Given two proteins with the following characteristics, which ion exchange condit

Question

Given two proteins with the following characteristics, which ion exchange conditions would likely separate the proteins (such that you can do with experiments with the purified proteins)? Assume that the buffer components and buffer concentration are compatible with both proteins?

MW pI

Protein A: 125,000 7.0

Protein B: 60,000 8.0

A) Cation exchange resin with a buffer pH of 6.5, and a gradient of 0-500mM NaCl

B) Anion exchange resin with a buffer pH of 7, and a gradient of 0-500mM NaCl

C) Anion exchange resin with a buffer pH of 6.5, and no gradient

D) Cation exchange resin with a buffer pH of 8, and a gradient of 0-500mM NaCl

Explanation / Answer

Ion Exchange Chromatography involves the separation of the ionizable molecules based on their total charge.

Changing pH of buffer on can change charge on molecule to be separated as per need.

A) In a Buffer with pH less than the Isoelectric point (pI) (pH < pI) protein of interest will be in its protonated form i.e. with net +ve charg and hence +vely charged cation exchange resin is chosen.

B) In a buffer with pH greater than the Isoelectric point (pI) (pH > pI) protein of interest will be in its ionized form and will have net –ve charge and a –vely charged anionic resin is chosen.

Hence for give Protein-A with pI = 7and B with pI=8.

Cation exchange resin with a buffer pH of 6.5 (<7 i.e. pH<pI) and a gradient of 0.500 mM NaCl is likely to separate. Option-A

At pH = 7 which is equal to pI of Protein-A. Protein-A will be in uncharged state and will not be able to get exchanged on resin. Option-B rulled out.

At pH = 6.5 one can cannot use anion exchange resin (see above explanation) hence option C rulled out.

At pH = 8 Protein B will be in neutral state and Protein A will be in anionic state (Ionized ) hence now cationic exchange cannot be used.

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