Answer

(b) Glycogen is a branched chain polymer of $\alpha D$ glucose units in which chain is formed by $\mathrm{C} 1-\mathrm{C} 4$ glycosidic linkage whereas branching occurs by the formation of $\mathrm{C} 1-\mathrm{C} 6$ glycosidic linkage. Structure of glycogen can be shown below similar to the structure amylopectin.

Structure of amylopectine

Glycogen is also known as animal starch present in liver, muscles and brain.

• (a) Amylose: Amylose is a linear polymer of $\alpha-D$ glucose units connected by $\mathrm{C}1-\mathrm{C}4$ glycosidic linkages without any branching. Therefore, it does not have the branched structure characteristic of glycogen.

• (c) Cellulose: Cellulose is a linear polymer of $\beta-D$ glucose units connected by $\mathrm{C}1-\mathrm{C}4$ glycosidic linkages. The $\beta$ 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.

Answer

(d) Glycogen is a polymer of $\alpha-D$ glucose stored in the liver, brain and muscles of animals, also known as animal starch.

• (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.

• (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.

• (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.

Answer

(c) Sucrose (cane sugar) is a disaccharide. One molecule of sucrose on hydrolysis gives one molecule of glucose and one molecule of fructose.

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.

• (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.

• (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.

• (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.

Thinking Process

This problem is based on the concept of anomer. Saccharides which differ in configuration at $\mathrm{C}-1$ are known as anomers.

Answer

(c) Anomers have different configuration at $\mathrm{C}$-1. If $\mathrm{OH}$ is present at right side anomeric carbon is known as $\alpha$-form and if $\mathrm{OH}$ is present at left side of anomeric carbon is known as $\beta$-form

• 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.

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 $\alpha$ - helix and $\beta$ - pleated structure.

In $\alpha$ - helix structure, a polypeptide chain forms all possible hydrogen bonds by twisting into a right handed screw with $-\mathrm{NH}$ group of each amino acid rest hydrogen bonded to $\mathrm{C} \mathrm{C}=\mathrm{O}$ of adjacent amino acid, which form a helix.

• 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 $\alpha$-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 $\alpha$-helix secondary structure.

• Dipole-dipole interactions: Dipole-dipole interactions occur between polar groups, but they are not the main stabilizing force for the $\alpha$-helix. The $\alpha$-helix is specifically stabilized by hydrogen bonds between the $-\mathrm{NH}$ group of one amino acid and the $\mathrm{C}=\mathrm{O}$ group of another amino acid in the polypeptide chain.

Answer

(b)

This structure represents sucrose in which $\alpha-D$ glucose and $\beta$-D- fructose is attached to each other by $\mathrm{C} _{1}-\mathrm{C} _{2}$ glycosidic linkage.

Since, reducing groups of glucose and fructose are involved in glycosidic bond formation, this is considered as non-reducing sugar.

• 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.

• 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.

• 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.

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 $\mathrm{HNO} _{3}$.

• 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 $\mathrm{HNO}_3$, not a vitamin.

Answer

(a) Nucleoside Species formed by the attachment of a base to $1^{\prime}$ position of sugar is known as nucleoside. The sugar carbon are numbered as $1^{\prime}, 2^{\prime}, 3^{\prime}$,…to distinguish them from bases.

Structure of nucleoside

Nucleotide Species formed by attachment of phosphoric acid to nucleoside at 5 ’ position of sugar nucleotide.

Structure of nucleotide

Dinucleotides are formed by phosphodiester linkage between $5^{\prime}$ and $3^{\prime}$ carbon atoms of the pentose sugar.

• (b) $1^{\prime}$ and $5^{\prime}$: This option is incorrect because the phosphodiester linkage in a dinucleotide is not formed between the $1^{\prime}$ and $5^{\prime}$ carbon atoms of the pentose sugars. The $1^{\prime}$ 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 $5^{\prime}$ carbon atoms. The $5^{\prime}$ carbon of one nucleotide connects to the phosphate group, which then links to the $3^{\prime}$ 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 $3^{\prime}$ carbon atoms. The linkage involves the $3^{\prime}$ carbon of one nucleotide and the $5^{\prime}$ carbon of the next nucleotide.

Answer

(b) Nucleic acids are polymer of nucleotides in which nucleic acids are linked together by phosphodiester linkage.

e.g., DNA, RNA etc.

• (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.

• (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.

• (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.

Answer

(c) Glucose is a aldohexose having structural formula.

