Hello All,
This is regarding cellulase enzyme calculation!
As per NREL protocol (NREL/TP 510-42628),
0.0053 concentration of enzyme gives 70 FPU/mL
this 0.0053 concentration of enzyme has got from 188.4 times dilution of 1 mL of original enzyme from Trichoderma ressei
my question is that, if i do back calculation for this, i will get 13216 FPU/mL of original enzyme, is it correct or not?
my second question is, in NREL technical report-510-42610, page number 7, point number 8.10, having an example,
0.23 mL of cellulase enzyme gives 55 FPU/mL. in that they have also given that same volume of enzyme gives 25 FPU/g cellulose loading
so, here, how to convert these calculations?
I would be thankful to all in advance for helping me in this regards
Hi Mood
You are correct in your activity conversion considering dilution factor.
To determine the specific activity (U/mg), you need to quantify the protein concentration of your enzyme extract (mg/ml). Commonly Bradford or Biuret assay are best option. After that you divide U/ml into mg/ml.
Filter Paper Activity = 0.37/ concentration of enzyme that releases 2 mg glucose.
Therefore FPU = 0.37/0.0053 which is equivalent to 69.81 FPU or alternatively FPU = 0.37*188.68 which is equivalent to 69.81 FPU. Either you divide by concentration of enzyme or multiply with Dilution Factor of enzyme which releases 2 mg of glucose from 50 mg Whatman 1 Filter Paper after 60 min of incubation.
hello，I am afraid you have a little mistake，70 FPU/ mL is the activity of original enzyme, it had be multiplied by the dilution ratio, because it was unbelievable the activity of cellulase was more than 10000.
70FPU is still the concentration of enzyme in your original enzyme as pointed out by Zhang.
Can I add a query of mine here?
What if the FPU/g is taken as 'FPU/ gram of biomass'?
I have found a mention of this unit in the attached paper by Nathan Mosier. The meaning of this unit is not clear to me and seems to be a mistake.
Article Features of Promising Technologies for Pretreatment of Ligno...
Cellulase (Cellic CTec3, 210 U mg-1), α-amylase (originated from Aspergillus oryzae, CAS:9001-19-8, 30 U mg-1), and amyglucosidase (originated from Aspergillus niger, CAS:9032-08-0, 70 U mg-1). For enzymatic hydrolysis of biomass 10 u.g-1 enzymes were added in 50 g L-1 solid loading . Can we calculate the FPU value from this information and how?
If you have used a pure cellulase, it is necessary to measure its density (g/ml). Sometimes, the density of pure cellulase density has been registered on the main pot. Then by dividing the enzyme activity (FPU/ml) by density (g/ml), you can obtain enzyme activity (FPU/g). If you use the enzyme extract, you can act like above (measuring the density of enzyme extract), but you must mention (FPU/g (enzyme extract)).
Mina Karimi-Avargani Muinat Olanike Kazeem I m going to use cellulase cocktail enzyme(Sigma)density 1.15g/ml which hydrolysis unit is 1000U/g. Do I need to calculate FPU?
You need to determine the standard rate at which your cellulase enzyme will liberate glucose. Your aim is to determine the amount of enzyme to release 2mg of Glucose from 50mg of filter paper (4 % conversion) in 60 min. Simply put, this standard is determined by preparing known concentrations of the enzyme and use it do digest glucose from filter paper. Follow the NREL procedure herewith attached
Hello! It can depends of the context. FPU/mL refers the units in a 1 mL of solution. On the other hand, FPU/g can be related to units added for 1 g of cellulose to hidrolize (or 1 g of substrate, if you don't have a pure biomass). Therefore, to convert cellulase enzyme activity, multiply the FPU/mL value by solution/cellulose ratio (mL/g).
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