(FEB 2022-11) 31. What is the fold difference between v at [S] =Km and v at [S] = 1000 Km where v is the initial velocity of an enzyme catalyzed reaction, [S] is substrate concentration and Km is the Michaelis constant? (1) 1.998 (2) 1000 (3) 2.998 (4) 3.998

(FEB 2022-11)
31. What is the fold difference between v at [S] =Km and v at [S] = 1000 Km where v is the
initial velocity of an enzyme catalyzed reaction, [S] is substrate concentration and Km is the Michaelis constant?
(1) 1.998                                                    (2) 1000
(3) 2.998                                                    (4) 3.998

The correct answer is (1) 1.998.


Introduction

Michaelis-Menten kinetics describe the relationship between the initial velocity of an enzyme-catalyzed reaction and substrate concentration. A key question is how the initial velocity v0 changes when substrate concentration varies greatly relative to the Michaelis constant (Km). This article calculates the fold difference in initial velocity between substrate concentrations of Km and 1000×Km, providing insights into enzyme saturation kinetics and reaction efficiency.


Michaelis-Menten Equation Recap

v0=Vmax[S]Km+[S]

Where:

  • v0 = initial velocity

  • Vmax = maximum velocity

  • [S] = substrate concentration

  • Km = Michaelis constant


Step 1: Calculate v at [S]=Km

vKm=Vmax×KmKm+Km=Vmax×Km2Km=Vmax2

Initial velocity at substrate concentration Km is half maximal velocity.


Step 2: Calculate v at [S]=1000Km

v1000Km=Vmax×1000KmKm+1000Km=Vmax×1000Km1001Km=10001001Vmax

Initial velocity approaches Vmax but slightly less due to substrate not being infinitely saturating.


Step 3: Calculate Fold Difference

Fold difference=v1000KmvKm=10001001VmaxVmax2=10001001×2≈1.998


Interpretation

  • Increasing substrate from Km to 1000Km nearly doubles initial velocity.

  • Saturation kinetics cause diminishing returns beyond substrate concentration near Vmax.

  • The fold difference being just under 2 shows that substrate concentration has to increase massively to fully saturate enzymes.


Practical Implications

  • Understanding these relationships guides substrate concentration selection in enzyme assays.

  • It elucidates enzyme efficiency limits; enzymes reach near-max velocity even at relatively moderate substrate concentrations.

  • Useful for drug discovery and metabolic control analysis.


Summary Table

Substrate Concentration Initial Velocity v0 Relative to Vmax
Km 12Vmax 0.5
1000Km 10001001Vmax≈0.999Vmax ≈ 1

Conclusion

In Michaelis-Menten kinetics, raising substrate concentration from Km to 1000Km approximately doubles the initial velocity from 50% to nearly 100% of Vmax, yielding a fold difference of about 1.998. This exemplifies the dramatic effect of enzyme saturation and is relevant for enzymology study and practical applications.

20 Comments
  • Aakansha sharma Sharma
    September 12, 2025

    the fold difference between v at [S] =Km and v at [S] = 1000 Km
    1) v at [S] =Km so we know Vo=VmaxS/Km+S so here S=Km
    Vmax Km/Km+Km=Vmax/2
    2)v at [S] = 1000 Km so here S=1000Km
    Vmax1000Km/Km+1000Km=Vmax1000/1001
    So we have to calculate Fold increase= S2/S1=Vmax1000/1001/Vmax/2= 2000/1001=1.998 is answer

  • Varsha Tatla
    September 13, 2025

    Very 1st we calculate the V at S =km
    #2 we calculate the V at S=1000km
    Now we calculate -fold increase by s2÷s1
    By this way we find answer is1.998
    So,here option 1st will be correct answer

  • Mohd juber Ali
    September 14, 2025

    (A). S=km
    V = vmax .km/km+km
    V = v max/2
    (B). S = 1000km
    V = Vmax.km /km +1000km
    V = Vmax Km/1001Km
    V= Vmax/1001km
    # S2/S1
    1000Vmax/1001km ✖️2km/Vmax
    2000/1001 ~2

  • Kanica Sunwalka
    September 14, 2025

    V at (S) = Km ……………………. let it be S1
    V at (S) = 1000 Km …………….. let it be S2

    by Vo = Vmax S/ Km + S
    we cal the value
    then divide S2 by S1 = > for fold increase
    ans is 1.998
    option 1 is correct

  • Kirti Agarwal
    September 14, 2025

    1.998

  • Anurag Giri
    September 14, 2025

    Coorect ans is 1.998

  • Khushi Agarwal
    September 14, 2025

    The correct answer is (1) 1.998
    S1= vmax/2
    S2= 1000vmax/1001
    S2/S1 = 1000vmax/1001 * 2/ vmax = 1.998

  • Rishita
    September 14, 2025

    1.998.

    • Neha Yadav
      September 14, 2025

      1.998 ans

  • Tanvi Panwar
    September 14, 2025

    V at s= Km = Vmax.km/2km=Vmax/2
    V at s =1000km= Vmax.1000km/km=1000km=Vmax.1000/1001
    so, fold increase is S2/S1= 1.998.

  • Payal Gaur
    September 14, 2025

    1.998

  • Pallavi Ghangas
    September 14, 2025

    1.998

  • Nilofar Khan
    September 15, 2025

    Correct answer is 1.998

  • Ayush Dubey
    September 15, 2025

    1.998

  • Asha Gurzzar
    September 15, 2025

    1.998answer

  • Minal Sethi
    September 16, 2025

    1. S=km
    V = vmax .km/km+km
    V = v max/2
    2. S = 1000km
    V = Vmax.1000km /km +1000km
    V = Vmax 1000Km/1001Km
    V= 1000Vmax/1001
    fold increase = S2/S1
    1000Vmax/1001 * 2/Vmax
    1.98

  • Palak Sharma
    September 16, 2025

    The correct answer is (1) 1.998
    V1= vmax/2
    V2= 1000vmax/1001
    V2/V1 = 1000vmax/1001 * 2/ vmax = 1.998

  • Arushi Saini
    September 16, 2025

    The correct answer is (1) 1.998
    S1= vmax/2
    S2= 1000vmax/1001
    S2/S1 = 1000vmax/1001 * 2/ vmax = 1.998

  • Muskan Yadav
    September 17, 2025

    fold increase is S2/S1
    The result is approximately 1.998, which is close to 2
    Therefore, option 1 is the correct answer

  • Deepika sheoran
    September 18, 2025

    1.998

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