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Biomolecules

NCERT Class 12 · Chemistry Based on NCERT Class 12 Chemistry textbook · Free CBSE study kit

Chapter Notes

BIOMOLECULES — CLASS 12 CBSE COMPREHENSIVE NOTES

10.1 CARBOHYDRATES

**Definition**: Carbohydrates are optically active polyhydroxy aldehydes or ketones, or compounds that hydrolyse to produce such units. They generally follow the formula Cx(H2O)y, though exceptions exist (e.g., rhamnose C6H12O5, acetic acid C2H4O2 is not a carbohydrate).

**Key Point**: The general formula Cx(H2O)y is NOT definitive for classification. Chemical structure and functional groups determine classification.

10.1.1 Classification of Carbohydrates

Carbohydrates are classified based on **hydrolysis behaviour**:

**1. Monosaccharides**: Single polyhydroxy aldehyde or ketone units that cannot be hydrolysed further. ~20 monosaccharides occur naturally. Examples: glucose, fructose, ribose, galactose.

**2. Oligosaccharides**: Hydrolyse to yield 2-10 monosaccharide units.

  • **Disaccharides** (most common): Yield 2 monosaccharide units
  • Examples: Sucrose (glucose + fructose), Maltose (glucose + glucose), Lactose (galactose + glucose)
  • **3. Polysaccharides**: Hydrolyse to yield many monosaccharide units. Non-sweet, called "non-sugars."

  • Examples: Starch, cellulose, glycogen
  • Classification as Reducing vs Non-Reducing Sugars

    **Reducing Sugars**: Reduce Fehling's solution and Tollens' reagent because they have free aldehyde or ketone groups. ALL monosaccharides (both aldoses and ketoses) are reducing sugars. Disaccharides like maltose and lactose are reducing (one anomeric carbon is free).

    **Non-Reducing Sugars**: Do NOT reduce Fehling's or Tollens' reagent. Example: Sucrose (both anomeric carbons are linked in glycosidic bond).

    10.1.2 Monosaccharides — Structure and Nomenclature

    Monosaccharides are named based on:

  • **Number of carbon atoms**: Triose (3C), Tetrose (4C), Pentose (5C), Hexose (6C), Heptose (7C)
  • **Functional group**: Aldose (contains —CHO group), Ketose (contains C=O group)
  • Example: Glucose = Aldohexose (6 carbons, aldehyde group)

    Example: Fructose = Ketohexose (6 carbons, ketone group)

    #### 10.1.2.1 GLUCOSE — Structure and Properties

    **General Formula**: C6H12O6 (Aldohexose)

    **IUPAC Name**: D-(+)-Glucose (D indicates stereochemistry relative to D-glyceraldehyde; (+) indicates dextrorotatory)

    **Preparation of Glucose**:

    **(a) From Sucrose (Cane Sugar)**:

    Sucrose + dilute HCl/H2SO4 → Glucose + Fructose (in equal amounts, in alcoholic solution)

    C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ + C₆H₁₂O₆

    (Sucrose) (Glucose) (Fructose)

    **(b) From Starch** (Commercial preparation):

    Starch + dilute H2SO4 → Glucose

    Conditions: 393 K, 2-3 atm pressure

    (C6H10O5)n + nH2O → nC6H12O6

    **Open-Chain Structure (Fischer Projection)**:

    Glucose was assigned a straight-chain structure based on:

    1. Molecular formula C6H12O6

    2. Prolonged heating with HI produces n-hexane (proves 6 carbons in straight chain)

    3. Forms oxime with hydroxylamine and cyanohydrin with HCN (proves aldehyde group —CHO)

    4. Oxidation by mild oxidizing agent (Br2 water) gives gluconic acid (proves —CHO is aldehyde, not ketone)

    5. Acetylation with acetic anhydride gives glucose pentaacetate (proves 5 —OH groups on different carbons)

    6. Oxidation by HNO3 gives saccharic acid (dicarboxylic acid), proves primary alcohol —CH2OH at end

    **Fischer Projection**:

    ```

    CHO

    |

    H—C—OH

    |

    HO—C—H

    |

    H—C—OH

    |

    H—C—OH

    |

    CH2OH

    ```

    Lowest asymmetric carbon (C5) has —OH on RIGHT side, correlating with D-(+)-glyceraldehyde, hence D-configuration.

