A 0.410 g sample of amylopectin was analyzed to determine the fraction of the total glucose residues that are branch points in the structure. The sample was exhaustively methylated and then digested, yielding 111 μmol of 2,3-dimethylglucose and 0.891 μmol of 1,2,3,6-tetramethylglucose.

What percentage of the total residues are branch points?

(The molecular weight of a glucose residue is C₆H₁₂O₆ minus H₂O = 162 g/mol.)

= ----- %

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The number of reducing ends in amylopectin is given by the moles of 1236tetramethylglucose So we wil... View the full answer