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Non-Rationalised Science NCERT Notes and Solutions (Class 12th)
Physics Chemistry Biology

Class 12th (Chemistry) Chapters
1. The Solid State 2. Solutions 3. Electrochemistry
4. Chemical Kinetics 5. Surface Chemistry 6. General Principles And Processes Of Isolation Of Elements
7. The P-Block Elements 8. The D-And F-Block Elements 9. Coordination Compounds
10. Haloalkanes And Haloarenes 11. Alcohols, Phenols And Ethers 12. Aldehydes, Ketones And Carboxylic Acids
13. Amines 14. Biomolecules 15. Polymers
16. Chemistry In Everyday Life



Chapter 14 Biomolecules



Carbohydrates

Carbohydrates are a large class of natural organic compounds primarily produced by plants. Common examples include sugar, glucose, and starch. Historically, they were thought to be hydrates of carbon with the general formula $\textsf{C}_x\text{(H}_2\text{O})_y$. While glucose ($\textsf{C}_6\text{H}_{12}\text{O}_6$) fits this ($\textsf{C}_6\text{(H}_2\text{O)}_6$), some compounds fitting the formula (like acetic acid, $\textsf{C}_2\text{(H}_2\text{O)}_2$) are not carbohydrates, and some carbohydrates (like rhamnose, $\textsf{C}_6\text{H}_{12}\text{O}_5$) do not fit the formula.

Chemically, carbohydrates are defined as **optically active polyhydroxy aldehydes or ketones** or compounds that yield such units upon hydrolysis. Many carbohydrates are sweet and called sugars (saccharides, from Greek 'sakcharon'). Common sugars include sucrose (cane sugar) and lactose (milk sugar).


Classification Of Carbohydrates

Carbohydrates are classified based on their hydrolysis behaviour:

Carbohydrates can also be classified based on their ability to reduce Fehling's solution and Tollens' reagent:

Monosaccharides are further classified by the number of carbon atoms and the functional group:

Combinations are used for naming, e.g., aldotriose (3C aldose), ketohexose (6C ketose).


Monosaccharides

Important monosaccharides include Glucose and Fructose.


Glucose

Glucose is a common monosaccharide, an aldohexose ($\textsf{C}_6\text{H}_{12}\text{O}_6$). It exists freely in nature (sweet fruits, honey, ripe grapes) and in combined form (in disaccharides and polysaccharides).

Preparation of Glucose:

Structure of Glucose: Glucose is also known as dextrose. It is the monomer unit for starch, cellulose, and glycogen. Its open-chain structure ($\textsf{C}_6\text{H}_{12}\text{O}_6$) was determined through various reactions:

The exact spatial arrangement of the –OH groups (configuration) was determined by Fischer. D-glucose is represented with the –OH on the lowest asymmetric carbon (C-5) on the right side, correlating it to D-(+)-glyceraldehyde. The (+) indicates dextrorotatory nature, which is not directly related to the D/L configuration notation.

Fischer projection of D-(+)-Glucose, D-(+)-Glyceraldehyde, Gluconic acid, and Saccharic acid.

Cyclic Structure of Glucose: The open-chain structure did not explain all properties of glucose, such as the absence of Schiff's test, no reaction with $\textsf{NaHSO}_3$ to form an addition product, and the existence of two crystalline forms ($\alpha$ and $\beta$) with different melting points and optical rotations.

This was explained by proposing that glucose exists primarily in cyclic hemiacetal forms formed by the reaction between the aldehyde group (C-1) and the hydroxyl group at C-5. This forms a six-membered ring, analogous to pyran (a cyclic ether with five carbons and one oxygen). The two cyclic forms, $\alpha$ and $\beta$, differ in the configuration at C-1 (the original aldehyde carbon), which is called the anomeric carbon. These two isomers ($\alpha$-anomer and $\beta$-anomer) are called anomers and exist in equilibrium with a small amount of the open-chain form in aqueous solution.

Cyclic hemiacetal structures of alpha-D-Glucose and beta-D-Glucose in equilibrium with open chain form.

The cyclic structures are more accurately represented by **Haworth structures**.

Haworth structures of alpha-D-Glucopyranose and beta-D-Glucopyranose.

