Biomolecules
Multiple Choice Questions (MCQs)
1. Glycogen is a branched chain polymer of
(a) amylose
(b) amylopectin
(c) cellulose
(d) glucose
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Answer
(b) Glycogen is a branched chain polymer of
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Structure of amylopectine
Glycogen is also known as animal starch present in liver, muscles and brain.
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(a) Amylose: Amylose is a linear polymer of
glucose units connected by glycosidic linkages without any branching. Therefore, it does not have the branched structure characteristic of glycogen. -
(c) Cellulose: Cellulose is a linear polymer of
glucose units connected by glycosidic linkages. The configuration and lack of branching make its structure different from that of glycogen. -
(d) Glucose: Glucose is a simple monosaccharide and not a polymer. It does not have a branched or linear polymeric structure like glycogen.
2. Which of the following polymer is stored in the liver of animals?
(a) Amylose
(b) Cellulose
(c) Amylopectin
(d) Glycogen
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Answer
(d) Glycogen is a polymer of
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(a) Amylose: Amylose is a polysaccharide and a form of starch found in plants, not animals. It is composed of long, unbranched chains of glucose molecules and is stored in plant tissues.
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(b) Cellulose: Cellulose is a polysaccharide that serves as a structural component in the cell walls of plants. Animals do not store cellulose; instead, it is a dietary fiber that some animals can digest with the help of symbiotic microorganisms.
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(c) Amylopectin: Amylopectin is another form of starch found in plants. It is a highly branched polymer of glucose, but it is not stored in animals. It is stored in plant tissues along with amylose.
3. Sucrose (cane sugar) is a disaccharide. One molecule of sucrose on hydrolysis gives…… .
(a) 2 molecules of glucose
(b) 2 molecules of glucose +1 molecule of fructose
(c) 1 molecule of glucose +1 molecule of fructose
(d) 2 molecules of fructose
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Answer
(c) Sucrose (cane sugar) is a disaccharide. One molecule of sucrose on hydrolysis gives one molecule of glucose and one molecule of fructose.
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Note Sucrose is a dextro-rotatory sugar on hydrolysis produces a laevorotatory mixture so, known as invert sugar. Sucrose is a non-reducing sugar while maltose and lactose are reducing sugar.
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(a) 2 molecules of glucose: This option is incorrect because sucrose is composed of one molecule of glucose and one molecule of fructose, not two molecules of glucose. Hydrolysis of sucrose does not produce two glucose molecules.
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(b) 2 molecules of glucose + 1 molecule of fructose: This option is incorrect because it suggests that hydrolysis of one molecule of sucrose yields three molecules (two glucose and one fructose), which is not possible. Sucrose hydrolysis yields only one glucose and one fructose molecule.
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(d) 2 molecules of fructose: This option is incorrect because sucrose is composed of one molecule of glucose and one molecule of fructose. Hydrolysis of sucrose does not produce two fructose molecules.
4. Which of the following pairs represents anomers?
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Show Answer
Thinking Process
This problem is based on the concept of anomer. Saccharides which differ in configuration at
Answer
(c) Anomers have different configuration at
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- Option (a) is incorrect because the two structures do not differ at the anomeric carbon (C-1); they differ at other carbon atoms.
- Option (b) is incorrect because the two structures are not cyclic forms of the same sugar; they represent different sugars.
- Option (d) is incorrect because the two structures are not stereoisomers differing at the anomeric carbon; they differ at multiple carbon atoms.
5. Proteins are found to have two different types of secondary structures viz
(a) peptide bonds
(b) van der Waals, forces
(c) hydrogen bonds
(d) dipole-dipole interactions
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Answer
(c) Secondary structures of protein denotes the shape in which a long polypeptide chain exists. The secondary structure exist in two type of structure
In
-
Peptide bonds: Peptide bonds are responsible for linking amino acids together in a polypeptide chain, forming the primary structure of proteins. They do not play a direct role in stabilizing the secondary structures like the
-helix. -
Van der Waals forces: While van der Waals forces can contribute to the overall stability of a protein’s tertiary structure by providing weak interactions between nonpolar side chains, they are not the primary stabilizing force for the
-helix secondary structure. -
Dipole-dipole interactions: Dipole-dipole interactions occur between polar groups, but they are not the main stabilizing force for the
-helix. The -helix is specifically stabilized by hydrogen bonds between the group of one amino acid and the group of another amino acid in the polypeptide chain.
