NCERT Solutions for Class 12 Science Chemistry Chapter 14 - Biomolecules

Answer:

Amylopectin is a made up of branched chain polymer of a - D glucose units. Here, C1-C4 glycosidic linkage are forming linear chains whereas branching occurs due to the formation of C1-C6 glycosidic linkage. Hence, structure of glycogen is similar to amylopectin.
Hence, the correct answer is option (ii).

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Glycogen is a branched biopolymer made up of linear chains of glucose joined together by glycosidic bonds a(1-4). It is stored in the liver of the animals.
Hence, the correct answer is option (iv).

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Sucrose is a disaccharide that is formed by the condensation reaction of one molecule of glucose and one molecule of fructose. The hydrolysis of sucrose yields one molecule of glucose and one molecule of fructose. 

Hence, the correct answer is option (iii).

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Anomers are cyclic monosaccharide which differs in configuration only at the acetal or hemiacetal carbon or at the anomeric carbon atoms.
Only the C1 carbon atoms of the cyclic monosaccharides differs in configuration in option (iii).
Hence, the correct answer is option (iii).

Answer:

Proteins have two different types of secondary structures:- α-helix and β-pleated sheet structure. Protein - helix structure is stabilised by hydrogen bonds. In the peptide bond, hydrogen bonds are formed between the backbone N-H group of an amino acid and the backbone C= O group of every fourth amino acid.
Hence, the correct answer is option (iii).

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Any carbohydrate structure that has Aldehydic or Hemiacetal group is a reducing sugar. This free aldehydic or hemiacetal group is easily oxidised when the oxidising agent is reduced.
In option no. (ii) the glycosidic linkage is formed by the aldehyde group of glucose and the keto group of fructose. So, this sugar lacks free aldehyde or hemiacetal group. Therefore, this is a non reducing sugar.
Hence, the correct answer is option (ii).

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The chemical name of vitamin C is Ascorbic acid. The others are not vitamins.
Vitamin is usually found in food and used as a dietary supplement.
Hence, the correct answer is option (ii).

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Dinucleotide is formed by joining two nucleotides together by phosphodiester linkage. These linkage exist between the 5' and 3' carbon atoms of nucleotide pentose sugars.
The 5' carbon atoms of pentose sugar in one nucleotide is joined to the 3' carbon atom atom of pentose sugar of the second nucleotide.
Hence, the correct answer is option (i).

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Nucleic acids are polymers of nucleotides that are linked together by the phosphodiester linkage.


e.g., DNA, RNA etc.
Hence, the correct answer is option (ii).

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Glucose is found in the pyranose form, which has a five - membered ring structure.

Glucose does not give 2, 4 - DNP test as the aldehyde group is not free in the cyclic structure.

Glucose is an aldohexose having structural formula.

While heating with HI, glucose produces n hexane

Hence, the correct answer is option (iii).

Answer:

Each polypeptide in a protein contains amino acids that are linked in a specific sequence. The primary structure of proteins is defined as this sequence of amino acids. In this case, amino acids are linked together by a peptide bond or peptide linkage.
Hence, the correct answer is option (i).

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Both DNA and RNA have four bases. The four nitrogenous bases found in DNA are adenine, guanine, thymine and cytosine.
The four nitrogenous bases found in RNA are adenine, uracil, guanine and cytosine. Thymine does not exist in RNA.
Hence, the correct answer is option (iii).

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The B group vitamins are known as water soluble vitamins as they are easily excreted from the body.
Vitamin B12 can be stored in our body because it is insoluble in water.
Hence, the correct answer is option (iv).

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DNA has four bases
(a) adenine (A)
(b) thymine (T)
(c) guanine (G)
(d) cytosine (C)
It does not contain uracil.
Hence, the correct answer is option (iv).

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Anomers are cyclic structures of monosaccharides that differ in structure at carbon - 1.
Here, I and II are anomers because they differ only at carbon-I.
Hence, the correct answer is option (i).

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The pentaacetate of glucose doesn't react with hydroxylamine. This shows that the aldehyde in glucose is not free and is involved in the formation of the cyclic structure. As a result, this reaction of glucose can only be explained by its cyclic structure.
Hence, the correct answer is option (iii).

Answer:

All the three monosaccharides have D configuration as the hydroxyl group on the lowest asymmetric carbon is on the right hand side. 

Hence, the correct answer is option (i).

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Anomeric carbon is carbon that is adjacent to the oxygen atom in the cyclic structure of glucose or fructose 'a' and 'b' are present next to an oxygen atom, as shown in the structure above.
The hydroxyl group configuration of both carbons differ.
Hence, the correct answer is option (iii).

