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Biomolecules Multiple Choice Questions (Type-I)
- Glycogen is a branched chain polymer of α-D-glucose units in which chain is formed by C 1 —C 4 glycosidic linkage whereas branching occurs by the formation of C 1 -C 6 glycosidic linkage. Structure of glycogen is similar to __________. (i) Amylose (ii) Amylopectin (iii) Cellulose (iv) Glucose Ans. (ii) Explanation: 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 C 1 -C 4 glycosidic linkage whereas branching occurs by C 1 -C 6 glycosidic linkage.
- Which of the following polymer is stored in the liver of animals? (i) Amylose (ii) Cellulose (iii) Amylopectin (iv) Glycogen Ans. (iv) Explanation: 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.
- 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 Ans. (iii) Explanation: One of the common disaccharides is sucrose which on hydrolysis gives equimolar mixture of D-(+)-glucose and D-(-) fructose. 12 22 11 2 6 12 6 6 12 6 Sucrose D ( ) Glu cos e D ( ) Fructose
C H O H O C H O C H O
− + − − − −
- Which of the following pairs represents anomers?
Ans. (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.
- 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 Ans. (iii)
(i) Aspartic acid (ii) Ascorbic acid (iii) Adipic acid (iv) Saccharic acid Ans. (ii) Explanation: Vitamin C is also known as Ascorbic acid.
- 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′ Ans. (i) Explanation: Nucleotides are joined together by phosphodiester linkage between 5’ and 3’ carbon atoms of the pentose sugar.
- Nucleic acids are the polymers of ______________. (i) Nucleosides (ii) Nucleotides (iii) Bases (iv) Sugars Ans. (ii) Explanation: Nucleic acids are long chain polymers of nucleotides, so they are also called polynucleotides.
- 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. Ans. (iii) Explanation: Fructose is present in furanose form.
- 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. Ans. (i) Explanation: Sequence of amino acids is said to be primary structure of proteins.
- 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) Explanation: 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).
- Which of the following B group vitamins can be stored in our body? (i) Vitamin B 1 (ii) Vitamin B 2 (iii) Vitamin B 6 (iv) Vitamin B 12 Ans. (iv) Explanation: Water soluble vitamins must be supplied regularly in diet because they are readily excreted in urine and cannot be stored (except vitamin B 12 ) in our body.
- Which of the following bases is not present in DNA? (i) Adenine (ii) Thymine
Explanation: The pentaacetate of glucose does not react with hydroxylamine indicating the absence of free —CHO group.
- 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 Ans. (i) Explanation: 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.
- 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. Ans. (iii)
Explanation: Two monosaccharides are held together by a glycosidic linkage between Cl of α -glucose and C2 of β -fructose.
- Three structures are given below in which two glucose units are linked. Which of these linkages between glucose units are between C 1 and C 4 and which linkages are between C 1 and C 6?
(i) (A) is between C1 and C4, (B) and (C) are between C1 and C (ii) (A) and (B) are between C1 and C4, (C) is between C1 and C (iii) (A) and (C) are between C1 and C4, (B) is between C1 and C (iv) (A) and (C) are between C1 and C6, (B) is between C1 and C Ans. (iii)
(i) 2
NH
CH CH − CH − COOH
(ii) 2
(^2 2) | NH
HOOC − CH − CH − CH − COOH
(iii) H 2 N—CH 2 —CH 2 —CH 2 —COOH (iv) 2
(^2) | NH
HOOC − CH − CH − COOH
Ans. (ii), (iv) Explanation: Amino acids are classified as acidic, basic or neutral depending upon the relative number of amino and carboxyl groups in their molecule. Equal number of amino and carboxyl groups makes it neutral; more number of amino than carboxyl groups makes it basic and more carboxyl group as compared to amino groups makes it acidic.
24. Lysine,
2
2 (^2 ) 4 |
NH
H N − CH − CH − COOH is _______________.
(i) α-Amino acid (ii) Basic amino acid (iii) Amino acid synthesised in body (iv) β-Amino acid Ans. (i), (ii)
25. Which of the following monosaccharides are present as five membered cyclic structure (furanose structure)? (i) Ribose (ii) Glucose (iii) Fructose (iv) Galactose Ans. (i), (iii) Explanation: Fructose: It also exists in two cyclic forms which are obtained by the addition of —OH at C5 to the group. The ring, thus formed is a five-membered ring and is named as furanose with analogy to the compound furan. Furan is a five-membered cyclic compound with one oxygen and four carbon atoms. 26. In fibrous proteins, polypeptide chains are held together by ___________. (i) van der Waals forces (ii) disulphide linkage (iii) electrostatic forces of attraction (iv) hydrogen bonds Ans. (i), (ii) Explanation: In fibrous proteins, main forces which stabilize structures of proteins are disulphide linkages and van der Waals. 27. Which of the following are purine bases? (i) Guanine
(ii) Adenine (iii) Thymine (iv) Uracil Ans. (i), (ii)
28. Which of the following terms are correct about enzyme? (i) Proteins (ii) Dinucleotides (iii) Nucleic acids (iv) Biocatalysts Ans. (i), (iv) Explanation: Life is possible due to the coordination of various chemical reaction in living organisms. This occurs with the help of certain biocatalysts called enzymes. Almost all the enzymes are globular Protein.