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?
(a)
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(b)
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(c)
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(d)
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Show Answer
Answer
(b)
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This structure represents sucrose in which
Since, reducing groups of glucose and fructose are involved in glycosidic bond formation, this is considered as non-reducing sugar.
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Option (a): This structure represents maltose, which consists of two glucose units linked by an α(1→4) glycosidic bond. The reducing end of one glucose unit is free, making maltose a reducing sugar.
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Option (c): This structure represents lactose, which consists of one galactose and one glucose unit linked by a β(1→4) glycosidic bond. The reducing end of the glucose unit is free, making lactose a reducing sugar.
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Option (d): This structure represents cellobiose, which consists of two glucose units linked by a β(1→4) glycosidic bond. The reducing end of one glucose unit is free, making cellobiose a reducing sugar.
7. Which of the following acids is a vitamin?
(a) Aspartic and
(b) Ascorbic acid
(c) Adipic acid
(d) Saccharic acid
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Answer
(b) Ascorbic acid is the chemical name of vitamin C. While others are not vitamins aspartic acid is an amino acid. Adipic acid is a dicarboxylic acid having 8 carbon chain. Saccharic acid is a dicarboxylic acid obtained by oxidation of glucose using
- Aspartic acid is an amino acid, not a vitamin.
- Adipic acid is a dicarboxylic acid with an 8-carbon chain, not a vitamin.
- Saccharic acid is a dicarboxylic acid obtained by oxidation of glucose using
, not a vitamin.
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?
(a) 5’ and 3'
(b)
(c) 5’ and 5'
(d) 3’ and 3'
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Answer
(a) Nucleoside Species formed by the attachment of a base to
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Structure of nucleoside
Nucleotide Species formed by attachment of phosphoric acid to nucleoside at 5 ’ position of sugar nucleotide.
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Structure of nucleotide
Dinucleotides are formed by phosphodiester linkage between
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(b)
and : This option is incorrect because the phosphodiester linkage in a dinucleotide is not formed between the and carbon atoms of the pentose sugars. The carbon is where the nitrogenous base attaches to the sugar, not where the linkage between nucleotides occurs. -
(c) 5’ and 5’: This option is incorrect because the phosphodiester linkage in a dinucleotide is not formed between two
carbon atoms. The carbon of one nucleotide connects to the phosphate group, which then links to the carbon of the next nucleotide. -
(d) 3’ and 3’: This option is incorrect because the phosphodiester linkage in a dinucleotide is not formed between two
carbon atoms. The linkage involves the carbon of one nucleotide and the carbon of the next nucleotide.
9. Nucleic acids are the polymers of
(a) nucleosides
(b) nucleotides
(c) bases
(d) sugars
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Answer
(b) Nucleic acids are polymer of nucleotides in which nucleic acids are linked together by phosphodiester linkage.
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e.g., DNA, RNA etc.
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(a) Nucleosides: Nucleosides are composed of a nitrogenous base attached to a sugar molecule (ribose or deoxyribose) but lack the phosphate group. Nucleic acids are not polymers of nucleosides because they require the phosphate group to form the phosphodiester bonds that link nucleotides together.
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(c) Bases: Bases refer to the nitrogenous bases (adenine, thymine, cytosine, guanine, and uracil) that are part of nucleotides. However, nucleic acids are not polymers of just the bases; they are polymers of nucleotides, which include a base, a sugar, and a phosphate group.
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(d) Sugars: Sugars (ribose in RNA and deoxyribose in DNA) are components of nucleotides, but nucleic acids are not polymers of sugars alone. The sugar must be part of a nucleotide, which also includes a nitrogenous base and a phosphate group, to form the nucleic acid polymer through phosphodiester linkages.
10. Which of the following statements is not true about glucose?
(a) It is an aldohexose
(b) On heating with
(c) It is present in furanose form
(d) It does not give 2, 4- DNP test
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Answer
(c) Glucose is a aldohexose having structural formula.
