I have Laccase from trametes versicolor with 0.5U/mg activity.
I need enzymatic activity of 50U/g in solution. What should be the amount of enzyme I add to obtain enzymatic activity of the above said?
The amount of solid enzyme you need to dissolve depends on the volume of the solution you require and the enzyme activity required per ml, not per g.
The question is not very clear .This is what I have understood
Kindly clarify you have activity of 0.5 U per mg and you need to have 50 U per gram .
You need 100 mg of enzyme to have 50 units of it. You can dissolve your enzyme 100 mg in minimum volume of Buffer and add to one gram
If you have a solution with 0.5 U/mg is the same of 500 U/g, so if you wish 50 U/g you must dilute 10 times that solution in the appropriate solvent or activity buffer.
Do the 50 U/g refer to the concentration (activity per weight of solution, which would be unusual as the concentration is normally given per volume), or does it refer to the specific activity (U per g of total protein, irrespective in which volume the protein is dissolved)? In the former case, you'd dilute with buffer, in the latter with the solution of an innocuous carrier protein (usually serum albumin, which is relatively cheap).
Btw., the enzyme unit (1 U = 1 µmol/min) is an outdated unit of activity, in the SI the correct way to express activities is in katal (1 kat = 1 mol/s, 1 U = 16.7 nkat).
- Badges
- Science method