A 64.3 mg sample of a protein (MW = 58,600) was treated with 2.00 mL of...

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A 64.3 mg sample of a protein (MW = 58,600) was treated with 2.00 mL of 0.0487 M sodium periodate (NaIO4) to selectively react with all of the serine and threonine residues. The resulting solution was then treated with excess iodide ion to convert the unreacted periodate into triiodide ion (13). 10 +31 + H₂O → 103 + 13 + OH A microtitration of the triiodide ion required 823 µL of 0.0988 M thiosulfate. 2 S₂03²1331+S406² Calculate the total number of serine + threoine residues per molecule of protein. A 64.3 mg sample of a protein (MW = 58,600) was treated with 2.00 mL of 0.0487 M sodium periodate (NaIO4) to selectively react with all of the serine and threonine residues. The resulting solution was then treated with excess iodide ion to convert the unreacted periodate into triiodide ion (13). 10 +31 + H₂O → 103 + 13 + OH A microtitration of the triiodide ion required 823 µL of 0.0988 M thiosulfate. 2 S₂03²1331+S406² Calculate the total number of serine + threoine residues per molecule of protein.

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