The Question is from neet-botany biomolecules-biology enzymes, and it was created by the author D VINOD

Question

A novel enzyme is discovered that catalyzes the hydrolysis of a specific peptide bond. Kinetic studies reveal a $K_m$ of 10 μM and a $V_{max}$ of 50 μmol/min.  A competitive inhibitor with a $K_i$ of 2 μM is added at a concentration of 5 μM. What will be the apparent $K_m$ and $V_{max}$ in the presence of the inhibitor?

D VINOD · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
	**Primary Concept**: Enzyme Kinetics - Competitive Inhibition
	**Related Topics**: Michaelis-Menten equation, Lineweaver-Burk plot
	**Prerequisites**: Understanding of $K_m$, $V_{max}$, and $K_i$.
## SOLUTION PATHWAY
**Given Information**:
	$K_m$ = 10 μM
	$V_{max}$ = 50 μmol/min
	$K_i$ = 2 μM
	[I] = 5 μM
**Method & Steps**:
	1. Calculate the apparent $K_m$ ($K_m'$) in the presence of the competitive inhibitor.
		Reasoning: Competitive inhibitors increase the apparent $K_m$ without affecting $V_{max}$.
		Formula/Rule: $K_m' = K_m (1 + \frac{[I]}{K_i})$
	2. Substitute the given values into the formula.
		Working: $K_m' = 10 \mu M (1 + \frac{5 \mu M}{2 \mu M}) = 10 \mu M (1 + 2.5) = 10 \mu M * 3.5 = 35 \mu M$
		Key Point:  $V_{max}$ remains unchanged in competitive inhibition.
	3. Determine the apparent $V_{max}$ ($V_{max}'$).
		Result: $V_{max}' = V_{max} = 50$ μmol/min
		Verification: The calculated $K_m'$ and $V_{max}'$ match option A.
## OPTION ANALYSIS
[A] ✓: $K_m' = 35$ μM, $V_{max}' = 50$ μmol/min. This is the correct answer, as calculated above. Competitive inhibitors increase the apparent $K_m$ while $V_{max}$ remains the same.
[B] :x: $K_m' = 10$ μM, $V_{max}' = 25$ μmol/min. This is incorrect. While a decrease in $V_{max}$ might be observed with other types of inhibition (e.g., non-competitive), competitive inhibitors do not affect $V_{max}$.
[C] :x: $K_m' = 25$ μM, $V_{max}' = 50$ μmol/min. This is incorrect. The calculation for $K_m'$ is flawed.
[D] :x: $K_m' = 35$ μM, $V_{max}' = 25$ μmol/min. This is incorrect. While the $K_m'$ is correctly calculated, the $V_{max}$ should not change with competitive inhibition.
## LEARNING AIDS
	**Common Mistakes**:
		1. Confusing the effects of different types of enzyme inhibition on $K_m$ and $V_{max}$.
			How to avoid: Carefully review the characteristics of each type of inhibition.
		2. Incorrectly applying the formula for $K_m'$ calculation.
			How to avoid: Double-check the formula and ensure correct substitution of values.
	**Quick Tips**:
		1. Remember that competitive inhibitors 'compete' with the substrate, effectively increasing the apparent $K_m$.
		2. $V_{max}$ can always be reached at a sufficiently high substrate concentration, even in the presence of a competitive inhibitor.

D VINOD · Answer

The correct Option is "Km' = 35 μM, Vmax' = 50 μmol/min"

D VINOD · Answer

Km' = 10 μM, Vmax' = 25 μmol/min

D VINOD · Answer

Km' = 25 μM, Vmax' = 50 μmol/min

D VINOD · Answer

Km' = 35 μM, Vmax' = 25 μmol/min

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