Chemistry 25 (Spring term 2016) Final Examination for NON-SENIORS GOOD LUCK AND HAVE A GREAT SUMMER!!!

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1 Name ANSWER KEY Distributed Thursday, June 2 Due Chemistry 25 (Spring term 2016) Final Examination for NON-SENIORS GOOD LUCK AND HAVE A GREAT SUMMER!!! by Thursday, June 2 at 1 pm to 362 Broad a drop box will be left outside 362 Broad to return finals during the week ***late penalties enforced*** Conditions Open this examination when you are ready to take it. This is a 3 hour examination that must be taken in one continuous stretch. You may use your Ch25 lecture notes, problem sets and solutions, the textbook (Kuriyan, Konforti and Wemmer), the course web site and a calculator. You may not use any other books, exams and problem sets from previous years of Ch25 (unless they are posted on the course web site), other web sites, non-calculator applications of Mathematica and related programs, discuss the exam with others, etc. This exam should have 20 pages total. Show your work! All work must be completed in the provided space. Getting the right answer is not enough the intermediate steps are needed for credit. Do not discuss this exam with other Ch25 students until after 1pm on June 9. useful relationships 1 µm = 10-6 m; 1 nm = 10-9 m; 1 Å = 10-8 cm = m; 1 µ = 10-6 m = 10 3 nm. N A = x mol -1 = Avogadro's constant force/energy 1 J = 1 N m = 1 C V; 1 pn = N; 1 pn nm = J = kj mol -1 ; J = 1 cal. pressure units: 1 pascal (Pa) = 1 N m -2 = 1 J m -3 = 10-5 bar 1atm = Pa = pn nm -2 = 760 mm Hg = lb/sq. in. = J cm -3 = kj m -3 electrostatics F = kj mol -1 V -1 = 96,485 C mol -1 ; q = x Coulombs = F /N A 4πε o = x C V -1 m -1 gas constant R J mol -1 K -1 = cal mol -1 K -1 = liter atm mol -1 K J K -1 (= R/N A = k B, the Boltzmann constant) unless otherwise stated: you may assume T = 298 K, P = 1 atm RT = 2.48 kj mol -1 and k B T = 4.11 x J at T = 298 K. for water, molecular weight = kg mol -1, liquid density = 1000 kg m -3, ε = 80 liquid heat capacity = 4.18 J gm -1 K -1 = 75.4 J mol -1 K -1 ; H fusion = 6.0 kj mol -1 ; H vap = kj mol -1 1

2 Name problem points 1a 5 1c 5 2a 3 2b 7 3a 7 3b 8 3c 5 4a 10 4b 4 4c 4 5a 5 5b 5 5c 5 5d 5 6a 10 6b 4 6c 8 total 100 extra credit +i 2

3 1 Realizing potential 1a (5 pts) Calculate the standard reduction potential E for the (½)O 2 /H 2 O half cell at 298 K: (½)O 2 + 2H + + 2e - H 2 O (l) using the free energy of formation for H 2 O (l) = kj mol -1 ; the free energies of formation of all other components (H +, O 2 and e - ) are 0 at ph 0. Describe the relationship between your result and the value listed in Table 11.2 of the text for this redox half cell (E = V). ( G f,products G f,reactants ) G = = G 1 f,h2 O 2 G O 2 2G H + 2G e = kj mol = 2FE E = 237 2F =1.229V in the text E ' = 0.816; E = E ' =1.230V The text gives E (reduction potential at ph 7) while this problem is based on free energies of formation at ph 0, and hence yields E (reduction potential at ph 0). The values should be the same after correction for 0.414(m/n), but they differ by 1 mv; for some reason the G f and E values are not exactly consistent. 3

4 1b (5 pts) A biological membrane has a surface charge density of 1 negative charge per 100 Å 2. The surrounding aqueous solution is at 298 K, with a dielectric constant = 78.5, a bulk ionic strength = 0.1 M and ph = What is the voltage at the surface of the membrane? σ = C m 2 = 0.16Cm 2 κ = Φ( 0) = σ m = m 1 ( )( ) 78.5 κε 0 ε = m 1 ( ) = 0.222V 4

5 2 Cellular redox problems A bioenergetically important redox reaction is the [NAD + ]/[NADH] couple: NAD + + H + + 2e - NADH E ' = V 2a (3 pts) If the ratio [NAD + ]/[NADH] is 10 inside a cell, what is the reduction potential at 298 K and ph 7 under these conditions? E = E!' RT nf [ ] [ ] ln red ox = [ ] log NADH "NAD # + $ % = ( 1) = 0.29V 5

