ANSWER KEY. Chemistry 25 (Spring term 2017) Final Examination GOOD LUCK AND HAVE A GREAT SUMMER!!

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1 Name ANSWER KEY Distributed Thursday, June 1, 2017 Due Chemistry 25 (Spring term 2017) Final Examination GOOD LUCK AND HAVE A GREAT SUMMER!! Thursday, June 8, 2017 by 1 pm to the drop box outside 362 Broad ***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 the Ch25 online lecture notes, problem sets and solutions, the course web site and a calculator. You may use handwritten notes you have made from thermodynamics texts, including EC and KKW. You may not use any books (including EC and KKW), exams and problem sets from previous years of Ch25 (unless they are posted on the course website), other web sites, non-calculator applications of Mathematica and related programs, discuss the exam with others, etc. This exam should have 24 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. 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 molecules mol -1 = Avogadro's number 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 = 96,487 C mol -1 ; q = x Coulombs = F /N A, 1 ev = x J, 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 = 6.01 kj mol -1 ; H vap = kj mol -1 1

2 Name problem points 1a 5 1b 2 1c 5 1d 5 2a 5 2b 5 3a 4 3b 4 3c 4 3d 2 4a 5 4b 5 4c 5 5a 5 5b 5 5c 2 6a 5 6b 4 6c 5 7a 5 7b 4 7c 5 5 (no problem 7c ) 8 4 total 100 2

3 1 (17 pts) All charged up The Donnan potential is relevant to the membrane potentials of cells, since there are differences in the concentrations of various ions between the two sides of the membrane. The following table lists the ionic concentrations for Na +, K + and Cl - in skeletal muscle: ion extracellular conc. (mm) intracellular conc. (mm) Na K Cl Equilibrium potential (mv) 1a (5 pts) Excitable cells can open ion-selective channels so that their membranes are permeable to one type of ion, which then sets the membrane potential as described by the Donnan potential. Calculate the equilibrium potential difference at 298 K across a membrane (defined as Φ inside Φ outside ) for each of the ions in the above table. Eq. 5.5 (with extracellular = outside and intracellular = inside) ( ) ( ) Φ inside Φ outside = RT zf ln c outside c inside 1b (2 pts) All excitable cells have negative membrane potentials when they are at rest; to which ion (or ions) are their membranes likely permeable? (a short sentence is all that is necessary) the membrane could be permeable to either (or both) K+ and Cl- since their concentrations are consistent with a negative membrane potential. 3

4 1c (5 pts) The solubility of the protein lactoglobulin increases by a factor of 100 as the ionic strength of the solution is increased by the addition of NaCl from 0 to 0.1 M in aqueous solution at 298 K. This is an example of the salting-in effect. Since the chemical potentials for lactoglobulin in the crystal (solid) and solution phases are equal in a saturated solution at equilibrium, this requires that the activity, a = γc, of lactoglobulin dissolved in solution remains constant under these conditions (assuming that the chemical potential of the solid lactoglobulin is unaffected by the changes in ionic strength). Consequently, the change in concentration of dissolved lactoglobulin with ionic strength must be compensated by the ionic strength dependence of the activity coefficient, such that the product γc is constant. z 2 Using the Debye-Huckel limiting law: lnγ i = i e 2 8πε o εk B T κ = 1.18z 2 i I, water 298 K, calculate the net charge on lactoglobulin from the solubility data. Neglect the contribution of the protein to the ionic strength. (NOTE: in the exam, the factor of 4πε o was omitted, but the numerical value is still correct) z 2 lnγ i = i q 2 8πε o εk B T = 1.18z 2 i ( γc) I=0 M = ( γc) I=0.1M ( c) I=0.1M =100 c γ I=0.1M ( ) I=0 M = γ I=0 M 2 ln[ γ I=0 M ] ln[ γ I=0.1M ] = z i ln[ 100] z 2 i = =12.3 z i = ± = ln[ 100] I for water, 298K This information was not provided in the problem, but the charge will be negative at ph 7, based on the isoelectric point ~5.2 4