Mutarotation: The specific rotation of glucose in water changes over time until it reaches an equilibrium value. This phenomenon, called mutarotation, is due to the interconversion between the $\alpha$ and $\beta$ cyclic forms via the open-chain intermediate.


Fructose

Fructose is an important **ketohexose** ($\textsf{C}_6\text{H}_{12}\text{O}_6$). It is obtained along with glucose from the hydrolysis of sucrose and is found naturally in fruits, honey, and vegetables. It is a laevorotatory compound and is designated as D-(-)-fructose. Its open-chain structure has a ketonic group at C-2 and a straight chain of six carbons.

Fischer projection of D-(-)-Fructose open chain.

Fructose also exists in cyclic forms due to the addition of the –OH group at C-5 to the ketone group (C-2). This forms a five-membered ring, analogous to furan, and the cyclic forms are called furanose structures. Like glucose, it exists as $\alpha$ and $\beta$ anomers, differing in configuration at the anomeric carbon (C-2).

Haworth structures of alpha-D-Fructofuranose and beta-D-Fructofuranose.


Disaccharides

Disaccharides yield two monosaccharide units upon hydrolysis. The two units are linked by a **glycosidic linkage**, an oxide linkage formed by the loss of a water molecule between the anomeric carbon of one monosaccharide and a hydroxyl group of another. If the anomeric carbons of both monosaccharides are involved in the glycosidic linkage, the disaccharide is non-reducing (e.g., sucrose). If one anomeric carbon is free, it's a reducing sugar (e.g., maltose, lactose).


Polysaccharides

Polysaccharides are polymers of many monosaccharide units joined by glycosidic linkages. They serve primarily as food storage or structural materials.


Importance Of Carbohydrates

Carbohydrates are vital for life:

Intext Question 14.1. Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six membered ring compounds) are insoluble in water. Explain.

Answer:

Glucose and sucrose are soluble in water because they contain multiple hydroxyl (–OH) groups. These –OH groups are polar and can form strong **hydrogen bonds** with water molecules. The formation of these hydrogen bonds overcomes the intermolecular forces holding the carbohydrate molecules together and allows them to disperse in water.

Cyclohexane and benzene, despite being six-membered ring compounds, are hydrocarbons. They consist only of carbon and hydrogen atoms, with no highly polar functional groups like hydroxyl groups. They are non-polar molecules and cannot form hydrogen bonds with water. Water is a polar solvent. Non-polar solutes do not dissolve significantly in polar solvents because the energy gained from solute-solvent interactions (weak van der Waals forces) is not sufficient to overcome the strong hydrogen bonding between water molecules and the intermolecular forces within the non-polar solute.

Intext Question 14.2. What are the expected products of hydrolysis of lactose?

Answer:

Lactose is a disaccharide found in milk. Upon hydrolysis with dilute acid or enzyme (lactase), it breaks down into two monosaccharide units. As discussed in the Disaccharides section, lactose is composed of $\beta$-D-galactose and $\beta$-D-glucose.

The expected products of hydrolysis of lactose are **$\beta$-D-galactose and $\beta$-D-glucose**.

$\textsf{Lactose} \xrightarrow{\textsf{H}_2\text{O / H}^+ \text{ or Lactase}} \beta\textsf{-D-Galactose} + \beta\textsf{-D-Glucose}$

Intext Question 14.3. How do you explain the absence of aldehyde group in the pentaacetate of D-glucose?

Answer:

D-glucose exists primarily in two cyclic hemiacetal forms ($\alpha$ and $\beta$) in equilibrium with a small amount of the open-chain aldehyde form. In the cyclic hemiacetal structure, the aldehyde group (at C-1) has reacted with the hydroxyl group at C-5 to form a ring. The carbon atom C-1 in the cyclic structure is a hemiacetal carbon, not a free aldehyde group.

When glucose is acetylated with acetic anhydride, all the free hydroxyl groups (including the hydroxyl group at the anomeric carbon, C-1, which is part of the hemiacetal) are acetylated. The molecule becomes glucose pentaacetate. Since all five hydroxyl groups (including the hemiacetal –OH at C-1) are esterified, the cyclic structure is "locked" in place by the ester linkages, and the ring cannot easily open to regenerate the free aldehyde group that would be present in the open-chain form.

