# Question #9056d

Feb 4, 2016

I count four different trisaccharides that can be produced by the incomplete hydrolysis of amylopectin.

#### Explanation:

Amylopectin is a highly branched polymer of glucose. Its partial structure is

I have labelled five of the glucose units as $\text{A, B, C, D}$, and $\text{E}$.

The possible trisaccharide combinations are $\text{ABC, DEB, EBC}$, and $\text{ABE}$.

α-Maltotriose (Rings $\text{A, B, C}$)

α-Maltotriose is formed from rings $\text{A, B}$, and $\text{C}$.

It consists of three glucose units joined by α-(1→4) linkages.

α-Isomaltotriose

Isomaltose is formed from rings $\text{D, E}$, and $\text{B}$.

It consists of three glucose units joined by an α-(1→4) and an α-(1→6) linkage.

α-Panose

α-panose is formed from rings $\text{E, B}$, and $\text{C}$.

α-Isopanose is formed from rings $\text{A, B}$, and $\text{E}$.
The $\text{A}$ and $\text{B}$ rings are joined by an α-(1-4) linkage, while the $\text{E}$ and $\text{B}$ rings are joined by an α-(1→6) linkage.