email: [email protected]: 453-6479

LEARNING RESOURCES:

ESTIMATED WORK TIME: 40 hours.

A. This study guide is provided in two forms: printed and electronic.I strongly encourage you to obtain the electronic form as a pdf fileand install it on your computer so that it can be read using AdobeAcrobat Reader. See Appendix II for an introduction on how toview a pdf file. The pdf file can be downloaded from the biochemis-try server (http://www.siu.edu/departments/biochem) and AcrobatReader can be downloaded without charge from Adobe’s web page(http://www.adobe.com/acrobat). They should also be installed onthe student computers. There are a number of advantages to usingthe electronic version including color, a hypertext index, and hyper-text links within the text. Hypertext links in the text body are in blueunderlined characters (such as this). Clicking on these will lead to ajump to the linked material for further details. The destination mate-rial is highighted with red underlined characters (such as this) tomake it easy for you to find on the page. The red underlined text isnot a hyperlink - only a destination.

This and other study guides are provided to help you focus on thetopics that are important in the biochemistry curriculum. These aredesigned to guide your studying and provide information that maynot be readily available in other resources. They are not designed toreplace textbooks, and are not intended to be complete. They areguides for starting your reading. The pdf electronic versions shouldbe especially useful for quick reviews at a later date. The hypertextNomenclature and Vocabulary sections should permit rapid scanningof the key points.

B.Textbooks: 1.Devlin, Textbook of Biochemistry with Clinical Correla-tions, Thomas. Core text for Medical Biochemistry.2.Champ & Harvey, Lippincotts Illustrated Reviews of Bio-chemistry, current ed., Lippincotts. Efficient presentation ofbasic principles. 3.Murray et al., Harper's Biochemistry, (23rd ed.) ('93), Pren-tice-Hall, Inc. An excellent review text for examinations.


http://www.siu.edu/departments/biochem

http://www.adobe.com/acrobat



Most textbooks of biochemistry contain sections on pH and dissocia-tion and protein structure; some are more extensive than others. Anybiochemistry textbook that covers the subject in sufficient detail soyou can answer the questions in the Problem Sets and Practice Examshould be sufficient. Additional material can be found on the web atthe National Institutes of Health (http://www.nih.gov), theNational Library of Medicine (http://www.nlm.nih.gov), and thefree MEDLINE PubMED Search system at the National Library ofMedicine (http://www3.ncbi.nlm.nih.gov/PubMed/).

C.Lecture/Discussions Especially recommended for those who have not had biochemistryand for those who have questions.

EVALUATION CRITERIA: A written examination will be scheduled. Answers to questions and

the solving of problems will be judged against the learning resources.Examples of exam questions are given in the Problem Sets. The passlevel is 70%.

Module 1: Acid/Base Chemistryof BiomoleculesINTRODUCTION: Water makes up about 70% of a typical cell by weight. It is one of

two solvents in which most of biochemistry occurs, the second beingthe lipids of membranes. Water is a very unusual substance and playsa central role in defining life as we know it. Its large dipole momentmeans that it is a highly polar liquid (at 37°C) and thus serves as anexcellent solvent for other polar (and hydrophilic) molecules. Apo-lar molecules are not easily dissolved in water and are referred to ashydrophobic. Hydrophobic molecules are excluded from an aque-ous environment because they cannot interact well with water andtherefore lead to a structuring of water in their vicinity (an unfavor-able process). Since hydrophobes generally mix well, they separate tominimize their interface with water and form a second distinct envi-ronment - the greasy, oily environment of lipids (lipophilic). Thebiochemical system can be viewed as two different environments: the


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http://www.nlm.nih.gov/

http://www3.ncbi.nlm.nih.gov



aqueous, polar environment (e.g. cytoplasm); and the hydrophobic,or lipophilic, non-aqueous environment (e.g. membranes).

Hydrophobic compounds are uncharged, nonpolar species and gener-ally contain largely aliphatic and aromatic organic groups. Hydro-philic compounds are polar and include sugars, salts, acids and bases,and polar organic groups such as amino, carboxyl, and alcoholgroups.

Many molecules become charged (i.e. they become ions) when dis-solved in water. Most notable of these are the acids and bases. Posi-tive ions are referred to as cations, and negative ions are anions. Thepredominant cations in blood plasma is Na+ (making up about 150meq/L out of a total of 170). The predominant anions are Cl- andbicarbonate (HCO3

-). In contrast the predominant cations in cyto-plasm are K+ and Mg++, and the anions are inorganic and organicphosphates and negatively charged proteins.

Acids and bases become charged in water through release or accep-tance of a proton. Acid/base balance in a living organism is criticalsince it defines the relative charge on many molecules including pro-teins important in cellular function. In many clinical situations,acid/base balance must be modified and controlled by the physicianto ensure the health of a patient.

Many of the properties of proteins and other biomolecules have theirorigin in the acidic and basic character of functional groups on thebiomolecule.

Common biological phenomenon, as well as experimental techniquesused in both clinical and research laboratories, make use of the acid/base properties of biomolecules. These include such common tech-niques such as ion exchange chromatography and electrophoresis.The establishment and maintenance of pH gradients in membranes,and the partitioning and compartmentation of biomolecules anddrugs in cells and subcellular particles are at least in part dependenton the acid/base properties of the molecules that are involved. Inorder to understand the function of these molecules, it would be bestto obtain a working knowledge of pH and proton dissociation. Sev-eral concepts and terms must be understood at a level sufficient towork problems that require calculating for example, pH, conjugateacid, conjugate base concentrations, pKa , isoelectric points, andbuffering capacity.

OBJECTIVES: You will need to understand pH, H+ ion concentration, the Hender-son-Hasselbalch equation, Ka, pKa, ionization, protonation-depro-




tonation, and conjugate acid and conjugate base. An understandingof chemical equilibria will be required. Examples of the types ofquestions and problems to be solved are included in the ProblemSets. More specific objectives that are part of this objective are as fol-lows:

a. When given the molarity or normality of a strong acid orbase, calculate the hydrogen ion concentration and the pH.b. When given the molarity of a weak acid or base and itspKa(s), calculate its percent dissociation, the hydrogen ionconcentration and the pH.c. When given the pH of a solution and the pKa values, calcu-late the concentration of the conjugate acid and conjugate baseand determine the net charge on the molecule.

Be prepared to sketch (in a qualitative fashion) titration curves formolecules or ions with single or multiple pKa's and answer questionsusing the titration curve concerning pH, titration (e.g., percent titra-tion and/or fraction of conjugate acid and conjugate base at a speci-fied pH), isoelectric point, buffering strength, ionic species present ata particular pH and the charge on the molecule.

Understand the role of electrostatic interactions and hydrophobicinteractions in determining solubility in aqueous and lipid solvents.When given a chemical compound, characterize it as hydrophilic orhydrophobic, and ionizable or unionizable. Be able to predict solu-bilities (qualitatively) when given molecular structures and pH.

Define the terms in the NOMENCLATURE and VOCABULARYlist and use them properly in answering questions concerning thismodule.

NOMENCLATURE and VOCABULARY:

Amino groupApolarBufferBuffering capacityCarboxyl groupConjugate acid Conjugate base EquilibriumEquivalentsHenderson Hasselbalch equationHydrophilicHydrophobicIsoelectric point (pI)Kw




LipophilicNeutralitypHpKa, pKbpOHPolarSaltStrong acid or baseTitrationWeak acid or base

STUDY GUIDE-1I. Equilibria The concept of equilibrium is central to all of biochemistry. Any

chemical change can be reversed, and the relative amounts of thestarting and final species are determined by their relative energies -the more stable species will predominant, but the least stable will alsoalways exist, even if only in a minute amount. No reaction ever goesto total completion, although sometimes the reaction can be viewedas essentially complete for all practical purposes. An example is thedissociation of NaCl in water:

NaCl <=> Na+ + Cl-

The solvent water is not explicitly written since it does not directlyparticipate in the reaction; it merely provides a medium. The arrowsare written in both directions to emphasize that the reaction proceedsin both directions. It is important to realize that a reaction is adynamic process with both forward and reverse reactions occurring,even at equilibrium. The charged sodium and chloride ions aremuch more polar than the NaCl so that the energetically preferredspecies in water is the dissociated ionic species. Most salts essentiallycompletely dissociate when dissolved in water, i.e. although Na+ andCl- ions can reassociate, they rarely do so.

Another compound which essentially completely dissociates in wateris hydrochloric acid (HCl):

HCl <=> H+ + Cl-

This is an acid because it contributes a H+ (i.e. a proton) upon disso-ciation. (Strictly speaking, the proton is taken up by a water mole-




cule to give H3O+). Since HCl completely dissociates, it is referredto as a strong acid. NaOH completely dissociates as follows:

NaOH <=> Na+ + OH-

This is a base because it contributes an OH- ( i.e. an hydroxide ion)upon dissociation. Since NaOH completely dissociates, it is referredto as a strong base.

Not all compounds become ions when dissolved in water, e.g. glu-cose. Some compounds ionize partially when dissolved in water, andthese are of central importance here. An example is acetic acid:

CH3COOH <=> CH3COO- + H+

which contains a carboxyl group. Another is ethylamine:

CH3CH2NH2 + H2O <=> CH3CH2NH3+

+ OH-

which contains an amino group. Note that water is an explicit reac-tant in the last reaction and the amine strips a proton off and leavesbehind a hydroxide ion. Since neither acetic acid or the amine pro-ceed to essentially complete ionization upon dissolving in water, theyare referred to as a weak acid and a weak base, respectively.

In fact, water itself can ionize to some extent:

H2O <=> H+ + OH-

In pure water the concentration of H+ is extremely low at 10-7 molar.Instead of working with such small numbers, we typically translatethem into different units by taking the negative logarithm of the con-centration. This is called the pH:

pH ≡ -log [H+] = -log( 10-7 ) = 7

The pH of pure water is 7. A pH of 7 indicates neutrality, i.e. theconcentration of H+

and OH- are the same. Increasing concentra-tions of H+ cause the pH to decrease; thus a pH less than 7 indicatesan acidic solution. A pH greater than 7 indicates a basic, or alka-line, solution.

There is no way to know if a compound is a strong or weak acid orbase by looking at it without some previous knowledge (and memori-




zation). Strong acids include hydrochloric, sulfuric, and phosphoricacids. Strong bases include sodium and potassium hydroxide. Weakacids include an organic compound containing a carboxyl group, andweak bases include any organic compound containing an aminegroup. For example, consider aspirin:

ASPIRIN

Is this an acid or base? Strong or weak? (Answer). What aboutprocaine? (Answer)

What about tyrosine?

(Answer)

OH

CH3

C

O

O

C

O

O CH2CH2 N

CH2CH3

CH2CH3

H2N

OH

COOH

H

H2N

CH2

C




II. Equilibrium constants Often we need to be more precise about the degree to which a reac-

tion progresses, and this is accomplished through an equilibriumconstant. Note that the position of the equilibrium is established bythe relative energies of the reactants and products - thus the "posi-tion" of the equilibrium is fixed by nature and can be described by afundamental "constant".

For any reaction, K is equal to the product of the concentrations ofthe products divided by the product of the concentrations of the reac-tants. For example, for the following reaction

A + B <=> C + D

the equilibrium constant, K, is given by K = [C] [D] / [A] [B]

where the brackets indicate that we are using concentrations. Forreactions which go to essential completion, such as the dissociation ofHCl in water, we do not normally discuss an equilibrium constantsince it is essentially infinite. However, for the ionization of a weakacid or base the equilibrium constant is very useful. For example, foracetic acid

We stress that K is a constant. Thus if we start with 1 M acetic acidor with 0.001 M acetic acid, the reaction will proceed until the aboveratio is achieved; the equilibrium will be reached and the product ofthe reactants divided by the reactants will be 1.74 x 10-5 M (althoughthe actual concentrations will be different in the two cases).

Again, similar to what we did with H+ concentrations above, theequilibrium constant can be translated into different units by takingthe negative logarithm of the equilibrium constant and this is referredto as a pK (in analogy to the pH):

pK ≡ - log K = - log (1.74 x 10-5) = 4.76

The pK for the dissociation of acetic acid is 4.76. The K and pK aretwo different ways of expressing the same thing. They are equivalentand you may use, or encounter, either.

K = [CH3COO−] [ H+]

[CH3COOH] = 1.74 x 10−5 M




The equilibrium constant for the dissociation of water is very small.

Pure water has a concentration of 55.5 molar, so the product of theconcentration of hydroxide and proton concentrations is 10-14. Thisis commonly referred to as Kw. Therefore,

pKw = pH + pOH = 14

where pOH ≡ -log [OH-]. In other words, if we know the pH, wealso know the hydroxide ion concentration, since pH = 14 - pOH.

III. Acids, Bases, and Salts

An acid is any substance that can donate a proton, and a base is anysubstance that can accept a proton (strictly speaking, we are using theLewis definition here). Thus HCl and acetic acid are both acids,whereas ammonia and ethylamine are both bases. What is glycine?

GLYCINE

If ethylamine is a base, when it accepts a proton it becomes an acidsince it can now donate that proton, and the resulting protonatedspecies is the conjugate acid. The ethylammonium ion is the conju-gate acid of ethylamine. If acetic acid donates a proton, the resultingacetate ion is a base since it can accept a proton. Acetate is the conju-gate base of acetic acid. As we will see below, conjugate acid/basepairs of weak acids and bases are essential in defining a buffer.

If we define the pK for an acid as pKa and the pK for a base as pKb,then it is possible to show that for conjugate acids and bases, pKa +

K = [H+] [OH−]

[H2O] = 1.8 x 10−16 M

COOHNH2

H

C

H




pKb = 14. For example, for acetic acid:

The addition of equivalent amounts of an acid and a base yields asalt. For example, addition of equivalent amounts of hydrochloricacid and sodium hydroxide yields NaCl and water. Addition ofequivalent amounts of sodium hydroxide to acetic acid yields sodiumacetate:

NaOH + CH3COOH <=> CH3COO- + Na+ + H2O

IV. pH and strong acids

If we dissolve HCl in water, the pH of the resulting solutions isstraightforwardly given by the negative log of the concentration ofHCl since all of the HCl is assumed to dissociate. Thus, a 0.001 Msolution of HCl has a pH of

- log [ 0.001 ] = 3

V. pH and strong bases

Likewise, if we dissolve NaOH in water, the pOH is straightfor-wardly given by the negative log of the concentration of the NaOH.BE CAREFUL: The pH is given by 14 - pOH (since pH + pOH =14).

VI. pH and weak acids The pH of a solution of a weak acid is determined not only by theconcentration of the acid but also by its pK since it does not com-

Ka = [acetate] [H+][acetic acid]

Kb = [acetic acid] [OH−]

[acetate]

KaKb = [H+] [OH−]

or

pKa + pKb = pH + pOH = 14


SIU School o f Medic ine BIOCHEMISTRY pH and Structural Biology

pletely dissociate. If we dissolve acetic acid in water we obtain thefollowing equilibrium:

CH3COOH <=> CH3COO- + H+

For every acetic acid which dissociates (not dissolves) we obtain anequivalent amount of acetate and protons. Thus, if we start with0.001 M acetic acid, we can write:

x is obtained by solving the quadratic equation.

VII. pH and salts of weak acids

When a salt of a weak acid is added to water, the salt dissociates com-pletely. The anion is now the conjugate base of the weak acid, andpartial re-association with a proton from water will occur to establishthe appropriate equilibrium for the base. An example should help toclarify these points. Consider dissolving sodium acetate in water:

CH3COO- + Na+ + H2O <=> CH3COOH + Na+ + OH-

Clearly, since acetic acid is a weak acid, the acetate cannot be com-pletely dissociated, and will pick up some protons from water inorder to establish the appropriate equilibrium. This leaves behindhydroxide ion, so the pH must increase. The pKb for the acetate is14 - 4.76 = 9.24, so if we start with 0.001 M acetate, we obtain:

and [OH-] is obtained by solving the quadratic equation. (Again,note that the pH is obtained by getting the pOH and subtracting thisfrom 14.)