Therefore, the absence of a free aldehyde group in glucose pentaacetate indicates that glucose reacted in its cyclic (hemiacetal) form, and all the hydroxyl groups present in that cyclic structure were acetylated, preventing the molecule from reverting to the open-chain aldehyde form under typical conditions for aldehyde tests.



Proteins

Proteins are the most abundant biomolecules in living systems, crucial for growth, maintenance, structure, and function. They are polymers of **$\alpha$-amino acids**.


Amino Acids

Amino acids contain both an **amino group (–NH$_2$)** and a **carboxyl group (–COOH)**. In $\alpha$-amino acids (the building blocks of proteins), both functional groups are attached to the same carbon atom, called the $\alpha$-carbon. The $\alpha$-carbon also carries a hydrogen atom and a variable side chain (R group).

Structure of an alpha-amino acid.

Over 20 different $\alpha$-amino acids are commonly found in proteins. They have trivial names (e.g., Glycine for its sweet taste, Tyrosine first from cheese). They are represented by three-letter (e.g., Gly, Ala) or one-letter (e.g., G, A) symbols.

Examples of common natural amino acids:

Name Side Chain (R) Three-letter symbol One-letter code
GlycineH–GlyG
Alanine$\textsf{CH}_3\text{–}$AlaA
Valine*$\textsf{(CH}_3)_2\text{CH–}$ValV
Leucine*$\textsf{(CH}_3)_2\text{CH–CH}_2\text{–}$LeuL
Isoleucine*$\textsf{CH}_3\text{–CH}_2\text{–CH(CH}_3)\text{–}$IleI
Serine$\textsf{HO–CH}_2\text{–}$SerS
Threonine*$\textsf{CH}_3\text{–CH(OH)–}$ThrT
Cysteine$\textsf{HS–CH}_2\text{–}$CysC
Methionine*$\textsf{CH}_3\text{–S–CH}_2\text{–CH}_2\text{–}$MetM
Aspartic acid$\textsf{HOOC–CH}_2\text{–}$AspD
Glutamic acid$\textsf{HOOC–CH}_2\text{–CH}_2\text{–}$GluE
Asparagine$\textsf{H}_2\text{N–CO–CH}_2\text{–}$AsnN
Glutamine$\textsf{H}_2\text{N–CO–CH}_2\text{–CH}_2\text{–}$GlnQ
Lysine*$\textsf{H}_2\text{N–(CH}_2)_4\text{–}$LysK
Arginine*$\textsf{HN=C(NH}_2)\text{–NH–(CH}_2)_3\text{–}$ArgR
Histidine*
Histidine side chain structure.
HisH
Phenylalanine*$\textsf{C}_6\text{H}_5\text{–CH}_2\text{–}$PheF
Tyrosine$\textsf{(p)HO–C}_6\text{H}_4\text{–CH}_2\text{–}$TyrY
Tryptophan*
Tryptophan side chain structure.
TrpW
Proline
Proline side chain structure.
(cyclic)
ProP

* denotes essential amino acids.


Classification Of Amino Acids

Amino acids are classified based on the net charge of their side chain (R group) or the ratio of amino to carboxyl groups:

Based on nutritional requirements:

Amino acids are typically colourless, crystalline, water-soluble solids with high melting points, behaving like salts rather than simple amines or carboxylic acids. This is due to the formation of a **zwitter ion** (dipolar ion) in aqueous solution. The carboxyl group loses a proton, which is accepted by the amino group, resulting in a molecule with both positive and negative charges but overall neutral.

Formation of zwitter ion in amino acid.

Zwitter ions exhibit **amphoteric behaviour**, reacting with both acids (accepting a proton at the carboxylate group) and bases (donating a proton from the ammonium group).

Except for glycine (where R=H), all other $\alpha$-amino acids have an asymmetric $\alpha$-carbon atom and are therefore **optically active**, existing as enantiomers (D and L forms). Most naturally occurring amino acids have the **L-configuration**, represented with the –NH$_2$ group on the left in a Fischer projection.

