# Enzyme Kinetics/ Biochemistry

You do a set of enzyme assays, plot a Lineweaver-Burk plot and estimate a Vmax reaction rate of 20 μmol.L-1.s-1 when the enzyme concentration is 10 ng.mL-1. Assuming that your enzyme is pure, that you have 20 mg of it and its Mr is 50,000:- - how many enzyme units do you have in total ?
- what is the specific activity ?
- what is the kcat ?

Answers
Okay, so number one:You have 20mg and know the molecular weight (50 kDa); this is a simple application of NCM, which statesNumber of moles * molecular weight = total massNumber of moles * 50 000 = 20 * 10^-3Number of moles = ( 20 * 10^-3 ) / 50 000Number of moles = 4 * 10^-7 molesNumber two:The specific activity gives a measure of enzyme catalytic capacity per unit mass (rather than per molar quantity, as the kcat does). Conveniently, the enzyme concentration is state in mass concentration so converting to useful units10 ng/mL = 10 000 ng/L = 10 µg/LWe now divide Vmax by this value to give 20/10 = 2 µmol. µg^-1. s^-1Number three:By the Michaelis-Menten equation, Vmax = kcat * E0, where kcat is the turnover number or number of moles turned over in unit time by a single enzyme unit, and E0 is the initial enzyme concentration. Thus,20 µmole. L^-1. s^-1 = kcat * 10/50 000 (I have divided by 50 000 because we now work with molar quantities)Thus kcat = 50 000 * 20 / 10= 100 000 s^-1.I hope this helps; any further questions, just ask :)
marenfrew
11 February 2015