chapter 03 solutions

8/10/2019 Chapter 03 Solutions

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DEFINITIONS

31 Quaternary structure

32 a helix

33 Primary structure

34 Binding site

35 Tertiary structure

36 Polypeptide backbone

37 b sheet

38 Protein domain

39 Secondary structure

TRUE/FALSE310 True. In both statesstretched like a string and properly foldeda protein

has a highly ordered arrangement of its atoms. A folded protein is stable at anear entropy minimum because the entropic cost is more than balanced bythe contributions of weak bonds. A stretched out protein, however, is notstable at this entropy minimum and will assume a more disordered state;that is, it will maximize its entropy.

311 True. In a b sheet the amino acid side chains in each strand are alternatelypositioned above and below the sheet. This relationship can be seen in Fig-ure 338 (see Answer 319), which shows that the carbonyl oxygens alternatefrom one side of the strand to the other. Thus, each strand in a b sheet canbe viewed as a helix in which each successive amino acid is rotated 180.

312 True. Chemical groups on such protruding loops can often surround amolecule, allowing the protein to bind to it with many weak bonds.

THOUGHT PROBLEMS

313 Free amino acids have an amino group and a carboxylate group, both ofwhich are charged at neutral pH. In proteins these groups are involved inpeptide bonds, which are uncharged. Thus, the hydrophobicity/hydro-philicity of a free amino acid is not the same as that of its side chain in a pro-tein.

To measure the hydrophobicity/hydrophilicity of the side chains, it iscommon to assess the properties of side-chain analogs. Thus, for alanine

one would use methane, for threonine, ethanol, for aspartic acid, acetic

THE SHAPE AND STRUCTURE OF PROTEINS In This Chapter

THE SHAPE AND A4STRUCTURE OFPROTEINS

PROTEIN FUNCTION A5

A

Chapter 3

3Proteins


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A46 Chapter 3: Proteins

acid, and so on. To assess hydrophilicity, one can measure the solubility ofthe side-chain analogs in water. In general, this is done by determining howthe side-chain analog partitions between a vapor of the analog and water.Hydrophobicity can be measured in an analogous way by assessing how aside-chain analog partitions between water and a nonpolar solvent such ascyclohexane. One might imagine that the rank order for hydrophilicitywould be the reverse of that for hydrophobicity. From the rank order lists inFigure 334, that is mostly true, but there are differences, most notably tyro-sine (Y) and tryptophan (W ).

Reference:Creighton TE (1993) Proteins, 2nd ed, pp 153155. New York: WHFreeman.

314 Hydrogen bonds, electrostatic attractions, van der Waals attractions, and the

hydrophobic force.

315

A. Heating egg-white proteins denatures them, allowing them to interact withone another in ways that were not possible at the lower temperature of thehens oviduct. This process forms a tangled meshwork of polypeptide chains.In addition to these interactions, interchain disulfide bonds also form, sothat hard-boiled egg white becomes one giant macromolecule.

B. Dissolving hard-boiled egg white requires a strong detergent to overcomethe noncovalent interchain interactions and mercaptoethanol to break thecovalent disulfide bonds. Together, but not separately, the two reagentseliminate the bonds that hold the tangled protein chains in place. Try it foryourself!

316 In an a helix, the carbonyl oxygen of the first amino acid hydrogen bonds tothe amide nitrogen of amino acid 5 (Figure 335). Thus, there can be no ahelix shorter than five amino acids. The single hydrogen bond that would beformed with five amino acids gives too little stability to the structure for anyhelicity to be detected. Only with six amino acidstwo hydrogen bondsdo you begin to detect some. Helicity becomes increasingly apparent asmore amino acids, hence more hydrogen bonds, are added.

317 The ends of a helices, like polar amino acids, are almost always found at thesurface of a protein where they can interact with polar water molecules. Inaddition to their partial charge, the backbones of the four amino acids ateither end of the helix carry hydrogen-bonding groups that are unsatisfiedby hydrogen-bonding within the helix (Figure 336). These groups also add

to the polarity of the termini of a helices.

LIVFMWACGYTSHQKNEDR

3.983.983.102.041.411.390.870.340.001.083.514.345.606.486.497.587.759.66

15.86

GLIVAFCMTSWYQKNEHDR

0.000.110.240.400.453.153.633.877.277.458.278.50

11.7711.9112.0712.6312.6613.3422.31

hydrophobicity(dark bars)

aminoacid

kcal/mole

hydrophilicity(light bars)

aminoacid

kcal/mole

increasinghydroph

obicity

increasinghydrophilicity

kcal/mo

le

20

10

0

10

20

hydrophilicity

hydrophobicity

GC

DEN

KQHST

R

Y

AWMFVIL

Figure 334 Measurements of

hydrophilicity and hydrophobicity ofside-chain analogs (Answer 313).

Partition coefficients are listed for each

amino acid side-chain analog. A partition

coefficient is the equilibrium constant for

the solubility of a solute in two differentphases, and is expressed as kcal/mole.

The rank order for hydrophobicity is

listed in descending order, while that forhydrophilicity is listed in ascending order.

If there were a perfect negativecorrespondence between hydrophobicity

and hydrophilicity, the amino acids

would appear in the same order in the

two lists. The data are plotted with the

amino acids arranged in order ofdecreasing hydrophobicity (dark bars)

from left to right. Hydrophilicity is

indicated by the overlaid gray bars.

N7

N3

N4

O1

O2

O3

O4

O5

O6

N5

N6

N8

N9

N2

N1

O7

O8

O O

+H3

Figure 335 Schematic of an a helix,showing the pattern of hydrogen bondsbetween carbonyl oxygens and amide

nitrogens within the helix (Answer 316

).


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THE SHAPE AND STRUCTURE OF PROTEINS A

L

S

S

V

M

E

W

T

F

ER

PEPTIDE 1

VI

F

I

L

K

D

1815

4

11

18

714

310 17

6

132

9

16

5

12

PEPTIDE 2

K

R

R

R

RF

V

V

L

LLF

I

T

S

1815

4

11

18

714

310 17

6

132

9

16

5

12

PEPTIDE 3

R

R

R

RV

V

L

I

LF

F

I

I

A

S18

15

4

11

18

7

14

310 17

6

13

2

9

16

5

12

318 The first two peptide sequences, but not the third, would give amphiphilichelices, as shown in Figure 337.

319 As illustrated in Figure 338, the first three strands of the sheet are antipar-allel to their neighbors, whereas the fourth strand is parallel to the third.

320

D. Because the side chains of the amino acids alternately project above andbelow the sheet, a sequence that could form a strand in an amphiphilic bsheet should have alternating hydrophobic and hydrophilic amino acids.Only choice D satisfies this condition.

321 None of these folds would give a knot when stretched out. This is a general

principle: proteins fold without forming knots. One might imagine that itwould be difficult to thread the end of a protein through an interior loop toform a knot. A folding pathway to such a knotted form might not be achie-veable through random motions in a reasonable time.

322 Antiparallel strands are commonly formed by a polypeptide chain that foldsback on itself. Thus, only a few amino acids are required to allow thepolypeptide chain to make the turn. By contrast, parallel strands must beconnected by a polypeptide chain that is at least as long as the strands in the

N

N

N

N

O

O

O

O

d+

d-

Figure 336 Representation of an

a helix showing dipole andunsatisfied hydrogen-bonding

groups at its ends (Answer 317).Non-hydrogen-bonded Os and Ns

are labeled.

Figure 337 Arrangement of amino

acids of the three peptide

sequences around helix wheels

(Answer 318).

N

N

N

N

N

N

N

N

N

N

NN

NN

N

N

N

N

N

N

N

Figure 338 A segment of b sheetshowing the polarity (N to C) of the

individual strands (Answer 319

).


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sheet. For a long peptide, a common solution for satisfying backbone hydro-gen-bonding requirements is an a helix (Figure 339).

