1. The following rate equation represents _____________ inhibition.
\(V=\frac{V_{max} [S]}{K_m (1+\frac{I}{K_i})+[S]}\)
a) competitive
b) mixed
c) non-competitive
d) uncompetitive
Explanation: Competitive inhibition occurs when both substrate and inhibitor compete for binding to the active site of the enzyme. In this kind of inhibition, Ki is much greater than total inhibitor concentration and ESI complex is not formed. This kind of inhibition is mostly seen when the substrate concentration are low. This inhibition can be overcome at sufficiently high substrate concentrations as Vmax remain unaffected. Hence the rate equation is represented by \(V=\frac{V_{max} [S]}{K_m (1+\frac{I}{K_i})+[S]}\) Where,
Km = Miachelis Menton constant, Ki = inhibitor constant, I = Inhibitor concentration and S = Substrate concentration.
2. Uncompetitive inhibition is most noticeable at low substrate concentration and can be overcome at high substrate concentration.
a) True
b) False
Explanation: Uncompetitive inhibitions occurs during multi-substrate reactions where the inhibitor is competitive to one substrate and uncompetitive to the other. These inhibition occur when one substrate is already bound to the enzyme. Ki is much greater than the total inhibitor concentration and EI complexes are not formed. Hence the above statement is false. This is true in case of competitive inhibition.
3. Inhibition of invertase by sucrose falls into which category of inhibition?
a) Substrate inhibition
b) Non-competitive inhibition
c) Product inhibition
d) Competitive inhibition
Explanation: Invertase inhibition by sucrose is kind of substrate inhibition. It is a special case of uncompetitive inhibition which occurs at high substrate concentrations. Product inhibition is case of competitive inhibition wherein the substrate and the inhibitor have structural similarity. Competitive inhibition is one wherein the substrate and the inhibitor compete for the active site of the enzyme. Non-competitive inhibition occurs when the inhibitor binds at the site away from the binding site, causing a reduction in catalytic activity.
4. ______________ inhibition represent the following rate equation.
\(V=\frac{\left (\frac{V_{max}}{1+\frac{[I]}{K_i}}\right )[S]}{K_m+[S]}\)
a) Substrate inhibition
b) Competitive inhibition
c) Product inhibition
d) Non-competitive inhibition
Explanation: Non-competitive inhibition is a special case of mixed inhibition. This kind of inhibition occurs when the inhibitor binds to the enzyme, far away from the catalytic site. In this case both EI and ESI complexes are equally formed. Hence the rate equation is given by the equation
\(V=\frac{\left (\frac{V_{max}}{1+\frac{[I]}{K_i}}\right )[S]}{K_m+[S]}\)
Where,
Km = Miachelis Menton constant, Ki = inhibitor constant, I = Inhibitor concentration and S = Substrate concentration.
5. The __________ inhibition gives the following rate equation.
\(V=\frac{V_{max} [S]}{K_m+[S](1+\frac{[S]}{K_s})}\)
a) Substrate
b) Mixed
c) Non-competitive
d) Competitive
Explanation: Substrate inhibition is a special case of uncompetitive inhibition which occurs in about 20% of all known enzymes. It is primarily caused by binding of parts of the substrate molecule to the subsites in the active site. At high substrate concentration, the resultant complex may become inactive causing reduction in the rate of reaction. Assuming the ESS complex may not form product, the rate equation may be given by \(V=\frac{V_{max} [S]}{K_m+[S](1+\frac{[S]}{K_s})}\)
Where,
Km = Miachelis Menton constant, Ks = Dissociation constant for substrate and S = Substrate concentration.
6. In competitive inhibition, what happens to Vmax and Km if [I] = Ki?
a) Lowers to 0.5 Vmax and 0.5 Km
b) Vmax is unchanged and Km increases 2Km
c) Lowers to 0.5 Vmax and Km remains unchanged
d) Lowers to 0.67 Vmax and Km increases to 2Km
Explanation: Competitive inhibition is one wherein the inhibitor and the substrate compete for the active site. Inhibitor and substrate are said to be structurally similar. Thus, the rate equation for competitive inhibition is given by \(V=\frac{V_{max} [S]}{K_m (1+\frac{I}{K_i})+[S]}\). According to this equation, Vmax remains unchanged and Km increases 2Km.
7. Which of these is correct for mixed inhibition?
a) Lowers to 0.5 Vmax and 0.5 Km
b) Vmax is unchanged and Km increases 2Km
c) Lowers to 0.5 Vmax and Km remains unchanged
d) Lowers to 0.67 Vmax and Km increases to 2Km
Explanation: Mixed inhibition is said to occur when both EI and ESI complexes are formed. The rate equation is this case is given by \(V=\frac{V_{max} [S]}{K_m (1+\frac{[I]}{K_i})+[S](1+\frac{[I]}{K_i’})}\)
Where
Km = Miachelis Menton constant, Ki = Dissociation constant for EI complex, [I] = Inhibitor concentration, S = Substrate concentration and Ki‘ = Dissociation constant for ESI complex
Now, if Ki = [I] = 0.5 Ki‘, then Vmax lowers to 0.67 and Km increases to 2Km.
8. What happens to Vmax and Km in case of uncompetitive inhibition when I = Ki‘?
a) Lowers to 0.5 Vmax and 0.5 Km
b) Vmax is unchanged and Km increases 2Km
c) Lowers to 0.5 Vmax and Km remains unchanged
d) Lowers to 0.67 Vmax and Km increases to 2Km
Explanation: Uncompetitive inhibition occurs during multi-substrate reaction where the inhibitor is competitive to one substrate and uncompetitive to the other. The inhibition cannot be overcome as Vmax and Km are equally reduced. The rate equation is given by, \(V=\frac{\left (\frac{V_{max}}{1+\frac{[I]}{K_i’}}\right ) [S]}{\left (\frac{K_m}{1+\frac{[I]}{K_i’}}\right )+[S]}\).
When I = Ki‘ in this equation, the Vmax and Km is reduced to half.
9. If I = Ki = Ki‘, then what will happen to Vmax and Km when inhibitor acts non-competitively?
a) Lowers to 0.5 Vmax and 0.5 Km
b) Vmax is unchanged and Km increases 2Km
c) Lowers to 0.5 Vmax and Km remains unchanged
d) Lowers to 0.67 Vmax and Km increases to 2Km
Explanation: Non-competitive inhibitors are those inhibitors which bind at site away from the binding site, causing reduction in catalytic activity. It is a rare case mixed inhibition and the rate equation is given by, \(V=\frac{\left (\frac{V_{max}}{1+\frac{[I]}{K_i}}\right ) [S]}{K_m+[S]}\). When If I = Ki = Ki‘, Vmax is lowered to 0.5 Vmax and Km remains unchanged.
10. Mercury causes irreversible inhibition.
a) True
b) False
Explanation: Irreversible inhibition is one in which loss of activity cannot be restored by the removal of inhibitor. Irreversible inhibition behaves as time-dependent loss of enzyme concentration, when the enzyme is totally inactive. In cases involving incomplete activation, there may be time dependent changes in both Km and Vmax. Heavy metal ions such as mercury, lead, etc., cause irreversible inhibition because they strongly bind to the amino acid backbone. Hence they should not come in contact with enzymes.