Ebook Physical chemistry for the life sciences (2nd edition) Part 1

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(BQ) Part 1 book Physical chemistry for the life sciences has contents: The first law, the second law, phase equilibria, chemical equilibrium, thermodynamics of ion and electron transport, the rates of reactions, accounting for the rate laws, complex biochemical processes. This page intentionally left blank Physical Chemistry for the Life Sciences Library of Congress Number: 2010940703 © 2006, 2011 by P.W Atkins and J de Paula All rights reserved Printed in Italy by L.E.G.O S.p.A First printing Published in the United States and Canada by W H Freeman and Company 41 Madison Avenue New York, NY 10010 www.whfreeman.com ISBN-13: 978-1-4292-3114-5 ISBN-10: 1-4292-3114-9 Published in the rest of the world by Oxford University Press Great Clarendon Street Oxford OX2 6DP United Kingdom www.oup.com ISBN: 978-0-19-956428-6 Physical Chemistry for the Life Sciences Second edition Peter Atkins Professor of Chemistry, Oxford University Julio de Paula Professor of Chemistry, Lewis & Clark College W H Freeman and Company New York This page intentionally left blank Contents in brief Prolog Fundamentals xxi PART Biochemical thermodynamics 21 The First Law The Second Law Phase equilibria Chemical equilibrium Thermodynamics of ion and electron transport 23 69 94 135 181 PART The kinetics of life processes 217 The rates of reactions Accounting for the rate laws Complex biochemical processes 219 243 273 PART Biomolecular structure Microscopic systems and quantization 10 The chemical bond 11 Macromolecules and self-assembly PART Biochemical spectroscopy 12 Optical spectroscopy and photobiology 13 Magnetic resonance 311 313 364 407 461 463 514 Resource section Atlas of structures Units Data 546 558 560 Answers to odd-numbered exercises Index of Tables Index 573 577 579 This page intentionally left blank Full contents Prolog xxi The structure of physical chemistry xxi (a) The organization of science xxi (b) The organization of our presentation xxii Applications of physical chemistry to biology and medicine (a) Techniques for the study of biological systems xxii 1.4 The measurement of heat 32 (a) Heat capacity 33 (b) The molecular interpretation of heat capacity 34 Internal energy and enthalpy 1.5 The internal energy (a) Changes in internal energy xxii 34 35 35 (b) Protein folding xxiii Example 1.1 Calculating the change in internal energy (c) Rational drug design xxv (b) The internal energy as a state function 37 (d) Biological energy conversion xxv (c) The First Law of thermodynamics 38 1.6 The enthalpy 36 38 Fundamentals (a) The definition of enthalpy 39 F.1 Atoms, ions, and molecules (b) Changes in enthalpy 39 (c) The temperature dependence of the enthalpy 41 (a) Bonding and nonbonding interactions (b) Structural and functional units (c) Levels of structure F.2 Bulk matter (a) States of matter (c) F.3 Energy (a) Varieties of energy (b) The Boltzmann distribution Checklist of key concepts Checklist of key equations Discussion questions Exercises Projects (a) Bomb calorimeters (b) Physical state Equations of state In the laboratory 1.1 Calorimetry Example 1.2 Calibrating a calorimeter and measuring the energy content of a nutrient (b) Isobaric calorimeters (c) Differential scanning calorimeters 10 11 13 17 17 18 18 19 Physical and chemical change 1.7 21 The First Law 23 The conservation of energy 23 1.1 Systems and surroundings 24 1.2 Work and heat 25 44 46 46 46 Thermochemical properties of fuels The combination of reaction enthalpies Example 1.4 Using Hess’s law Standard enthalpies of formation Example 1.5 Using standard enthalpies of formation 49 51 52 55 57 58 58 59 25 1.12 Enthalpies of formation and computational chemistry 61 (b) The molecular interpretation of work and heat 26 1.13 The variation of reaction enthalpy with temperature 62 (c) The molecular interpretation of temperature 26 (a) Exothermic and endothermic processes Case study 1.1 Energy conversion in organisms 1.3 44 47 Case study 1.2 Biological fuels 1.11 43 (b) Enthalpies of vaporization, fusion, and sublimation Example 1.3 Using mean bond enthalpies 1.10 42 (a) Phase transitions 1.8 Bond enthalpy 1.9 PART Biochemical thermodynamics Enthalpy changes accompanying physical processes 42 27 The measurement of work 29 (a) Sign conventions 29 (b) Expansion work 30 (c) Maximum work 31 Example 1.6 Using Kirchhoff’s law 63 Checklist of key concepts Checklist of key equations Discussion questions Exercises Projects 64 65 65 65 68 viii FULL CONTENTS The Second Law 69 Entropy 70 2.1 The direction of spontaneous change 70 2.2 Entropy and the Second Law 71 (a) The definition of entropy 71 2.3 2.5 112 (c) The chemical potential of a solute 114 Example 3.2 Determining whether a natural water can support aquatic life 116 Case study 3.2 Gas solubility and breathing 117 73 (c) The entropy change accompanying a phase transition 75 (d) Real solutions: activities 118 (d) Entropy changes in the surroundings 77 Case study 3.3 The Donnan equilibrium 119 Absolute entropies and the Third Law of thermodynamics 77 Example 3.3 Analyzing a Donnan equilibrium 121 78 The molecular interpretation of the Second and Third Laws 80 (a) The Boltzmann formula 80 (b) The relation between thermodynamic and statistical entropy 81 (c) The residual entropy 82 Entropy changes accompanying chemical reactions 82 (a) Standard reaction entropies 82 (b) The spontaneity of chemical reactions 83 The Gibbs energy 2.6 112 (b) The chemical potential of a solvent (b) The entropy change accompanying heating In the laboratory 2.1 The measurement of entropies 2.4 (a) The chemical