You do an enzyme kinetic experiment and calculate a Vmax of

You do an enzyme kinetic experiment and calculate a Vmax of 176 mol per minute. If each assay used 0.20 mL of an enzyme solution that had a concentration of 0.20 mg/mL, what would be the turnover number if the enzyme had a molecular weight of 128,000 g/mol? (Enter your answer to two significant figures.)

turnover number = __________________ sec-1

Solution

Ans. Part 1. Given,

            Vmax = 176 µmol/ min = (176 /60) µmol/ s = 2.93 µmol/ s

            Total volume of reaction mixture = 0.20 mL

            Concertation of enzyme = 0.20 mg/ mL

            Molar mass of enzyme = 128000 g / mol

So,

            Mass of enzyme in 1.0 mL reaction mixture = 0.2 mg

                                                            = 2.0 x 10^-4 g                                       ; [1 mg = 10^-3 g]

Number of moles of enzyme in 1.0 mL reaction mixture = Mass / molar mass

                                                = (2.0 x 10^-4 g) / (128000 g / mol)

                                                = 1.5625 x 10^-9 mol

                                                = 1.5625 x 10^-3 µmol                          ; [1 mol = 10⁶ µmol ]

Now,

1.0 mL reaction mixture contains 1.5625 x 10^-3 µmol enzyme. So, amount of enzyme in 0.2 mL reaction mixture is-

                                    0.2 mL x (1.5625 x 10^-3 µmol / mL)

                                    = 3.125 x 10^-4 µmol   

Part 2. Turnover number of Kcat is given by-

K_cat = V_max / Eo

            We have,

                                    Vmax = 176 µmol/ min

                                    [Eo] = 3.125 x 10^-4 µmol = Amount of enzyme

SO,

Kcat = (2.93 µmol/ s) / (3.125 x 10^-4 µmol)

            = 9.376 x 10³ s^-1

Hence, turnover number = 9.376 x 10³ s^-1