6 2b (7 pts) Another important intracellular redox species is the peptide glutathione. Glutathione can be found in either the oxidized form GSSG (where G denotes a tripeptide) that contains an oxidized disulfide linkage, or in the reduced form GSH, where the disulfide is reduced to two sulfhydryl groups. The glutathione half-cell reaction may be written: GSSG + 2H + + 2e - 2GSH E ' = V Assume that the total concentration of glutathione, defined as 2[GSSG] + [GSH], is 10 mm. What is E (in Volts) at ph 7 when 80% of the glutathione pool is reduced, ie [GSH] = 8 mm? E = E!' RT nf [ ] [ ] ln red ox = [ ] 2 [ ] log GSH GSSG.008 = log.001 [ ] 2 [ ] = 0.205V 6

7 3 Decisions, decisions In the presence of O 2, micro-organisms can oxidize NH 4 + to either N 2 or NO 3 -. The following half-cell reactions will be useful for this problem. O 2 + 4H + + 4e - 2 H 2 O E = V 2 NO H e - N 2 (g) +6 H 2 O E = V N 2 (g) + 8H + + 6e - 2NH 4 + E = V 3a (7 pts) Using the half-cell reactions, write the balanced reaction for the oxidation of NH 4 + by O 2 to yield 1/2N 2. Calculate G for this reaction in kj mol -1. for the reaction at ph 7 2 x {2NH + 4 N 2 (g) + 8H + + 6e - } G = -12F(+0.277) 3 x {O 2 + 4H + + 4e - 2 H 2 O} G = -12F(+0.816) 4NH O 2 2N 2 (g) + 4H + + 6H 2 O G = kj mol -1 x ¼ = NH (¾)O 2 (½)N 2 (g) + H + + (3/2)H 2 O G = kj mol -1 it is ambiguous whether this was to be done at ph 7 (where I should have used G ) or ph 0 (more traditionally G ). It is OK if the student uses the ph 0 reference state. The relevant E values are (in order of the reactions listed above) , , V and the G for the balanced reaction is kj mol -1 7

8 3b (8 pts) Using the half-cell reactions, write the balanced reaction for the oxidation of NH 4 + by O 2 to yield NO 3 -. Calculate G for this reaction in kj mol -1. Per mole of NH 4 +, will ammonia oxidation to N 2 or NO 3 - release the greater amount of free energy at ph 7 and 1 atm O 2? 2NH + 4 N 2 (g) + 8H + + 6e - G = -6F(+0.277) N 2 (g) + 6 H 2 O 2 NO H e - G = -10F(-0.749) 2NH H 2 O 2 NO H e - G = kj mol -1 4 x {O 2 + 4H + + 4e - 2 H 2 O} G = -16F(0.816) = NH O 2 2 NO H H 2 O G = x ½ = NH O 2 NO H + + H 2 O G = kj mol -1 For the ph 0 reference state, G = kj mol -1. For the ph 7 reference state, NH 4 + oxidation to NO 3 - is the more energetically favorable per mole of NH 4 + ( for NO 3 - vs for N 2 ), but this is switched for the ph 0 reference state ( for NO 3 - vs for N 2 ). This is because the NO 3 - product has 2H + released, whereas N 2 has 1H +, so the former is more ph sensitive than the latter. 8

9 3c (5 pts) Given that the pka of NH 4 + is 9.25, calculate E for the reduction of N 2 to NH 3, using the value E = V for the reaction N 2 (g) + 8H + + 6e - 2NH 4 + : N 2 (g) + 6H + + 6e - 2NH 3 E =?? K a = 10 -pka = = x NH 4 + NH 3 + H + G = -RTlnK a = kj mol -1 N 2 (g) + 8H + + 6e - + 2NH 4 G = -6F(0.275) = kj mol -1 2NH + 4 2NH 3 + 2H + G = -2RTlnK a = kj mol -1 N 2 (g) + 6H + + 6e - 2NH 3 G = kj mol -1 E = /(-6F) = V 9