Intext Question 14.4. The melting points and solubility in water of amino acids are generally higher than that of the corresponding halo acids. Explain.

Answer:

Amino acids exist as **zwitter ions** (dipolar ions) in their crystalline state and in aqueous solution. The zwitter ion is a highly polar species with distinct positive and negative charges within the same molecule. These strong electrostatic attractions between the positive and negative centers of different zwitter ions lead to strong intermolecular forces (ionic interactions, specifically salt bridges) in the solid state. These strong forces require significant energy to overcome, resulting in high melting points.

Furthermore, the charged nature of the zwitter ion allows amino acids to form strong **ion-dipole interactions** with polar water molecules, making them highly soluble in water. Hydrogen bonding is also involved.

Corresponding halo acids (e.g., $\textsf{CH}_2\text{ClCOOH}$) are polar due to the C–Cl bond and the carboxyl group, and can form hydrogen bonds through the carboxyl group. However, they do not form distinct positive and negative charges within the same molecule to the extent that amino acids do in their zwitterionic form. The intermolecular forces in solid halo acids are primarily dipole-dipole interactions and hydrogen bonds, which are generally weaker than the ionic interactions/salt bridges present in crystalline amino acids.

Similarly, while halo acids can form hydrogen bonds with water, they lack the strong ion-dipole interactions that contribute to the high solubility of zwitterionic amino acids.

Thus, the salt-like zwitterionic nature of amino acids accounts for their higher melting points and greater water solubility compared to simple halo acids.


Structure Of Proteins

Proteins are polymers of $\alpha$-amino acids linked by **peptide bonds** (or peptide linkages). A peptide bond is an amide linkage formed between the carboxyl group of one amino acid and the amino group of another, with the elimination of a water molecule.

Formation of a peptide bond between two amino acids.

Linking two amino acids forms a **dipeptide**. Three amino acids linked by two peptide bonds form a **tripeptide**, and so on. When many amino acids (more than ten) are linked, the product is called a **polypeptide**. A protein is generally defined as a polypeptide with more than 100 amino acid residues and a molecular mass above 10,000 u, although this distinction isn't always strict (e.g., insulin has 51 amino acids but is a protein). Amino acid residues in a polypeptide chain are linked sequentially, starting from the N-terminus (free amino group) to the C-terminus (free carboxyl group).

Proteins are classified by molecular shape:

Protein structure is described at four levels of complexity:


Denaturation Of Proteins

A protein in its biologically active, three-dimensional structure is called a **native protein**. Denaturation occurs when a protein's native structure is disrupted by physical (heat, radiation) or chemical (acids, bases, organic solvents) agents.

Denaturation primarily affects the **secondary, tertiary, and quaternary structures** (if present) by breaking the hydrogen bonds, ionic bonds, and van der Waals forces that stabilise them. The polypeptide chains unfold, and the helical/folded conformations are lost. The **primary structure (amino acid sequence) remains intact**.

Upon denaturation, the protein loses its specific three-dimensional shape and, consequently, its biological activity. Common examples include the coagulation of egg white upon boiling (albumins denature) and the curdling of milk due to lactic acid (casein denatures).

Intext Question 14.5. Where does the water present in the egg go after boiling the egg?

Answer:

The egg white consists mainly of water and dissolved proteins (like albumin). In the raw egg white, these proteins are folded into their globular, soluble native state, surrounded by water molecules. When the egg is boiled, the heat causes the proteins to undergo **denaturation**. The protein molecules unfold, disrupting their secondary and tertiary structures. As they unfold, hydrophobic regions of the protein molecules that were previously tucked inside become exposed. These exposed regions interact with each other and with other protein molecules, causing the proteins to aggregate or coagulate into a solid mass.

The water molecules that were surrounding the individual soluble protein molecules become trapped within the network of these coagulated, denatured protein aggregates. The water is not chemically changed or removed; it is physically immobilised within the solid protein matrix. This process is similar to how gels form, where a liquid is held within a solid network.

So, the water doesn't "go away"; it gets **trapped within the coagulated protein matrix** as the egg white solidifies.



Enzymes

Enzymes are **biocatalysts** that significantly speed up chemical reactions within living organisms under mild conditions (physiological temperature and pH). They are essential for processes like digestion, energy production, and synthesis of molecules.