5 1d (5 pts) The ion product of water, K W, is defined as the product of the activities of H + and OH -! in pure water. At 298 K, K W = a H + a OH = In pure water (ionic strength ~ 0), activities and concentrations are equivalent. Due to charge balance considerations, the concentration of [H + ] must equal the concentration of [OH - ] in pure water, so that H + [ ] = a H + = K w! =10 7 M. The concentration constant, K W, is defined as the product of the concentrations of H + and OH - : K W = H + [ ][ OH ] = K! W /( γ H +γ OH ) What is the concentration of [H + ] in an aqueous solution of 0.3 M NaCl at 298 K? Assume that the Debye-Huckel limiting law is valid for the activity coefficients of H + and OH - ions. The contributions of H + and OH - ions to the ionic strength may be neglected. K! W K W = γ H +γ OH = # $ H + % & # $ OH % & = # $ H + % & ±1 γ H + = γ OH = e ( )2 0.3 = ( 0.524) = = # $ H + % #H + $ % & = = log 10 # $ H + % & (not requested) & 2 5

6 2 (10 pts) Electrostatics, the Debye Huckel model and protein stability As a rough generalization, proteins are most stable at their isoelectric ph where the net charge Q on the protein is zero. As the ph changes in either direction from the isoelectric point, the absolute magnitude of the charge on the protein will increase due to the titration of ionizable groups. This results in an net electrostatic repulsion between like charges that contributes to a destabilization of the native state (N) of the protein relative to the denatured state (D), since the charges will be farther apart on average in the denatured state. A simplified model for the dependence of protein stability on the net charge Q of a protein may be developed based on the Born model for the energy of a spherical charge. Q 2 G Born = 8πε 0 εa = z2 q 2 8πε 0 εa where Q is the charge of the protein in Coulombs (with Q = z q, where z is the valence charge and q is the charge of the proton) and a is the radius of the sphere. Note that this expression gives the energy of a single ion (protein) (not a mole). A simple model for protein stability describes the free energy of unfolding, G u, as the sum of two terms, a non-electrostatic term, G non-el, and the electrostatic Born energy: G u = G non el G Born This model assumes that the Born energy destabilizes only the native state when Q 0, which is clearly an oversimplification. We will further assume that G non-el is independent of the protein charge Q. G#(kJ#mol *1 )# 70" 60" 50" 40" 30" 20" 10" G#unfolding#vs#z 2# y"=73"&"0.16"z 2 "" 0" 0" 100" 200" 300" 400" z 2# 2a (5 pts) For lysozyme, the free energy of unfolding (in kj mol -1 ) has been experimentally measured at different phs at 298 K in water and is plotted above as a function of the protein charge squared. If a = 15 Å and ε = 78.5, calculate the slope anticipated from the Born charging model and compare to the observed slope (-0.16 kj mol -1 charge -2 ). Do you expect this electrostatic model to over-, under- or exactly estimate the observed value? Explain briefly (1-2 sentences). Use an ionic strength of 0 (ie ignore activity coefficients). 6

7 q 2 ( ) 2 G Born = 8πε 0 εa z2 = ( ) 78.5 = J charge -2 z 2 N A to convert to moles = 589 J mol 1 charge -2 z 2 = kj mol 1 charge -2 z 2 ( ) z2 ( ) compared to the experimental value (0.16 kj mol -1 charge -2 ), this model will overestimate the significance of electrostatic effects since they are assumed to only destabilize the N state, while the contributions to the D state are completely neglected. Note: data came from Yang and Honig (JMB 231, 459 (1993)); titration curves from Tanford & Roxby (Biochem 11, 2192 (1972)); with calorimetry data from Privalov and Pfeil (including some info from Pfeil's book on protein stability) 7

8 2b (5 pts) Of course, it is usually a bad approximation to neglect activity coefficients when considering the thermodynamics of charged species. Because of the surrounding counterions, the chemical potential of a charged species will be lower in the presence of electrolytes by an amount given by the Debye-Huckel expression: µ γ = k B T lnγ = z2 q 2 κ 8πε 0 ε 1+κa (Notes: (i) because of the large size of the protein, we include the term 1+κa in the denominator; (ii) this expression is for a single ion (ie one protein), and not for a mole of ions). Including this term, our expression for the unfolding free energy becomes G u = G non el G Born + µ γ ( ) where the term in parentheses approximates the electrostatic energy for the N state. This again makes the dubious approximation that the electrostatic interactions affect only the activity of the N state (ie we ignore the contribution of both charge and ionic strength to the chemical potential of the D state, but we justify this approximation by simplifying the calculations for Ch 25). For an ionic strength of 0.2 M, evaluate the numerical factor multiplying z 2 for the sum of G Born + µ γ. Does the addition of this term improve or worsen the agreement with the experimental value (-0.16 kj mol -1 charge -2 )? κ = 0.2 ( ) = m -1 µ γ = q2 κ 8πε 0 ε 1+κa z2 = = J charge -2 z 2 N A to convert to moles = -405 J mol 1 charge -2 z 2 ( ) = kj mol 1 charge -2 z 2 ( ) 78.5 ( ) ( 1+ ( )( )) z 2 ( ) = vs 0.16 observed experimentally - this agreement is amazingly close - too close, given the crudeness of the electrostatic model - the strongest statement that can probably be made is that the terms most likely have the correct sign. 8

