Question

The kinetics of chymotrypsin were studied and yielded a KM of 5 mM and a Vmax of 0.2 mM/min. The enzyme concentration was 0.20 μg/ml. Calculate the turnover number, assuming one binding site per enzyme.

Answer 1

Ans. Given,

            Vmax = 0.2 mM / min                                 ; [1 min = 60 s]

= 0.00333 mM/ s                              ; [1 mM = 1000 uM]

= 3.33 uM/ s

            [E] = 0.20 ug/ mL

# Step 1: Convert [E] in terms of uM

Molar mass of Chymotrypsin = 25.6 kDa = 25600 g/ mol

Now,

            [E] = 0.20 ug/ mL

                        = 2.0 x 10^-7 g / mL

                        = (2.0 x 10^-5 g / 25600.0 g mol^-1) / mL

                        = 7.8125 x 10^-12 mol / mL                                       ;[1 mL = 0.001 L]

                        = 7.8125 x 10^-12 mol / 0.001 L

                        = 7.8125 x 10^-9 M

                        = 7.8125 x 10^-3 uM

Therefore, 0.2 ug/ mL chymotrypsin is equivalent to 7.8125 x 10^-3 uM chymotrypsin.

# Now,

            Turnover number, Kcat = Vmax / [E]

                                                = 3.33 uM s^-1 / (7.8125 x 10^-3 uM)

                                                = 426.24 s^-1

Hence, turnover number = 426.24 s^-1

# Note: The molar mass of chymotrypsin = 25.6 kDa (Wilcox 1970)

Change in molar mass of the enzyme would cause change in Kcat value, too.