Enzymes. Single Ubiquitin Molecule. Polypeptide products formed with LysC a seri

Question

Enzymes. Single Ubiquitin Molecule. Polypeptide products formed with LysC a serine protease.

Please help me solve this. I can't figure out where to even begin.

Please help me!

You have purified a protein (which you call protein X) that you suspect to be covalently modified by the addition of a single ubiquitin molecule. You also know that your purified protein is very unique, in that it only contains one lysine amino acid in its entire primary sequence. The molecular weight of the unmodified form of protein X is 60,000 Daltons. Suppose you also know that the c-terminal sequence of ubiquitin is HOOC-Gly-Gly-Lys-Ile-Gln-etc. where the indicated lysine at position 3 is also the only lysine amino acid in the ubiquitin sequence. You are now trying to confirm that your protein does indeed carry this ubiquitin modification. Suppose you add to your purifed protein X the enzyme LysC, a serine protease that specifically recognizes the protonated amine group on lysine side chains, and hydrolyzes the peptide bond on the c-terminal side of lysine. Assume the solution is buffered to pH 7 and the pKa of the lysines in question is > 10.Based on the information above, answer the following two questions: i) If protein X is modified by ubiquitin, how many polypeptide products would you expect to create if it is treated with LysC? Briefly explain why. ii) What would be the approximate molecular weight of the different peptide products?

Explanation / Answer

a) There will be only two polypeptide products as you suspect to be covalently modified by the addition of a single ubiquitin molecule also purified protein only contains one lysine amino acid in its entire primary sequence.

so that lysine must be of due to ubiquitin

the chain breakage will be due to LysC which cleaves towards the C-terminal side of lysine the left over peptide will be

HOOC-Gly-Gly-Lys-Ile-Gln-etc   ---> HOOC-Gly-Gly-NH2 + HOOC-Lys-Ile-Gln-etc

                                                                    I                             II

b) the molecular weight of protein = 60000 daltons

Molecular weight of one glycine amino acid = 75Da

So the molecular weight of fragment I = 150 - 18(mol weight of water) = 132 Da

Molecular weight of fragment II = 60,000-120= 59880 Da

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