Almost all enzymes are **globular proteins**. They are highly **specific** for the reaction they catalyse and the substrate they act upon. Enzymes are typically named after the substrate they act on or the type of reaction they catalyse, with the ending '-ase' (e.g., maltase catalyses maltose hydrolysis, oxidoreductase catalyses oxidation-reduction reactions).

Enzymes work by **reducing the activation energy** of a reaction, similar to chemical catalysts. For example, the activation energy for sucrose hydrolysis is much lower when catalysed by sucrase than by acid.


Mechanism Of Enzyme Action

The mechanism of enzyme action involves the enzyme binding to its specific substrate(s) at a region called the **active site**, forming an enzyme-substrate complex. The active site has a specific shape that fits the substrate (like a lock and key or induced fit model). Within the enzyme-substrate complex, the enzyme facilitates the chemical transformation of the substrate into product(s). The product(s) are then released from the active site, and the enzyme is free to catalyse the reaction again.

Key aspects of the mechanism:

Enzymes achieve catalysis through various means, including orienting substrates correctly, providing a favourable microenvironment, straining substrate bonds, and participating directly in the reaction via amino acid side chains at the active site.

Only small quantities of enzymes are needed because they are not consumed in the reaction.



Vitamins

Vitamins are organic compounds required in the diet in small amounts for normal growth, health, and maintenance of biological functions. They are considered **essential food factors** because the body generally cannot synthesise them (though some gut bacteria produce certain vitamins). Deficiency of vitamins causes specific diseases.

The term "Vitamine" was originally coined from 'vital' and 'amine' because early identified vitamins contained amino groups, but this is not true for all vitamins. The final 'e' was dropped, resulting in the term 'vitamin'.


Classification Of Vitamins

Vitamins are classified based on their solubility:

Examples of important vitamins, their sources, and deficiency diseases:

Sl. No. Name of Vitamin Sources Deficiency diseases
1.Vitamin AFish liver oil, carrots, butter, milkXerophthalmia (eye cornea hardening), Night blindness
2.Vitamin B$_1$ (Thiamine)Yeast, milk, green vegetables, cerealsBeri beri (loss of appetite, retarded growth)
3.Vitamin B$_2$ (Riboflavin)Milk, egg white, liver, kidneyCheilosis (fissuring at mouth corners), digestive disorders, skin burning sensation
4.Vitamin B$_6$ (Pyridoxine)Yeast, milk, egg yolk, cereals, gramsConvulsions
5.Vitamin B$_{12}$Meat, fish, egg, curdPernicious anaemia (haemoglobin deficient RBCs)
6.Vitamin C (Ascorbic acid)Citrus fruits, amla, green leafy vegetablesScurvy (bleeding gums)
7.Vitamin DSunlight exposure, fish, egg yolkRickets (bone deformities in children), Osteomalacia (soft bones, joint pain in adults)
8.Vitamin EVegetable oils (wheat germ oil, sunflower oil)Increased RBC fragility, muscular weakness
9.Vitamin KGreen leafy vegetablesIncreased blood clotting time

Excess vitamin intake can also be harmful; supplementation should be done under medical advice.



Nucleic Acids

Nucleic acids are biomolecules found in the nucleus of living cells, responsible for transmitting inherent characters (heredity) from one generation to the next. They are long chain polymers of repeating units called **nucleotides**, and are hence called **polynucleotides**. The two main types are **deoxyribonucleic acid (DNA)** and **ribonucleic acid (RNA)**.


Chemical Composition Of Nucleic Acids

Complete hydrolysis of DNA or RNA yields three components:

1. A pentose sugar: A five-carbon sugar. In DNA, it is $\beta$-D-2-deoxyribose. In RNA, it is $\beta$-D-ribose.

Structures of beta-D-Ribose and beta-D-2-Deoxyribose.

2. Phosphoric acid: $\textsf{H}_3\text{PO}_4$.

3. Nitrogen-containing heterocyclic compounds (Bases):

Structures of the DNA and RNA bases: Adenine, Guanine, Cytosine, Thymine, Uracil.

Note: DNA contains A, G, C, T. RNA contains A, G, C, U.