323 Proteins obviously cant search all possible conformations on their way tofinding the correct one. Thus, there must be defined pathways to simplifythe search. It is now thought that weak interactions rapidly cause the proteinto collapse into a molten globule, in which bonding interactions are tran-sient and chains maintain fluidity. Within the molten globule, very weak sec-

ondary structures form and disappear, as do tertiary interactions. The for-mation of small elements of correct secondary structure, stabilized byappropriate tertiary interactions, then appears to nucleate formation of thefinal structure. This general folding pathway represents a fight between themaximization of entropy, which tends to keep the protein as random as pos-sible, and the minimization of enthalpy through formation of weak bonds.Increasing numbers of weak interactions pull the structure through a suc-cession of increasingly well defined states to the final conformation. Thisconceptualization of folding has been likened to a funnel, and is commonlyreferred to as the folding funnel, with multiple routes of progress down thefunnel accompanied by an increase in nativelike structures.

Reference:Fersht A (1999) Structure and Mechanism in Protein Science, pp575600. New York: WH Freeman.

324 Many different strings of amino acids can give rise to identical protein folds.The many amino acid differences between the homeodomain proteins fromyeast and Drosophilaare among the many possible ones that do not alter fold-ing and function. This question could have been framed in another way;namely, how many amino acid changes are required to convert, say, an a helixinto a b sheet? The answer is: surprisingly few. These two answers underscorethe difficulty in predicting protein structures from amino acid sequences.

325 The statement is correct. The length of evolutionary separation does notdepend on how long yeast and Drosophilahave been around, but rather onwhen their last common ancestor existed. Evolutionary separation is calcu-lated as twice the time from when the last common ancestor existed; that is,

counting backwards from one kind of organism to the common ancestor andthen forward to the other kind of organism.

326

A. The protein in Figure 35 is composed of two domains. The protein can becleaved in the exposed peptide segment that links the two domains (Figure340). Fragments that correspond to individual domains are likely to foldproperly. It is common experience that isolated domains are easier to crys-tallize than the entire protein.

B. The ability to form a crystal depends on the surface characteristics of theprotein because it must be able to interact with itself in a repeating patternto form a crystal. Homologous proteins from different species, which foldthe same (like the homeodomain proteins in Problem 324), differ subtly intheir surface characteristics. As a result, the protein from one species may

crystallize readily, while that from another species may not crystallize at all.A single amino acid change sometimes makes all the difference.

327 Perhaps this is so. Nevertheless, it seems likely that new, and useful, proteinfolds have been invented during evolution by the chance fusion of genes. Adistribution of protein folds within the tree of life would be informative. Pro-tein folds that are distributed in all divisions of the treearchaea, eubacteria,and eucaryoteswere very likely present in their last common ancestor.More recently invented protein folds would likely be confined to a single divi-sion, or a few branches. Even if all protein folds were present in the last com-mon ancestor, it seems unlikely that there would be a one-to-one correspon-dence between folds and genes. Surely, the evolution that led up to the lastcommon ancestor would already have exploited some of the benefits of gene

duplication and refinement of function that lead to families of related genes.

Figure 339 Connections between

antiparallel and parallel strands of a

b sheet (Answer 322).

from lysozyme

from alcohol dehydrogenase

ANTIPARALLEL STRANDS

PARALLEL STRANDS

Figure 340 Catabolite activator protein

(Answer 326). The arrowshows the site

of cleavage in the exposed peptide

segment linking the two domains.


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The limited number of protein folds raises a more fundamental questionabout the total number of protein folds that are possible. It may be that evo-lutionary processes have already exploited most of the stable folds that arepossible. Alternatively, it may be that the number of possible stable foldsgreatly exceeds those currently used on Earth. The simple cataloguing ofnatural protein folds will not address this more basic question.

328 Generally speaking, an identity of at least 30% is needed to be certain that a

match has been found. Matches of 20% to 30% are problematical and diffi-cult to distinguish from background noise. Searching for distant relativeswith the whole sequence usually drops the overall identity below 30%because the less conserved portions of the sequence dominate the compar-ison. Thus, searching with shorter, conserved portions of the sequence givesthe best chance for finding distant relatives.

329 The close juxtaposition of the N- and C-termini of this kelch domain identi-fies it as a plug-in type domain. In-line type domains have their N- and C-termini on opposite sides of the domain.

330 As shown in Figure 341, the three protein monomers have distinctly differ-ent assembly properties because of the three-dimensional arrangement oftheir complementary binding surfaces. Monomer A would assemble into a

sheet; monomer B would assemble into a long chain; monomer C wouldassemble into a ring composed of four subunits.

331 Choice B () is the only arrangement of DNA-binding sites that matchesthe arrangement of subunits in the head-to-head Cro dimer. The DNA thatcorresponds to such an arrangement is known as a palindrome:

ATCG.CGAT

TAGC GCTA

Rotation of this sequence 180 about the central dot gives an identicalsequence, just as does rotation of the arrows (). This demonstrates thatthe Cro dimer and its recognition sequence have the same symmetry, asexpected.

332 Head-to-tail dimers have unsatisfied binding sites at each end, whichwould lead to the formation of chains (see Figure 341B).

333 Proteins 1, 3, 4, and 5 can form head-to-head dimers, as illustrated for pro-tein 1 in Figure 342A. All binding surfaces that allow proteins to interact arecomplementary. The binding surfaces that allow two copies of a protein toform a head-to-head dimer must be self-complementary because theybind to themselves. To be self-complementary, one half of the binding sitemust be complementary to the other half. This means that the two halvescan be folded on top of one another, with properly matched binding, acrossa line drawn through the center of the binding site, as illustrated for the pro-tein 1 binding surface in Figure 342B. There is no line across which proteins2 and 6 can be folded to make their binding partners match. Inclusion of

protrusions and invaginations would not have altered this general principle:complementary binding surfaces can be folded so that a protrusion on oneside inserts into an invagination on the other side.

334 The coil 1A segment of nuclear lamin C matches the heptad repeat at 9 of 11positions (Figure 343), which is very good. The match need not be perfect

(A)

(B)

(C)

Figure 341 Assembly of proteinmonomers (Answer 330).

+

d

a

+

d

a

+

da

(A)

(B)

* * ***** * * * * * * **

DLQELNDRLAVYIDRVRSLETENAGLRLRITESEEVV

-A--D---A--D---A--D---A--D---A--D---A

+ + + + + + - + + - +

hydrophobic amino acids

match with heptad repeat

Figure 342 Self-complementarity in

proteins (Answer 333). (A) Head-to-hea

dimer formation. (B) Self-complementa

binding surface.

Figure 343 Heptad repeat motif in the

coil 1A region of nuclear lamin C (Answ334). Hydrophobic amino acids are

marked with an asterisk (*). When a

hydrophobic amino acid occurs at an A

D in the heptad repeat, it is assigned a

The start of the heptad repeat was

positioned to maximize matches.


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to allow formation of a coiled-coil. The matches to the heptad repeat in theother two marked segments (coil 1B and coil 2, see Figure 39) are not asgood, but they are still acceptable for the formation of a coiled-coil.

Reference:McKeon FD, Kirschner MW & Caput D (1986) Homologies inboth primary and secondary structure between nuclear envelope and inter-

mediate filament proteins. Nature31, 463468.

335

A. The final fiber will be 70 nm in length and will contain 7 of the 10-nm sub-units and 5 of the 14-nm subunits (Figure 344). Assemblies that are anyshorter will have unoccupied binding sites at one end or the other.

B. It makes no difference how the original pair come together. So long as theunoccupied binding sites of the initial pair are filled in by other subunits, thefiber will always grow to the same length.

CALCULATIONS

336 At equilibrium there would be 1 unfolded protein for every 107 folded pro-teins. This ratio comes from substituting values for DG (9.9 kcal/mole), R,and Tinto the equation and solving for logK:

logK = (9.9)/[( 2.3) (1.98 103) (310)] = 107

Since K= [U]/[F],

logK= log ([U]/[F]) = 7

Taking the log of both sides,

[U]/[F] = 107, or [U] = 107 [F]

337

A. The stability of lysozyme at 37C is 10 kcal/mole. From your measurements

Gunfolded = 128 119 = 9 kcal/mole

and

Gfolded = 75 76 = 1 kcal/mole

Thus,

DG = Gunfolded Gfolded = 9 + 1 = 10 kcal/mole

B. This problem illustrates the essence of the protein folding problem: theoverall stability of a protein derives from small differences between largenumbers. For lysozyme, for example, not only must the hundreds of inter-actions in the folded state be evaluated, but so also must the hundreds ofinteractions in the unfolded state. In each state the sum of the enthalpiccontributions is large, as is the sum of the entropic contributions. As shownby the calculation above, these large numbers are subtracted from oneanother to give a small number. Thus, the enthalpic and entropic values ofthe individual interactions must be known with exquisite precision in orderto predict the stability of the folded protein. Yet individual weak bonds can

70 nm

Figure 344 Vernier assembly of 10-nm

and 14-nm subunits (Answer 335).


