(JUNE 2018)
36. The Vmax and Km from a Lineweaver-Burk plot of an enzyme reaction where 1/v = 40µM-1
min at 1/[S] = 0 and 1/[S] = -1.5 x 102 mM-1 at 1/v = 0 are
(1) 0.025 µM min-1 and 0.67 x 10-2mM
(2) 0.025 µM-1 min and 0.67 x 10-2
(3) 0.025 µM min-1 and 1.5 x 102 mM-1
(4) 0.038 µM min-1 and 0.67 x 10-2mM
The correct answer is (1) 0.025 µM min−1 and 0.67 \times 10^{-2}$$ mM.
Introduction
The Lineweaver-Burk plot, a double reciprocal graph of the Michaelis-Menten equation, is a classical method used to estimate kinetic parameters such as maximum velocity Vmax and Michaelis constant Km of enzyme reactions. Given intercepts from the plot, these parameters can be calculated directly. This article clarifies how to derive Vmax and Km given specific plot intercepts, tailored for academic rigor and practical enzyme kinetics understanding.
Given Data from the Lineweaver-Burk Plot
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1/v=40 μM−1 min at 1/[S]=0 (Y-intercept)
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1/[S]=−1.5×102 mM−1 at 1/v=0 (X-intercept)
Understanding the Lineweaver-Burk Equation
The Lineweaver-Burk plot plots 1v against 1[S]. The equation is:
1v=KmVmax1[S]+1Vmax
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Y-intercept =1Vmax
-
X-intercept =−1Km
Calculation of Vmax
From the Y-intercept:
1Vmax=40 μM−1 min
Calculate Vmax:
Vmax=140=0.025 μM min−1
Calculation of Km
From the X-intercept:
−1Km=−1.5×102 mM−1
Solve for Km:
Km=11.5×102=6.67×10−3 mM=0.0067 mM
Equivalent to 0.67×10−2 mM.
Interpretation and Significance
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Vmax=0.025 μM min−1 represents the maximum reaction velocity when enzyme is saturated.
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Km=6.67×10−3 mM indicates that half-maximal velocity occurs at this substrate concentration, reflecting substrate affinity.
Summary Table
| Parameter | Calculated Value | Unit |
|---|---|---|
| Vmax | 0.025 | μM min−1 |
| Km | 0.0067 (or 0.67×10−2) | mM |
Conclusion
By analyzing the intercepts of a Lineweaver-Burk plot, Vmax and Km can be precisely calculated. For the given data, the maximum velocity is 0.025 μM min−1, and the Michaelis constant is approximately 0.0067 mM. This process is fundamental in enzymology for characterizing enzyme behavior and substrate interactions accurately.



9 Comments
Aakansha sharma Sharma
September 14, 2025For Vmax
Given 1/v =40uM-min ,1/S = o means for Y intercept
So we have to calculate Vmax = 1/40uM-min = 0.025 uM- min
For Km
Given 1/S=1.5×10(2)mM-, 1/V=0 means for X-intercept
We have to calculate Km = 1/1.5×10(2)mM-=0.0067 so it can be written as 0.67× 10(-2) mM-
The correct answer is (1) 0.025 µM min−1 and 0.67 ×10(-2)mM
Kanica Sunwalka
September 14, 2025for Km
1/s = 1.5×10^2
On x axis – intercept i.e. 1/v =0
so km = 1/ 1.5 x 10 ^2 = 0.0067 = 0.67x 10^-2 mM
for Vmax
1/v = 40
On y axis – intercept 1/s =0
so Vmax = 1/40 =0.025 uM min ^-1
ans is 1 option
Kirti Agarwal
September 14, 2025Op 1
Khushi Agarwal
September 14, 2025The correct answer is (1) 0.025 µM min−1−1 and 0.67 \times 10^{-2}m
Vmax = 1/40uM-1
= 0.025 uM-1
Km = 1/1.5×10(2)mM-=0.0067 = 0.67× 10(-2) mM-
Rishita
September 14, 2025for Km
1/s = 1.5×10^2
On x axis – intercept i.e. 1/v =0
so km = 1/ 1.5 x 10 ^2 = 0.0067 = 0.67x 10^-2 mM
for Vmax
1/v = 40
On y axis – intercept 1/s =0
so Vmax = 1/40 =0.025 uM min ^-1
Manisha choudhary
September 14, 2025Option 1
Tanvi Panwar
September 14, 2025for Vmax. ; 1/v=0+1/Vmax =.025uM min ^-1
for Km; 0=Km/Vmax.(1/S)+1/Vmax. = 0.67x 10^-2 mM
Palak Sharma
September 16, 2025km= 0.67 x 10-2 mM
Vmax= 0.025 uM/min
Minal Sethi
September 18, 20251/v=0+1/Vmax =.025uM min -1
for Km; 0=Km/Vmax.(1/S)+1/Vmax. = 0.67x 10^-2 mM