To calculate the Vmax and Km of the reaction you will need to run various concentrations of the methyltransferase to be tested and calculate the activity at each concentration.
The following are example numbers and should not be used with the assay. The optimal starting amount of enzyme will need to be determined empirically
An Example: The following concentrations of a methyltransferase were used in the SAM510 assay and the activites were calculated as per the protocol.
|
[Methyltransferase] (mM) |
Activity |
|
0.1 |
0.04855 |
|
0.125 |
0.05664 |
|
0.167 |
0.06827 |
|
0.25 |
0.08395 |
|
0.5 |
0.11227 |
|
1 |
0.133 |
Convert both columns of data in the above table to the inverse. i.e Divide 1 by each number in the table above.
|
1/[Methyltransferase] (mM) |
1/Activity |
|
10 |
20.59836979 |
|
8 |
17.65574554 |
|
6 |
14.64772963 |
|
4 |
11.91231024 |
|
2 |
8.907403153 |
|
1 |
7.518796578 |
Plot the above as a chart and determine the linear equation for a linear trendline:
1/Vmax is the y-intercept. This is 6.0373 in the above example. So Vmax = 1/6.0373 = 0.1656 µmol/min/ml
The x intercept is -1/km. So in the above equation. 0=1.4527(-1/Km)+6.0373. So Km=-1.4525/-6.0337 or 4.5812mM
