MCB100A/Chem130 MidTerm Exam 2 April 4, 2013

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1 MCBA/Chem Miderm Exam 2 April 4, 2 Name Student ID rue/false (2 points each).. he Boltzmann constant, k b sets the energy scale for observing energy microstates 2. Atoms with favorable electronic configurations gain stability by forming covalent bonds. F Conformational entropy favors the folded state over the unfolded state 4. F QM potential energies are tractable for calculating the heat capacity of a protein 5. F he highest potential energy is the most probable energy state at constant NV 6. Statistical entropy can be applied to isothermal ideal gas expansion and ideal DNA pulling 7. F he natural log of multiplicity, ln(w), is not extensive 8. F he work done in an equilibrium process is less than work done in a non-equilibrium process 9. Chemical potential for ideal gas is only dependent on entropy changes with molecule numbers. he heat capacity at constant pressure is greater than the heat capacity at constant temperature. Cv is inversely proportional to the variance of the Gaussian distribution for energy at constant 2. F he units of entropy are kj/mol. he Helmholtz free energy is available energy to do mechanical or chemical work 4. he chemical potential, µ, is the tendency of system to realize concentration changes. 5. F A reaction will go forward if the reaction quotient, Q is greater than equilibrium constant K 6. Henderson-Hasselbach relates the ph to the acid dissociation constant Ka 7. F he standard enthalpy and standard entropy are always independent of temperature, pressure 8. F he chemical potential is related to the mechanical (expansion) work of Gibbs free energy 9. F If protein A has a larger partition function than protein B then B has the higher heat capacity 2. emperature signifies how multiplicity of bath grows with energy

2 Multiple Choice (5 points each). 2. Molecular interactions are classified as long-ranged when the power law exponent of r -n is (a) n < (b) n = (c) n < 2 (d) n > What is the value of k (a) ~2.5 kj/mol (c) ~.6 kcal/mol (b) ~2 cm - 2. What is the probability of observing system with energy E at constant N,V, (a) exp(-βe 2 )/Q (b) /Q (c) exp(-βe)/q 24. If a covalent bond vibrational excitation is ~25kJ/mole, is it significantly populated at 298K? (a) yes (b) no (c) maybe 25. For an isolated system (constant N,V,E) all energy microstates (a) have different Boltzmann probabilities (b) are equally probable (c) have different multiplicities 26. emperature signifies how multiplicity of bath (surroundings) grows with energy is embodied in (a) (de/ds) N,V (c) (de/d) N,V (b) (dw/de) N,V 27. he condition for the multiplicity to be at an extremum or maximum for large N is (a) dw/dn= (b) d(lnw)/dn= (c) both (a) and (b) (d) none of the above 28. Which state function(s) predict spontaneous change? (a) ds> (b) da< (c) dg< 29. Direction of spontaneous change is when dn particles move from regions of (a) low chemical potential to high (b) equal chemical potential (c) high chemical potential to low. he extent of a chemical reaction with large and negative values of ΔG correspond to (a) K eq >> (b) K eq ~ (c) K eq << (d) none of the above

3 Short roblems (5 points each). he basis of proton NMR is that the hydrogen atom has a magnetic moment, so that in a magnetic field it can populate two states: spin up (S ) and spin down (S ). he energy difference in a MHz NMR spectrometer is.5 x -2 kj/mol. For N= hydrogen atoms and = K (a) Compute the relative population difference N S N S / (N S +N S ) Q = e /k + e.5/2.5 = (N S ) = e.986 (N S ) = e.5/ N S = 5.5 N S = N S N S / (N S + N S ) =.7 (b) How does the population difference change with increasing temperature? It gets smaller 2. Identify all relevant energetic interactions and their functional forms found for the following amino acid constituents: (a) Intermolecular interaction between CH 4 and CH 4 (model for alanine-alanine) No net charge, no other permanent poles, and unfavorable electron pairing (same electronic configuration) E total = 4ε ij σ ij r ij 2 σ ij r ij 6 (b) Intermolecular interaction between NH + and COO - (model for zwitter ionic termini) Net charge, no other permanent poles, and unfavorable electron pairing (same electronic configuration) E total = 4ε ij σ ij r ij 2 σ ij r ij 6 + q iq j r ij

