Penicillin is hydrolyzed and thereby rendered inactive by penicillinase, an enzy

Question

Penicillin is hydrolyzed and thereby rendered inactive by penicillinase, an enzyme present in some resistant bacteria. The mass of this enzyme is 27500 g/mol. The amount of penicillin hydrolyzed in 3.0 minutes in a 100.0 ml solution containing 210^-7 g of purified penicillinase was measured as a function of the concentration of penicillin. Assume that the concentration of penicillin does not change appreciably during the assay. If the Km of the reaction is 1 M and the Vmax is 0.033 M/min, and assuming that each enzyme molecule has two active sites, what is the turnover number (kcat) in units of 1/s?

Explanation / Answer

Ans. Part A: Molar mass of penicillinase = 27500 g/mol

Moles of penicillinase = Mass / Molar mass

                                    = 7.0 x 10-7 g / (27500 g mol-1)

                                    = 7.273 x 10-12 mol                        

Given, volume of enzyme reaction medium = 100.0 mL = 0.100 L

Now, Enzyme concertation = Moles of enzyme / Volume of solution in liters

                                                = 7.273 x 10-12 mol / 0.100 L

                                                = 7.273 x 10-11 M

                                                = 7.273 x 10-5 uM

Part B: Using Turnover number, Kcat = Vmax / [Et]     - equation 1

                        Where, Vmax = maximum velocity            ;[Et] = concentration of enzyme

We have, Vmax = 0.033 uM/ min = 0.033 uM / 60s = 0.00055 uM s-1

Now,

            Kcat = 0.00055 uM s-1 / 7.273 x 10-5 uM = 7.653 s-1

Therefore, Kcat under given conditions = 7.653 s-1

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