Gradebo Priht Periodic Table ion 24 of 26 Map ninger Principles of Biochemistry

Question

Gradebo Priht Periodic Table ion 24 of 26 Map ninger Principles of Biochemistry son Cox sEVENT MHE/Freeman presented by Sapling Learning EDIT DNP-Val a A sample (435 mg) of an oligomeric protein of M 187,000 was treated COOH with an excess of 1-fluoro-2,4-dinitrobenzene (Sanger's reagent) under slightly alkaline conditions until the chemical reaction was complete. The peptide bonds of the protein were then completely hydrolyzed by heating it with concentrated HCI. The hydrolysate was found to contain 3.29 mg of DNP-Val (shown at the right) 2,4-Dinitrophenyl derivatives of the o- amino groups of other amino acids could not be found H3C 3 Calculate the number of polypeptide chains in this protein. Give the answer as a whole number. Number (b) 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 (M, 60,000, indicating three distinct polypeptides. SDS electrophoresis 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? Exit Previous Give Up & View Solution 0 Check Answer 0 Next Hint

Explanation / Answer

Ans. #A. Given-

            Mass of protein = 435 mg = 0.435 g

MW of protein = 187000 g/ mol

Now, moles of protein = Mass / MW = 0.435 g / (187000 g/ mol) = 2.326 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

                                                            = 0.00329 g/ (283.24 g/ mol)

                                                            = 1.162 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 = 1.162 x 10-5 mol / 2.326 x 10-6 mol

                                                                        = 4.993 = 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, the protein has 5 peptide chains.

#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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