Question

In an enzyme-catalyzed reaction, Vmax = 75 nmoles x liters-1 x min-1 and Km = 2.5 x 10-5 M.

a). What would v be at [S] = 2.5 x 10-5 M and at [S] = 5.0 x 10-5 M?

b). What would v be at [S] = 5.0 x 10-5 M if the enzyme concentration were doubled?

Answer 1

Michaelis-Menton equation can be used to solve this problem
V=Vmax[S]/Km+[S]

where V is velocity,Km is michaelis constant related to the enzyme substrate
affinity
{s} is substrate concentation

Vmax is the maximum velocity or rate.

a.vmax=75*10^(-9)mole/l/min [because nanomoles=10(-9)moles]
[s]=2.5*10^(-5)

Km=2.5*10(-5).

V=75*10(-9)*2.5*10(-5)/2.5*10(-5)+2.5*10(-5)
=75*2.5*10(-14)/5*10(-5)
v=37.5*10^(-9)moles/l/min

b.[s]=5*10^[-5],km=2.5*10^[-5}]

then v=75*10^[-9]*5*10^[-5]/[2.5+5]*10^[-5]
=[375/7.5]*10^[-9]
=50*10^[-5]moles/l/min