IT PTP1B Assay Kit; Colorimetric

Stock Code: 3586523
Manufacturer Part No: 539736-1KIT

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Descriptions

Analysis Note

Positive ControlRecombinant PTP1B

Conversion of A₆₂₀ to nmol of Phosphate with a Standard Curve.1. Plot standard curve data as A₆₂₀nm versus nmol PO₄^2-.2. Obtain a line-fit or fits to the data using an appropriate routine. Note that the plot may not be linear over the entire span from 0 to 3 nmol of phosphate. As shown in figure below, two linear fits, one for 0 to 1 nmol phosphate and a second for 1 to 3 nmol, can produce a more accurate correlation of A₆₂₀ to amount of phosphate.3. Use the slope and Y-intercept of the appropriate plot to calculate amount of phosphate released for a particular data point.4. Example (sample calculation with the standard curve in the figure below):a) Measured A₆₂₀ = 0.180. Since this A falls below that of the 1 nmol point in the figure, choose the 0 to 1 nmol fit, i.e. y = 6.58x - 0.386.b) Replacing x with A₆₂₀ and y with nmol of PO₄^2- produces:nmol of PO₄^2- = 6.58(A₆₂₀) - 0.386nmol of PO₄^2- = 6.58(0.180) - 0.386nmol of PO₄^2- = 0.798NOTE: For highest accuracy, a standard curve should be performed for each new set of assay data. This will normalize for variations in free phosphate in samples, time of incubation with the Green Reagent, and other experimental factors.Time Course and Rate Calculations1. Using a standard curve, convert A₆₂₀ measurements for a series of time points to nmol of phosphate, as described above.2. Plot nmol of phosphate versus time and obtain the slope in nmol/min. If the slope is decreasing at later times, restrict the time points used in the slope determination to the earliest, linear part of the plot.3. It may be useful to calculate the total amount of phosphopeptide per well contained at the chosen substrate concentration. For example, at 75 M IR5 peptide in 100 l reaction that is:75 x 10^-6 mol/L x 100 x 10^-6 L = 7500 x 10^-12 mol = 7.5 nmol4. Although when possible it is best to obtain initial rate data from the first few percent of substrate converted to product, this may not generate a sufficient signal for an accurate rate estimate. In practice, with PTP1B at 75 M IR5 substrate, time course plots remain linear for at least the first 15% of substrate hydrolyzed (~1 nmol of phosphate).Test Sample/Inhibitor Data1. It is important to obtain a "time zero" measurement and to subtract this value, expressed as nmol of phosphate, from both the control and test sample/inhibitor values. Again, a convenient way to obtain an accurate t=0 measurement is to add and mix the 25 l of Red Reagent into the 50 l of 2X Enzyme and then mix in the 50 l of 2X Substrate (see Time Course instructions).2. Calculate activity as a % of Control3. Add additional controls as necessary. For example, while an extremely low reading will generally indicate that the test sample is a potent inhibitor, there is the possibility that the compound interfered with the color development. In such a case, an appropriate control would be to compare the color reaction obtained from 1 nmol of phosphate (10 l of 100 M Phosphate Standard) with and without added test compound [buffer alone with no enzyme or PTP1B Substrate (IR5)]; see Phosphate Standard Curve instructions).