NCERT Solution Class 12th Chemistry Chapter – 14 Biomolecules
| Textbook | NCERT |
| class | Class – 12th |
| Subject | Chemistry |
| Chapter | Chapter – 14 |
| Chapter Name | Biomolecules |
| Category | Class 12th Chemistry Question & Answer |
| Medium | English |
| Source | last doubt |
NCERT Solution Class 12th Chemistry Chapter – 14 Biomolecules
?Chapter – 14?
✍Biomolecules✍
?Question & Answer?
Answer: For any compound to be water soluble, it should develop dipoles (partial negative and partial positive charges) at the two ends of compound. The development of charges results in the formation of hydrogen bond between the water molecule and the compound. The development of charges at two ends is due to the difference in the electronegativity between two atoms. The atom with higher electronegativity will acquire negative charge while the atom with lower electronegativity will acquire positive charge. A glucose molecule contains five –OH (highly electronegative) groups while a sucrose molecule contains eight −OH groups. Thus, glucose and sucrose undergo extensive H-bonding with water. Hence, these are soluble in water.But cyclohexane and benzene do not contain −OH groups. They contain only carbon and hydrogen atoms, as a result the dipole developed is very weak in nature and hence the hydrogen bond formed is not strong. Hence, they cannot undergo H-bonding with water and thus are insoluble in water.
Answer: Lactose is composed of β-D galactose and β-D glucose.
Thus, on hydrolysis, it gives β-D galactose and β-D glucose.
Answer: D-glucose reacts with hydroxylamine (NH2OH) to form an oxime because of the presence of aldehydic (-CHO) group or carbonyl carbon. This happens as the cyclic structure of glucose forms an open chain structure in an aqueous medium, which then reacts with NH2OH to give an oxime.
But pentaacetate of D-glucose does not react with NH2OH. This is because pentaacetate does not form an open chain structure.
Answer: Both acidic (carboxyl) as well as basic (amino) groups are present in the same molecule of amino acids. In aqueous solutions, the carboxyl group can lose a proton and the amino group can accept a proton, thus giving rise to a dipolar ion known as a zwitter ion.
Due to this dipolar behaviour, they have strong electrostatic interactions within them and with water. But halo-acids do not exhibit such dipolar behaviour.
For this reason, the melting points and the solubility of amino acids in water is higher than those of the corresponding halo-acids.
Answer: Denaturation of proteins is a process that changes the physical and biological properties of proteins without affecting the chemical composition of protein. In an egg, denaturation of protein is the coagulation of albumin present in the white of an egg. When an egg is boiled in water, the globular proteins present in it change to a rubber like insoluble mass which absorbs all the water present in the egg by making hydrogen bond with it.Answer: Vitamin C (Ascorbic acid) is a water soluble versatile vitamin. Humans cannot synthesize vitamin C due to the deficiency of a enzyme L- gulonolactone oxidase and also vitamin C is rapidly absorbed from the intestine. Because it is water soluble vitamin it is not stored in the body to a significant extent. As a result, it is readily excreted in the urine. Vitamin C is excreted as such or as its metabolite diketogulonic acid and oxalic acid in urineAnswer: The three dimensional structure of DNA (deoxyribose nucleic acid) was given by James Watson & Francis crick in 1953. They proposed that DNA polymer forms a duplex structure consisting of 2 strands of polynucleotide chains coiled around each other in the form of a double helix. The nucleotide form the backbone of DNA structure.When a nucleotide from the DNA containing thymine is hydrolysed, thymine β-D-2-deoxyribose and phosphoric acid are obtained as products.
Answer: RNA (Ribose nucleic acid) has a structure similar to DNA except that it is a single strand structure.A DNA molecule is double-stranded in which the pairing of bases occurs. Adenine always pairs with thymine, while cytosine always pairs with guanine. Therefore, on hydrolysis of DNA, the quantity of adenine produced is equal to that of thymine and similarly, the quantity of cytosine is equal to that of guanine.
But when RNA is hydrolysed, there is no relationship among the quantities of the different bases obtained. Hence, RNA is single-stranded.
Text solution
Answer: Monosaccharides are carbohydrates that cannot be hydrolysed further to give simpler units of poly-hydroxy aldehyde or ketone. Monosaccharides are classified on the bases of number of carbon atoms and the functional group present in them.Monosaccharides containing an aldehyde group are known as aldoses and those containing a -keto group are known as ketoses. Monosaccharides are further classified as trioses, tetroses, pentoses, hexoses, and heptoses according to the number of carbon atoms they contain. For example, a ketose containing 3 carbon atoms is called ketotriose and an aldose containing 3 carbon atoms is called aldotriose.
Answer: Reducing sugars are carbohydrates that reduce Fehling’s solution and Tollen’s reagent. All monosaccharides and disaccharides, excluding sucrose, are reducing sugars.Answer: Two main functions of carbohydrates in plants are:(i) Polysaccharides such as starch serve as storage molecules.
(ii) Cellulose, a polysaccharide, is used to build the cell wall.