    **Problem with Open-Chain Structure**:

    1. Glucose does NOT give Schiff's test despite aldehyde group

    2. Does NOT form NaHSO3 addition product

    3. Pentaacetate does NOT react with hydroxylamine (no free —CHO)

    4. Exists in TWO crystalline forms (α and β) unexplained by open-chain structure

    **Cyclic Structure (Haworth Projection)**:

    The —OH group at C5 adds to the aldehyde carbon (C1), forming a **hemiacetal**. This creates a **6-membered pyranose ring**. Two cyclic forms exist:

  • **α-D-Glucose**: —OH at C1 (anomeric carbon) is BELOW the ring plane
  • **β-D-Glucose**: —OH at C1 is ABOVE the ring plane
  • These are **anomers** (differ only in configuration at anomeric carbon C1).

    The cyclic structure explains:

  • Absence of free —CHO group in pentaacetate
  • Existence of two crystalline forms (α-form m.p. 419 K, β-form m.p. 423 K)
  • Non-reactivity with Schiff's reagent
  • Reducing property still present (free —CHO can form transiently from anomeric carbon in solution)
  • **Haworth Structure**: Oxygen in pyranose ring is written at back-right corner. Substituents above plane are on wedges, below plane are on dashes.

    #### 10.1.2.2 FRUCTOSE — Structure and Properties

    **General Formula**: C6H12O6 (Ketohexose)

    **IUPAC Name**: D-(−)-Fructose (Laevorotatory)

    **Properties**:

  • Natural ketohexose found in fruits, honey, vegetables
  • Contains ketone group (C=O) at C2
  • Also exists as two anomers (anomers differ at C2, the anomeric carbon in fructose)
  • **Cyclic Structure**:

    Fructose forms a **5-membered furanose ring** (not 6-membered) because the —OH at C5 adds to the ketone group at C2. This creates:

  • **α-D-Fructose**: —OH at C2 below ring plane
  • **β-D-Fructose**: —OH at C2 above ring plane
  • **Difference from Glucose**:

    | Feature | Glucose | Fructose |

    |---------|---------|----------|

    | Functional group | Aldehyde (—CHO) | Ketone (C=O) |

    | Ring size | Pyranose (6-membered) | Furanose (5-membered) |

    | Optical rotation | Dextrorotatory (+) | Laevorotatory (−) |

    10.1.3 Disaccharides

    Disaccharides are two monosaccharide units joined by a **glycosidic linkage** (C—O—C bond formed by loss of H2O between anomeric carbons or between anomeric carbon and hydroxyl of another sugar).

    #### (i) SUCROSE

    **Composition**: D-(+)-Glucose (1 molecule) + D-(−)-Fructose (1 molecule)

    **Structure**: C1 of **α-D-glucose** linked to C2 of **β-D-fructose** via 1,2-glycosidic linkage

    **Classification**: **NON-REDUCING SUGAR** (both anomeric carbons are involved in glycosidic bond, so no free aldehyde or ketone to reduce Fehling's)

    **Optical Rotation**:

  • Sucrose is dextrorotatory: (+) 66.5°
  • On hydrolysis:
  • Glucose: (+) 52.5°
  • Fructose: (−) 92.4°
  • Net mixture: (−) [because fructose's laevorotation > glucose's dextrorotation]
  • **Hydrolysis Products Called "Invert Sugar"**: Sign of rotation changes from (+) to (−), hence "invert."

    Hydrolysis equation:

    C12H22O11 + H2O → C6H12O6 (glucose) + C6H12O6 (fructose)

    #### (ii) MALTOSE

    **Composition**: Two molecules of α-D-glucose

    **Structure**: C1 of glucose (I) linked to C4 of glucose (II) via **1,4-α-glycosidic linkage**

    **Classification**: **REDUCING SUGAR** (C1 of second glucose unit has free anomeric carbon, can open and show aldehyde property)

    **Example**: Produced during starch digestion by amylase enzyme.

    #### (iii) LACTOSE

    **Composition**: β-D-galactose + β-D-glucose

    **Structure**: C1 of galactose linked to C4 of glucose via **1,4-β-glycosidic linkage**

    **Classification**: **REDUCING SUGAR** (free anomeric carbon at C1 of glucose)

    **Source**: Milk sugar (found in mammalian milk)

    **Lactose Intolerance**: Caused by deficiency of lactase enzyme needed to hydrolyse lactose in small intestine.

    10.1.4 Polysaccharides

    #### (i) STARCH

    **Source**: Plant storage polysaccharide (cereals, roots, tubers)

    **Composition**: Polymer of α-D-glucose units (unbranched + branched chains)

    **Two Components**:

    **(a) Amylose** (15-20% of starch):

  • Unbranched, water-soluble polymer
  • 200-1000 glucose units linked by **1,4-α-glycosidic bonds** (C1 of one glucose to C4 of next)
  • Helical/coiled structure in solution (blue colour with I2)
  • **(b) Amylopectin** (80-85% of starch):

  • Branched, water-insoluble polymer
  • **Main chains**: 1,4-α-glycosidic linkages
  • **Branch points**: 1,6-α-glycosidic linkages (—OH at C6 of one glucose links to C1 of branching unit)
  • Red/brown colour with I2
  • Branches occur approximately every 20-25 glucose units
  • **Hydrolysis of Starch**: Produces glucose (used as energy source)

    #### (ii) CELLULOSE

    **Source**: Plant cell walls (most abundant organic compound on Earth)

    **Composition**: Unbranched polymer of **β-D-glucose units**

    **Linkage**: **1,4-β-glycosidic bonds** (C1 of one glucose to C4 of next)

    **Structure**: Straight chain, forms extended/linear conformation (unlike coiled amylose)

    **Solubility**: Insoluble in water (due to strong hydrogen bonding within and between chains)

    **Uses**: Wood, paper, cotton fabric, cellulose derivatives (cellophane, nitrocellulose)

    **Key Difference from Starch**: β-glycosidic bonds (cellulose) vs α-glycosidic bonds (starch). Humans lack cellulase enzyme, so cannot digest cellulose; ruminants have cellulose-digesting bacteria.

    #### (iii) GLYCOGEN

    **Source**: Animal storage polysaccharide (liver, muscles, brain)

    **Composition**: Polymer of α-D-glucose units

    **Structure**: Highly **branched** (more branched than amylopectin)

  • Main chains: 1,4-α-glycosidic bonds
  • Branch points: 1,6-α-glycosidic bonds (more frequent than in amylopectin, every 8-12 glucose units)
  • **Function**: Stored energy source. When body needs glucose, glycogen phosphorylase breaks glycogen → glucose

    **Also Found In**: Yeast and fungi

    **Branching Advantage**: Highly branched structure provides multiple branch points for simultaneous enzyme attack, allowing rapid glucose mobilization when needed.

    10.1.5 Importance of Carbohydrates

    **Biological Functions**:

    1. **Energy source**: Glucose provides immediate energy (ATP synthesis)

    2. **Energy storage**: Starch in plants, glycogen in animals

    3. **Structural component**: Cell walls (cellulose in plants, chitin in fungi/insects)

    4. **Structural support**: Wood from cellulose

    5. **Textiles and clothing**: Cotton fiber (cellulose)

    6. **Industrial uses**: Textiles, paper, lacquers, brewing

    7. **In nucleic acids**: Ribose (C5) in RNA, deoxyribose (C5) in DNA

    8. **Conjugates**: Form glycoproteins (protein + carbohydrate) and glycolipids (lipid + carbohydrate)

    **Instant Energy**: Honey contains glucose and fructose; traditionally used in Ayurveda for immediate energy.

    ---

    10.2 PROTEINS

    **Definition**: Proteins are the most abundant biomolecules in living systems, serving as fundamental basis for structure and function of life. All proteins are **polymers of α-amino acids**.

    **Sources**: Milk, cheese, pulses, peanuts, fish, meat, eggs

    **Functions**:

  • Structural proteins (muscles, skin, hair, tendons)
  • Enzymatic proteins (catalyze reactions)
  • Transport proteins (hemoglobin carries O2)
  • Storage proteins (egg albumin)
  • Immune proteins (antibodies)
  • Hormonal proteins (insulin)
  • Support proteins (collagen, keratin)
  • **Etymology**: "Proteios" (Greek) = primary, of prime importance

    10.2.1 AMINO ACIDS — Structure and Classification

    **Basic Structure**:

    ```

    R

    |

    H2N—C—COOH

    |

    H

    ```

    Where:

  • —NH2 = amino group
  • —COOH = carboxyl group (carboxylic acid)
  • R = side chain (variable group)
  • The carbon bearing both amino and carboxyl groups is **α-carbon**
  • **Only α-amino acids** are obtained on hydrolysis of proteins. (Other amino acids like β, γ, δ exist but are not found in proteins.)