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vary considerably in strength. For example, individual hydrogen bonds inproteins vary from a few tenths of a kcal/mole to over 1 kcal/mole. Theseconsiderations make prediction of protein structure from these sorts of cal-culations virtually impossible.

Reference: Creighton TE (1993) Proteins, 2nd edn, pp 297300. New York:WH Freeman.

338 Since there are 20 possible amino acids at each position in a 300 amino acidlong protein, there are 20300 (which is 10390) possible proteins. The mass ofone copy of each possible protein would be

Thus, the mass of protein would exceed the mass of the observable universe(1080 g) by a factor of about 10290!!

339 The fraction of correctly synthesized proteins will be 0.91 for a 1000 aminoacid protein, 0.37 for a 10,000 amino acid protein, and 0.00005 for a 100,000

amino acid protein. The calculation is shown below for a 1000 amino acidprotein.

PC = (fC)n

= (0.9999)1000

= 0.91

340

A. As calculated in the previous problem, synthesis of one 10,000 amino acidprotein would be expected to occur correctly 37% of the time. By contrast,synthesis of each 200 amino acid subunit would be expected to occur cor-rectly 98% of the time [PC = (fC)n= (0.9999)200 = 0.98]. Assembly of the mix-ture of subunits into correct ribosomes follows the same equation [PC = (fC)n

= (0.98)50 = 0.37]. Thus, making a ribosome from subunits gives the samefinal fraction of correct ribosomes, 37%, as making them from one long pro-tein.

B. The assumption in part A that correct and incorrect subunits are assembledinto ribosomes with equal likelihood is not true. Any mistake that interfereswith the correct folding of a subunit, or that interferes with the ability of thesubunit to bind to other subunits, would eliminate the subunit from assem-bly into a ribosome. As a result, the fraction of correctly assembled ribo-somes would be higher than calculated in part A. Thus, the value of subunitsynthesis lies not in more accurate synthesis, but rather in permitting qual-ity control mechanisms to reject incorrect subunits efficiently.

DATA HANDLING341 If unfolding of the protein simply reflected the titration of a buried histidine,

it should require 2 pH units to go from 9% to 91% completion. The actualunfolding curve takes only 0.3 pH units to span this range. This sharp tran-sition indicates a highly cooperative process; when the protein starts tounfold, it completes the process rapidly. For example, it might be that sev-eral buried histidines can ionize when the chain starts to unfold, so thatwhen one goes they all go together. Note also that as soon as a buried histi-dine (pKof 4 in this example) becomes accessible to solvent its pKwill shifttoward its normal value of 6, significantly steepening its titration curve.

Reference:Creighton TE (1993) Proteins, 2nd ed, pp 288289. New York: WHFreeman.

110 d 300 aa gmass =aa protein 6 1023d

mass = 5.5 10370g

10390proteins


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342 As expected, hydrophobic amino acid side chains are most frequentlyburied and hydrophilic side chains are least commonly buried. Perhaps thebiggest surprise in this list is the high proportion of cysteine (C) side chainsthat are buried. Cysteine is generally grouped with polar amino acidsbecause of its SH group, but its hydrophobic/hydrophilic properties (seeAnswer 313) indicate that it is, at best, weakly polar. Tyrosine (Y ) alsodeserves comment. Tyrosine is usually grouped with polar amino acidsbecause of its hydroxyl moiety; however, its measured hydrophobic/

hydrophilic properties are ambiguous (see Answer 313), indicating that it isonly weakly polar. By the criterion of buriedness, it clearly behaves likeother polar amino acids.

343

A. These data are consistent with the hypothesis that the springlike behavior oftitin is due to the sequential unfolding of Ig domains. First, the fragmentcontained seven Ig domains and there are seven peaks in the force-versus-extension curve. In addition, the peaks themselves are what you mightexpect for sequential unfolding. Second, in the presence of a protein denat-urant, conditions under which the domains will already be unfolded, thepeaks disappear and the extension per unit force increases. Third, when thedomains are cross-linked, and therefore unable to unfold, the peaks disap-

pear and extension per unit force decreases.B. The spacing between peaks, about 25 nm, is almost exactly what you would

calculate for the sequential unfolding of Ig domains. The folded domainoccupies 4 nm, but when unfolded, its 89 amino acids would stretch toabout 30 nm (89 0.34 nm), a change of 26 nm.

C. The existence of separate, discrete peaks means that each domain unfoldswhen a characteristic force is applied, implying that each domain has a definedstability. The collection of domains unfolds in order from least stable to moststable. Thus, it takes a little more force each time to unfold the next domain.

D. The sudden collapse of the force at each unfolding event reflects an impor-tant principle of protein unfolding; namely, its cooperativity. Proteins tendto unfold in an all-or-none fashion (see Problem 341). A small number ofhydrogen bonds are crucial for holding the folded domain together (Figure

345). The breaking of these bonds triggers cooperative unfolding.Reference: Rief M, Gutel M, Oesterhelt F, Fernandez JM & Gaub HE (1997)Reversible folding of individual titin immunoglobulin domains by AFM.Science276, 11091112.

344

A. None are detected in this experiment. Treating first with radiolabeled NEMshows that many cytosolic proteins have cysteines that are not linked bydisulfide bonds. Treating first with unlabeled NEM to block these sites, fol-lowed by DTT to break disulfide bonds, should expose any SH groups thatwere linked by disulfide bonds. These newly exposed SH groups should belabeled by subsequent treatment with radiolabeled NEM. The absence oflabeling indicates that no cysteines were involved in disulfide bonds.

B. BSA and insulin are labeled extensively only after their disulfide bonds havebeen broken by treatment with DTT. In the absence of DTT treatment, BSAis weakly labeled. Since BSA has an odd number of cysteines, at least onecannot be involved in disulfide bonds. Structural analysis confirms that oneof its 37 cysteines is not involved in a disulfide bond.

C. Because the ER is the site where disulfide bond formation is catalyzed inpreparation for export of proteins, it is expected that lysates from cells thathave internal membranes would have many proteins with disulfide bonds.

N

C

Figure 345 Hydrogen bonds that lock

the domain into its folded conformation(Answer 343). The indicated hydrogen

bonds (gray lines), when broken, trigger

unfolding of the domain. If you compare

this topological diagram with the three-

dimensional structure in Figure 312A,you can pick out the two short b strandsthat are involved in forming these

hydrogen bonds.


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PROTEIN FUNCTION A

PROTEIN FUNCTION

DEFINITIONS

345 Scaffold protein

346 Feedback inhibition

347 Antibody

348 Active site

349 Enzyme

350 Protein phosphatase

351 Linkage

352 Protein kinase

353 Transition state

354 SCF ubiquitin ligase

355 Allosteric protein356 Proteomics

357 Coenzyme

TRUE/FALSE

358 False. The pKvalues of specific side-chain groups depend critically on theenvironment. On the surface of a protein, in the absence of surroundingcharged groups, the pKof a carboxyl group is usually close to that of the freeamino acid. In the neighborhood of negatively charged groups, the p Kof acarboxyl group is usually higher; that is, the proton dissociates less readilysince the increase in local density of negative charge is not favored. Theopposite is true in a positively charged environment. In hydrophobic sur-roundings the dissociation of a proton can be substantially suppressed,since the presence of a naked charge in such an environment is highly dis-favored. It is this ability to alter the reactivities of individual groups thatallows proteins to fine-tune their biological functions.