4 Long roblems (25 points each). E. coli DNA polymerase introduces about incorrect base in 4 internucleotide linkages during replication in vitro. However, in vivo the proofreading function of E. coli DNA polymerase is much better, making mistake in 8 nucleotide polymerization events. (a) If an average E. coli gene is ~ bases long, what is the probability of introducing an error per gene in vitro? p(in vitro)= -2 mutations per gene (b) What is the probability of introducing an error per gene in vivo? p(in vivo)= -6 mutations per gene (c) Given the large number of nucleotide polymerization events, N, in E. Coli, we will assume the number of errors, Err, is distributed according to a Gaussian distribution (Err) = σ 2 2π e (Err <Err>) /2σ 2 ; where <Err> is the mean error and σ is the standard deviation. What is <Err> and σ in vivo for 8 gene polymerization events? What is <Err> and σ in vitro for 8 nucleotide polymerization events? <Err> in vivo =p x N = -6 x 8 = 2 <Err> in vitro =p x N = -2 x 8 = 6 σ in vivo =[p x (-p) x N] /2 =[ -6 x (~) x 8 ] /2 = σ in vitro = [ -2 x ~ x 8 ] /2 =~ (d) What is the probability that the number of errors will be standard deviation larger than the mean in vivo? () = 2 2π e ( ) /2() 2 = =.24 (e) What is the probability that the number of errors in vivo will be 5 times smaller than the average number of errors in vitro? ( ) = Err=,,/5=2 2 2π e 2 ( ) 2 2( ) 2 = =

5 4. he variation in the constant pressure heat capacity, C, of a protein as a function of temperature is known as the melting curve, as it measures the relative populations of the folded and unfolded proteins in water. We would like to use C measurements made along this curve to determine a single equation that gives the standard Gibbs free energy for unfolding, ΔG unfold ( ), at any temperature. (a) Draw a typical heat capacity curve for a protein in water from 25C to 9C. Name intramolecular energy interactions that break and therefore new energy levels that become populated as the protein goes from the folded to the unfolded state. hydrogen bonds, van der Waals interactions (dispersion), salt bridges (b) Express ΔG unfold ( ) in terms of ΔH unfold ( ) and ΔS unfold ( ) ΔG unfold ( ) = ΔH unfold ( ) ΔS unfold ( ) (c) Express ΔH unfold ( ) in terms of C unfold ; express ΔH fold ( ) in terms of C fold ΔH unfold 2 ( ) = C unfold d ΔH fold 2 ( ) = C fold d (d) Express ΔS unfold ( ) in terms of C unfold ; express ΔS fold ( ) in terms of C fold ΔS unfold 2 unfold ( ) = C d ΔS fold 2 fold ( ) = C d

6 (continued) You are given the following thermodynamic cycles, where F corresponds to folded and U to unfolded ΔH F ( M ) U ( M ) ΔH 2 ΔH4 F( ) ΔH U ( ) ΔS F ( M ) U ( M ) ΔS 2 ΔS4 F( ) ΔS (e) Which leg of the following thermodynamic cycle allows us to determine ΔH unfold ( ) and ΔS unfold and how would I determine it from the other three legs of the cycle? ΔH unfold ( ) = ΔH 2 ( ) ΔH F U ( ) F ( ) ΔS U ( ) U ( ) ( ) + ΔH ( ) + ΔH 4 ( ) ΔS unfold ( ) = ΔS 2 ( ) + ΔS ( ) + ΔS 4 ( ) ( ) (f) Which leg of the thermodynamic cycle measures ΔH unfold M these quantities from experiment and what equations would you use? ΔH = M +Δ ( ) ΔH F ( m ) = C unfold d = M Δ U ( ) F ( ) and ΔS unfold ( M )? How would you get ( ) ΔS U ( ) area under peak of melting curve ΔG ( ) = ΔH ( ) ΔS ( ) = ΔS ( ) = ΔH ( m ) m m m m (g) What are the expressions for ΔH n and ΔS n for the other two legs of the cycle? ΔH 2 M ( ) = C fold d = C fold M ( ) ΔH 4 ( ) = C unfold d = C unfold M M ( ) ΔS 2 ( ) = M fold C d (h) Write the expressions for = C fold ln M ΔG unfold ΔS 4 ( ) = M C unfold ( ) and show your work d = C unfold ln M