10 4 Electrostatics and cooperativity The pk a s of succinic acid are p = 5.61 and p = 4.16 for the dissociations of a proton from singly and doubly protonated succinic acid, respectively. 4a (10 pts) What is the ph when an average of one proton is bound by succinate? =10 p = =10 p = $H % + & ' + 2 $ H + % & K ν = 1 =1 $ % 1+ H + & ' $ + H + % & ' 2 $ % 1+ H + & ' $ + H + % & $ = H + % & ' + 2 $ H + % & ' 2 ' 2 ' 2 $ % 1= H + & $H % + & ' = = ' 2 ph = log$ % H + & ' =

11 4b (4 pts) Following our analysis of the interaction energy for the binding of the 4 th O 2 relative to the 1 st O 2 by hemoglobin, calculate the interaction energy (in kj mol -1 ) associated with the binding of the 2 nd proton relative to the 1 st proton by succinate. ΔG int = RT ln 4 = RT ln = 4.8 kj 4( mol-1 ) 11

12 4c (4 pts) For the binding of protons to succinate, electrostatic interactions make a significant contribution to this interaction energy, since the electrostatic repulsion between the two negative charged carboxylate groups in succinate is significantly reduced by protonation. Using Coulomb s law, calculate the electrostatic repulsion in kj mol -1 between two points of charge -1 separated by 4 Å in aqueous solution (ε = 80). How does this value compare to interaction energy derived in 4b? ( ) 2 ( )( 80) ( ) G ES = q 19 1q 2 4πε 0 εr = = J molecule -1 = N A = 4339 J mol -1 =4.4 kj mol -1 the results are surprisingly similar given the crudeness of the electrostatic model. This illustrates the dominance of Coulomb interactions and the consequences of dampening the ES interactions in the high dielectric constant of water. 12

13 5 Hemoglobin and allostery The binding of oxygen to the four binding sites of hemoglobin under certain experimental conditions is characterized by the following intrinsic dissociation constants, κ i, for the binding to a single site of the i th molecule of O 2. i κ i (mm Hg) K i (mm Hg) a (5 pts) Calculate the values of the macroscopic dissociation constants K i corresponding to the κ i. List your results in the above table. need to correct for the statistical factors between the microscopic and macroscopic constants = 1 4 κ 1 = 2 3 κ 2 K 3 = 3 2 κ 3 K 4 = 4κ 4 13

14 5b (5 pts) With the above dissociation constants, calculate the fractions of hemoglobin having 0, 1, 2, 3 and 4 molecules of bound O 2 at half saturation (ν = 2) at a partial pressure of oxygen, p, = 4.90 mm Hg. List your results in the table below. Hint: the terms in the binding polynomial Z have the following values when p = 4.9 mm Hg: Z =1+ p + p2 + p 3 K 3 + p 4 K 3 K 4 = = i fraction with i bound O f 0 = 1 Z = = f 1 = Z f 2 = Z f 3 = Z f 2 = Z = = = =

15 5c (5 pts) For a hypothetical protein with four identical and non-interacting O 2 binding sites, calculate the fractions of protein with 0, 1, 2, 3 and 4 molecules of bound O 2, under conditions where the binding sites are on average half-occupied (ν = 2). i fraction with i bound O 2 0 1/16= /16= /16= /16= /16=0.063 For identical and non-interacting sites, the distribution of states is given by the binomial distribution. For half-occupancy (the probability p of occupying a site = ½ (which also equals 1-p)), the fraction with i bound ligands of N = 4 is given by N! " 1 $ % 4 i! ( N i)! # 2 & ' 15

16 5d (5 pts) For a hypothetical O 2 binding protein with 4 subunits obeying perfect all-ornone positive cooperativity, calculate the fractions of protein with 0, 1, 2, 3 and 4 molecules of bound O 2, under conditions where the binding sites are on average halfoccupied (ν = 2). i fraction with i bound O 2 0 ½ ½ For all-or-none cooperativity, the only states that are occupied have either 0 or 4 bound ligands. The fractional occupancy of these two states when ν = 2 can be found as follows: ν = 0 f f 4 = 4 f 4 f 4 = ν 4 = 1 2 f 0 =1 f 4 =

17 6 Diffusion and transport 6a (10 pts) A photon of light in the interior of the sun undergoes a random walk to reach the sun's surface, due to collisions with protons and free electrons. If a photon's step size in this process is approximately 1 cm, and the sun's radius is 7 x 10 8 m, estimate how long it takes, in seconds and in years, for a photon to travel from the center to the surface of the sun? What is the corresponding diffusion constant for this process, in units of m 2 s -1? The speed of light, c ~ 3 x 10 8 m s -1 and assume a one-dimensional diffusion process. x 2 = ( ) 2 = nl 2 = n( 10 4 m 2 ) n = steps each 1 cm step takes τ = s = s total time = nτ = ( )( ) = s = 5200 years x 2 = 2Dt D = x2 2t ( ) 2 ( ) = m 2 s =