VIII. Buffering If we add increasing amounts of a strong base to a strong acid, thebase will neutralize the acid (forming a salt). The pH will be deter-mined by the acid concentration until it is completely converted tosalt, and then the pH will rather abruptly change to a high value since

K = [x] [x]

[0.001 − x] = 1.74 x 10−5

Kb = [CH3COOH] [OH−]

[CH3COO−] =

[x][x][0.001−x]

= 5.75 x 10−10



it will be dictated by the increasing amount of base. This titration ofan acid with a base can be presented in pictorial form as followswhere we plot pH as a function of the equivalents of base added:

In contrast, if we add base to a weak acid, we will obtain somethingquite different. As an example, the titration of acetic acid might looklike the following:

In this case there is a broad plateau in the titration curve at elevatedpH over which the pH changes negligibly. This occurs in the rangewhere the concentration of base (and therefore acetate ion) is equiva-lent to the concentration of acid. In this range the pH is "buffered"and the acetate is referred to as a buffer. This elevated plateau onlyoccurs for weak acids and bases. The pH region of buffering is at thepK of the acid.

pH

1

7

14

[base]

pH

1

7

14

[base]



IX. Henderson - Hasselbalch Equation The concept of buffering is a key concept in biochemistry, so we will

be a little more precise here. If the weak acid is designated as HB,the equilibrium we are interested in is:

HB <=> H+ + B-

and

The last equation is known as the Henderson-Hasselbalch equation,and allows us to know the pH of any solution given its pKa and therelative amounts of conjugate base and acid. The titration curve inthe figure above is a plot of this equation. When [B-] and [HB] areequal, pH = pK (since log(1) = 0). This occurs in the middle of theplateau. Thus the plateau is centered at the pK, or maximal bufferingoccurs at the pK.

X. Buffer capacity It is important at this point to mention buffer strength or bufferingcapacity. It is clear that the buffering efficiency of a weak acid or baseis maximal at the pKa of the compound. That is, this is the pH atwhich there is the greatest resistance to pH change with the additionof acid or base. It follows that the buffer strength of a solution of aweak electrolyte depends upon two factors: the proximity of the pHto the pK of the compound and the concentration of the compound.

Ka = [H+] [B−]

[HB]

pKa = − log[H+] [B−]

[HB]

pKa = −log [H+] − log [B−][HB]

or

pH = pKa + log [B−][HB]



The greater the concentrations of conjugate acid and conjugate base,the greater the resistance to pH change. A solution of 0.01 molaracetic acid buffer at pH 5.0 has less buffer capacity than a 0.1 molaracetic acid buffer at pH 5.0.

XI. Charge on an ionizable group at aspecific pH We can use the Henderson-Hasselbalch equation to calculate the

effective charge on an ionizable group at a given pH. For example,what is the effective charge of acetate at pH 7.4? Rearranging theHenderson-Hasselbalch equation we have

Thus, the acetic acid is essentially completely ionized at this pH. Ifthe pH were 4.76, the acetic acid would be half ionized and the effec-tive charge would be -0.5 (i.e. half of a charge). Half of a charge doesnot exist, but the ensemble of all of the acetic acid/acetate ionsbehave as if there were an effective chage of -0.5. What about the ε-amino groups of lysine at pH 7.4? (The pK of these groups is about10.8.) What can you conclude from this about the normal charge onthe side chains of lysine, glutamate, and aspartate at physiologicalpH?

XII. Multiple equilibria Many compounds contain multiple ionizable groups. For example,

glycine has both a carboxyl group and an amino group, and thereforehas two pK's:

pK1 = 2.34 (COOH)

pK2 = 9.60 (NH3+)

Phosphoric acid has three ionizable hydroxyl groups and thereforehas three pK's:

pK1 = 2.0 (H3PO4)

pK2 = 6.7 (H2PO4-)

log[B−][HB] = pH − pKa = 7.4 − 4.76 = 2.64

B-[ ]HB[ ]

------------- 436=



pK3 = 12.5 (HPO4-2)

You should be able to predict from this information at what pHphosphate would make a good buffer.

Note that because compounds can have more than one ionizablegroup, it is often useful to refer to the concentration of equivalentsrather than the concentration of the compound itself. Thus a 0.001molar ( 1 mmolar or 1 mM) solution of phosphoric acid is a 3 mil-liequivalent (3 meq.) solution.

XIII. Titration of an Amino Acid

Let's now construct an accurate titration curve for glycine-hydrochlo-ride. (Glycine comes in three crystalline forms: GlyHCl, the com-pletely protonated form; glycine, the zwiterionic form; and Na-glycinate, the completely deprotonated form.)

The ionization processes which occur with glycine are described inthe following equations.

Since glycine-HCl is completely protonated and has two ionizablegroups it will take two equivalents of base [OH-] to titrate oneequivalent of glycine-HCl. Furthermore, because the pKa's arewidely separated, we will titrate the first group (carboxylic acid) com-pletely before beginning to substantially titrate the second group(amine).

With these ground rules we can construct the titration curve usingappropriate graphical coordinates: ordinate = [OH-] equivalents vs.abscissa = pH.



We begin with glycine in it's fully protonated form. The pH atwhich this occurs is around pH 1.0. Thus, we are starting our titra-tion in the lower left corner of the graph. Now, let's add sufficient[OH-] such that 10% of the molecules have their carboxyl groupstitrated, (0.1 equivalents titrated). To bring about this amount oftitration, 0.1 equivalents of base must be added. Putting this information into the Henderson-Hasselbalch equationwe can calculate the pH which will be achieved on addition of 0.1equivalents of base. The pH at which glycine will be 90% in the(0,+) ionic form and 10% in the (-,+) form is:

pH = 2.35 - 0.95pH = 1.40 for 0.1 equivalents of base added.

Upon adding 0.5 equivalents of base we will have titrated half of thetotal glycine carboxyl groups. This will require the use of 0.5 equiva-lents of base. By the Henderson-Hasselbalch equation we have:

pH = 2.35 + log (0.5/0.5)pH = 2.35 for 0.5 equivalents of base added.

Similarly, when enough base is added so that glycine is 90% (-,+) and10% (o,+) the pH becomes:

pH = 2.35 + log(0.9/0.1)pH = 3.30 for 0.9 equivalents of base added.

To a first approximation, 90%/10% is roughly 10/1. Therefore, as arule of thumb: one pH unit on either side of a pK represents a 90/10or 10/90 ratio of salt to acid. Thus, the buffering range of a buffergenerally extends one pH unit on either side of the pKa.

Plotting these values and drawing a smooth sigmoid curve throughthem gives a rather accurate pH titration curve for the carboxylgroup. The same can be done for the alpha amino group of glycine.

By rearranging the H-H equation and taking the antilog of both sideswe get the following variation:

(salt/acid) = 10pH- pKa

When pH = pKa + 1;(salt/acid) = 101 = 10/1

pH 2.35 0.100.90----------log+=

pH pKasalt[ ]acid[ ]

---------------log+=



When pH = pKa - 1;(salt/acid) = 10-1= 1/10

Though the ratios are actually 10/1 and 1/10, respectively to a goodapproximation they could be considered ca. 90/10 and 10/90.

The principle ionic species existing in the plateau regions are indi-cated on the graph. It is easy to identify the region in which theamino acid has a zero net charge, i.e. (-,+). The exact pH at whichthe ionic compound has a zero net charge is called the isoelectricpoint or the pI. To calculate the isoelectric point, one has only toidentify the pH which is exactly halfway between the two pKa's flank-ing the point of zero net charge.

pI = ( pKa1 + pKa2) / 2pI = (2.35 + 9.78) / 2pI = 6.06

XIV. Physiological Buffering

Normal arterial plasma pH is 7.4. A pH range of 6.8 to 7.8 is accept-able for life. The intracellular pH of an erythrocyte is about 7.2.Most other cells are around 7.0. Heavily exercised muscle pH candrop to 6.0.

As indicated above, the predominant anions in blood plasma arechloride and bicarbonate, and in cytoplasm are phosphates and pro-teins.

Intracellular pH is buffered by organic phosphates such as the sugarphosphates (pK's 6.5 to 7.6) and protein side chains (e.g. histidine(pK 5.6 to 7.0)). The inorganic phosphate concentration is low, giv-ing it an insignificant buffering capacity.

Carbonic acid-bicarbonate is an important extracellular buffering sys-



tem in mammals and is partially responsible for maintaining the pHof blood at 7.4, but maintenance of blood pH is largely controlledphysiologically rather than chemically since the pK of carbonic acid ismore than 1 unit from the extracellular pH.

CO2 is a gas which is hydrated by carbonic anhydrase in red bloodcells to form carbonic acid. The concentration of the acid species(H2CO3) can be controlled by respiratory regulation, i.e. yourbreathing rate. The equations involved are:

CO2 + H2O ↔ H2CO3; K1 = [H2CO3]/[CO2]

H2CO3 ↔ H+ + HCO3-; K2 = ([H+] [HCO3

-]) / [H2CO3]

CO2 + H2O ↔ H+ + HCO3-

By convention, the solvent [H2O] which is also a reactant in this pro-cess, is not included in the equilibrium constant. One can write anapparent dissociation constant for the total process as:

Kaapp = K1K2 = ([H+] [HCO3

-] / [CO2]

pKaapp = 6.1

Operationally, [CO2] includes both CO2 dissolved and H2CO3 .However, CO2 dissolved exceeds H2CO3 by 1000 fold at equilib-rium. Thus, the concentration of H2CO3 is negligible by compari-son. The corresponding Henderson-Hasselbalch expression becomes:

pH = 6.1 + log ([HCO3-] / [CO2 ])

The concentration of CO2 in the blood is commonly referred to interms of its partial pressure, e.g. PCO2 = 40 mm Hg. Partial pressureis converted to concentration with the conversion factor 0.03 meqliter-1 mmHg-1 at 37°C, so that

pH = 6.1 + log ([HCO3-] / 0.03 PCO2 )

with [HCO3-] expressed in milliequivalents per liter.

Note that total CO2 refers to CO2 + H2CO3 + HCO3- , since it is

measured by acidifying the solution with a strong acid to converteverything to CO2.

It should be noted that the human body is an open system and the



level of CO2 in the blood is regulated through respiration. A detailedconsideration of acid/base balance in the blood therefore requiresconsideration not only of the pK of carbonic acid but also the physi-ologically regulated level of CO2. This topic will be treated in muchgreater detail elsewhere.

PROBLEM SET-11. What is the (a) H+ ion concentration, (b) OH- ion concentration,(c) pH, and (d) pOH of a 0.001 M solution of HCl? (answer)

2. What is the pH of 0.004 M KOH solution? (answer)

3. The Ka for a weak acid, HA, is 1.6 X 10-6. What is the (a) pH and

(b) degree of ionization of the acid in a 10-3 M solution? (c) Calcu-late the pKa. (answer)

4. Given the following:

H2CO3 <=> H+ + HCO3- pKa = 6.1

0.03 meq/liter = 1.0 mm Hg

Normal blood concentrations of [HCO3-] and [H2CO3] are 24

meq./liter and 1.2 meq./liter respectively at pH 7.40

a.Knowing that the pH of blood can drop as low as 6.8 andstill be compatible with life, how many meq. of acid must beadded to plasma to achieve a pH = 6.8?

b.The upper limit to which blood plasma pH can be raisedfrom the normal pH = 7.4 and still be compatible with life is

equivalent to the addition of 29 meq./liter of HCO3- to the

normal blood values of HCO3- and H2CO3 (CO2 dis-

solved) (given above) without altering the normal H2CO3concentration. Calculate the upper limit of the blood pHcompatible with life. (answer)

5. Phosphate is an important body buffer. The dissociation processeswhich take place are:



a. At the pH of blood (pH 7.4), which species predominate?

b. What are their approximate proportions (to the nearest 1 %)at this pH? (show calculations) (answer)

6. Consider a patient who excretes one liter of urine, pH 5.6 in 24hrs. The principal buffer system in urine is phosphate, the total con-centration of which for the patient is 46.6 mM. This means that theconcentration of each of the four phosphate species (see above) addedtogether amount to 46.6 mM. Each of these ionic species will beassociated with its equivalent amount of the counterion Na+. Thekidney, by taking the phosphate from the blood at pH 7.4 andexcreting it at pH 5.6 achieves a significant conservation of Na+.

a. If the 46.6 mM phosphate was at pH 7.4 identify the majorconjugate acid and conjugate base species and calculate theiractual concentrations.

b. What would be the sum total concentration (mM) of Na+

needed to act as a counterion for the conjugate acid and basespecies?

c. When the 46.6 mM phosphate becomes a liter of urine at pH5.6, different proportions of conjugate acid and base exist. Cal-culate the actual concentrations of the acid and base species inthe liter of urine.

d. By the process of excreting the urine at pH 5.6 instead of 7.4the body has saved (retained) how much sodium per liter?(answer)

7. A lab report which you have just received is partially obliterated.You are able to make out the following data.

PCO2 (H2CO3 + CO2 dissolved) = 65mm Hg or



1.95 meq/L. Total CO2 = 35 meq/L. Knowing that with this buffer at pH 7.4 one has a conjugate base/conjugate acid ratio of 20/1, what is the pH of the patient's blood?(answer)

8. Use the following information for the next series of questions:

The pK for CO2/HCO3- system is 6.1; and 0.030 meq/L CO2 = 1

mm Hg. Remember, total [CO2] is assumed to mean the concentra-tion of dissolved CO2 plus the concentration of H2CO3 .

The blood of an individual who was breathing deeply and rapidly,(hyperventilating) was found to have a pH of 7.6 and a PCO2 of 20.7mm Hg.

a.Calculate the [HCO3-]/[CO2 ] ratio in the blood of this indi-

vidual.

b.Calculate the total [CO2 ] in the blood in meq/L.

c.The hyperventilation has produced a condition of respiratoryalkalosis. If the total CO2 concentration does not change as theindividual resumes normal breathing, what would be the PCO2in mm Hg when the blood returns to a pH of 7.4? (answer)

Answers-11.a) 0.001 M = 10-3 M HCl. HCl, hydrochloric acid, is a strong

acid therefore the [H+] = 10-3 M

b) 10-14 = [H+] [OH-] = Kw

Thus [OH-] = 10-11 M

c) pH = -log [H+] = 3

d) pOH = -log [OH-] = 11

What is the pH of 0.02 M HCl? 10-1 M HNO3? and 5 x 10-4 NHCl? (Ans.: 1, 7, 1.0, 3.3)

2. [KOH] = 0.004M The [OH-] concentration will be 0.004M for astrong base such as KOH which completely dissociates in water. Kw

= 10-14 = [H+] [OH-] which is a property of water. Thus, the [H+] is



2.5x10-12M and the pH is 11.6. What is the pOH of this solution?(2.4)

3.

Ka = 1.6 x 10-6

let X = [H+] thus

X = [A-]

[HA] = 10-3M since very little will dissociate if the pKa is

1.6x10-6.

; X=4 x 10-5 = [H+]

a. pH = 4.4

b. Degree of ionization is the concentration of [A-] divided by the

total acid concentration [A-]+[HA] times 100.

[A-]/total acid x 100.= degree of ionization

4x10-5/10-3 x 100 = (4x10-2) (100) = 4%

c. pKa = -log Ka

= - log 1.6x10-6

= - (-5.796)= 5.796

4.a. At pH = 7.4 , [HCO3-] = 24 meq and [H2CO3] (really dis-

solved CO2) = 1.2 meq.

[HCO3-] + [H2CO3] = 25.2 meq

If we add acid, we’ll convert some HCO3- to H2CO3 but the total

amount of these two components will not change. The question ishow many meq of acid must be added to drop the pH to 6.8?

6.8 = 6.1 + log([HCO3- - X] / [H2CO3 + X])

0.7 = log([HCO3- - X] / [H2CO3 + X])

5.01 = ([24 - X] / [1.2 + X]) Solving for: X we get 3.0 meq of acid added.

b. pH = 6.1 + log((24 + 29) / 1.2)

HA H+

A-

+=

KaH

+[ ] A-[ ]

HA[ ]-----------------------=

1.6 106× X2

10 3–----------=



thus pH = 6.1 + 1.68 = 7.78

5. At pH = 7.4, only the H2PO4- and HPO4

2- species will bepresent at appreciable quantities. You can solve for them using theH/H equation.

a. Of these two species, HPO42- will be most abundant because the

pH is above the pKa

b.

7.4 = 6.7 + log([ HPO42- ] / [ H2PO4

- ])

0.7 = log([ HPO42- ] / [ H2PO4

- ])

and ([ HPO42- ] / [ H2PO4

- ]) = 5.01

[HPO42- ] + [ H2PO4

- ] = 100% Thus we have two unknowns and two equations.