359 False. Assuming that the three-dimensional structure of at least one familymember is known, it would be possible to use evolutionary tracingfittingthe primary sequence to the structureto determine where the conservedamino acids cluster on the surfaces of the proteins. Clusters of conservedamino acids are likely to correspond to important regions such as thoseinvolved in binding to specific ligands or other proteins. Knowing wheresuch binding sites reside on the surface does not identify the proteins func-

tion. You would not know whether the protein was an enzyme or a structuralprotein, or what it bound to. Some other approach, usually biochemical,would be required to elucidate the function.

360 False. The turnover number is constant since it is Vmax divided by enzymeconcentration. For example, a 2-fold increase in enzyme concentrationwould give a 2-fold higher Vmax, but it would give the same turnover num-ber: 2 Vmax/2 [E] = k3.

361 True. The term cooperativity embodies the idea that changes in the confor-mation of one subunit are communicated to the other identical subunits inany given molecule, so that all of these subunits are in the same conformation.

362 True. Each cycle of phosphorylationdephosphorylation hydrolyzes one

molecule of ATP; however, it is not wasteful in the sense of having no benefit.


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Constant cycling allows the regulated protein to switch quickly from onestate to another in response to stimuli that require rapid adjustments of cel-lular metabolism or function. This is the essence of effective regulation.

363 False. Although many of the conformational changes induced by ligandbinding are relatively small, in some instances these local changes are prop-agated through a molecule to give rise to changes of more than a nanome-ter. The conformational change triggered by hydrolysis of GTP by EF-Tu, forexample, allows two domains of the protein to separate by 4 nm.

THOUGHT PROBLEMS

364 Antifreeze proteins function by binding to tiny ice crystals and arrestingtheir growth, thereby preventing the fish from freezing. Ice crystals that formin the presence of antifreeze proteins are abnormal in that their surfaces arecurved instead of straight. The various forms of the antifreeze proteins inthese fishes are all composed of repeats of a simple glycotripeptide (Thr-Ala/Pro-Ala) with a disaccharide attached to each threonine. The genes forthese antifreeze proteins were apparently derived by repeated duplication ofa small segment of a protease gene.

References:Cheng CHC & Chen L (1999) Evolution of an antifreeze protein.Nature401, 443444.

Jia Z & Davies PL (2002) Antifreeze proteins: an unusual receptorligandinteraction. Trends Biochem. Sci. 27, 101106.

365 To bind to valine, the valyl-tRNA synthetase uses a binding pocket of theproper shape that is lined with hydrophobic residues. Such a binding sitepermits valine to bind well but does not fully exclude threonine, which hasthe same shape and a single polar hydroxyl group (Figure 346). The secondbinding site is much more specific for threonine because it contains anappropriately positioned hydrogen-bond acceptor that makes a specifichydrogen bond with threonine but not with valine. Even though valine canfit into the site, it cannot bind tightly and is thus a very poor substrate for thehydrolysis reaction.

366Of all the possible pairs, only proteins 2 and 6 can interact in a way that sat-isfies all the binding groups on their binding surfaces (Figure 347). This sortof complementary arrangement of binding moieties is characteristic of sur-facesurface interactions between proteins.

367 The problem is that the off rate for the antibodyenzyme complex is too slow.In order for the peptide to displace the enzyme from the column, the enzymemust first dissociate from the antibody. The antibody binding sites wouldthen be quickly bound by the peptide, whose high concentration would pre-vent the enzyme from reattaching to the antibody (any newly exposed anti-body binding site would be bound by peptide). In principle, you could soakthe column with peptide for several days (for several dissociation half-times,see Problem 390), but this usually has adverse consequences for the qualityand activity of the enzyme preparation. In general, high-affinity antibodies

have slow off rates and are unsuitable for affinity chromatography.

tRNA

O

site 1 site 2

O

CH

CH

H3C

+H3N

CH3

COO

CH

CH

H3C

+H3N

OH

C O

Figure 346The two binding sites for

valine and threonine on valyl-tRNAsynthetase (Answer 365).

Figure 347 Binding of protein 2 with

protein 6 (Answer 366).

d

+

a

+

a

d

2

+ a

d

6 6

+ d

a

2


11/23

PROTEIN FUNCTION A

Special procedures have been devised for preparing or identifying anti-bodies that work in such experiments. Usually, lower-affinity antibodies areused, or chromatography is carried out under special conditions that reducethe affinity of the antibody.

Reference:Thompson NE & Burgess RR (1996) Immunoaffinity purificationof RNA polymerase II and transcription factors using polyol-responsivemonoclonal antibodies. Methods Enzymol. 274, 513526.

368 The reaction rate for the altered enzyme would be substantially slower than forthe normal enzyme. The reaction rate is related to the activation energy, whichis the difference in energy between the trough labeled ES in Figure 316 and thetransition state: the larger the activation energy, the slower the rate. If thealtered enzyme bound the substrate with higher affinity (a lower ES trough),then the activation energy would increase and the reaction would slow down.

369

D. Because an enzyme has a fixed number of active sites, the rate of the reac-tion cannot be further increased once the substrate concentration is suffi-cient to bind to all the sites. It is the saturation of binding sites that leads toan enzymes saturation behavior. The other statements are all true, but noneis relevant to the question of saturation.

370

A. Since k1 corresponds to the on rate and k1 corresponds to the off rate,

Kd = [E][S]/[ES] = koff/kon = k1/k1

B. Km is approximately equal to Kd when kcat is much less than k1; that is tosay, when the ES complex dissociates much more rapidly than substrate isconverted to product. This is true for many enzymes, but not all.

C. Because kcat is in the numerator of the expression, Km will always be some-what larger that Kd. Since lower values of Kd indicate higher binding affinity,Kmwill always underestimate the binding affinity. When kcat is much less thank1, the underestimate will be slight and Kmwill essentially equal Kd.

371 All explanations have at their heart the idea that the quantity of activeenzyme per total protein (the specific activity of the enzyme) is 10-fold lessin bacteria. Such a situation could arise for a number of reasons. 90% of theenzyme may fold incorrectly in bacteria. An essential cofactor of theenzyme, which is normally tightly bound, may be limiting in bacteria so thatonly 10% of the enzyme molecules acquire it. These explanations, whichpropose that there are 10% normally active enzymes among otherwise deadmolecules, account naturally for the observation that the Km is identical (Kmis independent of the concentration of active enzyme) while Vmax is lower(Vmaxis dependent on the concentration of active enzyme).

(One common suggestion is that the enzyme in bacteria folds so that eachmolecule has 10% of the normal activity. This possibility can be ruled outbecause the lower activity of each molecule would show up as a change in

Km as well as Vmax.)

372

A. An enzyme composed entirely of mirror-image amino acids would beexpected to fold stably into a mirror-image conformation; that is, it wouldlook like the normal enzyme when viewed in a mirror.

B. A mirror-image enzyme would be expected to recognize the mirror image ofits normal substrate. Thus, D hexokinase would be expected to add a phos-phate to L-glucose and to ignore D-glucose.

This experiment has actually been done for HIV protease. The mirror-image protease recognizes and cleaves a mirror-image substrate.

Reference:Milton RC, Milton SC & Kent SB (1992) Total chemical synthesisof a D-enzyme: the enantiomers of HIV-1 protease show reciprocal chiral

substrate specificity. Science256, 14451448.


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373 Phosphoglycolate is a transition-state analog for the triosephosphate iso-merase reaction. It has the two characteristics that define a transition-stateanalog: it resembles the reaction intermediate and it binds more tightly(here about 15 times more tightly) than the substrates.

References:Kyte J (1995) Mechanism in Protein Chemistry, pp 207208. NewYork: Garland Publishing.

Pauling L (1948) Chemical achievement and hope for the future.Am. Sci. 36,5058.

374

A. Amino acid side chains in proteins often have quite different pKvalues fromthose in solution. Glu 35 is uncharged because its local environment is non-polar, which makes ionization less favorable (raises its pK). The local envi-ronment of Asp 52 is more polar, permitting ionization near its solution pK.

B. As the pH drops below 5, Asp 52 picks up a proton and becomes nonionized,interfering with the mechanism. As the pH rises above 5, Glu 35 begins torelease its proton, also interfering with the mechanism.