potential of a gas 84 Focusing on the system 84 (a) The definition of the Gibbs energy 84 (b) Spontaneity and the Gibbs energy Case study 2.1 Life and the Second Law 85 85 (e) The thermodynamics of dissolving Colligative properties 3.9 The modification of boiling and freezing points 3.10 Osmosis 121 122 123 125 In the laboratory 3.1 Osmometry 127 Example 3.4 Determining the molar mass of an enzyme from measurements of the osmotic pressure 127 Checklist of key concepts Checklist of key equations Further information 3.1 The phase rule Further information 3.2 Measures of concentration 128 129 129 130 Example 3.5 Relating mole fraction and molality 131 132 132 134 2.7 The hydrophobic interaction 86 2.8 Work and the Gibbs energy change 88 Discussion questions Exercises Projects Example 2.1 Estimating a change in Gibbs energy for a metabolic process 89 Chemical equilibrium 135 Case study 2.2 The action of adenosine triphosphate 90 Thermodynamic background 135 Checklist of key concepts Checklist of key equations Discussion questions Exercises Projects 90 91 91 91 92 Phase equilibria 94 The thermodynamics of transition 94 4.1 The reaction Gibbs energy 135 4.2 The variation of ΔrG with composition 137 (a) The reaction quotient 137 Example 4.1 Formulating a reaction quotient (b) Biological standard states Example 4.2 Converting between thermodynamic and biological standard states 4.3 142 94 (a) The significance of the equilibrium constant 3.2 The variation of Gibbs energy with pressure 95 (b) The composition at equilibrium 3.3 The variation of Gibbs energy with temperature 98 (b) The location of phase boundaries 101 (c) Characteristic points 103 (d) The phase diagram of water 105 Phase transitions in biopolymers and aggregates 3.5 The stability of nucleic acids and proteins Example 3.1 Predicting the melting temperature of DNA 3.6 99 100 Phase transitions of biological membranes Case study 3.1 The use of phase diagrams in the study of proteins The thermodynamic description of mixtures 106 106 140 140 The condition of stability (a) Phase boundaries 139 Reactions at equilibrium 3.1 3.4 Phase diagrams 138 143 Example 4.3 Calculating an equilibrium composition 143 (c) The molecular origin of chemical equilibrium 144 Case study 4.1 Binding of oxygen to myoglobin and hemoglobin 144 4.4 The standard reaction Gibbs energy Example 4.4 Calculating the standard reaction Gibbs energy of an enzyme-catalyzed reaction 146 146 107 (a) Standard Gibbs energies of formation 147 108 (b) Stability and instability 149 The response of equilibria to the conditions 109 4.5 The presence of a catalyst 4.6 The effect of temperature 149 150 150 110 3.7 The chemical potential 110 3.8 Ideal and ideal–dilute solutions 111 Coupled reactions in bioenergetics 151 Case study 4.2 ATP and the biosynthesis of proteins 152 296 COMPLEX BIOCHEMICAL PROCESSES Fig 8.20 The mechanism of action of H+-ATPase, a molecular motor that transports protons across the mitochondrial membrane and catalyzes either the formation or hydrolysis of ATP The yellow shapes represent the species ADP, ATP, and Pi and the new L site is ready for the cycle to begin again The proton flux drives the rotation of a g subunit, and hence the conformational changes of the ab segments, as well as providing the energy for the condensation reaction itself Several key aspects of this mechanism have been confirmed experimentally For example, the rotation of a g subunit has been portrayed directly by using single-molecule spectroscopy (In the laboratory 12.6) Electron transfer in biological systems We saw in Case studies 4.2 and 4.3 that exergonic electron transfer processes drive the synthesis of ATP in the mitochondrion during oxidative phosphorylation Electron transfer between protein-bound co-factors or between proteins also plays a role in other biological processes, such as photosynthesis (Section 5.11 and Case study 12.3), nitrogen fixation, the reduction of atmospheric N2 to NH3 by certain microorganisms, and the mechanisms of action of oxidoreductases, which are enzymes that catalyze redox reactions We begin by examining the features of a theory that describes the factors governing the rates of electron transfer Then we discuss the theory in the light of experimental results on a variety of systems, including protein complexes We shall see that relatively simple expressions can be used to predict the rates of electron transfer between proteins with reasonable accuracy 8.9 The rates of electron transfer processes Electron transfer is of crucial importance in many biological reactions, and we need to see how to use the strategies we have developed to discuss them quantitatively Consider electron transfer from a donor species D to an acceptor species A in solution The net reaction, the observed rate law, and the equilibrium constant are D + A → D+ + A− v = kobs[D][A] K= [D+][A−] [D][A] Electron transfer (8.24) 8.9 THE RATES OF ELECTRON TRANSFER PROCESSES The proposed mechanism is: Electron transfer D + A DA ka, k a′ DA → D+A− vet = ket[DA] D A → DA vret = k′et[D+A−] D+A− → D+ + A− vd = kd[D+A−] + − KDA = kac [DA]c = k a′ [D][A] In the first step of the mechanism, D and A must diffuse through the solution and encounter to form a complex DA, in which the donor and acceptor are separated by a distance comparable to r, the distance between the edges of each species Next, electron transfer occurs within the DA complex to yield D+A− The D+A− complex has two possible fates One is the regeneration of DA The other is to break apart and for the ions to diffuse through the solution We show in the following Justification that kobs in eqn 8.24 is given by 1 k′ A k′ D = + a + et kobs