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Glucose on heating with
Glucose does not give 2, 4, DNP test due to its existence as cyclic structure shown below
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It is present in pyranose form, as shown below
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Pyranose means pyran (membered ring containing oxygen) like structure.
- Option (a) is incorrect because it is actually true that glucose is an aldohexose.
- Option (b) is incorrect because it is actually true that glucose on heating with HI forms n-hexane.
- Option (d) is incorrect because it is actually true that glucose does not give the 2,4-DNP test due to its cyclic structure.
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
(a) primary structure of proteins
(b) secondary structure of proteins
(c) tertiary structure of proteins
(d) quateranary structure of proteins
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Answer
(a) In primary structure of proteins when each polypeptide in a protein has amino acids linked with each other in a specific sequence. This type of structure is known as primary structure of proteins.
-
Secondary structure of proteins: This refers to the local folded structures that form within a polypeptide due to interactions between atoms of the backbone. The most common types of secondary structures are alpha helices and beta sheets, which are stabilized by hydrogen bonds. It does not describe the specific sequence of amino acids.
-
Tertiary structure of proteins: This refers to the overall three-dimensional structure of a single polypeptide chain, resulting from interactions between the side chains (R groups) of the amino acids. It includes various types of bonds and interactions such as hydrogen bonds, disulfide bridges, hydrophobic interactions, and ionic bonds. It does not describe the specific sequence of amino acids.
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Quaternary structure of proteins: This refers to the structure formed by the assembly of multiple polypeptide chains into a single functional protein complex. These polypeptide chains, also known as subunits, interact through various types of bonds and interactions. It does not describe the specific sequence of amino acids within a single polypeptide chain.
12. DNA and RNA contain four bases each. Which of the following bases in not present in RNA?
(a) Adenine
(b) Uracil
(c) Thymine
(d) Cytosine
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Answer
(c) DNA contain four bases adenine, guanine, thymine and cytosine. While RNA contain four bases adenine, uracil, guanine and cytosine. Thus, RNA does not contain thymine.
Hence, statement (c) is the correct choice.
- (a) Adenine: Adenine is present in both DNA and RNA. Therefore, it is not the correct answer.
- (b) Uracil: Uracil is present in RNA but not in DNA. Therefore, it is not the correct answer.
- (d) Cytosine: Cytosine is present in both DNA and RNA. Therefore, it is not the correct answer.
13. Which of the following B group vitamins can be stored in our body?
(a) Vitamin
(b) Vitamin
(c) Vitamin
(d) Vitamin
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Answer
(d) Vitamin
- Vitamin B1 (Thiamine) is incorrect because it is water-soluble and not stored in significant amounts in the body.
- Vitamin B2 (Riboflavin) is incorrect because it is water-soluble and not stored in significant amounts in the body.
- Vitamin B6 (Pyridoxine) is incorrect because, although it can be stored in the liver to some extent, it is still considered water-soluble and not stored in significant amounts compared to Vitamin B12.
14. Which of the following bases is not present in DNA?
(a) Adenine
(b) Thymine
(c) Cytosine
(d) Uracil
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Answer
(d) DNA contains following four bases
(a) adenine (A)
(b) thymine (T)
(c) guanine (G)
(d) cytosine (C)
It does not contain uracil.
- Adenine (A) is incorrect because it is one of the four bases present in DNA.
- Thymine (T) is incorrect because it is one of the four bases present in DNA.
- Cytosine (C) is incorrect because it is one of the four bases present in DNA.
15. There cyclic structures of monosaccharides are given below which of these are anomers.
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(i)
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(ii)
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(iii)
(a) I and II
(b) II and III
(c) I and III
(d) III is anomer of I and II
Show Answer
Answer
(a) Anomers Cyclic structures of monosaccharides which differ in structure at carbon-1 are known as anomers.
Here, I and II are anomer because they differ from each other at carbon- 1 only.
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Option (b) II and III: This option is incorrect because II and III do not differ only at carbon-1. They have differences in other parts of their structures as well, which means they are not anomers.
-
Option (c) I and III: This option is incorrect because I and III do not differ only at carbon-1. They have differences in other parts of their structures as well, which means they are not anomers.
