a) Write the reaction that occurs (pay attention to and label ends correctly) 5 AGCTG CAGCT > 5 AGCTG 3 3 TCGAC 5

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1 Chem 315 Applications Practice Problem Set 1.As you learned in Chem 315, DNA higher order structure formation is a two step process. The first step, nucleation, is entropically the least favorable. The second step, zippering, caused by the interactions of nearest neighbors, is rapid. The overall folding process 2sS à ds can be considered a two part process 2sS à NsS à ds where NsS is the nucleated intermediate Consider the folding of the following short sequence AGCTG with its compliment. a) Write the reaction that occurs (pay attention to and label ends correctly) 5 AGCTG CAGCT > 5 AGCTG 3 3 TCGAC 5 strand 1 + strand > double strand b) Express the free energy of the folding mathematically (equation) DG o folding = DG o ds (DG o s1 + DG o s2) c) Given the data in the table provided calculate the ΔG o, ΔH o, and ΔS o for the formation of the double stranded sequence. The folding process is two step initiation (nucleation) and H bond formation (zippering) therefore DG overall can be broken down into a broad two step process and the zippering step can be further broken down into formation of each H bond H bonding is free energy favorable nucleation is not Since this this strand does have at least one CG pair from the table the nucleation value is 1.8 kcal/mol The sum of the hydrogen bond values are for AG GC CT and TG (leading strand) = kcal/mol overall = kcal/mol (highly favorable) Likewise the DH o is kcal/mol and the ds o is cal/molk The entropy is unfavorable as expected and this interaction is enthalpy driven.

d) Write the equilibrium constant expression and calculate the value of the equilibrium constant at room temperature Keq = ds/ss 2 At 298 (values in the table used for part c) DG o folding = -

2 d) Write the equilibrium constant expression and calculate the value of the equilibrium constant at room temperature Keq = ds/ss 2 At 298 (values in the table used for part c) DG o folding = - RTlnKeq cal/mol = 1.985cal/mol K (298 K) lnkeq Keq = 3958 (products highly favored as expected by DG<0) e) Given an initial single strand concentration of 0.1mM, calculate the equilibrium concentrations of the single strands and double strands. Keq = ds/ss = x/ (0.0001M- x) 2 solve quadratic or recognize that given the large equilibrium constant the concentration of product will essentially be 0.1 mm and the reactants 0.

3 2. Draw a reaction coordinate diagram for enzyme catalyzed single substrate reaction that obeys the Michaelis- Menten Model. Label axes, all components and all energies appropriately. Comment on the overall favorability of the thermodynamics of the reaction you have illustrated. MM Kinetics E + S < == > E S < == > E + P (second step rate determining) This is basically it y axis of free energy Ea1 is DG # for step 1 and Ea2 is DG # for step 2 What is labeled DH here is DG. As shown this is a favorable reaction as DGrxn <0 3. In an assay for kinetics measurement for inhibition, okadaic acid (250 mm stock solution) was diluted. In each sample, 250 ul of the stock was and the final volume was adjusted to 3.0 ml. The experiment was conducted at room temperature. Assuming ideal behavior calculate: a) ΔH, ΔS and ΔG of dilution. With the assumption of ideality DH =0 DG = RT ln (C2/C1) C1= 250 mm C2 = 250mM*250uL/3000uL = mm DG = 8.314J/molK * 298* ln (20.83mM/250 mm) = J/mol = kj/mol DG = DH TDS J/mol = 0 298K (DS) DS = J/molK

