This module is designed to be used by instructors teaching undergraduate courses in which the concept of protein separation by size/shape is important (e.g. Biochemistry, Molecular biology, Biotechnology, Immunology, Bioanalytical chemistry, etc.). The students should have some basic knowledge of chemistry and biology. The in class activity portion of this module is designed to be completed by a small group of students in about a 1 hour time period.
After completing the module, the students will be able to:
1. Explain how size exclusion chromatography works
2. Illustrate (by drawing) the SEC separation process
3. Predict elution order of proteins
4. Interpret size exclusion chromatograms
5. Recognize when SEC would be useful
6. Estimate molecular mass using SEC
Inquiry-based in class questions:
(1) Most students will immediately answer this question correctly. They are aware that there is a large difference in size or mass between monosaccharides and proteins. If problems arise, the instructor may decide to ask them to find and compare the structures.
(2) Students will probably make the connection between glucose and diabetes. Therefore, they may say to get rid of the protein so that glucose can be measured in the blood. They should probably know what immunoglobulins do in the body. Guide them to think about what immunoglobulins are and what role they play in the body. Ask them if there may be a need for pure immunoglobulin (examples include use as a pharmaceutical or for development of an immunoassay).
(3) Students will know that small molecules move faster than larger molecules in solution. So, they may say something like put the sample in a flow system and the glucose molecules will move faster and can be separated first. Suggest to them what might happen if the sample is placed in a column which contains only buffer and the sample is added to the top and the solution allowed to flow through the column using a pump to continuously add buffer. They will most likely say that the glucose will separate out first. Now, draw some particles in the column with holes a little larger that glucose, but a lot smaller than an antibody. They will get that glucose can enter the holes (use the term pores) and that the antibody cannot. Now ask them what will happen to the rate at which glucose is able to exit (use elute) from the column. They will understand that glucose will be held up inside the pores. Now ask them what will happen to the antibody. They will probably get that the antibody will not be held up much due to the presence of the particles and will elute before the glucose molecules. It is worth emphasizing at this point that SEC is a method that depends on the molecules having no attractive forces with the surface of the solid phase. Attractive forces, which are exploited in virtually all other chromatographic methods but SEC, could be different for glucose and an immunoglobulin and if present, could interfere with the separation scheme.
(4) The smallest molecule can diffuse through the entire pore. The largest molecule does not fit into any of the pore. The intermediate-sized one can only diffuse through part of the pores. Because of this, the smallest one should take the longest to come through the column. The largest one should come through fastest. Of course, this requires that the molecules have no attractive interaction with the surface of the solid particles.
(5) Dissolving the compounds being separated in an especially good solvent (one they are highly soluble in) and using a solid particle with a surface with properties very different from the nature of glucose and the proteins (so a nonpolar surface) creates an environment where specific interactions with the surface can be eliminated.
(6) Students will generally know that larger molecules will produce broader peak profiles. So, they will most likely understand that peak 1 is most likely to belong to glucose and peak 2 is most likely to belong to an immunoglobulin.
(7) Students will probably not have too much difficulty in realizing that placing a smaller peak and larger peak in the same amount of space has less overlap (like resolution) than placing two larger peaks in the same amount of space. If there is any question, the instructor may have the students perform an exercise to illustrate this concept.
Out-of-class problem set:
(1) Students will refer to the schematic representation of the operation of an SEC column to draw this chromatogram. Ensure that they correctly show protein #1 eluting last and protein #3 eluting first. Based on the information in the module, sephadex G100 would be a good packing material.
(2) Students will look up the mass of insulin and realize that it is a lot smaller in mass (~6000 Da) than the other proteins depicting on the chromatogram, and therefore it will elute a lot later than other proteins depicted. Based on use of G100, insulin would have some accessibility to the pores.
(3) They will understand that for peaks to be well separated (resolved), there must be a two-fold difference in molecular mass. So, a peak of 40,000 Da eluting between proteins 1 and 2 may not be completely separated.
(4) Students should be able to extrapolate the corresponding Log MW associated with 3 mL retention volume:
Retention volume = 3mL yields a corresponding Log MW ~ 4.8
MW ~ 60,000Da
Synder, L.R.; Kirkland, J.J.; Glajch, J.L.; Practical HPLC Method Development; Wiley : New York (1997).