C. The native protein has 16 free SH groups per molecule. If you treat the nativ
ID: 14322 • Letter: C
Question
C. The native protein has 16 free SH groups per molecule. If you treat the native molecule with urea and mercaptoethanol, the 18,000 molecular weight component that you get has 6 free SH groups per 18,000. (Assume for this part that there is only 1 type of subunit in this protein.) If you treat the native molecule gently with mercaptoethanol alone, you do not change the molecular weight, but you get a protein with 32 free SH groups per molecule. Now if you add urea AFTER the gentle mercaptoethanol treatment, the molecular weight is reduced from 144,000 to 18,000. The 18,000 component obtained in this way has 4 free SH groups per molecule.Given all information so far, answer the following:
C-1. How many subunits (polypeptide chains) are present in each native molecule? _____________
C-2. How many cysteine residues are present in each subunit? ___________
C-3. How many S S bonds are present in each native molecule? _____________
C-4. How many S S bonds are formed within each polypeptide chain (connecting different parts of the SAME chain)?______________.
Explanation / Answer
C-1. There are 8 subunits in the native molecule. (144,000 mol. wt total /18,000 mol. wt per subunit = 8 subunits.) C-2. Each subunit has 6 SH groups, so there must be 6 cysteines. C-3 and C-4: There are 16 S-S bonds per native molecule and one S-S bond within each polypeptide. In each native molecule, there are 8 S-S bonds within chains and 8 S-S bonds between chains. Each subunit or chain has 2 SH groups that are not linked up in S-S bonds, 2 SH groups linked to each other to form an intrachain S-S bond and 2 SH groups linked to the subunits on either side. The gentle mercaptoethanol breaks the S-S bonds between chains but not the S-S bonds within chains. The S-S bonds within the chains are relatively resistant to breaking by mercaptoethanol because they are buried inside the protein. The S-S bonds between the chains are sensitive to breakage because they are exposed on the surface of the protein.
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