Glucose on heating with $\mathrm{HI}$ produces $n$ hexane.

$$ \mathrm{C} _{6} \mathrm{H} _{12} \mathrm{O} _{6} \xrightarrow{\mathrm{HI}} \mathrm{C} _{6} \mathrm{H} _{14} $$

Glucose does not give 2, 4, DNP test due to its existence as cyclic structure shown below

It is present in pyranose form, as shown below

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.

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.

• 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.

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.

Answer

(d) Vitamin $B_{12}$ can be stored in our body belongs to $B$ group vitamins, because it is not water soluble.

• 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.

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.

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.

• 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.

Answer

(c) “Pentaacetate of glucose does not react with hydroxylamine” showing absence of free $\mathrm{CHO}$ group. This can not be explained by open structure of glucose. While all other properties are easily explained by open structure of glucose.

Hence, option (c) is the correct choice.

• (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.

Answer

(a) $D$ and $L$ configuration are relative configuration decided by relating structure of given saccharide with $D$ or $L$ glyceraldehyde.

When $\mathrm{OH}$ an lowest asymmetric carbon is written at right hand side, it is represented as $D$ configuration and when $\mathrm{OH}$ is written on left hand side, it is represented as $L$ configuration.

• Option (b) II, III:

  • Compound I also has the $D$ configuration because the $\mathrm{OH}$ group on the lowest asymmetric carbon is on the right-hand side. Therefore, excluding Compound I makes this option incorrect.

• Option (c) I, II:

  • Compound III also has the $D$ configuration because the $\mathrm{OH}$ group on the lowest asymmetric carbon is on the right-hand side. Therefore, excluding Compound III makes this option incorrect.

• Option (d) III:

  • Compounds I and II also have the $D$ configuration because the $\mathrm{OH}$ group on the lowest asymmetric carbon is on the right-hand side. Therefore, excluding Compounds I and II makes this option incorrect.

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 ’ $a$ ’ and ’ $b$ ’ are present at adjacent to oxygen atom. Both carbons differ in configurations of the hydroxyl group.

• (a) ’ $a$ ’ carbon of glucose and ’ $a$ ’ carbon of fructose: This option is incorrect because the ’ $a$ ’ carbon of fructose is not adjacent to the oxygen atom in the cyclic structure, and thus it is not an anomeric carbon.

• (b) ’ $a$ ’ carbon of glucose and ’ $e$ ’ carbon of fructose: This option is incorrect because the ’ $e$ ’ carbon of fructose is not adjacent to the oxygen atom in the cyclic structure, and thus it is not an anomeric carbon.

• (d) ’ $f$ ’ carbon of glucose and ’ $f$ ’ carbon of fructose: This option is incorrect because the ’ $f$ ’ carbon of both glucose and fructose are not adjacent to the oxygen atom in the cyclic structure, and thus they are not anomeric carbons.

Answer

(c) Numbering of glucose starts from adjacent carbon of $\mathrm{O}$-atom to the other carbon atom ending at last $\mathrm{CH}_{2} \mathrm{OH}$ group as shown below

In this way, numbering for the disaccharides can be done as

(ii)

(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.

Answer

$(b, d)$

Sucrose on hydrolysis produces equimolar mixture of $\alpha-\mathrm{D}(+)$ glucose and $B-D(-)$. fructose. Since in sucrose $C-1$ of glucose and $C-2$ of fructose are linked with each other So, they are non-reducing in nature.

• (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.

Answer

$(a, c)$

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.

Answer

$(b, d)$

Amylopectin and glycogen have almost similar structure in which glucose are linked linearly to each other by $\mathrm{C} _{1}-\mathrm{C} _{4}$ glycosidic linkage and branched at $\mathrm{C} _{1}-\mathrm{C} _{6}$ glycosidic linkage.

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.

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 $\mathrm{COOH}$ group, and basic $-\mathrm{NH} _{2}$ group amino acid, proteins can be classified as

(i) If number of $\mathrm{COOH}$ groups = number of $\mathrm{NH} _{2}$ groups, amino acid is neutral.

(ii) If number of $\mathrm{COOH}$ groups $>$ number of $\mathrm{NH} _{2}$ groups, amino acid is acidic.

(iii) If number of $\mathrm{COOH}$ group < number of $\mathrm{NH} _{2}$ group, amino acid is basic.

