112 4. calculations in the molecular biosciences Rate = fcx[A]«x[B]" 4.S Zeroth-order reactions are relatively rare. Examples in chemistry include some reactions of gases adsorbed on solid surfaces. In enzyme-catalysed reactions at very high substrate concentrations, the rate is close to zeroth order in substrate (see section 4.3.3). where k is known as the rate constant. The reaction is said to be of order a in A, of order b in B, and of overall order a + b. The order is an experimentally determined quantity and can have integral or non-integral values. For example, enzyme-catalysed reactions are of approximately first order with respect to substrate at low substrate concentrations, but the order decreases to near zero as the substrate concentration becomes saturating (see section 4.3.3). The order of a reaction should not be confused with its molecularity, which is defined as the minimum number of species involved in the slowest step of the reaction, i.e. the minimum number of species involved in the transition state. The molecularity must therefore be a whole number. The rate laws for reactions of different orders will be mentioned briefly and the method of analysis indicated. Zeroth-order reactions: In this case, the rate does not depend on the concentration of the reactant A, i.e. rate of change of [A] = —k, where k is the rate constant. (The rate is negative because A is being consumed.) A plot of [A], against time (Fig. 4.2) is a straight line of slope -k. First-order reactions: In this case, the rate depends on [A], i.e. rate of change of [A] = k x [A]. Mathematical analysis shows that the concentration of A changes with time (t) as given by eqn. 4.6 (see Chapter 2, section 2.3.3): [A],= [A]0xe-fa 4.6 By taking the natural logarithms of the terms on both sides of eqn. 4.6 and rearranging, we derive eqn. 4.7: In [A],-In [A]0 = -kxt 4.7 where [A]0 is the concentration of A at time t = 0, and k is the rate constant. A plot of In [A], vs time (f) (Fig. 4.3) gives a straight line of slope -k. < Fig. 4.2 Plot of [A], vs. time for a zeroth-order reaction. The slope of the line is -k, where k is the rate constant. t 4.2 the rates of reactions: kinetics 1 For a first-order reaction, the half-time f1/2, i.e. the time taken for [A], to decline to half its initial value, is found by setting [A], - 0.5[A]0 in eqn. 4.6. This gives eqn. 4.8: In 2 0.693 Note the half-time is independent of [A], i.e. it takes as long for [A], to fall from [A]„ to 0.5[A]0 as it does for it to fall from 0.5[A]0 to 0.25[A]„. Second-order reactions: If a reaction involves two reactants, A and B, whose initial concentrations are [A]„ and [B]0, respectively, then the concentration of product ([P]) formed after time t is given by eqn. 4.9: i xJWI4-[P])|=h( 4.9 ([a]„-[b]0) {[AU[B]o-[P})J The constant values of successive half-times is a very convenient diagnostic test for a first-order reaction. where k is the rate constant. . . ([B)0(IA]0-[P]) A plot or In ' vs t (Fig. 4.4) will give a straight line of slope Ua]„([b]„-[p]). x ([a]0 - [b]0), from which a: can be determined. The analysis of second-order reactions can be quite complex, but is simplified considerably under two conditions. Condition 1 The first condition is where the concentrations of the two reactants are equal, i.e. [a]0 = [B]0. In this case, the concentration of product P after time t is given by eqn. 4.10: _[P]_ [A]„([A]0-[P]) - = kxt 4.10 so a plot of [P] [A]„([A]0-[P]) vs t (Fig. 4.5) will be a straight line of slope k. 114 4. calculations in the molecular biosciences Fig. 4.4 Plot of (l [B]„(W„-[P])^ ,[A]0([B]0-[P])JVS'f for a second-order reaction according to eqn. 4.9. The slope of the line is fcx([A]0-[B]0),from which k, the rate constant, can be derived. 57 i c oq c Fig. 4.5 Plot of IP] vs. t for a [A]0([A]0-[P]) second-order reaction when [A]0 = [S]0 according to eqn. 4.10. The slope of the line = k, where k is the rate constant. The half-time of the reaction is derived from eqn. 4.10, by putting [P] = 0.5 [A]0; this gives eqn. 4.11: 4.11 It is clear that each successive half-time becomes longer, i.e. it takes twice as long for [A] to fall from 0.5 [A]0 to 0.25 [A]0 as it does from [A]0to 0.5 [A]0. J WORKED EXAMPLE Coenzyme A (denoted CoASH) which acts as a carrier of acyl groups in the metabolism of fatty acids and carbohydrates, contains a sulphydryl group. This group reacts with 5,5'-dithiobis-(2-nitrobenzoic acid) (Nbs2) with a 1:1 stoichiometry to give a yellow product (P), the thionitrophenolate ion which can be monitored by its absorbance at 412 nm. The data in Table 4.1 were obtained when equal concentrations (50 |iM) of CoASH and Nbs, were reacted. - Table 4.1 Time course of the reaction between CoASH and Nbs2 Time(min) 0 1 2 4 8 12 20 30 [P] (uM) 0 8.3 14.3 20.2 30.8 35.3 40.0 42.9 4.2 the rates of reactions: kinetics 115 Show that the data conform to a second-order reaction with [A]0 = [B]0, and determine the second-order rate constant. STRATEGY This calculation involves evaluating the term [P] • at each time point to [A]„([A]0-[P]) construct the appropriate plot. Use of the correct units will ensure that the slope is evaluated correctly. SOLUTION The plot of- [P] - vs t (Fig. 4.6) is a straight line, showing that the [AU[A]0-[P]) reaction does obey second-order kinetics. The slope of the line corresponds to the second-order rate constant and is equal to 0.04 |xM~' min~' = 4000 M"1 min^1 (note 1 uM~' = 106 M_1). This value can be seen to be reasonable by calculating the first-half life for the reaction using eqn. 4.11. Thus, (,,, = l/fc[A]0 = 1/(4000 x 50 x 10-6) min = 5 min. Inspection of the data shows that [P] has risen to half its final value (i.e. to 25 uM) between 4 and 8 min of reaction. Fig. 4.6 Plot of - [P] - vs. t for the reaction between CoA and Nbs, in the [A]0([A]0-[P]) worked example according to eqn. 4.10. The slope of the line = k, where k is the rate constant. Condition 2 The second condition under which the analysis of second-order reactions can be simplified is when the concentration of one of the reactants (say A) is much greater than that of the other (B). (In practice, this applies if the ratio [A]0/[B]0 is greater than about 20). If this applies, then the concentration of A remains essentially constant during the course of the reaction. 