Solving for [ H2PO4- ] :

5.01[ H2PO4- ] + [ H2PO4

- ] = 100%

6.01[ H2PO4- ] = 100%

H2PO4- = 16.6% and HPO4

2- = 83.4%

6. [HPO42- ] + [ H2PO4

- ] = 46.6mM Using the Henderson-Hasselbalch equation:

7.4 = 6.7 + log([ HPO42- ] / [ H2PO4

- ])

and 0.7 = log([ HPO42- ] / [ H2PO4

- ])

5.01 = ([ HPO42- ] / [ H2PO4

- ])

and 5.01 [ H2PO4- ] = [ HPO4

2- ]Substituting into the first equation for this problem:

5.01 [ H2PO4- ] + [ H2PO4

- ] = 46.6mM

6.01 [ H2PO4- ] = 46.6 mM

[ H2PO4- ] = 7.8 mM and [ HPO4

2- ] = 38.8 mM

b.For each H2PO4- one Na+ is released and for each HPO4

2- two

Na+ are needed. Thus, 85.4 mM Na+ excreted at pH 7.4. c. Repeat the calculation as in a. except that the pH = 5.6



H2PO4- = 43.2 mM

HPO42- = 3.4 mM

Na+ excreted = 50 mMd.Calculate as in b.Na+ saved = 35.4 mM

7.a. 7.4 = pKa + log(20 / 1) , pKa = 7.4 - log 20 = 7.4 - 1.3 = 6.1b.pH = 6.1 + log(33.05 / 1.95) = 6.1 + 1.25 = 7.33

8.a. 7.6 = 6.1 + log([HCO3-] / [CO2]) ,

1.5 = log([HCO3-] / [CO2]) and ([HCO3

-] / [CO2]) = 31.62

b.pCO2 = 20.7 Thus [CO2] = (20.7) x 0.030 = 0.621meq/L.

Therefore, [HCO3-] = 31.62 * [CO2] = 19.64.

The total [CO2] = [HCO3-] + [CO2] = 19.64 + 0.621 = 20.26

meq/L.

c.The [HCO3-] can be calculated from the ratio and [CO2] since the

total remains constant.

7.4 = 6.1 + log ([HCO3-] / [CO2])

1.3 = log([HCO3-] / [CO2])

([HCO3-] / [CO2]) = 19.95

and [HCO3-] = 19.95[CO2]

Substituting in the equation above for total CO2.

19.95[CO2] + [CO2] = 20.26meq20.95[CO2] = 20.26meq [CO2] = 0.967 meq

or in mm Hg:pCO2 = 0.967/0.030 = 32 mm Hg

Answers to acid and base questions in text (pages 3 - 5)Aspirin is a weak acid.Procaine is a weak base with two basic groups.Tyrosine is both a weak acid and weak base, with two acidic

groups and one basic group.



s
Module 2: Amino Acids, Peptideand Proteins
ESSENTIAL CONCEPTS: 1.Proteins are composed of amino acids connected in a linear

sequence via peptide bonds.

2.Amino acids form zwitterions and can behave as acids and bases.

3.The characteristics of amino acids determine the properties ofpolypeptides.

4.Individual proteins can be isolated from mixtures containing otherproteins for analysis of their structure and function.

5.The primary sequence of a protein is encoded in the DNA anddetermines the final three-dimensional form adapted by the proteinin its native state.

6.The amino acid sequence of a protein determines its shape andconformation which are critical for its function.

7.The amino acid sequence of a protein can be determined andsequence information has been used to elucidate the molecular basisof biological activity, to determine the cause of abnormal function ordisease and to trace molecular events in evolution.

8. At this point we are not able to predict the structure of a proteinfrom its amino acid sequence with any confidence. The ability to dothis is key to using molecular biology in a rational manner in medi-cine. The area of protein design and engineering is one of the fron-tiers in modern molecular biology.

OBJECTIVES: The purpose of this problem unit is to provide you with a basicunderstanding of proteins including how they are isolated, purified,and sequenced. It is a foundation upon which a great deal of bio-chemistry and cellular and molecular biology has been built.



These objectives are designed to serve as guidelines for studying thismaterial using the learning resources.

1. Give an example of a protein involved in: a. enzymatic catalysisb. transport and storagec. coordinated motiond. mechanical supporte. immune protectionf. generation and transmission of nerve impulsesg. control of growth and differentiation.

2. Recognize the structure and three-letter abbreviation of each of the20 common protein amino acids and categorize the amino acids asnon-polar, polar, uncharged polar (at pH 7.0), sulfur containing, aro-matic, aliphatic, acidic, and basic.

3.a. Using a specific amino acid as an example, describe the proper-ties it exhibits that are shared by all amino acids, e.g., α-amino group,α-carboxyl group, side chain and ionic form.

b. Be able to recognize an amino acid from its structure.

4. Define pKa and pI. Estimate the pKa of the α-amino and α-car-boxyl group of an amino acid. Which amino acid side chains canionize in proteins, and what is the approximate pKa for these ioniz-able groups? When given the pKa's for an amino acid or peptide, cal-culate the pI. Relate pI to electrophoretic behavior of an amino acid,peptide or protein.

5.a. Given the structure of an amino acid side chain which can existin protonated and unprotonated forms, draw both forms.

b. Given the pKa's for the functional groups in an amino acid, showthe structure of the ionic form that would be most abundant at pH =pKa, at pH = pKa + 2 pH units, and at pH = pKa - 2 pH units.

c. What would be predominant ionic form of a particular amino acidat pH = 7? What charge would the amino acid have at, for example,pH = 7, 5, 9, or pI? What ionic species would predominate at pH =pI for amino acids such as His, Glu, Arg, Lys, Gly etc.?

d. Given the pKa's for an amino acid, draw a titration curve for theamino acid. Estimate the net charge on the amino acid at any point



on the curve. Show the points on the curve that correspond to thepKa's and pI. From the information in your titration curve, predictthe direction of migration of the amino acid in an electrophoreticfield at any pH.

6. Given the structure of an organic molecule (i.e., an amino acid ora drug) and the pH, predict if it will be water soluble (hydrophilic),lipid soluble (hydrophobic), positively charged, or negativelycharged.

7. Which of the 20 common amino acids is not an α-amino acid? Afew amino acids are not coded for in DNA but are derived from oneor another of the 20 fundamental amino acids after they have beenincorporated into the protein chain. This process is called post-trans-lational modification. Some of these derived amino acids are hydrox-yproline, 5-hydroxylysine, 6-N-methyl lysine, 3-methyl histidine,gamma-carboxyglutamate, and desmosine. Give examples of proteinscontaining each of these derived amino acids and describe the specialfunctions that they play in these proteins.

8. Draw a peptide bond and describe its relevant three-dimensionalfeatures (i.e. planar atoms, Φ and Ψ angles, bonds with freedom ofrotation, bonds without freedom of rotation.)

9. Give examples of techniques for purifying proteins which frac-tionate on the basis of size, solubility, charge and specific bindingaffinity. Understand the molecular basis for each of these separationtechniques.

10. Both direct protein sequencing using the Edman degradationprocedure and indirect protein sequencing using recombinant DNAtechniques have been used to determine the primary sequence ofmany proteins. In fact, more proteins have been sequenced usingrecombinant DNA techniques than by direct protein sequencing.Describe why it is important to determine the primary sequence of aprotein.

11. Describe the process of protein folding. What forces areinvolved in driving the folding process. How does folding differ invivo from that for the purified protein in vitro?

12.Recognize the terms in the NOMENCLATURE and VOCABU-LARY list and be able to use them properly in answering ques-tions.Be able to answer questions such as those in the Problem Sets and thePractice Exam.



NOMENCLATURE and VOCABULARY:

Affinity chromatographyAliphatic side chainAmino acidAmino acid compositionAmino acid sequenceα−amino groupAmphoteric moleculesAnionAnodeAromatic side chainβ-mercaptoethanolC-terminus (carboxy terminus)Carboxyl groupCathodeCationChaperoninChymotrypsinCNBrDenaturantDenaturedDialysisDipolar ions (zwitterions)Disulfide bondDithiothreitolDnaJDnaKEdman degradationElectrophoresisElectrophoretic mobilityGel filtrationGroELGroESHsp70HydrophilicHydrophobicImino acid Ion-exchange chromatographyIsoelectric focusingIsoelectric pointLigandMacromolecular crowdingMolecular chaperoneNative foldN-terminus (amino terminus)Peptide bondpI



pKaPolarPolyacrylamide gelResidueSalting inSalting outSodium Dodecyl Sulfate (SDS)Side chainTrypsinUrea Zwitterion

STUDY GUIDE-2I. Introduction Proteins are one of the essential macromolecular components of liv-

ing systems. They are typically large molecules composed of a hun-dred or more amino acids (residues) arranged in a linear sequence,i.e. they are polymers of amino acids. There are 20 naturally occur-ring amino acids. A typical one, alanine is shown below. All, exceptproline (which is actually an imino acid), contain a basic aminogroup and an acidic carboxyl group and are therefore amphoteric.They can contain both positively and negatively charged groups, i.e.they can be zwitterions. The form expected for alanine at acid pH isshown here. This is an α-amino acid in that the amino group is anα-amino group attached to the central α-carbon. The amino acids

differ from each other in the side chain attached to the central alphacarbon. In alanine, the side chain is a methyl group. A generic sidechain is sometimes symbolized with an R. Since the four groupsattached to the alpha carbon are different, the alpha carbon is anasymmetric center. The diagram of alanine shown above is a cartoonthat does not accurately represent the stereochemistry. All naturallyoccurring amino acids are in the L configuration. Sighting along thealpha carbon - alpha proton bond, reading clockwise the CO - R - Ngroups spell one of the major crops of Illinois (thus the CORN crib)

Cα

H

C

O

OH

H3N

CH3

+



and an accurate depiction of the L configuration is shown here.

Switching the position of the R and amino groups gives the D iso-mer, which is not observed in nature. Why the L isomer was chosenis one of the great enigmas of nature and will probably never beunderstood.

There are two commonly used abbreviations for the amino acids -three letter abbreviations such as Ala for alanine, and single lettercodes such as A. A complete listing of the symbols can be found inany biochemistry text.

Linkage of the amino acids occurs through dehydration (release of aH2O molecule) on the ribosome leading to an amide or peptidebond or linkage. For example, a tripeptide composed of alanine,phenylalanine, and glycine is shown below (in the form expected atneutral pH).

The arrows indicate the amide bonds. The peptide bond is notstrictly a single bond due to delocalization of electrons from the C=Oto the lone pair of the amide nitrogen. In fact it is about 40% double

H

COOHR

NH2

N C C N C C N C C

O

O

OH H

H H

H H

H

CH3

CH2

O-

H



bond. This partial double bond leads to restricted rotation. Theamide bond is therefore planar and typically it is found in the transconfiguration as shown above. Essentially free rotation is allowed forthe backbone bonds before and following the alpha carbon, and theseare referred to as the phi ( Φ) and psi (Ψ) bonds.

Note that the linear linkage leaves a free amino or N- terminus and afree carboxy or C-terminus.

The amino and carboxy termini are ionizable along with some of theside chains. Approximate pKa’s of the amine, carboxyl, and sidechains of various amino acids are shown below. In peptides and pro-teins, the amino terminus has a pK of about 9.5 and the C-terminusabout 2.2.

II. Modified Amino Acids

Some amino acids are not expressed by the DNA code directly, butare derived from one of the 20 natural amino acids after synthesis ofthe protein, i.e. they are formed post-translationally. The crosslink-ing of two adjacent cysteines to form a disulfide bond is the mostcommon post-translational modification. The disulfide bond is easilybroken by reducing it with dithiothreitol or mercaptoethanol.Other post-translationally modified amino acids include 4-hydrox-yproline, 5-hydroxylysine, ε-N-methyl lysine, 3-methyl histidine, γ-carboxyglutamic acid, and desmosine. Structures of these can befound in most biochemistry texts and will not be reproduced here.Other post-translationally modified amino acids include phospho-serine, phosphothreonine, and phosphotyrosine. These last threemodifications are reversible, and are often used for control purposes,

Table 1: pK a’s of amino acids

Amino acid carboxyl amino side chain

glutamate 2.1 9.5 4.1

aspartate 2.0 9.9 3.9

lysine 2.2 9.l 10.5

histidine 1.8 9.3 6.5

arginine 1.8 9.0 12.5

cysteine 1.9 10.7 8.4

threonine 2.1 9.1 13

tyrosine 2.2 9.2 10.5



i.e. activating or deactivating an enzyme or signal protein.

III. Protein Folding Each amino acid in a protein is linked to the next in a defined man-ner specified by the sequence of three base codons within a specificgene in the DNA of the chromosome (see http://www.ncbi.nlm.nih.gov/SCIENCE96/). In a sense, the DNA codedefines the structure of a protein through its sequence. The sequencedefines how the protein will fold. The amino acid sequence, i.e. theprimary structure of the protein, defines both the structure and thefunction of the protein. A protein, in general, has a specific uniquestructure with a defined role. The protein nearly always folds to thesame structure (amyloid and prion proteins are exceptions; see Mod-ule III). For example, the sequence for cro protein always folds to thestructure shown below. These structures are complicated but are nor-mally composed of well defined motifs. The more common motifswill be discussed in Module 3.

At this time, the native fold or structure of a protein cannot be pre-dicted with confidence from its sequence. The protein folding prob-lem is one of the most difficult challenges facing biochemists today.If we are to use sequence information in a truly rational manner andengineer proteins to do specific tasks, it will be necessary to be able topredict with high accuracy the structure of a protein from itssequence. At the present time, sequence information can beextremely useful in identifying specific functional motifs. Sequenceanalysis can often go a long way to identifying an unknown protein’sfunction (see http://www.nlm.nih.gov/databases/data-bases.html). In some cases when the sequence is quite similar to aprotein of known structure, the structure of the unknown protein can


http://www.ncbi.nlm.nih.gov/SCIENCE96/

http://www.ncbi.nlm.nih.gov/SCIENCE96/

http://www.nlm.nih.gov/databases/databases.html

http://www.nlm.nih.gov/databases/databases.html


be predicted with a high level of confidence.

The native fold of a protein is often dictated in part by the need forhydrophobic (aliphatic and aromatic) side chains to move out of thewater environment. Thus, hydrophobic residues will often form theinterior of a protein, and hydrophilic residues will coat the outside.In addition, the fold is dictated by the need to at least maintain thesame number of hydrogen bonds with the amide NH’s and carboxyloxygens. Moving the hydrophobic side chains into an oily interiorwill force segments of the protein backbone that these are attached tointo the interior as well. The only way to maintain hydrogen bondswith these segments of the backbone is to form secondary structuressuch as α-helix and β-sheet.

Proteins can be unfolded. This is often referred to as denaturation,and the unfolded protein is denatured. Clearly, when a protein isbeing synthesized within the cell, it is initially denatured and mustfold after release from the ribosome. All proteins must be able to foldfrom a random chain. However, denaturation is not always revers-ible. For example, a fried egg is denatured. Denaturation can beinduced by heat or by the addition of denaturants, i.e. compoundswhich weaken the forces that normally stabilize the folded protein.Denaturants include β-mercaptoethanol which breaks disulfidebonds, urea which preferentially binds to the unfolded protein, andsodium dodecyl sulfate (SDS) which binds to hydrophobic sidechains. Removal of denaturants often leads to renaturation, orrefolding.

IV. Folding in vivo Protein folding in vivo is complicated by two factors: the large num-ber of macromolecular components leads to crowding in the cell, andsequential synthesis of the polypeptide chain on the ribosome leads toexposure of hydrophobic residues that cannot collapse into a properlyfolded structure due to either sequestering of the rest of the sequenceon the ribosome or the absence of necessary domains due to incom-plete synthesis. The macromolecular concentration within the cell ison the order of 340 gm/liter. This highly crowded or restricted envi-ronment results in an effective decrease in the amount of water avail-able, and therefore an effective increase in the local concentration ofall components that may be many orders of magnitude greater thanexpected given their actual concentration in grams/liter. Macromo-lecular crowding leads to a greater chance for hydrophobic patchesand domains to collide and coalesce than would occur in a test tubeat the same concentration. Thus, the high effective concentrationleads to non-productive aggregation and kinetically trapped mis-folded polypeptide chains. The problem with misfolding polypep-



tides as they are being synthesized on the ribosome is similar. Thehydrophobic patches on the protein chain can collapse into misfoldedstructures during synthesis and prior to release due to the absence ofthe full chain. There is little evidence that proper folding can occuron the ribosome.

