7841 PrintPeriodic T 2018 11:59 PM Jump to... (0) | Leah Dollard- 0 of 26 er Pri

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7841 PrintPeriodic T 2018 11:59 PM Jump to... (0) | Leah Dollard- 0 of 26 er Principles of Biochemistry ox by Sapling DNP-Val A sample (543 mg) of an oligomeric protein of M. 185,000 was treated with an excess of 1-fluoro-2,4-dinitrobenzene (Sanger's reagent) under slightly alkaline conditions until the chemical reaction was complete. The NO2 bonds of the were then completely h heating t with concentrated HCI. The hydrolysate was found to contain 4.15 mg of DNP-Val (shown at the right). 2 amino groups of other amino acids could not be found. derivatives of the a- HC Calculate the number of polypeptide chains in this protein. Give the answer as a whole number Number A second oligomeric protein of M, 230,000 was shown by a similar endgroup analysis to consist of five polypeptide chains. SDS polyacrylamide gel electrophoresis in the presence of a reducing agent shows three bands: (M 30,000), (M 40,000) and y (M 60,000), indicating three distinct polypeptides sos is without reducing agent also yields three bands, with M of 30,000, 40,000, and 120.000 Which of the following oligomeric structures is consistent with this data? Check Answer NextExit

Explanation / Answer

Ans. #A. Given-

            Mass of protein = 543 mg = 0.543 g

MW of protein = 185000 g/ mol

Mass of DNP-Val adduct = 4.15 mg = 0.00415 g

# Moles of protein = Mass / MW = 0.543 g / (185000 g/ mol) = 2.935x 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

                                                            = 0.00415 g/ (283.24 g/ mol )

                                                            = 1.464 x 10-5 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-Val / Moles of Protein = 1.464 x 10-5 mol / 2.935 x 10-6 mol

                                                                        = 4.992 = 5 (nearest whole number)

Since, each 2,4-DNP-N-term adduct is given by one N-terminal residue, there are 5 N-terminal residues per molecule of protein (same as 5 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 = 5

#B. Given-

Molar mass of protein = 230000

Mass of a-subunit = 30000

Mass of b-subunit = 40000

Mass of y-subunit = 60000

# SDS provides uniform negative charge along the length of peptide chain and disrupts non-covalent interactions (H-bonds, van der Waal’s interactions, etc.).

A reducing reagent cleaves the disulfide bond in a peptide or between two different peptide chains.

# In either case, the sum of mass of all subunits must be equal to the molar mass of the protein-

            Mass of ab2y2 = 30000 + (2 x 40000) + (2 x 60000) = 230000

            Mass of a2by2 = (2 x 30000) + 40000 + ( 2 x 60000) = 220000

            Mass of a2b2y = (2 x 30000) + (2 x 40000) + 60000 = 200000

            Mass of a2b2y2 = (2 x 30000) + (2 x 40000) + (2 x 60000) = 260000

Hence, correct constitution of protein = ab2y2 Mr = 230000

So, correct option is – A. ab2y2

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