    **Nomenclature**: Amino acids named based on:

    1. Source (tyrosine from Greek "tyros" = cheese)

    2. Property (glycine from Greek "glykos" = sweet, because it tastes sweet)

    **Three-Letter and One-Letter Codes**: Used for brevity in protein sequences

  • Example: Glycine = Gly (3-letter), G (1-letter)
  • Example: Alanine = Ala (3-letter), A (1-letter)
  • #### 20 NATURALLY OCCURRING AMINO ACIDS (From Table 10.2):

    **Classification by Side Chain R**:

    **1. Non-Polar (Hydrophobic) Amino Acids**:

  • Glycine (Gly, G): R = H (only amino acid without stereoisomerism)
  • Alanine (Ala, A): R = CH3
  • Valine* (Val, V): R = (CH3)2CH—
  • Leucine* (Leu, L): R = (CH3)2CH—CH2—
  • Isoleucine* (Ile, I): R = CH3—CH2—CH(CH3)—
  • Proline (Pro, P): Imino acid; R = cyclic (5-membered ring including N)
  • Phenylalanine* (Phe, F): R = C6H5—CH2—
  • Methionine* (Met, M): R = CH3—S—CH2—CH2—
  • Tryptophan* (Trp, W): R = indole group (benzene + pyrrole fused)
  • **2. Polar Uncharged (Hydrophilic) Amino Acids**:

  • Serine (Ser, S): R = HO—CH2—
  • Threonine* (Thr, T): R = CH3—CHOH—
  • Asparagine (Asn, N): R = H2N—CO—CH2—
  • Glutamine (Gln, Q): R = H2N—CO—CH2—CH2—
  • Cysteine (Cys, C): R = HS—CH2— (contains disulfide-forming sulfur)
  • Tyrosine (Tyr, Y): R = p-HO—C6H4—CH2—
  • **3. Acidic Amino Acids** (negatively charged at physiological pH):

  • Aspartic acid (Asp, D): R = HOOC—CH2—
  • Glutamic acid (Glu, E): R = HOOC—CH2—CH2—
  • **4. Basic Amino Acids** (positively charged at physiological pH):

  • Lysine* (Lys, K): R = H2N—(CH2)4—
  • Arginine* (Arg, R): R = HN=C(NH2)—NH—(CH2)3—
  • Histidine* (His, H): R = imidazole ring (5-membered aromatic with 2 nitrogens)
  • **Asterisk (*)**: Denotes **essential amino acids** (cannot be synthesized by human body; must be obtained from diet). 10 amino acids are essential.

    #### Classification by Functional Groups and Charge

    **Neutral Amino Acids**: Equal —NH2 and —COOH groups

  • Examples: Glycine, Alanine, Valine, Leucine, etc.
  • **Acidic Amino Acids**: More —COOH than —NH2

  • Aspartic acid (1 extra —COOH)
  • Glutamic acid (1 extra —COOH)
  • Charge at physiological pH: Negative (—COO−)
  • **Basic Amino Acids**: More —NH2 than —COOH

  • Lysine (1 extra —NH2)
  • Arginine (1 extra —NH2)
  • Histidine (imidazole with basic nitrogen)
  • Charge at physiological pH: Positive (—NH3+)
  • 10.2.2 ISOELECTRIC POINT AND AMINO ACID BEHAVIOR

    **Isoelectric Point (pI)**: pH at which amino acid exists in **zwitterionic form** (equal positive and negative charges).

    **Zwitterion** (Dipolar Ion):

    ```

    R

    |

    H3N+—C—COO−

    |

    H

    ```

    At this pH:

  • —NH2 is protonated to —NH3+
  • —COOH is deprotonated to —COO−
  • Net charge = 0
  • Amino acid shows minimum solubility and maximum migration resistance in electric field
  • **pH Effects**:

  • **pH < pI**: Amino acid gains a proton, becomes positively charged (moves toward cathode in electrophoresis)
  • **pH > pI**: Amino acid loses a proton, becomes negatively charged (moves toward anode in electrophoresis)
  • **pH = pI**: Zwitterionic form, no net migration
  • 10.2.3 PEPTIDE BOND FORMATION

    **Definition**: The C—N bond formed between the carboxyl group (—COOH) of one amino acid and the amino group (—NH2) of another amino acid with elimination of water (H2O).

