Question

Is condition 2 competitive or noncompetitive inhibition? Condition 2: y = 10 x + 10 Condition 1: y = 5 x + 5

Answer 1

Ans is Non competitive

So we use condition 1as untreated and condition 2 as inhibitor traeted.So to know whether it is competitive or non competitive we hve to find vmax and km of both the condition by using formula given below as it follows straight line equation in the condition shown above.

1/V=(km/Vmax)1/S+1/Vmax refer to Y=mx+c

So if use the condition 1 here than 1/Vmax is equal to 5 because it is intercept of graph and hence Vmax= 0.2

Use this value for km/Vmax=5 and keep vmax value 0.2 so we will get Km =1.0

Same do for condition 2 here than 1/Vmax is equal to 10 so Vmax=0.1

use this value ofor Km/Vmax=10 and hence Km will be 1.0

This results shows that addition of inhibitor leads to change in Vmax but unchanged Km so this clearly relate to Non competitive inhibitor because  in the presence of the inhibitor because some of the enzyme molecules will always be “out of commission.Where as in case of Competitive Vmax is the maximum velocity of the enzyme. Competitive inhibitors can only bind to E and not to ES. They increase Km by interfering with the binding of the substrate, but they do not affect Vmax because the inhibitor does not change the catalysis in ES because it cannot bind to ES.