Answer

$(b, d)$

(b) $\mathrm{HOOC}-\mathrm{CH} _{2}-\mathrm{CH} _{2}-\underset{\mathrm{NH} _{2}}{\underset{|}{\mathrm{CH}}}-\mathrm{COOH}$

Number of $\mathrm{COOH}$ groups $=2$

Number of $\mathrm{NH} _{2}$ group $=1$

Since, number of $\mathrm{COOH}$ groups (2) $>$ number of $\mathrm{NH} _{2}$ group (1). Therefore, this amino acid is acidic amino acid.

(d) $\mathrm{HOOC}-\mathrm{CH} _{2}-\underset{\mathrm{NH} _{2}}{\underset{|}{\mathrm{CH}}}-\mathrm{COOH}$

Number of $\mathrm{COOH}$ groups $=2$

Number of $\mathrm{NH} _{2}$ groups $=1$

Since, Number of $\mathrm{COOH}$ groups (2) $>$ Number of $\mathrm{NH} _{2}$ groups (1). Therefore, amino acid is acidic. While other two are neutral amino acid as number of $\mathrm{NH} _{2}$ group in equal to number of $\mathrm{COOH}$ group in then.

• (a) $\left(\mathrm{CH} _{3}\right) _{2} \mathrm{CH} -\underset{\mathrm{NH} _{2}}{\underset{|}{\mathrm{CH}}}-\mathrm{COOH}$

  Number of $\mathrm{COOH}$ groups $=1$

  Number of $\mathrm{NH} _{2}$ groups $=1$

  Since the number of $\mathrm{COOH}$ groups (1) is equal to the number of $\mathrm{NH} _{2}$ groups (1), this amino acid is neutral, not acidic.

• (c) $\mathrm{H} _{2} \mathrm{~N}-\mathrm{CH} _{2}-\mathrm{CH} _{2}-\mathrm{CH} _{2}-\mathrm{COOH}$

  Number of $\mathrm{COOH}$ groups $=1$

  Number of $\mathrm{NH} _{2}$ groups $=1$

  Since the number of $\mathrm{COOH}$ groups (1) is equal to the number of $\mathrm{NH} _{2}$ groups (1), this amino acid is neutral, not acidic.

Answer

$(a, b, c)$

Lysine whose structural formula is written below as

$\mathrm{H} _{2} \mathrm{~N}-\left(\mathrm{CH} _{2}\right) _{4}-\underset{\mathrm{NH} _{2}}{\underset{|}{\mathrm{CH}}}-\mathrm{COOH}$

(a) It is an $\alpha$ amino acid.

(b) It is a basic amino acid because number of $\mathrm{NH} _{2}$ groups (2) is greater than number of $\mathrm{COOH}$ group (1).

(c) It is a non-essential amino acid. Because it is synthesised in our body.

• Option (d) is incorrect because lysine is not a $\beta$-amino acid. In a $\beta$-amino acid, the amino group is attached to the $\beta$-carbon (the second carbon from the carboxyl group), whereas in lysine, the amino group is attached to the $\alpha$-carbon (the first carbon from the carboxyl group).

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.

Answer

$(b, d)$

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.

• 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.

• 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.

Answer

$(a, b)$

Purines consist of six membered and five membered nitrogen containing ring fused together.

Guanine and adenine are purine bases whose structures are

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.

Answer

$(a, d)$

Enzymes are proteins which acts as biocatalyst having specific role/action in specific biochemical reaction.

e.g., (i) Maltase decomposes maltose to glucose.

$$ \underset{\text { Maltose }}{\mathrm{C} _{12} \mathrm{H} _{22} \mathrm{O} _{11}} \xrightarrow{\text { Maltase }} \underset{\text { Glucose }}{2 \mathrm{C} _{6} \mathrm{H} _{12} \mathrm{O} _{6}} $$

• (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.

• (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.

Answer

Sugar present in milk is known as lactose sugar. Two units of monosaccharides $\beta$-D-galactose and $\beta$-D-glucose are linked together.

Hence, are known as disaccharides.

Answer

Glucose on heating with HI produces $n$-hexane.

This suggests that all the six carbon atoms of glucose are linked in a straight chain.

Answer

Phosphoric acid is linked at 5’- position of sugar moiety of nucleoside to give a nucleotide.

Answer

Glycosidic linkage connects monosaccharide units in polysaccharides.

Glycoside linkage in lactose

Answer

Glucose on oxidation with $\mathrm{Br} _{2} / \mathrm{H} _{2} \mathrm{O}$ produces gluconic acid (six carbon carboxylic acid).