Biomolecules Assertion and Reason Type
Note: In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (i) Assertion and reason both are correct statements and reason explains the assertion. (ii) Both assertion and reason are wrong statements. (iii) Assertion is correct statement and reason is wrong statement. (iv) Assertion is wrong statement and reason is correct statement. (v) Assertion and reason both are correct statements but reason does not explain assertion.
60. Assertion: D (+) – Glucose is dextrorotatory in nature. Reason: ‘D’ represents its dextrorotatory nature. Ans. (iii) Explanation: Glucose is correctly named as D(+)-glucose. ‘D’ before the name of glucose represents the configuration whereas ‘(+)’ represents dextrorotatory nature of the molecule. The letters ‘D’ or ‘L’ before the name of any compound indicate the relative configuration of a particular stereoisomer. 61. Assertion: Vitamin D can be stored in our body. Reason: Vitamin D is fat soluble vitamin. Ans. (i) Explanation: Fat soluble vitamins are soluble in fat and oils but insoluble in water. They can be stored in liver and adipose (fat storing) tissues. 62. Assertion: β-glycosidic linkage is present in maltose,
Reason: Maltose is composed of two glucose units in which C–1 of one glucose unit is linked to C–4 of another glucose unit. Ans. (iv) Explanation:
Maltose is composed of two α -D-glucose units in which C1 of one glucose (I) is linked to C4 of another glucose unit (II).
63. Assertion: All naturally occurring α-amino acids except glycine are optically active. Reason: Most naturally occurring amino acids have L-configuration. Ans. (v) Explanation: Except glycine, all other naturally occurring α -amino acids are optically active, since the α -carbon atom is asymmetric. These exist both in ‘D’ and ‘L’ forms. Most naturally occurring amino acids have L-configuration. L-Amino acids are represented by writing the —NH 2 group on left hand side. 64. Assertion: Deoxyribose, C 5 H 10 O 4 is not a carbohydrate. Reason: Carbohydrates are hydrates of carbon so compounds which follow C x (H 2 O)y formula are carbohydrates. Ans. (ii) Explanation: Complete hydrolysis of DNA (or RNA) yields a pentose sugar, phosphoric acid and nitrogen containing hetero-cyclic compounds (called bases). In DNA molecules, the sugar moiety is β-D-2- deoxyribose. 65. Assertion: Glycine must be taken through diet. Reason: It is an essential amino acid. Ans. (ii) Explanation: The amino acids, which can be synthesised in the body, are known as non- essential amino acids. Glycine is an example of non-essential amino acid. 66. Assertion: In presence of enzyme, substrate molecule can be attacked by the reagent effectively. Reason: Active sites of enzymes hold the substrate molecule in a suitable position. Ans. (i) Explanation: In presence of enzyme, substrate molecule can be attacked by the reagent effectively because active sites of enzymes hold the substrate molecule in a suitable position.
Ans. Fructose is an important ketohexose. It is obtained along with glucose by the hydrolysis of disaccharide, sucrose. Fructose also has the molecular formula C 6 H 12 O 6 and on the basis of its reactions it was found to contain a ketonic functional group at carbon number 2 and six carbons in straight chain as in the case of glucose.
35. The letters ‘D’ or ‘L’ before the name of a stereoisomer of a compound indicate the correlation of configuration of that particular stereoisomer. This refers to their relation with one of the isomers of glyceraldehyde. Predict whether the following compound has ‘D’ or ‘L’ configuration.
Ans. The letters ‘D’ or ‘L’ before the name of any compound indicate the relative configuration of a particular stereoisomer. This refers to their relation with a particular isomer of glyceraldehydes. D’ before the name of glucose represents the configuration whereas ‘(+)’ represents dextrorotatory nature of the molecule. It may be remembered that ‘D’ and ‘L’ have no relation with the optical activity of the compound. For assigning the configuration of monosaccharides, it is the lowest asymmetric carbon atom which is compared. The given compound has L-configuration.
