Question

Advanced Biochemistry or andvanced Physical Chemistry.

A specific enzymatic catalyzed reaction acts on a substrate with a Km of 10 mM and Vmax of 20 μM/s.

How much must [S] increase for V to rise from 5% of Vmax to 10% of Vmax?

How much must [S] increase for V to rise from 90% of Vmax to 95% of Vmax?

From your above calculations, what can you say about the [S] as V approaches Vmax?

Answer 1

v = 5% to 10% of Vmax

   = 0.05 x 0.02 mM to 0.1 x 0.02 mM

   = 0.001 to 0.002 mM/s

Feed values in,

v = Vmax/1+ Km/[S]

when v = 0.001 mM/s

0.001 = 0.02/1 + 10/[S]

0.001 + 0.01/[S] = 0.02

[S] = 0.53 mM

when v = 0.002 mM/s

0.002 = 0.02/1 + 10/[S]

0.002 + 0.02/[S] = 0.02

[S] = 1.11 mM

So [S] increase from 0.53 to 1.11 mM

v = 90% to 95% of Vmax

   = 0.90 x 0.02 mM to 0.95 x 0.02 mM

   = 0.018 to 0.019 mM/s

Feed values in,

v = Vmax/1+ Km/[S]

when v = 0.018 mM/s

0.018 = 0.02/1 + 10/[S]

0.018 + 0.18/[S] = 0.02

[S] = 90 mM

when v = 0.019 mM/s

0.019 = 0.02/1 + 10/[S]

0.019 + 0.19/[S] = 0.02

[S] = 190 mM

So [S] increase from 90 to 190 mM

From the above calculation we can say that, as v approaches Vmax [S] increases exponentially.