Answer:

The highest priority is given to the carbon containing carbon which is adjacent to O-atom to the other carbon atom ending at last CH₂OH group. In this order, numbering of the disaccharides can  be done as follows:



Hence, the correct answer is option (c).

Answer:

On hydrolysis, sucrose produces an equimolar mixture of $\mathrm{\alpha }$-D(+) glucose and $\mathrm{\beta }$-D(–) fructose. So, it is a disaccharide.

Also, in sucrose, these two monosaccharides are held together by a glycosidic linkage between C₁ of $\mathrm{\alpha }$-D(+) glucose and C₂ of $\mathrm{\beta }$-D(–) fructose as shown in the figure above.  Since the reducing groups of glucose and fructose are involved in glycosidic bond formation, sucrose is a non reducing sugar.

Hence, the correct answers are options (ii) and (iv).

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The proteins having a structure in which the chain of polypeptides coil around to give a spherical shape are termed as globular protein. Examples globular proteins are insulin and albumin.
Hence, the correct answer are option (i) and (iii).

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Amylopectin and glycogen have almost similar structure as those of glucose. Amylopectin is highly branched polymer of $\mathrm{\alpha }$-D-glucose unit. Also, glycogen is a multi branched polysacchoride of glucose.
Hence, the correct answer are option (ii) and (iv).

Answer:

Since, the structure (ii) and (iv) have more –COOH group attached to the protein unit as compared to –NH₂ group and are acidic in nature.
Hence, the correct answer are option (ii) and (iv).

Answer:

Lysine is an amino acid synthesized in the body, also it contains excess of –NH₂ group in its structure, therefore it is a basic $\mathrm{\alpha }$-amino acid.
Hence, the correct answer are option (i), (ii) and (iii).

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Ribose and fructose consists of  5 carbon atom containing polyhydroxy carbonyl compounds forming a five membered cyclic furanose structure.
Hence, the correct answer are option (i) and (iii).

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In the structure of fibrous protein both hydrogen and disulfide bond run in parallel to each other, giving it a fiber like structure.
Hence, the correct answers are options (ii) and (iv).

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In purine bases, a six membered and a five membered nitrogen ring are fused together. Guanine and adenine are purine bases whose structures are shown below.

Hence, the correct answers are options (i) and (ii).

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Enzymes are biocatalysts which play a major role in biochemical reactions. They are generally made up of proteins.
Hence, the correct answers are options (i) and (iv).

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Lactose is commonly known as milk sugar. It consists of two monosaccharide units- β-D-galactose and β-D-glucose. Such oligosaccharides are called disachharides. The linkage is between C1 of galactose and C4 of glucose.

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On prolonged heating with HI, glucose given n-hexane



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

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

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In polysaccharides, a large number of monosaccharide units are joined together by glycosidic linkages. It is an oxide linkage formed by the loss of a water molecule. For example,  β-glycosidic linkage in a polysachharide cellulose is shown below.

Answer:

Glucose is converted to gluconic acid by oxidization with bromine water and to saccharic acid when oxidation is carried by conc. HNO₃.
(i)

(ii)

Answer:

When aldehyde group is present, the monosaccharides are known as aldoses. When ketone group is present, the monosaccharides are known as ketoses. Fructose has molecular formula C₆H₁₂O₆ containing 6 carbon atoms and a keto group and is classified as ketohexose.

Answer:

Glyceraldehyde contains one asymmetric carbon atom and exists in two enantiomeric forms as shown below. 


 For assigning the configuration of monosaccharides, it is the lowest asymmetric carbon atom which is compared.


The given compound has L configuration due to the presence of OH on the left side as in the case of  ‘L’ (–) isomer of glyceraldehyde.

Answer:

For assigning the configuration of monosaccharides, it is the lowest asymmetric carbon atom which is compared with the configuration of that particular stereoisomer. This refers to their relation with one of the isomers of glyceraldehyde. 

The configuration assigned to ribose and 2-deoxyribose is D because their lowest asymmetric carbon atom, i.e., C3 has OH on right side in open chain or on lower side in cyclic form. Thus, ribose is β, D-ribose deoxyribose and  β-D-2 deoxyribose.

Answer:

Sucrose is dextrorotatory but after hydrolysis gives dextrorotatory glucose and levorototory fructose. Since the laevorotation of fructose (–92.4°) is more than dextroration of glucose (+52.5°), the mixture is laevorotatory. Thus, the hydrolysis of sucrose brings about a change in the sign of rotation, from dextro(+) to laevo(–), and the product is named as invert sugar.

Answer:

α-amino acids form polypeptide chains by elimination of water molecules.

Peptide linkage is the amide formed between −COOH group and −NH₂ group of two amino acid molecules.