k a kaket C kd F Electron transfer rate constant (8.25) Justification 8.4 The rate constant for electron transfer in solution We begin by equating the rate of the net reaction (eqn 8.24) to the rate of formation of separated ions, the reaction products: v = kobs[D][A] = kd[D+A−] Next, we apply the steady-state assumption to the intermediate D+A−: d[D+A−] = ket[DA] − k′et[D+A−] − kd[D+A−] = dt It follows that [D+A−] = ket [DA] k′et + kd However, DA is also an intermediate, so we apply the steady-state approximation again: d[DA] = ka[D][A] − k a′[DA] − ket[DA] + k′et[D+A−] = dt Substitution of the initial expression for the steady-state concentration of D+A− into this expression for [DA] gives, after some algebra, a new expression for [D+A−]: [D+A−] = kaket [D][A] k a′k′et + k a′kd + kd ket When we multiply this expression by kd, we see that the resulting equation has the form of the rate of electron transfer, v = kobs[D][A], with kobs given by kobs = kdka ket k a′k′et + k a′kd + kdket To obtain eqn 8.25, we divide the numerator and denominator on the righthand side of this expression by kdket and solve for the reciprocal of kobs 297 298 COMPLEX BIOCHEMICAL PROCESSES To gain insight into eqn 8.25 and the factors that determine the rate of electron transfer reactions in solution, we assume that the main decay route for D+A− is dissociation of the complex into separated ions, or kd >> k′et It then follows that 1 A k′ D ≈ 1+ a kobs ka C ket F (8.26) There are two limits to consider: • When ket >> k a′, kobs ≈ ka and the rate of product formation is controlled by diffusion of D and A in solution, which fosters formation of the DA complex • When, ket > | DrG |, kobs may be estimated by a special case of the Marcus cross-relation: kobs = (kDDkAAK)1/2 Marcus cross-relation (8.34) where K is the equilibrium constant for the net electron transfer reaction (eqn 8.24) and kDD and kAA (in general, kii ) are the experimental rate constants for the electron self-exchange processes (with the colors distinguishing one molecule from another): D + D+ → D+ + D A− + A → A + A− kDD kAA Justification 8.5 The Marcus cross-relation To derive the Marcus cross-relation, we use eqn 8.33 to write the rate constants for the self-exchange reactions as kDD = ZDD e−D G ‡ DD kAA = ZAA e−D G ‡ /RT AA /RT For the net reaction and the self-exchange reactions, the Gibbs energy of activation may be written from eqn 8.29 as D‡G = D rG 32 + D rG + 14 l 4l 3 = DrG AA = and hence D‡GDD = 14 lDD and For the self-exchange reactions D rG DD D‡GAA = 14 lAA It follows that kDD = ZDD e−l kAA = ZAA e−l /4RT DD /4RT AA To make further progress, Marcus assumed that the reorganization energy of the net reaction is the arithmetic mean of the reorganization energies of the self-exchange reactions: l = 12 (lDD + lAA) Provided l >> DrG for the net reaction, the first term in D‡G may be neglected and the Gibbs energy of activation of the net reaction is D‡G = 12 DrG + 18 lDD + 18 lAA Therefore, the rate constant for the net reaction is kobs = Ze−D G /2RT e−l r /8RT −lAA/8RT DD e We can use eqn 4.10 (ln K = −DrG 3/RT) in the form K = e−D G /RT to write r kobs = (kDDkAAK)1/2 f where f= Z (ZAAZDD)1/2 In practice, the factor f is usually set to and we obtain eqn 8.34 The rate constants estimated by eqn 8.34 agree fairly well with experimental rate constants for electron transfer between proteins, as we see in the following example 301 302 COMPLEX BIOCHEMICAL PROCESSES Example 8.4 Using the Marcus cross-relation The following data were obtained for cytochrome c and cytochrome c551, two proteins in which heme-bound iron ions shuttle between the oxidation states Fe(II) and Fe(III): cytochrome c cytochrome c551 kii /(dm3 mol−1 s−1) 1.5 × 102 4.6 × 107 E 9/V +0.260 +0.286 Estimate the rate constant kobs for the process cytochrome c551(red) + cytochrome c(ox) → cytochrome c551(ox) + cytochrome c(red) Then compare the estimated value with the observed value of 6.7 × 104 dm3 mol−1 s−1 Strategy We use the standard potentials and eqns 5.16 (ln K = nFE 3cell /RT ) and 5.17a (E 3cell = E 3R − E 3L ) to calculate the equilibrium constant K Then we use eqn 8.34, the calculated value of K, and the self-exchange rate constants k ii to calculate the rate constant kobs Solution The two reduction half-reactions are Right: cytochrome c(ox) + e− → cytochrome c(red) E 3R = +0.260 V Left: cytochrome c551(ox) + e− → cytochrome c551(red) E 3L = +0.286 V The difference is E 3cell = (0.260 V) − (0.286 V) = −0.026 V It then follows from eqn 5.16 with v = and RT/F = 25.69 mV that ln K = − 0.026V 2.6 =− 25.69 × 10−3 V 2.569 Therefore, K = 0.36 From eqn 8.34 and the self-exchange rate constants, we calculate kobs = {(1.5 × 102 dm3 mol−1 s−1) × (4.6 × 107 dm3 mol−1 s−1) × 0.36}1/2 = 5.0 × 104 dm3 mol−1 s−1 The calculated and observed values differ by only 25 per cent, indicating that the Marcus relation can lead to reasonable estimates of rate constants for electron transfer Self-test 8.6 Estimate kobs for the reduction by cytochrome c of plastocyanin, a protein containing a copper ion that shuttles between the +2 and +1 oxidation states and for which kAA = 6.6 × 102 dm3 mol−1 s−1 and E 3cell = +0.350 V Answer: 1.8 × 103 dm3 mol−1 s−1 CHECKLIST OF KEY EQUATIONS 303 Checklist of key concepts Catalysts are substances that accelerate reactions but undergo no net chemical change A homogeneous catalyst is a catalyst in the same phase as the reaction mixture Enzymes are homogeneous, biological catalysts The Michaelis–Menten mechanism of enzyme kinetics accounts for the dependence of rate on the concentration of the substrate A Lineweaver–Burk plot is used to determine the parameters