The correct answer is (2) E+S ⇄ ES → E+P; ES+I ⇄ ESI.
Introduction
Noncompetitive inhibition is a fundamental model in enzyme kinetics, describing how inhibitors can reduce enzyme activity regardless of substrate concentration. Unlike competitive inhibitors, which compete for the active site, noncompetitive inhibitors bind elsewhere—often to both the free enzyme and the enzyme-substrate complex—resulting in unique effects on kinetic parameters. Recognizing the reaction mechanism for noncompetitive inhibition is crucial for students, researchers, and professionals interpreting enzyme assays or designing pharmaceuticals.
Mechanism of Noncompetitive Inhibition
-
Noncompetitive inhibitors bind to the enzyme or enzyme-substrate complex at a site distinct from the active site, creating both EI and ESI complexes.
-
Inhibition does not change substrate binding (Km), but decreases the overall reaction rate (Vmax).
-
Substrate and inhibitor can bind independently; inhibition cannot be overcome by increasing substrate concentration.
Analyzing the Schemes
-
Option 1:
-
Shows competitive inhibition, where inhibitor competes for the free enzyme (EI), but ES cannot bind inhibitor—not noncompetitive.
-
-
Option 2:
-
Shows normal catalysis (E + S ⇄ ES → E + P) and inhibition via the ES complex (ES + I ⇄ ESI).
-
Indicates the inhibitor can bind both free enzyme and ES, a hallmark of noncompetitive inhibition.
-
-
Option 3:
-
Describes a multisubstrate, multi-phase enzyme activity—irrelevant for classic noncompetitive inhibition.
-
-
Option 4:
-
Shows both competitive and noncompetitive binding modes but mixes multiple mechanisms, making it atypical and overly complex for pure noncompetitive inhibition.
-
Why Option 2 is Correct
-
The key feature: inhibitor binding to ES forms ESI, reducing product formation.
-
Noncompetitive inhibition does not affect substrate binding (Km), only product formation (Vmax), and can occur even when substrate is bound.
-
The scheme in Option 2 matches textbook noncompetitive inhibition reactions.
Summary Table
Mechanism | Binding Sites | Effect on Kinetics |
---|---|---|
Competitive | Active site (E) | Increases Km, no change Vmax |
Noncompetitive (Option 2) | E and ES | No change Km, lowers Vmax |
Uncompetitive | ES only | Lowers Km and Vmax |
Mixed | Both, but complex | Both Km and Vmax can change |
Biological and Clinical Importance
-
Noncompetitive inhibition is exploited in drug design for specific regulation regardless of substrate levels.
-
Examples include certain metabolic enzyme inhibitors and allosteric regulators.
Conclusion
Option 2 accurately represents noncompetitive enzyme inhibition, where the inhibitor can bind to the ES complex to inactivate the enzyme. This results in decreased overall catalytic rate, unaffected by substrate concentration, a distinct and diagnostically important kinetic signature.
36 Comments
Santosh Saini
September 12, 2025E+S➡️ EScomplex, ES+I➡️ESI (reversible reaction)
Kirti Agarwal
September 12, 20251. Competitive
2 non compititive
3 multisubstrate
4 mixed competitive
Mahima Sharma
September 17, 20252nd
Khushi Vaishnav
September 12, 2025E+S – ES – E+P
ES+I – ESI
Kajal
September 12, 2025Option 2nd is correct as in non competitive enzyme inhibition inhibitor have equal affinity for both the enzyme n enzyme substrate complex…
yashika
September 12, 2025Non competitive inhibitor
Bharti yadav
September 13, 2025E+S ⇄ ES → E+P
ES+I ⇄ ESI.
In non competitive inhibition inhibitor have equal affinity for enzyme and enzyme substrate complex.
yashika
September 13, 2025Binds to both e and es complex
anjani sharma
September 13, 2025In non competitive inhibition, inhibitor have equal Affinity for enzyme and ES complex
So answer b
Kanica Sunwalka
September 13, 2025in non competitive inhibiton ,
inhibitor have equal affinity for ezyme and ES complex
option 2 is correct
Aakansha sharma Sharma
September 13, 2025The correct answer is (2) E+S ⇄ ES → E+P; ES+I ⇄ ESI.
Rishita
September 14, 20252 nd is right answer
Pratibha Jain
September 14, 2025correct answer is option (2)
E+S ⇄ ES → E+P; ES+I ⇄ ESI.
Santosh Saini
September 14, 2025In non competitive inhibition the inhibitor has equal affinity for enzyme and ES complex
Dharmpal Swami
September 14, 2025E+S=ES=E+P
ES+I=ESI
Konika Naval
September 14, 2025Option 2
Palak Sharma
September 14, 2025E+S ⇄ ES → E+P
ES+I ⇄ ESI.
In non competitive inhibition inhibitor have equal affinity for enzyme and enzyme substrate complex.
Sakshi yadav
September 14, 2025E+S ⇄ ES → E+P; ES+I ⇄ ESI.
Non competative inhibitor have both equal affinity for enzyme and enzyme substrate
Ankita Pareek
September 14, 2025Inhibitor of non competitive inhibition has affinity of binding for both enzyme and enzyme substrate so that option 2nd is correct
Ajay Sharma
September 14, 2025Second is correct
Priya dhakad
September 14, 2025E+S ⇄ ES → E+P; ES+I ⇄ ESI.
Sakshi Kanwar
September 14, 20252nd as inhibitor binds to the ES complex
Soniya Shekhawat
September 15, 2025Indicates the inhibitor can bind both free enzyme and ES, this is of noncompetitive inhibition.
Aafreen Khan
September 15, 2025In non competitive, inhibitor binds to both free enzyme and ES complex
E+S–ES –E+P
ES+I– ESI.
Vanshika Sharma
September 15, 2025Reaction 2 is correct
Bhawna Choudhary
September 15, 2025The correct answer is (2) E+S ⇄ ES → E+P; ES+I ⇄ ESI.
Mohd juber Ali
September 15, 2025In non conpetative inhibition
The affinity of E equal to ES
Inhibitor bind with active site of enzyme and also bund with ES complex and form ESI option 2 right
Divya rani
September 16, 2025In non competitive inhibition inhibitor has same Affinity for enzyme and ES complex so 2nd option is correct .
Nilofar Khan
September 16, 2025correct answer is (2) E+S ⇄ ES → E+P; ES+I ⇄ ESI.
In noncompetitive inhibition, inhibitor binds with both enzyme and enzyme sub. Complex. Same affinity for both
Lokesh Kumawat
September 16, 2025Option 2nd
Tanvi Panwar
September 16, 20252nd reaction indicates non competitive inhibition.
Khushi Agarwal
September 17, 2025The correct answer is (2) E+S ⇄ ES → E+P; ES+I ⇄ ESI
Priya khandal
September 17, 20252 is right sir
Avni
September 17, 2025(2) E+S ⇄ ES → E+P; ES+I ⇄ ESI
Muskan Yadav
September 19, 2025E+S ⇄ ES → E+P; ES+I ⇄ ESI option 2nd reaction indicates non competitive inhibition.
Kajal
September 25, 2025The correct answer is (2) E+S ⇄ ES → E+P; ES+I ⇄ ESI