Match List-$I$ with List-$II$.
List-$I$ (Biopolymer) List-$II$ (Monomer) $A$. Starch $I$. nucleotide $B$. Cellulose $II$. $\alpha$-glucose $C$. Nucleic acid $III$. $\beta$-glucose $D$. Protein $IV$. $\alpha$-amino acid
Choose the correct answer from the options given below:
| List-$I$ (Biopolymer) | List-$II$ (Monomer) |
|---|---|
| $A$. Starch | $I$. nucleotide |
| $B$. Cellulose | $II$. $\alpha$-glucose |
| $C$. Nucleic acid | $III$. $\beta$-glucose |
| $D$. Protein | $IV$. $\alpha$-amino acid |
Choose the correct answer from the options given below:
- A$A-II, B-I, C-III, D-IV$
- B$A-IV, B-II, C-I, D-III$
- C$A-I, B-III, C-IV, D-II$
- D$A-II, B-III, C-I, D-IV$
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The correct observation in the following reactions is:
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$\text{Sucrose}$ $\xrightarrow[\text{Cleavage (Hydrolysis)}]{\text{Glycosidic bond}} A + B$ $\xrightarrow[\text{reagent}]{\text{Seliwanoff's}} ?$
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View SolutionIdentify the correct conversion during acidic hydrolysis from the following $:$
$(A)$ Starch gives galactose.
$(B)$ Cane sugar gives an equal amount of glucose and fructose.
$(C)$ Milk sugar gives glucose and galactose.
$(D)$ Amylopectin gives glucose and fructose.
$(E)$ Amylose gives only glucose.
Choose the correct answer from the options given below $:$
$(A)$ Starch gives galactose.
$(B)$ Cane sugar gives an equal amount of glucose and fructose.
$(C)$ Milk sugar gives glucose and galactose.
$(D)$ Amylopectin gives glucose and fructose.
$(E)$ Amylose gives only glucose.
Choose the correct answer from the options given below $:$
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