Penicillin is hydrolyzed by penicillinase, an enzyme present in some resistant b

Question

Penicillin is hydrolyzed by penicillinase, an enzyme present in some resistant bacteria. The MW of this enzyme is 29000 g/mol. The amount of penicillin hydrolyzed in 2.0 minutes in a 80.0 ml solution containing 2.410-7 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, the Vmax is 0.05 M/min, the enzyme concentration is 1.03410-4 M, and assuming that each enzyme molecule has three active sites, how long would it take to hydrolyze two penicillin molecules (in units of s)?

Explanation / Answer

Turn over number Kcat = Vmax / Etot

Vmax = 0.05M/min = 5 *10-8 M/min

Etot = 2.4 *10-7g / 80ml

= 2.4 *10-7 g /29000 g/mol = 8.275 * 10-12 mol

80 ml = 0.08 L; So Etot = 8.275 * 10-12 mol / 0.08 L = 1.03 * 10 -10 M

Kcat = Vmax / Etot

Kcat = 5 *10-8 M/min / 1.03 * 10 -10 M

Kcat  = 485.4/min

Kcat = 8.09/s

Time to hydrolyze two penicillin molecules = 2/Kcat = 2/8.09/s = 0.24s

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