4 b) What problems (if any) are inherent in the assumption made? Comment on the likelihood that the assumption is valid and explain why or why not. The assumption made is that the solution is ideal. This gives us DH=0. This solution is dilute so the solute solute interactions are minimal. However, there will be interactions between the solvent and solute. The more dilute the solution the more ideal the behavior so the diluted solution acts more ideally than the more concentrated solution (250mM). 4. In a regulator holding reservoir, gases from three 1 L tanks are being mixed to feed into a breathing tube. The gasses are N2, O2 and NO2, 14.0, 8.0 and 6.25g, respectively. The volume of the new tank is 2L and room temperature is maintained. Assuming the gases exhibit no reactivity or interactions with each other calculate the ΔH, ΔS and ΔG of mixing of the gases. Again DH = 0 (see problem 3) DGmixture = DG N2 + DG O2 + DGNO2 Concentrations 14.0g N2 = 0.5 mol; 8.0 g O2 = 0.25 mol and 6.25 g NO2 = mol Each initially in 1 L gives concentrations of 0.5, 0.25 and M, respectively Concentrations in the new tank are halved to 0.25, and M, respectively DG = RT ln (C2/C1) + RT ln (C2/C1) + RT ln (C2/C1) = RT (ln 0.25/0.5) + ln(0.125/0.25) + ln(0.0679/0.125) (or you can recognize that the dilution is ½ and occurring 3x) = 8.314J/molK (298K) *3ln J/mol Spontaneous (as expected for expansion of each gas into larger container) DS= 0 (- 5125J/mol)/298K = J/mol K (DS>0 for mixing) 5. What is the osmotic pressure created by a 1 M solution of glucose at human physiological temperature? What pressure would be generated by a 1M salt solution (NaCl). Explain the difference and compare your answer to the pressure you experience walking across campus (thermodynamic pressure not social or academic pressure!) pi = crt (C molar concentration) pi = 1mol/L ( Latm/molK)(310K) pi = atm (very large but this is a very concentrated solution at an elevated temperature!) for the salt since the dissolution results in two ions the effective concentration (assuming no ion pairing) is 2M and the pressure would be almost 51 atm!

5 In solid sugar, individual sugar molecules interact with each other by van der Waals forces. A salt is made up of anions and cations, and these interact electrostatically. Equimolar solutions of a salt and a sugar therefore do not have the same number of solute particles. Consequently, the osmotic pressure of the salt will ordinarily be greater than that of the sugar. Subtle solvent-interaction effects are important too, but the underlying cause of osmotic pressure is a soluble solute in a region of solution which can undergo expansion and which is separated from a region of higher concentration solvent by a semi-permeable membrane. 6. a) Calculate the osmotic pressure created by g of a 10kDa protein in 10 ml of dialysis tubing suspended in 200 ml water. The membrane cutoff is 1kDa. b) The tubing also contains 2.0g GdHCl salt used in unfolding the protein. What is the osmotic pressure created by the salt? c) At what concentration of salt will the system come to equilibrium (wrt salt)? d) Assuming the tubing can swell 3 fold what will be the pseudo equilibrium concentration of the protein? e)how much salt will remain in the swelled tubing at equilibrium? f) How much work will be done by the protein in tubing? a) pi = crt pi =(0.0015g / 10,000g/mol)/0.01L * Latm/molK * 298 K pi = atm (very small pressure because very few particles) b) pi = crt pi =(2.0g / 95.53g/mol)/0.01L * Latm/molK * 298 K pi = atm (much higher pressure more particles!) c) Equilibrium will occur when the concentration of salt throughout is constant. Total mass salt 2.0g with total volume (both sides of membrane) 210 ml (2.0g/95.53 g/mol)/0.210 L = = 0.10 M d) Equilibrium will occur when the concentration of protein throughout is constant. Initial concentration of protein: (0.0015g/10,000 g/mol)/0.010 L = 1.5E- 5 M Total mass protein g with total volume (both sides of membrane) 210 ml (0.0015g/10,000 g/mol)/0.210 L = 7.1E- 7 M (if protein could distribute throughout the entire solution) But the 10 kd protein cannot pass through the 1 kd cutoff membrane so water will move in. Since the tubing can swell 3 fold the final concentration of the protein (pseudo equilibrium) (0.0015g/10,000 g/mol)/0.03 L = 5E- 6 M pseudo equilibrium concentration

6 The initial GdHCl concentration inside membrane is 2.09 M. The true equilibrium concentration is 0.1 M. Salt can pass through the membrane so it will reach equilibrium. With a 30 ml membrane 0.1 mol/l * 0.03L * g/mol = 0.257g salt inside membrane at equilibrium e) assuming constant external pressure of atm w=- pdv = atm*(0.03L- 0.01L) * * J/Latm = J 7. Glutamine is an important biomolecule made from glutamate. Calculate the equilibrium constant for the reaction: Glutamate + NH3 <==> Glutamine + H2O a) Use standard free energies of formation data to obtain the free energy change for the reaction. DG = DGproducts DG reactants DG = (Glutamine + H2O) (Glutamate + NH3) DG o formation (kj/mol) Glutamate NH Glutamine H2O ( ) ( ) = 13.2 kj/mol (not spontaneous as written) b) Under physiological conditions, the concentration of ammonia is near 10 mm. Calculate the ratio of [glutamine]/[glutamate] at equilibrium. Assume the concentration of water is invariant. DG o = - RTlnKeq 13200J/mol = J/molK 298 K ln [glutamate][0.01m]/1[glutamine] = ln [glutamate][0.01m]/1[glutamine] e = [glutamate][0.01m]/1[glutamine] [glutamine] = [glutamate][0.01m] [glutamine]/[glutamate] = 2.04 c) Physiologically glutamine is synthesized by coupling with ATP hydrolysis Is this necessary or not - explain? Yes this is necessary as written the reaction is not spontaneous and will require an energetic driving force. d) Write the reaction for the coupled process. Glutamate + NH3 + ATP <==> Glutamine + H2O +ADP + Pi