A complicated molecular chaperone machinery has evolved thatassists protein folding in vivo. At the present time there appears to betwo ubiquitous systems that are utilized: the Hsp70 system and thechaperonin system. Both function in most cells including bacteriaand eukaryotes. These were originally thought to be associated withstress or heat shock, thus the name Hsp (heat shock protein). It isnow clear that they are essential for the proper folding of most, if notall, proteins within the cell.

The Hsp70 and chaperonin systems do not contain specific informa-tion to direct the folding process. Rather, they sequester theunfolded chains to decrease their effective concentrations and preventthe unfolded chain from aggregating with other hydrophobic chainsor misfolding.

Much of our understanding of the cellular protein folding machinerycomes from the E. coli proteins. The Hsp70 analogues in E. coli areDnaK and DnaJ that function to maintain an unfolded protein in asoluble, monomeric state. The chaperonin system in E. coli is theGroEL/GroES system which forms a cavity into which the unfoldedchains can be inserted. GroEL proteins (MW 57,000) form twostacked seven membered rings approximately 140 Å across and 150Å high with an internal cavity about 50Å in diameter. (See http://bioc09.uthscsa.edu/~seale/Chap/struc.html.) The lining of thecavity contains hydrophobic patches that assist in binding theunfolded chain. It now appears that folding of the polypeptide canproceed within the cavity. In a sense, the GroEL/GroES complexprovides a “cage” within which the protein is free to fold withoutinterference. Both the Hsp70 and chaperonin systems containATPase activities. The purpose of the ATP hydrolysis is to modulatethe affinities of the systems for hydrophobic residues, thus providinga timing mechanism for binding and release. The systems cyclethrough high and low affinity states as the ATP is bound, hydrolyzedto ADP and phosphate, and then the ADP is released. If thepolypeptide has not folded within this time window prior to openingof the cage and release, it is free to bind again. Thus, the chaperoninsdo not direct the folding process, they simply provide a temporaryhiding place.


http://bioc09.uthscsa.edu/~seale/Chap/struc.html


V. Protein Purification Characterization of a new or unknown protein often begins with

sequencing. In some cases amino acid composition (i.e. the relativeamounts of the various amino acids) may be of interest, but in gen-eral the sequence and if possible the structure are necessary. Biologicalsamples, e.g. cytoplasm, usually contain hundreds if not thousands,of proteins. Thus the protein of interest must be isolated and puri-fied.

Purification and separation of a protein from other proteins, or fromsmaller molecules, is achieved by applying a combination of severalmethods. These methods take advantage of the specific properties ofthe protein such as solubility, molecular size, molecular charge, orbinding of the protein to a specific substance.

Some proteins require inorganic ions for water-solubility, and addi-tion of these in low concentrations (e.g. less than 1 M) can often leadto salting-in of the protein (i.e. an increase in solubility). Furtherincreases in salts, e.g. ammonium sulfate, can lead to loss of solubilityand precipitation, called salting-out. The concentration of saltrequired to precipitate a protein varies with each protein. Thus, acrude mixture can be initially fractionated by progressively increasingthe ammonium sulfate concentration in stages (e.g. 10%, 20%, 30%ammonium sulfate), and collecting the precipitate after each stage bycentrifugation.

Small molecules and ions can be removed from protein solutions bydialysis through a semipermeable membrane. Dialysis membranescan be purchased with pore sizes that will permit molecules withmolecular weights less than, for example, 1000, 3500, or 12000 topass freely into or out of the bag while retaining molecules larger thanthe molecular weight cut off. The protein solution is put into thedialysis bag, the ends sealed with clamps, and the bag is immersed inthe desired buffer. The pores allow water and small molecules to passthrough, but retain the protein molecules. The buffer on the outsideof the bag freely moves in, and the buffer on the inside moves out andbecomes diluted.

Proteins are most commonly purified using various column chroma-tography methods. The protein solution is passed through a glasscolumn containing the chromatographic medium of choice.

Gel filtration chromatography (molecular exclusion chromatogra-phy or molecular sieving) uses a column of insoluble, but highlyhydrated, polymers such as Sephadex, agarose or polyacrylamide.These materials are made in the form small porous beads. Small mol-



ecules can enter the pores but larger molecules cannot. Therefore,the volume of solvent available (the distribution volume) for thesmall molecules is greater than for the larger molecules. Thus, thesmaller molecules flow through the column more slowly and a mix-ture can be separated by size.

Gel filtration can be used to estimate the molecular weight of a pro-tein. In this case the gel column is standardized with proteins ofknown molecular weights. The shape of a molecule influences its dis-tribution volume and therefore its rate of passage through the col-umn. In reality these columns separate proteins based on theiraverage radii (i.e. Stokes' radii), not their molecular weights. There-fore, long fibrous proteins elute from the gel filtration column earlierthan would be expected based on their actual molecular weights.These columns are therefore most suitable for estimating the molecu-lar weights of globular proteins. Ion-exchange chromatography separates proteins and other mole-cules by charge. Both cation and anion exchange resins are availablefor protein purification. For example, a cation exchange column ofinsoluble ion-exchange material carrying carboxy methyl groups,(carboxylate groups) such as carboxy methyl cellulose (CM cellulose)can be used. At neutral pH, these groups are negatively charged andwill bind protein molecules carrying a net positive charge (or con-taining regions on their surfaces that have a net positive charge.) Thebound proteins are retained on the column material, or retarded inflow rate. They can be eluted from the exchanger by washing with asolution containing positive ions, e.g., Na+ salts, which will exchangeplaces with the positively charged protein bound to the carboxylategroups. Phosphocellulose is another type of cation exchange mate-rial. Probably the most often used ion exchange material for proteinpurification is DEAE (diethylaminoethane) linked to either cellulose(DEAE Cellulose) or Sephadex (DEAE-Sephadex). These ionexchange materials are positively charged at neutral pH and bind neg-atively charged groups on the protein's surface. The proteins areeluted from the columns by increasing the concentration of negativeions such as Cl- or by changing the pH. (Would you raise or lowerthe pH to elute proteins from an anion exchange column?)

Cation exchange column chromatography (containing small sul-fonated polystyrene beads) is used in the automated analysis of aminoacid mixtures. These mixtures can be obtained either from proteinsby acid hydrolysis (6 N, HCl, 24 hrs, 110°C), or from body fluidssuch as urine or plasma (existing as free amino acids).

Affinity (adsorption) chromatography is based on the property



that some proteins bind strongly to other molecules (called ligands)by specific, non-covalent bonding. The ligand is covalentlyattached to the surface of large, hydrated particles of a porous mate-rial such as cellulose, Sephadex beads, agarose particles, polyacryla-mide particles or porous glass beads. These are then used to make achromatographic column. If a solution containing several proteins ispoured down the column, the protein to be selectively adsorbed willbind tightly to the ligand molecules, whereas, the other proteins willpass through the column unhindered. After traces of the other pro-teins are washed off the column, the adsorbed protein is eluted byadding a strong solution of pure ligand. The unbound ligand com-petes for the protein with the ligand that was attached to the columnsupport material.

Antibodies to a specific protein can often be prepared (after the pro-tein has been purified once) and can be used to purify the desiredprotein from mixtures of proteins (such as a tissue extract or bodyfluid). The interaction of protein and antibody may produce anantigen-antibody complex large enough to be centrifuged out of solu-tion, allowing recovery of the protein. However, it is often necessaryto create a larger complex by first adding rabbit anti-gamma globulin(anti-IgG) to the antibody-protein mixture and then recovering thetriple complex.

The antibody can be linked to a column support material to make avery specific affinity column for purifying individual proteins fromcomplex protein mixtures. The proteins that are bound by the anti-body can be eluted by changing the ionic conditions.

Many of the techniques used in the purification of proteins havefound wide usage in clinical laboratory practice including automatedkits that can be easily used by technicians with little training. Plasmaprotein patterns, for example, are routinely examined by gel electro-phoresis, and a wide range of affinity binding assays (including radio-immunoassays) for hormones and drugs make use of the specificbinding of one substance to another.

VI. Demonstration of purity

The most common method of documenting the purity of a proteinpreparation is electrophoresis. In an electrical field, proteinsmigrate in a direction determined by the net charge on the molecule.(See http://www.rit.edu/~pac8612/electro/E_Sim.html for aninteresting demonstration). The net charge on a protein is deter-mined by the nature of the ionizing groups on the protein and theprevailing pH. For each protein there is a pH, called the isoelectricpoint (pl), at which the molecule has no net charge and will not


http://www.rit.edu/~pac8612/electro/E_Sim.html


move in an electrical field. At pH values more acid than the pI, theprotein will bear a net positive charge and, behaving as a cation, willmove toward the negatively charged pole (the cathode). At pH valuesabove the pI, the protein will have a net negative charge and willbehave as an anion, moving toward the positively charged pole (theanode).

Zone electrophoresis utilizes paper, starch, or gel blocks saturatedwith buffer to separate proteins with different electrophoretic mobili-ties. This type of electrophoresis is often used to fractionate plasmaproteins for diagnostic purposes.

Electrophoresis is most often done on a cross-linked polyacryla-mide gel (i.e. polyacrylamide gel electrophoresis, PAGE). PAGEcan be done without detergent (native protein separation) or with adetergent such as sodium dodecyl sulfate (SDS PAGE). The separa-tion of native proteins on PAGE is based on a combination of theircharge and their molecular weight. However, SDS PAGE separatesproteins only on the basis of their molecular weight. The SDS dena-tures the protein, thereby minimizing the effects of the protein'sshape on the molecular weight determination. The SDS also dissoci-ates quaternary structures into monomers. The subunits of proteinsstabilized by interchain disulfide bonds can also be separated if areducing agent such as 2-mercaptoethanol or dithiothreitol (Cle-land's reagent) is added with the SDS. Because the SDS forms nega-tively charged micellar particles with the protein, the effect ofprotein's own charge is lost. The SDS protein micelles migrate to thepositive pole of the electrophoresis chamber since they are coatedwith the negatively charged SDS molecules. The cross-linked poly-acrylamide acts as a molecular sieve. The electrophoretic mobility isdetermined by the size or molecular weight, and large polypeptidesremain near the origin, small polypeptides migrate farther into thegel).

Isoelectric focusing is a special form of electrophoresis that is espe-cially useful in analysis of proteins. In isoelectric focusing polyamino-polycarboxylic acids (amphoteric molecules) with known isoelectricpoints are used to establish a pH gradient in an electrical field. Aprotein will migrate to that part of the gradient where it has no effec-tive net charge, i.e. its isoelectric point or pI, and focus into a nar-row band. This technique is probably the most effective for resolvingproteins which have very similar pI values.



VII. Cleavage of peptide bonds The diagram below gives a summary of the cleavage sites of some fre-

quently used reagents and enzymes (i.e. proteases) that cleave peptidebonds. Cleavage can occur on either side of the amino acid with sidechain R2. Cleavage position 1 corresponds to the amino side, andcleavage position 2 corresponds to the carboxyl side. In general, fora given protease or reagent the identity of side chain R2 determineswhether or not cleavage will occur, and if so, which side the cleavagewill occur on. Cyanogen bromide (CNBr) cleaves on the carboxyside (position 2) of methionine residues only, i.e. it is methioninespecific. Trypsin cleaves on the carboxy side of positively chargedresidues, i.e. lysine and arginine. Chymotrypsin cleaves on the car-boxy side of aromatic residues tyrosine, phenylalanine, and tryp-tophan. Pepsin cleaves on either side of tyrosine, tryptophan,phenylalanine, and leucine. Both subtilisin and pronanse are nonspe-cific - i.e. the side chain has no effect on the site of attack. Carbox-ypeptidases hydrolyze the C-terminal amino acid.

VIII. Sequencing Protein sequencing is commonly done by a process known as Edmandegradation. The actual chemistry can be found in any biochemis-try textbook and will not be reproduced here. The important point isthat a reagent (phenylisothiocyanate) is used that reacts specificallywith the amino terminal residue. Treatment of the product with HClresults in release of the modified amino terminal residue leaving aprotein that is shorter by one amino acid. The modified amino acid,i.e. the phenylthiohydantoin derivative, can be identified so that theidentity of the original unmodified N-terminal amino acid can beknown. Repeating this process leads to successive identification ofthe amino acid sequence from the amino terminus one residue at atime. This process has been automated and is now performed bypeptide sequencers. The process is capable of sequencing a peptide of

N C C N C C N C C

O

O

OH H

H H

H H

R1

R2

R3

1 2



about 30 to 40 residues before the error rate becomes too great.Large proteins are therefore sequenced by sequencing fragments cre-ated using the specific cleavage reagents described above (sectionVII).

PROBLEM SET - 21. Which of the following amino acids alters polypeptide folding insuch a way that when it occurs in a peptide chain, it interrupts the α-helix and creates a rigid kink or bend?

a.Phe d.Trpb.Lys e.Hisc.Pro f.Cys (answer)

2. Which of the following amino acids has a lone electron pair at oneof the ring nitrogens which makes it a potential ligand important inbinding the iron atoms in hemoglobin?

a.Lys d.Prob.Tyr e.Hisc.Trp f.Arg (answer)

3. Which of the following amino acids plays a crucial role in stabiliz-ing the structure of many different proteins by virtue of the ability oftwo such residues on different (or the same) polypeptides to form acovalent linkage between their side chains?

a.Cys d.Lysb.Gly e.Metc.Tyr f.Glu (answer)

4. What is the minimum number of pKa values for a single aminoacid and for a peptide?

a.1,1 d.2,1b.1,2 e.2,3 (answer)c.2,2

5. Proteins can act as buffers. A buffer is a solution that resistschanges in pH when acid or base is added. The pH range over whicha buffer is effective is called the buffering range, usually defined aspKa + 1 to pKa - 1. Indicate the buffering range (or ranges) for theside chains of His, Asp, and Lys. (answer)

6. Which of the following amino acids have side chains that, whenthey are in a protein under normal physiological conditions (near pH7), are almost entirely positively charged?

a.Glu d.Trp



b.His e.Lysc.Arg f.Cys (answer)

7. Match the following reagents, often used in protein chemistry,with one or more of the given tasks for which it is best suited.Tasks:

a. reversible denaturation of a protein devoid of disulfide bondsb. hydrolysis of peptide bonds on the carboxyl side of aromaticresiduesc. cleavage of peptide bonds on the carboxyl side of methionined. hydrolysis of peptide bonds on the carboxyl side of lysine andarginine residuese. two reagents needed for reversible denaturation of a proteinwhich contains disulfide bonds

Reagents:1. CNBr2. urea3. 2-mercaptoethanol4. trypsin5. 6 N HCl6. chymotrypsin (answer)

8. Predict the direction of migration {i.e., stationary (0), towardcathode (C) or toward anode (A)} of the peptide Lys - Gly - Ala - Glyduring electrophoresis at pH 1.9, pH 3.0, pH 6.5, and pH 10.0: (answer)

9-15. Match the following proteins with their physiological functionbelow:

a. hemoglobinb. chymotrypsinc. acetylcholine receptor proteind. myosine. collagenf. gammaglobulinsg. nerve growth factor

9. catalysis (answer)10. transport and storage (answer)11. generation and transmission of nerve impulses (answer)12 .immunity (answer)13. coordinated action (answer)14. control of growth differentiation (answer)15. mechanical support (answer)

16-21 Match the following:



a. Lysb. Gluc. Leud. Cyse. Phef. Ser

16. nonpolar aliphatic (answer)17. nonpolar aromatic (answer)18. basic (answer)19. acidic (answer)20. sulfur containing (answer)21. hydroxyl containing (answer)

22. Which of the following are true?a. pI is defined as the pH where a molecule has no net chargeb. pKa of an ionizable group is defined as the pH where 1/2 ofthe groups are ionized and 1/2 are notc. at a pH equal to its pI, a protein will not move in the electricfield in an electrophoresis experimentd. the pKa of a charged group on a protein depends on thatgroups local environment (answer)

23-28. Match the following methods of purifying proteins withtheir corresponding molecular basis.

a. size separationb.chargec. specific bindingd. solubility (answer)

23. electrophoresis (answer)24. gel-filtration (answer)25. salting-out (answer)26. immunoprecipitation (answer)27. affinity chromatography (answer)28. isoelectric focusing (answer)

29. Which of the following might be considered important reasonsfor determining the amino acid sequence of a protein?

a. Knowledge of sequence helps elucidate the molecular basis ofbiological activity.b. Amino acid alteration may cause abnormal function and dis-ease.c. Amino acid sequence allows one to trace molecular events inevolution.d. Rules of folding of polypeptides into three-dimensional struc-tures may be deduced from amino acid sequences. (answer)



30. Which of the following are true concerning the peptide bond?a. The peptide bond is planar because of the partial double bondcharacter of the bond between the carboxyl carbon and nitrogen.b. There is relative freedom of rotation of the bond between thecarboxyl carbon and the nitrogen.c. The hydrogen bonded to the nitrogen atom is trans to theoxygen of the carboxyl carbon.d. There is no freedom of rotation in the bond between thealpha-carbon and the carboxyl carbon. (answer)

31. You have a mixture of proteins with the following properties:(Mr = molecular weight)

a.Mr = 12,000,pI = 10b.Mr = 62,000, pI = 4c.Mr = 28,000, pI = 7d.Mr = 9,000,pI = 5

Other factors aside, what order of emergence would you expect fromthese proteins to elute from:

a) an anion exchange resin such as DEAE-cellulose with a linearsalt gradient elution andb) a Sephadex G-50 gel exclusion column (answer)

32. Which of the following α-amino acids is a diamino-monocar-boxylic acid?

a. Leucineb. Lysinec. Glutamic acidd. Glycinee. Proline (answer)

33. The peptide bond has a "backbone" of atoms in which of the fol-lowing sequences?

a. C-N-N-Cb. C-C-C-Nc. C-C-N-Cd. N-C-C-Ce. C-O-C-N (answer)

34-35 Each question below contains four suggested answers of whichone or more is correct. Choose answer.

