From the abbreviated name of the compound Gal($\beta$1 $\rightarrow$ 4)Glc, we know that:
A) C-4 of glucose is joined to C-1 of galactose by a glycosidic bond.
B) the compound is a D-enantiomer.
C) the galactose residue is at the reducing end.
D) the glucose is in its pyranose form.
E) the glucose residue is the $\beta$ anomer.

Question

From the abbreviated name of the compound Gal($\beta$1 $\rightarrow$  4)Glc, we know that:
A) C-4 of glucose is joined to C-1 of galactose by a glycosidic bond.
B) the compound is a D-enantiomer.
C) the galactose residue is at the reducing end.
D) the glucose is in its pyranose form.
E) the glucose residue is the $\beta$ anomer.

Quizplus · Accepted Answer

The abbreviation Gal(F1F1F1S1F1F1F104)Glc indicates that the galactose residue is linked to the glucose residue via a glycosidic bond between carbons 1 and 4, respectively. The notation F1F1F1S1 indicates the directionality of the glycosidic bond, with the reducing end pointing towards the S1 position. The abbreviation F1F1F1S1F1F1F10 indicates the anomeric configuration of the glucose residue, which is not relevant to the location of the glycosidic bond. The enantiomeric configuration and pyranose/furanose form of the monosaccharides are not indicated by the abbreviation.

From the Abbreviated Name of the Compound Gal β\betaβ 1 →\rightarrow→ 4)Glc, We Know That: A) C-4 of Glucose Is Joined

From the Abbreviated Name of the Compound Gal $\beta$ 1 $\rightarrow$ 4)Glc, We Know That:
A) C-4 of Glucose Is Joined