(MODEL PAPER) 65. Which one of the following enzyme reaction represents noncompetitive initiation?

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

  1. Option 1:

    • Shows competitive inhibition, where inhibitor competes for the free enzyme (EI), but ES cannot bind inhibitor—not noncompetitive.

  2. 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.

  3. Option 3:

    • Describes a multisubstrate, multi-phase enzyme activity—irrelevant for classic noncompetitive inhibition.

  4. 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, 2025

    E+S➡️ EScomplex, ES+I➡️ESI (reversible reaction)

  • Kirti Agarwal
    September 12, 2025

    1. Competitive
    2 non compititive
    3 multisubstrate
    4 mixed competitive

    • Mahima Sharma
      September 17, 2025

      2nd

  • Khushi Vaishnav
    September 12, 2025

    E+S – ES – E+P
    ES+I – ESI

  • Kajal
    September 12, 2025

    Option 2nd is correct as in non competitive enzyme inhibition inhibitor have equal affinity for both the enzyme n enzyme substrate complex…

  • yashika
    September 12, 2025

    Non competitive inhibitor

  • Bharti yadav
    September 13, 2025

    E+S ⇄ ES → E+P
    ES+I ⇄ ESI.
    In non competitive inhibition inhibitor have equal affinity for enzyme and enzyme substrate complex.

  • yashika
    September 13, 2025

    Binds to both e and es complex

  • anjani sharma
    September 13, 2025

    In non competitive inhibition, inhibitor have equal Affinity for enzyme and ES complex
    So answer b

  • Kanica Sunwalka
    September 13, 2025

    in non competitive inhibiton ,
    inhibitor have equal affinity for ezyme and ES complex
    option 2 is correct

  • Aakansha sharma Sharma
    September 13, 2025

    The correct answer is (2) E+S ⇄ ES → E+P; ES+I ⇄ ESI.

  • Rishita
    September 14, 2025

    2 nd is right answer

  • Pratibha Jain
    September 14, 2025

    correct answer is option (2)
    E+S ⇄ ES → E+P; ES+I ⇄ ESI.

  • Santosh Saini
    September 14, 2025

    In non competitive inhibition the inhibitor has equal affinity for enzyme and ES complex

  • Dharmpal Swami
    September 14, 2025

    E+S=ES=E+P
    ES+I=ESI

  • Konika Naval
    September 14, 2025

    Option 2

  • Palak Sharma
    September 14, 2025

    E+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, 2025

    E+S ⇄ ES → E+P; ES+I ⇄ ESI.
    Non competative inhibitor have both equal affinity for enzyme and enzyme substrate

  • Ankita Pareek
    September 14, 2025

    Inhibitor of non competitive inhibition has affinity of binding for both enzyme and enzyme substrate so that option 2nd is correct

  • Ajay Sharma
    September 14, 2025

    Second is correct

  • Priya dhakad
    September 14, 2025

    E+S ⇄ ES → E+P; ES+I ⇄ ESI.

  • Sakshi Kanwar
    September 14, 2025

    2nd as inhibitor binds to the ES complex

  • Soniya Shekhawat
    September 15, 2025

    Indicates the inhibitor can bind both free enzyme and ES, this is of noncompetitive inhibition.

  • Aafreen Khan
    September 15, 2025

    In non competitive, inhibitor binds to both free enzyme and ES complex
    E+S–ES –E+P
    ES+I– ESI.

  • Vanshika Sharma
    September 15, 2025

    Reaction 2 is correct

  • Bhawna Choudhary
    September 15, 2025

    The correct answer is (2) E+S ⇄ ES → E+P; ES+I ⇄ ESI.

  • Mohd juber Ali
    September 15, 2025

    In 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, 2025

    In non competitive inhibition inhibitor has same Affinity for enzyme and ES complex so 2nd option is correct .

  • Nilofar Khan
    September 16, 2025

    correct 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, 2025

    Option 2nd

  • Tanvi Panwar
    September 16, 2025

    2nd reaction indicates non competitive inhibition.

  • Khushi Agarwal
    September 17, 2025

    The correct answer is (2) E+S ⇄ ES → E+P; ES+I ⇄ ESI

  • Priya khandal
    September 17, 2025

    2 is right sir

  • Avni
    September 17, 2025

    (2) E+S ⇄ ES → E+P; ES+I ⇄ ESI

  • Muskan Yadav
    September 19, 2025

    E+S ⇄ ES → E+P; ES+I ⇄ ESI option 2nd reaction indicates non competitive inhibition.

  • Kajal
    September 25, 2025

    The correct answer is (2) E+S ⇄ ES → E+P; ES+I ⇄ ESI

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