Biomolecules and Polymers

Hydrolysis of sucrose gives :

Which one of the following compounds contains $\beta$-C₁-C₄ glycosidic linkage?

Given below are two statements. Statement I : Maltose has two $\alpha$-D-glucose units linked at C₁ and C₄ and is a reducing sugar. Statement II : Maltose has two monosaccharides : $\alpha$-D-glucose and $\beta$-D-glucose linked at C₁ and C₆ and it is a non-reducing sugar. In the light of the above statements, choose the correct answer from the options given below :

A polysaccharide 'X' on boiling with dil H₂SO₄ at 393 K under 2-3 atm pressure yields 'Y'. 'Y' on treatment with bromine water gives gluconic acid. 'X' contains $\beta$-glycosidic linkages only. Compound 'X' is :

Glycosidic linkage between C₁ of $\alpha$-glucose and C₂ of $\beta$-fructose is found in

Animal starch is the other name of

Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Amylose is insoluble in water. Reason R: Amylose is a long linear molecule with more than 200 glucose units. In the light of the above statements, choose the correct answer from the options given below.

Match List I with List II .tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-7btt{border-color:inherit;font-weight:bold;text-align:center;vertical-align:top} .tg .tg-amwm{font-weight:bold;text-align:center;vertical-align:top} .tg .tg-0pky{border-color:inherit;text-align:left;vertical-align:top} List I List II Test Functional group / Class of Compound A. Molisch's Test I. Peptide B. Biuret Test II. Carbohydrate C. Carbylamine Test III. Primary amine D. Schiff's Test IV. Aldehyde Choose the correct answer from the options given below :

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Ketoses give Seliwanoff's test faster than Aldoses. Reason (R) : Ketoses undergo $\beta$-elimination followed by formation of furfural. In the light of the above statements, choose the correct answer from the options given below :

Match List I with List II .tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-c3ow{border-color:inherit;text-align:center;vertical-align:top} .tg .tg-7btt{border-color:inherit;font-weight:bold;text-align:center;vertical-align:top} .tg .tg-0pky{border-color:inherit;text-align:left;vertical-align:top} List - I List - II (A) Glucose/$\mathrm{NaHCO_3/\Delta}$ (I) Gluconic acid (B) $\text { Glucose } / \mathrm{HNO}_3$ (II) No reaction (C) $\text { Glucose } / \mathrm{HI} / \Delta$ (III) n-hexane (D) Glucose/Bromine water (IV) Saccharic acid Choose the correct answer from the options given below: