Question

An enzyme-catalyzed reaction to the presence of 5 nM of reversible inhibitor yields a V_max value that is 80% of the value in absence of the inhibitor. The KMvalue is unchanged.

a) what type of inhibition is occurring?

b) what proportion of the enzyme molecule will have bound inhibitor?

c) Draw the Lineweaver-Burk (known as double-reciprocal plot) for uninhibited and inhibited reaction.

SHOW ALL YOUR WORK PLEASE

Answer 1

Ans. #A. Given- Vmax is decreased but Km remains unaffected in presence of a reversible inhibitor.

A decrease in Vmax but Km remaining unaffected is the characteristic of a non-competitive (Mixed) inhibitor. So, the given inhibitor is a non-competitive inhibitor.

#B. For a specified set of reaction parameters (with [E]_total and Km as the two variables), the reaction velocity depends on the enzyme concertation and Km value. Note that Km remains unaffected.

For a set of reaction parameters with Km remaining constant, the rate of reaction is directly proportional to the enzyme (active or uninhibited enzyme molecules) – i.e. greater is the [E]_active, greater would be reaction velocity.

# Given than V_apparent = 80% of Vmax = 0.8 Vmax. That is, the apparent reaction velocity (V,app) is 80% of the Vmax (uninhibited) – indicating 80% of the total enzyme molecules are in active state of catalysis; the rest being inhibited by the inhibitor.

So, the proportion of inhibitor-bound enzyme molecules = 100% - 80% = 20%    

#C.

Non-Compatitive (Mixed) Inhibitor No I Vmax Km Increases A common X-axis intercept indicates the same value of Km for both the uninhibited and inhibited kinetics