-
Option (d) III is anomer of I and II: This option is incorrect because III is not an anomer of either I or II. Anomers differ only at the carbon-1 position, and III has differences in other parts of its structure compared to I and II.
16. Which of the following reactions of glucose can be explained only by its cyclic structure?
(a) Glucose forms pentaacetate
(b) Glucose reacts with hydroxylamine to form an oxime
(c) Pentaacetate of glucose does not react with hydroxyl amine
(d) Glucose is oxidised by nitric acid to gluconic acid
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Answer
(c) “Pentaacetate of glucose does not react with hydroxylamine” showing absence of free
Hence, option (c) is the correct choice.
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(a) Glucose forms pentaacetate: This reaction can be explained by the presence of five hydroxyl (OH) groups in the open-chain form of glucose, which can react with acetic anhydride to form pentaacetate.
-
(b) Glucose reacts with hydroxylamine to form an oxime: This reaction involves the aldehyde group (CHO) in the open-chain form of glucose reacting with hydroxylamine to form an oxime, which can be explained by the open structure of glucose.
-
(d) Glucose is oxidised by nitric acid to gluconic acid: This reaction involves the oxidation of the aldehyde group (CHO) in the open-chain form of glucose to a carboxylic acid group (COOH), forming gluconic acid. This can be explained by the open structure of glucose.
17. Optical rotations of some compounds alongwith their structures are given below which of them have
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(a) I, II, III
(b) II, III
(c) I, II
(d) III
Show Answer
Answer
(a)
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When
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Option (b) II, III:
- Compound I also has the
configuration because the group on the lowest asymmetric carbon is on the right-hand side. Therefore, excluding Compound I makes this option incorrect.
- Compound I also has the
-
Option (c) I, II:
- Compound III also has the
configuration because the group on the lowest asymmetric carbon is on the right-hand side. Therefore, excluding Compound III makes this option incorrect.
- Compound III also has the
-
Option (d) III:
- Compounds I and II also have the
configuration because the group on the lowest asymmetric carbon is on the right-hand side. Therefore, excluding Compounds I and II makes this option incorrect.
- Compounds I and II also have the
18. Structure of disaccharide formed by glucose and fructose is given below. Identify anomeric carbon atoms in monosaccharide units.
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(a) ’
(b) ’
(c) ’
(d) ’
Show Answer
Answer
(c) Carbon adjacent to oxygen atom in the cyclic structure of glucose or fructose is known as anomeric carbon. As shown in the structure above ’
-
(a) ’
’ carbon of glucose and ’ ’ carbon of fructose: This option is incorrect because the ’ ’ carbon of fructose is not adjacent to the oxygen atom in the cyclic structure, and thus it is not an anomeric carbon. -
(b) ’
’ carbon of glucose and ’ ’ carbon of fructose: This option is incorrect because the ’ ’ carbon of fructose is not adjacent to the oxygen atom in the cyclic structure, and thus it is not an anomeric carbon. -
(d) ’
’ carbon of glucose and ’ ’ carbon of fructose: This option is incorrect because the ’ ’ carbon of both glucose and fructose are not adjacent to the oxygen atom in the cyclic structure, and thus they are not anomeric carbons.
19. Three structures are given below in which two glucose units are linked. Which of these linkages between glucose units are between
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(i)
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(ii)
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(iii)
(a)
(b) (A) and (B) are between
(C)
(d) (A) and (C) are between C1 and C6, (B) is between
Show Answer
Answer
(c) Numbering of glucose starts from adjacent carbon of
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In this way, numbering for the disaccharides can be done as
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(ii)
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(iii)
-
Option (a) is incorrect because:
- It states that (A) is between C1 and C4, and (B) and (C) are between C1 and C6. However, (B) is actually between C1 and C4, not C1 and C6.
-
Option (b) is incorrect because:
- It states that (A) and (B) are between C1 and C4, and (C) is between C1 and C6. However, (C) is actually between C1 and C4, not C1 and C6.
-
Option (d) is incorrect because:
- It states that (A) and (C) are between C1 and C6, and (B) is between C1 and C4. However, (A) and (C) are actually between C1 and C4, not C1 and C6.
Multiple Choice Questions (More Than One Options)
20. Carbohydrates are classified on the basis of their behaviour on hydrolysis and also as reducing or non-reducing sugar. Sucrose is a…… .
