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276  Chapter 10: Biochemical Reactions: Enzyme Kinetics

                           where


                                                 V max,app  =  V,,,,,/U   +   cd&)  <   Vm,,  (10.4-14)


                           Equations 10.4-13 and -14 illustrate that, relative to the uninhibited case, V,,, changes
                           (decreases), but K,,,  remains the same.
                             If we do not make the assumption leading to K2 = K3,  then the four-step mechanism
                           above also represents mixed (competitive and noncompetitive) inhibition, and both K,,,
                           and  vmlu  change. In this case, equation 10.4-12 may be written as

                                                              V man,appCS
                                                         rp  =                              (10.4-15)
                                                             cs  +  Km,a*p


                           where


                                                    V max,app  =  V,,,/(l   + cIIK3)        (10.4-16)
                                                 K     =  K,(l +  cIIK2)l(1  +  cIIK3)      (10.4-17)
                                                  maPP


                             Treatment of the full six-step kinetic scheme above with the SSH leads to very cum-
                           bersome expressions for cn,   cm, etc., such that it would be better to use a numerical
                           solution. These can be simplified greatly to lead to a rate law in relatively simple form,
                           if we assume (1) the first four steps are at equilibrium, and (2) k,., =  kr2:


                                                                V??UlXCS                    (10.4-18)
                                                rp =  cs  +  K,(l +  c,lK,)l(l  +  cIIK3)



                           This represents competitive inhibition in the sense that V,,,,, is unchanged (relative to
                           the uninhibited reaction), but Km  is changed.


      10.5  PROBLEMS FOR CHAPTER 10

                           10-1  If the activation energy for the decomposition of  Hz.02  in aqueous solution catalyzed by the
                                enzyme catalase is 50  kJ  mol-‘,  and that for the uncatalyzed reaction is 75  kJ  mol-‘,  calculate
                                the ratio of the rate of the catalyzed reaction to that of the uncatalyzed reaction at 300 K. What
                                assumptions have you made in your calculation?
                                                  equation, 10.2-9, is developed in Section 10.2.1 from the point of view
                           10-2
                                The  Michaelis-Menten
                v
                                of homogeneous catalysis and the formation of an intermediate complex. Use the  Langmuir-
             7O-v
            0              10-3  Hinshelwood model of surface catalysis (Chapter  8),  applied to the substrate in liquid solution
                                and the enzyme as a “colloidal particle” with active sites, to obtain the same form of rate law.
                                Ouellet et al. (1952) have reported a kinetics analysis of the enzymatic diphosphorylation of
                                adenosine triphosphate  (ATP). Because of the suggestion that myosin might be the transducer
                                which, in muscles, converts the free energy of ATP into external mechanical work, the sys-
                                tem chosen for study was the hydrolysis of ATP (S) in the presence of myosin to give ADP
                                (adenosine diphosphate) and phosphate. Their initial-rate data obtained at 25°C are as follows
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