Structure Of Nucleic Acids

A **nucleoside** is formed by the attachment of a base to the 1' position of the pentose sugar. The carbons in the sugar ring are numbered 1', 2', 3', 4', 5' to distinguish them from the atoms in the base.

A **nucleotide** is formed when a nucleoside is linked to a phosphoric acid molecule at the 5' position of the sugar moiety (via a phosphodiester bond). It consists of a base, a sugar, and a phosphate group.

Structure of a nucleoside and a nucleotide.

Nucleotides link together to form the polynucleotide chain of nucleic acids. They are joined by **phosphodiester linkages** between the 5' carbon of one nucleotide's sugar and the 3' carbon of the next nucleotide's sugar.

Formation of a dinucleotide showing phosphodiester linkage.

The sequence of nucleotides in the nucleic acid chain constitutes its **primary structure**. Nucleic acids also have secondary structures.

DNA Secondary Structure: James Watson and Francis Crick proposed the **double helix** structure for DNA. Two polynucleotide strands coil around each other, held together by hydrogen bonds between specific pairs of bases on opposite strands. The strands are **complementary**: Adenine (A) always pairs with Thymine (T) via two hydrogen bonds (A=T), and Cytosine (C) always pairs with Guanine (G) via three hydrogen bonds (C$\equiv$G). The sugar-phosphate backbone forms the outside of the helix, and the bases are stacked inside. The two strands run in opposite directions (antiparallel).

Double strand helix structure of DNA showing base pairing.

RNA Secondary Structure: RNA is typically a **single-stranded** molecule. However, it can fold back on itself to form regions of double helix within the single strand, stabilised by complementary base pairing (A=U, C$\equiv$G). This folding creates complex 3D structures essential for RNA function.

There are three main types of RNA with different functions:


Biological Functions Of Nucleic Acids

Nucleic acids are fundamental for life processes:

**DNA fingerprinting** is a technique that uses the unique sequence of bases in an individual's DNA for identification purposes in forensics, paternity testing, identifying remains, and studying evolutionary relationships. It is highly reliable as DNA sequence is unique and stable.



Hormones

Hormones are signaling molecules that act as **intercellular messengers** in the body. They are produced by endocrine glands and transported via the bloodstream to target tissues where they exert specific effects. Hormones help to maintain **homeostasis** (balance of biological activities) and regulate various physiological processes.

Chemically, hormones are diverse:

Functions of hormones include:

Dysfunction of endocrine glands or hormone production can lead to various diseases (e.g., hypothyroidism/hyperthyroidism from thyroid issues, Addison's disease from adrenal cortex dysfunction).

Intext Question 14.6. Why cannot vitamin C be stored in our body?

Answer:

Vitamin C (Ascorbic acid) is a **water-soluble vitamin**. Water-soluble vitamins are readily absorbed into the bloodstream and circulate in the body's aqueous environment. They are not significantly stored in fatty tissues or the liver (unlike fat-soluble vitamins).

Because they are soluble in water, excess amounts of vitamin C (beyond what the body needs and can utilise immediately) are efficiently filtered by the kidneys and excreted in the urine. This continuous excretion prevents significant accumulation or storage in the body.

Therefore, vitamin C needs to be consumed regularly in the diet to maintain adequate levels in the body.

Intext Question 14.7. What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?

Answer:

A nucleotide from DNA consists of three components: a base, a pentose sugar, and a phosphate group. In DNA, the sugar is $\beta$-D-2-deoxyribose. The base is either Adenine (A), Guanine (G), Cytosine (C), or Thymine (T). The question specifies a nucleotide containing thymine (T) from DNA.

Complete hydrolysis breaks the bonds linking these components.

  • The bond between the base (Thymine) and the sugar (2-deoxyribose) is a glycosidic bond.
  • The bond between the sugar (2-deoxyribose) and the phosphate group is a phosphodiester bond (specifically a phosphoester bond in a single nucleotide unit).

Therefore, hydrolysis of a DNA nucleotide containing thymine would break these bonds, yielding the individual components:

1. **Thymine (T)** (the base)

2. **$\beta$-D-2-deoxyribose** (the pentose sugar specific to DNA)

3. **Phosphoric acid ($\textsf{H}_3\text{PO}_4$)** (the phosphate group)

Intext Question 14.8. When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA?

