Sunday 1 November 2009

Deriving Lineweaver-Burk Reciprocal Plot from Michaelis Menten Equation I

After a recent seminar, a number of students have asked how to derive the Lineweaver-Burk Reciprocal Plot equation from the Michaelis Menten equation... so, here goes.

Blog Post Bonus: Download a PDF of this blog post for your class notes.
The enzymatic reaction can be viewed as:

Enzymes equation

Where E = enzyme; S = substrate; ES = Enzyme substrate complex; and P = product

The Michaelis-Menton Equation describing the reaction is:

Michaelis menton

Where v = rate (initial velocity); Vmax = maximum velocity (100% of enzyme catalytic sites occupied); Km = Michaelis constant (concentration of substrate to achieve half Vmax); S = substrate concentration

Now, if the equation you are starting with does not look like the one above then have a look at another post (apparently I do it the 'old fashioned way’!)

In the lab, we can change the substrate concentration (S) in a reaction and measure the rate (v). As you know, a plot of substrate concentration (S) against rate (v; initial velocity) gives a curve, which plateaus at Vmax, with Km the concentration at half-Vmax:

Substrate concentration against rate

Plot of Substrate Concentration against Initial Velocity (rate)

The direct measurement for Vmax can never be achieved in the lab as the concentration of the substrate needed would never be reached. Also, in the lab, we would use a computer to calculate the values and determine Vmax, and Km. However, you should be able to calculate Vmax, and Km yourself, just so you know the computer is right, and it is possible to calculate these values from experimental data by using a linear plot and an equation derived from the Michaelis-Menton Equation.

As you know, the equation for a straight line is:

Equation 1

Equation for a straight line, where m = the gradient and c = the intercept on the y-axis

So, the problem is, how do we get:

Equation 2

to look like equation 1? The answer is, we rearrange....


We need to 'extract' our substrate concentration S and our rate v so we can plot them on a straight-line graph. Basically, we need to get equation 2 to look like equation 1.

The equation for a straight line is:

Equation 1

And our Michaelis-Menton Equation:

Equation 2

So, the first thing we need to do is invert equation 2 to get:

Equation 3

If you didn’t understand that ‘mathematical move’ consider this:

Equation 4

The above is true. That is, 2 over 1 = 2, 4 over 1 = 4 and 2 times 4 is 8. If I simply invert (flip) all the parts:

Equation 5

It is also true.

Equation 3 is getting closer to what we want. However, the Vmax over v is a problem as we don’t know Vmax and can only measure v and S in the lab. Therefore, we need to separate out the terms we can measure so we can have x and y as in equation 1.

To ‘remove’ the Vmax from the lefthand side we need to divide both sides by Vmax:

Equation 6

Note the new Vmax term on both sides of the equation - compare to equation 3

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As we now have Vmax over v multiplied by Vmax we can cancel out both Vmax:

Equation 7

to give:

Equation 8

If that bit of maths has lost you, consider this:

Equation 9

if you divide both sides by 2 it is still correct:

Equation 10

However, on the righthand side the two 2s can be cancelled to give:

Equation 10a

which is still correct.

In equation 5 we now have 1/v and this is our y in equation 1. All we need to do now is ‘extract’ x (which is our substrate concentration) from equation 5.

If you consider the following it is true:

Equation 11

we can ‘separate terms’ on this and express it in several other ways:

Equation 12

That is, once we find a ‘common’ element (in the above example 1/2, and in equation 5 1 over [S]Vmax) we rearrange, so:

Equation 13

extracting 1 over [S]Vmax we get:

Equation 14

multiplying through with 1 over [S]Vmax gives:

Equation 15

As you can see in equation 7 we have two terms after the + that can cancel out, and our experimental variable (S) can be separated, so:

Equation 16
Equation 17

Finally, separating out 1/[S] gives:

Equation 18

Which when compare to equation 1:

Equation 19

It can be seen that:
  • y = 1/v
  • x = 1/[S]
  • c, the y-intercept = 1/Vmax
  • m, the gradient = Km/Vmax
And, the intercept for the x axis, (i.e. when rate (y) = 0) is 1/-Km.

Hence, the final graph is:


Blog Post Bonus: Download a PDF of this blog post for your class notes.
If you would like to test your skills at performing enzyme kinetics calculations then you might like to look at:


  1. Thanks for this derivation Nick, superbly explained and very easy to follow!

  2. Just what I needed - Thanks

  3. great explanation

  4. Thank you, this is great

  5. Thank you, this is great

  6. Very helpful thank you

  7. WONDERFUL! it couldn't be more clear, great job!