Re: How to calculate the enzyme Kcat from its Vmax and M.W.

Hi Sam,

It’s nice to hear from a fellow “McGillian” (I’m B.Eng. 2000). You’re so close; k_cat is really just a V_Max expressed in units of turnovers per enzyme molecule per second.

All you need to do here is some unit conversion. 2000 micromoles (of product molecule) is really just a number. It’s 2000 x 10^-6 x Avogadro’s number (6.02 x 10²³). Divide by 60 to convert the min^-1 to sec^-1. Now you just need to convert the mg in the denominator to the number of enzyme molecules in one mg. 60 kDa is 60,000 g/mol, or 1/60,000 mol/g, or 1/60,000,000 mol/mg. Use Avogadro’s number again to get molecules/mg (1/60,000,000 mol/mg x 6.02 x 10²³ molecules/mol). You’ll see that the two instances of Avogadro’s numbers cancel, and that your boss’ advice is not quite right. Use your eventual mastery of this conversion to show him/her why.

I don’t want to finish solving this for you. It will be good for you to get comfortable with these calculations by practicing them. Most biochemistry textbooks will have a section on these types of calculations. If you write the units of every quantity next to the numbers as you multiply or divide them, you can follow the unit conversion process and “cancel out” units that appear in both the numerator and denominator. See the excellent tutorial at http://www.youtube.com/watch?v=XKCZn5MLKvk.

Regards,
Alex Tobias