that occur in the Michaelis–Menten mechanism In sequential reactions, the active site binds all the substrates before processing them into products In ‘ping-pong’ reactions, products are released in a stepwise fashion In competitive inhibition of an enzyme, the inhibitor binds only to the active site of the enzyme and thereby inhibits the attachment of the substrate In uncompetitive inhibition, the inhibitor binds to a site of the enzyme that is removed from the active site but only if the substrate is already present In noncompetitive inhibition, the inhibitor binds to a site other than the active site, and its presence reduces the ability of the substrate to bind to the active site 10 Fick’s first law of diffusion states that the flux of molecules is proportional to the concentration gradient 11 Fick’s second law of diffusion (the diffusion equation) states that the rate of change of concentration in a region is proportional to the curvature of the concentration in the region 12 Diffusion is an activated process 13 The flux of molecules through biological membranes is often mediated by carrier molecules 14 Protons migrate by the Grotthus mechanism, Fig 8.16 15 Electrophoresis is the motion of a charged macromolecule, such as DNA, in response to an electric field Important techniques are gel electrophoresis, isoelectric focusing, pulsed-field electrophoresis, two-dimensional electrophoresis, and capillary electrophoresis 16 According to the Marcus theory, the rate constant of electron transfer in a donor–acceptor complex depends on the distance between electron donor and acceptor, the standard reaction Gibbs energy, and the reorganization energy, l Checklist of key equations Property Equation Comment Michaelis–Menten rate law v = vmax[S]0 /([S]0 + KM) vmax = k b[E]0; assumes S is in excess Lineweaver–Burk plot 1/v = 1/vmax + (KM/vmax)(1/[S]0) Based on Michaelis–Menten mechanism Fick’s first law J = −Ddc/dx Fick’s second law ∂c/∂t = D∂ 2c/∂x Temperature-dependence of D D = D0e Mobility of an ion u = ez/6pha Marcus expression ket ∝ e−bre−D G/RT ‡ with D‡G = (DrG + l)2/4l Marcus cross-relation Also known as the diffusion equation −E a /RT k obs = (kDDkAAK )1/2 Assumes the validity of Stokes’ law 304 COMPLEX BIOCHEMICAL PROCESSES Further information Further information 8.1 Fick’s laws of diffusion Fick’s first law of diffusion Consider the arrangement in Fig 8.22 In an interval Dt the number of molecules passing through the window of area A from the left is proportional to the number in the slab of thickness l and area A, and therefore volume lA, just to the left of the window where the average (number) concentration is c(x − 12 l), and to the length of the interval Dt: number coming from left ∝ c(x − 12 l)lADt On writing the constant of proportionality as D (and absorbing l into it), we obtain eqn 8.11 Fick’s second law Consider the arrangement in Fig 8.23 The number of solute particles passing through the window of area A located at x in an infinitesimal interval dt is J(x)Adt, where J(x) is the flux at the location x The number of particles passing out of the region through a window of area A at x + dx is J(x + dx)Adt, where J(x + dx) is the flux at the location of this window The flux in and the flux out will be different if the concentration gradients are different at the two windows The net change in the number of solute particles in the region between the two windows is net change in number = J(x)Adt − J(x + dx)Adt = {J(x) − J(x + dx)}Adt The calculation of the rate of diffusion considers the net flux of molecules through a plane of area A as a result of arrivals from on average a distance 12 l in each direction Fig 8.22 Likewise, the number coming from the right in the same interval is number coming from right ∝ c(x + l)lADt The net flux is therefore proportional to the difference in these numbers divided by the area and the time interval: J∝ c(x − 12 l)lADt − c(x + 12 l)lADt = {c(x − 12 l) − c(x + 12 l)}l ADt We now express the two concentrations in terms of the concentration at the window itself, c(x), as follows: dc dx dc c(x − 12 l) = c(x) − 12 l × dx c(x + 12 l) = c(x) + 12 l × From which it follows that dc D A dc D # !A J ∝ c(x) − 12 l l − c(x) + 12 l F C @C dx dxF $ ∝ −l dc dx To calculate the change in concentration in the region between the two walls, we need to consider the net effect of the influx of particles from the left and their efflux toward the right Only if the slope of the concentrations is different at the two walls will there be a net change Fig 8.23 Now we express the flux at x + dx in terms of the flux at x and the gradient of the flux, dJ/dx: J(x + dx) = J(x) + dJ × dx dx It follows that net change in number = − dJ × dx Adt dx The change in concentration inside the region between the two windows is the net change in number divided by the volume of the region (which is Adx), and the net rate of change is obtained by dividing that change in concentration by the time interval dt Therefore, on dividing by both Adx and dt, we obtain 305 EXERCISES dJ A dc D rate of change of concentration = =− C dt F dx Finally, we express the flux by using Fick’s first law (at this point we need to acknowledge that c depends on both x and t and therefore use partial differential notation): ∂c ∂ A ∂c D ∂ 2c =− −D =D ∂t ∂x C ∂x F ∂x which is eqn 8.12 Discussion questions 8.1 Discuss the features and limitations of the Michaelis–Menten mechanism of enzyme action 8.2 Prepare a report on the application of the experimental strategies described in Chapters and to the study of enzyme-catalyzed reactions Devote some attention to the following topics: (a) the determination of reaction rates