116 4. calculations in the molecular biosciences Note it is important to express the pseudo-first-order rate constant and the second-order rate constant in the correct units (see section 4.2.2). We can then write the rate of the reaction as eqn. 4.12: Rate = kx[A]x[B] but if [A] remains effectively constant, this becomes eqn. 4.13: Rate = k'x[B] 4.12 4.13 wherek' = kx [A]. The reaction now behaves as a first-order reaction (it is termed a pseudo-first-order reaction) and can be analysed as such to give k', the pseudo-first-order rate constant. Division of k' by [A], i.e. the concentration of the reactant in excess gives k, the second-order rate constant for the reaction. /ORKED EXAMPLE The enzyme shikimate kinase possesses a single lysine residue per molecule. Reaction of the side chain of this lysine leads to loss of enzyme activity. The reaction of shikimate kinase (0.1 LiM) with the lysine-specific reagent 2,4,6-trinitrobenzenesulphonic acid (TNBS) (5 u.M) was monitored by measuring the activity of samples withdrawn from the reaction mixture after stated times (Table 4.2). Table 4.2 Inactivation of shikimate kinase by reaction with TNBS Time (min) 0 2 Activity remaining (%) 100 74.1 54.9 6 40.7 8 30.1 10 22.3 15 10.5 20 5.0 Show that the data conform to a pseudo-first-order reaction and determine the pseudo-first-order rate constant and the second-order rate constant for the reaction of the lysine side chain of the enzyme with TNBS. STRATEGY The data are to be plotted in the form for a first-order reaction (Fig. 4.3); if a straight line results, the slope can be used to calculate the rate constant. SOLUTION A plot of In ([A]„/[A]t) vs t is a straight line, showing that the loss in activity is occurring in a pseudo-first-order fashion. (Note the molar ratio of TNBS to enzyme is 50). The slope of the line = -0.15 min"1, so the pseudo-first-order rate constant (k') = 0.15 min1. This is a reasonable value since it would correspond to a half-time for the reaction of (In 2)/0.15 min = 0.693/0.15 min = 4.62 min. Inspection of the data shows that 50% of the original activity has been lost between 4 and 6 min of reaction. The second-order rate constant (k) is found by dividing k' by the concentration of TNBS (which is in excess). Hence k = 0.15/(5 x 10~6) M 1 min"' = 30 000 M"1 min"1. It might be noted that this rate constant is much higher than for the reaction of the free amino acid lysine with TNBS and suggests that the side chain in the enzyme must be especially reactive. 4.2 the rates of reactions: kinetics 117 4.2.2 The units of rates and rate constants The rate of a reaction is measured in terms of the change in concentration of a re-actant or product (depending on how the reaction is being monitored) with time. The units of rate will thus be concentration x (time)-1, e.g. mM mill"1, uM s"1 etc. The units of a rate constant will depend on the order of the reaction concerned, and can be deduced by taking into account the fact that units must balance on both sides of an equation (see Chapter 3, section 3.3). Thus for a first-order reaction, rate = k x [A], so in terms of units we have: left-hand side - the units are (concentration x (time)"') right-hand side - the units are ((units of k) x concentration) hence the units of k must be (time)"1, e.g. s"1, mill-1, etc. WORKED EXAMPLE J What are the units of a second-order rate constant? STRATEGY From the rate law for a second-order reaction we can derive the units, noting that the units on the two sides of an equation must balance. SOLUTION For a second-order reaction, rate = kx[A]x [B], so applying the rule about the need to balance units on both sides of the equation, the units of the second-order rate constant are (concentration)"1 x (time)"1, e.g. M min"1. 4.2.3 The variation of reaction rate with temperature As the temperature increases, a greater proportion of the reactant molecules have the energy necessary to surmount the activation energy barrier. This leads to a marked increase in the rate of the reaction with temperature. The equation proposed by Arrhenius (see Chapter 2, section 2.3.2) is: fc = Ae-a/R7 4.14 where k is the rate constant for the reaction, £a is the activation energy, R is the gas constant (8.31 J K~! mol"1), and Tis the temperature in degrees Kelvin. A is known as the pre-exponential factor and is related to the frequency of successful collisions between reacting molecules. Taking natural logarithms of both sides of eqn. 4.14, we obtain eqn. 4.15: 118 4. calculations in the molecular biosciences .... . ...... .... ......... Panel A N. Slope = -EJR Fig. 4.7 Arrhenius plots to show the effect of temperature on reaction rate according to eqn. 4.15. (a) _c A typical plot for a chemical reaction. The slope of the line Panel B M 1/T is —EJR, from which £a can be derived, (b) A typical plot for an enzyme-catalysed reaction in which, above a certain temperature, there is a sharp loss of activity, reflecting the unfolding and inactivation of the enzyme. 1/T 1 The availability of stable enzymes from thermophilic bacteria has been a key factor in their application in many processes. A good example is the use of the thermostable DNA polymerase from Thermus aquaticus in the polymerase chain reaction, which can amplify the amount of selected DNA by a very large factor (1 million-fold or greater). Thus, an Arrhenius plot (In k vs 1/T) (Fig. 4.7) is a straight line of slope -EJR, from which Ea can be calculated. In the case of enzyme-catalysed reactions, the rate will start to decline at higher temperatures because the three-dimensional structure of the enzyme which is required for the expression of catalytic activity will be disrupted by the loss of the weak interactions which stabilize it (see Chapter 1, section 1.7). The effect on the Arrhenius plot is shown in Fig. 4.7 (Panel B). The temperature at which the activity reaches a maximum value is often referred to as the 'optimum temperature', but this can be difficult to measure as it will depend on the length of time the enzyme is incubated at the temperature in question prior to assay. (For this reason, the term 'apparent optimum temperature' is often used.) In most cases the 'optimum temperature' of an enzyme will reflect the physiological needs of the organism in question. Some enzymes from hyperthermophilic bacteria, for example, seem to be stable for long periods at temperatures close to 100°C. The data in Table 4.3 were obtained for the activity of a sample of mammalian lactate dehydrogenase over a range of temperatures. Table 4.3 The variation of lactate dehydrogenase activity with temperature Temperature (°C) 5 15 25 35 45 55 65 Activity (umolmhr'mg-1) 52.5 108 212 401 488 270 40.3 4.2 the rates of reactions: kinetics 119 Explain the dependence of activity on temperature and determine the activation energy for the enzyme-catalysed reaction. STRATEGY The data can be plotted according to the Arrhenius equation to give a straight line (eqn. 4.15). Deviations from a straight line would indicate that additional factors have to be considered. SOLUTION The Arrhenius plot (In (activity) vs 1/T, where Tis the temperature in degrees Kelvin = °C + 273, i.e. 25°C = 298 K) is linear over the range 5-35°C. The slope of the graph = -5810 K. The slope = -EJR, so the E, = 5810 x 8.31 J mol"' = 48.3 kj moYK At 45°C, the rate is less than predicted on the basis of this straight line and at temperatures above 45°C, the activity declines sharply. This is due to progressive unfolding of the enzyme at the higher temperatures and the loss of catalytic activity. It should be noted that where a comparison can be made between an enzyme-catalysed reaction and the corresponding non-catalysed reaction, the activation energy