    **Formation Mechanism**:

    ```

    R1 R2

    | |

    H2N—C—COOH + H2N—C—COOH →

    | |

    H H

    (Amino acid 1) (Amino acid 2)

    R1 R2

    | |

    H2N—C—CO—NH—C—COOH + H2O

    | |

    H H

    (Dipeptide)

    ```

    **Peptide Bond Structure**:

    ```

    O

    —CO—NH— (amide linkage)

    ```

    **Properties of Peptide Bond**:

    1. **Partial double bond character**: Due to resonance, C—N bond has ~40% double bond character, preventing free rotation

    2. **Planar structure**: The six atoms (C, O, N, H, and the two α-carbons) lie in same plane

    3. **Trans configuration preferred**: Due to steric reasons, —CO and —NH usually in trans arrangement across peptide bond

    4. **pH stability**: Stable under physiological pH; hydrolysed under extreme pH with heat

    **Polypeptide Formation**: Multiple amino acids join through peptide bonds:

  • **Dipeptide**: 2 amino acids
  • **Tripeptide**: 3 amino acids
  • **Oligopeptide**: Few (2-20) amino acids
  • **Polypeptide/Protein**: Many (>100) amino acids
  • **Naming Convention**: Read from **N-terminus (free —NH2)** to **C-terminus (free —COOH)**

  • Example: Ser-Ala-Gly (Serine at N-end, Glycine at C-end)
  • N-terminus usually written on left, C-terminus on right
  • 10.2.4 PROTEIN STRUCTURE — Four Levels of Organization

    #### 1. PRIMARY STRUCTURE

    **Definition**: The linear sequence of amino acids joined by peptide bonds, from N-terminus to C-terminus.

    **Determination by**:

  • Sanger's method (sequential hydrolysis and chromatography)
  • Modern: Mass spectrometry and DNA sequencing
  • **Importance**:

  • Determines overall 3D shape
  • Determines biological function
  • **One change** in primary sequence can cause disease (e.g., sickle cell anemia: valine instead of glutamic acid at position 6 in hemoglobin)
  • **Notation**:

  • Using 3-letter code: Gly-Ala-Val-Pro
  • Using 1-letter code: GAVP
  • **Number of Possible Proteins**: With 20 amino acids, a protein of n amino acids can have 20^n possible sequences (enormous diversity).

    #### 2. SECONDARY STRUCTURE

    **Definition**: Regular, repeating, local structures in the polypeptide chain maintained by **hydrogen bonds between backbone atoms** (not side chains).

    **Types**:

    **(a) α-Helix**:

  • Spiral/coil structure
  • Hydrogen bonds between C=O of residue (n) and N—H of residue (n+4)
  • **3.6 amino acids per turn**
  • **Pitch** (rise per turn) = 5.4 Å
  • Stabilized by hydrogen bonds in the polypeptide backbone
  • Common in proteins: Keratin (hair, nails), myoglobin (muscle oxygen storage)
  • Right-handed helix (natural form)
  • **(b) β-Sheet (β-Pleated Sheet)**:

  • Extended conformation of polypeptide chains
  • Hydrogen bonds between adjacent extended polypeptide chains (can be parallel or antiparallel)
  • **Parallel**: N-termini on same side, polypeptides run in same direction
  • **Antiparallel**: Chains run in opposite directions (more stable, more common)
  • Side chains alternate above and below the plane of sheet
  • Common in proteins: Silk fibroin (β-keratin), immunoglobulins
  • Sheet-like, extended structure
  • **(c) Random Coil (Loop/Turn)**:

  • Irregular structure connecting α-helices or β-sheets
  • No regular hydrogen bonding pattern
  • Rich in Pro and Gly (Pro breaks helix due to constrained ring; Gly flexible due to no side chain)
  • **Stabilization**: All secondary structures stabilized by **hydrogen bonds (H-bonds)** between backbone C=O and N—H, NOT involving side chains.

    #### 3. TERTIARY STRUCTURE

    **Definition**: Overall 3D shape of polypeptide chain, resulting from interactions between side chains (R groups) of amino acids.

    **Stabilizing Forces** (between R groups of amino acids far apart in primary sequence):

    1. **Hydrogen Bonds**: Between polar side chains

  • Example: —OH of Ser, —COOH of Asp with —NH2 of Lys
  • 2. **Ionic Interactions (Salt Bridges)**: Between oppositely charged R groups

  • Example: —COO− of Asp/Glu with —NH3+ of Lys/Arg
  • 3. **Disulfide Bonds (Covalent)**: Between —SH groups of two Cysteine residues

  • 2 Cys—SH → Cys—S—S—Cys (most strong; only covalent bond in tertiary structure)
  • Particularly important in extracellular proteins (pancreatic hormones, antibodies)
  • 4. **Hydrophobic Interactions**: Non-polar R groups cluster in protein interior (away from aqueous environment)

  • Example: Phe, Met, Leu side chains associate
  • 5. **Van der Waals Forces**: Weak interactions between atoms in close proximity

