Exam 1 Key - Tesmer Keys - Fall 2007

Name:_______________________ CB501 Exam I ? Oct 4, 2007 Protein Stability & Folding (potentially the most important exam of your life) Dr. John Tesmer TTH 9-11am 1) You must show your work to get full credit 2) Box your answers when appropriate 3) The test is closed book, closed notes 4) If your answers lack units, you will not get points 5) Explain every answer PLEASE READ THE QUESTIONS CAREFULLY. IF YOU DON?T UNDERSTAND A QUESTION?PLEASE ASK ME FOR CLARIFICATION Useful Constants: R = 8.315 J/?K?mol k b = 1.38 x 10 -23 J/?K 1 cal = 4.184 J h = 6.626 x 10 -34 J?s N A = 6.022x10 23 Good luck! Page 2 1) Permanent dipoles (as found in an asymmetric molecule like water) have partial positive and negative charges associated with the dipole moment. However, the electrostatic interaction energy of such dipoles are not worth as many kJ/mol per unit charge as is the interaction between two point charges. Why? (5 pts) Dipoles have both positive and negative charges, and while one of these point charges will attract the opposite charge on another dipole, it will also be repelled by the same charge on another dipole. 2)Hydrogen bonds can in theory provide up to ~20 kJ/mol of free energy, as calculated from model compounds. However, in proteins, mutations that eliminate hydrogen bonds (without perturbing surrounding structure) often only contribute 2-5 kJ/mol to the stability of the protein. Explain why. (5 pts) Stability is determined by the relative free energy of the unfolded versus the folded state. Groups that can hydrogen bond interact favorably through hydrogen bonds with water in the unfolded state, nearly as well as they can with another side chain or backbone in the folded protein. Thus the stabilization from two groups that hydrogen bond is simply the difference in free energy between these two states. 3)For each of the following reagents/conditions, explain the basis for how they destabilize the native state of a protein (4 pts each part). a)Temperature As temperature goes up, the entropy term in ?G = ?H ? T?S begins to dominate. Because the folded state has lower entropy than the unfolded, the protein will thus tend to unfold as the temperature goes up. b)Urea Denaturants such as urea have both nonpolar and polar character, and they can therefore interact favorably with both polar and nonpolar groups. At high concentrations, they can compete for the non- covalent interactions holding the protein core together. c) If one monitored denatured by circular dichroism (CD), draw the expected thermal denaturation curve for a typical single domain ?-helical protein. Be sure to label the axes. Here is a denaturation curve from class. Units on the Y axis should be molar ellipicity at 220 nM (1 pts) and on the X axis should be increasing temperature. There should be a region of smooth cooperative transition, increasing from low values to high as the protein denatures (2 pts). Before and after denaturation, the curve should be linear (but not necessarily slope of 0) (1 pt). Page 3 (problem 3 continued) d)Would you expect the melting point (T m ) to be the same or different if one monitored the thermal denaturation by fluorescence? Explain. The same. The melting point is dependent on the thermodynamics of the protein, not the method used to measure it. 4) When one puts a nonpolar amino acid into water, it has a significantly higher partial specific volume (cm 3 /g) than an equivalently sized polar amino acid. a) What is the molecular basis for this effect, and what ramifications does it have for the stability of a protein as a function of temperature? (4 pts) It reflects the fact that water tends to form hydrogen-bonded ?cages? around the non-polar portions of the amino acid (2 pts). It amounts to a large change in heat capacity as the protein unfolds. Both ?H and ?S are not invariant with temperature in folding reactions, but change as a function of ?Cp. (1 pt). This dependence of ?G on the heat capacity leads to both cold and heat denaturation of the protein, and a temperature of maximal stability (1 pt). b) Does this effect stabilize the folded or unfolded state of a protein? Explain your reasoning. (3 pts) It stabilizes the unfolded state by maximizing the non-covalent interactions between solvent and exposed hydrophobic groups. (3 pts) There is of course an entropic price to do this, and thus the ability of water to do this ?favor? for unfolded proteins is a function of temperature. Page 4 5)The rate of any reaction (k) is related to the free energy barrier of its transition state. The Eyring equation relates the rate to the thermodynamic parameters ?H ? and ?S ? (the activation enthalpy and entropy). This is very similar to the Arrhenius equation. It is not important here that you understand how this equation is derived, but note that a plot of ln(k/T) versus 1/T (Eyring plot) is expected to give a straight line, with slope and intercept determined by ?H ? and ?S ? , respectively. The photosynthetic reaction center (PSII) is an integral membrane protein that transfers