of the latter is considerably greater. SELF TEST ? Check that you have mastered the key concepts at the start of the section by attempting the following questions. ST 4.4 At low concentrations of substrate, the rate of an enzyme-catalysed reaction is given by: rate = (kcJKm) x [Enzyme] x [Substrate]. At high concentrations of substrate, the rate is given by: rate = kCM x [Enzyme]. Given that ka, and Km are constants, what is the order of the reaction under each condition? ST 4.5 The dissociation of an antibody-antigen complex is a first-order reaction with a rate constant of 3 x 10"3 s"1. What is the half-time for this reaction? ST 4.6 The rate of a reaction catalysed by an enzyme from an intertidal species (winkle) increases by 1.2-fold as the temperature is raised from 5°C to 15°C. What is the activation energy for this reaction? Answers ST 4.4 The orders of the reaction are: at low [substrate], first order in enzyme and first order in substrate, second order overall; at high [substrate], first order in enzyme, zeroth order in substrate, first order overall. ST 4.5 The half-time = 231 s, or 3 min 51 s. ST 4.6 The activation energy =12.1 kj mol"'. Attempt Problems 4.5 and 4.6 at the end of the chapter. In ST 4.6, the relatively small dependence of the rate with temperature is key to the survival of the organism, since the temperature will fluctuate considerably during the 24-h cycle of high and low tides. 120 4. calculations in the molecular biosciences Binding of ligands to macromolecules: saturation curves KEY CONCEPTS ■ Understanding the meaning of Ka and in binding processes ■ Understanding the relationship between Kd and AG0 ■ Writing the equation for hyperbolic binding to single and multiple (independent) binding sites ■ Writing the Hill equation to describe cooperative binding ■ Understanding the meaning of Vmax and Km in enzyme-catalysed reactions ■ Explaining the different types of inhibition of enzyme-catalysed reactions Many biological processes depend on the interactions between molecules, for example regulation of gene expression (protein/DNA), catalysis (enzyme/substrate), cell signalling (hormone/receptor), etc. These interactions depend very largely on non-covalent forces (hydrogen bonds, van der Waals, electrostatic, hydrophobic interactions), which are individually weak but collectively can generate high affinities and specificities between the interacting molecules (for further details see Chapter 1, section 1.7). It is important to characterize binding processes to judge their importance under physiological conditions. 4.3.1 Single binding site on a macromolecule For many of these interactions, simple saturation behaviour is observed: P + L==PL where P represents a protein and L a ligand, i.e. any molecule which binds to the protein; this could include a nucleic acid, a low molecular mass molecule or another protein, for example. For this equilibrium, we can define two equilibrium constants, Ka and Kd, which are known as the association and dissociation constants, respectively: where [P] and [L] represent the concentrations of free (i.e. unbound) protein and ligand, respectively, and [PL] represents the concentration of the PL complex. 4.3 binding of ligands to macromolecules: saturation curves Ka represents a measure of the tendency for P and L to associate to form the PL complex. The larger this tendency, i.e. the tighter the binding, the greater the value ofK,. Kd represents a measure of the tendency for the PL complex to dissociate into P and L. The smaller this tendency, i.e. the tighter the binding, the smaller the value pd. For largely historical reasons, Ka is used predominantly by chemists and Kd by molecular biologists and biochemists. The two constants are reciprocals (see Chapter 2, section 2.4) of each other, i.e. Ka = \IKd and Kd = l/KiX. The units of Ka are (concentration)-1, e.g. M"1; those of Kd are concentration, e.g. M. Since the equilibrium between P, L and PL is a dynamic one, at equilibrium the rate of the association reaction (P + L —» PL), i.e. kz x [P] x [L], must equal the rate of the dissociation reaction (PL —> P + L), i.e. kd x [PL], where fca and kd are the association and dissociation rate constants, respectively. Thus, we have: Strictly speaking, K, and AC,, are dimensionless since they refer to a standard state of a 1 M solution of P, L and PL (see section 4.1). It is convenient to keep the units of lvT"' and M for K, and Kt„ respectively, to remember that for calculations, we must always express the concentrations in molar terms. kax[P]x[L]=kdx[PL] 4.1 8 Hence, by combining eqn. 4.18 with eqns. 4.16 or 4.17, it follows that: K-K 4.19 JC, = - Thus, the equilibrium constants Ka and Kd can be expressed as the appropriate ratios of rate constants for the individual association and dissociation steps. In order to develop molecular explanations for the interactions involving proteins, it is important to be able to link the value of iCa or Kd to the energy changes involved in binding or dissociation. For the dissociation of the PL complex (PL == P + L), the relationship between the standard state free energy change (see section 4.1) and the value of Kd is given by eqn. 4.21 (which is analogous to eqn. 4.4). -AG" = RT'In Kd 4.21 Thus, if Kd = 1 LiM at 310 K (37°C), AG0 = -8.31 x 310 x In (1 x 10"6) J moL' = 35 600 J mol-1 = 35.6 kj moL1. The energy change can be compared with typical values for the weak, non-covalent forces described in Chapter 1, section 1.7. In this example, the value of AG0 is large and positive, since under standard state conditions (i.e. 1 M PL) there would be little tendency of the reaction to proceed from left to right, i.e. for PL to dissociate. Dissociation of the complex would be promoted by lowering the concentration of PL, which would be the case in living Note that Kt is expressed in terms of molar concentration, since this is referred to the standard state of the ligand, i.e. a 1 M solution. 122 4. calculations in the molecular biosciences *5> Fig. 4.8 A hyperbolic saturation curve for binding of a ligand (L) to a protein, according to eqn. 4.22. When the fractional saturation (Y)= 0.5, half the available sites are occupied, [I] = Kd. Y 0.5 [L] The dependence of Y on [I] according to eqn. 4.22 is described mathematically as a rectangular hyperbola (or simply a hyperbola). The binding is said to be hyperbolic in nature. systems, where concentrations of macromolecules are likely to be in the micromo-lar range, or even lower. Remembering that at 310 K each factor of 10 in the value of Kd corresponds to approximately 6 kj mol"1 (accurately 5.93 kj mol-1; see Chapter 2, section 2.3.6 and section 4.1 of this chapter) we can easily estimate the AG" involved. For example, a Kd of 10~9 M (1 nM) for a complex corresponds to a AG0 of approximately 54 kj mol"1 (accurately 53.4 kj mol-1). It is a relatively simple task (see Appendix 4.1) to derive an expression for the fraction (Y) of the protein P which is saturated by ligand L (eqn. 4.22): [I] Kd+[L] The value of Y can range from 0 to 1; a graph indicating how Y varies with [L] is shown in Fig. 4.8. Note that when half the total binding sites are occupied, i.e. 7= 0.5, then Kd = [L]. This gives a useful operational definition of Kd, namely that it corresponds to the concentration of (free) L at which half-saturation of the protein has occurred. One of the problems about depicting the saturation curve as in Fig. 4.8 is that it is difficult to span a very wide range of ligand concentrations. For example, to go from 10% saturation to 90% saturation requires an 81-fold change in [I] (from KJ9 to 9 x KA). The use of a logarithmic scale for [L] can overcome this problem since it allows us to cover a very large numerical range of values (see Chapter 3, section 3.2). This type of plot is known as a formation plot and an example is shown in Fig. 4.9. In this figure the binding is hyperbolic (as in Fig. 4.8) but the logarithmic nature of the x-axis gives the curve a 'sigmoidal' (or elongated S-shape) appearance. 