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The maths of compartmentation
Let's imagine a Thingy, X, which is split between two compartments A and B. There are therefore two forms of X, XA and XB. The amount of X in some tube (for instance a sample we've collected from a gradient) is the sum of XA and XB.
If that particular tube contains 30% of the total amount of compartment A, then it will contain 30% of the total amount of XA, and 30% of any marker-thing (for instance a marker enzyme) . Rather than talking in percentages, we will use fractions (which means we don't have to worry about multiplying and dividing things by 100). Our 30% corresponds to a fraction of 0.3 We'll use the letter A for the amount of marker enzyme in compartment A. Therefore the fraction of compartment A that we've got in our tube is:
And since the amount of XA in our tube is equal to that fraction of the total XA.
Something very similar applies to B, so the sum of X in our tube is given by:
If we divide through by the total amount of X everywhere, we get:
But, of course, XA,Tot/XTot is the fraction of Thingy X that is in compartment A, exactly the number we would like to calculate. This has all got a bit scary-looking because of all the subscript "tubes" and "Tots", so we can simplify things by promising to work in fractions of the total amount of marker recovered. So, if we found 200 units of marker enzyme A in our gradient, we will divide all measurements of marker enzyme A by 200, and just call these numbers A. Also for the sake of simplicity, we'll use the letter "a" for the fraction of X that is found in compartment A. Therefore:
Now we need to solve this to find a and b. There are various methods.
Simultaneous equationsThe easiest way is to get two, different versions of this equation and solve them. To take a concrete example, you might have something like
This is what would happen if you had measurements from a crude extract containing all the compartments, and for a fraction enriched in A (20% yield) relative to B (1% contamination). "0.181" and "1.000" are your yields of the Thingy X in two tubes, and the other numbers (0.2, 0.01, and 1) are the yields of the marker enzymes. You can solve these simultaneous equations and get an answer for a and b.
Of course you can do the algebra and come up with a general solution for two compartments rather than solving the specific case above.
If you have three compartments, then you need three simultaneous equations to solve, which means you need three different tubes to assay. In theory for four compartments you could do it with four tubes.
But what happens if you have more than the right number of tubes? You might, for instance, have two compartments but a gradient separated into four fractions. In this case your data are "over defined", which is not a bad thing, because it gives you some built-in quality control of your data. However, it does make things more difficult. For instance, with our two compartments and four measurements, we could arive at several different answers depending on which measurements we use (none of the measurements being exactly right, in the real world).
The graphical method
For two compartments there is a very convenient graphical approach. This takes our original equation
And modifies it by dividing through by a marker enzyme
Now this is the equation of a straight line. If we plot X/A versus B/A we should get a straight line with gradient "b" and intercept "a". And if our data are over-defined (more than two points) we simply do a linear regression (fit a line).
There are two downsides to this method. The first is that it only works for two compartments (unless you happen to have a sheet of n-dimensional paper handy, and an (n-1) dimensional ruler to put through the points!). The other is worth knowing about: this method only works if all your fractions contain a reasonable amount of A and B (i.e. you have a very poor gradient with bad separation). It is ideal for non-aqueous fractionations, where mixing of compartments is virtually certain.
The reason it needs a bad gradient is that you have a term B/A. If a fraction contains no B, then the answer is zero, which is accurate but not very informative. In fractions where there is no A, then the answer is huge, tending to infinity. And in a nice clean bit between two bands in the gradient there is nothing of anything, and the answer is totally random.
The general least-squares method
The principle is to minimise squared differences between the real measurements of unknown enzymes and what they ought to have been using the marker enzymes. For an unknown enzyme X, the amount measured in a fraction ought to be...
where a is the proportion of X in compartment "A", and A is the amount of the marker enzyme for compartment "A" found in the fraction we are measuring (as above)
The error is the difference between measured X and theoretical X
Squared error is given by (from now on just say "X", not Xtheo.
I'm only writing out the bits for 2 compartments, but it is easy enough to envisage C, D, E... etc.
For all fractions, the sum of error-squared is:
This can be minimised in the normal way with respect to a, b, etc (the proportions of X in each compartment). Differentiate with respect to a, b, etc., and make each dErr2/dx = 0.
For a:
You can get similar equations for all the other compartments. The other one in this case is:
If you have 3 compartments, you will get three equations, each with three things on the right hand side, and so on. These can be solved by any reliable numerical method. It is important you use a reliable method, because solving a set of simultaneous equations can be a tricky business if two of them are nearly the same (i.e. two of the compartments separated very badly on the gradient). In a case like this you need to solve the equations in the right order, which is what all the reliable numerical methods guarantee.