375 Water from rusty pipes provides iron, which is essential for all forms of life.Egg white contains a special protein, ovotransferrin, which binds iron very

tightly, analogous to the binding of biotin by avidin. Washing eggs in rustywater allows iron to enter in sufficient quantities to exceed the bindingcapacity of ovotransferrin, thereby making free iron available to themicroorganisms.

376 This simple question required decades of research to provide a completeand satisfying answer. At the simplest level, hemoglobin binds oxygen effi-ciently in the lungs because the concentration (partial pressure) of oxygen ishighest there. In the tissues the concentration of oxygen is lower because itis constantly being consumed in metabolism. Thus, hemoglobin will tend torelease (bind less) oxygen in the tissues. This natural tendencyan effect onthe binding equilibriumis enhanced by allosteric interactions among thefour subunits of the hemoglobin molecule. As a consequence, much moreoxygen is released in the tissues than would be predicted by a simple bind-

ing equilibrium.

377 Although the rateof diffusion cannot be altered by changes to the enzymes,the average distance over which a molecule must diffuse can be manipu-lated. Linking the two enzymes together decreases the average distance fordiffusion of the first product to the second enzyme. A decrease in the dis-tance reduces the timefor diffusion and, thus, increases the overall rate ofthe reactions catalyzed by the pair of enzymes.

378 When [S] >> Km, the enzyme will be virtually saturated with substrate at alltimes and capable of operating at maximum rate, independent of small fluc-tuations in substrate concentration. For many enzymes that use ATP and asecond substrate, as protein kinases do, the Km for ATP is usually very low (afew mM for most protein kinases) relative to the concentration of ATP in the

cell (12 mM). This situation allows the kinases to operate effectively regard-less of the typical fluctuations in ATP concentration. Under these conditionsthe rate of phosphorylation depends solely on the concentration of theother substrate.

When [S] Km, the rate of the enzyme-catalyzed reaction will vary in pro-portion to the changes in substrate concentration. This is the typical situa-tion for most enzymes involved in metabolic pathways, which allows themto keep up with the changing flow through the pathway.

When [S]


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PROTEIN FUNCTION A

body use hexokinase, which has a low Km, to add phosphate to glucose to ini-tiate its metabolism. By contrast, liver cells use glucokinase, which has a highKm, to carry out this reaction. Between meals the circulating glucose is routedmainly to nonliver cells, which use the low Km enzyme hexokinase. Aftermeals when the circulating concentration of glucose is much higher, the livercaptures a much larger fraction (because glucokinase activity increases muchmore with higher substrate concentration than does hexokinase activity).The liver uses much of that glucose to make glycogen, which serves as a glu-cose reserve for use between meals.

379

C. Cells cannot influence rates of diffusion, which are limited by physicalparameters beyond a cells control. As discussed in Answer 377, cells can

decrease the time it takes for substrates to reach an enzyme by increasingthe concentration of enzyme or by linking related enzymes in multienzymecomplexes.

380 One reasonable proposal would be for excess AMP to feedback inhibit theenzyme for converting Eto F, and excess GMP to feedback inhibit the stepfrom Eto H. Intermediate E, which would then accumulate, would feedbackinhibit the step from R5P to A. Some branched pathways are regulated injust this way. Purine nucleotide synthesis is regulated somewhat differently,however (Figure 348). AMP and GMP regulate the steps from E to F andfrom Eto H, as above, but they also regulate the step from R5P toA. Regula-tion by AMP and GMP at this step might seem problematical since it sug-gests that a rise in AMP, for example, could shut off the entire pathway evenin the absence of GMP. The cell uses a very clever trick to avoid this problem.Individually, excess AMP or GMP can inhibit the enzyme to about 50% of itsnormal activity; together they can completely inhibit it.

381 In resting muscle ATP usage is at a minimum; hence, the group of ATP-likesignal metabolites accumulate. Specific members of this group inhibitglycogen phosphorylase and stimulate glycogen synthase, ensuring thatglycogen reserves are maintained or increased.

In exercising muscle ATP usage is high and AMP-like signal metabolitesincrease. Specific AMP-like signal metabolites stimulate glycogen phospho-rylase and inhibit glycogen synthase, ensuring a breakdown of glycogen toprovide glucose units for ATP production.

382 The substrate, phosphate, and the activator, AMP, both bind to the rarer con-

formation of glycogen phosphorylase, thereby shifting the conformationalequilibrium in favor of the more active species. This makes good biologicalsense because phosphate and AMP concentrations both rise when the cellincreases its rate of ATP hydrolysis, and activation of glycogen phosphory-lase is one way to provide metabolic substrates for the synthesis of addi-tional ATP. In both cases the overall activity of the enzyme increases becausethe fraction of enzymes in the high-activity conformation is increased.

383 The first MWC postulate, which states that the subunits are arranged sym-metrically, rules out all arrangements except those shown in the leftmostand rightmost columns of the diagram. If ligand binds much more tightly tocircles, then the allowed arrangements are those shown in Figure 349. If theligand binds equally to both subunit conformations, then all the arrange-ments in the leftmost and rightmost columns are allowed, consistent with

MWC postulate 1.

R5P A B C D

F G AMP

H I GMP

E

50%

50% 100%

100%

Figure 348 Pattern of inhibition in the

metabolic pathway for purine nucleotidsynthesis (Answer 380).

Figure 349 Arrangements of subunit

conformations that are consistent with

the MWC postulates (Answer 383).

Shadedarea indicates those

arrangements that are excluded by the

MWC postulates for a ligand with affinit

for one conformation of subunit (circle)


14/23


Detailed studies on a few cooperative enzymes such as aspartate transcar-bamoylase have found no evidence for intermediate, nonsymmetrical con-formations.

384 The rate of the metabolic reaction depends on the population of enzymemolecules, not on an individual enzyme molecule. While an individualmolecule is either on or offdepending on whether it is phosphorylatedthe activity of the population of enzyme molecules depends on the propor-

tionof these molecules that are phosphorylated. As the proportion of phos-phorylated molecules increases from 0% to 100%, the activity of the popula-tion of enzymes (the rate of the metabolic reaction) will decrease smoothlyfrom 100% to 0%. The phosphorylation state of a population of enzymemolecules is controlled by the balance between the opposing activities ofprotein kinases, which attach phosphates, and protein phosphatases, whichremove them.

385

E. A mutation that decreases the rate of GTP hydrolysis by Ras would prolongits activated state, leading to excessive stimulation of cell proliferation.Indeed, many cancers contain just such a mutant form of Ras. The mutantRas proteins in all other choices would lead to a decreased ability to trans-mit the downstream signal, thus decreasing cell proliferation. For example,a mutation that increased the affinity of Ras for GDP (choice B) would pro-long the inactivestate of Ras, thereby interfering with the growth signal anddecreasing cell proliferation.

386 A nonfunctional GAP (choice A) or a permanently active GEF (choice D)would allow Ras to remain in the active state (with GTP bound) longer thannormal, and thus might cause excessive cell proliferation.

387

A. In the absence of ATP a motor protein would stop moving. The conforma-tional shifts that are required for movement are triggered by ATP bindingand hydrolysis. In the absence of ATP the motor protein would be stuck inits lowest-energy conformation.

B. If the free-energy change for the hydrolysis of ATP by the motor protein werezeroconditions under which ATP is as easily made as hydrolyzedthemotor protein would wander back and forth. With zero free-energy changethere would be no barrier between conformations.

CALCULATIONS

388

A. The equilibrium constant, K, equals 106 M1.

B. The same calculation as above, when each component is present at 109 M,gives an equilibrium constant of 109 M1.

C. This example illustrates that interacting cellular proteins present at low con-centrations need to bind to one another with high affinities if a high pro-portion of the molecules are to be bound together. A three-order of magni-tude decrease in the equilibrium constant corresponds to a free-energy dif-ference of about 4.2 kcal/mole. Thus, effective binding at the lower con-centration would require the equivalent of 45 extra hydrogen bonds. Thefree-energy difference between the two equilibrium constants can be calcu-lated. For an equilibrium constant of 106 M1,

DG = 2.3 RT log K

K=[AB] 106M

[A][B] (106M) (106M)

K= 106M1

=


15/23

PROTEIN FUNCTION A

Substituting,

DG = 1.41 kcal/mole (6)DG = 8.46 kcal/mole

For an equilibrium constant of 109 M1,

DG = 1.41 kcal/mole (9)

DG = 12.69 kcal/mole

Thus, the higher equilibrium constant corresponds to a free-energy differ-ence that is 4.2 kcal/mole more negative. To supply this amount of bindingenergy with hydrogen bonds (about 1 kcal/mole) would require about 45extra hydrogen bonds.

