Which one of the following compounds contains $$\beta - C_1 - C_4$$ glycosidic linkage?

A glycosidic linkage joins the anomeric carbon $$C_1$$ of one monosaccharide to some carbon (commonly $$C_4$$ or $$C_6$$) of a second monosaccharide after the elimination of water.

The notation $$\beta - C_1 - C_4$$ means: the anomeric carbon $$C_1$$ of the first sugar is in the $$\beta$$-configuration and it is linked to the $$C_4$$ carbon of the second sugar.

Let us recall the main disaccharides/polysaccharide units mentioned in the options and the type of link each contains:

Option A (Sucrose): $$\alpha\text{-}D\text{-glucose}\,(C_1) - \beta\text{-}D\text{-fructose}\,(C_2)$$ linkage, i.e. $$\alpha\,1 \rightarrow 2\,\beta$$. No $$C_4$$ carbon is involved, so it is not $$\beta\,1 \rightarrow 4$$.

Option B (Amylose): the unbranched component of starch. Each glucose unit is joined by $$\alpha\,1 \rightarrow 4$$ glycosidic bonds (not $$\beta$$).

Option C (Lactose): composed of $$\beta\text{-}D\text{-galactose}$$ and $$\beta\text{-}D\text{-glucose}$$. The linkage is $$\beta\,1 \rightarrow 4$$, joining $$C_1$$ of galactose in the $$\beta$$-form to $$C_4$$ of glucose.

Option D (Maltose): two $$\alpha\text{-}D\text{-glucose}$$ units connected by an $$\alpha\,1 \rightarrow 4$$ bond (not $$\beta$$).

Only Option C (Lactose) contains the required $$\beta - C_1 - C_4$$ glycosidic linkage.

Hence, the correct choice is Option C (Lactose).