( 1 vρ) = ( sec)( N sec m 1 )

Size: px

Start display at page:

Download "( 1 vρ) = ( sec)( N sec m 1 )"

Clarence Tate
5 years ago
Views:

1 Page 1 of 5 + formula sheet 1. You have isolated a novel protein that has been implicated in a disease and you would like to use what you have learned in Chem/Biochem 471 to deduce some of its properties. a) You subject the protein to a centrifugation experiment under nondenaturing conditions and observe the following: Sedimentation coefficient s = 6.73 x sec, partial specific volume = cm 3 g -1, frictional coefficient f = 1.01 x N sec m -1, density ρ = g cm -3, What can you deduce about the protein? ( 1 vρ)m s = f sf m = ( 1 vρ) = ( sec)( N sec m 1 ) 1 (0.739cm 3 g 1 )(0.9982cm 3 g) Molecular weight of the protein is 156 kda mol 103 g kg b) You conduct an SDS-PAGE experiment (denaturing conditions) and obtain the results shown in the diagram (single band in right lane). Label the bands in the left lane, which contains proteins of known molecular weights of 40 kda, 80 kda, 70 kda, and 120 kda. From top to bottom, the markers are 120, 80, 70, and 40. Based on this gel, what is the molecular weight of the protein? 80 kda How can you reconcile the results of a) and b)? Protein is a dimer of two 78 kda subunits. Because of the denaturing conditions of b we see the monomer molecular weight. =156,000g / mol =156kDa c) You concentrate your protein as much as you can and then estimate the concentration using the absorbance at 280 nm. You can t look up an extinction coefficient for this newly discovered protein so you use the protein sequence to predict its extinction coefficient. Your protein contains 1 Trp and 3 Tyr; for each Trp ε 280 = 5690M 1 cm 1 and for each Tyr ε 280 =1280M 1 cm 1. 1Trp Calculate ε protein 280 = 5690M 1 cm 1 3Tyr M 1 cm 1 = 9530M 1 cm 1 protein Trp protein Tyr Is it reasonable to use only these two amino acids to estimate the extinction coefficient at 280nm? Explain. Yes, because the other amino acids do not absorb at 280 nm: Tyr and Trp are the only ones with a large enough box of extended conjugation to shift their absorption out as far as 280 nm. For your final protein solution in a 1cm cuvette, A 280 = 0.2. What is your protein concentration? A = εcl, so c = A εl = 0.2 (9530M 1 cm 1 )(1cm) = M = 21µM d) A fellow student in your lab tells you that your protein concentration is quite low, but you cannot increase its concentration any further. Why is this a potential problem if you want to use NMR spectroscopy to obtain some structural information about your protein? Relate your answer to one of the formulas on your formula sheet. NMR has very low sensitivity, because E<<kT so Ni/Nj 1. Tiny population difference between energy levels, means the signal is tiny, so the low sample concentration would be a problem.

2 Page 2 of 5 + formula sheet 2. Suppose that you analyze another protein of interest (protein B) using experiments similar to those of problem 1 and conclude that this protein contains 3 identical subunits. a) You discover that ATP copurifies with protein B and hypothesize that each subunit has a binding site for this ligand. You put 1 x 10-4 M protein into a dialysis bag and put the bag into a solution containing ATP. After waiting to reach equilibrium, you find the following: [ATP] = 3 x 10-4 M inside the bag and [ATP] = 1 x 10-4 M outside the bag Assuming that the binding sites are identical, what is the binding association constant for L binding to one site? [ATP] bound =[ATP] in [ATP] out = 2 x 10-4 M ν = NK [L] A 1+ K A [L] = [L] bnd [P] tot 3K A [L] = 2(1+ K A [L]) K A [L] = 2 2 K A = M = M 1 = M M = 2 b) What can you learn about the structure of your protein by measuring its circular dichroism spectrum? An estimate of the percent alpha helix and percent beta sheet. c) If the single Trp residue is near the ligand binding site, which do you think is a better choice to monitor ligand binding to the protein, Trp fluorescence or circular dichroism? Explain your choice. Trp fluorescence because it monitors the local environment of the Trp which could change upon ligand binding. CD will only change if there is a significant change in secondary structure upon ligand binding, which is less likely. d) You use your chosen method to monitor ligand binding. Sketch the results you would expect if (i) the binding sites are identical or (ii) if they have positive cooperativity. Indicate the concentration of 50% binding for case (i). 50% binding at [L]=Kd =1/Ka=5x10-5 M

