please I want to know if my calculation is true or not
my enzyme assay consist of To 200 μL resuspended cells (sample), added 287.5 μL phosphate buffer ,225 μL 5M NaCl, and 37.5 μL N-succinyl-Ala-Ala-Pro-Phe-p-nitroanilide incubated for 30 min.
• Stop the reaction by the addition of 175 μL of 80% (v/v) acetic acid.
• Centrifugation for 5 min at 13,000 g
• Measuring the release of pNA at 410 nm.
One unit of protease activity defined as 1 nmol of p-nitroanilide released per minute under the conditions of the assay.
if I have for example absorbance 0.065
Units /ml enzyme=
=(0.065*(925/1000)*1000)/(30*8.8*(200/1000))
925/1000=Total volume (in milliliters) of assay
1000 = Factor for converting μmoles to nanomoles as per the Unit Definition
30=assay time per minutes
8.8 = Millimolar extinction coefficient of p-nitroaniline at 410 nm under the assay condition I am confused on it as in one article I read (The release of p-nitroaniline was followed at 410 nm ([E 410]=8800 M-1cm -1) in another one I read molar absorption coefficient ( 8.800/M/cm) so I have to use 8.8 or 0.0088 as millimolar??
200/1000=Volume (in milliliters) of enzyme sample used
The specific protease activity calculated by division of result on the protein content in 0.2ml (sample volume)
Thanks
To avoid the confusion of stringing together a bunch of numbers and parentheses without any units, I like to break the calculation down into steps.
The extinction coefficient is 8800 M-1cm-1 or, equivalently, 8.8 mM-1cm-1. (In some places, a dot (.) is used to separate the digits in large numbers instead of being used as a decimal point.)
The absorbance is 0.065, which corresponds to 7.39 µM (=7.39 x 10-6 moles/L) product based on the extinction coefficient, and assuming a pathlength of 1 cm, which you didn't mention.
The volume of the solution in which the measurement was made was 925 µl = 0.000925 L.
The amount of product formed was 7.39 x 10-6 moles/L x 0.000925 L = 6.84 x 10-9 moles = 6.84 nmoles.
The reaction time was 30 min, so the average rate of product formation was 6.84/30 = 0.228 nmoles/min = 0.228 units (U).
The volume of enzyme added was 0.2 ml, so the activity of the preparation was 0.228/0.2 = 1.14 U/ml.
For the specific enzyme activity I have to divide the final number on the protein content in 0.2 ml , but I wonder there are a reverse relationship between the protein content and specific enzyme activity, I mean the sample with higher protein content would be with lower specific enzyme activity?
Yes the specific activity calculated is Activity/ mg of total protein determined seperately by standard methods.
Specific activity of enzyme with high protein concentration means enzyme of relatively lower purity as total proteins includes all proteins other than enzyme proteins
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