(DEC 2006)
20. The steady state hypothesis for enzyme suggest that
(1) Rate of formation of ES complex is equal to rate of breakdown of ES complex
(2) Rate of formation of ES complex is equal to rate of formation of products
(3) Rate of formation of ES complex and its dissociation into E and S are equal
(4) Enzyme are steadily consumed in the reaction
The correct answer is (1) Rate of formation of ES complex is equal to rate of breakdown of ES complex.
Introduction
The steady-state hypothesis is a fundamental concept in enzyme kinetics, providing a basis for the quantitative description of enzymatic reaction rates. It assumes that during the initial phase of an enzyme-catalyzed reaction, the concentration of the enzyme-substrate (ES) complex remains relatively constant. This allows simplification of complex kinetic equations and aids in deriving the Michaelis-Menten equation. This article explains the steady-state hypothesis and why it assumes equal rates of formation and breakdown of the ES complex.
Background: Enzyme and Substrate Interaction
In an enzymatic reaction:
E+S⇌k−1k1ES→k2E+P
-
E: free enzyme
-
S: substrate
-
ES: enzyme-substrate complex
-
P: product
The ES complex forms as a transient intermediate.
Definition of Steady-State Hypothesis
The steady-state hypothesis states that:
d[ES]dt=0
During the initial reaction phase, the concentration of the ES complex does not change significantly over time. This means:
Rate of formation of ES=Rate of breakdown of ES
Why Is This Assumed?
-
The formation of ES complex is rapid.
-
Breakdown of ES to product and free enzyme is relatively slower.
-
After a brief transient phase, the formation and breakdown rates balance, making ES concentration steady.
-
This assumption helps simplify kinetic analyses, making quantitative description of reaction velocity feasible.
Mathematical Expression
-
Rate of ES formation:
vformation=k1[E][S]
-
Rate of ES breakdown (both to substrate and product):
vbreakdown=k−1[ES]+k2[ES]=(k−1+k2)[ES]
Setting these equal at steady state:
k1[E][S]=(k−1+k2)[ES]
From this, the concentration of ES can be solved in terms of measurable quantities.
Incorrect Options
-
(2) Rate of formation of ES complex equals rate of product formation: Incorrect because product formation only accounts for one ES breakdown route, ignoring dissociation back to enzyme and substrate.
-
(3) Rate of formation of ES complex equals rate of its dissociation into E and S: Incomplete, overlooks product formation pathway; steady-state includes both breakdown routes.
-
(4) Enzymes are steadily consumed: False; enzymes are catalysts and are regenerated, not consumed.
Significance in Enzyme Kinetics
The steady-state hypothesis forms the base for the Michaelis-Menten kinetic model, enabling calculation of:
-
Initial reaction velocity.
-
Michaelis constant (Km), indicating substrate affinity.
-
Maximum reaction velocity (Vmax).
This model guides experimental design, pharmaceutical research, and understanding metabolic regulation.
Summary Table
| Option | Correct/Incorrect | Explanation |
|---|---|---|
| (1) | Correct | Rates of ES formation and breakdown are equal at steady state |
| (2) | Incorrect | Ignores reversible substrate dissociation |
| (3) | Incorrect | Omits ES breakdown to products |
| (4) | Incorrect | Enzymes are catalysts, not consumed |
Conclusion
The steady-state hypothesis assumes that during an enzyme-catalyzed reaction, the concentration of the enzyme-substrate complex remains constant because the rate of formation equals the rate of breakdown. This simplification is pivotal for understanding and modeling enzyme kinetics effectively.
This knowledge is essential for students, researchers, and professionals involved in enzymology, molecular biology, and biomedical sciences aiming for a thorough comprehension of biochemical reaction mechanisms.
46 Comments
Sakshi yadav
September 12, 2025Rate of formation of ES complex is = to rate of breakdown of ES complex.
Mohd juber Ali
September 12, 2025Optioan a is right
Mansukh Kapoor
September 12, 2025The correct answer is option 1st
Rate of formation of ES complex is equal to rate of breakdown of ES complex
Aakansha sharma Sharma
September 12, 2025Rate of formation of ES complex = rate of breakdown of ES complex.
Priya dhakad
September 12, 2025Rate of formation of Es complex is equal to rate of breakdown of ES complex.
Varsha Tatla
September 12, 2025Steady state at where formation of ES complex=breakdown of ES complex
Tanvi Panwar
September 12, 2025Rate of formation of ES is equal to rate of break down of ES into E and S.
