Lineweaver–Burk Plot: Finding Km and Vmax and Spotting Enzyme Inhibition

By Aniket Bhardwaj · 11 November 2026 · Physical Chemistry

The Michaelis–Menten equation gives a curve, and reading K_m and V_max from a curve by eye is unreliable. The Lineweaver–Burk plot turns the same equation into a straight line, which is easy to fit and easy to compare for different inhibitors. This page shows how to build the plot, read the constants and recognise competitive and non-competitive inhibition from the line pattern.

From Michaelis–Menten to a straight line

v = V_max [S] / (K_m + [S])
Take the reciprocal of both sides:
1/v = (K_m / V_max) × (1/[S]) + 1/V_max

This has the form y = mx + c, so plot 1/v on the y-axis against 1/[S] on the x-axis.

Slope = K_m / V_max
y-intercept = 1/V_max, so V_max = 1 / intercept
x-intercept = −1/K_m, which is where the line crosses the x-axis on the negative side
K_m = slope ÷ intercept (slope × V_max)

Here v is the initial rate, [S] the substrate concentration, V_max the maximum rate when the enzyme is saturated, and K_m the substrate concentration at which v = V_max/2.

Worked examples

Example 1: finding V_max and K_m. A practice data set for an enzyme gives these initial rates:

[S] (mM)	v (µM min⁻¹)	1/[S] (mM⁻¹)	1/v (µM⁻¹ min)
0.5	10.0	2.000	0.1000
1	16.7	1.000	0.0599
2	25.0	0.500	0.0400
4	33.3	0.250	0.0300
8	40.0	0.125	0.0250

A least-squares straight-line fit of 1/v against 1/[S] gives slope = 0.0400 and intercept = 0.0200.

V_max = 1 / 0.0200 = 50 µM min⁻¹

K_m = slope / intercept = 0.0400 / 0.0200 = 2.0 mM

Quick check with two points: using (2.000, 0.1000) and (0.125, 0.0250), slope = (0.1000 − 0.0250) / (2.000 − 0.125) = 0.0750 / 1.875 = 0.0400, and intercept = 0.1000 − 0.0400 × 2.000 = 0.0200. ✓

Check with the original equation: at [S] = 2 mM = K_m, v should be V_max/2 = 25.0, which matches the table. ✓

Example 2: competitive inhibitor. With the same enzyme, an inhibitor at [I] = 2.0 mM has K_i = 1.0 mM. A competitive inhibitor multiplies the slope by α = 1 + [I]/K_i and leaves V_max unchanged.

α = 1 + 2.0/1.0 = 3.0

New slope = 0.0400 × 3.0 = 0.120. Intercept stays 0.0200 (V_max = 50 µM min⁻¹).

Apparent K_m = 0.120 / 0.0200 = 6.0 mM, which equals 2.0 × 3.0. ✓

The lines with and without inhibitor cross on the y-axis, because the intercept is the same.

Example 3: pure non-competitive inhibitor. For a pure non-competitive inhibitor with [I]/K_i = 1.0, both slope and intercept are multiplied by α = 1 + 1.0 = 2.0.

New slope = 0.0400 × 2.0 = 0.0800; new intercept = 0.0200 × 2.0 = 0.0400.

Apparent V_max = 1 / 0.0400 = 25 µM min⁻¹, half the original.

K_m = 0.0800 / 0.0400 = 2.0 mM, unchanged. ✓

The x-intercept stays at −1/K_m = −0.50 mM⁻¹, so the lines cross on the x-axis.

Reading the line pattern

Type	V_max	K_m	Where the lines meet
No inhibitor	Reference	Reference	—
Competitive	Unchanged	Apparent K_m rises	On the y-axis
Pure non-competitive	Falls	Unchanged	On the x-axis
Uncompetitive	Falls	Apparent K_m falls	Parallel lines, no crossing

Common mistakes

Plotting v against [S] and calling it a Lineweaver–Burk plot. You must take reciprocals of both quantities.
Reading the y-intercept as V_max. It equals 1/V_max.
Taking K_m as the slope. The slope is K_m/V_max; divide by the intercept to get K_m.
Forgetting the minus sign: the x-intercept is −1/K_m, so K_m is positive.
Trusting the points at very low [S]. Their reciprocals are large, so small errors in v stretch the line badly. Many modern workers fit the Michaelis–Menten curve directly, or use other linear forms, for this reason.

Quick revision: how to plot and fit correctly

Start by making a table with four columns: [S], v, 1/[S] and 1/v. Do the reciprocals carefully, keeping at least three significant figures, because small rounding errors in 1/v at low [S] change the slope. Plot 1/[S] on the horizontal axis and 1/v on the vertical axis, then draw the best straight line through the points, or fit it by least squares.

Extend the line backwards until it meets the y-axis (this gives 1/V_max) and then further until it meets the x-axis on the negative side (this gives −1/K_m). If your line passes through the points almost perfectly, the enzyme follows simple Michaelis–Menten behaviour in the range you studied. If the points curve, there may be substrate inhibition, cooperativity or experimental error, and a straight line is the wrong model.

When two lines are drawn on one graph, always state which one is the control and which has the inhibitor. Then compare the intercepts first: the same y-intercept means competitive inhibition, and the same x-intercept means pure non-competitive inhibition.

Where this is used in exams

Exam topic	How the plot is used
Enzyme kinetics (biochemistry and physical chemistry units)	Finding K_m and V_max from tabulated data.
Inhibition questions	Identifying the inhibitor type from changes in slope and intercepts.
Data-fitting practice	Straight-line least-squares fitting of reciprocal data.

Check the current official syllabus of your exam to see whether enzyme inhibition is included and at what depth.

Enter your 1/[S] and 1/v pairs to get the slope and intercept of the best straight line, then work out V_max and K_m.

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