Regarding enzyme kinetics

Question

Expressed as a multiple of its KM, how much substrate is required in order for an enzyme to be 50 % saturated? (i.e. what is [S]/KM?). And 98 % saturated?

Answer 1

From the Michaelis Menten equation, you should be able to calculate these values.  

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

Set Vmax = 1 and solve for [S].  The answer for [S] at 50% saturation will be the Km.  Thus,

For 50% saturation, v = 0.5 Vmax = 0.5 and

0.5 = (1)(S)/Km + (S)

[S] = Km

 

For 98% saturation, v = 0.98 Vmax = 0.98

0.98 = (1)(S)/Km + (S)

[S] = 50Km or 50 times the Km value.