Question

You have purified the peotide hormone that binds to the receptor below. To determine its amino acid sequence you digested the polypeptide with trypsin and in a separate reaction you cleaved the polypeptide with cyanogen bromide. Cleavage with trypsin yielded 5 peptides that were sewuences by Edman degradation as shown in the following:

1. Ser-Leu

2. Asp-Val-Arg

3. Val-Met-Glu-Lys

4. Ser-Gln-Met-His-Lys

5. Ile-Phe-Met-Leu-Cys-Arg

the fragments were purified using high-performance liquid chromatography, HPLC, with a hydrophobic C18 column. PREDICT which fragment eould elute last from the column, providing the chemical basis for your prediction

Answer 1

Answer- 2.Asp-Val-Arg elute last from column

Explanation-

1. In hydrophobic column of HPLC, hydrophobic peptides get bind and their movement is retarded so that highly hydrophobic peptide will elute lastly. In similar way hydrophilic (polar) peptide will elute firstly as it will not interact with hydrophobic column and its movement not retarded by hydrophobic interaction. According to this above polypeptides shows elution pattern as follows-

first elute- Ser-Gln-Met-His-Lys = this peptide is most hydrophilic as it contains two polar amino acids serine and glutamine along with one positive charge due to histidine and one positive charge due to positively charged N-terminal. Positive charge of lysine is neutralized as it is present at negatively charged C-terminal. So that this peptide will elute firstly in hydrophobic HPLC column.

second elute-Ser-Leu= this dipeptide elute secondly because it is second most polar peptide as it has positive charge due to its N-terminal and negative charge due to C-terminal and it has one polar amino acid serine. So that it elute in second elute.

third elute-Val-Met-Glu-Lys = this peptide has one positive charge due to N-terminal and one negative charge due to negatively charged glutamate. The positive charge of lysine is neutralized by negatively charged C-terminal. So that this will elute in third elution.

fourth elute-Ile-Phe-Met-Leu-Cys-Arg = this peptide elute in fourth elution as it is fourth polar polypeptide. It contains four nonpolar amino acids, one polar amino acid cysteine and one positive charge due to positively charged N-terminal. Negative charge of C-terminal is neutralized by positively charged arginine.

fifth elute-Asp-Val-Arg = this peptide has negatively charged aspartate at N-terminal and positively charged arginine at C-terminal so that both charged terminal of peptide gets neutralized makes this tripeptide is most hydrophobic. Also this peptide has nonpolar valine so that overall this peptide does not has any charge and will elute lastly in hydrophobic column of HPLC.