1Biochemistry I Fall Term, 2004September 22, 2004Lecture 10: Biochemical Energetics 2Assigned reading in Campbell: Chapter 12.6-12.10.(Omit section 12.11)Key Terms:LiposomesMetabolismAnabolismCatabolismOxidation-reduction reactionsCouplingLinks:(I) Review Quiz on Lecture 10 concepts(O) Protein Folding-Unfolding Transition: a Shockwave movie of an "αα structure"(O) Some Amino Acid Properties12.6 Hydrophobic Interactions: A Case Study in ThermodynamicsThe descriptions in Campbell are a good introduction to the concepts involved in describing theformation of liposomes and the folding of proteins into their native conformation. The followingsections add to those concepts some experimental results that use for their interpretation therelationships between ∆G, ∆H, ∆S, and the equilibrium constant.A. Measurement of protein folding/unfolding equilibriaFor the reversible protein unfolding reaction,N <=> U with Keq = [U]/[N]the observed "transition curve" between native (N) and unfolded (U) states is a steep function oftemperature, reflecting the cooperativity of the stabilizing forces. The examples below show theunfolding transitions of Protein G plotted as the fraction unfolded vs. the temperature. Curves forthe wild type and two different mutant proteins are shown.2Both the enthalpy (∆H) and entropy (∆S) of unfolding are available from these data. To obtainthese values the experimental data are analyzed using the following approach.:The van't Hoff equation relates the equilibrium constant to temperature and allows us to extract∆H and ∆S for the transition and to calculate protein stability at any temperature:∆G = -RT lnKeq = ∆H - T∆Srearrange:lnKeq = -∆H/RT + ∆S/RA plot of lnKeq vs 1/T yields a straight line with slope = -∆H/R.Three steps in using the van't Hoff equation, starting from a transition curve:1. Determine the spectroscopic signal (e.g. A280nm or fluoresence) for 100% native protein(N) and 100% unfolded protein (U). Measure the signal as a function of temperature (T)to produce a transition curve. Determine the fraction of unfolded (or native) protein as afunction of temperature.2. Calculate K as a function of T:K = Fu / Fn Note that K = 1 and ∆G = 0 at Tm.Tm is the midpoint of the transition curve where [N] = [U]; it also called the "meltingtemperature".3. Plot lnK vs 1/T (K-1), a "van't Hoff plot"a. Determine ∆H of unfolding: = -R*slopeb. Determine ∆S of unfolding: = ∆H/Tmc. Calculate the protein stability at any other temperature using: ∆G = ∆H - T∆S.3B. Example of Protein G unfolding (and some typical calculations)These calculations are based on the curves shown above.For the unfolding reaction of wild type Protein G (Thr 53):∆H = 50.4 kcal/mol from a van't Hoff plot (not shown).We calculate ∆S at Tm:∆S = ∆H/Tm = 50,400/342 = 147.4 cal/mol/K.For the Ala 53 mutant Protein G:∆H = 43.7 kcal/mol.∆S = 132.4 cal/mol/K.Now compare the wild type protein to the Ala 53 mutant at the same temperature.1. At 69°C (the Tm for wild type):∆Gwt = 0∆Gmutant = ∆H - T∆S = 43,700 - 342*132.4 = -1.58 kcal/mol.Thus, ∆Gwt - ∆Gmutant = ∆∆G = 0 - (-1.58) = 1.58 kcal/mol.Conclusion: Unfolding of wild type protein is less favorable than the Ala 53 mutant by1.58 kcal/mol.2. At 27°C:∆Gwt = 50,400 - 300*147.4 = 6.2 kcal/mol.∆Gmutant = 43,700 - 300*132.4 = 4.0 kcal/mol.∆∆G = 6.2 - 4.0 = 2.2 kcal/mol.The equilibrium constant at 27°C for wild type protein unfolding is calculated from:∆G = -RTlnKeqKeq = exp(-6.2/0.6) (RT = 0.6 kcal/mol/K at 300K)Keq = 3.25*10-5The fraction of wild type protein in the unfolded state at this temperature is:Fu = Keq/(1 + Keq)Fu = 3.3*10-54There are two remarkable and general features of these protein stability results:1. The ∆G for Protein G folding is modest. -- only -6.2 kcal/mol at room temperature. Thisis also the case for other proteins where the ∆G values range from about -5 to -15kcal/mol.2. The modest -∆G for protein folding in general, represents the small difference between alarge favorable ∆H and an almost-as-large unfavorable T∆S contribution.C. The hydrophobic effect can be measured as a ∆∆∆∆G of transfer.During protein folding, the transition from the countless unfolded states to a single nativestate is accompanied by the burial of solvated nonpolar sidechains (and polar peptideunits) into the nonsolvated core of the protein. The reduction in solvent-accessible area ofthese groups favors the folded state. An empirical correlation that appears to havepredictive value also provides insight into the energetics of the hydrophobic effect.12.7 The Nature of Metabolism12.8 The Nature of Oxidation and Reduction12.9 Coenzymes in Biological Important Oxidation-Reduction ReactionsThese three sections cover topics that were mostly covered in introductory biology. Review theconcepts in the Key Terms list. Knowledge of the structures of nucleotides and coenzymes arenot required at this point in Biochemistry I.12.10 Coupling of Production and Use of EnergyThe first step in the glycolytic pathway is the phosphorylation of glucose to make glucose-6-phosphate:Glucose + ATP <=> ADP + G-6-PThe reaction is catalyzed by the enzyme, hexokinase.Use the values in Table 12.1 (Campbell, p. 422) to calculate the following for the hexokinasereaction:1. ∆G°'2. Keq3. Will the above values be different in the cell, where the substrate and productconcentrations are in the mM range?Compare your answers to those provided in lecture (on