<p> Name: ______MCB 111/211 Midterm 2 2005</p><p>DG = - RT ln K k = (kT/h) e-DG=/RT R = 1.987 cal/deg-mole</p><p>1. Find the letter below that best matches the following statements. Use a letter only once. (10 pts.)</p><p>A. Dead-End Elimination B. ribosome display C. PHD D. m value E. often largely an effect on the entropy change of a reaction F. 110 M G. 1 mM H. protein stability curve I. Tm J. Cm K. pKa changes between folded and unfolded states L. hydrophobic effect M. assumption in Chou-Fasman method for predicting secondary structure N. assumption in the 3D-1D profile (threading) method for predicting structure O. free energy of transfer from water to octanol P. heat required to change the temperature by 1 °K Q. Polar interactions i) __L___ accounts for much of the temperature dependence of protein stability ii) __P___ Cp iii) __H___ small difference between large, nearly balancing effects iv) __E___ effective concentration v) __A___ finds the global energy minimum sequence in computational protein design vi) __N___ the protein folds vii) __J___ midpoint of solvent-induced denaturation curve viii) __O___ measure of the hydrophobic effect ix) __M___ interactions with residues nearby in the sequence determine structure x) __K___ measure of the contribution of electrostatic interactions to protein stability</p><p>2. HIV and RSV are related viruses that each encode a protease that cleaves their respective Gag polyproteins. The sizes of the proteases are 98 (HIV) and 124 residues (RSV). BLAST reveals that they contain a segment of 43 residues with 19 identities. The E-value for the alignment is 6.5. </p><p>2a. What is the definition of the E-value? (4 pts.)</p><p>The number of hits of that quality expected using a random sequence to search the database.</p><p>1/5 Name: ______</p><p>2b. Do you conclude that the HIV and RSV proteases are homologous or not? Why or why not? (5 pts.)</p><p>No. The sequence match is likely to be random. The e-value is >0.001.</p><p>Or</p><p>Yes. HIV and RSV are related.</p><p>2c. Suggest a way to test your answer to 2b. Using that method, what results would expect if the folds are similar? (6 pts.)</p><p>Threading. E-value <0.001 or Z-score >7.</p><p>Solve the structures. Ca rmsd <2.5 Å.</p><p>3. Dahiyat and Mayo (Science 278, 82 (1997)) designed a sequence (called FSD-1) that adopted a zinc-finger fold without binding zinc. The zinc-binding residues (Cys2His2) in the natural protein (Zif268) were replaced by Ala, Lys, Phe and Phe. More hydrophobic surface area was buried in the folded FSD-1 structure than in the wild-type protein (which had zinc in the core). </p><p>3a. Although the design calculations examined ~2 x1027 sequences, over 1062 structural variants were ranked computationally. What accounts for the much larger number of structural variants than sequence variants? (6 pts.)</p><p>Most residues were tried in multiple rotamers.</p><p>3b. Assume that unfolding the natural zinc finger causes release of the Zn2+ ion. How do you expect the replacement of the Zn2+ site in the natural protein by hydrophobic residues in FSD-1 to influence the Cp of unfolding? Justify your answer. (6 pts.)</p><p>Cp goes up because the core of FSD-1 is more hydrophobic. The hydrophobic effect has a large, positive Cp.</p><p>2/5 Name: ______</p><p>3c. Briefly describe two ways you would measure Cp of unfolding. (6 pts.)</p><p>DSC. Cp corresponds to the offset of the folded and unfolded baselines.</p><p>Measure H at different temperatures. H (T) = H (T0) + Cp (T-T0) Cp = (H (T) - H (T0)) / (T-T0)</p><p>3d. (9 pts.) Considering only the effects of replacing the Zn2+ site in the natural protein by hydrophobic residues in FSD-1, do you expect Zif268 or FSD-1 to be more destabilized (or less stabilized) in: i) 20% ethanol (vs. no ethanol) FSD-1 ii) 1 M ammonium sulfate (vs. low salt) Zif268 iii) 0 °C (low temperature vs. room temperature) FSD-1</p><p>3e. FSD-1 and the natural zinc finger are both 28 amino acids long. Assuming each residue gains 12 degrees of freedom upon unfolding, estimate the Gconf for unfolding at 37 °C. (8 pts.)</p><p>28 S = R lnWu = R ln12 = R x 28 ln12 cal/mol-deg = 138.25 cal/mol-deg</p><p>O Gconf = -TSconf = -310 K x 138.25 cal/deg-mole = -42.858 kcal/mol</p><p>3f. At 20 °C, the stability (Gu) of FSD-1 was 1.2 kcal/mole. Assuming a two-state model, what is the ratio of unfolded protein to folded protein at this temperature? (8 pts.)</p><p>Gu = -RT ln K, where K = [U]/[F]</p><p>K = e -Gu/RT = e -1200/(1.987 x 293) = 0.127</p><p>3/5 Name: ______</p><p>NOT COVERED IN 2006 4. Bolon and Mayo (Proc Natl Acad Sci Figure 1. A. Mechanism of ester USA 98, 14274 (2001)) used the same hydrolysis. Nu = nucleophile. B. design method to make a mutant of the Transition state analog involving a protein thioredoxin that catalyzed histidine nucleophile. hydrolysis of an ester substrate. They created the design by calculating the location in the protein that would accept a model for the transition state of the reaction in which a histidine attacked the ester bond (Fig. 1). The best designed mutant contained the changes Phe12Ala, Leu17His and Tyr70Ala. </p><p>4a. What is the logic of using a model of the transition state in this way to create an enzyme? (6 pts.)</p><p>1. The enzyme binds the transition state tighter than the ground states.</p><p>2. The enzyme changes the chemical mechanism of the the reaction (His nucleophile vs. H2O).</p><p>4b. The rate of the catalyzed reaction was 180-fold faster than the uncatalyzed reaction at 20 °C. By how much did the mutant thioredoxin lower the activation energy for the reaction? (8 pts.)</p><p>≠ ≠ ≠ ≠ kcat/kuncat = 180 = exp ((-G cat - -G uncat)/RT) = exp ((G uncat - G cat)/RT)</p><p>≠ ≠ RT ln 180 = (G uncat - G cat) = 3.023 kcal/mole</p><p>5. A 23-residue fragment of the transcriptional activator CREB folds upon binding to the coactivator CBP. Ser133 of CREB is phosphorylated. At 25° C, the pKa for protonation of the di-anionic phosphoryl group is 5.4 in free CREB and 4.3 in the CREB/CBP complex. </p><p>5a. What is the free energy of the electrostatic interaction of CBP with the Ser133-phosphoryl group? (Mestas & Lumb, Nature Structural Biology 6, 613 (1999)). (6 pts.) pKa = -log Ka = 2.303 ln Ka, 10-pKa = Ka</p><p>G = -RT (ln Ka(free) – ln Ka(bound))</p><p>= -RT ln (Ka(free)/ Ka(bound))</p><p>= -RT ln (10-5.4/10-4.3) = -1.5 kcal/mole</p><p>4/5 Name: ______</p><p>5b. Does phosphorylation of CREB favor or oppose CBP binding? (6 pts.)</p><p>Favor</p><p>5c. Suggest two possible mechanisms by which CREB phosphorylation could regulate CBP binding. (6 pts.) NOT COVERED IN 2006</p><p>1. CBP directly contacts the PO4 in the complex.</p><p>2. Long range electrostatic intereactions with CBP</p><p>3. PO4 promotes the folding of CREB to favor CBP binding (allosteric interaction)</p><p>5/5</p>