Q.19 Which one of the following parameters changes upon doubling the enzyme concentration?
(A) KM (B) V max (C) k cat (D) Keq
Vmax changes when enzyme concentration doubles in enzyme kinetics.
This is a standard Michaelis-Menten kinetics question relevant for CSIR NET Life Sciences preparation.
Correct Answer
(B) Vmax
Option Analysis
KM (A) remains unchanged. It represents substrate concentration at half Vmax and reflects enzyme-substrate affinity, independent of total enzyme amount.
Vmax (B) doubles proportionally with enzyme concentration. More enzyme molecules mean more active sites available for catalysis at saturating substrate levels.
kcat (C) stays constant. This turnover number measures substrate molecules converted per enzyme per second, an intrinsic enzyme property unaffected by concentration.
Keq (D) does not change. Equilibrium constant depends on reaction thermodynamics (ΔG), not catalyst amount or kinetics.
Introduction to Enzyme Kinetics Parameters
In enzyme kinetics, understanding how doubling enzyme concentration affects KM, Vmax, kcat, and Keq is crucial for CSIR NET Life Sciences. The Michaelis-Menten model describes reaction velocity (v) as v=Vmax[S]KM+[S]. Vmax determines maximum rate, while other parameters reflect intrinsic properties.
Why Vmax Changes with Enzyme Concentration
Vmax equals kcat×[E]t, where [E]t is total enzyme concentration. Doubling [E]t doubles available active sites, thus doubling Vmax at saturating substrate. This linear relationship holds under standard conditions where [S]≫[E]t.
Parameters That Remain Unchanged
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KM: Substrate concentration yielding half Vmax; measures affinity. Independent of enzyme amount as it normalizes relative to Vmax.
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kcat: Turnover number (Vmax/[E]t); ratio keeps it constant despite proportional Vmax increase.
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Keq: Thermodynamic equilibrium constant; enzymes accelerate rates but don’t alter ΔG or position.
| Parameter | Effect of Doubling [E] | Reason | CSIR NET Relevance |
|---|---|---|---|
| KM | No change | Affinity constant | Competitive inhibition questions |
| Vmax | Doubles | ∝ [E]_t | Correct answer |
| kcat | No change | Vmax/[E] ratio | Efficiency calculations |
| Keq | No change | Thermodynamic | Equilibrium vs kinetics |
CSIR NET Exam Tips
Practice Lineweaver-Burk plots: 1/v vs 1/[S] shows Vmax as y-intercept, shifting up with higher [E]. Remember: only Vmax scales with enzyme amount; others are intrinsic.