7 ΔH unfold ( ) = C fold M ( ) + ΔH ( M ) + C unfold ( M ) ΔS unfold ( ) = C fold ln M + ΔS ( M ) + C unfold ln M ΔG unfold ( ) = C fold M ( ) + ΔH ( M ) + C unfold ( M ) % C fold # $ ln M + ΔS ( M ) + C unfold ln M & ( ' ΔG unfold ( ) = ΔH unfold ( M ) + ΔC M # ( ) % ΔS ( M ) + ΔC ln $ M & ( '

8 4. For the folding of a residue protein, containing 6 glycines and 4 prolines, at K: (i) When unfolded, each residue except glycines and prolines can take conformations of equal energy; glycine can take 4 conformations; proline can take 2 conformations. (ii) he conformation of the folded state has every residue in possible conformation except glycine which has 2 conformations. (iii) A His-Asp ion pair, whose interaction energy is -5kJ/mol in vacuum, is found in the interior of a folded protein with dielectric constant ε p =4. he ion pair remains intact when unfolded in water whose dielectric constant is ε W =8. (iv) Assume water molecules have 6 possible configurations when surrounded by other waters, protein backbone or polar sidechains, otherwise they have only 2 possible configurations. (v) Every hydrophobic sidechain in unfolded state interacts with 2 water molecules, and all hydrophobic sidechains are buried in folded state. (v) Assume water-protein interaction energy is zero. (a) Calculate the Helmholtz free energy for the protein-protein interactions only (ΔA protein ). Show your work. Does ΔA protein favor folding? ΔU protein = = 8.25 kj/mol ΔS protein = 2.5 kj/mol ( ln ln ) = 2.5 kj/mol ( ) = kj/mol ΔA protein = ΔU protein ΔS protein = 8.25 kj/mol kj/mol=8.4 kj/mol no does not favor folding (b) Calculate the free energy for water,non-hydrophobic interactionss (ΔA non-hphobe,h2o ). Show your work. Does ΔA non-hphobe,h2o favor folding over nonfolding? ΔU non hphobe,h 2O = ΔS non hphobe,h 2O = 2.5 kj/mol ( ln6 # nonhphobe water ln6 # nonhphobe water ) = ΔA non hphobe,h 2O = A fold A unfold = ΔU non hphobe,h 2O ΔS non hphobe,h 2O = no does not favor folding (c) Calculate the free energy for water,hydrophobic interactions (ΔA will be per hydrophobic residue). Show your work. What is minimum number of hydrophobic sidechains required to ensure that folding in water is spontaneous at room temperature ( K)? ΔU hphobe,h 2O = ΔS hphobe,h 2O = 2.5 kj/mol ( ln6 2 ln2 2 ) = 5.49 kj/mol/#hphobe n hphobe ΔA hphobe,h 2O = A fold A unfold = ΔU hphobe,h 2O ΔS hphobe,h 2O = 5.49n hphobe kj/mol ΔA protein + ΔA non hphobe,h 2O + n hphobe ΔA hphobe,h 2O = 8.4 kj/mol=5.49n hphobe kj/mol n hphobe =

9 (Extra Credit). o find the probability, p, that both minimizes the Gibbs free energy and maximizes the entropy, we start with the following equation L = k b p v ln p v + α p v E v + λ p v V v + γ p v (a) What is the first term on the right-hand side of the equation? he probabilistic version of entropy (b) What is the meaning of the second term on the right-hand side of the equation? he system exchanges energy with the bath (energy fluctuates) (c) What is the meaning of the third term on the right-hand side of the equation? Volume fluctuations (d) What is the meaning of the fourth term on the right-hand side of the equation? he probability summed over all microstates adds up to (e) What is the meaning of L (the Lagrangian)? It is the maximization of the entropy (2 nd Law) under the constraints that energy and volume fluctuate (f) How do I determine the probability, p? dl dp = ; dl dp = k b + ln p ( ) + αe + λv + γ = p = exp αe + λv + γ k b k b (g) Can you solve for the α Lagrange multiplier? S = k b p ln p k b ln p = ( αe + λv + γ k b ) S = p ( αe + λv + γ k b ) S = α E λ V + γ k b ( ) p S E N, V = = α

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