electrons across membranes. The Eyring plot for the rate of electron transfer has a surprising non-linear form, consisting of two distinct lines. This transition between the lines is thought to represent a change in rate limiting step (occurring at T thermo or T meso ) for the electron transfer. At higher (ambient) temperatures, the ln(k/T) term is constant. This allows the rate of electron transfer to be relatively constant over the expected range of temperatures that the organism is expected to exist at, which is important for the health of these organisms. Under the range of temperatures shown, the protein does NOT denature. A) What features/properties of a protein structure (or transition state) determine (1) ?H and (2) ?S? (6 points) ?H is determined by the relative enthalpy of the non-covalent interactions of the folded vs. the unfolded state (when considering stability), or between the ground state and transition state (when considering transition state). 3 pts ?S is determined by the relative disorder/conformational freedom of these two states. 3 pts B) Comparing the plots for the thermophilic and mesophilic PS II enzymes, how different are their non-covalent interactions in the transition state? (3 pts) The two Eyring plots have identical slopes, and thus they have the same ?H ? and the same net enthalpic stabilization in the transition state. C) At temperatures below the transition temperature for each protein (please note that the x-axis is 1/T), is the transition state of the thermophile more or less dynamic than that of the mesophile (here you might want to pull out your calculator)? Is this what you would expect? Explain. (3 pts) ?S ? is determined by the y-intercept when 1/T is 0. ?S ? = R(ln(k/T) ? ln(k B /h)) = R(ln(k/T) ? 23.8). The ln(k/T) term will intercept the y axis around ~2.8 for the thermophile, and ~3.5 for the mesophile, meaning that ?S ? will negative for this system, and that the thermophile will have a more negative ?S ? than the mesophile. That is, there is a bigger entropic penalty for the thermophile to enter the transition state, which is presumably even more rigid than its ground state. One might have expected a mesophile, with a less entropically constrained ground state, to incur a larger penalty upon entering the transition state (3 pts) To be technically correct, one cannot really answer the question ?which is more dynamic? because ?S only tells you the relative difference between states, and not the final entropy, of the transition state. I?ll give you credit if you answer this way too (but I still expect the math!). Filled circles: mesophile Open circles: thermophile T thermo T 0 T meso T 0 Page 5 Problem 5 continued D) To try to understand the basis for the difference in the transition temperature between the mesophile and thermophile PSII, the authors of this study decided to target a GXXXG-like motif found in one of the transmembrane helices in the dimer interface of PSII (which is a constitutive dimer) by site directed mutagenesis. They chose to substitute a serine found at the ?G? position of this motif with either an alanine or a valine. What effect do you predict each of substitutions will have on the stability of PSII? Explain. Consider both enthalpy and entropy. (5 pts) GXXXG motifs are found in transmembrane helices to optimize packing/noncovalent interactions between them. The S->A mutation will enable closer packing (decreasing ?H of folding) and should also increase ?S of folding (favoring folding) because one does not have to give up conformation freedom of the alanine side chain when it packs. The opposite trends expected for the S->V substitution. E) Would the transition temperature T meso shift to lower or higher temperatures as a consequence of each of these substitutions? Why? (3 pts) T meso might be expected to shift to higher temps with the S->A, because the protein is more stable and it may take higher temperatures to reach the transition point. The opposite is expected for S->V. F) At the physiological temperatures of each organism (above the transition temperature), the slope of the Eyring plot is essentially 0. What does this imply about the mechanism by which the rate of electron transfer is controlled at these temperatures? (3 pts) This is a striking example of non-Arrhenius behaviour. The slope is zero, implying that ?H ? is zero (but the rate is of course still temperature dependent). The rate is solely determined by the relative entropy of the transition state to that of the ground state?.BTW: one way of thinking about this is that at high temperatures, the noncovalent interactions of both the ground and transition state are the same, perhaps because the ground state has less optimized packing (is more ?fluid?) at higher temperatures. That is, PSII appears to have evolved a mechanism to keep its transition state structure as ?fluid/non-optimal? as its ground state. In this example, PSII appears to do it by having conserved cavities inside the transmembrane domains. G) At lower temperatures than that of the transition temperature, the