Answer to Question #97245 in Biochemistry for Viswanath Vittaladevaram

Question

180 µL of medium containing (200 µg/mL) Ampicillin were treated with 20 µL of ꞵ-lactamase enzyme to give finial concentrations ranging from 1:1 to 1:10 ratio of ampicillin to B-lactamase enzyme. The ꞵ-lactamase activity is 400 Unit/mg (Unit = 1 µmol / min hydrolysis of benzylpenicillin at pH 7 at 25C). The Ampicillin mass = 349.406 g/mol. Explain how to prepare the ꞵ-lactamase stock solution concentrations that can be used to give a final concentration ratio from 1:1 to 1:10 (ampicillin to enzyme) in the medium?

Answer 1

There is 36µg of ampicillin in the sample. Therefore, 36 to 360µg of enzyme ( representing 14.4 to 144 Units) are required. The enzyme molecular weight is 29kDa and assuming that the preparation is pure. The final concentration of ampicillin in the sample is (36µg/349.406)/200µL=516.16µM. So, the concentration of enzyme needed for the 1:1 ratio and 5.16mM for the 1:10 ratio corresponds to 0.103µmol and 1.03µmol of enzyme in the 200µL sample, respectively. If the molecular weight is 29kDa then; 0.103*10^-6*29000=2.987mg and 29.87mg respectively will be required.