Use the data in the following table for questions 29 and 30

Use the data in the following table for questions 29 and 30 Estimate the amount of substrate one needs to add to fill half the active sites: 4 mu M 20 mu M 50 mu M 828 mu M 1600 mu M For the above enzyme, if the enzyme concentration [E total] was 25uM, what was the approximate turnover number of per second? 300 30 3 0.003 cannot tell

Solution

The plot of initial velocity to substrate concentration for an enzyme that obeys Michaelis - Menten kinetics the KM or the Michaelis constant is the substrate concentration where all the sites are half filled. KM = Vmax/2 ([S] = KM, then V0 = Vmax/2) . Vmax is the maximum concentration where the saturation of the enzyme conversion to substrate saturation is reached and half the maximum velocity is the KM. So here the saturation of substrate is at 1656 uM and the KM or half saturation is at 828 uM.

Answer is 828uM or d

The catalytic turnover number is calculated from the equation which is \"number of substrate molecules converted into product by an enzyme molecule in a unit time when the enzyme is fully saturated with substrate\" which means the VMax is 7498 uM /min divided by the enzyme substrate concentration i.e., 25 uM so 7498 divided by 25 gives 299.92 and the inverse or 1/299.92 is 0.003. This is K2 = Vmax/ enzyme concentration and 1/k2 is the turnover number. Answer is 0.03 or d.