polypeptide B consists of nine amino acid residues Complete hydrolysis of polype

Question

polypeptide B consists of nine amino acid residues Complete hydrolysis of polypeptide B yields Alanine Arginine Cysteine Glutamic acid Leucine Methionine Serine Tyrosine Valine Treatment of polypeptide B with elastase yields a tripeptide and a hexapeptide Treatment of polypeptide B with cyanogen bromide yields an intact tetrapeptide along with one or more other compounds Partial hydrolysis of polypeptide B yields the dipeptide Val-Tyr, the tripeptide Ala-Ser-GIu and other peptides Treatment of polypeptide B with chymotrypsin yields a tripeptide and a hexapeptide Treatment of polypeptide B with trypsin yields a tetrapeptide and a pentapeptide Treatment of polypeptide B with lithium aluminum hydride in dry diethyl ether, followed by treatment with 6 M aqueous hydrochloric acid at 100 degree C for 24 hours yields compound C along with one or more other compounds

Explanation / Answer

your primary sequence from N-terminus to C-terminus would be LEU-VAL-TYR-ARG-MET-CYS-ALA-SER-GLU From treatment with Trypsin, we now know that ARG should be towards left of MET. As, Trypsin cleaves the peptide bond following the positively charged amino acids like ARG, and to get a tetrapeptide and pentapeptide it should be placed towards left of MET. From partial hydrolysis we know that VAL-TYR and ALA-SER-GLU are the sub group polypeptides of our primary polypeptide chain. From chymotrypsin treatment, which cleaves peptide bonds selectively at hydrophobic amino acids like TYR and it should be placed at the third position or the sixth position., and is along with VAL From last reaction given, we can deduce that LEU should be at the end of n-terminus. That is, it should be the starting amino acid. This is so because LAH reduces and then hydrolyses in the presence of HCL and the product C is that of reduced LEU. If Leu is at the end it would not have been hydrolysed as product C. From Elastase treatment, we know that it cleaves the bond at AL or VAL and gives a tri and a hexapeptide, and ALA is preceeded or followed by SER and GLU so it can not be placed in the third position and it is placed at 6th position. Hence, TYR is now placed at third position. Finally we have our sequence. Which is given above. I hope this helps.

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