(a) monosaccharide
(b) disaccharide
(c) reducing sugar
(d) non-reducing sugar
Show Answer
Answer
Sucrose on hydrolysis produces equimolar mixture of
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(a) monosaccharide: Sucrose is not a monosaccharide; it is composed of two monosaccharides, glucose and fructose, linked together. Monosaccharides are single sugar molecules, such as glucose or fructose, not combinations of them.
-
(c) reducing sugar: Sucrose is not a reducing sugar because the glycosidic bond between glucose and fructose involves the anomeric carbon of both sugars, preventing them from acting as reducing agents. Reducing sugars have free anomeric carbons that can participate in redox reactions, which sucrose does not have due to its specific linkage.
21. Proteins can be classified into two types on the basis of their molecular shape, i.e., fibrous proteins and globular proteins. Examples of globular proteins are
(a) insulin
(b) keratin
(c) albumin
(d) myosin
Show Answer
Answer
The structure of protein which results when the chain of polypeptides coil around to give a spherical shape are known as globular protein. These proteins are soluble in water, e.g., insulin and albumin are globular protein.
Hence, (a) and (c) are correct choices.
-
(b) keratin: Keratin is a fibrous protein, not a globular protein. It forms long, filamentous structures and is insoluble in water. Keratin is primarily found in hair, nails, and the outer layer of skin.
-
(d) myosin: Myosin is also a fibrous protein. It is a motor protein involved in muscle contraction and other motility processes within cells. Myosin molecules form long, fibrous structures and are not spherical in shape.
22. Which of the following carbohydrates are branched polymer of glucose?
(a) Amylose
(b) Amylopectin
(c) Cellulose
(d) Glycogen
Show Answer
Answer
Amylopectin and glycogen have almost similar structure in which glucose are linked linearly to each other by
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Structure of amylopectin
Glycogen are carbohydrates stored in animal body. The structure to similar to amylopectin and is rather more highly branched.
-
Amylose: Amylose is a linear polymer of glucose. It consists of glucose units linked by α(1→4) glycosidic bonds without any branching.
-
Cellulose: Cellulose is also a linear polymer of glucose, but it consists of glucose units linked by β(1→4) glycosidic bonds. It does not have any branching and has a different type of glycosidic linkage compared to amylopectin and glycogen.
23. Amino acids are classified as acidic, basic or neutral depending upon the relative number of amino and carboxyl groups in their molecule. Which of the following are acidic?
(a)
(b)
(c)
(d)
Show Answer
Thinking Process
This problem is based on concept of nature of amino acid, that either it is acidic, basic or neutral.
Depending upon the number of acidic
(i) If number of
(ii) If number of
(iii) If number of
Answer
(b)
Number of
Number of
Since, number of
(d)
Number of
Number of
Since, Number of
-
(a)
Number of
groupsNumber of
groupsSince the number of
groups (1) is equal to the number of groups (1), this amino acid is neutral, not acidic. -
(c)
Number of
groupsNumber of
groupsSince the number of
groups (1) is equal to the number of groups (1), this amino acid is neutral, not acidic.
24. Lysine,
(a)
(b) basic amino acid
(c) amino acid synthesised in body
(d)
Show Answer
Answer
Lysine whose structural formula is written below as
(a) It is an
(b) It is a basic amino acid because number of
(c) It is a non-essential amino acid. Because it is synthesised in our body.
- Option (d) is incorrect because lysine is not a
-amino acid. In a -amino acid, the amino group is attached to the -carbon (the second carbon from the carboxyl group), whereas in lysine, the amino group is attached to the -carbon (the first carbon from the carboxyl group).
25. Which of the following monosaccharides are present as five membered cyclic structure (furanose structure)?
(a) Ribose
(b) Glucose
(c) Fructose
(d) Galactose
Show Answer
Answer
(a, c)
Ribose and fructose has five membered cyclic furanose structure because it include 5 carbon atom containing polyhydroxy carbonyl compound.
Hence (a) and (c) are correct choice.
- Glucose: Glucose typically forms a six-membered cyclic structure known as a pyranose, not a five-membered furanose structure.