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In α-helix, a polypeptide chain is stablised by the formation of hydrogen bonds between –NH – group of amino acids in one turn with the group of amino acids belonging to adjacent turn.

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Oxidoreductase enzymes are the class of enzymes which catalyse the oxidation of one substrate with simultaneous reduction of another substrate.

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When milk is curdled, its sugar get oxidized to form lactic acid by a bacteria. It is an example of denaturation of protein, where the hydrogen bonds of a protein are disturbed when it is subjected to some physical or chemical change.

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Glucose on acetylation with acetic anhydride forms a pentaderivative which confirms the presence of five –OH groups in glucose molecule.

Answer:

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, without which it does not react with NH₂OH group and hence doesn't form the oxime.

Answer:

It is recommended to take vitamin C on a daily basis or regularly because being water soluble, it gets flushed out of the body through urine. So, the amount of vitamin C does not get stored in the body and thus, needs to be supplied regularly in diet.

Answer:

Sucrose is dextrorotatory but after hydrolysis gives dextrorotatory glucose and levorototory fructose. Since the laevorotation of fructose (–92.4°) is more than dextroration of glucose (+52.5°), the mixture is laevorotatory.

Answer:

Amino acids behave like salts rather than simple amines or carboxylic acids due to the presence of both acidic (–COOH) and basic (–NH₂) groups. In solution, –COOH group can lose a proton and an amine group can accept a proton, giving rise to a dipolar ion called zwitter ion.

Answer:

In glycylalanine, the carboxyl group of glycine combine with the amino group of alanine.

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When a protein in its native form, is subjected to physical change like change in temperature or chemical change like change in pH, the hydrogen bonds are disturbed. Due to this, globules unfold and helix get uncoiled and protein loses its biological activity.

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Enzymes are biological catalysts required in need only in small quantities for the progress of a reaction. Similar to the action of chemical catalysts, enzymes are said to reduce the magnitude of activation energy to increase the rate of conversion of reactants into products. For example, the 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 hydrolyzed by the enzyme sucrase.

Answer:

Glucose reacts with hydroxylamine (NH₂OH) to form a monoxime and adds one molecule of hydrogen cyanide (HCN) to give a cyanohydrin. Thus, glucose contains a carbonyl group, which may be aldehydic or ketonic in nature.

Answer:

Nucleotides are joined together by phosphodiester linkage between 5′ and 3′ carbon atoms of the pentose sugar.
The word 'diester' in this linkage indicates that two –OH groups of phosphoric acid are involved in the formation of two ester linkages.
Phosphoric acid is involved in the formation of phosphodiester linkage.

Answer:

When two monosaccharides are linked to each other with an oxide linkage formed by the loss of a water molecule, the bond is a glycosidic linkage. For example, the glycosidic linkage between C1 of α-D-glucose and C2 of β-D-fructose in sucrose is shown below.




They are present in carbohydrates (disaccharides, oligosaccharides and polysaccharides).

Answer:

Starch consists of two components— Amylose and Amylopectin.
In amylose, α-D-(+)-glucose units are held together by C1–C4 glycosidic linkage while amylopectin is a branched chain polymer of α-D-glucose units in which chain is formed by C1–C4 glycosidic linkage whereas branching occurs by C1–C6 glycosidic linkage.
Glycogen is known as animal starch because its structure is similar to amylopectin.
Cellulose is composed of only β-D-glucose units which are joined by glycosidic linkage between C1 of one glucose unit and C4 of the next glucose unit.

Answer:

Enzymes contain cavities which are of characteristic shape and possess active groups such as -NH₂ , -COOH, -SH, -OH, etc. These are actually the active centres on the surface of enzyme particles. The molecules of the reactant (substrate), which have complementary shape, fit into these cavities just like a key fits into a lock.  Enzymes hold the substrate in a suitable position so that it can be attached by the reagent effectively. On account of the presence of active groups, an activated complex is formed which then decomposes to yield the products.

Answer:

All naturally occurring α-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 (–NH₂) group is present towards the right hand side while L-form shows the presence of –NH₂ group on the left hand side.

Answer:

On oxidation with nitric acid, glucose yields 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. Also, since one primary alcoholic (–OH) group is present, the other four –OH groups are secondary hydroxyl groups.

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When it is subjected to physical change like change in temperature, i.e., on boiling, coagulation of egg white occurs because hydrogen bonds get disturbed due to which globules unfold and helix gets uncoiled and protein losses its biological activity.

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Here, D represents relative configuration of glucose with respect to glyceraldehyde. D (+) – Glucose is dextrorotatory in nature because ‘(+)’ represents dextrorotatory nature of the molecule.

Hence, the correct answer is option (iii).

Answer:

Vitamin D can be stored in our body because vitamin D is fat soluble vitamin.

