How Does Penicillinase Affect Penicillin Kinetics in Staphylococcus Aureus?

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Homework Statement

Penicilline is hydrolyzed and thereby rendered inactive by penicillinase , an enzyme present in some resistant bacteria. The mass of this enzyme in staphylococcus aureus is 29.6kd. The amount of penicilline hydrolysed in 1 minute in a 10ml solution containing 10^-9g 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.The following data was collected: Penicillin in micro mole:1, 3, 5, 10, 30, 50 and amount hydrlysed in nanomole: 0.11, 0.25, 0.34, 0.45, 0.58, 0.61 respectively.
1. plot V0 vs and 1/V0 vs 1/ for these data and find the value of KM
2. what is the value of Vmax
3. What is the turnover number of penicillinase under the experimental conditions. Assume one active site per enzyme molecule.

Homework Equations

The Attempt at a Solution

 

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[Blueshift5]

1. To plot V0 vs , we first need to calculate the initial velocity (V0) for each concentration of penicillin. This can be done by dividing the amount of penicillin hydrolyzed (in nanomoles) by the time (1 minute) and the enzyme mass (29.6kd). The results are as follows: 0.0037, 0.0085, 0.0115, 0.0152, 0.0196, 0.0206.

Next, we plot V0 vs on a graph with V0 on the y-axis and on the x-axis. We can then use the data points to draw a curve and determine the value of KM, which is the substrate concentration at which the reaction rate is half of Vmax.

To plot 1/V0 vs 1/, we first need to calculate 1/V0 by taking the reciprocal of each V0 value. The results are as follows: 270, 118, 87, 66, 51, 48.

Next, we calculate 1/ by taking the reciprocal of each concentration of penicillin. The results are as follows: 1, 0.33, 0.20, 0.1, 0.03, 0.02.

We then plot 1/V0 vs 1/ on a graph with 1/V0 on the y-axis and 1/ on the x-axis. We can then use the data points to draw a straight line and determine the value of KM, which is equal to the x-intercept of the line.

2. Vmax is the maximum reaction rate, which can be determined by finding the highest V0 value from the graph in part 1. In this case, the highest V0 value is 0.0206 nanomoles/min.

3. The turnover number is the number of substrate molecules converted to product per enzyme active site per unit time. In this case, we can calculate the turnover number by dividing the Vmax value (0.0206 nanomoles/min) by the amount of enzyme used (10^-9g), which is equal to 2.06 x 10^10 molecules of substrate converted to product per enzyme active site per minute.