- Galactose: Galactose also typically forms a six-membered cyclic structure known as a pyranose, not a five-membered furanose structure.
26. In fibrous proteins, polypeptide chains are held together by…… .
(a) van der Waals forces
(b) disulphide linkage
(c) electrostatic forces of attraction
(d) hydrogen bonds
Show Answer
Answer
In fibrous proteins, polypeptide chains are held together by hydrogen and disulphide bond, in parallel manner. Due to which fibre-like structure is obtained. Such proteins are generally known as fibrous proteins. These proteins are generally insoluble in water. e.g., Keratin, myosin.
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van der Waals forces: These are weak intermolecular forces that occur between molecules due to temporary dipoles. While they do contribute to the overall stability of protein structures, they are not the primary forces holding polypeptide chains together in fibrous proteins.
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electrostatic forces of attraction: These forces, also known as ionic bonds or salt bridges, occur between oppositely charged side chains of amino acids. While they can play a role in the stabilization of protein structures, they are not the main forces responsible for holding the polypeptide chains together in fibrous proteins.
27. Which of the following are purine bases?
(a) Guanine
(b) Adenine
(c) Thymine
(d) Uracil
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Answer
Purines consist of six membered and five membered nitrogen containing ring fused together.
Guanine and adenine are purine bases whose structures are
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While thymine and uracil are pyrimidene bases.
Hence (a) and (b) are correct choices.
- Thymine and uracil are pyrimidine bases, not purine bases. Pyrimidines consist of a single six-membered nitrogen-containing ring, unlike purines which have a fused ring structure.
28. Which of the following terms are correct about enzyme?
(a) Proteins
(b) Dinucleotides
(c) Nucleic acids
(d) Biocatalysts
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Answer
Enzymes are proteins which acts as biocatalyst having specific role/action in specific biochemical reaction.
e.g., (i) Maltase decomposes maltose to glucose.
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(b) Dinucleotides: Dinucleotides are molecules consisting of two nucleotides joined together. They are not proteins and do not function as enzymes. Enzymes are typically composed of amino acids, not nucleotides.
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(c) Nucleic acids: Nucleic acids, such as DNA and RNA, are macromolecules that store and transmit genetic information. They are not proteins and do not act as enzymes. Enzymes are specific types of proteins that catalyze biochemical reactions.
Short Answer Type Questions
29. Name the sugar present in milk. How many monosaccharide units are present in it? What are such oligosaccharides called?
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Answer
Sugar present in milk is known as lactose sugar. Two units of monosaccharides
Hence, are known as disaccharides.
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Answer
Glucose on heating with HI produces
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This suggests that all the six carbon atoms of glucose are linked in a straight chain.
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Answer
Phosphoric acid is linked at 5’- position of sugar moiety of nucleoside to give a nucleotide.
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Answer
Glycosidic linkage connects monosaccharide units in polysaccharides.
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Glycoside linkage in lactose
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Answer
Glucose on oxidation with
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Glucose on oxidation with nitric acid produces saccharic acid. (dicarboxylic acid)
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Answer
Monosaccharides contain carbonyl group.
Hence, are classified as aldose or ketose.
When aldehyde group is present, the monosaccharides are known as aldose.
When ketone group is present, the monosaccharides are known as ketose. Fructose has molecular formula
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Thinking Process
This problem is based on relative configuration i.e.,
If
If
Answer
Here,
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Answer
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In case of cyclic structure of saccharide, if
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Answer
Sucrose is dextrorotatory but sucrose on hydrolysis gives dextrorotatory glucose and laevorotatory fructose. Hence, mixture becomes laevorotatory.
This sugar which on hydrolysis changes its sign of rotation from dextro to laevo is known as invert sugar.
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Answer
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Answer
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Answer
Oxidoreductase enzymes A class of enzymes which catalyses the oxidation of one substrate with simultaneous reduction of another substrate is known as oxidoreductase enzymes.
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Answer
Curdling of milk is caused due to formation of lactic acid by the bacteria present in milk. It is an example of denaturation of protein, i.e., when a protein is subjected is some physical or chemical chages. Hydrogen bond get disturbed. Globules unfold and helix uncoil and protein loss its biological acturty.
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Answer
Glucose on reaction with acetic anhydride produces glucose pentaacetate.
