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NCERT Exemplar · Class 12 · Chemistry

Chapter 14: Biomolecules

Practice each question, then open the answer or worked solution.

II. Multiple Choice Questions (Type-II) Note: In the following questions, two or more options may be correct.

31 questions
Q30Written answer

How do you explain the presence of all the six carbon atoms in glucose in a straight chain?

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Worked solution

When glucose is heated for a prolonged time with HI, it forms n-hexane, suggesting that all the six carbon atoms are linked in a straight chain.

Q31Written answer

In nucleoside, a base is attached at the 1C position of the sugar moiety. A nucleotide is formed by linking the phosphoric acid unit to the sugar unit of a nucleoside. At which position of the sugar unit is the phosphoric acid linked in a nucleoside to give a nucleotide?

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Worked solution

A nucleoside is formed when a nitrogenous base is attached to a 1’ position of a five-carbon sugar. Phosphoric acid is linked to the 5’ carbon of the sugar in a nucleoside molecule to give a nucleotide molecule.

Q32Written answer

Name the linkage connecting monosaccharide units in polysaccharides.

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Worked solution

The monosaccharide units in polysaccharides are linked by glycosidic bonds. A glycosidic linkage is when an oxide linkage is formed between two monosaccharide units with the loss of a water molecule.

Q33Written answer

Under what conditions is glucose converted to gluconic and saccharic acid?

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Worked solution

Glucose is converted to gluconic acid, which is a six-carbon carboxylic acid, on treatment with a mild oxidizing agent like Br2 water. Glucose is converted to saccharic acid, which is a dicarboxylic acid, on treatment with nitric acid.

Q34Written answer

Monosaccharides contain carbonyl group and hence are classified as aldose or ketose. The number of carbon atoms present in the monosaccharide molecule is also considered for classification. In which class of monosaccharide will you place fructose?

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Worked solution

The molecular formula of fructose is C6H12O6. It also contains a ketone group; hence it is placed in the class of ketohexoses. Monosaccharides are classified depending on the number of carbon atoms the molecule contains.

Q35Written answer

The letters ‘D’ or ‘L’ before the name of a stereoisomer of a compound indicates

the correlation of the configuration of that particular stereoisomer. This refers to

their relationship with one of the isomers of glyceraldehyde. Predict whether the

following compound has ‘D’ or ‘L’ configuration.

NCERT Exemplar Solutions of Class 12 Chemistry Chapter 14 Biomolecules-9

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Worked solution

The –OH group is linked on the left side of the C5 carbon atom. Hence, the given compound has ‘L’ configuration.

Q36Written answer

Aldopentoses named as ribose and 2-deoxyribose are found in nucleic acids. What is their relative configuration?

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Worked solution

The configuration of both the aldopentoses is D-configuration. Ribose is named β-D-ribose, and 2-deoxyribose is β-D-2-deoxyribose.

Q37Written answer

Which sugar is called invert sugar? Why is it called so?

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Worked solution

Sucrose is also known as invert sugar. It is a naturally occurring sugar derived from sugarcane and sugarbeet. The hydrolysis of sucrose brings about a change in the sign of rotation, from Dextro (+) to laevo (–), and thus the product is named as invert sugar.

Q38Written answer

Amino acids can be classified as α-, β-, -, δ- and so on depending upon the relative position of the amino group concerning the carboxyl group. Which type of amino acids forms polypeptide chain in proteins?

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Worked solution

α-amino acids, alpha-amino acids, where the amino acid is linked to the α-carbon in the molecule, are the type of amino acids which form a polypeptide chain.

Q39Written answer

α-Helix is a secondary structure of proteins formed by twisting the polypeptide chain into right-handed screw-like structures. Which type of interaction is responsible for making the a-helix structure stable?

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Worked solution

A stable α-Helix is formed as a right-handed screw helix structure as the –NH group of each amino acid residue hydrogen is bonded to the –C=O of an adjacent turn of the helix.