9 9

10 3 (14 pts) Cooperativity - the more, the merrier. Hemoglobin must be able to pick up oxygen in the lungs at a partial pressure of 100 mm Hg, while releasing it throughout the rest of the body where the oxygen pressure may be closer to 20 mm Hg. 3a (4 pts) An individual R state subunit in hemoglobin may be modeled by a single myoglobin subunit that binds oxygen with an intrinsic association constant of 2 mm Hg -1. For myoglobin with this intrinsic association constant, calculate the moles O 2 bound per mole myoglobin at oxygen partial pressures of 100 mm Hg and 20 mm Hg. Based on this analysis, would myoglobin provide an efficient mechanism to deliver oxygen from the lungs to actively respiring tissue? for myoglobin ν = K p 1+ K p p =100 mm Hg, ν = = p = 20 mm Hg, ν = = myoglobin would not be so effective since it will only release ~2% of the bound O 2. 10

11 3b (4 pts) Paradoxically, by incorporating an inactive T state into hemoglobin, as well as making hemoglobin an oligomer, cooperative oxygen binding may be achieved. The balance between oxygenated and deoxygenated states of hemoglobin is reflected in the equilibrium constant L between unliganded T and R states. For the simplified MWC model for hemoglobin we discussed in class, the binding curve may be expressed as: ( )n 1 y = ν n = α 1+ α 1+ α ( ) n + L where y is the fractional occupancy, n is the number of subunits and α is the normalized concentration = K p, where K is the intrinsic R state association constant (2 mm Hg -1 ) and p is the partial pressure of O 2 in mm Hg. For a tetramer (n = 4), what value of L would be required so that y = 0.5 when p = 60 mm Hg (the average of 20 and 100 mm Hg), with K = 2 mm Hg -1? at p = 60 mm Hg, α = K.p = α ( ) 3 ( ) 4 + L y = 1 2 = α 1+α 1+α (( ) 4 + L) = α ( 1+α) 3 L = 2α ( 1+α) 3 ( 1+α) 4 ( ) 3 ( α 1) solve for L at α = 120 ( ) = 1+α = =

12 3c (4 pts) Using the value for L obtained in problem 4b, calculate the fractional occupancies of hemoglobin at oxygen partial pressures of 20 and 100 mm Hg. Relative to myoglobin, does hemoglobin have better or worse oxygen binding characteristics for delivering O 2 from the lungs to actively respiring tissue? If you are unsure of your answer for L from 4b, use L = L = 2.11 x 10 8 at p = 100 mm Hg, α = 200 ( ) 3 ( ) 4 + L y = α 1+α 1+α ( ) = L = at p = 20 mm Hg, α = 40 ( ) y = L = (with L = 10 8, y = and 0.027, respectively) hemoglobin is much better than myoglobin at delivering O 2 from lungs and releasing it in actively respiring tissue. 3d (2 pts) The binding of small organic phosphate molecules such as diphosphoglycerate (DPG) to hemoglobin provides a physiological adaptation to adjust the oxygen binding properties of hemoglobin to high altitudes. DPG increases the cooperativity of oxygen binding to hemoglobin. In terms of the MWC model, does the increased cooperativity of hemoglobin with bound DPG reflect an increase or decrease in the value of L? Explain briefly (one sentence). increased cooperatively implies increased L (conversely, decreased L implies decreased cooperativity - in the limit L = 0, the oxygen binding sites are all identifical and non-interacting) 12