over a long time scale, (b) the determination of the rate constants and equilibrium constant of binding of substrate to an enzyme, and (c) the characterization of intermediates in a catalytic cycle Your report should be similar in content and extent to one of the Case studies found throughout this text 8.3 A plot of the rate of an enzyme-catalyzed reaction against temperature has a maximum, in an apparent deviation from the behavior predicted by the Arrhenius relation (eqn 6.19) Provide a molecular interpretation for this effect 8.4 Describe graphical procedures for distinguishing between (a) sequential and ping-pong enzyme-catalyzed reactions and (b) competitive, uncompetitive, and noncompetitive inhibition of an enzyme 8.5 Some enzymes are inhibited by high concentrations of their own products (a) Sketch a plot of reaction rate against concentration of substrate for an enzyme that is prone to product inhibition (b) How does product inhibition of hexokinase, the enzyme that phosphorylates glucose in the first step of glycolysis, provide a mechanism for regulation of glycolysis in the cell? Hint: Review Case study 4.3 8.6 Provide a molecular interpretation for the observation that mediated transport through biological membranes leads to a maximum flux Jmax when the concentration of the transported species becomes very large 8.7 Discuss the mechanism of proton conduction in liquid water For a more detailed account of the modern version of this mechanism, consult our Quanta, matter, and change (2009) 8.8 Discuss how the following factors determine the rate of electron transfer in biological systems: (a) the distance between electron donor and acceptor, and (b) the reorganization energy of redox active species and the surrounding medium Exercises 8.9 As remarked in the text, Michaelis and Menten derived their rate law by assuming a rapid pre-equilibrium of E, S, and ES Derive the rate law in this manner, and identify the conditions under which it becomes the same as that based on the steady-state approximation (eqn 8.1) 8.10 Equation 8.4a gives the expression for the rate of formation of product by a modified version of the Michaelis–Menten mechanism in which the second step is also reversible Derive the expression and find its limiting behavior for large and small concentrations of substrate 8.11 For many enzymes, such as chymotrypsin (Case study 8.1), the mechanism of action involves the formation of two intermediates: E + S → ES v = ka[E][S] ES → E + S v = ka′[ES] ES → ES′ v = kb[ES] ES′ → E + P v = kc[ES′] Show that the rate of formation of product has the same form as that shown in eqn 8.1, written as: v= vmax + KM/[S]0 but with vmax and KM given by vmax = kbkc[E]0 kb + kc and KM = kc(ka′ + kb) ka(kb + kc) 8.12 The enzyme-catalyzed conversion of a substrate at 25°C has a Michaelis constant of 0.045 mol dm−3 The rate of the reaction is 1.15 mmol dm−3 s−1 when the substrate concentration is 0.110 mol dm−3 What is the maximum velocity of this reaction? 8.13 Find the condition for which the reaction rate of an enzyme- catalyzed reaction that follows Michaelis–Menten kinetics is half its maximum value 8.14 Isocitrate lyase catalyzes the following reaction: isocitrate ion → glyoxylate ion + succinate ion The rate, v, of the reaction was measured when various concentrations of isocitrate ion were present, and the following results were obtained at 25°C: [isocitrate]/(mmol dm−3) V/(pmol dm−3 s−1) 31.8 70.0 46.4 97.2 59.3 116.7 118.5 159.2 222.2 194.5 Determine the Michaelis constant and the maximum velocity of the reaction 306 COMPLEX BIOCHEMICAL PROCESSES 8.15 The following results were obtained for the action of an ATPase on ATP at 20°C, when the concentration of the ATPase was 20 nmol dm−3: [ATP]/(mmol dm−3) v/(mmol dm−3 s−1) 0.60 0.81 0.80 0.97 1.4 1.30 2.0 1.47 3.0 1.69 Determine the Michaelis constant, the maximum velocity of the reaction, the turnover number, and the catalytic efficiency of the enzyme 8.16 Enzyme-catalyzed reactions are sometimes analyzed by use of the Eadie–Hofstee plot, in which v/[S]0 is plotted against v (a) Using the simple Michaelis–Menten mechanism, derive a relation between v/[S]0 and v (b) Discuss how the values of KM and vmax are obtained from analysis of the Eadie–Hofstee plot (c) Determine the Michaelis constant and the maximum velocity of the reaction from Exercise 8.14 by using an Eadie–Hofstee plot to analyze the data 8.17 Enzyme-catalyzed reactions are sometimes analyzed by use of the Hanes plot, in which [S]0/v is plotted against [S]0 (a) Using the simple Michaelis–Menten mechanism, derive a relation between [S]0/v and [S]0 (b) Discuss how the values of KM and vmax are obtained from analysis of the Hanes plot (c) Determine the Michaelis constant and the maximum velocity of the reaction from Exercise 8.14 by using a Hanes plot to analyze the data 8.18 An allosteric enzyme shows catalytic activity that changes on noncovalent binding of small molecules called effectors For example, consider a protein enzyme consisting of several identical subunits and several active sites In one mode of allosteric behavior, the substrate acts as effector, so that binding of a substrate molecule to one of the subunits either increases or decreases the catalytic efficiency of the other active sites Consequently, reactions catalyzed by allosteric enzymes show significant deviations from Michaelis–Menten behaviour (a) Sketch a plot of reaction rate against substrate concentration for a multi-subunit allosteric enzyme, assuming that the catalytic efficiency changes in such a way that the enzyme with all its active sites occupied is more efficient than the enzyme with one fewer