Regarding enzyme kinetics

Question

The KI of a competitive inhibitor is 600 nM. If the KM of the un-inhibited enzyme is 80 nM,

 

a. What is the KM when 400 nM of inhibitor is present?

b. What concentration of inhibitor is needed to increase the KM of the reaction to twice its original value? And to 10 times its original value?

Answer 1

Making certain assumptions about the conditions used to conduct the analysis, one can use 

K_m' = K_m(1 + [I]/K_i] where K_m' is the apparent K_m and [I] the concentration of inhibitor.

(a)  K_m' = (80 nM)(1 + 400nM/600 nM) = (80)(1.67) = 133 nM 

 

For apparent K_m (K_m') to be 2x the original value, it would be 2x80 = 160 nM.  Use this value and solve for [I]. The answer should be the same as the K_i (600 nM) since this is the definition (essentially) of K_i (or IC₅₀)

(b)  160 nM = (80 nM)(1 + [I]/600)

160 =80 + 80[I]/600

80 = 80[I]/600

[I] = 600 nM (same as K_i) ... no big surprise

 

To calculate [I] to increase K_m to 10x original, do same calculation using 800 nM as K_m' and solve for [I]. I think you'll get around 5400 nM = 5.4 uM