# DBT JRF 2022 — Question 104

Exam: DBT JRF 2022 · Section: B · Topic: Biochemistry › Enzyme Kinetics - Substrate Concentration · Difficulty: Medium

## Question

In an enzyme catalyzed reaction following Michaelis-Menten kinetics, both the enzyme amount and substrate (S) are doubled and we observe that the reaction rate is also doubled. From this, we can conclude that:

## Options

- **A.** Enzyme amount has no effect on enzyme kinetics
- **B.** The substrate is toxic to enzyme
- **C.** Initial substrate (S) concentration << $K_m$
- **D.** Initial substrate (S) concentration >> $K_m$

## Answer

Option D

## Explanation

Zero-order kinetics: doubling enzyme doubles rate: The Michaelis-Menten equation: $v = \frac{V_{max}[S]}{K_M + [S]}$ If $[S] \gg K_M$ , then $K_M + [S] \approx [S]$ : $v \approx \frac{V_{max}[S]}{[S]} = V_{max}$ This is ZERO-ORDER kinetics in [S]: the rate is independent of substrate concentration, depending only on $V_{max} = k_{cat}[E]_T$ . Consequences: • Doubling [S] at this regime has NO effect on rate (rate already maximal at $V_{max}$ ). • Doubling enzyme concentration [E]ᵀ doubles $V_{max}$ → doubles the rate. The observation that 'doubling S had no effect, but doubling enzyme doubled the rate' confirms we are at zero-order in S (i.e., $S \gg K_M$ ). Answer: $S \gg K_M$ (zero-order kinetics in S); rate is proportional to enzyme concentration, not substrate.

Source: https://mypyqbuddy.com/questions/in-an-enzyme-catalyzed-reaction-following-michaelis-menten-kinetics-6a104cd5920734408bc0c777