Gel filtration chromatography (also known as size exclusion chromatography, molecular sieve chromatography, or gel permeation chromatography) is based on the differential distribution of the components in a sample between the mobile and stationary phases.
Specifically, in gel filtration chromatography, this differential distribution depends on the size and shape of the components.
Here, we take a more in-depth look at how the column fractionates the components in your sample and how the gel filtration chromatogram is used to determine molecular weight.
The Gel Filtration Column is not Your Typical Sieve
In gel filtration chromatography, the stationary phase is comprised of porous beads packed into a column. The mobile phase is the running buffer or other solvent. Sample components partition between the stationary and mobile phases based on their size-based accessibility to the pores of the matrix beads.
The smaller the size of the molecule or particle, the greater access it has to the pores of the matrix (or the accessible stationary phase), and the slower it moves through the column. Very large particles, which are completely excluded from the pores, elute first from the column at the void volume (V0). Analytes of intermediate sizes somewhat permeate into the pores and elute next. Very small molecules with high access to the pores, elute last just before one column volume of buffer has passed through the column (called total accessible volume [VT]). Therefore, although a gel filtration column separates particles by size, it is not a conventional sieve, which has pores of a uniform size.
The separated components are visualized as a plot of the volume of the mobile phase eluted through the column versus the detector signal. This plot, called a chromatogram, shows the location of the individual peaks and the quality of resolution of these peaks.
Take a Look at an Example
In Figure 1, there are two well-separated peaks corresponding to the two components (A and B) in the sample. Component A (red) requires solvent of volume Ve (A) to elute at its maximum concentration. Given that component A is completely excluded from the stationary phase (beads’ pores), Ve (A) = V0. Component B (green) has near complete access to the pores of the matrix. Therefore, its flow through the column is retarded. It elutes just before one column volume worth of the solvent can pass through the column (i.e., its elution volume, Ve (B), is almost equal to VT).
The location of the peaks, their heights, and widths convey a great deal of information about the sample components as well as the column.
Importantly, you can use peak locations to determine the molecular weights of the sample components. Let’s see how this is done.
How to Estimate the MW using a Gel Filtration Chromatogram
The distribution of an analyte between the stationary and mobile phases depends on its size. It is described by a parameter called its equilibrium distribution coefficient (Kd). Kd is the ratio of the concentrations of the component in the stationary and the mobile phases. For component A, Kd = 0, since it cannot access the pores of the stationary matrix. For component B, Kd can be written as
where Vi is the total solvent accessible volume of the beads’ pores.
The smaller the particle, the more it permeates into the beads’ pores, and therefore, the larger the Kd value. So, Kd depends on particle size; therefore, it is essential to determine Kd to estimate particle size.
It is difficult to determine the concentrations of an analyte in the stationary and mobile phases. Therefore, you express Kd in terms of the easily measurable parameters, Ve and Vo. We know that the flow of component B through the column is retarded and it elutes at volume Ve (B). The volume (Ve(B) – V0) is the ‘extra’ stationary phase volume that B is able to access by virtue of its small size. This ‘extra’ volume is a component of the internal pore volume Vi, and the partition coefficient of B can be expressed as a fraction of Vi. Therefore,
VT is the total solvent accessible volume, a parameter that is difficult to measure. Therefore, substitute VT by the term Vc, which is the total geometric volume of the column:
where Kav (B) is not a true partition coefficient but is easily determined.
That was not too difficult, was it? Nothing works like quantification to help clear concepts!
From Kav to Molecular Weight
There is a linear relationship between the Kav of molecules and the logarithms of their molecular weights over a considerable size range. You can exploit this relationship to determine the molecular weight of an unknown sample.
First, determine the Kav values for a set of standards of known molecular weights. Second, plot these Kav values against the logarithmic values of the corresponding molecular weights to get a “selectivity curve.” For example, Figure 2 shows a selectivity curve for a popular, pre-packed gel filtration column determined using a set of commercially available protein standards. You can then use the calibration curve to determine the molecular weight of your protein of interest.
Note: you must choose the molecular weight standards carefully; take the range of pore sizes in your matrix beads into account.
Additionally, bear in mind, this assumes that the molecular shapes of the standards and your samples are the same. Commercially available protein standards are largely globular in shape. If your protein has an elongated shape, then the determined molecular weight is likely to be inaccurate. In such cases, use analytical ultracentrifugation or gel filtration coupled with multiple angle light scattering for molecular weight determination.
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