(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

(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 KmVmax, 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

  • Km=10 mM

  • Vmax=100 μmol/min

  • 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

  • Since substrate concentration equals Km, the initial velocity is exactly half of Vmax.

  • This is consistent with the definition of Km as the substrate concentration at half-maximal velocity.


Practical Implications

  • Understanding how to compute V0 helps in designing enzyme assays.

  • Enables interpretation of enzyme efficiency and substrate affinity.

  • 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, 2025

    50umol/min

  • Aakansha sharma Sharma
    September 12, 2025

    We 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, 2025

    The correct answer is option 1st
    50

    • Anju
      September 14, 2025

      Ans:vo=100×10/10+10= 1000/20
      Vo=50

  • Khushi Vaishnav
    September 13, 2025

    50umol/min

  • Kanica Sunwalka
    September 13, 2025

    Vo = Vmax S / Km+S
    ans is 50umol/min

  • Sakshi Kanwar
    September 13, 2025

    Vo = Vmax . S / Km + S
    = 100. 10 / 10+10
    = 1000/20
    = 50

  • Mohd juber Ali
    September 14, 2025

    Option 1 right
    Initial valocity= 50

  • Sakshi yadav
    September 14, 2025

    V=vmax.s/km+s = 100×10/10+10=50umol/min.

  • Aafreen Khan
    September 14, 2025

    V= Vmax S/ Km+S
    = 100×10 / 10+10
    = 50umol/min

  • Soniya Shekhawat
    September 14, 2025

    We know that v =vmax *s/km+s so by put the value then answer is come is 50 micromol/min.

  • Pooja
    September 14, 2025

    Vo = Vmax ×S / Km + S
    = 100× 10 / 10+10
    = 1000/20
    = 50

  • Dharmpal Swami
    September 14, 2025

    100×10/10+10=50umol/min.

  • Bhawna Choudhary
    September 14, 2025

    50umol/min is correct answer

  • Bhawna Choudhary
    September 14, 2025

    50umol/min is correct

  • Kirti Agarwal
    September 14, 2025

    50

  • Anurag Giri
    September 14, 2025

    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

  • Tanvi Panwar
    September 14, 2025

    Vo= Vmax.S/Km+S
    =50umol/min is answer

  • Neha Yadav
    September 14, 2025

    Vo = Vmax .S / Km +S
    = 100 .10 / 10+10
    Vo = 50 umol/min

  • Rishita
    September 14, 2025

    Done sir 👍🏻

  • Anjali
    September 14, 2025

    100×10/10+10=50umol/min.

  • Vanshika Sharma
    September 14, 2025

    50 is correct answer

  • Avni
    September 14, 2025

    The correct answer is (1) 50 µmol/min.

  • Ayush Dubey
    September 14, 2025

    50 µmol/min

  • Payal Gaur
    September 14, 2025

    50 micromol/min

  • Pallavi Ghangas
    September 14, 2025

    50 micromol per min.

  • Heena Mahlawat
    September 14, 2025

    50

  • Nilofar Khan
    September 15, 2025

    correct answer is (1) 50 µmol/min.
    Vo = Vmax . S/Km+S

  • Mitali saini
    September 15, 2025

    V= Vmax S/ Km+S
    = 100×10 / 10+10
    = 50umol/min

  • Khushi Agarwal
    September 15, 2025

    Correct 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, 2025

    50is correct answer

  • Simran Saini
    September 15, 2025

    Vo=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, 2025

    Vo = Vmax . S / Km + S
    = 100. 10 / 10+10
    = 1000/20
    = 50

  • Sonal nagar
    September 16, 2025

    Option 1
    50 µmol/min.

  • Minal Sethi
    September 16, 2025

    V = Vmax*S/Km+S
    V = 50 microlitre

  • Muskan Yadav
    September 17, 2025

    The initial velocity at substrate concentration 10 mM is 50 µmol/min is the correct answer .

  • Deepika sheoran
    September 18, 2025

    Option 1 st
    50 umol/min.

  • Khushi Singh
    September 25, 2025

    50 is correct

  • Juber Khan
    April 6, 2026

    Given Vmax = 100
    Km( V0 is half of Vmax)
    V0 = 100/2

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