Question

You have mixture of proteins with the following properties. Which of these proteins would be eluted last from an anion-exchange column when the column is developed with 1.0M NaCl in pH 7.0 buffer. Note, that the anion-exchange column was equilibrated with the pH 7.0 buffer before use.

A. MW = 18,500, pI = 10.2

B. MW = 28,000, pI = 7.0

C. MW = 62,000, pI = 3.9 CORRECT ANSWER

D. MW = 9,000, pI = 5.8

E. MW = 30,000, pI = 7.8

Please show how C. would be the correct answer

Answer 1

Ans. Given,

            pI of protein A = 10.2

            pI of protein B = 7.0

            pI of protein C = 3.9

pI of protein D = 5.8

            pI of protein C = 7.8

At pH = pI, the protein is in Zwitter ion form, has equal number of positive and negative charges on it, and it electrically neutral.

At pH > pI, there protein molecule donates H⁺ ions to restore the pH of the medium (here, electrophoresis medium of buffer) in attempt to restore the pH towards its pI.

At pH < pI, the protein accepts proton and become positively charged.

# Anion exchange column exchanges anion. That is, the matrix is positively charged to which negatively charged ligands (here, protein) bind.

* A pH 2 units greater than pI is considered sufficient to generate maximum negative charge on the protein.

# The pH of anion-exchange resin in 7.0 because it’s pre-treated with a buffer of pH 7.0

# Protein would A would be eluted first because it is positively charged at pH 7.

# As the pH is lowered (more acidic), the protein would gradually become neutral when pH becomes equal to the pI of specified protein.

At pH = pI, the protein can’t be retained in the positively charged matrix of anion-exchange column.

# At pH 3.9, the lowest pH in series, protein C would be in form of Zwitter ion form, thus electrically neutral. So, protein C would be eluted when pH of elution buffer gradually becomes equal to 3.9.

# Since H 3.9 would be achieved at the last in elution series, protein C would be elulted last.