Eyring plot has a negative slope. What does this imply about the mechanisms by which the rate of electron transfer is limited at these temperatures? (3 pts) As discussed above, the rate is dependent on both ?H ? (disruption of more optimal non-covalent interactions on the way to the transition state) as well as the change in order of the transition state (?S ? ). At higher T, no additional heat needs to be put into break the limiting non-covalent interactions as the transition state is approached. As a footnote, in PSII and enzymes in which hydrogen tunneling is expected to occur, the rate appears dependent on ?islands? of local flexibility or non-optimal packing in the protein, which allow the transfer to take place. Changes in the protein structure that impact the flexibility of these islands thus impact the rates, as shown in this study. Page 6 6)A student is studying the stability and folding pathway of a small, well-behaved protein domain. A) By comparing sequences, the student notices that there is an invariant glutamate and lysine in the protein. By mutating either residue to alanine, the stability of the protein is decreased, as measured by chemical denaturation. Is this the expected result if these residues play a functional role in the domain? Explain. (5 pts) Functional residues (i.e. catalytic) are often destabilizing. The fact that stability is lost upon mutation of these residues implies that they are not functional, but rather structural. B) The protein is so interesting that the student decides to determine its structure by NMR. Looking at the spectrum, the student identifies the proton that residues on the glutamate side chain at low pH. By changing the pH, the student notes that the proton signal disappears at a pKa of 2. When the student does the same experiment with the Lys to Ala mutant, the glutamate pKa is observed to be 4. Explain a likely molecular basis for this result (5 pts) A pKa of 4 is in the normal range for a glutamate side chain. When the glutamate is protonated, it is neutral. Thus a likely molecular basis is that these residues are forming an electrostatic interaction, possibility a salt bridge, such that it creates a chemical environment that favors deprotonation/ionization of glutamate (hence a drop in observed pKa). C) Based on the answers above, where are these residues likely found in the domain and how are they interacting, if at all? Explain. (3 pts) They are most likely buried, because salt bridges/electrostatic interactions on the exterior of proteins do not tend to be stabilizing. D) The protein is determined to have two subdomains, one consisting of a helical bundle and the other a ? sheet. The student also sees that there is a lag phase in the folding curve (as measured by CD signal versus time), but not in the unfolding curve (and thus can be modeled by a single exponential). How is this result best interpreted? (5 pts) The lag phase in the folding curve indicates the presence of folding intermediates (2 pts). The fact that unfolding can be represented by a simple exponential suggests that the transition state is similar to the native state (2 pts). Thus, one might expect the transition state to be similar to that of barnase, with the rate of folding determined by pre-folding of the subdomains (making folding intermediates) and then the association of these subdomains in the transition state. (1 pts) Page 7 (Question 6 continued) E) Assume the glutamate and lysine interact between the two subdomains of the protein. What ? F values would you predict for these residues if the rate limiting step for folding involved bringing the two subdomains together? Explain. (5 pts) A value close to 1, because in the transtion state they will likely be forming similar interactions as they do in the native state. F)What ? F values would you predict for these two residues if the rate limiting step involved nucleation condensation of the two individual subdomains? (3 pts) A value close to 0, because in the transition state they are not yet interacting. Mutation of either residue is not expected to lead to a change in reaction rate. G) If the ? F values for all the core residues of the protein were 0.5, what are two plausible folding mechanisms consistent with this observation? (5 pts) 1) There are two parallel folding pathways, one in which the core residues are in near native environments, and the other in which they are not 2) All the residues are progressing towards their native state, but are not necessarily in their native configuration, in the transition state. This is more of a nucleation-condensation model. 7) Baker and colleagues demonstrated that their ultra stable protein, Top7, does not fold cooperatively. They suggest that cooperative folding is a property selected for in nature. Why might cooperative folding be beneficial to life? (6 pts) If proteins did not fold cooperatively, then they would be prone to side pathways such as aggregation or the formation of amyloids. Other good answers will be considered. John Tesmer Microsoft Word - Exam I_key.doc