Chapter 7, Problem 27P

(a1)

Summary Introduction

To explain: The basis of the procedure for determining the number of $\left(\alpha 1\to 6\right)$ branch proteins in amylopectin.

Introduction:

Carbohydrates are present in different forms. The carbohydrate made up of only one kind of sugar is monosaccharide, if formed of two types of sugars it is called disaccharide and if made up of many types of sugar it is polysaccharide.

Explanation of Solution

The carbon number 6 of glucose residue present in $\left(\alpha 1\to 6\right)$ branch proteins of amylopectin will be protected from methylation and hydrolysis process. This is because hydroxyl group present in carbon 6 is involved in joining another molecule of glucose present in amylopectin through glycosidic linkage. Thus, methylation process will only occur at carbon 2 and 3 of branched amylopectin unit. Hence, the branched residue of amylopectin would yield 2,3-di-O-methylglucose.

Pictorial representation: Fig 1:  represents the structure of 2,3-di-O-methylglucose.

Fig 1: Structure of 2,3-di-O-methylglucose.

(a2)

Summary Introduction

To explain: What happens to the unbranched glucose residues in amylopectin during the methylation and hydrolysis procedure.

Introduction:

Carbohydrates are present in different forms. The carbohydrate made up of only one kind of sugar is monosaccharide, if formed of two types of sugars it is called disaccharide and if made up of many types of sugar it is polysaccharide.

Explanation of Solution

The polysaccharides are large polymer molecules composed of monosaccharides unit. Starch is an example of polysaccharide that are made up of amylose and amylopectin molecule. Amylose is a linear chain of starch that are compose of glucose units. Each glucose unit in amylose are joined through α-glycosidic bond. Whereas, amylopectin is branched structure of starch that contain β-glycosidic bond.

In case of unbranched residue, methylation and hydrolysis process would yield 2,3,6-tri-O-methylglucose, as the hydroxyl group at carbon 2, 3, and 6 are free.

(b)

Summary Introduction

To calculate: The percentage of glucose residues in amylopectin contained $\left(\text{α1}\to \text{6}\right)$ branch.

Introduction:

The polysaccharides are large polymer molecules composed of monosaccharides unit. Starch is an example of polysaccharide that are made up of amylose and amylopectin molecule. Amylose is a linear chain of starch that are compose of glucose units. Each glucose unit in amylose are joined through α-glycosidic bond. Whereas, amylopectin is branched structure of starch that contain β-glycosidic bond.

Given: 258 mg or $258\times {10}^{-3}\text{g}$ of amylopectin yield 12.4 mg or $12.4\times {10}^{-3}\text{g}$ of 2,3-di-O-methylglucose.

Explanation of Solution

The average molecular weight of glucose is 162g/mol, then $258\times {10}^{-3}\text{g}$ of amylopectin will contain:

$\begin{array}{c}\text{Total}\text{weight}=\frac{258\times {10}^{-3}}{162\text{g}/\text{mol}}\\ =1.59\times {10}^{-3}\text{g}/\text{mol}\text{of}\text{glucose}\end{array}$

Thus, the total weight of amylopectin in $258\times {10}^{-3}\text{g}$ is $1.59\times {10}^{-3}\text{g}/\text{mol}\text{of}\text{glucose}$.

The weight of branched residue of 2,3-di-O-methylglucose in $12.4\times {10}^{-3}\text{g}$

$\begin{array}{c}\text{Weight}\text{of}\text{branched}\text{residue}=\frac{12.4\times {10}^{-3}}{208\text{g}/\text{mol}}\\ =5.96\times {10}^{-5}\text{g}/\text{mol}\text{of}\text{glucose}\end{array}$

Thus, the average molecular weight of glucose in 2,3-di-O-methylglucose is $5.96\times {10}^{-5}\text{g}/\text{mol}\text{of}\text{glucose}$.

The percentage of glucose residue glucose residue present in amylopectin having $\left(\alpha 1\to 6\right)$ branch and yield 2,3-di-O-methylglucose can be calculated as:

$\text{Percentage}\text{of}\text{glucose}=\frac{\text{weight}\text{of}\text{branched}\text{residue}}{\text{Total}\text{weight}\text{of}\text{amylopectin}}\times 100$

The weight of branched residue is $5.96\times {10}^{-5}\text{g}/\text{mol}\text{of}\text{glucose}$, and total weight of amylopectin is $1.59\times {10}^{-3}\text{g}/\text{mol}\text{of}\text{glucose}$.

Now, substitute the value in the above given equation and calculate the percentage of glucose residue in amylopectin.

$\begin{array}{c}\text{Percentage}\text{of}\text{glucose}=\frac{\left(5.96\times {10}^{-5}\right)}{\left(1.59\times {10}^{-3}\right)}\times 100\\ =0.0375\times 100\\ =3.75\%\end{array}$

Thus, the percentage of glucose residue in amylopectin having $\left(\text{α1}\to \text{6}\right)$ branch is 3.75 %.

Conclusion

The percentage of glucose residue in amylopectin is 3.75 %.