A. if 1,2, and 3 are correctB. if 1 and 3 are correctC. if 2 and 4 are correctD. if 4 is correctE. if 1,2,3, and 4 are correct



34. Amino acids found in proteins that are formed by post-transla-tional modification of one of the common amino acids include whichof the following?

1) Isoleucine2) Glutamic acid3) Threonine4) 4-Hydroxyproline (answer)

35. Separation of one protein from other proteins on the basis ofmolecular size can be achieved by:

1) electrophoresis on polyacrylamide gels containing sodiumdodecyl sulfate (SDS)2) affinity chromatography3) gel filtration (molecular exclusion chromatography)4) ion-exchange chromatography (answer)

36. Given the following tripeptide:

acetyl-lys-gln-his

(where acetyl- indicates acetylation of the N-terminus)

a. Construct a titration curve for the tripeptide. Label axesclearly. b. Determine the numerical value of the isoelectric point for thepeptide. (Show your work.) c. State the pH ranges over which the peptide would be consid-ered a good buffer. d. Identify each ionic form of the peptide which exists at pH7.4. (answer)

37. A drop of a solution containing a mixture of glycine (pKa's =2.34 and 9.6), alanine (pKa's = 2.34 and 9.69), glutamic acid (pKa's= 2.19, 9.67 and 4.25), lysine (pKa's = 2.18, 8.95 and 10.53) andhistidine (pKa's = 1.82, 9.17 and 6.0) was placed in the center of apaper strip and dried. The paper was moistened with a buffer of pH6.0 and an electric current was applied to the ends of the strip.

a. Which amino acid(s) moved toward the anode? (Rememberanions move toward anodes in electrophoresis chambers.)b. Which amino acid(s) moved toward the cathode?c. Which amino acid(s) remained at or near the origin? (answer)

Answers-21. c



2. e3. a4. c5. As calculated by the Henderson-Hasselbach equation, a bufferingrange of pKa ± 1 encompasses 82% of the total buffering capacity ofan ionizable group.

His 0.8 to 2.8 8.2 to 10.2 5.0 to 7.0Asp 1.1 to 3.1 8.8 to 10.8 2.9 to 4.9Lys 1.2 to 3.2 8.0 to 10.0 9.5 to 11.5

6. c, e7. a-2, b-6, c-1, d-4, e-2 and 38. a. moves toward the cathode at all pH values

9. b

10. a

11. c.

12. f

13. d

14. g

15. e

16. c

17. e

18. a

19. b

20. d

21. f

22. a,b,c,d

23. a,b



24. a

25. d

26. c

27. c

28. b

29. a,b,c,d

30. a,c

31.(a) a,c,d,b(b) b,c,a,d

32. b

33. c

34. c

35. b36.a. There are three ionizable groups in this peptide with pKa's of1.8, 6.0 and 10.8. Thus, it will require 3 equivalents of base (X axis)to titrate the proton from each of these functional groups. Plateauswill occur at pH ( Y axis) 1.8 (0.5 equivalent of base), 6.0 (1.5 equiv-alents), and 10.8 (2.5 equivalents) and inflection points will occur at1 and 2 equivalents of base. b. pI = 6.0 + 10.82 = pH 8.4 c. One pH unit on either side of each of the pKa's. d. For class discussion if needed.

37.a. Glu b. Lys and His c. Gly and Ala


Module 3: Structural Biology andDiseaseESSENTIAL CONCEPTS: 1. The amino acid sequence of a polypeptide determines its three
dimensional structure in solution.

2. Noncovalent interactions are primarily responsible for maintain-ing protein conformation.

3. Native proteins in aqueous solutions have most of their nonpolarside chains inside, and most of their polar side chains outside.

4. Proteins contain common recurring folding patterns.

5. Many proteins contain prosthetic groups.

6. An individual protein can contain one or more subunits.

7. An alteration in a single nucleotide base coding for an amino acidin a protein can alter the function or stability of that protein causingdisease.

8. The folded protein is a dynamic structure that can exist in differ-ent conformations with altered biological activity.

9. Many soluble, cellular proteins are globular - e.g. IgG antibodies,myoglobin, hemoglobin.

10. An IgG antibody molecule is composed of four polypeptidechains – two identical light chains, and two identical heavy chains -with two identical antigen-binding sites.

11. There are 5 different classes of H chains in antibody, each withdifferent biological activity.

12. Myeloma proteins are homogeneous antibodies made by plasma-cell tumors.

13. Collagen is the major protein of the extracellular matrix.



14. Collagen chains have an unusual amino acid composition andsequence.

15. The final functional form of a protein often involves post-trans-lational modifications of the protein. For example, after procollagenmolecules are secreted from fibroblasts, they are cleaved by specificproteases and then self-assemble into collagen fibrils.

16. Once formed, collagen fibrils are greatly strengthened by cova-lent cross-linking.

17. Elastin is a cross-linked, random-coil protein that gives tissuestheir elasticity.

18. Fibrous proteins can consist of twisted α-helices (e.g. fibrin), β-sheets (e.g., silk protein) or collagen triple helices. Filamentousstructures can also be assembled from globular protein subunits (e.g.,F-actin).

19. Proteins can have multiple conformations, some of which caninduce disease states, e.g. amyloidogenic proteins and prions.

20. Prion diseases appear to be the first example of a disease causedby the transmission of a misfolded protein. Infection does notrequire tranmission through genetic material.

OBJECTIVES: 1.Differentiate between the primary, secondary, tertiary and quater-nary structure of a protein.

2. Describe the types of noncovalent and covalent bonds that deter-mine primary, secondary, tertiary and quaternary structure of a pro-tein. Be able to identify examples of hydrogen bonds, ionic bonds,and hydrophobic interactions

3.Differentiate between the common recurring protein chain foldingpatterns: α-helix, antiparallel β-sheet, parallel β-sheet.

4.Compare the structure of myoglobin to that of hemoglobin and listfunctional differences between these two proteins.

5.Describe the symptoms of sickle-cell anemia and explain theirmolecular origin.

6.Define prosthetic group.



7.Some proteins have covalently attached carbohydrates. Describethe linkages that are known to couple carbohydrate to proteins. Glu-cose can attach to hemoglobin. What significance can you draw fromthe amount of glycosylated hemoglobin?

8.Describe what is meant by "protein denaturation." Recognize thatheat, vigorously shaking a protein solution, urea, guanidine hydro-chloride, high pH, low pH, and SDS can denature proteins. Is itpossible for a denatured protein to regain biological activity?

9.Using hemoglobin as an example, describe how conformationalchanges affect biological activity. What are "allosteric" proteins?

10.Outline the subunit polypeptide chain structure of the immuno-globin IgG.

11.Describe how proteins can be quantitatively measured and local-ized by highly specific antibodies.

12.Distinguish between fibrous and globular proteins with regard totheir solubilities and shapes. Compare and contrast the structuralorganization of the fibrous proteins: collagen, fibrin, keratin, silkprotein, and F-actin.

13.Describe the distinctive amino acid composition of collagen,name the most abundant amino acid in collagen and suggest reasonsfor its high frequency.

14.Explain the relationship between tropocollagen and collagen.

15.Describe the relationship of scurvy to the hydroxylation of col-lagen. What is the role of ascorbic acid (Vitamin C)?

16.Describe the relationship of procollagen to tropocollagen.Explain how defects in the conversion of procollagen to tropocol-lagen can lead to Ehlers-Danlos syndromes.

17.Describe the spatial relationship of tropocollagen to the collagenfiber. Explain how the collagen fibers are stabilized. Relate lathyrismto the cross-linking of collagen microfibrils. Explain how homocysti-nuria could affect this process.

18. Describe the differences and similarities of the molecular basis ofsickle cell anemia, amyloidosis, and prion disease.

19. Describe the molecular origin of the effects of aspirin, ibuprofen



and naproxen. Indicate the importance of isozymes in drug designstrategies involving cyclooxygenase.

20.Recognize the terms in the NOMENCLATURE and VOCABU-LARY list and use them properly when answering questions such asthose in the Problem Set, Practice Exam and at the end of the chap-ters in Stryer.

21.After reading a given passage from a primary resource, a medicaljournal or a textbook that describes the structure and function offibrous proteins, connective tissue biochemistry, the pathology ofconnective tissue, heritable disorders of connective tissue, the bio-chemistry of wound healing, or any of the principle molecular com-ponents of connective tissue, answer questions about the passage(which may involve the drawing of inferences or conclusions) or usethe information given to solve a problem.

NOMENCLATURE and VOCABULARY

Actin Allosteric protein Alpha helix AmyloidAmyloidogenicAntibody specificity Antibody Antigen Antiparallel sheetApoprotein Ascorbic acidBeta-bendBeta-sheet Collagen Constant region Disulfide bridges Drug designEhlers-Danlos syndrome Elastin EpitopeFibrinFibrinogenFibrous protein Globular protein Glycoprotein Glycosylation Heavy chain Heme



HemoglobinHemoglobin A Hemoglobin S Hemoglobinopathies Heterogeneity Hybridoma cellHydrogen bond Hydrophobic interactions Hydroxyproline IgA, IgD, IgE, IgG, IgM Ionic interactionsIsozymesImmunoglobulin KeratinLathyrism Light chain Monoclonal antibodiesMonomerMyeloma MyoglobinNuclear Magnetic Resonance (NMR)Oligomeric Parallel sheetPlasma proteins PrionsProtien engineeringPrPPolyclonal antibodiesPrimary structureProsthetic group Protomers Quaternary structure Ribonuclease ARandom coil Scurvy Secondary structureSickle cell anemiaStructural Biology Subunit Tertiary structure Triple helix Tropocollagen Three-dimensional structure Variable regionWestern blottingX-ray Crystallography



STUDY GUIDE-3I. Introduction Proteins serve many important functions in cells. They are involved

in basic metabolic catalysis (e.g. hexokinase in sugar metabolism),digestion (e.g. chymotrypsin), ion transport across membranes (e.g.Na+, K+ ATPase), motility (e.g. myosin, dynein), mechanical support(e.g. actin, tubulin), immune response (e.g. immunoglobulins), nerveimpulse generation (e.g. ion channels), and control and differentia-tion (growth hormone, adenylate cyclase).

II. Structural Biology The three dimensional structures of many proteins have been deter-mined in the last ten years or so and constitute the field of structuralbiology. These structures have been obtained using X-ray crystal-lography and nuclear magnetic resonance (NMR) spectroscopy. Inthis section we will use three dimensional structural information toexplain a few properties of proteins and the basis of some diseases.The explosion of structural information in the last decade along withthe promise of performing genetic engineering means that this typeof information will become much more prevalent in the future ofmedicine.

The molecular graphics files used here are contained in the ProblemUnit 1 Folder on the Biochemistry server (http://www.siu.edu/departments.biochem). These are “kinemage” files and must beviewed with the freeware program Kinemage that can be downloadedfrom the Biochemistry server also or from the Protein Science web set(http://prosci.org/Kinemage/). A brief guide to the use ofKinemage is provided as an Appendix to this study guide. Anotherfreeware molecular graphics program is also available called RasMolwhich can be used to view any of the protein (and other) structuralfiles located in the Brookhaven Protein Databank (http://pdb.pdb.bnl.gov).

III. SecondaryStructure As mentioned in Module 2, protein folding is initially driven by

hydrophobic collapse, i.e. the hydrophobic side chains coalesce intoan oily droplet to avoid interaction with water and enhance hydro-phobic interactions. Removal of the associated backbone fromwater leads to loss of hydrogen bonds between water and the amideprotons and carboxyl oxygens. These must be compensated for byforming hydrogen bonds between the amide protons and the car-boxyl oxygens. In fact, hydrogen bonding between the amide pro-


http://www.siu.edu/departments.biochem

http://www.siu.edu/departments.biochem

http://prosci.org/Kinemage/

http://pdb.pdb.bnl.gov

http://pdb.pdb.bnl.gov


tons and the carboxyl oxygens may be stronger than with watermolecules. Two of the most energy efficient ways to accomplish max-imization of internal hydrogen bonding is to form either the α-helixor the β-sheet secondary structures. Secondary structure is the nexthigher order structure above the linear sequence (or the primarystructure).

Open the file helix.kin with Kinemage. The image of an α-helixcan be moved with the mouse to help visualize it from various angles.The 3D effect in molecular graphics programs work to its fullestextent only if the molecule is moving. Oxygens are colored red,nitrogens blue, carbons green, and hydrogens yellow. Various displayoptions can be toggled off and on using the menu at the right byclicking in the associated boxes. For example, the hydrogen bondscan be displayed by clicking in the H-bond box. Note that all of thecarboxyl C=O bonds are aligned and point in the same direction:towards the C-terminus. The N-H’s point in the opposite directionand are oriented to make hydrogen bonds with the carboxyl oxygens4 residues away in the linear sequence. Thus, all NH’s and C=O’s areinvolved in H-bonds. The side chains are splayed outwards aroundthe helix. The helix has 3.6 residues per turn. It has a pitch of 5.4 Å(i.e. the rise per turn). All of the phi and psi angles are the same: phi= -57° and psi = 47°. In a Ramachandran plot the α-helix falls in atight region in the lower left quadrant. (The α in α-helix presumablycomes from α-keratin, a protein rich in α-helix.)

Next open the Kinemage file bsheet.kin. This displays a segment ofβ−sheet from an actual protein. (The β in β-sheet presumably comesfrom β-keratin, a protein rich in β-sheet.) The image shows only onestrand in the extended conformation when opened. Note that con-secutive oxygens are pointing in opposite directions. Click on thebox next to the “three chains” label to display the three-strandedsheet. The planar sheet is in the central portion of the image, withconnecting loops on the periphery. The adjacent strands are orientedto permit maximal hydrogen bonding of antiparallel strands. A turnin the backbone leading from one strand to the neighboring antipar-allel strand is a β-bend or β-turn. The phi and psi angles are approx-imately -139 and 135 degrees and the sheet falls in the upper leftquadrant of a Ramachandran plot. The hydrogen bonds can be dis-played in purple by clicking the H-bond box. Hydrogen bondingcan also be accomplished between parallel strands giving parallel β-sheet. Click on the side chains box to show that side chains in a β-sheet fall on the faces of the sheet and do not disrupt the H-bonding.

Secondary structure as well as higher order structures are stabilized bynot only hydrogen bonding, but also electrostatic interactions, van



der Waals forces, S-S (disulfide) cross-bridges, and hydrophobicinteractions.