Dharmpal Swami
September 13, 2025Rate of formation of ES complex is equal to break down in to E And S
Sakshi Kanwar
September 13, 2025steady-state includes both breakdown routes that is ES complex remains constant over time because its rate of formation equals to its rate of breakdown.
Kanica Sunwalka
September 13, 2025steady state means rate of formation of ES complex = rate of breakdown of ES complex (E+S)
option A is correct
Kirti Agarwal
September 13, 2025In steady state the rate of formation of ES complex is equal to the rate of breakdown of ES complex
Neha Yadav
September 13, 2025In steady state rate of formation of ES complex is equal to the rate of breakdown of ES complex
Bhawna Choudhary
September 13, 2025The rate of formation of ES complex is equal to the breakdown of ES complex
Santosh Saini
September 13, 2025Rate of formation of ES complex is equal to rate of breakdown of ES complex
Nilofar Khan
September 14, 2025correct answer is (1)
Rate of formation of ES complex is equal to rate of breakdown of ES complex.
Soniya Shekhawat
September 14, 2025Steady State hypothesis is rate of ES formation is equal to ES breakdown.
Aafreen Khan
September 14, 2025The correct answer is 1
Rate of formation of ES complex is equal to rate of breakdown of ES complex
Khushi Agarwal
September 14, 2025The correct answer is A
Rate of formation of ES complex is equal to rate of breakdown of ES complex
Pooja
September 14, 2025Option A is correct
Rate of formation of ES complex is equal to rate of breakdown of ES complex.
Ayush Dubey
September 14, 2025Rate of formation of ES complex is equal to rate of breakdown of ES complex.
Vanshika Sharma
September 14, 2025Rate of formation of ES complex is equal to rate of breakdown of ES complex
Payal Gaur
September 14, 2025Option A Rate of formation of ES complex is equal to the rate of break down of ES complex
Kajal
September 14, 2025Steady state means rate of formation of ES complex is equal to rate of dissociation of this complex
Anurag Giri
September 14, 2025The correct answer is (1) Rate of formation of ES complex is equal to rate of breakdown of ES complex.
Mitali saini
September 14, 2025The correct answer is (1) Rate of formation of ES complex is equal to rate of breakdown of ES complex.
Rishita
September 14, 2025Rate of formation of ES complex is equal to rate of breakdown of ES complex.
Avni
September 14, 2025The correct answer is (1) Rate of formation of ES complex is equal to rate of breakdown of ES complex.
Anju
September 14, 2025Ans:1
Rate of formation of ES = rate of breakdown of ES COMPLEX
Asha Gurzzar
September 14, 2025A is correct answer
Arushi Saini
September 14, 2025In steady state the rate of formation of ES complex is equal to the rate of breakdown of ES complex
HIMANI FAUJDAR
September 14, 2025Ans 1 ,Rate of formation of ES complex = rate of breakdown of ES complex.
Heena Mahlawat
September 14, 2025Rate of formation of ES complex is equal to rate of breakdown of ES complex
Aartii sharma
September 14, 2025Rate of formation of ES complex is equal to rate of breakdown of ES complex
Pallavi Ghangas
September 14, 2025Rate of formation of ES complex is = to rate of breakdown of ES complex.
anjani sharma
September 14, 2025Answer 1
Rate of formation of ES complex is equal to rate of breakdown of ES complex.
Deepika sheoran
September 15, 2025Option A Rate of formation of ES complex is Equal to Rate of breakdown of ES complex.
Priyanshi sharma
September 15, 2025The Rate of formation of ES complex is equal to rate of breakdown of ES complex
yashika
September 15, 2025Rate of fornation of es complex is equal to rate of breakdown of es complex
Simran Saini
September 15, 2025Rate of formation of ES complex is equal to rate of breakdown of ES complex.
Khushi Vaishnav
September 15, 2025Rate of formation of ES complex is equal to rate of breakdown of ES complex.
Muskan singodiya
September 15, 2025Option 1 is correct because rate of formation of Es complex is equal to rate of bareckdown of es complex
Anjana sharma
September 16, 2025Rate of formation of ES complex is equal to rate of breakdown of ES complex.
Sonal nagar
September 16, 2025Option 1.
Rates of ES formation and breakdown are equal at steady state.
Minal Sethi
September 16, 2025rate of formation of ES complex is equal to rate of breakdown of ES complex
Muskan Yadav
September 17, 2025Rate of formation of ES complex is equal to rate of breakdown of ES complex .
Yogita
September 17, 2025Rate of formation of ES complex= rate of breakdown of ES complex