Glucose on oxidation with nitric acid produces saccharic acid. (dicarboxylic acid)

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 $\mathrm{C} _{6} \mathrm{H} _{12} \mathrm{O} _{6}$ containing 6 carbon and keto group and is classified as ketohexose.

Thinking Process

This problem is based on relative configuration i.e., $D$ and $L$ configuration. This can be done by relating structure of monosaccharides with structure of glyceraldehyde.

If $\mathrm{OH}$ is present at right side of second last carbon of monosaccharide is considered as D configuration.

If $\mathrm{OH}$ is present at left side of second last carbon of monosaccharide is considered as Lconfiguration.

Answer

Here, $\mathrm{OH}$ group present on second last carbon in at left side hence this has L configuration.

Answer

$\beta$-D-ribose 5

$\beta$-D-2-deoxyribose

In case of cyclic structure of saccharide, if $-\mathrm{OH}$ group present at second last carbon is present at bottom side, then it is considered as $D$ configuration (as shown above)

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.

Answer

$\alpha$ amino acid forms polypeptide chain by elimination of water molecules.

Answer

$\alpha$-helix is a secondary structure of proteins formed by twisting of polypeptide chain to right handed screw like structure. Hydrogen bonds formed between - $\mathrm{NH}-$ group of amino acids in one turn with the $>\mathrm{C}=\mathrm{O}$ groups of amino acids belonging to adjacent turn is responsible for making the $\alpha$-helix structure stable.

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.

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.

Answer

Glucose on reaction with acetic anhydride produces glucose pentaacetate.

This reaction explain presence of five $-\mathrm{OH}$ groups.

Answer

Compound $(A)$ does not form an oxime on reaction with $\mathrm{NH} _{2} \mathrm{OH}$ due to absence of $\mathrm{CHO}$ group or $>\mathrm{C}=\mathrm{O}$ group.

Answer

Vitamin $\mathrm{C}$ is water soluble hence, they are regularly excreted in urine and can not be stored in our body, so, they are supplied regularly in diet.

Answer

Sucrose is dextrorotatory. On hydrolysis, it produces a mixture of glucose and fructose having specific rotation $+52.5^{\circ}$ and $-92.4^{\circ}$. Thus, the respectively net resultant mixture become laevorotatory.

Hence, the mixture is laevorotatory and product is known as invert sugar.

Answer

Amino acids have acidic $\mathrm{COOH}$ group as well as $\mathrm{NH} _{2}$ group hence, $\mathrm{COOH}$ looses its $\mathrm{H}$ to $\mathrm{NH} _{2}$, hence they exist as Zwitter ion.

Answer

Glycine and alanine on reaction with each other produces glycylalanine as

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.

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 $6.22 \mathrm{~kJ} \mathrm{~mol}^{-1}$ to $2.15 \mathrm{~kJ} \mathrm{~mol}^{-1}$.

Answer

Glucose on reaction with bromine water produces gluconic acid. This indicates the presence of $\mathrm{CHO}$ group.

Answer

(i) $5^{\prime}$ and $3^{\prime}$ carbon atoms of pentose sugar.

(ii) Most probably the resemblance of with 2 ester $(-\mathrm{COO})^{2-}$ groups joined together.

(iii) Phosphoric acid $\left(\mathrm{H} _{3} \mathrm{PO} _{4}\right)$.

Nucleosides are joined together by phosphodiester linkage between $5^{\prime}$ and $3^{\prime}$ carbon atoms of pentose sugar and a dinucleotide with phosphoric acid $\left(\mathrm{CH} _{3} \mathrm{PO} _{4}\right)$ is formed

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

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 $\alpha$-glucose in which two types of linkage are observed i.e., $\mathrm{C} _{1}-\mathrm{C} _{6}$ and $\mathrm{C} _{1}-\mathrm{C} _{4}$ glycosidic linkage.

(ii) Cellulose is a straight chain polysaccharide of $\beta$-D glucose in which glucose are linked together by $\mathrm{C} _{1}-\mathrm{C} _{4}$ glycosidic linkage.

(iii) Glucose is a monosaccharide.

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.

Answer

All naturally occurring $\alpha$-amino acids (except glycine) are optically active due to the presence of chiral carbon atom. These have either $D$ - or $L$-configuration. $D$-form means that, the amino $\left(-\mathrm{NH} _{2}\right)$ group is present towards the right hand side. L-form shows the presence of $-\mathrm{NH} _{2}$ group on the left hand side.

