I'm working on production of laccase enzyme by microorganism. But I
don’t know how can I calculate unite of enzyme??
My reaction mixture was 270µl supernatant (crude enzyme), 130 µl of
ABTS as substrate and 200µl of buffer. The formation of green dye was
followed spectrophotometrically at 420 nm for 1000 second by 30 s
intervals. The result of this assay is in attachment. Would you please
guide me about calculation of this unite with Beer and Lambert's Law?
Hi there,
1 Unit of enzyme is defined by the amount required for the transformation of 1µmol of substrate per minute. Despite the absence of the attached file, you should be able to calculate the rate of the reaction in terms of variation of absorbance per minute. Then knowing epsilon and the volume of reaction, the rate can be expressed as µmole converted per minute and in fine in U...
Hi there,
1 Unit of enzyme is defined by the amount required for the transformation of 1µmol of substrate per minute. Despite the absence of the attached file, you should be able to calculate the rate of the reaction in terms of variation of absorbance per minute. Then knowing epsilon and the volume of reaction, the rate can be expressed as µmole converted per minute and in fine in U...
I can't do much better than Dominique. I could add that (after you have converted your absorbance values to umoles/minyou have to specify the temperature and pH since these change things dramatically. For some reason we still use 25C as the standard temperature. If you want to be pedantic (since minutes are not an SI unit) you could use the 'katal as a measurement. This was officially adopted in 1999. One katal is the amount of enzyme that converts 1 mole of substrate per second, so
1 U = 1/60 micro katal = 16.67 nano katal.
Personally - I think that giving the Km and kcat - for a given pH and temperature is more acceptable definition of activity.
Best wishes
I just looked up the Sigma website (which is rather good for assays) - try
http://www.sigmaaldrich.com/technical-documents/protocols/biology/enzymatic-assay-of-laccase.html
this gives the standard assay for laccase based on absorbtion and the conversion to units.
you have to do a standard curve for the final product of the reaction to be able to calculate the amount of final product obtained from the reaction by the enzyme.
then you can calculate how much amount of final product obtained per minute. at this point you can calculate the activity of the enzyme.
Dear Rezvan ,
- This seems to be like a kinetic pattern. You can not use the data until 1000s because, every enzyme has its own activation energy this will be constant until certain time interval beyond that the activity will be decreased. In your case the end result is Greenish color. So take the reading upto 180s by 30 s intervals. If you take upto 1000s ab value respect to time interval you will get a Kinetic curve of enzyme of interest.
I agree the valid point of Mr.Naresh, see the time period where the data should be taken for the calculation. From he reference data given, the volume of the substrate is almost double of crude enzyme, pl check and optimize the concentration of the substrate, that too influence the activity.
I did know your reagents' concentration. So, there is a template, and I hope it may help you.
You just follow thiese links. Hope it will be helpful.
Best wishes
http://ac.els-cdn.com/S0141022905003030/1-s2.0-S0141022905003030-main.pdf?_tid=347083f0-6d4d-11e4-a1e5-00000aacb35d&acdnat=1416114088_8c58e0c20b993fb4d4d332eb83629e00
http://ac.els-cdn.com/S0141022901005221/1-s2.0-S0141022901005221-main.pdf?_tid=36c4a866-6d4d-11e4-b9b2-00000aacb35e&acdnat=1416114092_512cf9b0dbf36f70eb3a0f227a5239c2
Usually, 1 Unit of enzyme is defined by the amount required for the transformation of 1µmol of substrate per minute. Enzymatic activity is measured as Unit/liter by the following equation:
alfa = deltaA/t x (epsilon x d x (1000 mmol/1mol) x (V/v))-1
where alfa (activity), deltaA is the adsorption variation at around 414 nm, epsilon is the molar extinction coefficient (36000 M-1cm-1), d is the cuvette optical path, V is the total reaction volume (mL) and v is the enzymatic solution volume (mL).
In general, unit of enzyme is typically used to describe enzyme catalytic activity, where a unit (U) refers to the amount of enzyme that catalyzes the conversion of 1 micro mole (μmole) of substrate per minute. Thus, 1 enzyme unit (U) = 1 μmol/min, where μmol refers to the amount of substrate converted. Because each enzyme has a unique substrate, a unit of activity is different for one enzyme versus another.
Just go to this link for detailed explanation.
http://www.biomol.de/details/IN/technical-guides/Enzyme-Units-and-Specific-Activity-Explained.pdf
Unit Definition - One unit will oxidize 1.0 µmole of 2,2' azino bis (3 ethylbenzthiazoline-6-sulfonic acid) per minute at pH 5.0 at 25°C. - See more at: http://www.sigmaaldrich.com/technical-documents/protocols/biology/enzymatic-assay-of-peroxidase-abts-as-substrate.html#sthash.ztEXxFAT.dpuf
The answer is in the attached document below.
Please to see the established equation for calculation. Welcome if you have relative questions.
Laccase units calculation. Correction of an unintentional mistake:
For application of the given equation (answer/Nov 19, 014), consider (O.60/0.27) instead of (1/0.27). This is due to the final volume of the assay, 0.60 ml.
Please to read the attached document for your data discussion. Relative questions will be welcomed.
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