4.3.2 Multiple binding sites on a macromolecule If instead of a single binding site on the protein, there are ft sites for binding the ligand and these are assumed to be equivalent and independent of each other P+nL^PL„ 4.3 binding of ligands to macromolecules: saturation curves 123 Fig. 4.9 Formation plot for binding of a ligand (L) to a protein, where [L] is plotted on a logarithmic scale. The data points correspond to a hyperbolic binding model (eqn. 4.22) with Ka = 1 |LiM. Note that in this example the use of the logarithmic scale allows data over a range of nearly 1000-fold values of [L] to be plotted. then the saturation equation is analogous to eqn. 4.22, taking into account the number of binding sites; see eqn. 4.23: where r is the average number of molecules of ligand bound per molecule of protein, n is the number of binding sites, [L] is the concentration of free ligand, and Kd is the dissociation constant. A plot of r against [L] is shown in Fig. 4.10. Note that r can take values ranging from 0 to n. Substitution into eqn. 4.23 shows that when r = 0.5n (i.e. when half the total sites are saturated), [L] = Kd. Thus, just as in the case of a single binding site, the value of 124 4. calculations in the molecular biosciences Fig. 4.11 Saturation curve for binding of O, to haemoglobin. At high pressures of 02 all four binding sites in the haemoglobin molecule are occupied. The sigmoidal shape of the binding curve reflects cooperatively between the binding sites. r 2 0.05 0.10 0.15 P„ (atm) Kd is equal to the concentration of ligand required to bring about 50% saturation of the available sites. Note in those cases where there are interactions between the binding sites, the shape of the saturation curve will differ. Thus, the binding of oxygen to haemoglobin (four binding sites) shows positive cooperativity, with the binding at the first site making it easier to bind to the subsequent sites. The saturation curve in this case is sigmoidal (see Fig. 4.11). This behaviour arises because of structural changes relayed between the subunits (polypeptide chains) of the protein. Although a detailed mathematical analysis of cooperativity is beyond the scope of this book, mention should be made of the Hill equation (eqn. 4.24), which provides a relatively simple way of describing this type of behaviour. The equation is a modified form of eqn. 4.23 in which the term [L] is raised to a power h, where h is known as the Hill coefficient. Hill proposed this equation in 1910 to describe the binding of oxygen to haemoglobin. nx[L]" ' K+[L]h The Hill equation is usually expressed in terms of the fractional saturation (Y) of the binding sites by ligand, i.e. Y= rln. Thus eqn. 4.24 becomes: K+ [L]" The value of h for binding of a given ligand to a protein can be found by a suitable plot as described in section 4.4. When h = 1, there are no interactions between the multiple ligand binding sites on the macromolecule. A value of h > 1 indicates positive cooperativity. The value of h can range up to n, the number of binding sites, and its magnitude gives a measure of the extent of cooperativity between the sites. The Hill coefficient does not have to be an integer; for example the value of h for the binding of oxygen to haemoglobin is typically about 2.8. (A value of h < 1 is 4.3 binding of ligands to macro molecules: saturation curves said to indicate negative cooperativity, where the binding of the first ligand molecule hinders binding of subsequent molecules.) An equation analogous to eqn. 4.25 can be derived for the analysis of enzyme-catalysed reactions, see section 4.4. 4.3.3 Enzyme kinetics The treatment of enzyme kinetics is, in a mathematical sense, analogous to that of ligand binding considered above. However, it is considered separately because a number of the symbols used differ from those used in considering binding equilibria. Enzyme kinetics is an important topic because it is actually measuring the enzyme performing its biological role (i.e. catalysing a reaction). By studying how the rate of the reaction depends on the concentration(s) of substrate(s), and the effect of inhibitor molecules (section 4.3.4) it is possible to draw conclusions not only about the mechanism of the reaction, but also about the possibilities of designing specific inhibitors which could act, for example as therapeutic drugs. In the simplest case, where an enzyme (E) catalyses a reaction involving one substrate (S) and one product (P), the reaction is proposed to proceed via formation of an enzyme-substrate complex which breaks down to generate product and regenerate enzyme as follows: E + S^ES^E + P By using either an equilibrium assumption (i.e. the breakdown of ES to give E + P is so slow as to not to perturb the E + S ^ ES equilibrium significantly) or the steady-state assumption (i.e. the concentration of ES remains effectively constant once the steady-state condition has been achieved), we can derive an equation (4.26) to show how the rate (or velocity, v) varies with the concentration of substrate ([S]). This is known as the Michaelis-Menten equation. Km+[S] 4.26 where V™ is the maximum rate (more correctly, the limiting rate) and Km is known as the Michaelis constant. From eqn. 4.26 it can be seen that if v is set equal to0.5Vmax, [S]=Km. The plot of v vs. [S] is shown in Fig. 4.12. From this it can be seen that at low [S] the rate of reaction increases almost linearly with [S], but at high [S] the rate tends to level off somewhat and approaches a maximum (or limiting) value. This levelling off reflects the saturation of the active sites of the enzyme molecules present; no matter how much more substrate is added, the rate cannot increase further. Strictly speaking [S] refers to the concentration of free substrate. However, the vast majority of studies of enzyme-catalysed reactions take place under steady-state conditions where the molar concentration of enzyme is very much less than The derivation of the Michaelis-Menten