Question:easy

Glycosidic linkage in maltose is present between

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Maltose always has \(\alpha(1\rightarrow4)\) glycosidic linkage between two glucose units.
Updated On: Jun 15, 2026
  • C-1 of \(\alpha\)-D-glucose and C-4 of \(\alpha\)-D-glucose
  • C-1 of \(\alpha\)-D-glucose and C-4 of \(\beta\)-D-galactose
  • C-1 of \(\beta\)-D-glucose and C-4 of \(\alpha\)-D-glucose
  • C-1 of \(\beta\)-D-glucose and C-4 of \(\beta\)-D-glucose
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Identify the sugar units in maltose.
Maltose is a disaccharide built from two D-glucose units joined by a glycosidic bond formed by loss of water between two hydroxyl groups.
Step 2: Recall the anomeric forms.
In maltose both glucose units are in the $\alpha$-D-glucose form, unlike lactose which contains galactose, or cellobiose which uses the $\beta$ form.
Step 3: Locate the linkage carbons.
The bond joins the anomeric carbon C-1 of one $\alpha$-D-glucose to the C-4 hydroxyl of the second $\alpha$-D-glucose.
Step 4: Name the linkage.
This is an $\alpha(1\rightarrow4)$ glycosidic linkage between two $\alpha$-D-glucose units.
Step 5: Eliminate the wrong options.
Options mentioning galactose or a $\beta$-D-glucose anomeric carbon do not describe maltose, so they are rejected.
Step 6: Choose the option.
C-1 of $\alpha$-D-glucose and C-4 of $\alpha$-D-glucose matches option (1).
\[ \boxed{\text{C-1 of }\alpha\text{-D-glucose and C-4 of }\alpha\text{-D-glucose (Option 1)}} \]
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