Question

Vmax = 3.423 µM/sec and enzyme concentration is 250 ug/mL how do I Calculate Kcat? the enzyme is alkaline phosphatase and its MW is 140 kDa

Additional information:

Alkaline phosphatase:  500 µg/mL stock solution in reaction buffer.

#	[Enzyme] ug/mL	Alkaline phosphatase (µL)	Buffer (µL)	Total volume (µL)
1	100	200	1800	2000
2	125	250	1750	2000
3	150	300	1700	2000

Answer 1

Molar mass of enzyme = 140 kDa = 140 x 10³ Da = 140 x 10³ g/mol

mass concentration of enzyme = 250 ug/mL

we have to convert ug to grams and mL to Liter

1 g = 10⁶ ug

1 L = 1000 mL

mass concentration of enzyme = 250 ug/mL * (1 g / 10⁶ ug) * (1000 mL / 1 L)

mass concentration of enzyme = (250 * 1000 / 10⁶) g/L

mass concentration of enzyme = 0.25 g/L

molar concentration of enzyme = (mass concentration of enzyme) / (molar mass of enzyme)

molar concentration of enzyme = (0.25 g/L) / (140 x 10³ g/mol)

molar concentration of enzyme = 1.8 x 10^-6 mol/L

molar concentration of enzyme = 1.8 x 10^-6 M

molar concentration of enzyme = 1.8 x 10^-6 M * (10⁶ uM / 1 M)

molar concentration of enzyme = 1.8 uM

k_cat = (V_max) / (molar concentration of enzyme)

k_cat = (3.423 uM/s) / (1.8 uM)

k_cat = 1.917 s^-1