Helpful for Biochemistry as well as biotechnology students.

1. Lineweaver - Burk plot
Double reciprocal plot
BTH-3506 : ENZYMOLOGY
CE-1
-By
1) Gunali Chaudhari (8111)
2) Nikita Jedhe (218213)

2. Need of MM equation transformation
Fig-1 : Michaelis – Menten plot
Fig-1 : A typical curve of Michaelis – Menten
plot of enzyme kinetics is hyperbole curve.
The curve is not perfectly horizontal means
maximum velocity (Vmax) can be achieved only at
infinite substrate concentration.
Because of this there are errors in Vmax and Km
estimation.
Hence in 1934 Hans Lineweaver and Dean Burk
transformed Michaelis – Menten equation into
Lineweaver – Burk equation.
Lineweaver – Burk plot has the great
advantage of allowing a more accurate
determination of Vmax and Km.
V or V0 - Reaction velocity
[s] - Substrate concentration
Vmax - Maximum velocity
Km - Michaelis – Menten constant

3. How to calculate x-intercept?
Fig-3 : Lineweaver – Burk plot
X-intercept
at y = 0
y = 1/V0= 0
Put, y = 0 in Lineweaver – Burk equation –
0 = km . 1 + 1
Vmax [s] Vmax
- 1 = Km . 1
Vmax Vmax [s]
-1 . Vmax = 1
Vmax Km [s]
-1 = 1
Km [s]

4. 1 = Km+[s]
V0 Vmax [s]
Reciprocal of MM equation
Double reciprocal plot
Fig-2 : Lineweaver – Burk plot
v0 = Vmax [s]
Km + [s]
Michaelis – Menten equation
Inverting the equation
1 = Km + 1
V0 Vmax [s] Vmax
Lineweaver – Burk
equation
Comparing this equation with straight line equation
y = mx + c, we get –
y= 1/v0, m(slope)= Km/Vmax, x= 1/[s], c (y-intercept)= 1/Vmax

5. Practice Problem 1
 If the X-intercept on a Lineweaver – Burk plot or double reciprocal plot is
equal to -2 , what is the Km value ?
1) -2
2) 0.5
3) 1
4) -0.5
5) -1
Solution - According to Lineweaver – Burk plot,
x- intercept = -1/Km
-1/Km = -2
½ = Km
Km = 2) 0.5

6. Practice problem 2
 The following experimental data were collected during a study of the catalytic activity
of an intestinal peptidase with the substrate glycylglycine:
Glycylglycine + H2O peptidase 2 glycine
[S]
(Mm)
V
(µm/min)
1.5 0.21
2.0 0.24
3.0 0.28
4.0 0.33
8.0 0.40
16.0 0.45
1/[s]
(Mm)-1
1/v
(µm/min)-1
0.67 4.76
0.50 4.16
0.33 3.57
0.25 3.03
0.125 2.50
0.0625 2.22
y = 4.240x + 2.004
R² = 0.991
0
1
2
3
4
5
6
0 0.2 0.4 0.6 0.8
Reciprocal
of
Reaction
velocity
1/v
or1/
v
0
(1/µm/min)
Reciprocal of Glycylglycine concentration 1/[s] (1/Mm)
y= mx + c = y= 4.240x + 2.004
C = 1/Vmax = 2.004
Vmax = 1/2.004 = 0.49 µm/min
m = Km/Vmax = 4.240, 4.240= Km/0.49
Km= 4.240 × 0.49 = 2.07 Mm

7. Lineweaver – Burk plot of bisubstrate reactions
Sequential Mechanism Ping pong Mechanism
Fig-4 : This enzyme reaction involving a ternery
complex formation. For this mechanism,
Lineweaver –Burk plot at varying S1 and different
fixed values of S2 give a series of intersecting lines.
Fig-5 : This enzyme reaction not involving a ternery
complex formation. This is double displacement
pathway. For this mechanism, Lineweaver – Burk plot
at varying S1 and different fixed values of S2 give a
series of parallel lines.

8. Practice problem 3
 Consider the following set of data and answer the following questions –
[s] (mM) V
(mMol/min)
V (+inhibitor)
(mMol/min)
0.5 23.5 16.67
1.0 32.2 25.25
1.5 36.9 30.49
2.5 41.8 37.04
3.5 44.0 38.91
A) Plot the data on a Lineweaver – Burk plot and determine
Km and Vmax.
B) The 2nd set of velocities represents the rate of the reaction
when an inhibitor is added. Plot this data on the same graph
as above and determine new Km and Vmax and the type of
inhibitor (competitive, uncompetitive, non-competitive).
C) Can the effects of the inhibitor be over – ridden by adding
more substrate? Why?
1/[s] (1/mM) 1/V
(mMol/min)-1
1/v (+ inhibitor)
(mMol/min)-1
2 0.042 0.059
1 0.031 0.039
0.66 0.027 0.032
0.40 0.023 0.026
0.28 0.022 0.025

9. Lineweaver – Burk plot for Enzyme Inhibition
Fig-6 : Competitive
Inhibition
Slope – Change
Y-intercept- Same
Km – increased
Vmax- unaffected
Fig-7 : Non competitive
Inhibition
Slope – Change
Y- intercept - Change
Km – unaffected
Vmax- reduced
Fig-8 : Uncompetitive
Inhibition
Slope – Same
Y- intercept - Change
Km – reduced
Vmax- reduced

10. y = 0.011x + 0.018
R² = 0.997
y = 0.020x + 0.018
R² = 0.998
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0 0.5 1 1.5 2 2.5
1/[s]
(1/mM)
1/V (1/mM/min)
without inhibitor
with inhibitor
B) y= 0.020x + 0.018
After solving for Km and Vmax
We get, Vmax= 55.55 mM/min Km= 1.1 mM
m = change, c = same, Vmax= unaffected, Km = increased
The type of inhibitor is Competitive.
A) y= 0.011 + 0.018
After solving for Km and Vmax
We get,
Vmax= 55.55 mM/min Km=0.61mM

11. Limitations
Fig-9 : Limitations of Lineweaver – Burk plot
Fig-9 : all data found at large substrate
concentrations will be clustered near the
origin.
small experimental errors are found far
to the right of the y-axis calling for a
large extrapolation back to obtain x- and
y- intercepts.
To circumvent these problems,
experimental data is often plotted on a
Eadie-Hofstee plot, another linear form
of the Michaelis-Menten plot or by
computer programs that fit the data to the
Michaelis-Menten equation itself.

12. References
• Lehninger Principles Of Biochemistry- 4th edition and 7th
edition
• www.chem.libretexts.org
• www.sciencing.com
• www.ncbi.nlm.nih.gov