18 6b (4 pts) Due to the low solubility of ferric iron (Fe 3+ ) in aerobic environments, acquisition of this essential nutrient is a major challenge for many organisms. Human blood is a rich source of iron that is attractive to pathogens, and an elaborate set of strategies have evolved where pathogens try to get our iron and we try to keep it away from them. The major serum containing iron transport protein is transferrin (Tf), which binds Fe +3 with a dissociation constant estimated to be M: Tf-Fe +3 Tf + Fe +3 K = M What is the concentration of free Fe +3 in solution when transferrin is half saturated with ligand (ie when (Tf) = (Tf-Fe +3 ))? [ K = Tf ]! " Fe +3 # $ =10 22 M!Tf " Fe +3 # $ when Tf! " [ ] = Tf Fe +3 # $ at equilibrium,! " Fe+3# $ =10 22 M 18

19 6c (8 pts) An actively growing E. coli cell doubles and divides in ~1000 sec, which requires the uptake of 10 5 iron atoms during this time. Apply the "diffusion to capture" model and calculate the rate at which Fe +3 would reach the surface of the bacteria in a solution with the concentration of free Fe +3 equal to that determined in problem 6b. Can diffusion alone account for a sufficient uptake rate of iron to support the growth of this bacteria? For this problem, use D = 10-9 m 2 s -1 and assume the bacteria to be spherical with a constant radius 10-6 m (of course, this is not realistic for a growing cell, but it simplifies the analysis). If you are unsure of your answer to problem 6b, use [Fe +3 ] = M. for [Fe +3 ] = M, the number of Fe +3 per m 3 = (moles/liter) x 6.02 x (ions per mol) x 1000 (liters/m 3 ) = 6 x 10 4 ions/m 3 from the diffusion to capture problem, the number of Fe +3 reaching the surface of the bacteria per second = I = 4π DaC 0 = 4π 10 9 ( )( 10 6 )( ) = ions per second Over 1000 seconds, the total number of ions reaching the surface of the bacteria by diffusion alone would be ~8 x 10-7, which is insufficient to support the growth of this bacteria. In reality, bacteria can be motile (they swim around instead of waiting for diffusion to bring metabolites to them), they have active iron uptake systems and they will grow more slowly than a 1000 second doubling time under these conditions to get address this problem. 19

20 Extra credit: The catalytically active BS complex An enzyme E requires a bound activator, B, to form the EB complex that is then able to bind substrate S before converting it to product P. The overall mechanism is: 1/K B 1/K BS k E + B!!!!! EB + S!!!!! EBS!! E + B + P Only the EBS complex is catalytically active, and S can only bind to the EB complex (ie, the ES complex does not form). Derive the rate law using the rate-determining-step/preequilibrium treatment discussed in class on Tuesday, May 31 and determine the values of the exponents p, q and r (integers (positive, negative, or 0) in the following expression: v V max = p ( S)+ K BS ( S) q ( 1+ K B ( B) r ) [ ][ S] [ ] [ ][ B] [ EB] [ EBS] K BS = EB EBS K B = E v V max = [ EB] = EBS [ E] = EB [ EBS]+ [ EB]+ E [ S] [ ] K BS [ S] [ ] K B [ B] = [ EBS ] K B [ B] [ ] = 1 1+ K BS S [ ] + K B [ B] [ S] K BS S [ ] K BS S [ ] ( S) = = " K % [ S]+ K BS + K B BS $ ' [ S]+ K #[ B] BS 1+ K " % p q B ( S)+ K $ ' BS ( 1+ K B ( B) r ) & # [ B] & p = q = +1,r = 1 20

Chemistry 25 (Spring term 2015) Final Examination for NON-SENIORS GOOD LUCK AND HAVE A GREAT SUMMER!!!

Name Chemistry 25 (Spring term 2015) Final Examination for NON-SENIORS GOOD LUCK AND HAVE A GREAT SUMMER!!! Distributed Thursday, June 4 Due by Thursday, June 11 at 1 pm to 362 Broad Conditions Open this