    **Importance**:

  • Determines active site shape (for enzymes)
  • Determines biological activity
  • Denaturation (breaking of tertiary structure by heat, acid, base, detergent) causes loss of function
  • **Example**: Hemoglobin (carries O2), Myoglobin (oxygen storage), Enzymes

    #### 4. QUATERNARY STRUCTURE

    **Definition**: Arrangement of multiple polypeptide subunits (chains) in proteins containing more than one polypeptide chain. **NOT present in single-chain proteins.**

    **Stabilizing Forces**: Same as tertiary structure (H-bonds, ionic interactions, disulfide bonds, hydrophobic interactions, van der Waals forces) — but between different polypeptide chains.

    **Examples of Multi-Subunit Proteins**:

    **(a) Hemoglobin**:

  • 4 subunits (2α and 2β chains)
  • Quaternary structure allows cooperative binding of O2 (allosteric effect)
  • Abnormal hemoglobin (e.g., sickle cell) has altered quaternary structure
  • **(b) Immunoglobulins (Antibodies)**:

  • 4 polypeptide chains (2 heavy, 2 light)
  • Quaternary structure creates antigen-binding sites
  • **(c) Enzymes**:

  • Many have multiple subunits for catalytic activity
  • Example: Lactase (breaks lactose)
  • **Allosteric Effect**: Binding of substrate or regulator at one subunit affects binding affinity at other subunits (e.g., O2 binding to one hemoglobin subunit increases affinity of other subunits).

    10.2.5 CLASSIFICATION OF PROTEINS

    **Based on Composition**:

    1. **Simple Proteins**: Contain only amino acids

  • Examples: Albumin, Globulin, Keratin, Collagen
  • 2. **Conjugated Proteins**: Contain non-protein group (prosthetic group) + amino acid chains

  • **Glycoproteins**: Protein + carbohydrate (mucins, antibodies)
  • **Lipoproteins**: Protein + lipid (cell membranes, cholesterol transport)
  • **Metalloproteins**: Protein + metal ion (hemoglobin has iron-heme, cytochrome c)
  • **Nucleoproteins**: Protein + nucleic acid (ribosomes, chromatin)
  • **Phosphoproteins**: Protein + phosphate group (casein in milk)
  • **Based on Function**:

    1. **Enzymes**: Biological catalysts (e.g., amylase digests starch, pepsin digests protein)

    2. **Antibodies/Immunoglobulins**: Immune defense

    3. **Hormones**: Regulation (e.g., insulin for blood glucose)

    4. **Transport Proteins**: Carry molecules (hemoglobin carries O2)

    5. **Storage Proteins**: Store amino acids/iron (ferritin)

    6. **Structural Proteins**: Support (collagen in connective tissue, keratin in hair)

    7. **Motor Proteins**: Movement (myosin in muscle contraction)

    10.2.6 DENATURATION OF PROTEINS

    **Definition**: Loss of secondary, tertiary, and quaternary structures while primary structure (peptide bonds) remains intact. Results in loss of biological function.

    **Causes of Denaturation**:

    1. **Heat**: Breaks H-bonds, increases molecular motion

  • Example: Cooking egg white (albumin) → opaque, insoluble (originally transparent, soluble)
  • 2. **pH Change**:

  • Extreme pH (very acidic or basic) disrupts ionic interactions
  • May cause protonation/deprotonation of amino acids
  • 3. **Chemical Reagents**:

  • **Urea** or **Guanidinium chloride**: Disrupts hydrophobic interactions
  • **Heavy metal salts** (Pb²⁺, Hg²⁺): Form complexes with —SH, —COO−, —NH2 groups
  • **Detergents**: Break hydrophobic interactions
  • 4. **Mechanical Action**: Vigorous shaking breaks H-bonds

  • Example: Egg white beaten → foam (denatured albumin)
  • 5. **Organic Solvents**: Ethanol disrupts H-bonds and ionic interactions

    **Characteristics of Denatured Protein**:

  • Loss of solubility (in most cases)
  • Increase in viscosity (unfolded chains occupy larger volume)
  • Loss of enzymatic activity (active site shape changed)
  • Coagulation (proteins aggregate)
  • Precipitation
  • **Renaturation**: In some cases, denatured proteins can refold to native structure if denaturing conditions removed (reversible denaturation). Irreversible denaturation occurs when peptide bonds break or extensive aggregation occurs.