13 4 (15 pts) Binding agreements 4a (5 pts) Zinc is often found coordinated to the imidazole (Im) side chains of histidine residues in proteins. In model systems, zinc is found to coordinate up to 4 imidazoles, according to the following scheme: K K K K Zn + Im ZnIm + Im ZnIm2 + Im ZnIm3 + Im ZnIm4 K 1, K 2, K 3 and K 4 are the measured macroscopic association constants for this system. The experimental values for these constants are listed below. Extract the intrinsic microscopic constants, κ i, from the K i for each binding step, and indicate whether the binding process shows no cooperativity, positive cooperativity or negativity cooperativity for each step. K M -1 κ 1 69 K M -1 κ K M -1 κ K M -1 κ κ 1 = 1 4 K 1 κ 2 = 2 3 K 2 κ 3 = 3 2 K 3 κ 4 = 4K 4 since all the κ i s increase as i increases imidazole binding shows positive cooperativity at each step. 13

14 4b (5 pts) Aspartate transcarbamylase (ATCase) is an enzyme with 6 equivalent active sites that catalyzes the reaction of aspartate and carbamyl phosphate to form N- carbamyl aspartate, in the first committed step of pyrimidine biosynthesis. Succinate acts as a competitive inhibitor for aspartate, binding to the same location in the active site. The dependence of the enzyme velocity, v, on (aspartate) is shown below in Figure A (carbamyl phosphate is at saturating concentrations and can be ignored). The sigmoidal dependence of v on (aspartate) is as expected for an allosteric enzyme. In the second experiment, the concentration of (aspartate) is held constant at the low level indicated by the arrow in Figure A, and increasing amounts of succinate are added. The dependence of v on (succinate) is shown in Figure B, and reveals the surprising result that although succinate is a competitive inhibitor, it increases the activity of ATCase at low concentrations. Succinate cannot participate as a substrate in this reaction. Using the Monod-Wyman-Changeux model, briefly explain these results qualitatively. ATCase is an allosteric enzyme with 6 active sites. At low (succinate), binding of succinate to the active site shifts the equilibrium from T to R, thereby creating more R state enzyme and increasing the activity. Provided the concentration of succinate is low enough, the binding of succinate to only one active site in an active hexamer stabilizes be 5 empty active sites that can binding aspartate, leading to the increase in activity. As (succinate) increases, however, it occupies more and more active sites, which prevents aspartate from binding. At sufficiently high succinate concentration, succinate will bind to most of the active sites, thereby keeping aspartate from binding even thought ATCase will be predominately in the R-state (after all, succinate is a competitive inhibitor). 14

15 4c (5 pts) P and L interact to form a complex PL with association constant K. K P + L! (!!! PL with K = PL ) P ( )( L) In our analysis of ligand binding, we generally assumed that the ligand, L, is in vast excess over the protein, P, so that the binding of ligand to P to form PL does not appreciably impact the concentration of free L. Under these conditions, ( PL) Eq. A ν = ( P)+ ( PL) = K ( L ) 1+ K ( L) = 1 2 when ( L ) = 1 K. In the case of tight binding systems where the total ligand concentration comparable to the total protein concentration, however, the formation of PL will influence the amount of free L left in solution. The following example illustrates this point. Define the total concentrations of the protein and ligand as (P tot ) and (L tot ), respectively: ( P)+ ( PL) = ( P tot ) ( L)+ ( PL) = ( L tot ) Taking into account the effect of (PL) on free (L), calculate ν for the case where ( P tot ) = ( L tot ) = 1 K. Hint: incorporate the conservation of mass relations into the association constant expression and solve for (PL) when ( P tot ) = ( L tot ) = 1 K ( K = PL ) ( P) ( L) = ( PL) ( ) ( PL) ( ) ( PL) "# P $ tot % L tot "# $ % Explain briefly why ν for the tight binding case is less than that observed when (L) = 1/K for the weak binding case of Eq. A. 15

16 ( ) ( ) PL K = "# ( P tot ) ( PL) $ % "# ( L ) ( PL) $ = PL tot % " 1 K ( PL $ " ) # & % ' 1 K ( PL $ ) # & % ' " ( PL) = K 1 K ( PL $ ) # & % ' ( PL ) = K "( 1 + * ) K - #&, ( PL) = ( K PL )+ PL 0 = 1 K 3 ( PL )+ K ( PL) 2 3± 9 4 ( * 1 ) K 2K ( ν = PL ) P tot + -K, ( ) 2 = 3± 5 2K ( ) = K ( PL ) = 3± 5 2 $ %' = 0.38 and 2.61 since ν must be less than or equal to 1, in this case, ν = The tight binding case has a smaller value of ν since binding of L to P when (L tot ) = 1/K reduces the concentration of free L below 1/K which is required for ½ occupancy binding. 16

