MCQs Based On Class 12 Chemistry Chapter 14 Biomolecules
MCQ Questions for Class 12 Chemistry are prepared by the subjects experts according to the latest pattern. These MCQs are very important for students who wants to score well in CBSE Board Exam. These MCQs on chapter 14 Biomolecules will also help students in understanding the concepts very well.
Q.1. Glycogen is a branched chain polymer of α-D-glucose units in which chain is formed by C1—C4 glycosidic linkage whereas branching occurs by the formation of C1-C6 glycosidic linkage. Structure of glycogen is similar to _____________________ .
(i) Amylose
(ii) Amylopectin
(iii) Cellulose
(iv) Glucose
Answer
Answer: (ii) AmylopectinExplanation: Polysaccharides contain a large number of monosaccharide units joined together by glycosidic linkages. These are the most commonly encountered carbohydrates in nature. Amylopectin is insoluble in water and constitutes about 80-85% of starch. It is a branched chain polymer of alpha-D-glucose units in which chain is formed by C1-C4 glycosidic linkage whereas branching occurs by C1-C6 glycosidic linkage.
Q.2. Which of the following polymer is stored in the liver of animals?
(i) Amylose
(ii) Cellulose
(iii) Amylopectin
(iv) Glycogen
Answer
Answer: (iv) GlycogenExplanation: The carbohydrates are stored in animal body as glycogen. It is also known as animal starch because its structure is similar to amylopectin and is rather more highly branched. It is present in liver, muscles and brain.
Q.3. Sucrose (cane sugar) is a disaccharide. One molecule of sucrose on hydrolysis gives
(i) 2 molecules of glucose
(ii) 2 molecules of glucose + 1 molecule of fructose
(iii) 1 molecule of glucose + 1 molecule of fructose
(iv) 2 molecules of fructose
Answer
Answer: (iii) 1 molecule of glucose + 1 molecule of fructoseExplanation: One of the common disaccharides is sucrose which on hydrolysis gives equimolar mixture of D-(+)-glucose and D-(-) fructose.
C12H22O11 (Sucrose) + H2O → C6H12O6(D-(+)-glucose) + C6H12O6 (D-(-) fructose)
Q.4. Which of the following pairs represents anomers?
Answer
Answer: (iii)Explanation: The two cyclic hemiacetal forms of glucose differ only in the configuration of the hydroxyl group at Cl, called anomeric carbon (the aldehyde carbon before cyclisation). Such isomers, i.e., alpha-form and beta-form, are called anomers.
Q.5. Proteins are found to have two different types of secondary structures viz. α-helix and β-pleated sheet structure. α-helix structure of protein is stabilised by:
(i) Peptide bonds
(ii) van der Waals forces
(iii) Hydrogen bonds
(iv) Dipole-dipole interactions
Answer
Answer: (iii) Hydrogen bondsExplanation: α-helix and β-pleated sheet structure: These structures arise due to the regular folding of the backbone of the polypeptide chain due to hydrogen bonding between >C—O and —NH— group of the peptide bond.
α-Helix is one of the most common ways in which a polypeptide chain forms all possible hydrogen bond by twisting into a right-handed screw (helix) with the -NH group of each amino acid residue hydrogen bonded to the >C=O of an adjacent turn of the helix.
Q.6. In disaccharides, if the reducing groups of monosaccharides i.e. aldehydic or ketonic groups are bonded, these are non-reducing sugars. Which of the following disaccharide is a non-reducing sugar?
Answer
Answer: (ii)Explanation: One of the common disaccharides is sucrose which on hydrolysis gives equimolar mixture of D-(+)-glucose and D-(-) fructose. These two monosaccharides are held together by a glycosidic linkage between Cl of α-glucose and C2 of beta-fructose. Since the reducing groups of glucose and fructose are involved in glycosidic bond formation, sucrose is a non-reducing sugar.
Q.7. Which of the following acids is a vitamin?
(i) Aspartic acid
(ii) Ascorbic acid
(iii) Adipic acid
(iv) Saccharic acid
Answer
Answer: (ii) Ascorbic acidExplanation: Vitamin C is also known as Ascorbic acid.
Q.8. Dinucleotide is obtained by joining two nucleotides together by phosphodiester linkage. Between which carbon atoms of pentose sugars of nucleotides are these linkages present?