    ---

    10.3 NUCLEIC ACIDS

    **Definition**: Nucleic acids are polynucleotides (polymers of nucleotides) that store and transmit genetic information. The two main types are **DNA (deoxyribonucleic acid)** and **RNA (ribonucleic acid)**.

    **Biological Importance**:

  • Carry genetic information
  • Control protein synthesis
  • Direct all life processes
  • **Primary Components**:

  • **Nitrogenous bases** (purines and pyrimidines)
  • **Pentose sugar** (ribose in RNA, deoxyribose in DNA)
  • **Phosphoric acid** (H3PO4)
  • 10.3.1 NUCLEOTIDES — Building Blocks

    **Definition**: A nucleotide is a monomer consisting of three components:

    1. A pentose sugar (ribose or deoxyribose)

    2. A nitrogenous base (purine or pyrimidine)

    3. A phosphate group (or phosphoric acid)

    **Structure**:

    ```

    Base

    |

    O—P—O—Sugar

    |

    O

    (Phosphate)

    ```

    #### Pentose Sugars

    **(a) Ribose** (in RNA):

  • 5-carbon sugar (pentose)
  • Contains —OH group at C2 position
  • **(b) Deoxyribose** (in DNA):

  • 5-carbon sugar (pentose)
  • Contains —H (hydrogen, not —OH) at C2 position
  • Name: "deoxy" = lacking oxygen
  • **Difference**:

  • Ribose: —OH at C2
  • Deoxyribose: —H at C2
  • #### Nitrogenous Bases

    **Purines** (Double-ring structure, molecular formula C5H4N4):

    1. **Adenine (A)**

    2. **Guanine (G)**

    **Pyrimidines** (Single-ring structure, molecular formula C4H5N3):

    1. **Cytosine (C)** (in both DNA and RNA)

    2. **Thymine (T)** (in DNA only)

    3. **Uracil (U)** (in RNA only)

    **Memory Aid**: "PURines Are larger" (2 rings, larger molecule)

    **Base Pairing in Double Helix**:

  • **A—T** (or A—U in RNA): 2 hydrogen bonds
  • **G—C**: 3 hydrogen bonds
  • G—C pairs are stronger (more stable)
  • #### Nucleotide Formation

    Nucleotide forms when phosphate group links to C5

    MCQs — 10 Questions with Answers

    Q1. Which of the following compounds fits the general formula Cx(H2O)y but is NOT a carbohydrate?

    • A. Glucose (C₆H₁₂O₆)
    • B. Acetic acid (CH₃COOH) ✓
    • C. Sucrose (C₁₂H₂₂O₁₁)
    • D. Fructose (C₆H₁₂O₆)

    Answer: B — Acetic acid fits Cx(H2O)y as C₂(H₂O)₂ but lacks polyhydroxy aldehyde/ketone functional groups, so it is not classified as a carbohydrate.

    Q2. Glucose reacts with hydroxylamine to form an oxime and with HCN to form a cyanohydrin. This indicates the presence of which functional group in glucose?

    • A. Carboxylic acid (—COOH)
    • B. Ester (—COO—)
    • C. Carbonyl (>C=O) ✓
    • D. Ether (—O—)

    Answer: C — Both oxime formation and HCN addition are characteristic reactions of aldehydes and ketones (carbonyl groups), confirming the presence of >C=O in glucose.

    Q3. When glucose is oxidised with bromine water, gluconic acid is formed. What does this observation prove about glucose structure?

    • A. Glucose contains a ketone group at C1
    • B. Glucose contains an aldehyde group at C1 ✓
    • C. Glucose contains a primary alcohol at C1
    • D. Glucose contains an ester group at C1

    Answer: B — Mild oxidation with Br₂ water selectively oxidises aldehydes (—CHO) to carboxylic acids (—COOH); ketones are not oxidised by Br₂ water, so the formation of gluconic acid proves —CHO exists in glucose.

    Q4. Glucose pentaacetate is a stable compound formed by acetylation of glucose. How many hydroxyl (—OH) groups must glucose contain?

    • A. Three
    • B. Four
    • C. Five ✓
    • D. Six

    Answer: C — Pentaacetate means five acetyl groups are attached; since each —OH group is converted to —O—CO—CH₃, glucose must contain exactly five separate hydroxyl groups.

    Q5. When glucose is oxidised with nitric acid, saccharic acid (a dicarboxylic acid) is formed. This proves that glucose has which two functional groups at opposite ends of the carbon chain?

