Define still another number, Kcat, as Kcat = Vmax / [Etotal]. Justify the name for this value, the turnover number of an enzyme. Hints: Start with the equation for Vmax = k2 [ES] = k2 [Etotal], and compare that equation with the definition of Kcat. As [Etotal] increases, what happens to Vmax? Finally, is Kcat a constant? Other hints: begin with Vmax = k2 [Etotal]. Then, solve for k2 and compare this to the expression for Vmax. Our goal is to show that k2 = Kcat.