(FEB 2022-1)
28. The enzyme alkaline phosphatase was tested for its catalytic activity using the substrate para- nitrophenylphosphate. The KM obtained was 10 mM and Vmax was 100umol/min. Which one of the following options represents the initial velocity of the reaction at a substrate concentration of 10 mM?
(1) 50 µmol/min (2) 100 µmol/min
(3) 500 µmol/min (4) 20 µmol/min
The correct answer is (1) 50 µmol/min.
Introduction
The Michaelis-Menten equation serves as the cornerstone for enzyme kinetics, relating reaction velocity to substrate concentration. This article explains how to calculate initial velocity V0 from known Km, Vmax, and substrate concentration, using alkaline phosphatase’s activity toward para-nitrophenylphosphate as an example. This understanding is essential for students and researchers analyzing enzyme behavior in biochemical reactions and laboratory assays.
Given Data
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Km=10 mM
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Vmax=100 μmol/min
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Substrate concentration, [S]=10 mM
Michaelis-Menten Equation
The initial velocity V0 of an enzyme-catalyzed reaction is given by:
V0=Vmax[S]Km+[S]
Calculation
Substitute the given values:
V0=100×1010+10=100020=50 μmol/min
Therefore, the initial velocity at substrate concentration 10 mM is 50 µmol/min.
Explanation
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Since substrate concentration equals Km, the initial velocity is exactly half of Vmax.
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This is consistent with the definition of Km as the substrate concentration at half-maximal velocity.
Practical Implications
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Understanding how to compute V0 helps in designing enzyme assays.
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Enables interpretation of enzyme efficiency and substrate affinity.
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Assists in comparing enzyme activities and inhibitor effects.
Summary Table
| Parameter | Value | Description |
|---|---|---|
| Km | 10 mM | Substrate concentration at half Vmax |
| Vmax | 100 µmol/min | Maximum reaction velocity |
| Substrate concentration [S] | 10 mM | Experimental substrate concentration |
| Calculated V0 | 50 µmol/min | Initial velocity at given [S] |
Conclusion
Applying the Michaelis-Menten equation with the given values shows that at a substrate concentration equal to Km, the enzymatic reaction achieves half of its maximum velocity. Hence, for the alkaline phosphatase acting on para-nitrophenylphosphate, the initial velocity under these conditions is 50 µmol/min.
39 Comments
Varsha Tatla
September 12, 202550umol/min
Aakansha sharma Sharma
September 12, 2025We know that Vo=Vmax S/Km+S
Here Vmax=100umol/ min,S=10mM,Km= 10mM
So Vo= 100×10/10+10=1000/20 = 50umol/min is answer
Mansukh Kapoor
September 13, 2025The correct answer is option 1st
50
Anju
September 14, 2025Ans:vo=100×10/10+10= 1000/20
Vo=50
Khushi Vaishnav
September 13, 202550umol/min
Kanica Sunwalka
September 13, 2025Vo = Vmax S / Km+S
ans is 50umol/min
Sakshi Kanwar
September 13, 2025Vo = Vmax . S / Km + S
= 100. 10 / 10+10
= 1000/20
= 50
Mohd juber Ali
September 14, 2025Option 1 right
Initial valocity= 50
Sakshi yadav
September 14, 2025V=vmax.s/km+s = 100×10/10+10=50umol/min.
Aafreen Khan
September 14, 2025V= Vmax S/ Km+S
= 100×10 / 10+10
= 50umol/min
Soniya Shekhawat
September 14, 2025We know that v =vmax *s/km+s so by put the value then answer is come is 50 micromol/min.
Pooja
September 14, 2025Vo = Vmax ×S / Km + S
= 100× 10 / 10+10
= 1000/20
= 50
Dharmpal Swami
September 14, 2025100×10/10+10=50umol/min.
Bhawna Choudhary
September 14, 202550umol/min is correct answer
Bhawna Choudhary
September 14, 202550umol/min is correct
Kirti Agarwal
September 14, 202550
Anurag Giri
September 14, 2025Vo=Vmax S/Km+S
Here Vmax=100umol/ min,S=10mM,Km= 10mM
So Vo= 100×10/10+10=1000/20 = 50umol/min is answer
Tanvi Panwar
September 14, 2025Vo= Vmax.S/Km+S
=50umol/min is answer
Neha Yadav
September 14, 2025Vo = Vmax .S / Km +S
= 100 .10 / 10+10
Vo = 50 umol/min
Rishita
September 14, 2025Done sir 👍🏻
Anjali
September 14, 2025100×10/10+10=50umol/min.
Vanshika Sharma
September 14, 202550 is correct answer
Avni
September 14, 2025The correct answer is (1) 50 µmol/min.
Ayush Dubey
September 14, 202550 µmol/min
Payal Gaur
September 14, 202550 micromol/min
Pallavi Ghangas
September 14, 202550 micromol per min.
Heena Mahlawat
September 14, 202550
Nilofar Khan
September 15, 2025correct answer is (1) 50 µmol/min.
Vo = Vmax . S/Km+S
Mitali saini
September 15, 2025V= Vmax S/ Km+S
= 100×10 / 10+10
= 50umol/min
Khushi Agarwal
September 15, 2025Correct answer is 1
100× 10/10+10 = 50 micromol/ min
Nd in easy way to learn is Km enzyme/substrate concentration pe depend nahi karta, aur jab ES wapas E+S banne me fast ho (product banane se tez), tab Km ≈ dissociation constant.
Asha Gurzzar
September 15, 202550is correct answer
Simran Saini
September 15, 2025Vo=Vmax S/Km+S
Vmax=100umol/ min,S=10mM,Km= 10mM
Vo= 100×10/10+10=1000/20 = 50umol/min is answer
Arushi Saini
September 15, 2025Vo = Vmax . S / Km + S
= 100. 10 / 10+10
= 1000/20
= 50
Sonal nagar
September 16, 2025Option 1
50 µmol/min.
Minal Sethi
September 16, 2025V = Vmax*S/Km+S
V = 50 microlitre
Muskan Yadav
September 17, 2025The initial velocity at substrate concentration 10 mM is 50 µmol/min is the correct answer .
Deepika sheoran
September 18, 2025Option 1 st
50 umol/min.
Khushi Singh
September 25, 202550 is correct
Juber Khan
April 6, 2026Given Vmax = 100
Km( V0 is half of Vmax)
V0 = 100/2