I have a doubt in calculating the activity of immobilized cellulase i've used the formula Activity of cellulase (μmol/ml min) = 1000 w/Mvt; where, w is the amount of glucose produced, M is the molecular weight of glucose, v is the volume of
the sample and t is the reaction time. Here in the formula instead of ml (i.e volume) i have used grams (g) of beads taken, then the unit for cellulase would be (μmol/g min). The way i am calculating the activity is correct?
Yes it is correct.
This activity will be for 1gms of beads.
Can u provide me the complete details of cellulase activity testing procedure you followed including DNS prepration, glucose standard curve prepration and protein measurement? Actually we are facing some problem in testing.
Thanks
correct , but you may multiply by the activity of enzymes by bead in order to standardise your results
It is correct, you will have specific activity regarding to the support. If you need volumetric activity, you need to estimate the load of enzyme in the support.
Check this Sheldon and van Pelt. Enzyme immobilisation in biocatalysis: why, what and how. Chem. Soc. Rev. , 2013,42, 6223. DOI: 10.1039/c3cs60075k
your right , here you also calculate for free enzyme along with immobilized so that you can get the confirmation.
Dear Nisha S,
You will initially have to calculate the free enzyme's activity - IU/mL (the quantity that was immobilized). Later, post immobilization, please perform the assay ( I performed the IUPAC Ghose assay). I observed a decrease in enzyme activity after immobilization within calcium alginate beads.
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