The enzymatic reaction can be viewed as:
Where E = enzyme; S = substrate; ES = Enzyme substrate complex; and P = product
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. Both Dr Bevitt and Prof Lakey both start the rearrangement of the equation from the point shown in the other 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:
Plot of Substrate Concentration against Initial Velocity (rate)
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....
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 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:
Note the new Vmax term on both sides of the equation - compare to equation 3
As we now have Vmax over v multiplied by Vmax we can cancel out both Vmax:
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]Vmax) we rearrange, so:
extracting 1 over [S]Vmax we get:
multiplying through with 1 over [S]Vmax 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/Vmax
- m, the gradient = Km/Vmax
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.