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Chapter 7: Problem 8

A 0.2 -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 \(50 \mu \mathrm{mol}\) of 2,3 -dimethylglucose and \(0.4 \mu \mathrm{mol}\) of 1,2,3,6 tetramethylglucose. a. What fraction of the total residues are branch points? b. How many reducing ends does this sample of amylopectin have?

Short Answer

Expert verified
a) The fraction of the total residues that are branch points is approximately 32.5%. b) This sample of amylopectin has 0.4 µmol of reducing ends.

Step by step solution

01

Calculation of Total Residues

Determine the total number of glucose residues in the sample. Since 1 glucose molecule has a molar mass of about 162 g/mol and the sample consists of 0.2 g of amylopectin, we have a total of (0.2 g) / (162 g/mol) = 1.23 µmol of glucose residues.
02

Finding the fraction of Branch Points

Find out the fraction of total residues that are branch points. As given, there are \(0.4 \mu \mathrm{mol}\) of 1,2,3,6 -tetramethylglucose residues, which are the branch points. Therefore, the fraction of branch points is \(0.4 \mu \mathrm{mol}\) / 1.23 µmol = 0.325. This means that around 32.5% of the total residues are branch points.
03

Calculation of Reducing Ends

Identify the number of reducing ends. Since each branch point molecule contributes one reducing end, the number of reducing ends is equal to the number of branch points, which is 0.4 µmol.

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