4
$\begingroup$

I was wondering if anyone knows what the degree of curvature is or bond angle between the $\alpha$ glucose molecules in glycogen is. I know that glycogen/amylose/amylopectin have a general curved backbone (as well as branches in glycogen and amylopectin). This is a consequence of the 1,4-glycosidic bond between $\alpha$ glucose molecules. On the other hand cellulose has straight molceules, which is a consequnce of the 1,4-glycosiodic bond between $\beta$ molecules with alternating orientation. However I cannot find anywhere what the degree of this curvature is in glycogen backbone, or the bond angle between molecules in $\alpha$ 1,4-glycosidic bond?

$\endgroup$
3
$\begingroup$

Interesting question. Here, the question actually comes down to how the bond actually looks vs how we depict the bond. In detail, see this image of maltose from here:

maltose

Pay attention to the geometry of the glycosidic bond. From how it is depicted, one would conclude that it is straight in geometry. But how it actually looks is like this:

maltose actual

Now, the obvious conclusion is that the glycosidic bond does have an angle.

Now, with introduction over, lets come to the main point. After searching a lot, I found the value of this angle. It is 109°, much closer to the general tetrahedral angle (109.5°) of sp3 hybridized oxygen atom (deviation could be due to steric hindrance). See this article:

amylose and amylopectin

300 alpha-glucose units, bonded by $\alpha$-1,4 glycosidic linkages, making a 109 degree angle between 2 C-O bonds causing the polymer to twist into a helix. -OH group of C2 projects out into the middle and form H bonds with each other, stabilising the shape.

Yes, the glucose molecules in amylose form intermolecular hydrogen bonds which stabilize the molecule. See this article and in the diagram, notice that the hydrogen bond is formed between adjacent glucose units (to the left and right of the glycosidic oxygen atom):

amylose h-bonds

Although the $\alpha$-(1$\rightarrow$4) links are capable of relatively free rotation around the ($\Phi$) phi and ($\Psi$) psi torsions, hydrogen bonding between the O3' and O2 oxygen atoms of sequential residues tends to encourage a helical conformation. These helical structures are relatively stiff and may present contiguous hydrophobic surfaces.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.