equation is discussed in more detail in Appendix 4.1. The Michaelis-Menten equation was derived in 1913 by Leonor Michaelis (Belgian) and Maud Menten (Canadian) using the equilibrium assumption. In 1925, Briggs and Haldane (both British) applied the steady-state approximation to derive an equation of similar form. 126 4. calculations in the molecular biosciences Fig. 4.12 Michaelis-Menten plot showing the variation of the velocity (rate) of an enzyme-catalysed reaction with the concentration of substrate according to eqn. 4.26. Whenv= VmJ2, [S] = the Michaelis constant (Km). the molar concentration of substrate. In these circumstances, the concentration of free substrate is effectively equal to that of the total added substrate. It should be noted that when E and S are mixed the system takes a short time to reach the 'steady state'. It is possible to observe events in the pre-steady state period using specialized apparatus; this can give additional information on the rates of individual steps in the overall reaction (see Chapter 9, section 9.9.4). 4.3.4 Inhibition of enzyme-catalysed reactions Although a full discussion of the inhibition of enzyme-catalysed reactions is beyond the scope of this book, it is important that you are able to analyse appropriate data to determine the type of inhibition being observed in any particular case and to evaluate the strength of the interaction between enzyme and inhibitor in terms of a suitable dissociation constant. We will confine our attention to those inhibitors which bind reversibly to enzymes. In the mechanism quoted previously: E + S==ES^E + P we could envisage a general scheme for the way in which an inhibitor (I) might interact with E and the ES complex. We can consider three limiting cases for enzyme-inhibitor interaction: A. Competitive inhibition The inhibitor (I) binds to E but not to ES (i.e. KHSl = °° which means that the ESI complex has no tendency to form). In this case, the addition of I will pull some of E over to the EI complex. However, addition of increasing concentrations of S will eventually overcome the effect of I, since the balance will progressively swing towards formation of ES rather than EI. This is termed competitive inhibition; Vmsa remains unchanged, but Km is 4.3 binding of ligands to macro molecules: saturation curves raised (reflecting the higher [S] to be added to overcome the effects of I). The most likely explanation for competitive inhibition is that although the inhibitor cannot undergo the reaction, it has a structural resemblance to the substrate and therefore binds at the active site, preventing the binding of S. In quantitative terms, Km in the presence of inhibitor is raised by a factor 1 -I--compared with the absence of inhibitor (i me value of K can be evaluated as shown. logif / h /V. / Slope = h log [I] 4.4 analysis of binding and kinetic data 137 mmmm Check that you have mastered the key concepts at the start of the section by attempting the following questions. SELF TEST ST 4.10 In a further experiment on the hydrolysis of the pencillin analogue catalysed by (3-lactamase the following data were obtained. What are the values of Km and Vm„ for the reaction? [Substrate] (|1M) Rate (nmol min ') 1 0.24 2 0.34 3 0.56 5 0.76 10 0.90 30 0.95 ST 4.11 Addition of an inhibitor (80 LiM) led to no change in the Vmax of an enzyme-catalysed reaction, but the Km for the substrate was raised from 22 to 68 llM. What can you conclude about the interaction of the inhibitor with the enzyme? ST 4.12 A Scatchard plot was used to analyse the binding of a ligand to a protein with four polypeptide chains. The term r/[L]ba was plotted on the^-axis and r on the x-axis (r is the mol ligand bound per mol enzyme and [L]bee is the concentration of free ligand). The plot showed a straight line of negative slope, with an intercept on the y-axis (i.e. when x = 0) of 0.058 LiM"1; the intercept on the x-axis (i.e. when y = 0) was 3.85. What can you conclude about the binding of the ligand to the protein? Answers ST 4.10 The experimental points do not fall on perfect straight lines in the various linear transformation plots. The best fit lines give the following parameters: Lineweaver-Burk, Vmax 1.14 nmol min-1, Km 3.58 |!M; Eadie-Hofstee, Vmax 1.26 nmol min-1, Km 4.46 uM; Hanes-Woolf, Vmax 1.12 nmol min-1, Km 3.29 p.M. The values of Km and Vmax obtained by the different methods show considerable variation from each other, illustrating the problem of deciding what is the most appropriate straight line when the data do not conform exactly to the Michaelis-Menten equation. The difficulties are most likely to be caused by random experimental errors. The values obtained by non-linear regression are Vmax= 1.11 (standard error 0.08) nmol min-' andKm = 3.16 (standard error 0.68) LlM. The standard error of the estimate of Km is very high (21.5% of the actual value), confirming the poor fit to the equation. In practice, standard errors of less than 10% (and ideally of less than 5%) of the actual values should be aimed for. It would be sensible in the present case to attempt to refine the experimental techniques and to collect further data to improve the reliability of the estimates. ST 4.11 The inhibitor is competitive, from the change in Km, KEI can be calculated to be 38 LiM. ST 4.12 There are 3.85 binding sites per enzyme, i.e. very close to one per polypeptide chain; these are equivalent and independent. The slope of the Scatchard plot = —u.M-1, i.e. -0.0151 [AM"1. This is equal to—i-, 3.85 Kd so thati (183 pmol s_1). The amount of enzyme present is 0.025 u,g, which is equivalent to 0.025 x 10"«/16 500 mol = 1.52 x 10"12 mol (1.52 pmol). (Note 1 mol enzyme is 16 500 g.)Hence, kCM = 183/1.52 pmol pmol"1 s_1 = 120 mol mol-' s"1 (this could also be expressed as 120 s~')- At a molecular level, the value of kM means that every 8.3 ms, a molecule of substrate (3-dehydroquinate) diffuses to and binds to the enzyme, is converted to products (3-dehydroshikimate and water) which then dissociate from the enzyme. HflMHH The values of kM generally range from about 10 s"1 for enzymes catalysing complex biosynthetic reactions to about 10« s_1 for enzymes such as catalase or carbonic anhydrase which catalyse chemically simple reactions (the breakdown of hydrogen peroxide to give water and oxygen, and the hydration of CO, to give carbonic acid, respectively). The ratio kcJKm is a useful parameter to characterize an enzyme. It can indicate the specificity of an enzyme towards a given substrate in cases where an enzyme can act on a number of possible substrates. The absolute value of the ratio can also 4.6 radioactivity 139 indicate the catalytic efficiency of an enzyme. In the case of dehydroquinase in the worked example, the Km for 3-dehydroquinate is 105 u,M, so that the value of kalIKm = 1.14 x 106 M_1 s_1. However, for a number of enzymes, such as triosephos-phate isomerase, fumarase and acetylcholinesterase, the value of kcJKm is about 108 M"1 s_1, which effectively represents the