36. Aldopentoses named as ribose and 2-deoxyribose are found in nucleic acids. What is their relative configuration? Ans. configuration assigned is O. Thus, robose is β-D-ribose Deoxyribose is β-D-2-deoxyribose. 37. Which sugar is called invert sugar? Why is it called so? Ans. Sucrose is dextrorotatory but after hydrolysis gives dextrorotatory glucose and
laevorotatory fructose. Since the laevorotation of fructose ( 92.4 )− 0 is more than dextrorotation of glucose ( 52.5 )+ 0 , the mixture is laevorotatory. Thus, hydrolysis of sucrose brings about a change in the sign of rotation, from dextro (+) to laevo (-) and the product is named as invert sugar.
38. Amino acids can be classified as α-, β-, γ-, δ- and so on depending upon the relative position of amino group with respect to carboxyl group. Which type of amino acids form polypetide chain in proteins? Ans. α -amino acids form polypeptide chain in proteins. 39. α-Helix is a secondary structure of proteins formed by twisting of polypeptide chain into right handed screw like structures. Which type of interactions are responsible for making the α-helix structure stable?
Ans. In α -helix, apolypeptide chain is stabilized by the formation of hydrogen bonds between
—NH— group of amino acids in one turn with the >C —O groups of amino acids belonging to adjacent turn.
40. Some enzymes are named after the reaction, where they are used. What name is given to the class of enzymes which catalyse the oxidation of one substrate with simultaneous reduction of another substrate. Ans. Oxidoreductase is class of enzymes which catalyse the oxidation of one substrate with simultaneous reduction of another substrate. 41. During curdling of milk, what happens to sugar present in it? Ans. When milk is curdeled, its sugar get oxidize to form lactic acid. 42. How do you explain the presence of five —OH groups in glucose molecule? Ans. Acetylation of glucose with acetic anhydride gives glucose pentaacetate which confirms the presence of five —OH groups. Since it exists as a stable compound, five —OH groups should be attached to different carbon atoms. 43. Why does compound (A) given below not form an oxime?
Ans. Glucose pentaacetate (structure A) doesn’t have a free —OH group at C1 and so can’t be converted to the open chain form to give —CHO group and hence doesn’t form the oxime.
44. Why must vitamin C be supplied regularly in diet?
energy. For example, activation energy for acid hydrolysis of sucrose is 6.22 kJ mol-1, while the activation energy is only 2.15 kJ mol-1^ when hydrolysed by the enzyme, sucrase.
50. How do you explain the presence of an aldehydic group in a glucose molecule? Ans. Glucose gets oxidised to six carbon carboxylic acid (gluconic acid) on reaction with a mild oxidising agent like bromine water. This indicates that the carbonyl group is present as an aldehydic group. 51. Which moieties of nucleosides are involved in the formation of phosphodiester linkages present in dinucleotides? What does the word diester in the name of linkage indicate? Which acid is involved in the formation of this linkage? Ans. When nucleoside is linked to phosphoric acid at 5-position of sugar moiety, it forms a nucleotide. Nucleotides are joined together by phosphodiester link between 5’ and 3’ carbon atoms of the pentose sugar.
Phospheric acid is involved in the formation of this linkage.
52. What are glycosidic linkages? In which type of biomolecules are they present? Ans. The two monosaccharides are joined together by an oxide linkage formed by the loss of a water molecule. Such a linkage between two monosaccharides are held together by a glycosidic linkage between C1 of α -glucose and C2 of β-fructose. Since the reducing
groups of glucose and fructose are involved in glycosidic bond formation, between two monosaccharide units through oxygen atom is called glycosidic linkage.
53. Which monosaccharide units are present in starch, cellulose and glucose and which linkages link these units? Ans. In starch and glycogen, glycosidic α -linkage is present. Cellulose is a straight chain polysaccharide composed only of β-D-glucose units which are joined by glycosidic linkage between C1 of one glucose unit and C4 of the next glucose unit. 54. How do enzymes help a substrate to be attacked by the reagent effectively? Ans. Enzymes are highly specific. They react with substrate molecule and form intermediate complex. They reduce the magnitude of activation energy. 55. Describe the term D- and L- configuration used for amino acids with examples. Ans. Amino acid exists both in ‘D’ and ‘L’ forms. Most naturally occurring amino acids have L- configuration. L-Amino acids are represented by writing the —NH 2 group on left hand side and D-amino acids are represented by writing the —NH 2 group on right hand side. 56. How will you distinguish 1° and 2° hydroxyl groups present in glucose? Explain with reactions. Ans. On oxidation with nitric acid, glucose as well as gluconic acid both yield a dicarboxylic acid, saccharic acid. This indicates the presence of a primary alcoholic (—OH) group in glucose.
Acetylation of glucose with acetic anhydride gives glucose pentaacetate which confirms the presence of five —OH groups. Since it exists as a stable compound, five —OH groups should be attached to different carbon atoms.
