9. You are carrying out an enzyme reaction in the lab and obtain Km and Vmax values. While
repeating the experiment, you use only 1/10th of the enzyme. What would you observe?
a. You would get the same result because enzyme is the same and hence properties are
unchanged
b. Vmax would change but Km remains the same
c. Vmax would remain same and Km would be 1/10th of the original value
d. Both Vmax and Km would decrease
Reducing enzyme concentration to 1/10th in an enzyme reaction decreases Vmax to 1/10th of its original value while Km remains unchanged. This follows Michaelis-Menten kinetics where Vmax depends directly on total enzyme amount but Km reflects intrinsic enzyme-substrate affinity independent of enzyme quantity.
The correct answer is b. Vmax would change but Km remains the same.
Option Analysis
a. You would get the same result because enzyme is the same and hence properties are unchanged
Incorrect. While the enzyme’s intrinsic properties like catalytic constant remain identical, Vmax scales linearly with enzyme concentration since fewer enzyme molecules mean lower maximum turnover rate. Reaction velocity plots shift downward proportionally, altering measurable Vmax without affecting Km.
b. Vmax would change but Km remains the same
Correct. Vmax = kcat × [Etotal], so 1/10th enzyme reduces Vmax by 10-fold. Km = (k-1 + kcat)/k1 depends only on rate constants for binding and catalysis, unchanged by enzyme amount. Lineweaver-Burk plots show same x-intercept (-1/Km) but altered y-intercept (1/Vmax).
c. Vmax would remain same and Km would be 1/10th of the original value
Incorrect. Vmax cannot stay constant with less enzyme as saturation rate drops. Km unaffected by [Etotal]; reducing it would imply higher substrate affinity, which contradicts kinetics where affinity derives from microscopic rates.
d. Both Vmax and Km would decrease
Incorrect. Km independent of enzyme concentration; decreasing it alongside Vmax misrepresents affinity changes that actually occur with inhibitors or mutations, not quantity variation.
Michaelis-Menten Fundamentals
Michaelis-Menten equation v = Vmax[S] / (Km + [S]) models hyperbolic velocity-substrate curves. Vmax represents saturation when all active sites bind substrate, directly proportional to enzyme molecules available. Km equals substrate concentration at half Vmax, measuring binding efficiency via ratio of dissociation to association rates.
Experimental Implications
In lab repeats with 1/10th enzyme, velocity measurements across substrate concentrations yield identical Km from half-saturation point but scaled-down Vmax. Normalize by enzyme amount to compare specific activities. CSIR NET questions test this distinction, emphasizing proportional Vmax scaling for accurate kinetic parameter reporting.
| Parameter | Original Enzyme | 1/10th Enzyme | Reason |
|---|---|---|---|
| Vmax | Vmax | Vmax/10 | Fewer active sites |
| Km | Km | Km | Intrinsic affinity |
