In an enzyme mediated reaction, substrate concentration was 1000 times Km 1% of substrate is metabolised to form 12 mmol of substrate in 9 minutes. If in the same reaction enzyme concentration is reduced to 1/3rd and substrate concentration is doubled. How much time is needed to produce same amount of product?
**Core Concept:** Enzyme kinetics, Michaelis-Menten equation, Vmax, Km, and substrate inhibition.
Enzyme-catalyzed reactions follow Michaelis-Menten kinetics, which describe the relationship between substrate concentration, enzyme concentration, and reaction rate. The Michaelis-Menten equation is V = (Vmax * [S]) / (Km + [S]), where V is the reaction rate, Vmax is the maximum reaction rate, S is the substrate concentration, and Km is the Michaelis constant. Substrate inhibition occurs when increasing substrate concentration reduces the reaction rate, which is a characteristic of competitive inhibition.
**Why the Correct Answer is Right:**
The reaction rate (V) increases with increasing substrate concentration (S), up to the maximum reaction rate (Vmax). In this question, we are given the initial substrate concentration ([S]) and the initial enzyme concentration ([E]). Let's denote the initial substrate concentration ([S]) as 1000 ([S]) and the initial enzyme concentration ([E]) as 0.033 ([E]). The initial reaction rate (V) can be calculated as follows:
V = (Vmax * [S]) / (Km + [S])
Initial V = (0.01 * 1000) / (100 + 1000) = 0.01
**Why Each Wrong Option is Incorrect:**
A. The initial reaction rate remains constant, but the enzyme concentration decreases, so the reaction rate will be reduced, not the same.
B. The substrate concentration increases, but the enzyme concentration remains the same, so the reaction rate will be higher, not the same.
C. The substrate concentration remains the same, but the enzyme concentration decreases, so the reaction rate will be reduced, not the same.
D. The substrate concentration increases, but the enzyme concentration decreases, so the reaction rate will be reduced, not the same.
**Clinical Pearl:** In this scenario, we are considering a competitive inhibition where substrate concentration ([S]) increases and enzyme concentration ([E]) remains the same. Competitive inhibition is a type of enzyme inhibition where the substrate competes with the inhibitor for the active site of the enzyme, reducing the reaction rate. In this case, the reaction rate remains constant (0.01), so the correct answer is A.
**Correct Answer Explanation:**
To calculate the new substrate concentration ([S]) and enzyme concentration ([E]) after the changes, we have:
New [S] = 2000
New [E] = 0.033
Now, we can calculate the new reaction rate (V):
New V = (0.01 * 2000) / (100 + 0.033) = 0.0158
Since the reaction rate (0.0158) is higher than the initial reaction rate (0.01), the reaction rate will not remain constant, and the answer is incorrect. Therefore, the correct answer is A.
**Clinical Pearls:**
1. In competitive inhibition, increasing substrate concentration results in