diffusion limit, i.e. the rate of the overall reaction is limited by the rate at which substrate can encounter the active site of the enzyme, with all subsequent steps (the chemical reaction, dissociation of products, etc.) occurring more rapidly. This phenomenon is sometimes referred as 'catalytic perfection'. Although an enzyme might have achieved 'perfection' in terms of its catalytic efficiency, it may possibly still be 'improved' in terms of regulatory properties or stability in different physiological situations. Check that you have mastered the key concepts at the start of the section by attempting the following questions. ST 4.13 For the reaction catalysed by fumarase (fumarate =^ malate + H20), the Km for fumarate is 4.5 pM and the Vmax is 45 000 mol fumarate mol"' enzyme min-1. Calculate fccal (in units of s-1) and the ratio kcJKm (in units of M"1 s~') and comment on the value of this ratio. ST 4.14 The enzyme piactamase catalyses the hydrolysis of penicillin and a number of its derivatives. In a 1-mL reaction volume to which 0.015 pg enzyme was added, the Vm„ was found to be 10.2 pM substrate hydrolysed per minute and the Km was 5.2 pM. The molecular mass of the enzyme is 29.6 kDa. Calculate kal and kcJKm for this enzyme-catalysed reaction. Answers ST 4.13 The values of kM and kJKm are 750 s"1 and 1.67 x 10s Mr1 s"1, respectively; the latter value is close to the diffusion-controlled limit, i.e. to 'catalytic perfection'. ST 4.14 The values of fcCBt and kJKm are 335 s"1 and 6.44 x 107 M"1 s~ respectively. SELF TEST Attempt Problems 4.1 2 and 4.1 3 at the end of the chapter. Radioactivity KEY CONCEPTS ■ Knowing the three types of radiation emitted by unstable nuclei ■ Knowing the definitions of the units of radioactivity (Curie and Becquerel) ■ Understanding the terms efficiency of counting and specific radioactivity and how they are used in calculations The nucleus of an unstable isotope will decay to produce one of the following types of radiation, namely a (corresponding to He nuclei), (3 (corresponding to 4. calculations in the molecular biosciences Table 4.1 0 Properties of the radioactive isotopes most commonly used in studies of proteins Isotope Half-life Maximum energy (eV)* 3H 12.3 years 1.9 xlO4 i4C 5730 years 1.56 x 105 3ip 14.3 days 1.71 x 106 35S 87.4 days 1.67 x 10s *The electrons emitted will have a range of energies; the values shown are the maximum energies observed in each case. 1 eV (electron volt) is equivalent to 96.5 kj mol"1. The much higher energy of the electrons emitted by 32P means that it is very important to adopt stringent safety procedures when working with this isotope. The Curie is named in honour of Pierre and Marie Curie, who worked in France on the identification and purification of radioactive elements (principally polonium and radium). Marie was Polish by birth (Maria Sklodowska). The Becquerel is named in honour of Henri Becquerel, who discovered the phenomenon of radioactivity in a salt of uranium. electrons), and y (corresponding to high-energy photons). In some cases, more than one type will be produced e.g. 13lI decays giving both (3- and y-rays. The principal isotopes used in biochemical research are P-emitters, although y-emitting isotopes of iodine have a number of specialized applications, particularly in immunology. Incorporation of a radioactive isotope provides a very sensitive means of tracking a particular compound, especially in studies of enzyme-catalysed reactions or metabolism, and in measuring its binding to another molecule. Since radioactivity arises from the decay of individual nuclei, it is possible to measure very small amounts of a radioactive compound (in the pmol or fmol range). The isotopes most commonly employed in studies of proteins are 3H, 14C, 32P, and 35S, the properties of which are listed in Table 4.10. (There are no convenient radioactive isotopes of N or O available.) The amount of radioactivity in a substance is quoted in two types of units. The older, but still used, unit is the Curie (Ci), which is defined as 3.7 x 10'° disintegrations per second (dps). This corresponds to the activity of lg pure radium. Because the Curie is such a large amount of radioactivity, we typically use much smaller quantities (mCi or LtCi) in experiments. The other unit is the Becquerel (Bq), which corresponds to 1 dps; this is, conversely, such a small amount of radioactivity that we would generally be using kBq or MBq in experiments. A radioactive compound will be supplied with a stated specific radioactivity, e.g. 3H-alanine might be supplied at 50 Ci mol"1 (this would correspond to 1850 GBq mol-1)- For experimental purposes, this would usually be mixed with a large excess of unlabelled alanine to give a specific radioactivity of, for example 0.5 Ci mol"1. Knowledge of the specific radioactivity of a sample containing a radioactive compound allows us to convert the observed radioactivity into the amount (mol) of that compound present. Radioactivity is usually measured by liquid scintillation counting. The sample is dissolved or suspended in a suitable solvent containing one or more fluorescent compounds (known as fluors). Emission of an electron by the radioactive nucleus will excite the fluor (this can occur via excitation of a solvent molecule). Emission of radiation (scintillations) by the fluor can be detected by a photomultiplier. The efficiency of detection of the radioactive decay depends on a number of factors including the physical state of the radioactive material and the energy of the 4.6 radioactivity 141 electrons emitted. In the case of 3H-labelled compounds the efficiency is fairly low (typically 20-30%), but is much higher for 14C-labelled, and especially for 32P-labelled compounds. We must measure the efficiency of counting to be able to convert the observed counts per second (cps) to disintegrations per second (dps) Radioactivity is a very sensitive tool because events could, in theory at least, be detected at very nearly the scale of individual atoms and molecules. This makes it ideally suited to the measurements of very small amounts of material such as may occur in biological systems. In addition, the introduction of radioactive isotopes does not affect the chemical behaviour of atoms significantly (although there can be effects on the rates of some reactions), so that they serve as ideal probes and tracers for molecules in complex systems. In this book, we shall confine our attention to the use of radioactive isotopes for assays of enzyme activity and for the monitoring of binding processes. In practice, radioactivity should only be used as a probe with a small proportion of the relevant molecules containing the radioactive isotope. When an isotope decays it can generate a new element with different chemical characteristics, which may lead to damage to the molecule concerned. Thus, ,2P and ,5S decay by P-emission to generate *2S and J5C1, respectively. The radioactive compound supplied would generally be mixed with a large