17 5 (12 pts) Who needs O 2? 5a (5 pts) A anaerobic denitrification process termed anammox (for anaerobic ammonia oxidation) has been identified in microorganisms from the Black Sea. The overall reaction is: Eq. A NH NO 2 N 2 (g) + 2H 2 O From the thermodynamic data tabulated in Eqs B-C below, calculate G for this process as written above in Eq. A. If this value reflects the actual free energy change of the physiological process, and if it takes 50 kj/mole to synthesize ATP inside a cell, are the energetics of the anammox process sufficient to drive ATP synthesis? Eq. B 2NO H + + 6e - N 2 (g) + 4H 2 O E = V Eq. C N 2 (g) + 8H + + 6e - 2NH + 4 E = V 2NH + 4 N 2 (g) + 8H + + 6e - G = -nfe = -6F(+0.277) 2NO H + + 6e - N 2 (g) + 4H 2 O G = -nfe = -6F(+0.968) 2NH NO 2 2N 2 (g) + 4H 2 O G = -6(96.487)( ) = kj mol -1 for the reaction as written in Eq. A: NH NO 2 - N 2 (g) + 2H 2 O G = /2 = kj mol -1 Thermodynamically, this G is ample to drive the synthesis of multiple ATP molecules. 17

18 5b (5 pts) The mechanism of this reaction proceeds through the reduction of nitrite, NO - 2, to nitric oxide (NO) that is then condensed with ammonia (NH 3 ) to yield hydrazine (N 2 H 4 ). Hydrazine is quite reactive and toxic, and in the last step of the anammox process, hydrazine is oxidized to N 2. This last step is coupled to generation of a protonmotive force across the membrane of an unusual bacterial organelle (the anammoxosome), driving ATP synthesis through the reverse of the following reduction half cell Eq. D N 2 (g) + 4H e - N 2 H 4 If the free energy of formation of hydrazine is 149 kj mol -1 at 298 K, calculate E and E for Eq. D. (Hint: remember E = 0 V for 2H + + 2e - H 2 ). N 2 + 2H 2 N 2 H 4 G = 149 kj mol -1 4H + + 4e - 2H 2 G = -2F(0) = 0 N 2 (g) + 4H e - N 2 H 4 G = = 149 kj mol -1 = -4FE E = 149/(-4F) = 149/(-4x96.487) = V E = E m/n = = V 5c (2 pts) You should find that hydrazine is a stronger reductant than NADH and that it should be able to reduce H + to form H 2 briefly suggest a plausible explanation why this doesn t happen during the enzyme catalyzed anammox reaction. When bound to the active site of the enzyme, access of H+ to hydrazine is likely restricted. 18

19 6 (14 pts) Where was I going? 6a (5 pts) In our introduction to transport processes, we discussed how the step size δ and time interval τ appropriate for a random walk analysis of diffusion may be estimated for a protein from the one dimensional diffusion coefficient and the mean square velocity from the kinetic theory of gases: D = δ 2 k and B T 2τ 2 = mv 2 x 2 = m $ δ & ' 2 ) 2 % τ ( δ and τ were explicitly evaluated for lysozyme, a water-soluble protein of molecular weight gm mol -1, and found to be ~10-11 m and s, respectively. The diffusion constants for membrane proteins in a lipid bilayer are ~2 orders of magnitude smaller than for water-soluble proteins, with values for D = m 2 s -1 measured for a membrane protein with the same molecular weight of lysozyme at 298 K. (i) estimate the step size δ (in meters) and time interval τ (in sec) characterizing the random walk properties of a 14 kd membrane protein. (ii) Using the Stokes equation for the frictional coefficient of a sphere of radius 16 Å and D = m 2 s -1, estimate the corresponding viscosity of a membrane bilayer. How much more viscous than water is a membrane? (i) from the analysis on page 8-1 of the lecture notes, for a membrane protein of 14 kd 2 mv x 2 = k BT 2 v ~ v x 2 D = δ 2 1 2! = k BT $ # & " m % 1 2 = δ =13 m s 1 τ 2τ =10 12 m 2 s -1 for a membrane protein the same molecule weight as lysozyme δ = 2D v ~ m τ = δ v ~ sec (ii) D = k B T f η = = k B T 6πηR k B T 6π DR = 6π ( )( ) = 0.14 J m-3 s = 0.Pa s, or 140x the viscosity of water 19

