Question

Penicillin is hydrolyzed and thereby rendered inactive by penicillinase (also known as B- lactamase), an enzyme present in some resistant bacteria. The amount of penicillin hydrolyzed in 1 minute in a 10-ml solution containing 10-13 moles 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. [Penicillin] (PM) Amount hydrolyzed (nanomoles) 0.11 0.25 0.34 0.45 0.58 0.61 Plot Vo versus [S] and 1/Vo versus 1/[S] for these data.* Does penicillinase appear to obey Michaelis-Menten kinetics? [Select] * It is recommended that you use graph paper to plot the data, as this is what will be available to you during the exam. You may just use Microsoft Excel for this homework, but then this would not be a good practice for the exam. If so, what is the value of KM? [Select] What is the value of Vmax? [Select] What is the turnover number of penicillinase under these experimental conditions? Assume one active site per enzyme molecule.

Penicillin is hydrolyzed and thereby rendered inactive by penicillinase (also known as β-lactamase), an enzyme present in some resistant bacteria. The amount of penicillin hydrolyzed in 1 minute in a 10-ml solution containing 10^-13 moles 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.

I got the right answers for the first two parts but I can't figure out how to find the turnover number. Please give a step by step for the last part! Thank you

Answer 1

From the graph we can say that it follows Michaleis-Menton Kinetics.

9000 y = 7.6234x + 1.4638 in N/T 7 N O 0.2 0.4 0.8 1 1.2 0.6 1/[S]

Now, y=1/V

x= 1/S

for x=0, y=1.4638. Let this value of y be called as A.

A=1/Vmax = 1.4638

Thus, Vmax = 1/A = 1/1.4638 = 0.68315

For the equation y=mx+c, m is the slope.

Hence, Slope=7.6234

Now, Slope= KM / Vmax

So, 7.6234= KM / 0.68315

KM = 7.6234*0.68315 = 5.2079

Now, Turnover No. = Vmax / Total enzyme concentration

Here, Enzyme concentration = 10-13 moles

Turnover no. = 0.68315/ 10-13

= 683.15*1010