Question

Question 1

Carbonic anhydrase catalyzes the interconversion of bicarbonate and carbon dioxide in

water. Human carbonic anhydrase has a KM for bicarbonate of 20 mM. An enzyme

preparation was tested in a 300 nM bicarbonate solution, and found to convert 13% of

the bicarbonate to CO2 after 4 minutes. What percentage of the bicarbonate will have

been converted to CO2 after 8 minutes? (Hint: Consider the concentration of

bicarbonate compared to the KM of the enzyme, and therefore what kind of rate law

applies. You do not need the Michaelis-Menton equation to solve this problem.)

Answer 1

since V= Vmax*S/(KM+S)

given KM= 20mM= 20*10-3 M, S= substrate concentration= 300nM= 300*10-9 M

since KM>>S

V= VmaxS/KM

but V= -dS/dt= consumpton of substrate

-dS/dt= VmaxS/KM

-dS/S= (Vmax/KM)*dt

when integrated noting that at t=0, S= So and t= t , S= S

-ln(S/SO)= (Vmax/KM)*t

but conversion X= 1-S/SO

hence -ln(1-X)= (Vmax/KM)t

given X=0.13, t= 4min

-ln(1-0.13)= (Vmax/KM)*4

0.14/4= Vm/KMax = 0.035

for the second case of t=8 min

-ln(1-X)= 0.035*8, X= 1-0.76=0.24

24% conversion takes place

X= 0.