bound substrate molecule, and so on Compare your sketch with Fig 8.2, which illustrates Michaelis–Menten behavior (b) Your plot from part (a) should have a sigmoidal shape (S shape) that is typical for allosteric enzymes The mechanism of the reaction can be written as E + nS ESn → E + nP and the reaction rate v is given by vmax v= + K′/[S]n0 where K′ is a collection of rate constants analogous to the Michaelis constant and n is the interaction coefficient, which may be taken as the number of active sites that interact to give allosteric behavior Plot v/vmax against [S]0 for a fixed value of K′ of your choosing and several values of n Confirm that the expression for v does predict sigmoidal kinetics and provide a molecular interpretation for the effect of n on the shape of the curve 8.19 (a) Show that the expression for the rate of a reaction catalyzed by an allosteric enzyme of the type discussed in Exercise 8.18 may be rewritten as log v vmax − v = n log [S]0 − log K′ (b) Use the preceding expression and the following data to determine the interaction coefficient for an enzyme-catalyzed reaction showing sigmoidal kinetics: [S]0 /(10−5 mol dm−3) V/(mmol dm−3 s−1) 0.10 0.0040 0.40 0.25 0.50 0.46 [S]0/(10−5 mol dm−3) V/(mmol dm−3 s−1) 0.60 0.75 0.80 1.42 1.0 2.08 [S]0/(10−5 mol dm−3) V/(mmol dm−3 s−1) 1.5 3.22 2.0 3.70 3.0 4.02 For substrate concentrations ranging between 0.10 mmol dm−3 and 10 mmol dm−3, the reaction rate remained constant at 4.17 mmol dm−3 s−1 8.20 A simple method for the determination of the interaction coefficient n for an enzyme-catalyzed reaction involves the calculation of the ratio [S]90 /[S]10, where [S]90 and [S]10 are the concentrations of substrate for which the reaction rates are 0.90vmax and 0.10vmax, respectively (a) Show that [S]90 /[S]10 = 81 for an enzyme-catalyzed reaction that follows Michaelis–Menten kinetics (b) Show that [S]90 /[S]10 = (81)1/n for an enzyme-catalyzed reaction that follows sigmoidal kinetics, where n is the interaction coefficient defined in Exercise 8.19 (c) Use the data from Exercise 8.19 to estimate the value of n 8.21 Yeast alcohol dehydrogenase catalyzes the oxidation of ethanol by NAD+ according to the reaction CH3CH2OH(aq) + NAD+(aq) → CH3CHO(aq) + NADH(aq) + H+(aq) The following results were obtained for the reaction: [CH3CH2OH]0 /(10−2 mol dm−3) V/(mol s−1 (kg protein)−1) V/(mol s−1 (kg protein)−1) V/(mol s−1 (kg protein)−1) V/(mol s−1 (kg protein)−1) (a) (b) (c) (d) 1.0 0.30 0.51 0.89 1.43 2.0 0.44 0.75 1.32 2.11 4.0 0.57 0.99 1.72 2.76 20.0 0.76 1.31 2.29 3.67 where the concentrations of NAD+ are (a) 0.050 mmol dm−3, (b) 0.10 mmol dm−3, (c) 0.25 mmol dm−3, and (d) 1.0 mmol dm−3 Is the reaction sequential or ping-pong? Determine vmax and the appropriate K constants for the reaction 8.22 One of the key events in the transmission of chemical messages in the brain is the hydrolysis of the neurotransmitter acetylcholine by the enzyme acetylcholinesterase The kinetic parameters for this reaction are kcat = 1.4 × 104 s−1 and KM = 9.0 × 10−5 mol dm−3 Is acetylcholinesterase catalytically perfect? 8.23 The enzyme carboxypeptidase catalyses the hydrolysis of polypeptides, and here we consider its inhibition The following results were obtained when the rate of the enzymolysis of carbobenzoxy-glycyl-d-phenylalanine (CBGP) was monitored without inhibitor: [CBGP]0 /(10−2 mol dm−3) Relative reaction rate 1.25 0.398 3.84 0.669 5.81 0.859 7.13 1.000 (All rates in this exercise were measured with the same concentration of enzyme and are relative to the rate measured when [CBGP]0 = 0.0713 mol dm−3 in the absence of inhibitor.) When 2.0 mmol dm−3 phenylbutyrate ion was added to a solution containing the enzyme and substrate, the following results were obtained: [CBGP]0 /(10−2 mol dm−3) Relative reaction rate 1.25 0.172 2.50 0.301 4.00 0.344 5.50 0.548 In a separate experiment, the effect of 50 mmol dm−3 benzoate ion was monitored and the results were [CBGP]0/(10−2 mol dm−3) Relative reaction rate 1.75 0.183 2.50 0.201 5.00 0.231 10.00 0.246 EXERCISES Determine the mode of inhibition of carboxypeptidase by the phenylbutyrate ion and benzoate ion 8.24 Consider an enzyme-catalyzed reaction that follows Michaelis– 307 8.34 The viscosity of water at 20°C is 1.0019 × 10−3 kg m−1 s−1 and at 30°C it is 7.982 × 10−4 kg m−1 s−1 What is the activation energy for the motion of water molecules? Menten kinetics with KM = 3.0 mmol dm−3 What concentration of a competitive inhibitor characterized by KI = 20 mmol dm−3 will reduce the rate of formation of product by 50 per cent when the substrate concentration is held at 0.10 mmol dm−3? 8.35 The mobility of a Na+ ion in aqueous solution is 5.19 × 10−8 m2 s−1 V−1 at 25°C The potential difference between two electrodes placed in the solution is 12.0 V If the electrodes are 1.00 cm apart, what is the drift speed of the ion? Use h = 8.91 × 10−4 kg m−1 s−1 8.25 Some enzymes are inhibited by high concentrations of their own 8.36 It is possible to estimate the isoelectric point of a protein from substrates (a) Show that when substrate inhibition is important, the reaction rate v is given by its primary sequence (a) A molecule of calf thymus histone contains one aspartic acid, one glutamic acid, 11 lysine, 15 arginine, and two histidine residues Will the protein bear a net charge at pH = 7? If so, will the net charge be positive or negative? Is the isoelectric point of the protein less than, equal to, or greater than 7? Hint: See Exercise 4.45 (b) Each molecule of egg albumin has 51 acidic residues (aspartic and glutamic acid), 15 arginine, 20 lysine, and seven histidine residues Is the isoelectric point of the protein less than, equal to, or greater than 7? (c) Can a mixture of calf thymus histone and egg albumin be separated by gel electrophoresis with the isoelectric focusing method? v= vmax + KM/[S]0 + [S]0/KI where KI is the equilibrium constant for dissociation of the inhibited enzyme–substrate complex (b) What effect does substrate inhibition have on a plot of 1/v against 1/[S]0? 