Question

Three isoforms of haemoglobin are needed to be expressed and purified using the budding yeast expression system. These haemoglobins are very similar both in size and amino acid sequence. Briefly describe how would you design your experiments to express and purify these proteins to homogeneity.

Answer 1

It is given that isoforms are the same in size and amino acid sequences. Therefore, gel filtration chromatography can not be used. However, it might be possible that charge present on the isoforms are different. Theoretically, charge on any protein will be determined by the presence of charged amino acid and terminal groups on N and C terminal end. In this calculation, we are expecting that all the charged amino acids are exposed to the solvent. However, in the solution, this is not the case. It might be possible that some amino acids may not be exposed and therefore determination of the theoretical value of the charge is different from the practical value. Hence in solution, isoforms may have different charges.

This charge difference can be used to separate isoforms by ion exchange and isoelectric focusing chromatography. Either of the technique can be used. Genes for all these isoforms are cloned in the yeast vector and transformed into yeast cells for their expression. After expression, proteins will be separated out by using either ion exchange or isoelectric focusing chromatography.