6. (a) Please fill in the correct answer as deriving the Michaelis-Menten equation step by step, starting from Michaelis-Menten mechanism as following and [S]o is excess relative to enzyme. E+S E ES ka, ka' ESP+E kb V= (1) The rate of product formation according to the Michaelis-Menten mechanism is (2 points) (2) The concentration of the enzyme-substrate complex by invoking the steady-state approximation and writing Ka[E][S] - =O (4 points) d[ES] = dt (3) it follows that [ES]=- *[E][S] (2 points) (4) Now we define the Michaelis constant as Km=(ka'+kb)/ka. To express the rate law in terms of the concentrations of enzyme and substrate added, we note that [E]o = [E] + [ES). Therefore, the equation can be changed to be [E]o = [ES] *(1+ (2 points) (5) The free substrate concentration is approximately equal to the initial substrate concentration and we can write [S] = [S]o. It then follows that: [ES]= (2 points) Therefore, this mechanism leads to the Michaelis-Menten equation for the rate of product formation: v= Ko[E]o 1+KM/(S) (6) What kind of condition will be achieved to maximum velocity? and what is maximum velocity? (4 points) (b) The enzyme-catalyzed conversion of a substrate at 298K has a Michaelis constant of 90 umol dm and a maximum velocity of 20 umol dm's when the enzyme concentration is 2 nmol dm. Please calculate turnover rate keat and catalytic efficiency n of this reaction. (hint: kcat= Vmax/[E]o, n=kca/KM) (4 points)