Q4 Classify the following into monosaccharides and disaccharides.Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose
Answer: Monosaccharides:Ribose, 2-deoxyribose, galactose, fructose
Disaccharides: Maltose, lactose
Answer: Glycosidic linkage refers to the linkage formed between two monosaccharide units through an oxygen atom by the loss of a water molecule.For example, in a sucrose molecule, two monosaccharide units, ∝-glucose and β-fructose, are joined together by a glycosidic linkage.
Answer: Glycogen is a carbohydrate (polysaccharide). In animals, carbohydrates are stored as glycogen.Starch is a carbohydrate consisting of two components – amylose (15 – 20%) and amylopectin (80 – 85%).
However, glycogen consists of only one component whose structure is similar to amylopectin. Also, glycogen is more branched than amylopectin.
Answer: (i) On hydrolysis, sucrose gives one molecule of ∝-D glucose and one molecule of β – D-fructose.
(ii) The hydrolysis of lactose gives β-D-galactose and β-D-glucose.
Answer: Starch consists of two components – amylose and amylopectin. Amylose is a long linear chain of ∝-D-(+)-glucose units joined by C1-C4 glycosidic linkage (∝-link).
Amylopectin is a branched-chain polymer of ∝-D-glucose units, in which the chain is formed by C1-C4 glycosidic linkage and the branching occurs by C1-C6 glycosidic linkage.
On the other hand, cellulose is a straight-chain polysaccharide of β-D-glucose units joined by C1-C4 glycosidic linkage (β-link).
Answer: (i) When D-glucose is heated with HI for a long time, n-hexane is formed.
(ii) When D-glucose is treated with Br2 water, D- gluconic acid is produced.
(iii) On being treated with HNO3, D-glucose get oxidised to give saccharic acid.
Answer: 1) Aldehydes give 2, 4-DNP test, Schiff’s test, and react with NaHSO4 to form the hydrogen sulphite addition product. However, glucose does not undergo these reactions.
(2) The pentaacetate of glucose does not react with hydroxylamine. This indicates that a free -CHO group is absent from glucose.
(3) Glucose exists in two crystalline forms – ∝ and β. The ∝-form (m.p. = 419 K) crystallises from a concentrated solution of glucose at 303 K and the β-form (m.p = 423 K) crystallises from a hot and saturated aqueous solution at 371 K. This behaviour cannot be explained by the open chain structure of glucose.
Answer: Essential amino acids are required by the human body, but they cannot be synthesised in the body. They must be taken through food.For example: valine and leucine
Non-essential amino acids are also required by the human body, but they can be synthesised in the body.
For example: glycine and alanine
Q12 Define the following as related to proteins(i) Peptide linkage (ii) Primary structure (iii) Denaturation.
Answer: (i) Peptide linkage:The amide formed between -COOH group of one molecule of an amino acid and -NH2 group of another molecule of the amino acid by the elimination of a water molecule is called a peptide linkage.
(ii) Primary structure:
The primary structure of protein refers to the specific sequence in which various amino acids are present in it, i.e., the sequence of linkages between amino acids in a polypeptide chain. The sequence in which amino acids are arranged is different in each protein. A change in the sequence creates a different protein.
(iii) Denaturation:
In a biological system, a protein is found to have a unique 3-dimensional structure and a unique biological activity. In such a situation, the protein is called native protein. However, when the native protein is subjected to physical changes such as change in temperature or chemical changes such as change in pH, its H-bonds are disturbed. This disturbance unfolds the globules and uncoils the helix. As a result, the protein loses its biological activity. This loss of biological activity by the protein is called denaturation. During denaturation, the secondary and the tertiary structures of the protein get destroyed, but the primary structure remains unaltered.
One of the examples of denaturation of proteins is the coagulation of egg white when an egg is boiled.
Answer: There are two common types of secondary structure of proteins:(i) ∝-helix structure
(ii) β-pleated sheet structure
∝– Helix structure:
In this structure, the -NH group of an amino acid residue forms H-bond with the group of the adjacent turn
of the right-handed screw (∝-helix).
β-pleated sheet structure:
This structure is called so because it looks like the pleated folds of drapery. In this structure, all the peptide chains are stretched out to nearly the maximum extension and then laid side by side. These peptide chains are held together by intermolecular hydrogen bonds.
Answer: The H-bonds formed between the -NH group of each amino acid residue and the group of the adjacent turns of the ∝-helix help in stabilising the helix.Answer:
| Fibrous protein | Globular protein | ||
| 1. | It is a fibre-like structure formed by the polypeptide chain. These proteins are held together by strong hydrogen and disulphide bonds. | 1. | The polypeptide chain in this protein is folded around itself, giving rise to a spherical structure. |
| 2. | It is usually insoluble in water. | 2. | It is usually soluble in water. |
| 3. 4. 5. | Fibrous proteins are usually used for structural purposes. For example, keratin is present in nails and hair; collagen in tendons; and myosin in muscles. Fibrous proteins are made up of regular amino acid proteins. Fibrous proteins are less sensitive to any changes in pH or temperature. | 3. 4. 5. | All enzymes are globular proteins. Some hormones such as insulin are also globular proteins. The amino acid sequece is irregular in globular proteins. Globular proteins are sensitive to any changes in pH, temperature etc. |
Q 16. How do you explain the amphoteric behaviour of amino acids?
Answer:In the presence of water or aqueous solution, the carboxyl group of an amino acid can lose a proton and the amino group can accept a proton to give a dipolar ion known as zwitterion.
Therefore, the amino acid can act both as an acid and as a base, in the presence of zwitterionic form.
So. The amino acid show amphoteric behaviour.
Q 17. What are enzymes?
Answer:The protein that catalyses the biological reactions are called enzymes. They are very particular in nature and for some specific substrate, they catalyse particular reactions.
The enzymes are named after a particular reaction or in common bases, they are named after a particular class of substrate.
Example: Maltase is the enzymes which are used to catalyse the hydrolysis of maltose into glucose.
Also, oxidoreductase enzymes are those which are used to catalyse the oxidation of one substrate with the simultaneous reaction of another substrate.
The name of an enzyme ends with “ – ase ”
Q 18. What is the effect of denaturation on the structure of proteins?
Answer:The outcome of denaturation, helixes get uncoiled and globules get unfolded. There would be no change in the primary structure of the protein while the secondary and the tertiary structure gets destroyed. We can say that the secondary and the tertiary – structured proteins are changed into primary – structured proteins. Also, because of the loss of secondary and the tertiary structure the enzymes loses its activity.
Q 19. How are vitamins classified? Name the vitamin responsible for the coagulation of blood.
Answer: On the basis of their solubility in water or fat, vitamins are classified into two groups.
(i)Fat-soluble vitamins: Vitamins that are soluble in fat and oils, but not in water, belong to this group. For example: Vitamins A, D, E, and K
(ii)Water-soluble vitamins:Vitamins that are soluble in water belong to this group. For example: B group vitamins (B1, B2, B6, B12, etc.) and vitamin C
However, biotin or vitamin H is neither soluble in water nor in fat.
Vitamin K is responsible for the coagulation of blood.
Q 20. Why are vitamin A and vitamin C essential to us? Give their important sources.
Answer: These two vitamins are essential to us because the deficiency of these two vitamins causes us harmful disease like the deficiency of vitamin causes us xerophthalmia (hardens the cornea of the eye) night blindness. While the deficiency of vitamin C causes scurvy (bleeding gums).
The sources of these two vitamins are:
Vitamin A : Carrots, fish liver oil, milk and butter.
Vitamin C: amla, citrus fruits and green leafy vegetables.
Q 21. What are nucleic acids? Mention their two important functions.
Answer: It is a molecule which is found as one of the constituents of chromosomes which is found in the nuclei of all the living cells.
Nucleic acid can be categorised into two categories: ribonucleic acid (RNA) and deoxyribonucleic acid (DNA).
Nucleic acids are long-chain polymers of nucleotides, so they are also known as polynucleotides.
(i) It is responsible for heredity. In heredity, there is a transfer of inherent characters from one generation to another. This process is held by the DNA.
(ii) The protein cell synthesis is held by the Nucleic acid (both RNA and DNA). The protein synthesis is majorly done by the various RNA molecules in a cell while DNA contains the message for the synthesis of a specific protein.
Q 22. What is the difference between a nucleoside and a nucleotide?
Answer:A Nucleotide is formed by the combination of all the three basic components of nucleic acids (i.e., base, a pentose sugar, and phosphoric acid).
Therefore, Nucleotide = Base + Sugar + Phosphoric acid
On the other hand, A nucleoside is formed by the attachment of a base to 1’ position of the sugar.
Nucleoside = Sugar + Base
Q 23. The two strands in DNA are not identical but are complementary. Explain.
Answer: In the helical structure of DNA, the hydrogen bond holds the two strands between specific pairs of bases. Adenine forms a hydrogen bond with thymine, while cytosine forms a hydrogen bond with guanine. SO, as its result, the two strands acts as a complementary for each other.
Q 24. Write the important structural and functional differences between DNA and RNA.
Answer: The difference on the basis of their functions is:
| DNA | RNA | ||
| 1 | DNA is the chemical basis of heredity. | 1 | RNA is not responsible for heredity. |
The differences on the basis of their structures are as follows:
| DNA | RNA | ||
| 1 | The sugar moiety in DNA molecules is \beta -D-2 deoxyribose. | 1 | The sugar moiety in RNA molecules is \beta -D-ribose. |
| 2 | Bases are Adenine(A), Guanine(G), Cytosine(C), Thymine(T). | 2 | The bases are Adenine(A), Guanine(G), Cytosine(C), Uracil(U). |
| 3 | The helical structure of DNA is double-stranded. | 3 | The helical structure of RNA is single-stranded. |
Q25 What are the different types of RNA found in the cell?
Answer: (i) Messenger RNA (m-RNA)
(ii) Ribosomal RNA (r-RNA)
(iii) Transfer RNA (t-RNA)