B. Alberts, A. Johnson, J. Lewis и др. - Molecular Biology of The Cell (6th edition) (1120996), страница 65
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Why is it better to search with a short sequencethan with a long sequence? Do you not have more chancesfor a “hit” in the database with a long sequence?3–10 The so-called kelch motif consists of a fourstranded β sheet, which forms what is known as a β propeller. It is usually found to be repeated four to seven times,forming a kelch repeat domain in a multidomain protein.β2β4β33–11 Titin, which has a molecular weight of about3 × 106, is the largest polypeptide yet described.
Titinmoleculesextendfrom muscle thick filaments to theProblemsp3.06/3.06Z disc; they are thought to act as springs to keep the thickfilaments centered in the sarcomere. Titin is composed of alarge number of repeated immunoglobulin (Ig) sequencesof 89 amino acids, each of which is folded into a domainabout 4 nm in length (Figure Q3–2A).You suspect that the springlike behavior of titin iscaused by the sequential unfolding (and refolding) of individual Ig domains. You test this hypothesis using the atomicforce microscope, which allows you to pick up one end ofa protein molecule and pull with an accurately measuredforce. For a fragment of titin containing seven repeats of theIg domain, this experiment gives the sawtooth force-versus-extension curve shown in Figure Q3–2B.
If the experiment is repeated in a solution of 8 M urea (a protein denaturant), the peaks disappear and the measured extensionbecomes much longer for a given force. If the experimentis repeated after the protein has been cross-linked by treatment with glutaraldehyde, once again the peaks disappearbut the extension becomes much smaller for a given force.(A)NC(B)400force (pN)3–4An enzyme reaches a maximum rate at high substrate concentration because it has a fixed number ofactive sites where substrate binds.NFigure Q3–1 Thekelch repeat domain ofgalactose oxidase fromD. dendroides (Problem3–10). The seven individualβ propellers are colorcoded and labeled.
TheN- and C-termini areindicated by N and C.3002001000050100150200extension (nm)Figure Q3–2 Springlike behavior of titin (Problem 3–11). (A) Thestructure of an individual Ig domain. (B) Force in piconewtons versusextension in nanometers obtained by atomic force microscopy.Problems p3.12/3.11CHAPTER 3 END-OF-CHAPTER PROBLEMS3–12 Rous sarcoma virus (RSV) carries an oncogenecalled Src, which encodes a continuously active proteintyrosine kinase that leads to unchecked cell proliferation.Normally, Src carries an attached fatty acid (myristoylate)group that allows it to bind to the cytoplasmic side of theplasma membrane. A mutant version of Src that does notallow attachment of myristoylate does not bind to themembrane. Infection of cells with RSV encoding either thenormal or the mutant form of Src leads to the same highlevel of protein tyrosine kinase activity, but the mutant Srcdoes not cause cell proliferation.A.Assuming that the normal Src is all bound to theplasma membrane and that the mutant Src is distributedthroughout the cytoplasm, calculate their relative concentrations in the neighborhood of the plasma membrane.
Forthe purposes of this calculation, assume that the cell is asphere with a radius (r) of 10 μm and that the mutant Srcis distributed throughout the cell, whereas the normal Srcis confined to a 4-nm-thick layer immediately beneath themembrane. [For this problem, assume that the membranehas no thickness. The volume of a sphere is (4/3)πr3.]B.The target (X) for phosphorylation by Src resides inthe membrane. Explain why the mutant Src does not causecell proliferation.3–13 An antibody binds to another protein with anequilibrium constant, K, of 5 × 109 M–1.
When it binds toa second, related protein, it forms three fewer hydrogenbonds, reducing its binding affinity by 11.9 kJ/mole. Whatis the K for its binding to the second protein? (Free-energychange is related to the equilibrium constant by the equation ΔG° = –2.3 RT log K, where R is 8.3 × 10–3 kJ/(mole K)and T is 310 K.)3–14 The protein SmpB binds to a special species oftRNA, tmRNA, to eliminate the incomplete proteins madefrom truncated mRNAs in bacteria. If the binding of SmpBto tmRNA is plotted as fraction tmRNA bound versus SmpBconcentration, one obtains a symmetrical S-shaped curveas shown in Figure Q3–3.
