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Unformatted text preview: MCB124-Baldwin ANSWER SET 3 page 1 1. THERMODYNAMICS OF PROTEIN FOLDING i. Below is a temperature vs. stability curve for a globular single domain protein. Describe the significance of the four labeled parts. (A) Hot Tm-the point above Tmax at which the protein is half unfolded. (B) Cold Tm-the point below Tmax at which the protein is half unfolded . (C ) The temperature of maximum protein stability (Tmax) (D ) The temperature range where protein folding is favorable . ii . At temperatures below (B) what is believed to be the favorable driving force(s) for folding? Intramolecular interactions in the folded state, and the favorable entropy of solvent release from the hydrophobic effect. iii . At temperatures above (A) what is believed to be the favorable driving force(s) for folding? Intramolecular interactions in the folded state, and the favorable enthalpy change on burying hydrophobic groups- that hydrophobic water has poorer interactions than in liquid water, and the entropy of water release. iv . What would you expect to happen to ∆ G and the apparent ∆ H and ∆ S values of an interior Leu-Ala mutant and why? The free energy of folding would become more positive and less favorable ∆∆ G > 0. The ∆ H of folding would become more positive due to losses of vanderwaals interactions in the folded state. The ∆ H component of the hydrophobic effect would also contribute but although its sign would depend on the reference temperature, the contribution would expected to be smaller. The ∆ S of folding would decrease, due to the fewer number of solvent molecules released on folding from the reduction of hydrophobic surface in the unfolded state, however, there would also be a counteracting change in the ∆ S contribution from the chain, since the unfolded state has fewer degrees of side chain freedom to loose on folding. v. A certain Leu-Ala mutant has the opposite change in ∆ H and ∆ G to that you expected but the same ∆ S. What happened in this particular instance? A favorable ∆∆ H and ∆∆ G for this mutation can only result from a release of strain in the folded state. The ∆ S is primarily due to changes in the unfolded state, which are the same for any protein. Δ G f T o C C B A D MCB124-Baldwin ANSWER SET 3 page 2 2. Stability Measurements and Calculations First, do a van’t Hoff plot The van’t Hoff equation is lnK = - ∆ H/R(1/T) + ∆ S/R ˚C Kfold 1/T ˚K lnKf 62 2.6343601 2.985E-03 9.686E-01 64 1.38607031 2.967E-03 3.265E-01 66 0.7189229 2.950E-03 -3.300E-01 68 0.36773196 2.933E-03 -1.000E+00 We plot lnKf vs 1/T (in ˚K), fit a line to it and solve for all of the parameters in a)-c). Remember that the ∆ H and ∆ S values calculated are at the Tm only! The Van’t Hoff plot appears to be linear because the data cover a very narrow range of temperatures (if viewed over a larger temperature range, it would appear curved due to the effect of ∆ Cp) The linear equation is lnKf = 37487/T -110.92 The linear equation is lnKf = 37487/T -110....
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- Spring '09
- Glu, Hydrophobe, answer set