Regarding enzyme kinetics

Question

 

1. The KM of an enzyme/substrate combination is 125 mM, with a VMAX of 20 μM.s-1.  The concentration of enzyme is 45 nM.

 

a. At what rate is product being formed when the concentration of substrate is 10 mM?

b. At the same substrate concentration, how many molecules of product are catalyzed by each molecule of enzyme per second?

Answer 1

v = Vmax[S]/Km+[S]

 

(a)

v = (0.02 mM s^-1)(10 mM)/125 mM + 10 mM

v = 0.00148 mM/sec

 

(b)

 

0.00148 mM/sec = 1.48x10^-6 M/sec (note: M is moles/liter so number of molecules will ultimately depend on volume)

concentration of enzyme = 45 nM = 4.5x10^-8 M

molecules product in 1 liter/sec = 1.48x10^-6 moles x 6.02x10²³ molecules/mole = 8.9x10¹⁷ molecules

molecules of enzyme in 1 liter = 4.5x10^-8 moles x 6.02x10²³ molecules/mole = 2.7x10¹⁶ molecules of enzyme

molecules of product/molecule of enzyme = 8.9x10¹⁷/2.7x10¹⁶ = 33