Answer:

In DNA, there is a specific relationship between the quantities of bases: the amount of Adenine (A) equals the amount of Thymine (T), and the amount of Cytosine (C) equals the amount of Guanine (G) (Chargaff's rules, A=T and C=G). This is because DNA exists as a double helix where A on one strand always pairs with T on the other, and C on one strand always pairs with G on the other.

The observation that there is no specific relationship among the quantities of different bases obtained upon hydrolysis of RNA (i.e., A $\ne$ U and C $\ne$ G generally, although A=U and C=G pairing can occur in folded regions) suggests that **RNA is typically a single-stranded molecule**. Unlike the DNA double helix, RNA does not consistently exist as two complementary strands paired throughout its length. While RNA can fold and form double-helical regions through intramolecular base pairing, the overall structure is not a continuous double helix, so the base pairing rules (A=U, C=G) do not apply across the entire molecule in a fixed stoichiometric ratio.



Summary

Biomolecules are the complex organic molecules that constitute living systems. Key classes discussed are carbohydrates, proteins, nucleic acids, and vitamins, along with hormones.



Exercises

Questions covering definitions, classification, structures, preparation, properties, functions, and differences between various biomolecules discussed in the chapter.



Intext Questions



Question 14.1. Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six membered ring compounds) are insoluble in water. Explain.

Answer:

Question 14.2. What are the expected products of hydrolysis of lactose?

Answer:

Question 14.3. How do you explain the absence of aldehyde group in the pentaacetate of D-glucose?

Answer:

Question 14.4. The melting points and solubility in water of amino acids are generally higher than that of the corresponding halo acids. Explain.

Answer:

Question 14.5. Where does the water present in the egg go after boiling the egg?

Answer:

Question 14.6. Why cannot vitamin C be stored in our body?

Answer:

Question 14.7. What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?

Answer:

Question 14.8. When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA?

Answer:



Exercises



Question 14.1. What are monosaccharides?

Answer:

Question 14.2. What are reducing sugars?

Answer:

Question 14.3. Write two main functions of carbohydrates in plants.

Answer:

Question 14.4. Classify the following into monosaccharides and disaccharides.

Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose.

Answer:

Question 14.5. What do you understand by the term glycosidic linkage?

Answer:

Question 14.6. What is glycogen? How is it different from starch?

Answer:

Question 14.7. What are the hydrolysis products of

(i) sucrose and (ii) lactose?

Answer:

Question 14.8. What is the basic structural difference between starch and cellulose?

Answer:

Question 14.9. What happens when D-glucose is treated with the following reagents?

(i) HI

(ii) Bromine water

(iii) $HNO_3$

Answer:

Question 14.10. Enumerate the reactions of D-glucose which cannot be explained by its open chain structure.

Answer:

Question 14.11. What are essential and non-essential amino acids? Give two examples of each type.

Answer:

Question 14.12. Define the following as related to proteins

(i) Peptide linkage

(ii) Primary structure

(iii) Denaturation.

Answer:

Question 14.13. What are the common types of secondary structure of proteins?

Answer:

Question 14.14. What type of bonding helps in stabilising the $\alpha$-helix structure of proteins?

Answer:

Question 14.15. Differentiate between globular and fibrous proteins.

Answer:

Question 14.16. How do you explain the amphoteric behaviour of amino acids?

Answer:

Question 14.17. What are enzymes?

Answer:

Question 14.18. What is the effect of denaturation on the structure of proteins?

Answer:

Question 14.19. How are vitamins classified? Name the vitamin responsible for the coagulation of blood.

Answer:

Question 14.20. Why are vitamin A and vitamin C essential to us? Give their important sources.

Answer:

Question 14.21. What are nucleic acids? Mention their two important functions.

Answer:

Question 14.22. What is the difference between a nucleoside and a nucleotide?

Answer:

Question 14.23. The two strands in DNA are not identical but are complementary. Explain.

Answer:

Question 14.24. Write the important structural and functional differences between DNA and RNA.

Answer:

Question 14.25. What are the different types of RNA found in the cell?

Answer: