So let's suppose we have the beaker. Inside that beaker we have a mixture of some particular type of enzyme and the concentration, a total of that enzyme inside the mixture is given to us to be 0.1 molar now, let's suppose that experimentally, we calculate the V max value of that particular enzyme to be 60,000 molar per second. So at this concentration, when all the active sites of that enzyme mixture are filled with the substrate, we know that the maximum rate of the reaction is 60,000 molar every single second. So what that basically means is 60,000 of the substrate molecules are transformed into product every single second, when all the active sites are actually occupied, are actually saturated with that subtra. This is what 60,000 molar per second actually tells us. So to calculate K two, also known as Kcat, the turnover number, we simply take the VMAX.