A substrate is converted to a product via the reaction sequence(1) E +S ESk2(2) ES + S ES2k4(3) ES2ES +P(4) ES , E +P(a). Using pseudo-steady state hypothesis for various forms of enzyme active sites, obtain relationsbetween rates of individual steps in the mechanism.(b). Relating concentrations of vacant active sites to concentrations of occupied active sites, obtainexpressions for the rate of consumption of S, (-Rs), and the rate of formation of P, Rp.(c). What are the maximum values of (-Rs) and Rp and under which conditions are these attained?

Using the Michaelis-Menten equation is one of the best approaches to enzyme kinetics. Michaelis-Menten equation relating relates the velocity to substrate concentration for any reaction with a substrate S binds reversibly to an enzyme E to form an enzyme-substrate complex ES, which then reacts irreversibly to generate a product P and to regenerate the free enzyme E. This equation can be represented as: E+S⇌ ES→k2E+P Rate of disappearance of S = -k1[E] [S] / t Rate of breakdown of [ES] =- k-1 [ES] / t - k2 [ES] / t Rate of formation of P = k2 [ES] / tUnder the steady-state assumption, the rate of formation is equal to the rate of breakdown a. As per Pseudo steady state hypothesis , net rate of reaction of active intermediates is zero, i.e. reactive intermediates reacts as fast it is formed. Thus, net rate of formation of reactive intermediates is zero. Actually, active intermediates are highly reactive and disappears fast. rA = 0 In the given reaction steps, [ES] and [ES2] are the reaction intermediates, Rate of formation of [ES] + rate of formation of [ES2] = 0 1. Rate of formation of ES complex=( k1ES) /delta tRate of break down of ES complex=(-k6ES) /delta t= (-k2ES) /delta tSimalarly the rate of formation and breakdown for others reaction of the sequence are as follows2. Rate of formation of ES2 complex=(k3ESS ) /delta tRate of break down of ES2 complex=(-k5ES2) /delta t = (-k4 ES2 ) /delta tUnder psuedosteady state assumption, the rate of formation of intermediates is equal 0Rate of formation of [ES]+ Rate of formation of [ES2] = 0 k1ES+k3ESS=0Rearranging the above equation k1ES=-k3ESSk1[E] =-k3 [ES]Thus, k1 and k3 depends on [E] when [S] is very low.b. b. Rate of consumption ofsubstrate can be given by the:rate of formation of ES and ES2 complex as, substrate are occupied hereRate of consumption of S, Rs=(k1ES+k3ESS ) /delta tRate of formation of P can be given adding the rate of breakown of ES2 and ES complexRate of formation of P, Rp=(-k5ES2-k6ES) /delta t c. Maximum value of Rs can be when rate of product formation is high. Maximum value of Rp can be when rate of substrate consumption is high. This condition can be achieved when Rs = Rp k1 [E][S] + k3[ES][S] = -k5[ES2] - k6[ES] k1 [E][S] + k3[ES][S]+ k5[ES2] + k6[ES] = 0 Thus, the sum of rate of substrate consumption and rate of product formation is zero.