(i) 5′ and 3′
(ii) 1′ and 5′
(iii) 5′ and 5′
(iv) 3′ and 3′
Answer
Answer: (i) 5′ and 3′Explanation: Nucleotides are joined together by phosphodiester linkage between 5’ and 3’ carbon atoms of the pentose sugar.
Q.9. Nucleic acids are the polymers of
(i) Nucleosides
(ii) Nucleotides
(iii) Bases
(iv) Sugars
Answer
Answer: (ii) NucleotidesExplanation: Nucleic acids are long chain polymers of nucleotides, so they are also called polynucleotides.
Q.10. Which of the following statements is not true about glucose?
(i) It is an aldohexose.
(ii) On heating with HI it forms n-hexane.
(iii) It is present in furanose form.
(iv) It does not give 2,4-DNP test.
Answer
Answer: (iii)Explanation: Fructose is present in furanose form.
Q.11. Each polypeptide in a protein has amino acids linked with each other in a specific sequence. This sequence of amino acids is said to be–
(i) primary structure of proteins.
(ii) secondary structure of proteins.
(iii) tertiary structure of proteins.
(iv) quaternary structure of proteins.
Answer
Answer: (i) primary structure of proteins.Explanation: Sequence of amino acids is said to be primary structure of proteins.
Q.12. DNA and RNA contain four bases each. Which of the following bases is not present in RNA?
(i) Adenine
(ii) Uracil
(iii) Thymine
(iv) Cytosine
Ans. (iii) Thymine
Answer
Answer: (iii) ThymineExplanation: DNA contains four bases viz. adenine (A), guanine (G), cytosine (C) and thymine (T). RNA also contains four bases; the first three bases are same as in DNA but the fourth one is uracil (U).
Q.13. Which of the following B group vitamins can be stored in our body?
(i) Vitamin B1
(ii) Vitamin B2
(iii) Vitamin B6
(iv) Vitamin B12
Answer
Answer: (iv) Vitamin B12Explanation: Water soluble vitamins must be supplied regularly in diet because they are readily excreted in urine and cannot be stored (except vitamin B12) in our body.
Q.14. Which of the following bases is not present in DNA?
(i) Adenine
(ii) Thymine
(iii) Cytosine
(iv) Uracil
Answer
Answer: (iv) UracilExplanation: Uracil is present in RNA but not in DNA.
Q15. Three cyclic structures of monosaccharides are given below which of these are anomers.
(i) I and II
(ii) II and III
(iii) I and III
(iv) III is anomer of I and II
Answer
Answer: (i) I and IIExplanation: This behavior could not be explained by the open chain structure (I) for glucose. It was proposed that one of the —OH groups may add to the —CHO group and form a cyclic hemiacetal structure. It was found that glucose forms a six-membered ring in which —OH at C-5 is involved in ring formation. This explains the absence of —CHO group and also existence of glucose in two forms as shown below.
Q.16. Which of the following reactions of glucose can be explained only by its cyclic structure?
(i) Glucose forms pentaacetate.
(ii) Glucose reacts with hydroxylamine to form an oxime.
(iii) Pentaacetate of glucose does not react with hydroxylamine.
(iv) Glucose is oxidised by nitric acid to gluconic acid.
Answer
Answer: (iii) Pentaacetate of glucose does not react with hydroxylamine.Explanation: The pentaacetate of glucose does not react with hydroxylamine indicating the absence of free —CHO group.
Q.17. Optical rotations of some compounds along with their structures are given Below. Which of them have D configuration?
(i) I, II, III
(ii) II, III
(iii) I, II
(iv) III
Answer
Answer: (i) I, II, IIIExplanation: All those compounds which can be chemically correlated to (+) isomer of glyceraldehyde are said to have D-configuration whereas those which can be correlated to (-) isomer of glyceraldehyde are said to have L-configuration.
Q.18. Structure of a disaccharide formed by glucose and fructose is given below. Identify anomeric carbon atoms in monosaccharide units.
(i) ‘a’ carbon of glucose and ‘a’ carbon of fructose.
(ii) ‘a’ carbon of glucose and ‘e’ carbon of fructose.
(iii) ‘a’ carbon of glucose and ‘b’ carbon of fructose.
(iv) ‘f’ carbon of glucose and ‘f’ carbon of fructose.
Answer
Answer: (iii) ‘a’ carbon of glucose and ‘b’ carbon of fructose.Explanation: Two monosaccharides are held together by a glycosidic linkage between Cl of α-glucose and C2 of β-fructose.
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