389 The antibody binds to the second protein with an equilibrium constant, K,of 5 107 M1.

A useful shortcut to problems of this sort recognizes that DG is related tolog K by the factor 2.3 RT, which equals 1.4 kcal/mole at 37 C. Thus, a fac-tor of ten increase in the equilibrium constant (an increase in log Kof 1) cor-responds to a decrease in DG of 1.4 kcal/mole. A 100-fold increase in Kcor-

responds to a decrease in DG of 2.8 kcal/mole, and so on. For each factorof ten increase in K, DG decreases by 1.4 kcal/mole; for each factor of tendecrease in K, DG increases by 1.4 kcal/mole. This relationship allows aquick estimate of changes in equilibrium constant from free-energy changesand vice versa. In this problem you are told that DG increased by 2.8kcal/mole (a weaker binding gives a less negative DG). According to therelationship developed above, this increase in DG requires that Kdecreaseby a factor of 100 (a decrease by 2 in log K); thus, the equilibrium constantfor binding to the second protein is 5 107 M1.

The solution to the problem can be calculated by first determining thefree-energy change represented by the binding to the first protein:

DG = 2.3 RT log K

Substituting for K,

DG = 2.3 (1.98 103 kcal/(K mole)) (310 K) log (5 109)DG = 1.41 kcal/mole 9.7DG = 13.68 kcal/mole

The free-energy change associated with binding to the second protein isobtained by adding 2.8 kcal/mole to the free-energy change for binding tothe first protein, giving a value of 10.88 kcal/mole. Thus, the equilibriumconstant for binding to the second protein is

log K= (10.88 kcal/mole)/(1.41 kcal/mole)= 7.7

K = 5 107 M1

390

A. The equilibrium constants for the two reactions are the same, 108 M1.

K= [AbPr]/([Ab][Pr]) = kon/koff

For the first antibodyprotein reaction,

K= kon/koff= 105 M1 sec1/103 sec1

K= 108 M1


16/23


For the second reaction,

K= kon/koff= 103 M1 sec1/105 sec1

K= 108 M1

B. Since the first reaction has both a faster association rate and a faster disso-ciation rate, it will come to equilibrium more quickly than the second reac-tion.

C. The time it takes for half the complex to dissociate can be calculated fromthe relationship given in the problem:

2.3 log [AbPr]t/[AbPr]0 = kofft

Substituting 0.5 for [AbPr]t/[AbPr]0 and rearranging the equation,

For the first complex, with koff= 103 sec1,

t= 692 seconds, or 11.5 minutes

For the second complex, with koff= 105 sec1, the calculation gives

t= 6.9 104 seconds, or about 19 hours

Thus, the first complex, which falls apart relatively quickly, would be muchmore difficult to work with than the second complex, which falls apart moreslowly. Inappropriate reliance on the equilibrium constant, instead of the offrate constant, can lead an investigator astray in this sort of experiment.

391

A. At equilibrium, the rates of the forward and reverse reactions are equal. Thisis the definition of equilibrium. The overall reaction rate at equilibrium willbe 0.

B. The equilibrium constant equals 103. At equilibrium, the forward andreverse reactions are equal. Thus,

kf[A] = kr [B]

and

kf/kr = [B]/[A] = K

Thus,

K= 104 sec1/107 sec1 = 103

C. Enzyme catalysis does not alter the equilibrium for a reaction; it only speedsthe attainment of equilibrium. Thus, the equilibrium constant is 103. If theequilibrium is unchanged and kfis increased by a factor of 109, then kr mustalso be increased by a factor of 109.

392 At [S] = zero, the rate equals 0/Km and the rate is therefore zero. At [S] = Km,the ratio of [S]/([S] + Km) equals 1/2 and the rate is 1/2 Vmax. At infinite [S]the ratio of [S]/([S] + Km) equals 1 and the rate is equal to Vmax.

393 If Km increases, then the concentration of substrate necessary to give half-maximal rate also increases. At a concentration of substrate equal to the Km

of the unphosphorylated enzyme, the phosphorylated enzyme would have aslower rate; thus, phosphorylation inhibits this enzyme.

2.3 log 0.5t=

koff

2.3 log 0.5t=

koff


17/23

PROTEIN FUNCTION A

394

D. The substrate concentration must be increased by a factor of 16 to increasethe rate from 20% to 80% Vmax. Substituting a rate of 20% Vmax into theMichaelisMenten equation gives

0.2 Vmax= (Vmax) [S]/([S] + Km)

Cancelling Vmaxand multiplying both sides by ([S] + Km) gives

0.2 [S] + 0.2Km = [S]0.8 [S] = 0.2 Km

[S] = 0.25 Km at 20% Vmax

An analogous calculation shows that

[S] = 4 Km at 80% Vmax

Thus, [S] must increase by a factor of 16 (4 Km/0.25 Km) for the rate to gofrom 20% to 80% Vmax.

395 The turnover number for carbonic anhydrase is 6.1 107/min (or 1.0

106

/sec). For this calculation, it is necessary to express the amount of CO 2hydrated and the amount of the enzyme on the same molecular basis, eitheras molecules or moles. For CO2,

For carbonic anhydrase,

The turnover number

396 The numerical value of the product of the Kd values for the substrates is 3.0 107 [(27 106) (11 103)], which is 9-fold greater than the Kd for PALA (2.7 108), suggesting that PALA might be a transition-state analog. PALA, how-ever, is composed not of aspartate plus carbamoyl phosphate, but of succi-nate plus a close analog of carbamoyl phosphate. If one substitutes the Kd forsuccinate, 0.9 mM, into the calculation, the product of the Kd values is 2.4 108 [(27 106) (0.9 103)], which is very close to the Kd for PALA. Thus,PALA is likely to be a bisubstrate analog rather than a transition-state analog.

You may have wondered whether it is valid to compare Kd values in thisway. Recall that DG = 2.3RTlogK. Using this equation, we could have con-verted Kd values to DG values and compared the sum of DG values for

aspartate and carbamoyl phosphate with the DG for PALA. Since DG is pro-portional to logK, this is equivalent to comparing the products of the Kd val-ues for aspartate and carbamoyl phosphate with the Kd for PALA. Do the cal-culation for DG values and convince yourself that this is true.

Reference: Fersht A (1999) Structure and Mechanism in Protein Science, pp360361. New York: WH Freeman.

397

A. An understanding of this phenomenon comes from a consideration of Kd,which for the binding of PALA to aspartate transcarbamoylase (ATC) is

Since the concentration of enzyme is negligible relative to the concentration

0.90 g 6 1023d molecule

min mL g 44 d= 1.23 1022molecules/min mL

10 mg mmole 6 1017molecules

mL 30,000 mg mmole= 2.0 1014molecules/min mL

1.22 1022molecules/min mL

2.0 1014molecules/mL= 6.1 107/mink3= Vmax/[E] =

[PALA][ATC][PALAATC]

Kd= = 2.7 108M


18/23


of added PALA, we can use 2.7 106 as the equilibrium concentration ofPALA. Substituting this value into the Kd expression gives

Thus, in the presence of PALA only 1% of the enzyme is free. This is true for

normal cells and resistant cells. For resistant cells, which have 100 times asmuch enzyme, 1% free corresponds to the amount that is present in unin-hibited normal cells. Thus, they can grow perfectly well in the presence ofPALA.

B. Mutational resistance to an inhibitor requires a subtle change in the enzymethat allows it to decrease its binding affinity for the inhibitor, while not sig-nificantly altering its binding affinity for substrates. This may not be a feasi-ble response to PALA because it so closely resembles the two substrates. Achange that decreases PALA binding will likely decrease substrate binding aswell.

Reference: Wahl GM, Padgett RA & Stark GR (1979) Gene amplificationcauses overproduction of the first three enzymes of UMP synthesis in N-(phosphonacetyl)-L-aspartate (PALA)resistant hamster cells.J. Biol. Chem.254, 86798689.

398

A. The relative concentrations of the normal and mutant Src proteins areinversely proportional to the volumes in which they are distributed. Themutant Src is distributed throughout the volume of the cell, which is

Vcell = (4/3)pr3 = (4/3)p (10 106 m)3 = 4.1888 1015 m3

Normal Src is confined to the 4-nm layer beneath the membrane, which hasa volume equal to the volume of the cell minus the volume of a sphere witha radius 4 nm less than that of the cell:

Vlayer = Vcell (4/3)p (r 4 nm)3= Vcell (4/3)p [(10 106 m) (4 109 m)]3

= (4.1888 1015 m3) (4.1838 1015 m3)Vlayer = 0.0050 1015 m3

Thus, the volume of the cell is 838 times greater than the volume of a 4-nm-thick layer beneath the membrane (4.1888 1015 m3/0.0050 1015 m3).

Even allowing for the interior regions of the cell from which it would beexcluded (nucleus and organelles), the mutant Src would still be a couple oforders of magnitude less concentrated in the neighborhood of the mem-brane than the normal Src.

B. Its lower concentration in the region of its target X at the membrane is thereason why mutant Src does not cause cell proliferation. This notion can be

quantified by a consideration of the binding equilibrium for Src and X:

The lower concentration of the mutant Src in the region of the membranewill shift the equilibrium toward the free components, reducing the amountof complex. If the concentration is on the order of 100-fold lower, theamount of complex will be reduced up to 100-fold. Such a large decrease incomplex formation could readily account for the lack of effect of the mutantSrc on cell proliferation.

Src + XSrc X

K = [Src X][Src][X]

[ATC][PALAATC]

= 0.01


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PROTEIN FUNCTION A

DATA HANDLING

399

A. Your results support the idea that the PI 3-kinase interacts with the activatedPDGF receptor through its SH2 domains. The interaction is blocked specifi-cally by the phosphorylated pentapeptides 708 and 719. In their nonphos-phorylated forms these same pentapeptides do not block the association.

B. The common features of the seven peptides that can bind to PI 3-kinase are

a phosphotyrosine and a methionine (M) located three positions away in theC-terminal direction. (Although not shown explicitly here, there seems to beno requirement for specific amino acids on the N-terminal side of the phos-photyrosine.)

C. Recognition of a couple of amino acids in a short sequence is characteristicof a surfacestring interaction. Indeed, recognition of sequences by SH2domains is often cited as a prime example of such an interaction.

3100 The immunoblot shows that antibodies BPA1 and BPA2 react specificallywith a 220-kd protein, which is likely to be Brca1. By contrast, although anti-body C20 reacts with the same protein, it seems to react even more stronglywith a second protein of about 180 kd. Thus, a likely explanation for the con-tradictory cell-localization experiments is that C20 antibodies were identify-

ing the location of the 180-kd protein, whereas BPA1 and BPA2 were show-ing the location of Brca1. Brca1 is now thought to function in the nucleus.Additional experiments have identified the epidermal growth factor (EGF)receptor as the protein with which C20 cross-reacts. The EGF receptor has acouple of regions of similarity to the peptide that was used to generate theC20 antibody. Cross-reactivity of antibodies is not an uncommon problem.For this reason, cell-localization studies are usually performed with anti-bodies raised against more than one region of a protein. Agreement with dif-ferent antibodies decreases the likelihood that cross-reactivity is a problem.

References: Jensen RA, Thompson ME, Jetton TL, Szabo CI, van der Meer R,Helou B, Tronick SR, Page DL, King MC & Holt JT (1996) BRCA1 is secretedand exhibits properties of a granin. Nat. Genet. 12, 303308.

Thomas JE, Smith M, Rubinfeld B, Gutowski M, Beckmann RP & Polakis P(1996) Subcellular localization and analysis of apparent 180-kDa and 220-kDa proteins of the breast cancer susceptibility gene, BRCA1. J. Biol. Chem.271, 2863028635.

3101

A. The slopes (1/Kd) of the lines in Figure 326 can be estimated by taking thedifference between two points on the y axis divided by the differencebetween the corresponding points on the xaxis. Thus, the slope of line A is

1/Kd = (0.08 0.35)/(5 107 M 1 107 M)= 6.75 105 M1

Kd = 1.5 106 M

For line B,

1/Kd = (0.20 0.90)/(5 107 M 1 107 M)Kd = 5.7 107 M

The precise values are dependent on your estimate of the corresponding val-ues on the xaxis.

B. The lower the Kd, the tighter the binding; thus, the tighter IPTG-bindingmutant of the Lac repressor corresponds to line B (Kd = 5.7 107 M) and thewild-type Lac repressor corresponds to line A (Kd = 1.5 106 M). That alower value corresponds to tighter binding is apparent from the definition ofKd in the problem. Tighter binding will give more complex (PrL) and fewer


20/23


free components (Pr + L); thus, the ratio of concentrations, Kd, will besmaller.

References: Gilbert W & Muller-Hill B (1966) Isolation of the Lac repressor.Proc. Natl Acad. Sci. U.S.A. 56, 18911898.

Kyte J (1995) Mechanism in Protein Chemistry, pp 175177. New York: Gar-land Publishing, 1995.

3102A. When the concentrations of free and bound ligand are equal, their ratio

becomes 1 and the concentration of free protein is equal to Kd:

When [L] equals [PrL],

Kd = [Pr]

B. Visual inspection of the data in Figure 327 shows that the concentrations offree and bound tmRNA are approximately equal when the concentration of

SmpB is 18.8 nM. Thus, Kd is around 20 nM.Not all of the added SmpB protein is free, since some is obviously bound

to tmRNA. But because tmRNA was included at a concentration of 0.1 nM,the bound fraction at an SmpB concentration of 18.8 nM is only0.05 nM. Thus, the correction for bound SmpB is minuscule (less than 1%)and can be neglected.

C. It is critical in this kind of experiment that the tmRNA concentration be keptwell below Kd. If the concentration of tmRNA had been 100 nM, for example,the shift to 50% bound would have occurred at around 50 nM SmpB (andmost of the protein would have been in the bound, not the free, form). IftmRNA were included at 100 mM, the shift to 50% bound would haveoccurred at around 50 mM SmpB. Thus, if tmRNA were included at a con-centration above Kd, the point at which 50% was shifted to the bound form

would bear no relationship to Kd.

Reference: Karzai AW, Susskind MM & Sauer RT (1999) SmpB, a uniqueRNA-binding protein essential for the peptide-tagging activity of SsrA(tmRNA). EMBO J. 18, 37933799.

3103 The calculated values of fraction bound versus protein concentration areshown in Table 36. Also shown are rule-of-thumb values, which are easierto remember (see the answer to Question 254).

These relationships are useful not only for thinking about Kd, but also forenzyme kinetics, which we cover in other problems. The rate of a reactionexpressed as a fraction of the maximum rate is

which has the same form as the equation for fraction bound. Thus, when theconcentration of substrate, [S], is 10-fold above the Michaelis constant, Km,the rate is 90% of the maximum, Vmax. When [S] is 100-fold below Km, therate is 1% of Vmax.

The relationship also works for the fractional dissociation of an acidicgroup, HA, as a function of pH. When the pH is 2 units above pK, 99% of theacidic group is ionized. When the pH is 1 unit less than pK, 10% is ionized.

3104

A. Since the concentration of Lac repressor is 105 times the Kd for binding, youwould expect 99.999% of the sites in a bacterial population to be occupiedby the Lac repressor.

B. When inducer is present, the concentration of Lac repressor will be only

Table 36 Calculated values forfraction bound versus protein

concentration (Answer 3103).

[PROTEIN] FRACTION RULE-OF-

BOUND (%) THUMB

104 Kd 99.99 99.99

103 Kd 99.9 99.9

102 Kd 99 99

101 Kd 91 90

Kd 50 50

101 Kd 9.1 10

102 Kd 0.99 1

103 Kd 0.099 0.1

104 Kd 0.0099 0.01

[Pr][L][PrL]

Kd=

=V

max

[S]rate[S] + K

m


21/23

PROTEIN FUNCTION A

100 times more than the Kd, but you would still expect 99% of the sites to beoccupied.

