Chapter 7, Problem 8P

Answers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book.

Determining the Branch Points and Reducing Ends of Amylopectin 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-μmol of 2,3-dimethylgluetose and 0.4 μmol of 1,2,3,6- letramethylglucose.

1. What fraction of the total residues are branch points?
2. I low many reducing ends does this sample of amylopectin have?

Interpretation Introduction

(a)

Interpretation:

The fraction of the total residues that are branch points in the structure of amylopectin is to be calculated.

Concept Introduction:

The simplest hydrolyzed form that is obtained from the carbohydrates is known as monosaccharide. Polysaccharides are those types of sugars which contain more than ten units of monosaccharides.

Amylopectin is a polysaccharide and is one of the two forms of starch. It is composed of long chain of glucose attached by the $\alpha \left(1\to 4\right)$ glycosidic linkage and has $\alpha \left(1\to 6\right)$ branches with each $24$ to $30$ of glucose units.

Answer to Problem 8P

The fraction of the total residues that are branch points in the structure of amylopectin is $0.0407$ .

Explanation of Solution

The given mass of amylopectin sample is $0.2\text{g}$ .

The given amount of $2,3-\text{dimethylglucose}$ is $50\mu \text{mol}$ .

The conversion of $\mu \text{mol}$ into $\text{mol}$ is done as,

  $1\mu \text{mol}={10}^{-6}\text{mol}$

Thus, the moles of $2,3-\text{dimethylglucose}$ becomes $50\times {10}^{-6}\text{mol}$ .

The given amount of $1,2,3,6-\text{tetramethylglucose}$ is $0.4\mu \text{mol}$ or $0.4\times {10}^{-6}\text{mol}$ .

The molecular weight of glucose is $180\text{g/mol}$ .

The molecular weight of water is $18\text{g/mol}$ .

If glucose unit losses a water molecule while forming a glycosidic linkage in amylopectin, then, the molecular weight of glucose in amylopectin is $180-18=162\text{g/mol}$ .

The moles of a substance is calculated by the equation as,

  $\text{Moles}=\frac{\text{Mass}\left(\text{g}\right)}{\text{Molar mass}}$

Substitute the value of mass and molar mass of the sample in the above equation.

  $\begin{array}{c}\text{Moles}=\frac{0.2\text{g}}{162\text{g/mol}}\\ =0.00123\text{mol}\end{array}$

The moles of $2,3-\text{dimethylglucose}$ which are formed by $0.00123\text{mol}$ of glucose are the branch points of amylopectin sample. Thus, the fraction of the total residues that are branch points in the structure of amylopectin are calculated as,

  $\text{Fraction}=\frac{\text{Moles}\text{of}2,3-\text{dimethylglucose}}{\text{Moles}\text{of}\text{glucose}\text{in}\text{amylopectin}}$

Substitute the values of moles of $2,3-\text{dimethylglucose}$ and glucose in the above equation.

  $\begin{array}{c}\text{Fraction}=\frac{50\times {10}^{-6}\text{mol}}{0.00123\text{mol}}\\ =0.0407\end{array}$

Thus, the fraction of the total residues that are branch points in the structure of amylopectin is $0.0407$ .

Interpretation Introduction

(b)

Interpretation:

The total number of reducing ends possessed by the sample of amylopectin is to be calculated.

Concept Introduction:

The simplest hydrolyzed form that is obtained from the carbohydrates is known as monosaccharide. Polysaccharides are those types of sugars which contain more than ten units of monosaccharides.

Amylopectin is a polysaccharide and is one of the two forms of starch. It is composed of long chain of glucose attached by the $\alpha \left(1\to 4\right)$ glycosidic linkage and has $\alpha \left(1\to 6\right)$ branches with each $24$ to $30$ of glucose units.

Answer to Problem 8P

The total number of reducing ends possessed by the sample of amylopectin is $\text{2}\text{.41}\times {10}^{17}\text{molecules}$ .

Explanation of Solution

The given amount of $1,2,3,6-\text{tetramethylglucose}$ is $0.4\mu \text{mol}$ or $0.4\times {10}^{-6}\text{mol}$ .

The moles of the reducing ends is calculated by the moles of $1,2,3,6-\text{tetramethylglucose}$ which is formed by the methylation of carbon at 1 and $6$ of $2,3-\text{dimethylglucose}$ .

If one mole of compounds possesses $6.022\times {10}^{23}\text{molecules/mol}$ of the reducing ends then the number of molecules possessed by $0.4\times {10}^{-6}\text{mol}$ mole of $1,2,3,6-\text{tetramethylglucose}$ is,

  $\begin{array}{c}\text{Number}\text{of}\text{reducing}\text{ends}=0.4\times {10}^{-6}\text{mol}\times 6.022\times {10}^{23}\text{molecules/mol}\\ =\text{2}\text{.41}\times {10}^{17}\text{molecules}\end{array}$

Thus, the total number of reducing ends possessed by the sample of amylopectin is $\text{2}\text{.41}\times {10}^{17}\text{molecules}$ .