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rate the following proteins from mouse kidney in order to analyze the structure

ID: 143553 • Letter: R

Question

rate the following proteins from mouse kidney in order to analyze the structure of Suppose you wish to sepa Protein B, a potential cancer metastasis protein. The following is known about the prote the kidney sample: ins you isolated from Protein 46.1 32.5 14.3 44.9 * 1 Dalton (Da) 1 g/mol pl Molecular Weight (kDa) 6.0 4.8 12.1 9.4 Beginning with a mixture of all four proteins, design a purification scheme to separate Protein B from the contaminating proteins. For chromatography steps, indicate which elution fractions (early vs. late) will contain which protein(s). a. A colleague has developed an alternative purification method, but it ends with Protein B in a solution of 8 M urea, and the protein is not functional. What has happened to the protein? What specific purification method could you use to restore the function? b.

Explanation / Answer

a. We can use ion exchange chromatography to purify protein B.

We know that when pH>pI, proteins have a net negative charge

pH<pI proteins have net positive charge.

So, if we take a mobile phase with pH 5, proteins A,C, D will have net positive charge and protein B will have negative charge.

We can do anion exchange chromatography in which the negatively charged protein (i.e. B) will bind to the stationary phase and all the other will elute out on the basis of their size, larger eluting out first.

We can then modify the salt concentration of our protein B and elute it in the end.

Therefore, the sequence of elution in anion exhange chromatography will be: A, D, C, B.

b. Urea is a denaturant and has likely denatured the protein due to which the protein has become non functional.

It is known that the molecular mass of urea is 60.06 g/mol whereas the molecular mass of protein B is 32.5 g/mol. This difference in their molecular mass will enable us to separate urea and our protein B using Dialysis, in which a semi permeable membrane with pores of a chosen size allows the entry or exit of only certain sized molecules.

Based on the difference in size of our protein and urea, we can select an available dialysis membrane and separate the protein and urea.

Once separated from urea, the protein's function will be restored.