8.26 What is (a) the flux of nutrient molecules down a concentration gradient of 0.10 mol dm−3 m−1, (b) the amount of molecules (in moles) passing through an area of 5.0 mm2 in 1.0 min? Take for the diffusion coefficient the value for sucrose in water (5.22 × 10−10 m2 s−1) 8.27 How long does it take a sucrose molecule in water at 25°C to 8.37 We saw in Section 8.8 that to pass through a channel, the ion must first lose its hydrating water molecules To explore the motion of hydrated Na+ ions, we need to know that the diffusion coefficient D of an ion is related to its mobility u by the Einstein relation: diffuse (a) mm, (b) cm, and (c) m from its starting point? 8.28 The mobility of species through fluids is of the greatest importance for nutritional processes (a) Estimate the diffusion coefficient for a molecule that steps 150 pm each 1.8 ps (b) What would be the diffusion coefficient if the molecule traveled only half as far on each step? 8.29 The diffusion coefficient of a particular kind of t-RNA molecule is D = 1.0 × 10−11 m2 s−1 in the medium of a cell interior at 37°C How long does it take molecules produced in the cell nucleus to reach the walls of the cell at a distance 1.0 mm, corresponding to the radius of the cell? 8.30 The diffusion coefficients for a lipid in a plasma membrane and in a lipid bilayer are 1.0 × 10−10 m2 s−1 and 1.0 × 10−9 m2 s−1, respectively How long will it take the lipid to diffuse 10 nm in a plasma membrane and a lipid bilayer? 8.31 Diffusion coefficients of proteins are often used as a measure of molar mass For a spherical protein, D ∝ M−1/2 Considering only one-dimensional diffusion, compare the length of time it would take ribonuclease (M = 13.683 kg mol−1) to diffuse 10 nm to the length of time it would take the enzyme catalase (M = 250 kg mol−1) to diffuse the same distance 8.32 Is diffusion important in lakes? How long would it take a small pollutant molecule about the size of H2O to diffuse across a lake of width 100 m? 8.33 Pollutants spread through the environment by convection (winds and currents) and by diffusion How many steps must a molecule take to be 1000 step lengths away from its origin if it undergoes a one-dimensional random walk? D= uRT zF where z is the charge number of the ion and F is Faraday’s constant (a) Estimate the diffusion coefficient and the effective hydrodynamic radius a of the Na+ ion in water at 25°C For water, h = 8.91 × 10−4 kg m−1 s−1 (b) Estimate the approximate number of water molecules that are dragged along by the cations Ionic radii are given in Table 9.3 8.38 For a pair of electron donor and acceptor, ket = 2.02 × 105 s−1 for DrG = −0.665 eV The standard reaction Gibbs energy changes to DrG = −0.975 eV when a substituent is added to the electron acceptor and the rate constant for electron transfer changes to ket = 3.33 × 106 s−1 Assuming that the distance between donor and acceptor is the same in both experiments, estimate the value of the reorganization energy 8.39 For a pair of electron donor and acceptor, ket = 2.02 × 105 s−1 when r = 1.11 nm and ket = 2.8 × 104 s−1 when r = 1.23 nm (a) Assuming that DrG and l are the same in both experiments, estimate the value of b (b) Estimate the value of ket when r = 1.48 nm 8.40 Azurin is a protein containing a copper ion that shuttles between the +2 and +1 oxidation states, and cytochrome c is a protein in which a heme-bound iron ion shuttles between the +3 and +2 oxidation states The rate constant for electron transfer from reduced azurin to oxidized cytochrome c is 1.6 × 103 dm3 mol−1 s−1 Estimate the electron self-exchange rate constant for azurin from the following data: Cytochrome c Azurin kii /(dm3 mol−1 s−1) 1.5 × 102 ? E cell /V 0.260 0.304 308 COMPLEX BIOCHEMICAL PROCESSES Projects 8.41 Autocatalysis is the catalysis of a reaction by the products For example, for a reaction A → P it can be found that the rate law is v = k[A][P] and the reaction rate is proportional to the concentration of P The reaction gets started because there are usually other reaction routes for the formation of some P initially, which then takes part in the autocatalytic reaction proper Many biological and biochemical processes involve autocatalytic steps, and here we explore one case: the spread of infectious diseases (a) Integrate the rate equation for an autocatalytic reaction of the form A → P, with rate law v = k[A][P], and show that KES,a = ESH+2 ESH + H+ KES,b = [ES−][H+] [ESH] [ESH][H+] [ESH +2] in which only the EH and ESH forms are active (a) For the mechanism above, show that v= v′max + K M′ /[S]0 with [P] e at = (1 + b) [P]0 + be at where a = ([A]0 + [P]0)k and b = [P]0/[A]0 Hint: Starting with the expression v = −d[A]/dt = k[A][P], write [A] = [A]0 − x, [P] = [P]0 + x and then write the expression for the rate of change of either species in terms of x To integrate the resulting expression, the following relation will be useful: 1 D A = + ([A]0 − x)([P]0 + x) [A]0 + [P]0 C [A]0 − x [P]0 + x F (b) Plot [P]/[P]0 against at for several values of b Discuss the effect of autocatalysis on the shape of a plot of [P]/[P]0 against t by comparing your results with those for a first-order process, in which [P]/[P]0 = − e−kt (c) Show that for the autocatalytic process discussed in parts (a) and (b), the reaction rate