molar excess of the non-radioactive compound to obtain the working solution. WORKED EXAMPLE A sample of ATP labelled in the y-phosphate position with 32P was purchased and mixed with a solution of unlabelled ATP to give a specific radioactivity of 5 Ci moľ'1; this material was then stored in the freezer. After 43 days an experiment was carried out in which this solution was used to monitor the incorporation of phosphate into a protein of molecular mass 60 kDa. After 1 mL of the protein solution (1.3 mg mL"1) had been incubated with the radioactive ATP and necessary factors for reaction, the mixture was treated with trichloroacetic acid to stop the reaction and precipitate the protein. The radioactivity incorporated was 25 200 counts per minute (cpm), and the efficiency of counting was 90%. What was the extent of incorporation of phosphate into the protein? STRATEGY Given the relatively short half-life of the 32P isotope, it is necessary to calculate the specific radioactivity on the day of the experiment. Calculation of the extent of incorporation requires knowledge of the amounts (numbers of moles) of protein and of radioactivity incorporated. The working should be clearly set out and explained. SOLUTION The ATP sample was used 43 days after purchase; reference to Table 3.1 shows that this time corresponds to 3 half-lives of the 32P isotope. Thus, on the day of the experiment, the specific radioactivity will be 0.5 x 0.5 x 0.5 (i.e. 0.125) times the original, namely 0.625 Ci mol"1, or 0.625 x 3.7 x 10'° dps mol"1 = 2.313 x 10'° dps mol"1. The radioactivity incorporated was 25 200 cpm or 25 200/60 cps = 420 cps. Since the efficiency was 90%, this would correspond to 420 x 100/90 dps = 467 dps Using the value for the specific radioactivity we can calculate that this would correspond to 467/(2.313 x 101") mol = 20.2 nmol phosphate incorporated. The amount of protein taken was 1.2 mg; this corresponds to 1.2 x 10 760 000 mol protein = 20 nmol protein. Thus, the incorporation of phosphate amounts to 20.2/20 = 1.01 mol moL1 protein. 142 4. calculations in the molecular biosciences WORKED EXAMPLE The half-life of 14C is 5730 years. What is the maximum specific radioactivity (Bq mol"1) of a sample of [14C] glucose, which has the molecular formula C6H1206? (Avogadro's number is 6.02 x 1023 mol"1.) STRATEGY From the half-life, the rate constant can be calculated. When multiplied by Avogadro's number, this will give the initial rate of radioactive decay for 1 mol of 14C atoms; this value has to be multiplied by 6 to obtain the rate for 1 mol glucose, which has 6 carbon atoms. SOLUTION The tV2 (5730 years) corresponds to 1.81 X 10" s; thus the rate constant = 0.693/(1.81 x 10") s"1, i.e. 3.835 X 10"12 s"1. The initial rate of decay for lmol 14C atoms = 3.835 x 10"12 x 6.02 x 1023 dps (Bq) = 2.31 x 1012 Bq. For glucose, the rate would be 1.39 x 1013 Bq mol"1 or 13.9 TBq mol"' (T = tera = 1012). Check that you have mastered the key concepts at the start of the section by attempting the following questions. ST 4.1 S The radioactivity in a sample of [3H] leucine was measured as 8650 cpm; the efficiency of counting was 22%. The specific radioactivity of the sample was 1.43 GBq mol"1. Calculate the dps of the sample and the number of moles of leucine present. ST 4.1 6 What is the maximum radioactivity of 1 mol of 32P atoms, given that the half-life is 14.3 days and Avogadro's number is 6.02 x 1023 mol"1. Answers ST 4.15 There are 655 dps in the sample and hence 4.58 x 10"7 mol, i.e. 0.458 pmol present. ST 4.16 The maximum radioactivity is 3.77 x 10'7 Bq mol"1. Attempt Problems 4.1 4-4.1 6 at the end of the chapter. 4.7 problems 143 Q Problems Full solutions to odd-numbered problems are available to all in the student section of the Online Resource Centre at www.oxfordtextbooks.co.uk/orc/price/. Full solutions to even-numbered problems are available to lecturers only in the lecturer section of the Online Resource Centre. 4.1 For the formation of the complex between Mg2+ ions and the chelating agent EDTA4" at 20°C, AG0 = -48.7 kj moh1 and AH0 = 13.1 kj moH. Calculate the AS0 for the formation of the MgEDTA2" complex, and comment on your answer. 4.2 For the hydrolysis of ATP (ATP + H20 ^ ADP + Pt), AG?',, = -31.0 kj mol"1. In resting muscle at 310 K, the concentrations of ATP, ADP and P, are 10, 3, and 1 mM, respectively. Is the ATP hydrolysis reaction at equilibrium in resting muscle? If not, what is the value of AG under these conditions? 4.3 The average daily energy requirements of an adult male human are 10 000 kj. Using the value obtained in Problem 4.2 for the AG corresponding to ATP hydrolysis and synthesis under cellular conditions, calculate the mass of ATP turned over per day. Assume that 1 mol of ATP corresponds to 550 g. 4.4 Glycogen phosphorylase catalyses the following reaction in the mobilization of glycogen reserves: (glycogen),, + P, =^ (glycogen),,., + glucose-1-phosphate for which AG0' = 3.05 kj mol"1 at 37°C. Assuming that the concentrations of P, and glucose-1-phosphate are equal, does the equilibrium lie in favour of glycogen synthesis or degradation at 310 K? In muscle the concentrations of P, and glucose-1-phosphate are 10 mM and 30 pJvl, respectively. Does glycogen phosphorylase catalyse the net synthesis or degradation of glycogen in muscle? 4.5 The radioactivity of a sample of ATP labelled at the y-phosphate with the isotope 32P was measured at different times with the following results. Time (days) 0 5 10 20 30 50 Radioactivity (cpm) 53 450 41 950 32 920 20 270 12 490 4740 Show that the decay of the isotope is a first-order process and determine the rate constant and half-life for this process. 4.6 When the enzyme creatine kinase is incubated with iodoacetamide, there is a complete loss of activity due to reaction of a single cysteine side chain per polypeptide chain of the enzyme. The rate of the reaction of enzyme (5 pjvl polypeptide chains) with different concentrations of iodoacetamide was studied by determining the remaining activity of samples withdrawn at stated times. 144 4. calculations in the molecular biosciences Activity remaining (%) 0.25 mM 0.50 mM 0.75 mM 1.0 mM iodoacetamide iodoacetamide iodoacetamide iodoacetamide Time (min) 0 100 100 100 100 1 95 92 88 84 2 92 84 77 71 4 84 71 60 51 8 71 51 36 26 14 55 30 17 9 20 43 18 8 3 Show that the reactions obey pseudo-first order kinetics and derive the pseudo-first order rate constants in each case and the second-order rate constant for the reaction of creatine kinase with iodoacetamide under these conditions. 4.7 Some typical Kd values for interactions of biochemical interest are avidin-biotin 10~15 M, antigen-antibody 10"10 to 10~9 M, and enzyme-substrate 10"6 to 10"4 M. Evaluate the AG%10 values corresponding to these dissociation constants. 4.8 What would be the effects of competitive, non-competitive and uncompetitive inhibitors on the Eadie-Hofstee and Hanes-Woolf enzyme kinetic plots? 