t's just algebra... Below you will derive an equation to express the velocity of an enzyme catalyzed reaction in the presence of a competitive inhibitor *Write a Kinetic scheme for a single substrate enzyme catalyzed reaction in the presence of a competitive inhibitor. * Just as in the absence of a competitive inhibitor, the rate (velocity) of product formation with a competitive inhibitor present is (a) v= keat [ES] MaNs Damance, In the presence of substrate and competitive inhibitor, the enzyme can be in one of three forms (III In blanks) [E]an -[6] + [ENY + ( Divide. Divide equation (a) by the mass balance equation: (6) tots s loc Equilibria: ES and El complexes have dissociation constants that can be expressed as (III in blanks): (c) Km- []*[ ] (d) K = []*[ ] [ 1 [ ] Rearrange equation (C) and (d) to solve for ES and El respectively (it in blanks) (e) [ES] - [*] (O [BI] - [1] Substitute. Rewrite equation (a) with the terms ES and El substituted with the right hand side of (0) and (1) Factor & simplify. Now, on the right hand side of equation (8), I have factored out the [E] terms, so they cancel, to give (1) [S] /Km [E]or 1 + [S]/K + [l/K; Finally, the numerator and denominator on the right side of equation (1) can be multiplied by K., and factored to give a simplified form of the velocity equation: