An oligomeric protein has a molecular weight of 415000. A 740 mg sample was trea
ID: 54954 • Letter: A
Question
An oligomeric protein has a molecular weight of 415000. A 740 mg sample was treated with an excess of 1-fluoro-2,4-dinitrobenzene (Sanger's reagent) until the reaction was complete. Next the peptide bonds were hydrolyzed by heating with concentrated HCl. The product of the complete hydrolysis contained 1.8 mg of the compound shown below, and no other 2,4-dinitrophenyl derivatives of the ?-amino groups of other amino acids could be found. How many polypeptide chains are in this protein? (Enter your answer as a whole number). HINT: The dinitrophenyl derivative identifies the end groups of each polypeptide chain. Start by calculating the number of moles of protein in the sample. Then calculate the number of moles of endgroup obtained.
If someone could please show me the STEPS I would greatly appreciate it! Do not put this as an answer because it does not make any sense. "Calculate the mw of the protein that was provided. From there, you divide 1.8 so it will 1.8/mw of protein...then you divide 740mg/415000mmol= 0.0018...then you divide again the values that you got from your protein and enter the answer as a whole #. for example: you calculated 3.4 for your final answer you will only have to put 3." Seriously unhelpful.
Explanation / Answer
Sanger's reagent i.e. 1- fluoro- 2,4- dinitrobenzene tags the amino- terminal residue of a peptide & then hydrolysing the peptide into its constituent amino acid residues, the amino terminal amino acid can be identified.
In case of a oligomeric protein having different amino terminal residues, then subjecting it to this procedure can help us identify the amino-terminal residues & also know how many peptide chains are there (number of different amino -terminal amino acid labeled reveals number of peptide chains).
Here, molecular weight of the oligomeric protein is 415000.
740 mg of the protein was taken; hence, number of moles of the protein taken = 740/415000 = 0.00178
Therefore, no. of moles of polypeptide chains = .00178 * x (x be the number of polypeptide chains in the oligomeric protein)
Here, we get only one labeled amino acid (I think the compound here is 2, 4- dinitroalanine); this implies that all the polypeptides in the oligomeric protein has alanine in the amino-terminal end.
Molecular weight of the compound ~ 256
1.8 mg of the compound was obtained after completion of reation; hence number of moles of N- terminal alanine = 1.8/ 256 =.00703
Now, 0.00703 = 0.00178 * x
Hence, number of polypeptide chains in the oligomeric protein = 0.00703/ .00178 = 3.949 = 4 (approximately)