20 6b (4 pts) The movement of charged molecules in an electric field can be analyzed by the same approach we used for sedimentation with the following force balance equation (lecture note Eqn. 8.19): fv = qe u = v E = q f u is the electrophoretic or ionic mobility. It can be measured experimentally, or estimated for spherical ions using Stoke s equation for the frictional coefficient. Using the Stoke s equation for f, calculate u in m 2 s -1 V -1 for a spherical ion of +1 valence charge and a diameter of 3 Å in water (η = Pa s) at 298 K. q u = 6πηR = π 10 3 ( )( ) = m 2 s 1 V 1 20

21 6c (5 pts) The conductivity, G, of a solution is determined by the mobilities, u i, and concentrations, c i, of the of the constituent ions, i, through the relationship: G = F where F is the Faraday (this is for the specific case where the valence charges of the ions are either +1 or -1). The conductivity of water provides a sensitive test for purity (at least for ionic contaminations), since the conductance of pure water should only be determined by the Ionic mobilities of H + and OH -, which at 298 K are and x 10-8 m 2 s -1 V -1, respectively. Assuming the concentrations of (H + ) = (OH - ) = 10-7 M, calculate G for pure water at ph 7 and express the results in the SI units of S m -1, where S = the SI unit of conductance (siemen). Hint: resistivity is 1/conductivity. Water deionizing systems measure the resistivity of the water they produce, which for freshly deionized water is typically reported in the non-si units as ~18 megaohm cm. Ohm is the reciprocal of siemen, and working through the conversion factors, 18 megohm cm = ohm m = S -1 m. Fresh deionized water produced by our Physical Plant has a resistivity of megaohm cm, but as CO 2 is absorbed and becomes hydrated to carbonic acid, the ph and resistivity drop to ~ 2 megaohm cm when it comes out of the deionized water taps in our labs. G = F c i u i = 96487$ %( ) ( ) & ' i = Sm 1 $ 1 G =1.8 & 105 S 1 m = S 1 m =18 MΩcm % ) ' * not requested, just FYI i c i u i 21

22 7 (14 pts) K decay and hot bodies 7a (5 pts) A typical adult human contains about 160 gms of potassium. Approximately 0.01% of this is a radioactive isotope of potassium, 40 K. 40 K has a half life of 1.3 x 10 9 years. Calculate the number of decays per second (ie, Becquerels) expected from the decay of 40 K in this typical person. 160 gm 10 4 = gm 40 K gm 40 gm mol atoms mol 1 = atoms 40 K k = ln = t 1/ yr 365 day/year 24 hr/day 3600 sec/hr = s 1 dn dt = kn = s atoms 40 K = 4073 atoms/s = 4073 Bq 22

23 7b (4 pts) 40 K decays by two major pathways to form 40 Ca and 40 Ar in the relative ratio of 89 % : 11 %. (the β's represent emitted particles of no relevance to this particular problem) ( ) ( ) K k 1!! Ca + β 89% K k 2!! Ar + β + 11% Assuming these decay processes are all first order, determine the numerical values for the rate constants k 1 and k 2, in s -1. d 40 K ( ) = k 40 1 ( K) k 40 2 ( K) = s 1 ( 40 K) dt k 1 = = s 1 k 2 = = s 1 23

24 8 (4 pts) Last but not least Briefly explain (1 sentence each) the contributions of 2 of the following scientists to our understanding of bioenergetics and hydrophobicity: Peter Mitchell, Agnes Pockels, and Benjamin Franklin and Charles Tanford. Peter Mitchell chemiosmotic theory for bioenergetics Agnes Pockels early studies of hydrophobicity by inventing a device for measuring surface forces Benjamin Franklin early studies of hydrophobicity by measuring surface area of oil on the surface of a pond Charles Tanford quantitation of hydrophobicity by measurement of transfer free energies of solutes from non-aqueous to aqueous solutions. 24