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 concentrationIn 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)**

The direct measurement for V

_{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 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

So, the problem is, how do we get:

to look like equation 1 so we can plot a straight line using the terms we can measure, i.e v and S? The answer is, we rearrange....

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 as 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 3.

So, we have:

which is the same as:

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

extracting 1 over [S]V

multiplying through with 1 over [S]V

As you can see in equation 6 we have two terms after the + that can cancel out, and our experimental variable (S) can be separated, so:

/div>

so we get:

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.

Finally, separating out 1/[S] from 7 gives:

Which when compare to equation 1:

It can be seen that:

Hence, the final graph is:

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 as 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 3.

So, we have:

which is the same as:

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}in equation 4 we get:multiplying through with 1 over [S]V

_{max}gives:As you can see in equation 6 we have two terms after the + that can cancel out, and our experimental variable (S) can be separated, so:

/div>

so we get:

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.

Finally, separating out 1/[S] from 7 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:

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