7 e) Calculate the standard state free energy and the equilibrium constant for this reaction at STP. This can be done by using the value for part a of 13.2 kj/mol and adding the value for the ATP hydrolysis kj/mol overall kj/mol f) Assume NH3 and Pi are maintained at about 10mM and that the ration of ATP to ADP = 1. What is the ratio of glutamate to glutamine at equilibrium? DG o = - RTlnKeq J/mol= J/molK* 298 K*ln[glutamate][0.01M][ADP][Pi]/1[ATP][glutamine] 7.78 = ln [glutamate][0.01m] 2 /[glutamine] e 7.78 = [glutamate][0.01m] 2 /[glutamine] 2396 [glutamine] = [glutamate][0.01m] 2 [glutamine]/[glutamate] = 4.17E- 8 note large change with coupled reaction 8. ATP solutions are generally made fresh in the laboratory before use. If they are made ahead of time they must be stored at - 4 C. Explain haynie 5.5 Because ΔG < 0, ATP hydrolysis is spontaneous at room temperature. But ΔG = ΔH TΔS, so the driving force for hydrolysis is reduced by lowering the temperature, even if ΔH and ΔS are approximately constant in the temperature range of interest. Changes in ΔG are reflected in Keq, which as we have seen in Chapter 4 is the ratio of the rate of the forward reaction to the reverse reaction. The precise value of ΔG alone, however, does not tell us whether a reaction is fast or slow. The rate of the forward reaction of ATP hydrolysis depends on the correct orientation of a water molecule and the bond hydrolyzed. The rate of the reverse reaction depends on the diffusion of ADP and Pi, which must interact to produce ATP. At 4 C molecules move more slowly than at 25 C, slowing the rate of exploration of possible relative orientations of ATP and water and therefore reducing the rate of hydrolysis. The effect is much more pronounced at 20 C, because although there is still some movement of solvent, most of the water molecules are effectively fixed, because although there is still some movement of solvent, most of the water molecules are effectively fixed in a lattice and molecular reorientation is greatly reduced in comparison with the liquid state.

8 9. The synthesis of ATP under standard conditions requires 7.7 cal/mol. In vivo the synthesis is couplesd to the movement of 2 protons across a membrane. What ph difference exists across the mitochondrial membrane as a result of ATP synthesis? Haynie 5.37 The chemical equation describing the situation before us is: ATP + H20 + 2H + inside ADP + Pi + 2H + outside G = RTln([H + outside] 2 /[H + inside] 2 ) = RTln([H + outside]/[h + inside]) 2 = 2RT(2.303)log([H + outside]/[h + inside]) = 4.606RTlog([H + outside]/[h + inside]) = 4.606RT( ph). Solving for ph, ph = 7700 cal mol 1 /( cal mol 1 K K) = For the binding of a ligand to a protein at 25 o C and ph=7.0, K d = 1x10-7 M. What is the free energy change for association of the ligand with the protein under standard conditions? For the dissociation reaction, ΔG o' = RTlnK d ΔG o' = (8.315x10-3 kj/(molk))(298k)( 16.1) = kj/mol Dissociation is not favored under standard conditions. For the association reaction, ΔG o' = RTlnK a = RTln(1/K d ) ΔG o' = (8.314)(298)(16.1) = 39.9 kj/mol Association is favored under standard conditions.

reduction kj/mol

reduction kj/mol 1. Glucose is oxidized to water and CO 2 as a result of glycolysis and the TCA cycle. The net heat of reaction for the oxidation is -2870 kj/mol. a) How much energy is required to produce glucose from