The enzymatic reaction can be viewed as:

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

Where v = rate (initial velocity); V

_{max}= maximum velocity (100% of enzyme catalytic sites occupied); K_{m}= Michaelis constant (concentration of substrate to achieve half V_{max}); S = substrate concentrationNow, 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 V

_{max}, with K

_{m}the concentration at half-V

_{max}:

**Plot of Substrate Concentration against Initial Velocity (rate)**

_{max}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 V

_{max}, and K

_{m}. However, you should be able to calculate V

_{max}, and K

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

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

**Rearranging**

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:

And our Michaelis-Menton Equation:

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

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

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:

It is also true.

Equation 3 is getting closer to what we want. However, the V

_{max}over v is a problem as we don’t know V

_{max}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 V

_{max}from the lefthand side we need to divide

**both**sides by V

_{max}:

Note the new V

_{max}term on both sides of the equation - compare to equation 3

As we now have V

_{max}over v multiplied by V

_{max}we can cancel out both V

_{max}:

to give:

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

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

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

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:

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

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

_{max}) we rearrange, so:

extracting 1 over [S]V

_{max}we get:

multiplying through with 1 over [S]V

_{max}gives:

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:

Finally, separating out 1/[S] gives:

Which when compare to equation 1:

It can be seen that:

- y = 1/v
- x = 1/[S]
- c, the y-intercept = 1/V
_{max} - m, the gradient = K
_{m}/V_{max}

_{m}.

Hence, the final graph is:

If you would like to test your skills at performing enzyme kinetics calculations then you might like to look at: Maths4Biosciences.com.

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

ReplyDeleteJust what I needed - Thanks

ReplyDeletegreat explanation

ReplyDeleteThank you, this is great

ReplyDeleteThank you, this is great

ReplyDeleteVery helpful thank you

ReplyDeleteWONDERFUL! it couldn't be more clear, great job!

ReplyDelete