reaches a maximum at tmax = −(1/a) ln b (d) In the so-called SIR model of the spread and decline of infectious diseases, the population is divided into three classes: the susceptibles, S, who can catch the disease, the infectives, I, who have the disease and can transmit it, and the removed class, R, who have either had the disease and recovered, are dead, are immune, or are isolated The model mechanism for this process implies the following rate laws: dS = −rSI dt ESH ES− + H+ dI = rSI − aI dt dR = aI dt (i) What are the autocatalytic steps of this mechanism? (ii) Find the conditions on the ratio a/r that decide whether the disease will spread (an epidemic) or die out (iii) Show that a constant population is built into this system, namely that S + I + R = N, meaning that the timescales of births, deaths by other causes, and migration are assumed large compared to that of the spread of the disease 8.42 In general, the catalytic efficiency of an enzyme depends on the pH of the medium in which it operates One way to account for this behavior is to propose that the enzyme and the enzyme–substrate complex are active only in specific protonation states This proposition can be summarized by the following mechanism: EH + S ESH ka, ka′ ESH → E + P kb EH E− + H+ KE,a = [E−][H+] [EH] EH+2 EH + H+ KE,b = [EH][H+] [EH +2] v′max = vmax [H+] KES,a + 1+ KES,b [H+] [H+] KE,a + KE,b [H+] K M′ = KM [H+] KES,a + 1+ KES,b [H+] 1+ where vmax and KM correspond to the form EH of the enzyme (b) For pH values ranging from to 14, plot v′max against pH for a hypothetical reaction for which vmax = 1.0 mmol dm−3 s−1, KES,b = 1.0 mmol dm−3, and KES,a = 10 nmol dm−3 Is there a pH at which vmax reaches a maximum value? If so, determine the pH (c) Redraw the plot in part (b) by using the same value of vmax, but KES,b = 0.10 mmol dm−3 and KES,a = 0.10 nmol dm−3 Account for any differences between this plot and the plot from part (b) 8.43 Studies of biochemical reactions initiated by the absorption of light have contributed significantly to our understanding of the kinetics of electron transfer processes The experimental arrangement is that for time-resolved spectroscopy (In the laboratory 7.2) and relies on the observation that many substances become more efficient electron donors on absorbing energy from a light source, such as a laser With judicious choice of electron acceptor, it is possible to set up an experimental system in which electron transfer will not occur in the dark (when only a poor electron donor is present) but will proceed after application of a laser pulse (when a better electron donor is generated) Nature makes use of this strategy to initiate the chain of electron transfer events that leads ultimately to the phosphorylation of ATP in photosynthetic organisms (a) An elegant way to study electron transfer in proteins consists of attaching an electroactive species to the protein’s surface and then measuring ket between the attached species and an electroactive protein cofactor J.W Winkler and H.B Gray, Chem Rev 92, 369 (1992), summarize data for cytochrome c modified by replacement of the heme iron by a Zn2+ ion, resulting in a zinc–porphyrin (ZnP) moiety in the interior of the protein, and by attachment of a ruthenium ion complex to a surface histidine amino acid The edge-to-edge distance between the electroactive species was thus fixed at 1.23 nm A variety of ruthenium ion complexes with different standard reduction potentials were used For each rutheniummodified protein, either Ru2+ → ZnP+ or ZnP* → Ru3+, in which the zinc-porphyrin is excited by a laser pulse, was monitored PROJECTS This arrangement leads to different standard reaction Gibbs energies because the redox couples ZnP +/ZnP and ZnP +/ZnP* have different standard potentials, with the electronically excited porphyrin being a more powerful reductant Use the following data to estimate the reorganization energy for this system: DrG 9/eV ket /(106 s−1) 0.665 0.657 0.705 1.52 0.745 1.52 0.975 8.99 1.015 5.76 5.50 10.1 (b) The photosynthetic reaction center of the purple photosynthetic bacterium Rhodopseudomonas viridis is a protein complex containing a number of bound co-factors that participate in electron transfer reactions The table below shows data compiled by Moser et al., Nature 355, 796 (1992), on the rate constants for electron transfer 309 between different co-factors and their edge-to-edge distances (BChl, bacteriochlorophyll; BChl2, bacteriochlorophyll dimer, functionally distinct from BChl; BPh, bacteriopheophytin; Q A and Q B, quinone molecules bound to two distinct sites; cyt c559, a cytochrome bound to the reaction center complex.) Are these data in agreement with the behavior predicted by eqn 8.31? If so, evaluate the value of b Reaction r/nm ket /s−1 BChl−→BPh 0.48 1.58 × 1012 BPh−→Chl +2 0.95 3.98 × 109 Reaction r/nm ket /s−1 Q A−→QB 1.35 3.98 × 107 Q A−→BChl+2 2.24 63.1 BPh−→QA 0.96 1.00 × 109 cyt c559→Chl +2 1.23 1.58 × 108 This page intentionally left blank ... Exercises Project 209 211 212 212 212 215 PART The kinetics of life processes 217 The rates of reactions 219 Reaction rates 219 In the laboratory 6 .1 Experimental techniques 219 (a) The determination... temperature 14 9 15 0 15 0 11 0 3.7 The chemical potential 11 0 3.8 Ideal and ideal–dilute solutions 11 1 Coupled reactions in bioenergetics 15 1 Case study 4.2 ATP and the biosynthesis of proteins 15 2 FULL... 91 91 91 92 Phase equilibria 94 The thermodynamics of transition 94 4 .1 The reaction Gibbs energy 13 5 4.2 The variation of ΔrG with composition 13 7 (a) The reaction quotient 13 7 Example 4 .1 Formulating
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