Sapling Learning Ma oligomeric protein has a molecular weight of 445000. A 930 m
ID: 1087871 • Letter: S
Question
Sapling Learning Ma oligomeric protein has a molecular weight of 445000. A 930 mg sample was treated with an excess of 1-luoro-2,4-dinitrobenzene (Sanger's reagent) until were hydrolyzed of the compound shown below, and no other 2,4-dinitrophenyl derivatives of the r amino acids could be found. by h eating with concentrated HCI. The product of the complete hydrolysis contained 26 a-amino groups of NO2 Now nany panlyweprias ahehni nrg miberis protein? are in -COOH Number CHJ Previous Gvo up &View; Scubo. ) Check Answer @Net Exit Hint careesExplanation / Answer
Ans. #A. Given-
Mass of protein = 930 mg = 0.930 g
MW of protein = 445000 g/ mol
Mass of DNP-Met adduct = 2.6 mg = 0.0026 g
# Moles of protein = Mass / MW = 0.930 g / (445000 g/ mol) = 2.090 x 10-6 mol
# Sanger’s reagent forms covalent exclusively with the N-terminal residue of a peptide chain. Upon hydrolysis with HCl, the 2,4-DNP-N-term adduct is released from the peptide chain.
Now,
Moles of 2,4-DNP-N-term adduct produced = Mass/ Molar mass
= 0.00260 g/ (315.30 g/ mol)
= 8.246 x 10-6 mol
# 1 mol 2,4-DNP reacts with 1 mol N-terminal residue. So, the number of adducts must be equal to the number of polypeptide chains in the protein.
Now,
Moles of DNA-Met / Moles of Protein = 8.246 x 10-6 mol / 2.090 x 10-6 mol
= 3.946 = 4 (nearest whole number)
Since, each 2,4-DNP-N-term adduct is given by one N-terminal residue, there are 4 N-terminal residues per molecule of protein (same as 4 mol N-ter residues per mol protein). Also, one N-terminal represents one peptide chain (subunits/ monomer/ peptide chains).
Therefore, number of polypeptide chains in protein = 4
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