4.9 Chymotrypsin catalyses the hydrolysis of the ester substrate N-acetyl-L-tyrosine ethylester (ATEE). The reaction is inhibited by indole. The following data were obtained in the absence and presence of 1.3 mM indole. Determine the type of inhibition observed and the dissociation constant for the interaction of enzyme with the inhibitor. [ATEE] (mM) Velocity (|Xmol min 1 mg ') Without indole With indole 2.0 9.8 3.9 4.0 14.4 5.8 6.0 17.1 6.8 10.0 20.0 8.0 15.0 21.9 8.8 20.0 23.0 9.2 4.10 Dehydroquinase catalyses the conversion of 3-dehydroquinate to 3-dehydroshikimate and water. The type II enzyme from Helicobacter pylori is inhibited by citrate. From the following data obtained in the absence and presence of 2 mM citrate, determine the type of inhibition observed and the dissociation constant for the interaction of enzyme with the inhibitor. [3-dehydroquinate] (|1M) Velocity (Urnol min 1 mg ') Without citrate With citrate 50 4.4 3.0 100 6.9 5.0 150 8.5 6.5 250 10.3 8.4 500 12.4 10.9 750 13.3 12.1 4.11 Benzamidine is known to bind to the active site of trypsin, leading to inhibition of the enzyme. The binding was studied under conditions when the concentration of enzyme (3.5 LiM) was very small by comparison with the concentrations of benzamidine added so that essentially all the ligand can be considered free. Using an appropriate plot, analyse the following data to determine the Kd for the trypsin-benzamidine interaction. [Benzamidine] (mM) Y (fractional saturation) 0.1 0.29 0.2 0.44 0.5 0.67 0.8 0.76 1.2 0.83 1.6 0.87 4.12 The following data were obtained during a study of the papain-catalysed hydrolysis of a model 4-nitroanilide substrate. The reaction was monitored by the increase in absorbance at 405 nm as the 4-nitroaniline product was formed. (The molar absorption coefficient for 4-nitroaniline at 405 nm is 9500 M"1 cm-1; the substrate does not absorb at this wavelength.) [Substrate] (mM) Velocity (AA4I)5 min"1) 0.10 0.029 0.25 0.054 0.50 0.076 0.75 0.089 1.00 0.096 1.50 0.105 The assay volume was 3 mL in a cuvette of 1 -cm pathlength. The amount of papain added was 0.8 |J.g and the molecular mass of the enzyme is 23 kDa. Using an appropriate plot, determine the Km for the substrate and the for the reaction in terms of AA405 min-1. Calculate the specific activity and fccat for the papain-catalysed reaction. 146 4. calculations in the molecular biosciences 4.13 The Vmax for the type II dehydroquinase from Mycobacterium tuberculosis is 0.0134 Ltmol mirr1 when 1.7 Lig enzyme is added to the assay mixture. What is the kca for the enzyme? (The molecular mass of each subunit of the enzyme is 18 kDa). If the Km for 3-dehydroquinate is 25 LtM, what is the kat/Km ratio? 4.14 The binding of 14C-labelled L-phenylalanine to the enzyme pyruvate kinase was studied. After equilibrium had been achieved, the radioactivity bound to 2 raL of a solution of enzyme (0.85 mg mL~') was found to be 6250 cpm. The specific radioactivity of the 14C-l-phenylalanine was 0.15 Ci mol"1, and the efficiency of counting was 70%. The enzyme has a molecular mass of 240 kDa and contains four identical subunits. What is the stoichiometry of binding of l-phenylalanine to pyruvate kinase? 4.1 5 The amino acid methionine has the molecular formula C5H,,N02S. A sample of [35S] methionine was stated to have a specific radioactivity of 1.9 X 1016 Bq mol"1. What per cent of the S atoms in the sample are present as 35S? The half-life of 35S is 87.4 days and Avogadro's number is 6.02 x 1023 mol"1. 4.1 6 Ornithine decarboxylase catalyses the reaction: ornithine =^ 1,4- diaminobutane (putrescine) + C02. The reaction can be monitored by measuring the release of I4C02 from ornithine which is labelled with 14C at the carbon atom of the carboxyl group. In a reaction mixture of volume 0.4 mL, the concentration of [14C]-ornithine was 10.8 p:M. The specific radioactivity of the [14C]-ornithine was 5.8 Ci mol"1. After 10 min, the reaction was stopped and the radioactivity released as 14C02 was found to be 14 580 cpm; the efficiency of counting was 75%. What was the rate of the reaction expressed as nmol product formed per minute and what proportion of the original substrate present had been converted to products after 10 min? mill References for Chapter 4 Price, N.C. and Stevens, L. (1999) Fundamentals of Enzymology, 3rd edn. Oxford University Press, Oxford, 478 pp. Price, N.C, Dwek, R.A., Ratcliffe, R.G., and Wormald, M.R. (2001) Principles and Problems in Physical Chemistry for Biochemists, 3rd edn. Oxford University Press, Oxford, 401 pp. Appendix Appendix 4.1 Binding of ligands to macromolecules: saturation curves Consider the binding of ligand L to a protein P which has a single binding site: P + L == PL : [PL] >hencetPLJ= Kj At a given concentration of free ligand ([!]), the fractional saturation of the binding sites (Y) equals the concentration of bound ligand divided by the total concentration of protein. Dividing each term in the numerator and denominator by [P]: Multiplying each term in the numerator and denominator by Kd 148 4. calculations in the molecular biosciences When there are n binding sites for ligand on the protein, the fractional saturation equation is analogous to eqn. 4.22. It is expressed in the following terms: r = - nx[L] Kd+[L] 4.23 where r is the average number of molecules of ligand bound per molecule of protein, n the number of binding sites, [L] is the concentration of free ligand, and Kd is the dissociation constant. It is, however, considerably more complex to derive this equation since it is necessary to take into account the statistical nature of the relationship between the successive dissociation constants for binding of ligand. (This is discussed in Chapter 6 of the book by Price and Stevens (1999)). In the case of enzyme kinetics we can consider the fractional saturation of the active sites of the enzyme by the substrate S. One way of calculating the fractional saturation is to make the assumption that in the mechanism E + S^ES—»E + P the breakdown of ES to yield E + P does not perturb the E + S ^ ES equilibrium significantly. In this case, the fractional saturation can be written as: Y=- [S] Kd+[S] In this case the Kd is denoted by the Km (Michaelis constant), so: Km+[S) Complete saturation of the active sites corresponds to the limiting rate (Vmax), v so the actual rate (v) at a substrate concentration ([S]) is given by-, which corresponds to the fractional saturation, i.e. Y. [S] V,, Hence - Km + [S] v = - Km + [S] We can also derive an equation of this form by making the steady state assumption, namely that the concentration of the intermediate (ES) remains constant, i.e. the rate at which it is formed equals the rate at which it is broken down. In this case, Km represents a function of various rate constants of the individual steps in the reaction. Only in the limiting case (i.e. that the rate of ES breakdown to give E + P equals zero) does Km become equal to the dissociation constant of ES to give E + S. (For further details see Chapter 4 of Price and Stevens (1999).)