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This reaction explain presence of five
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(A)
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Answer
Compound
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Answer
Vitamin
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Answer
Sucrose is dextrorotatory. On hydrolysis, it produces a mixture of glucose and fructose having specific rotation
Hence, the mixture is laevorotatory and product is known as invert sugar.
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Answer
Amino acids have acidic
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Answer
Glycine and alanine on reaction with each other produces glycylalanine as
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Answer
Due to physical and chemical change, hydrogen bonds in proteins are disturbed. Due is this globules unfold and helix gets uncoiled and therefore, protein loses its biological activity. This is known as denaturation of proteins.
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Answer
Enzymes, the biocatalysts reduce the magnitude of activation energy by providing alternative path. In the hydrolysis of sucrose, the enzyme sucrase reduces the activation energy from
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Answer
Glucose on reaction with bromine water produces gluconic acid. This indicates the presence of
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Answer
(i)
(ii) Most probably the resemblance of with 2 ester
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(iii) Phosphoric acid
Nucleosides are joined together by phosphodiester linkage between
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Answer
Linkage between two monosaccharides due to oxide linkage formed by the loss of a water molecule, is known as glycosidic linkage as shown below
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Answer
Monosaccharides units present in starch, cellulose and glucose can be determined by knowing the product of their hydrolysis.
(i) Starch is a polysaccharide of
(ii) Cellulose is a straight chain polysaccharide of
(iii) Glucose is a monosaccharide.
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Answer
At the surface of enzyme, active sites are present. These active sites of enzymes hold the substrate molecule in a suitable position, so that it can be attacked by the reagent effectively. This reduces the magnitude of activation energy.
Enzymes contains cavities of characteristics shape and possessing active groups known as active centre on the surface. The molecules of the reactant (substrate) having complementary shape, fit into these cavities. On account of these active groups, an activated complex is formed which then decomposes to yield the products.
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Answer
All naturally occurring
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Answer
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Hence, one
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Answer
Denaturation of proteins Protein present in egg white has an unique three dimensional structure. When it is subjected to physical change like change in temperature. i.e., on boiling, coagulation of egg white occurs due to denaturation of protein.
During denaturation hydrogen bonds are disturbed due to this globules unfold and helix gets uncoiled and protein looses its biological activity.
Matching The Columns
58. Match the vitamins given in Column I with the deficiency disease they cause given in Column II.
Column I (Vitamins) |
Column II (Diseases) |
||
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A. | Vitamin A | 1. | Pernicious anaemia |
B. | Vitamin |
2. | Increased blood clotting time |
C. | Vitamin |
3. | Xerophthalmia |
D. | Vitamin C | 4. | Rickets |
E. | Vitamin D | 5. | Muscular weakness |
F. | Vitamin E | 6. | Night blindness |
G. | Vitamin K | 7. | Beri-beri |
8. | Bleeding gums | ||
9. | Osteomalacia |
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Answer
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Column I (Enzymes) |
Column II (Reactions) |
||
---|---|---|---|
A. | Invertase | 1. | Decomposition of urea into |
B. | Maltase | 2. | Conversion of glucose into ethyl alcohol. |
C. | Pepsin | 3. | Hydrolysis of maltose into glucose. |
D. | Urease | 4. | Hydrolysis of cane sugar. |
E. | Zymase | 5. | Hydrolysis of proteins into peptides. |
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Answer
A.
B.
C.
D.
E.
Column I (Enzymes) |
Column II (Reaction) |
|
A. | Invertase | Hydrolysis of cane sngar. |
B. | Maltase | Hydrolysis of maltose into glucose. |
C. | Pepsin | Hydrolysis of protein into peptides. |
D. | Urease | Decomposition of urea into |
E. | Zymase | Conversion of glucose into ethyl alcohol. |
Assertion and Reason
In the following questions a statement of Assertion (A) followed by a statement of Reason (
(a) Assertion and reason both are correct statements and reason explains the assertion.
(b) Both assertion and reason are wrong statements.
(c) Assertion is correct statement and reason is wrong statement.
(d) Assertion is wrong statement and reason is correct statement.
(e) Assertion and reason both are correct statements but reason does not explain assertion.