C. If 99% of the binding sites were occupied by the repressor even in the pres-ence of the inducer, you would expect that the genes would still be veryeffectively turned off. This sort of straightforward calculation, and its non-biological answerafter all, the genes are known to be turned on by theinducertells you that some critical information is missing.

D. Low-affinity, nonspecific binding of the Lac repressor is the missing infor-

mation suggested by the calculation in part C. Since there are 4 106 non-specific binding sites in the genome (a number equal to the size of thegenome), there is a competition for repressor between the multitude of low-affinity sites and the single high-affinity site. This competition reduces theeffective concentration of the repressor. As can be calculated, the competi-tion reduces repressor occupancy at the specific site to about 96% in theabsence of lactose, and to about 3% in the presence of lactose. These num-bers account nicely for the genes being turned off in the absence of lactoseand turned on in its presence.

3105

A. It is important that only a small quantity of product is made, because other-wise the rate of the reaction would decrease as the substrate was depleted

and product accumulated. Thus, the measured rates would be lower thanthey should be, and the kinetic parameters would be incorrect.

B. The MichaelisMenten plot, shown in Figure 350, is a rectangular hyper-bola, as expected if this enzyme obeys MichaelisMenten kinetics. To deter-mine values for Km and Vmax from this plot by visual inspection, you mustestimate the rate at infinite substrate concentration. From the curve of theline in the figure you might reasonably estimate Vmaxas anywhere from 1.8to 2.0 mmol/min. (As developed in the answer to Problem 3103, a usefulrule of thumb is that at a concentration of substrate 10-fold above Km, therate is about 90% of Vmax.) If you chose 2.0 mmol/min, then 0.5 Vmax(1.0 mmol/min) corresponds to a substrate concentration of 1.0 mM, which isthe value of Km. The visual uncertainty in this plot led early researchers totransform the equation into a straight-line form so a line could be fitted to

the data and the kinetic parameters could be more accurately determined.C. As indicated in the LineweaverBurk plot in Figure 350, theyintercept is 0.5(1/Vmax) and the xintercept is 1.0 (1/Km). Thus, Vmaxequals 2.0 mmol/minand Km equals 1.0 mM. Although this form of a straight-line plot is commonlydiscussed in textbooks, it is rarely used in practice because the data pointsthat are most reliable are tightly grouped at one end of the line. Conse-quently, the slope of the line is unduly influenced by the low (and usuallyless accurately determined) rates at low substrate concentration. Otherstraight-line transformations of the MichaelisMenten equation such as theEadieHofstee plot, which is analogous in form to the Scatchard plot shownin Problem 3101, are generally preferred. In this era of computers, however,the data can be fitted perfectly well to the nonlinear MichaelisMentenequation, although it is still common to present such data in a linear form.

3106

D. Since NAM cannot occupy site C, that site must normally be occupied byNAG; and since the cell-wall polysaccharide is an alternating polymer ofNAM and NAG, the NAM monomers must occupy sites B, D, and F. Becausecleavage occurs after NAM monomers, the site of cleavage must be betweensites B and C or between sites D and E. Since tri-NAG occupies sites AC butis not cleaved, whereas longer NAG polymers are, the catalytic groups forcleavage must lie between sites D and E.

3107

A. Binding of aspartate normally shifts the conformation of ATCase from thelow-activity to the high-activity state. At low aspartate concentrations notall of the ATCase will have been shifted to the high-activity conformation.

The peculiar activating effect of malate occurs because its binding helps

0 5.0 10

0.5

0

1.0

1.5

2.0

rate(mol/min)

S (M)

MichaelisMenten plot

1/rate(min/mol)

1/S (1/M)

2

4

6

8

5 10 15

1.0

0.5

0

LineweaverBurk plot

Figure 350 MichaelisMenten andLineweaverBurk plots of the data in

Table 33 (Answer 3105). Thexandyintercepts are indicated on the

LineweaverBurk plot.


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complete the shift of ATCase to the high-activity conformation. In the pres-ence of a low concentration of malate the number of active sites in the high-activity conformation increases; thus, enzyme activity increases.

B. This peculiar activating effect of malate is not observed at high aspartateconcentrations because ATCase is already entirely shifted to its high-activityconformation. Under these conditions each molecule of malate that bindsto an active site will reduce the total number of sites accessible to aspartateand thus reduce the overall activity of ATCase.

References: Cantor CR & Schimmel PR (1980) Biophysical Chemistry, pp944945. New York: WH Freeman.

Gerhart JH & Pardee AB (1963) The effect of the feedback inhibitor, CTP, onsubunit interactions in aspartate transcarbamylase. Cold Spring HarborSymp. Quant. Biol. 28, 491496.

3108 These changes are exactly what you would expect for an allosteric enzymesuch as ATCase. Because binding at one active site (one of six per ATCasemolecule) is sufficient to shift the conformation of a molecule of ATCase, thechange in global conformation (change in sedimentation) is expected tolead the change in occupancy of binding sites (change in spectral mea-surement).

References: Cantor CR & Schimmel PR (1980) Biophysical Chemistry, pp954956. New York: WH Freeman.

Kirschner MW & Schachman HK (1973) Local and gross conformationalchanges in aspartate transcarbamylase. Biochemistry12, 29973004.

3109

A. Both cyclin A and phosphorylation of Cdk2 are required to activate Cdk2 forefficient phosphorylation of histone H1 (see Figure 332, lane 5). Absence ofcyclin A (lane 1) or absence of phosphorylation of Cdk2 (lane 3) results inmuch reduced levels of histone H1 phosphorylation.

B. Cyclin A, which binds tightly to both forms of Cdk2 (Kd = 0.05 mM), dramat-ically improves the binding of both forms to histone H1. In the absence of

cyclin A, P-Cdk2, for example, binds histone H1 with a Kd of 100 mM, whereasin the presence of cyclin A, it binds histone H1 with a Kd of 0.7 mM, anincrease in the tightness of binding of more than a factor of 100. In addition,as shown in Figure 332, phosphorylation of Cdk2 activates its proteinkinase activity, allowing it to phosphorylate histone H1, when cyclin A is pre-sent to increase its ability to bind to histone H1.

C. Given that the intracellular concentrations of ATP and ADP are more than10-fold higher than the measured dissociation constants, the changes inaffinity for ATP and ADP are unlikely to be critical for the function of Cdk2.The binding sites for ATP will be nearly saturated regardless of the phospho-rylation state of Cdk2. ADP, which binds at the same site as ATP, is unlikely tointerfere significantly with ATP binding, because ADP has a higher Kd and itscellular concentration is generally lower than that of ATP.

Reference:Brown NR, Noble MEM, Lawrie AM, Morris MC, Tunnah P, DivitaG, Johnson LN & Endicott JA (1999) Effects of phosphorylation of threonine160 on cyclin-dependent kinase 2 structure and activity. J. Biol. Chem. 274,87468756.

3110

A. Mutant 2 is Asp-181Ala (D181A). It is the best candidate because it has aKm close to that of the wild-type enzyme, but a very low turnover number(kcat). With these kinetic parameters it might be expected to bind normallyto its target substrates but not remove phosphate from tyrosine. The nextmost likely candidate would be Arg-221 Lys, which has a slightly lower Kmthan wild-type PTP1B and turns over slowly, although about 20 times fasterthan D181A. (Further studies identified the band at 180 kd as the epidermal

growth factorEGFreceptor.)


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PROTEIN FUNCTION A

B. C215S showed no activity, as expected since the SH group of cysteine isrequired for catalysis. Because C215S was not active, it was not possible todetermine its Km, which might have been similar to that of the wild-typeenzyme. (Measurements of Kd were not made.) Thus, C215S was a reason-able candidate to test. Lack of success with C215S suggests that it bindsphosphotyrosine-containing proteins very poorly.

Reference: Flint AJ, Tiganis T, Barford D & Tonks NK (1997) Development of

substrate-trapping mutants to identify physiological substrates of proteintyrosine phosphatases. Proc. Natl Acad. Sci. U.S.A. 94, 16801685.

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