IV. Tertiary Structure It is commonly found that the positions of hydrophobic and hydro-philic residues in helices and sheet are such that these structures canhave hydrophobic faces. This results in packing of the secondarystructures to give higher order structure that is referred to as tertiarystructure. For example, open the kinemage RNaseA.kin. This is thestructure of ribonuclease A (molecular weight about 14,000), a pro-tein which breaks down RNA. This is a classic globular proteinwhich has been extensively studied to obtain an understanding ofprotein folding. It is highly water soluble and exists in solution as amonomer. When you initially open the file, only the backbone is dis-played, i.e. only the Cα carbon are shown, linked together with“pseudobonds”. Place the mouse on the backbone near the edge ofthe molecule and click once. Move the mouse to the opposite side ofthe protein, and click again. The distance between these two points isgiven in Angstroms at the bottom of the window and should be onthe order of 40Å. The kinemage is set up so that the various second-ary structural elements making up the tertiary structure are coloreddifferently and can be toggled on and off with the menu at the right.The three helices are green, the three stranded sheet blue, and the twoso called β-ribbons are red. A β-ribbon is a two stranded sheet.Structure which cannot be classified in a common motif is referred toas random coil, although in the protein structure it may be rigid andmay not be a coil in layman terms. Toggle off the Cα backbone, andturn on the main chain along with mc. This displays all of the mainchain Cα, CO, and N atoms. Turn on the H-bonds to see the stabi-lizing bonds in the secondary structure elements. Next turn off theH-bonds (to simplify the view) and check sidechains, cys, SS balls,and ss. This displays the four disulfide bonds that are important incrosslinking the structure. Turn on the hydrophobic side chains suchas phenylalanine, valine, isoleucine, leucine, methionine and notethat the are located predominantly within the core. Turn on thecharged side chains such as lysine, arginine, aspartate, and glutamateand note their location. Can you locate any oppositely charged sidechains that might form stabilizing ionic interactions on the surface,e.g. adjacent lysines and aspartates?

Open the kinemage file myoglobin.kin. Myoglobin (MW 17,200) isthe protein that serves as a reservoir for oxygen in muscle tissue. It islargely composed of alpha helices. A cleft within the structure formsa binding pocket for the non-proteinaceous group essential for theproteins function. This is the prosthetic group heme with its associ-



ated iron which binds oxygen (shown in green). The protein withoutthe prosthetic group is referred to as the apoprotein, or in the case ofmyoglobin, apomyoglobin.

V. Isozymes Aspirin, ibuprofen, naproxen, and other nonsteroidal anti-inflamma-tory drugs (NSAIDS) are important in relieving pain as well as reduc-ing inflammation. They act by binding to cyclooxygenase (COX)and inhibiting its function in the conversion of arachidonic acid toprostaglandin H2, a precursor in the pathway to a number of prostag-landins. These play an important role in inflammation, pain, labor,and other physiological processes. In 1991 it was shown that thereare actually two forms, i.e. isozymes, of COX, given the very originalnames COX-1 and COX-2. COX-2 appears to be more importantin inflammation and pain, while COX-1 is associated with some ofthe undesirable side-effects of NSAIDS such as upset stomach, ulcers,and kidney failure. The structures of COX isozymes have recentlybeen determined, both in the uninhibited form, and with boundligands. The structures of the two isozymes are virtually identical, aswould be expected given the similar amino acid sequences. TheNSAID ligand binding site on COX-2 differs from that on COX-1 inthat valine replaces isoleucine at residue 523. The larger binding sitecavity on COX -2 has permitted the design of new drugs specific forCOX-2 in the last few years by drug companies such as Merck andCo., Roche Bioscience, and G.D. Searle. It is hoped that this willpermit the treatment of pain and inflammation without the adverseside effects associated with COX-1 inhibition. The success ofendeavors such as this has led to a very large investment in structuralbiology and rational drug design by even some of the smaller drugcompanies. In the US alone more than $2 billion dollars is spentevery year on NSAIDs. The design of a better product could havesignificant commercial as well as medical benefits. The huge poten-tial commercial benefit to be reaped from NSAID/COX structuralwork clearly explains the ongoing competition between numerousdrug research centers.

VI. Quaternary Structure

Open the file coiledcoil.kin. The coiled-coil is composed of twohelices with hydrophobic faces shown in orange. The coalescence ofthe hydrophobic faces leads to a wrapping of one helix around theother to form the coiled coil (see also). The hydrophilic residues (inblue) are located on the outside of the coiled coil and help to solubi-lize the large structure. The two helices are separate molecules. Theformation of higher order structure from multiple protein subunits(monomers or protomers) is referred to as quaternary structure.There are a number of advantages to forming higher order multiunit



oligomeric structures. Perhaps most importantly, it permits theintroduction of cooperativity between subunits and a new level ofcontrol through allostery that cannot be accomplished with mono-meric units. In allosteric proteins, if one unit switches to an acti-vated (or deactivated) state the others have a tendency to follow. Thecooperativity creates an all-or-nothing switch.

Another classic example of quaternary structure is the packing of fourhemoglobin chains to give the hemoglobin tetramer. This is the oxy-gen carrying protein of the red blood cell, and the cooperativityallows for maximal loading of oxygen in the lungs and efficientdumping in the peripheral tissues. Each member (monomer) of thetetramer is very similar to myoglobin in its overall fold. However,differences of surface residues leads to potential interactions that sta-bilize the formation of the tetramer (see hemoglobin.kin). Myoglo-bin lacks these residues and cannot form a tetramer. As we will seebelow, one additional surface residue change in hemoglobin leads tohigher order structures that a composed of polymers of hemoglobinin sickle cell anemia.

VII.Sickle cell anemia In 1904 in Chicago a black medical student was admitted to hospitalwith weakness, dizziness, headaches, shortness of breath, enlargedheart, kidney damage. He was found to be anemic with a 50%reduction in red blood cell count. Many of his red cells were “sick-led”, i.e. they were not the normal doughnut shape, but were elon-gated and curved to look like a sickle. The disease was labeled sicklecell anemia. (See http://www.emory.edu/PEDS/SICKLE/). Epide-miological and genetic studies showed that 9% of American blackswere carriers of the gene for sickle cell anemia. Four out of 1000were homozygous. The disease can be fatal before age 30 due toinfections, renal failure, and cardiac failure. The sickle shape of thered cells apparently leads to clogging of the capillaries, increased sick-ling, and catastrophic organ damage. Linus Pauling showed in 1949that the pI of hemoglobin isolated from sickle cells (i.e. hemoglobinS or HbS) was different from normal hemoglobin A (HbA). Sicklecell anemia therefore results from a defect in the hemoglobin mole-cule and is referred to as a hemoglobinopathy. A peptide map ofHbS was obtained by fragmenting the protein with trypsin to give 28peptides. The mixture was chromatographed on paper in a solventmixture of pyridine, acetic acid and water to partially separate thepeptides according to hydrophobicity. The paper was then turned 90degrees and an electric field applied to separate according to charge.A two dimensional pattern of 28 resolved dots was observed corre-sponding to the 28 peptides. This provides a “fingerprint” that ischaracteristic for the protein and is referred to as a peptide map. One


http://www.emory.edu/PEDS/SICKLE/


of the spots was found to be different in HbS from that observed inHbA. The peptide corresponding to this spot was isolated andsequenced by Ingram in 1954. The peptide in HbA had the sequence

Val - His - Leu - Thr - Pro - Glu - Glu - Lys

In HbS the sequence was

Val - His - Leu - Thr - Pro - Val - Glu - Lys

Thus the only difference between the normal and diseased states wasthe substitution of valine for glutamate at position 6 in the A chain!This single mutation has dramatic consequences. Not all mutationshave such a pronounced effect. Some are totally benign, and otherscan lead to a protein unfolding. This one leads to aggregation.

The substitution of valine for glutamate creates a hydrophobic patchon the surface of hemoglobin. Open the kinemage for hemoglobin.Turn on both the normal hemoglobin and also the sickle cell hemo-globin by clicking the boxes “Sickle” and “sub1”. Note that the twostructures are virtually identical. Click on the boxes for Glu6 or Val6to see the sickle cell substitution. Most importantly, it is known thatnormal deoxy hemoglobin naturally has a hydrophobic patch that is notpresent in the oxygenated form. The valine substitution creates a addi-tional new patch that can interact with the normal patch that is cre-ated when the hemoglobin becomes deoxygenated. Click on “sub2”to see the binding of one hemoglobin to another when “sickling”occurs (you may want to zoom out using the slide bar at the right).Note that Val6 in HbS fits very nicely into a pocket created by ala70,ala76, and leu88 in the adjacent tetramer. Thus aggregation andpolymerization are promoted by deoxygenation (i.e. low oxygen)leading to elongated polymers of hemoglobin S and stretching of thered cell into the sickle shape. However, polymerization is normallyslow, so that under most conditions the deoxygenated blood cell canget through the capillary bed without sickling occurring. However,should partial blockage occur, the sickling occurs rapidly and cata-strophically. Thus, even in heterozygous individuals, stress (exercise,pneumonia, etc.) can lead to sickling of some cells which can spreadin the capillary bed. The mutation appears to have evolved to “kill”red cells infected with the malaria parasite. The sickling is designedto be limited to the infected cell. The parasite competes for the oxy-gen, leading to decreased oxygen tension, sickling, and lysis of the cellthrough breakage. Thus sickle cell anemia presumably leads to deathof the cells infected with the parasite as a defense mechanism. Thedefense is particularly brutal since it largely sacrifices the homozygousindividuals. With the structure of hemoglobin known and the locus



of the lesion characterized, it is know hoped that a combination ofmolecular biology and pharmacological targets can be used to treatthe disease. One possibility is to design peptides that bind to theHbS hydrophobic patch, preventing by competition the polymeriza-tion reaction.

VIII. Immunoglobu-lins

Blood plasma proteins can be separated into a number of groups byzone electrophoresis using a Tiselius Cell. This is a classical tech-nique that is no longer used since the advent of acrylamide and othersolid bed techniques. However, it led to the current nomenclaturefor plasma proteins, since each of the bands in the Tiselius Cell werelabelled with Greek letters such that we now have α-globulins, β-globulins, and γ-globulins. The latter are the immunoglobulins,synthesized by lymphocytes. These are the antibodies elicited byantigens (foreign molecules). A given antigen elicits a heteroge-neous mixture of immunoglobulins, each made by a specific B-cell.The mixture is polyclonal. The major antibody class is IgG. IgM isthe initial antibody class elicited about 1 day after introduction of anantigen. IgG requires about 10 days. IgA is another class that iscommonly found in mucosal secretions and colostrum and milk.IgD and IgE are two other classes, the latter important in allergies.The classic antibody is the IgG immunoglobulin, containing fourprotein chains: two light chains and two heavy chains organized ina Y structure:

The heavy chains are linked with disulfide crosslinks, and both lightchains are linked to the heavy chains by disulfide crosslinks. Anti-bodies are also glycoproteins in that they contain carbohydrateattached at specific sites. The protein is said to be glycosylated. Dif-ferences in the H-chain define the classes of immunoglobulins.

Papain, a protease, can cleave the IgG to release the two individual“heads” composed of the upper portion of the H-chain and the asso-

heavy chain

light chain

variable regions

Fc

Fab



ciated L-chain. Pepsin can cleave the IgG tetramer to release the two-headed Fab fragment. Each of the H and L chains are composed ofrepeating elements of approximately 110 residues - the H-chain has 4units and the L-chain 2. Each unit is a domain composed of animmunoglobulin fold which is a seven stranded β-barrel. The termi-nal units of each of the H and L chains which make up the variableregions are referred to as VH and VL domains, respectively. Theother domains are constant domains referred to as CH and CLdomains. The antigen binding crevice is located at the ends of theheads defined by the variable regions of the heavy and light chains. Itis this region which varies from one antibody to another and definesthe specificity of each. This is the specific binding site for theepitope (the actual site on the antigen that elicited the immuneresponse). This is defined by the hypervariable loops joining thestrands of the sheet in the immunoglobulin folds of these domains.The antibodies demonstrate clearly the exquisite selectivity and diver-sity that can be achieved by proteins using only 20 amino acids.

Antibodies have become a very useful reagent in molecular biologyand clinical biochemistry. See, for example, the ELISA assay for HIVdescribed in Devlin (page 171). Much of this stems from the abilityto obtain large quantities of an antibody raised to interact with a spe-cific region of the antigen surface, i.e. the epitope. These are mono-clonal antibodies (see page 74 in Voet and Voet) and are producedby a hybridoma, a hybrid myeloma which has been created by fusinga spleen cell producing a specific antibody with an immortalmyeloma cell. The hybridoma cells can be raised in cell culture, orinjected into a rat to induce tumors producing monoclonal antibod-ies. These can then be used for various purposes, e.g. detecting spe-cific proteins in clinical diagnostic kits, probing acrylamide gels forspecific proteins (Western blotting, see page 94 in Voet and Voet), orfor affinity chromatography.

IX. Fibrous Proteins The proteins discussed above are largely globular, highly soluble pro-teins found in the cytoplasm. We now move to fibrous proteins.These include the structural proteins collagen, elastin, actin, silk,tubulin, fibrin, and keratin as well as the motile proteins such asmyosin and dynein. These are large oligomeric structures composedof many subunits. Polymerized hemoglobin S is an example of afibrous protein.

X. Fibrin Blood clots are composed of fibrin, an insoluble matrix of proteincomposed of subunits derived from fibrinogen. Fibrinogen is com-posed of six subunits - two each of Aα, Bβ, and γ − with a total



molecular weight of about 340,000. The A and B portions are pep-tides removed by the protease thrombin to create the fibrin monomerα2β2γ2. This spontaneously aggregates to give what is called a softclot. Crosslinking of the fibrin units by “fibrin stabilizing factor”(Factor XIII) leads to the final blood clot which is an open mesh ofcrosslinked fibrin strands. The crosslinking occurs between lysineand glutamine side chains as shown in the figure below.

Fibrinogen (Aα)2(Bβ2)(γ2)

charge repulsionprevents aggregation -

thrombin

A and B peptides

these ends are removed by thrombin

Fibrin monomer(reduced for clarityrelative to drawingabove)

“soft clot”

spontaneous polymerization



XI. Collagen Collagen is another structural protein that is important in maintain-ing the structure of skin, tendons, bone, cornea, cartilage, and bloodvessels. Similar to fibrin, it is initially expressed in a form which can-not aggregate known as procollagen (MW 300,000). Procollagen is atriple helix with globular heads at both the amino and carboxy ter-minals. Removal of the globular heads by amino and carboxyl pro-collagen peptidases leads to formation of tropocollagen, the triplehelical portion of procollagen. This rapidly aggregates and assemblesspontaneously into collagen. The collagen chains are quite large,composed of over 1000 residues each. Approximately 1/3 of the resi-dues are glycine and another third are either proline (Pro) or hydrox-yproline (Hyp). This highly unusual amino acid composition isessential for the structure of collagen. The glycines are arranged tooccur in a regular pattern of every third residue, e.g.

---- Gly - Pro - Hyp - Gly - Pro - Ile - Gly- Pro - Ala ----

This sequence folds into an extended helix with 3 residues per turn(as opposed to 3.6 for the α-helix) and is referred to as a polyprolineType II helix. The glycines fall on one face of the helix. The absenceof a side chain on glycine permits close approach and wrapping ofthree such helices around each other to form a three stranded “rope”.The resulting structure is stabilized by van der Waals interactionsbetween the strands at the glycine interface and H-bonding crosslinksfrom the hydroxyl groups on hydroxyproline. Hydroxyproline isabsolutely essential for proper stabilization of the mature collagen. 2-

Lys

NH2

Gln

C

O NH2

NH3

Lys

N

CO

Gln

fibrin stabilizing factor



hydroxyproline is shown here. 3-hydroxyproline can also be formed.

Both of these are formed by post-translational modification of pro-line with prolyl hydroxylase, which requires ascorbic acid. Theessential role of hydrogen bonding and hydroxyproline is indicatedby scurvy. The lack of ascorbic acid (vitamin C) in the diet leads tothe inability to form hydroxyproline, and therefore the lack ofcrosslinks in collagen leading to scurvy.

Crosslinking of collagen also occurs via Schiff base linkages. Lysinecan be oxidized to allysine by lysyl oxidase which requires Cu++.Adjacent lysine and allysine side chains spontaneously react to form a

C

O

CαNC

O

N

OH

H

CN C

OHH

NH2

CN C

H

NH2

OH

CN C

OHH

NH2

CN C

H OH

C

OH

CN C

H OH

C

CN C

OHH

N

H

lysyl amino oxidase

Schiff base link

allysine



Schiff base linkage. β-aminoproprionitrile is found in sweet peas andspecifically inhibits lysyl oxidase leading to decreased crosslinking ofcollagen and abnormalities in the bones, joints, and blood vessels ofcattle eating sweet peas, a condition known as lathyrism. The occur-rence of such inhibitors gives us hope that we might be able to designspecific inhibitors targeted for proteins involved in clinical problems(e.g. HbS and COX-2) and that protein engineering will become areality.

There are a number of different types of collagens. Type I makes upabout 90% of the collagen in the body and is found in skin, tendon,bone, cornea, and internal organs. Type II is found in cartilage.Type III is found in skin and blood vessels. Type IV is found in thebasal lamina (a thin layer of extracellular matrix that lies underneathepithelia cells).

There are a large number of diseases associated with collagen in addi-tion to scurvy mentioned above. Menkes’ Syndrome results from alysyl oxidase deficiency due to abnormal copper metabolism.Marfan’s Syndrome results from a mutation in the gene for one of theprocollagen chains leading to a longer chain which leads to spideryfingers and toes and weak aorta and pulmonary arteries. Homocysti-nuria results from a defect in cysteine synthesis and high levels ofhomocysteine appear in the urine and blood. Homocysteine reactswith lysine aldehydes preventing crosslinking. Ehlers-Danlos Syn-dromes are characterized by hyperextensible joints and skin due toimproper processing of collagen.

XII. Keratin α-keratin is found in hair and nails. It is composed of coiled-coil,which was discussed above and an example can be found in thekinemage file coiledcoil.kin. Coiled-coils are characterized by twohelices wound around each other. The amino acid sequence shows acharacteristic 7 residue repeat:

a - b - c - d - e - f - g

where residues a and d are almost always nonpolar, e.g. valine, leu-cine, or isoleucine. The alternating separation of hydrophobic resi-dues by two, three, two, three, two .... residues leads to ahydrophobic face that winds around the outside of each α-helix.Open the kinemage coiledcoil.kin and turn off the “outer” sidechainsby clicking in the “outer” box. The remaining residues at the inter-face of the two helices are largely leucine and valine. Rotate the mol-ecule with the mouse so that it is viewed end on down the axes of thetwo helices. Reduce the z-slab (the thickness of the image viewed) at



the right by sliding the “slide bar” all the way to the top. Slowlyincrease the thickness of the viewing slab by pressing the increasearrow at the bottom of the slide. As the viewing thickness isincreased note that the hydrophobic sidechains at positions a and dalternate back and forth along hydrophobic face forming a knob andhole effect. Rotate the molecule 90° and note that the knobs fromone helix fit nicely into the holes of the other. This perfect mating ofthe two surfaces is like a lock and key and leads to stabilizationthrough not only hydrophobic interactions, but also van der Waalsinteractions. It is one of the best examples of molecular recognitionbetween biomolecules through matching of opposing faces.

Packing of the coiled-coils against each other leads to formation ofthe hair and other structures. α-keratin is also rich in cysteine anddisulfide crossbridges are formed between neighboring coiled coils inhair. Hard keratin found in hair and nails is much higher in cysteinethan soft keratin found in skin. Chemical reduction of -S-S- linksbetween neighboring coiled coils breaks these links. Resetting themby adding an oxidizing agent at new positions after bending the hairis the basis of a “perm”.

The coiled-coil is quite flexible and springy. In contrast, β-keratin iscomposed of β-sheet. It is a much more rigid structure. Silk is also aβ-sheet protein, composed of stacked sheets of fibroin. Fibroin formsβ-sheets with largely glycine on one face and alanine and serine onthe other. Again a knob and hole structure is formed that increasesthe packing efficiency and strength of the material. Silk is largelyunstretchable due to the nature of the β-sheet, but is quite flexibleand strong.

XIII. Elastin Elastin, as might be guessed from its name, is a very elastic proteinand is found in lungs, aorta, and ligaments. It is extensible and iscomposed largely of glycine (1/3), alanine and valine (1/3) and is alsorich in proline. There are few polar residues, making it insoluble.Most notably, it has no organized structure. It also contains a new



amino acid known as desmosine. Desmosine is formed from 3 all-

ysines and one lysine to form an aromatic link between four proteinbackbones. The positively charged aromatic ring gives desmosine(and the tissues it is found in) its yellow color. Since formation ofallysine requires lysyl oxidase (see above), a copper metabolism defectcan lead to reduced crosslinking and strength in elastin.

XIV. Actin Actin is the muscle protein which forms the substrate upon whichthe myosin ATPase moves. It is the thin filament of the muscle sar-comere. It is also an important cytoskeletal protein. In contrast tothe fibrous proteins described above which are composed of largelyelongated fiber-like protomers, actin is composed of globular sub-units with a molecular weight of 42,000. The protomers, G-actin,contain binding sites for other G-actin subunits such that they canform infinite fibers. Each fiber is composed of two strands of G-actinwound around each other to form a helical rope of beads.

XV. Amyloid A number of diseases have been shown to be associated with amyloidfibril formation in vivo. Amyloid is an abnormal assembly of proteinthat is fibrous in nature. It can be composed of quite different pro-teins, but they all form fibrils 60 to 100 Å in diameter and variablelength. The molecular structure is composed of cross-β repeated pat-terns where the β strands are oriented perpendicular to the axis of thefibril.

N+

(CH2)2(CH2)2

(CH2)3

(CH2)4

DESMOSINE



Table 2 below lists 10 different diseases associated with different pro-teins that form amyloid. These are all associated with the deposit ofamyloid fibril of very similar secondary and quaternary structure.There is little if any sequence (primary) or tertiary structural similar-ity among the soluble precursor proteins.

The formation of amyloid resembles the process of sickling of HbS.However, aggregation and fibril formation cannot occur with thenormal folded protein. For example, variant forms of TTR (tran-sthyretin) that cause familial amyloid polyneuropathy cannot formamyloid even when the protein is incubated at very high concentra-tions. Indeed, the structures of the variants that are capable of form-ing amyloid are virtually identical to the normal form. This is asexpected since the mutations leading to the amyloid susceptibleforms are conservative, e.g. Val -> Met. It is now becoming clear thatthe tendency to form amyloid results from the destabilization of thenative folded form relative to an intermediate, alternative form thatmay be on the normal folding pathway. Thus, the intermediate formis populated to a greater extent than in the normal protein. It is thisform that is a direct precursor necessary for amyloid formation. Thefigure below shows the currently held view.

Table 2: Amyloidogenic proteins and the amyloid diseases resulting from their assembly into fibrils

Clinical syndrome Precursor protein

Alzheimer’s diseases β-protein

Primary systemic amyloidosis Immunoglobulin light chains

Secondary systemic amyloidosis

Serum amyloid A

Senile systemic amyloidosis Transthyretin

Familial amyloid polyneuropathy I

Transthyretin

Hereditary cerebral amyloid angiopathy

Cystatin C

Type II Diabetes Islet amyloid polypeptide

Atrial amyloidosis Atrial natriuretic factor

Injection-localized amyloidosis Insulin

Hereditary renal amyloidosis Fibrinogen



The horizontal pathway is the normal folding/unfolding pathway forthe protein showing only one of many possible intermediates. Thisintermediate is important since a side reaction is possible that leads toalternative structure that differs from the native protein. A recentsummary of our current understanding of amyloidogenic proteinscan be found in a review article by Jeffrey Kelley entitled “Alternativeconformations of amyloidogenic proteins govern their behavior” inCurrent Opinion in Structural Biology, 6, 11-17 (1996).

It is becoming apparent that amyloid can be cleared, i.e. the layingdown of amyloid is reversible. Recent therapeutic efforts at slowingdown the deposition of amyloid seem promising since if the forma-tion is slowed, the existing amyloid may be dissolved. (See, for exam-ple, “Treatment of amyloidosis” by S.Y. Tan et al. Am. J. KidneyDisease (1995), 26, 267-85). Check out http://medicine.bu.edu/amyloid/amyloid1.htm.

XVI. Prions A prion is a protein that is an infectious particle that lacks nucleic

Native folded protein Amyloidogenic intermediate Random coil

Amyloid


http://medicine.bu.edu/amyloid/amyloid1.htm

http://medicine.bu.edu/amyloid/amyloid1.htm


acid. Prion diseases are associated with the conversion of a normallysoluble prion protein into an insoluble β-sheet aggregate, similar towhat is observed with amyloidogenic proteins, that leads to disordersin the central nervous system including dementia (Creutzfeldt-JakobDisease) and ataxic (Scrapie, Bovine Spongiform Encephalopathy) ill-nesses. The diseases may be genetic, infectious, or sporadic (sponta-neous). A number of prion diseases are listed in the table below.

There are significant differences between the mechanisms of amyloi-dosis and prion disease. Although there are different prion diseases,they all seem to be associated with PrP. PrP is constitutivelyexpressed in normal, adult, uninfected brain. Normal prion proteinis symbolized by PrPC, while the infectious PrP is indicated by PrPSc

(after the prion disease Scrapie found in sheep). Prion diseases areassociated with the conversion of the prion cellular protein (PrP)from its α-helical structure to a β-sheet structure of PrPSc. PrPC con-tains about 40% α-helix and little β-sheet, while PrPSc is about 30%α-helix and 45% β-sheet. The only apparent difference betweenPrPC and PrPSc is their structure. They have identical sequences andthere are no apparent post-translational modification differences.Similar to amyloidogenic proteins, PrP appears to be a clear violationof the commonly held view that every protein has only one stablefolded conformation. Current evidence indicates that pre-existingPrPSc provides a template that in a sense catalyzes the conversion of

Table 3: Prion Diseases

Disease HostMechanism of Pathogenesis

Kuru Humans Infection through canni-balism

Variant Creutzfeldt-Jakob Disease

Humans Infection from bovine prions

Familial Creutzfeldt-Jakob Disease

Humans Germline mutations in PrP gene

Sporadic Creutzfeldt-Jakob Disease

Humans Spontaneous conver-

sion of PrPC to PrPSc

Scrapie Sheep Infection in genetically susceptible sheep

Bovine spongi-form encephalopa-thy

Cattle Infection with prion contaminated meat or bone meal

Feline spongiform encephalopathy

Cats Infection with prion contaminated meat



PrPC to more PrPSc. The different strains of prion diseases appear tobe different PrPSc structures which are self-propagating. Thereappears to be no genetic or nucleic acid component to the propaga-tion or infection process. Propagation appears to very similar to theprocess of crystallization. There is some evidence that initiation andpropagation may require a chaperone protein referred to as Protein X,possibly similar to Hsp70. It is interesting to note that recombinantPrP can be folded into either the α-helical or β-sheet forms, but nei-ther are infectious. An additional agent, perhaps Protein X isrequired. Current efforts at designing therapeutic agents are focusingon stabilizing PrPC and modifying Protien X.

It should be noted that not all workers in the prion field believe thatprion infection can be explained by simply a proteinaceous infectiousparticle. A very nice, balanced review of the field can be found byPrusiner et al. entitled “Prion Protein Biology” in Cell (1998) 93,337-348. Additional material can be found at http://why-files.news.wisc.edu/012mad_cow/glossary.html and http://w3.aces.uiuc.edu/AnSci/BSE/.

PROBLEM SET - 3l. Which one of the following amino acids is likely to be found in theinterior of a globular protein?

(a) leucine (b) serine(c) glutamine (d) aspartic acid(e) arginine (answer)

2. Which one of the following factors is considered to be the majorforce which leads to the conformational stability of globular proteins?

(a) hydrogen bonding(b) hydrophobic interactions(c) ionic interactions(d) disulfide bonding.(answer)

3. Protein-carbohydrate linkages in the glycoproteins involvesugar residues and which of the following amino acids?

(a) Asparagine(b) Serine(c) 5-hydroxylysine(d) Cysteine(e) N-terminal valine of hemoglobin b-chains (answer)


http://w3.aces.uiuc.edu/AnSci/BSE/

http://w3.aces.uiuc.edu/AnSci/BSE/

http://whyfiles.news.wisc.edu/012mad_cow/glossary.html

http://whyfiles.news.wisc.edu/012mad_cow/glossary.html


4. All of the following bond types are significant for the maintenanceof the secondary, tertiary and quaternary structure of enzymes or pro-teins EXCEPT:

(a) hydrophobic interactions(b) disulfide bonds(c) ester bonds(d) hydrogen bonds(e) electrostatic interactions (answer)

5. At their isoelectric point, proteins have(a) no ionized groups(b) no positively charged groups(c) no negatively charged groups(d)equal numbers of positively and negatively charged groups(e) none of the above (answer)

6. The α-helical arrangements of amino acids in a polypeptide chainrepresents the:

(a) primary structure of the protein(b) secondary structure of the protein(c) tertiary structure of the protein(d) quaternary structure of the protein(e) none of the these (answer)

7. Which of the following features of hemoglobin is considered partof its quaternary structure?

(a) sequence of amino acids(b) α-helices(c) ligand binding properties(d) subunit interactions(e) electrophoretic mobility (answer)

8. All of the following are true EXCEPT:(a) Proteins that contain more than one polypeptide chain areconjugated proteins.(b) Hemoglobin is a conjugated protein.(c) Glycoproteins are conjugated proteins(d) Many simple proteins contain only one N-terminal aminoacid per molecule of protein.(e) Some proteins have more than one conformation or shape.(answer)

9. In aqueous solution at pH 7, most proteins are folded so that thenonpolar amino acid side chains are inside in a nonpolar environ-ment, whereas most of the polar side chains are outside, in contactwith water. Which of the following amino acids are likely to have



their side chains on the inside of a globular protein in solution?(a) Val (b) His(c) Ile (d) Pro(e) Phe (f ) Asp (answer)(g) Lys

10. The α-helix structure(a) is maintained by hydrogen bonding between amino acid sidechains.(b )makes up about the same percentage of all globular proteins.(c) can serve a mechanical role in forming stiff bundles of fibersin such proteins as keratin, myosin, and fibrin.(d) is stabilized by hydrogen bonds between the NH of one pep-tide bond and the carboxyl oxygen of the third amino acidbeyond it.(answer)

11. Which of the following are true concerning the way a polypep-tide may fold?

(a) A tightly coiled rod called an α-helix can be formed in whichhydrogen bonds within the rod stabilize the structure.(b) The α-helix is mainly found in collagen and tropocollagen.(c) An extended structure called a β-sheet can be formed inwhich hydrogen bonds between different portions of the samechain stabilize the structure.(d) Long segments of β-sheet structures are commonly found inmost proteins.(answer)

12. Which of the following are true?(a) The primary structure of a peptide refers to the way that adja-cent amino acids interact with one another.(b) Secondary structure refers to the steric relationships of aminoacids which are close to one another.(c) Tertiary structure refers to the interactions that could poten-tially involve 3 different amino acids in the same polypeptide.(d) Quaternary structure deals with the interactions of multiple molecules of a multimeric protein. (answer)

13. Hemoglobin is a tetrameric protein consisting of two α and twoβ polypeptide subunits. The structure of the α and β subunits isremarkably similar to that of myoglobin. However, at a number ofpositions, hydrophilic residues in myoglobin have been replaced byhydrophobic residues in hemoglobin.

(a) How can this observation be reconciled with the generaliza-tion that hydrophobic residues fold into the interior of proteins?(b) In this regard, what can you say about the interactions deter-



mining quaternary structure in hemoglobin? (answer)

14. (a) Discuss how the molecular structure of Hb S differs from thatof Hb A.(b) How does oxygenation and deoxygenation affect the structure ofHb S? (answer)

15. What is the most abundant serum class of immunoglobulin?(answer)

16. What class of immunoglobulins appear first in the serum afterinjections of an antigen? (answer)

17. The subunit structure of IgG (H specifying a heavy chain and L alight chain) is as follows:

a. HLb. H2Lc. HL2d. H2L2 (answer)

Which of the following statements (numbers 28-34) are true andwhich are false? If they are false, be sure that you understand whythey are false.

18. Both the heavy and light chains contain variable regions.(answer)

19. An immunoglobulin synthesized by a particular myelomapatient would exhibit a range of binding affinities for its antigen.(answer)

20. Each IgG contains one combining site. (answer)

21. IgA is found in external secretions. (answer)

22. Each IgG can precipitate its antigen because it contains asingle binding site. (answer)

23. Diagram an IgG molecule showing the location of the variousconstant and variable regions, the Fab and Fc regions, location of thecarbohydrate, antigen binding site, papain and pepsin cleavage sites.(answer)

24. Which of the following are qualities of elastin?



a. high "stretchability"b. high hydroxylysine contentc. high aliphatic side-chain amino acid contentd. cross-linked through complex lysine derivatives (answer)

25-29. Given the properties of collagen and elastin, predict whetheryou would expect their substantial presence in the following tissues.Use A for collagen, B for elastin, or C for neither.25. tendon (answer)26. liver (answer)27. ligament (answer)28. aorta (answer)29. bone (answer)

30. The collagen triple-helix structure is characterized by extensivesequences of (Gly-X-Pro)n or (Gly-X-HyPro)n in which X is anyamino acid.

a. Why must Gly be present every third residue?b. What are the principal bonds that hold the three helicestogether in the superhelix? (answer)

31. Which of the following residues can be acted upon by an enzymein which vitamin C is a cofactor?

a. Hydroxylysineb. Prolinec. Norleucined. Aspartatee. Desmosine (answer)

32. All of the following statements about collagen are correctEXCEPT:

a. It is a glycoprotein.b. Peptide bond cleavage is needed before the very large collagenmolecules of connective tissue can be formed.c. Each of the constituent chains is a typical α-helix.d. It is a nutritionally poor protein since it contains a high per-centage of simple amino acids and a low percentage of the morecomplex, essential amino acids.e. In ascorbic acid deficiency, collagen chains are produced withabnormally low content of hydroxyproline.(answer)

33. One effect of insufficient procollagen amino-peptidase activityis:

a. Lathyrism



b. one of the Ehlers-Danlos syndromesc. under-hydroxylation of collagend. insufficient number of alpha-beta-unsaturated aldol crosslinks(answer)

34. Chaperones are protein assemblies in the cell which are impor-tant in protein folding because

a. there is a unique chaperone for every protein that determinesthe correct fold.b. they break down incorrectly folded proteins into their substit-uent amino acids.c. they provide isolated enclosures to prevent aggregation of theunfolded protein.d. they specifically remove virus and prion particles. e. they exaggerate the immune response. (answer)

35. Prion diseases are generally believed to be associated witha. infection by a virus.b. infection by DNA.c. infection by protein.d. infection by RNA. e. spontaneous protein conversions. (answer)

Answers-3 1. a2. b3. a,b,c,e4. c5. d6. b7. d8. a9. a,c,d,e10. c,d11. a,c12. b,c13. a.Hydrophobic patches occur on the outside of the hemoglobinsubunits where the a and b chains fit together. Thus, these patchesare on the outside of the subunit, but on the inside of the multimericprotein.b.Hydrophobic interactions plan an important role.

14.a. Hb S differs from Hb A in its primary structure. At position b6, Hb S has a Val substituted for Glu. The difference makes HbS



have in increased affinity for deoxygenated HbA or HbS, which leadsto polymerization and sickling.b.Deoxygenated Hb S is 25 times less soluble than deoxygenated HbA.

15. IgG16. IgM17. d18. True19. False, the myeloma immunoglobulin would be a single molecularspecies from a single cell and not a mixture of different species frommany cells.

20. False, each have two

21. True

22. False, because it contains two combining sites, it can form net-works of antibody-antigen complexes which are insoluble.

23. Use your textbook to check your answer.

24. a,c,d25. a,b26. c27. a,b28. a,b29. a

30.a. Every third residue in each of the three helices falls in the inte-rior of the superhelix, too close to the other two polypeptides to haveany side chain other than a hydrogen atom.b.Each polypeptide folds in a helical structure, designated the pro-line helix. The amide hydrogens and the carboxyl oxygens of eachpeptide bond extend perpendicularly to the helix axis and form H-bonds to corresponding groups of the adjacent helices.

31. b32. c33. b34. c35. c



OVERALL PRACTICE EXAMThis exam contains questions that are similar to what you can expectin the scheduled two hour evaluation for Problem Unit 1.

1.Hydrogen bonding is involved in all of the following structural fea-tures in proteins EXCEPT:

A.Alpha-helix.B.Beta-sheet conformation.C.Reverse-turn (beta bend).D.Random coli.E.Collagen triple helix. (answer)

2.Which of the following sections of a polypeptide chain have aminoacid sidechains, all of which are capable of forming hydrogen bonds?

A.leu-val-pheB.cys-his-alaC.ile-ser-trpD.val-arg-proE.asp-lys-ser(answer)

3.To make an acetic acid/sodium acetate buffer at pH 4.1, what is therequired ratio of sodium acetate to acetic acid? (pKa = 4.7)

A.4/1B.5/1C.1/4D.1/5E.1/2 (answer)

4.All of the following statements concerning IgM are true EXCEPTwhich one?

A.It is the first class of antibodies detected in serum after antigenexposure.B.It can consist of two kappa or two lambda light chains.C.It exists in serum as pentameric glycoprotein.D.It has multiple hypervariable sites within variable regions ofboth H and L chains.E.It is a harmful mediator of allergic reactions. (answer)

5.Dialysis is a process which:A.depends primarily on molecular shape.B.is well-adapted for protein separations which depend primarilyon charge difference.



C.permits large adjustment of the salt content of a protein solu-tion without a large change in the volume of that protein solu-tion.D.is preferably carried out at -10 to -20 C.E.is limited because only microgram quantities of protein can beprocessed. (answer)

6.A major force that contributes to the conformation of proteins and,in globular proteins, occurs primarily in their interior is:

A.hydrogen bonds.B.charged dipoles.C.hydrophobic interactions.D.disulfide bridges.E.hydration by water. (answer)

7.Ascorbic acid has which of the following roles in collagen biosyn-thesis?

A.catalystB.inhibitorC.oxidizing agentD.reducing agentE.high energy compound (answer)

8.Which one of the following bonds is LEAST likely to break duringprotein denaturation?

A.HydrophobicB.HydrogenC.DisulfideD.Electrostatic (answer)

9.What is the pH of a solution consisting of 500 ml of 0.004 M HCl+ 500 ml of 0.002 M NaOH?

A.1.0B.2.0C.3.0D.3.3E.4.0 (answer)

10.An unknown organic acid which was isolated from the sweat andtears of a first year medical student was found to be an ineffectivebuffer at pH 7, but buffered well at pH 4.5. The acid:

A.is a strong acid, i.e., it completely dissociates in water.B.is completely dissociated around pH 4.5.C.possesses a pK near 7.D.All of the above.E.None of the above. (answer)



11.At what pH value would you expect the electrostatic attractionbetween the side chains of histidine (pK = 6.5) and glutamic acid (pK= 4.25) in a protein to be strongest?

A.pH 3.0B.pH 5.5C.pH 7.0D.pH 10.0E.should be the same at all pH values (answer)

12.The buffering capacity of a buffer:A.can be expressed as the mole equivalents of (H+) or (OH-)required to change the pH of 1 liter of buffer solution by 1.0 pHunit.B.is greatest at the pH where pH = pKa'.C.is directly proportional to the buffer concentration.D.All of the above statements are correct.E.Only two of the above statements are correct. (answer)

13.Adding an organic solvent to a protein solution may cause all ofthe following EXCEPT which one?

A.aggregation.B.denaturation.C.alteration of electrostatic interactions.D.rupture of covalent bonds.E.rupture of hydrophobic bonds. (answer)

14.At a pH of 8.6, serum proteins will move in an electrical fieldtoward the anode (+) at rate dependent upon their:

A.carbohydrate contentB.lipid contentC.charge and molecular weight.D.N-terminal amino acid.E.intramolecular disulfide content. (answer)

15.Defective collagen in scurvy is due to insufficient vitamin Cwhich:

A.is ordinarily incorporated into crosslinks between tropocol-lagen molecules.B.is usually involved in the hydroxylation of prolyl residues.C.inhibits the oxidative degradation of collagen.D.is required for the conversion of lysyl residues into aldehydes(answer)

.16.Hydroxyproline residues in collagen are the result of:

A.incorporation of hydroxyproline from hydroxyproline-tRNA.



B.hydroxylation of existing proline residues in the protein.C.deamination of histidyl residues in the proteinD.conversion of free proline to hydroxyproline and then incor-poration into collagen. E.synthesis from hydroxyglutamic acid. (answer)

17.All of the following statements regarding collagen are correctEXCEPT which one?

A.Collagen is the most abundant protein in the human body.B.Collagen has an amino acid composition typical of that foundin most soluble proteins.C.Collagen is a very inelastic protein in the native state.D.Collagen is a very insoluble protein in the native state.E.Collagen is made up of subunits termed tropocollagen.(answer)

18.What is the ratio of the acid to conjugate base forms of the car-boxylic acid side chain of aspartate-102 in alpha-chymotrypsin at pH6.0, if the pKa for the group is 4.0?

A.10 to 1B.1 to 10C.100 to 1D.1 to 100E.1 to 1000 (answer)

19.Collagen contains large amounts of:A.alpha-helix.B.beta-helix.C.cysteine (or its disulfide form, cystine).D.intrachain hydrogen bonds.E.proline and hydroxyproline. (answer)

20.Which of the following statements regarding collagen are correct?1)Glycosyltransferases attach galactose (and sometimes afterwards glucose) residues to hydroxylysine.2)Glycosyltransferases attach galactose (and sometimes after-wards glucose) residues to lysine.3)Glycosyltransferases attach galactose (and sometimes after-wards glucose) residues to hydroxyproline.4)One third of the amino acids are glycine.5)The intra-molecular crosslinks of collagen increase with age.

A.1, 3, 5B.1, 2, 4C.1, 4, 5D.2, 3, 4



E.3, 4, 5 (answer)

21.For a certain weak acid, the ratio of the acidic species to the conju-gate base is found by analysis to be 10 to 1 at pH 5.0. What is the pKof the acid?

A.1B.4C.5D.6E.10 (answer)

Answer the following questions using the key outlined below: A. If 1, 2, and 3 are correctB.If 1 and 3 are correctC.If 2 and 4 are correctD.If only 4 is correctE.If all four are correct

22.Hydrophobic interactions:1.arise in part as a consequence of the properties of water.2.are restricted to residues in alpha-helices or beta-pleated sheetregions.3.are often involved in formation of multi-subunit protein struc-tures.4.are possible only in oligomeric proteins. (answer)

23.One effect of insufficient procollagen peptidase is:1.lathyrism.2.Ehlers-Danlos syndrome.3.under-hydroxylated collagen.4.decreased tensile strength. (answer)

24.Which of the following do the alpha-helix, the beta-sheet and thecollagen triple helix have in common?

1.high glycine content2.high proline/hydroxyproline content3.a large net charge at pI4.hydrogen bonds between the amide hydrogen and the carboxyloxygen of the polypeptide backbone. (answer)

25.Which of the following diseases are associated with protein mis-folding?

1. sickle cell anemia2. myeloma3. scurvy



4. amyloidosis (answer

Answers for Practice Exam 1.D.

2.E.

3. C4.1 = 4.7 + log(Acetate/Acidic acid) -0.6 = log (Acetate/Acidic acid) (Acetate/Acidic acid) = 0.25

4.E.

5.C.

6.C.

7.D.8.C.9.C.10.E.

11.B.Both need to be charged. At pH lower than 6.5 his has a chargegreater than +0.5, and at pH above 4.25 glu has a charge greater than-0.5. The strongest attraction will occur at a pH between the pKavalues for these two amino acids.

12.D.13.D.14.C.15.B.16.B.17.B.18.D.19.E.20.C.21.D.Solu: 5.0 = pK + log(1/10).22.B.23.C.24.D.25.D.



APPENDIX I: Using KinemageKinemage is a freeware program that can be run on Macintosh, PCWindows, and UNIX computers. It can be downloaded from thebiochemistry server or from http://prosci.org/Kinemages/. Thislatter site also contains other files that can be viewed with Kinemage,including other protein tutorials.

Kinemage is opened by double clicking on the icon MAGE_4.3.This version sets the number of colors displayed by your monitor to256. If you use an earlier version, you must do this manually priorto starting the program by using the Monitor control panel. Afterthe program loads, click on the PROCEED button. Two windowsare opened: a TEXT:Kinemages window and a MAGE Color Graph-ics window. Pull down the FILE menu to OPEN FILE and select anyfile with the .kin suffix (Kinemage cannot display pdb files down-loaded from the Brookhaven Protein Databank. A free programPREKIN_4.0 allows for the conversion of pdb files to kin files if youdesire). Text associated with kinemage files is sometimes printed inthe TEXT window and the color image appears in the MAGE ColorGraphics window. Click anywhere in the Color Graphics window tobring it to the forefront, and drag it to the desired position in themonitor display for optimal viewing by placing the cursor in theupper title bar, hold down the mouse button, and drag the window toits desired location. The window can be resized by placing the cursorin the triangle in the lower right corner, holding down the mousebutton, and dragging the corner to its desired size.

The image can be rotated to any desired orientation by placing thecursor anywhere in the Color Graphics window and dragging themouse while holding down the mouse button. A bit of experimenta-tion will show you that the type of rotation is determined by theplacement of the cursor in the graphics window. The image can bemade larger by using the zoom slide bar at the right. The z-slab slidebar controls the thickness of the viewing slab. This can help to sim-plify complicated structures by removing overlying and underlyingatoms that are not of interest, e.g. in a binding site. Most kinemageshave been set up to allow the viewer to select various views, atoms,side chains, etc. by clicking on or off the boxes in the menu at theright side of the Color Graphics window. Most of these are self-explanatory and are designed for self-instruction and exploring.

One additional nice point about Kinemage is the ability to measuredistances. Clicking on any atom will result in the display of its label


http://prosci.org/Kinemage/


at the lower left of the graphics window (e.g. o glu 62 indicaes thatthis is the oxygen of glutamate 62). Clicking on any other atom willnot only give its label but the distance from the previous atom to thisatom is given in the lower center of the graphics window. Thisshould allow you to get a feel for the size of various biomolecules.

APPENDIX II: Using Acrobat Reader with pdf Files

Portable Document Format (PDF) files can be read by AcrobatReader, a free program which can be downloaded from the AdobeWeb site (http://www.adobe.com/acrobat). If Acrobat Reader isinstalled on your system, it will automatically open simply by double-clicking on the pdf file that you wish to read.

Acorbat Window The document will be displayed in the center of your window and anindex will appear at the left side of the screen. Each entry in theindex is a hypertext link to the associated topic in the text.

Using hypertext links in a pdf document is exactly like that in a webpage or html document. When you place the cursor over a hypertextlink, it changes to a hand with the index finger pointing to the under-lying text. Clicking the mouse causes the text window to jump tothat location. The index does not change. Magnification may needto be adjusted using the menu option in the lower part of the screento optimize the view and readibility. The best magnification is usu-ally around 125%.

Subheadings in the index can be viewed by clicking on the open dia-monds to the left of appropriate entries to cause them to point down-wards. Clicking again will close the subheadings lists.

Hypertext links Hypertext links in the text (not in the index) are indicated by blueunderlined text. The cursor should change to a hand with the indexfinger pointing to this text when it passes over it. Clicking will causethe text page to move to the associated or linked text which will behighlighted in red underlined text. Red underlined text is not ahyperlink, only a destination.


http://www.adobe.com/acrobat


How to back up to a previous window:

If you wish to return to a previous text window after following ahypertext link, use the double solid arrow key at the top of the Acro-bat window (or use the key equivalent “command - “). Acrobatkeeps a record of your last 20 or so windows so that multiple stepsback can be made my repeating the command.

Links to web sites A number of url links to web sites are located in the pdf file andappear in blue underlined type starting with http:// (e.g. http://www.som.siu.edu). Clicking on these should open a web browsersuch as Netscape and take you to those web sites. You may need toresize the Acrobat Window to view the web browser window dis-played underneath it.

COMMENTSI hope that you find this pdf file useful. Comments on how to makeit better would be greatly appreciated. Please notify me in person orby email ( [email protected]) of any errors so that they can beremoved. The online version on the Biochem server can be easilyupdated.


http://www.som.siu.edu

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