60. Assertion (A) D (+) - Glucose is dextrorotatory in nature.
Reason
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Answer
(c) Assertion is correct but reason is wrong statement
Here, D represents relative configuration of glucose with respect to glyceraldehyde.
Reason (R) Vitamin
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Answer
(a) Assertion and reason both are correct statements and reason explains assertion. Vitamin
Reason (R) Maltose is composed of two glucose units in which C - 1 of one glucose unit is linked to
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Answer
(d) Assertion is wrong statement and reason is correct statement.
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Because maltose is composed of two glucose unit in which
Reason (R) Most naturally occurring amino acids have L-configuration.
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Answer
(e) Assertion and reason both are correct and reason does not explain assertion. All naturally occurring
Glycine is optically inactive because glycine does not have all four different substituent as shown below.
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Reason (
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Answer
(b) Both assertion and reason are wrong statements. Deoxyribose
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Reason (R) It is an essential amino acid.
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Answer
(b) Both assertion and reason are wrong statements. Correct asserstion and reason are Glycine must not be taken through diet because it can be synthesised in our body and a non-essential amino acid.
Reason (R) Active sites of enzymes hold the substrate molecule in a suitable position.
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Answer
(a) Assertion and reason both are correct and reason explains assertion. In presence of enzyme, substrate molecule can be attacked by a reagent effectively because active sites of enzymes hold the substrate molecule in a suitable position. So, enzyme catalysed reactions are stereospecific reactions.
Long Answer Type Questions
67. Write the reactions of D-glucose which can’t be explained by its open chain structure. How can cyclic structure of glucose explain these reactions?
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Answer
Chemical reactions of D-glucose which can’t be explained by its open chain structure are
(i) Glucose does not give Schiff’s test and does not produce hydrogensulphite addition product with
(ii) The pentaacetate of glucose does not react with hydroxylamine.
In actual, glucose exist in two different crystalline form
Cyclic structure exist in equilibrium with open structure and can be represented as
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Due to formation of cyclic structure of glucose
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Answer
Evidences on the basis of which glucose was assigned the following structure are as follows
(i) Glucose on reaction with
(ii) Glucose on reaction with acetic anhydride produces glucose penta acetate which indicates presence of five
(iii) Glucose on oxidation with bromine water produces gluconic acid indicates presence of
(iv) Glucose on reaction with
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Answer
Carbohydrates that are used as storage molecules in plants and animals are as follows
(i) Plant contains mainly starch, cellulose, sucrose etc.
(ii) Animal contain glycogen in their body. So, glycogen is also known as animal starch. Glycogen is present in liver, muscles and brain when body needs glucose, enzyme breaks glycogen down to glucose.
(iii) Cellulose is present in wood, and fibre of clothes.
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Answer
Primary structure of proteins Proteins consist of one or more polypeptide chains. Each polypeptide is a protein contains amino acids joined with each other in a specific sequence. Secondary structure of proteins it refers to the shape in which a long polypeptide chain can exist.
A structure of twisting of all a polypeptide chain formed by possible H-bonds into a right handed screw (helix ) with the -NH froup of each amino acid,and residuw hydrogen binded to the -CO- of an adjacent tum of the helix Hence, called |
All peptude chains are tretched out to nearly maximum extension and then laid side by side which are held together by intermolecular hydrogen bonds. This structure resembles the hydrogen bonds. This structure resembles the pleated folds of the drapery . Hence, called |
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Answer
On complete hydrolysis of DNA, following fragments are formed a pentose sugar (
Structures
(i) Sugar
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(ii) Phosphoric acid
(iii) Nitrogen bases DNA contains four bases
Adenine (A), Guanine (G), Cytosine (C) and thymine (T).
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A unit formed by the attachment of a base to 1’-position of sugar is called nucleoside. When nucleoside links to phosphoric acid at 5’-position of sugar moiety, a nucleotide is formed. Nucleotides are joined together by phosphodiester linkage between
In DNA two nucleic acid chains are wound about each other and held together by hydrogen bonds between pairs of bases.
The two strands are complementary to each other because hydrogen bonds are formed between specific pair of base adenine form hydrogen bonds with thymine whereas cytosine form hydrogen bonds with guanine.
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