This curve is a visual display ofa very useful relationship between Kd and concentration,which has broad applicability. The general expression forfraction of ligand bound is derived from the equation forKd (Kd = [Pr][L]/[Pr–L]) by substituting ([L]TOT – [L]) for[Pr–L] and rearranging. Because the total concentration ofligand ([L]TOT) is equal to the free ligand ([L]) plus boundligand ([Pr–L]),fraction bound = [Pr–L]/[L]TOT = [Pr]/([Pr] + Kd)Figure Q3–3 Fractionof tmRNA bound versusSmpB concentration(Problem 3–14).1.0fraction boundA.Are the data consistent with your hypothesis thattitin’s springlike behavior is due to the sequential unfolding of individual Ig domains? Explain your reasoning.B.Is the extension for each putative domain-unfolding event the magnitude you would expect? (In anextended polypeptide chain, amino acids are spaced atintervals of 0.34 nm.)C.Why is each successive peak in Figure Q3–2B a little higher than the one before?D.Why does the force collapse so abruptly after eachpeak?1710.750.50.25010–1110–910–710–5concentration of SmpB (M)For SmpB and tmRNA, the fraction bound = [SmpB–tmRNA]/[tmRNA]TOT = [SmpB]/([SmpB] + Kd).
Usingthis relationship,the fraction of tmRNA boundProblemscalculatep3.28/3.24for SmpB concentrations equal to 104 Kd, 103 Kd, 102 Kd,101 Kd, Kd, 10–1 Kd, 10–2 Kd, 10–3 Kd, and 10–4 Kd.3–15 Many enzymes obey simple Michaelis–Mentenkinetics, which are summarized by the equationrate = Vmax [S]/([S] + Km)where Vmax = maximum velocity, [S] = concentration ofsubstrate, and Km = the Michaelis constant.It is instructive to plug a few values of [S] into theequation to see how rate is affected. What are the rates for[S] equal to zero, equal to Km, and equal to infinite concentration?3–16 The enzyme hexokinase adds a phosphate toD-glucose but ignores its mirror image, L-glucose. Supposethat you were able to synthesize hexokinase entirely fromD-amino acids, which are the mirror image of the normalL-amino acids.A.Assuming that the “D” enzyme would fold to a stable conformation, what relationship would you expect it tobear to the normal “L” enzyme?B.Do you suppose the “D” enzyme would add aphosphate to L-glucose, and ignore D-glucose?3–17 How do you suppose that a molecule of hemoglobin is able to bind oxygen efficiently in the lungs, and yetrelease it efficiently in the tissues?3–18 Synthesis of the purine nucleotides AMP andGMP proceeds by a branched pathway starting with ribose5-phosphate (R5P), as shown schematically in FigureQ3–4.
Using the principles of feedback inhibition, proposea regulatory strategy for this pathway that ensures an adequate supply of both AMP and GMP and minimizes thebuildup of the intermediates (A–I) when supplies of AMPand GMP are adequate.R5PABCDFGAMPHIGMPEFigure Q3–4 Schematic diagram of the metabolic pathway forsynthesis of AMP and GMP from R5P (Problem 3–18).Problems p3.19/3.18172Chapter 3: ProteinsREFERENCESGeneralBerg JM, Tymoczko JL & Stryer L (2011) Biochemistry, 7th ed. NewYork: WH Freeman.Branden C & Tooze J (1999) Introduction to Protein Structure, 2nd ed.New York: Garland Science.Dickerson RE (2005) Present at the Flood: How Structural MolecularBiology Came About. Sunderland, MA: Sinauer.Kuriyan J, Konforti B & Wemmer D (2013) The Molecules of Life:Physical and Chemical Principles.
New York: Garland Science.Perutz M (1992) Protein Structure: New Approaches to Disease andTherapy. New York: WH Freeman.Petsko GA & Ringe D (2004) Protein Structure and Function. London:New Science Press.Williamson M (2011) How Proteins Work. New York: Garland Science.The Shape and Structure of ProteinsAnfinsen CB (1973) Principles that govern the folding of protein chains.Science 181, 223–230.Caspar DLD & Klug A (1962) Physical principles in the construction ofregular viruses.
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