3 Page 3 of 5 + formula sheet 3. One mole of an ideal gas initially at 25 C and 1 atm. is heated at constant pressure until it expands to a volume of 35L. You are given that C P = 20.8 /K mol. a) What is the final temperature of the gas? PV=nRT T final = PV nr = (1atm)(35L) = 427K =153 C (1mol)( Latm / molk) b) Calculate the initial volume and the work done on the gas during this process. V initial = nrt (1mol)( Latm / molk)(298k) = = 24.5L P 1atm w = PΔV = 1atm( )L = 10.5Latm = 1063 = 1.1k Latm c) Calculate the entropy change of the gas. Can t use S=q/T because not at constant T. ΔS = C p ln T f 427K = 20.8 ln T i molk 298K = 29.8 molk d) Calculate ΔE for this process. 2 approaches give same answer: ΔE = q + w = C p ΔT + w = (1mol)(20.8 )(129K)+ w = 2.7k 1.1k =1.6k molk ΔE = 3 2 nrδt = 3 (1mol) molk (129K) =1609 =1.6k

4 Page 4 of 5 + formula sheet 4. The shells of marine organisms contain CaCO 3 largely in the crystalline form known as calcite. There is a second crystalline form known as aragonite. These forms convert between themselves according to Answer the following using the data provided. Calcite Aragonite Thermodynamic parameter (25 ºC) ΔH 0 f S 0 f ΔG 0 f V CaCO 3 (calcite) /K mol cm 3 /mol CaCO 3 (aragonite) /K mol cm 3 /mol a) What is G for the conversion of calcite to aragonite? G = ( ) = 1.05 b) Would you expect calcite in nature to convert spontaneously to aragonite at 1 atm. and 25 ºC? Explain briefly. No because G is positive. c) Over what temperature range (eg T > x or T < x) is the calcite to aragonite conversion spontaneous (at P = 1 atm)? Hint: at what value of G do reactions become spontaneous? H = ( ) = S = /K mol (92.88 /K mol) = /K mol Conversion becomes spontaneous when G=0= H-T S, so T S= H. T= H/ S = /-4.18 /K mol = 43K Spontaneous at T< 43K c) Suppose mineral deposits are found that contain the thermodynamically unfavorable form of CaCO 3 and this form persists for years under ordinary conditions (T = 25 C, P = 1 atm). What could you conclude about the free energy diagram for the calcite to aragonite reaction? There is a high activation energy barrier (Ea>>kT) for the conversion, so the kinetics are slow which accounts for not finding the thermodynamically predicted form.

5 Page 5 of 5 + formula sheet 5. In the TCA cycle, malate is converted to oxaloacetate by the reduction of NAD + to NADH. The electrons from NADH are subsequently fed into the electron transport chain, eventually to O2. Answer the following questions, assuming T = 37 C. reaction ε 0 (V) NAD + + H e - NADH O2 + 4 H e - 2H2O a) What is ε 0 for electron transfer from NADH to oxygen? NADH is being oxidized, so reverse its equation and sign of ε 0. ε o' = V V =1.136V b) What is G 0 for the oxidation of 2NADH by oxygen? ΔG o' = n e Fε o' = ( 4)(96,500 )(1.136V ) = 438k / mol molv c) This electron transfer is coupled to pumping protons across the mitochondrial membrane. Calculate the G 0 cost per mole of protons pumped out of the mitochondrion, given the following parameters: ph (in) = 7.0, ph (out) = 6.0, and membrane potential = -140 mv (outside positive). To help you get the correct value, note that there is an energy cost due to the concentration gradient and an energy cost due to the charge gradient (both terms are unfavorable). ΔG 0' = RT ln a final a initial + zfδφ = ( / molk)(310k)ln (+1)(96,500 / molv )(V out V in ) ΔG 0' = ( / molk)(310k)ln(10)+ (96,500 / molv )(0 ( 0.140))V ΔG 0' = 5935 / mol / mol =19445 / mol =19.4k / mol d) This proton gradient is then used to drive synthesis of ATP inside the mitochondrion, where [ATP] = 1 mm, [Pi] = 2.5 mm, [ADP] = 1 mm. Given G 0 = 30.5 for the synthesis of ATP from ADP and Pi, calculate the free energy per mole needed for this reaction in the mitochondrion. ΔG = ΔG o + RT lnq = ΔG o + RT ln [ATP] [ADP][P i ] ΔG = 30.5 k mol ( M ) (310K)ln molk ( M )( M ) ΔG = 30.5 k k k = 45.9 mol mol mol

Chemistry 425 September 29, 2010 Exam 1 Solutions

Chemistry 425 September 29, 2010 Exam 1 Solutions Name: Instructions: Please do not start working on the exam until you are told to begin. Check the exam to make sure that it contains exactly 6 different