$D$-alanine

$L$-alanine

Answer

$1^{\circ}$ and $2^{\circ}$ hydroxyl groups present in glucose can be identified by the reaction of glucose with nitric acid. Primary $\mathrm{OH}$ group present in glucose are easily oxidise to $-\mathrm{COOH}$ group while secondary $\mathrm{OH}$ group does not.

Hence, one $\mathrm{OH}$ is primary $\mathrm{OH}$ group.

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.

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Answer

A. $\rightarrow(4)$

B. $\rightarrow(3) \quad$

C. $\rightarrow(5)$

D. $\rightarrow(1) \quad$

E. $\rightarrow(2)$

Answer

(c) Assertion is correct but reason is wrong statement $D(+)$ glucose is dextrorotatory because it rotates the plane polarised light to right.

Here, D represents relative configuration of glucose with respect to glyceraldehyde.

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(a) Assertion and reason both are correct statements and reason explains assertion. Vitamin $\mathrm{D}$ can be stored in our body because vitamin $\mathrm{D}$ is fat soluble vitamin.

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(d) Assertion is wrong statement and reason is correct statement. $\alpha$-glycosidic linkage is present in maltose

Because maltose is composed of two glucose unit in which $\mathrm{C}-1$ of one glucose unit is linked to $\mathrm{C}-4$ of another glucose unit.

Answer

(e) Assertion and reason both are correct and reason does not explain assertion. All naturally occurring $\alpha$-amino except glycine are optically active.

Glycine is optically inactive because glycine does not have all four different substituent as shown below.

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(b) Both assertion and reason are wrong statements. Deoxyribose $\mathrm{C} _{5} \mathrm{H} _{10} \mathrm{O} _{4}$ is a carbohydrate because it follow $\mathrm{C} _{5}\left(\mathrm{H} _{2} \mathrm{O}\right) _{2}$ formula and exist as polyhydroxy carbonyl compound whose cyclic structure is as shown below

$\beta$-o-2 deoxyribose

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(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.

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(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.

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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 $\mathrm{NaHSO} _{3}$, despite having aldehyde group

(ii) The pentaacetate of glucose does not react with hydroxylamine.

In actual, glucose exist in two different crystalline form $\alpha$ form and $\beta$ form. It was proposed that one of the $\mathrm{OH}$ groups may add to the $-\mathrm{CHO}$ group and form cyclic hemiacetal structure. Glucose forms a 6 membered pyranose structure.

Cyclic structure exist in equilibrium with open structure and can be represented as

Due to formation of cyclic structure of glucose $\mathrm{CHO}$ group of glucose remain no longer free due to which they do not show above given reactions.

Answer

Evidences on the basis of which glucose was assigned the following structure are as follows

(i) Glucose on reaction with $\mathrm{HI}$ produces $n$ hexane which indicates presence of six carbon atom linked in a having straight chain.

$$ \mathrm{C} _{6} \mathrm{H} _{12} \mathrm{O} _{6} \xrightarrow{\mathrm{HI}} n \text { hexane } $$

(ii) Glucose on reaction with acetic anhydride produces glucose penta acetate which indicates presence of five $\mathrm{OH}$ groups.

$$ \mathrm{C} _{6} \mathrm{H} _{12} \mathrm{O} _{6} \xrightarrow{\mathrm{Ac} _{2} \mathrm{O}} \text { Glucose pentaacetate } $$

(iii) Glucose on oxidation with bromine water produces gluconic acid indicates presence of $-\mathrm{CHO}$ group.

$$ \text { Glucose } \xrightarrow{\mathrm{Br} _{2} / \mathrm{H} _{2} \mathrm{O}} \text { Gluconic acid } $$

(iv) Glucose on reaction with $\mathrm{HNO} _{3}$ produces saccharic acid which indicates presence of one primary $\mathrm{OH}$ group.

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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.

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.

$\alpha$-helix structure of proteins

$\beta$-pleated sheet structure of proteins

Answer

On complete hydrolysis of DNA, following fragments are formed a pentose sugar ( $\beta$-D-2-deoxyribose) phosphoric acid $\left(\mathrm{H} _{3} \mathrm{PO} _{4}\right)$ and bases (nitrogen containing heterocyclic compounds).

Structures

(i) Sugar

$\beta$-D-2-deoxyribose

(ii) Phosphoric acid

(iii) Nitrogen bases DNA contains four bases

Adenine (A), Guanine (G), Cytosine (C) and thymine (T).

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 $5^{\prime}$ and $3^{\prime}$ carbon atoms of the pentose sugar.

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.