Hence, the correct answer is option (i).

Answer:

α-Glycosidic linkage is present in maltose because maltose is composed of two α-D-glucose units in which C-1 of one glucose is linked to C-4 another glucose unit.



Hence, the correct answer is option (iv).

Answer:

All naturally occurring α-amino acids except glycine are optically active. Glycine is optically inactive because glycine does not have all four different substituents attached to α-carbon, i.e., α-carbon atom is not asymmetric as shown below.

Most naturally occurring amino acids have L-configuration. L-Amino acids are represented by writing the –NH₂ group on left hand side.

Hence, the correct answer is option (v).

Answer:

Most of the carbohydrates were hydrates of carbon so compounds which follow C_x(H₂O)_y formula were considered as carbohydrates. But all the carbohydrates do not fit into this formula. Deoxyribose is a carbohydrate because carbohydrates are optically active polyhydroxy aldehydes or polyhydroxy ketones or substances which give such units on hydrolysis. 

Hence, the correct answer is option (ii).

Answer:

Glycine must not be taken through diet because it can be synthesised in our body and is a non-essential amino acid.

Hence, the correct answer is option (ii).

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In presence of enzyme, substrate molecules can be attacked by a reagent effectively because active sites of enzymes hold the substrate molecules in a suitable position. So, enzymes catalysed reaction are stereospecific reactions.

Hence, the correct answer is option (i).

Answer:

Following reaction and facts could not be explained by open-chain structure of glucose.

• Despite having the aldehyde group, glucose does not form the hydrogen sulphide addition product with NaHSO₃.
• The pentaacetate of glucose does not react with hydroxylamine indicating the absence of a free – CHO group. 
• Glucose is found to exist in two different crystalline forms which are named as α and β. The α-form of glucose (m.p. 419 K) is obtained by crystallisation from concentrated solution of glucose at 303 K while the β-form (m.p. 423 K) is obtained by crystallisation from hot and saturated aqueous solution at 371 K. 

Answer:

Glucose is an aldohexose and is also known as dextrose. It was assigned the structure given below.

(i) Its molecular formula was found to be C₆H₁₂O₆.
(ii) On prolonged heating with HI, it forms n-hexane, suggesting that all the six carbon atoms are linked in a straight chain.
(iii) Glucose reacts with hydroxylamine to form an oxime and adds a molecule of hydrogen cyanide to give cyanohydrin. These reactions confirm the presence of a carbonyl group (>C= O) in glucose.
(iv) Glucose get oxidized to six carbon carboxylic acid (gluconic acid) on reaction with a mild oxidizing agent line bromine water. This indicates that the carbonyl group is present as an aldehydic group.
(v) 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.
(vi) 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.

Answer:

Starch is the main storage polysaccharide of plants. It is the most important dietary source for human beings. The high content of starch is found in cereals.
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. When the body needs glucose, enzymes break the glycogen down to glucose. Glycogen is also found in yeast and fungi
Cellulose occurs exclusively in plants and it is the most abundant organic substances in the plant kingdom. The cell wall of bacteria and plants is made up of cellulose. We build furniture, etc. from cellulose in the form of wood and clothe ourselves with cellulose in the form of cotton fibre. 

Answer:

Proteins may have one or more polypeptide chains. Each polypeptide in a protein has amino acids linked with each other in a specific sequence and it is this sequence of amino acids that is said to be the primary structure of that protein.
Secondary structure of protein refers to the shape in which a long polypeptide chain can exist and includes α-helix and β-pleated sheets structure.
α-Helix is one of the most common ways in which a polypeptide chain forms all possible hydrogen bonds by twisting into a right handed screw (helix) with the –NH group of each amino acid residue hydrogen bonded to the >C=O of an adjacent turn of the helix.

In β-pleated sheet structure all peptide chains are stretched out to nearly maximum extension and then laid side by side which are held together by intermolecular hydrogen bonds. The structure resembles the pleated folds of drapery and therefore is known as β-pleated sheet.

Answer:

On complete hydrolysis of DNA, following fragments are formed:
A pentose sugar (β-D-2-deoxyribose), phosphoric acid (H₃PO₄) and bases (nitrogen containing hetero cyclic compounds).
(i) Sugar

(ii) Phosphoric acid

(ii) 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 known as nucleoside. In nucleosides, the sugar carbons are numbered as 1′, 2′, 3′, etc. in order to distinguish these from the bases. When nucleoside is linked to phosphoric acid at 5′-position of sugar moiety, we get a nucleotide. Nucleotides are joined together by phosphodiester linkage between 5′ and 3′ carbon atoms of the pentose sugar.

The diagram to show pairing of nucleotide bases in double helix of DNA. is shown below.