    • A. Two aldehyde groups (—CHO)
    • B. One aldehyde (—CHO) and one primary alcohol (—CH₂OH) ✓
    • C. Two ketone groups (C=O)
    • D. One aldehyde (—CHO) and one secondary alcohol

    Answer: B — Strong oxidation by HNO₃ converts both —CHO (at C1) and —CH₂OH (at C6) to —COOH, forming saccharic acid with two carboxylic acid groups; this proves both functional groups are present in glucose.

    Q6. Which statement about D/L and (+)/(−) notation in carbohydrate nomenclature is correct?

    • A. D means dextrorotatory (+) and L means laevorotatory (−)
    • B. D and L denote stereochemical configuration; (+) and (−) denote optical rotation direction ✓
    • C. D and L are related to letter 'd' and 'l' used for optical activity
    • D. D(+)-glucose and L(−)-glucose have identical chemical structures

    Answer: B — D/L are stereochemical descriptors based on the highest-numbered chiral carbon; (+) and (−) indicate optical rotation; they are independent properties of glucose.

    Q7. Which of the following is NOT correct regarding reducing and non-reducing sugars?

    • A. All monosaccharides are reducing sugars
    • B. Maltose is a non-reducing sugar because it lacks a free carbonyl group ✓
    • C. Polysaccharides like starch and cellulose are non-reducing sugars
    • D. Reducing sugars can reduce Fehling's and Tollens' reagents

    Answer: B — Maltose IS a reducing sugar because it has a free anomeric carbon with a —CHO or —C(OH)— group that can exist as an aldehyde; starch and cellulose are non-reducing because no free carbonyl is available.

    Q8. When sucrose is hydrolysed with dilute HCl, the products obtained are glucose and fructose in a 1:1 molar ratio. This indicates that sucrose is composed of:

    • A. One glucose and one galactose unit
    • B. One glucose and one fructose unit ✓
    • C. Two glucose units
    • D. Two fructose units

    Answer: B — The 1:1 molar ratio of glucose to fructose in the hydrolysate directly shows sucrose = glucose + fructose; the hydrolysis reaction is C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ (glucose) + C₆H₁₂O₆ (fructose).

    Q9. Which of the following reactions would NOT occur with glucose?

    • A. Oxidation with Br₂ water to form gluconic acid
    • B. Oxidation with HNO₃ to form saccharic acid
    • C. Acetylation with acetic anhydride to form pentaacetate
    • D. Reduction with Tollens' reagent to form alcohol ✓

    Answer: D — Tollens' reagent is an oxidising agent that converts aldehydes to carboxylic acids (and gets reduced to Ag mirror); glucose is oxidised, not reduced, by Tollens' reagent.

    Q10. In the Fischer projection of D(+)-glucose, if the aldehyde group (—CHO) is at C1 and the primary alcohol (—CH₂OH) is at C6, how are the carbon atoms arranged in the straight-chain structure?

    • A. C1(—CHO) — C2 — C3 — C4 — C5 — C6(—CH₂OH) with four chiral carbons
    • B. C1(—CHO) — C2 — C3 — C4 — C5 — C6(—CH₂OH) with five chiral carbons ✓
    • C. C1(—CHO) — C2 — C3 — C4 — C5 — C6(—CH₂OH) with three chiral carbons
    • D. Branched chain with —CHO at the central carbon

    Answer: B — The straight chain has 6 carbons total; C1 (aldehyde carbon) and C6 (primary alcohol carbon) are not chiral; C2, C3, C4, C5 are chiral centres, giving five chiral carbons total.

    Flashcards

    What is the general molecular formula for carbohydrates, and why is it misleading?

    Cx(H2O)y; misleading because non-carbohydrates like acetic acid fit the formula, and some true carbohydrates like rhamnose do not.

    Define monosaccharide and give two examples.

    A carbohydrate that cannot be hydrolysed further to give simpler polyhydroxy aldehyde or ketone units; examples: glucose and fructose.

    What is the difference between aldose and ketose monosaccharides?

    Aldose contains an aldehyde group (—CHO) while ketose contains a ketone group (C=O); glucose is aldohexose, fructose is ketohexose.

    How are reducing sugars identified in the lab?

    Reducing sugars (all monosaccharides and some disaccharides like maltose) reduce Fehling's solution and Tollens' reagent to give a precipitate or mirror.

    Write the hydrolysis reaction of sucrose and name the products.

    C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ (glucose) + C₆H₁₂O₆ (fructose); sucrose gives equal moles of glucose and fructose.

    What evidence proves glucose has a straight-chain structure with 6 carbon atoms?

    Prolonged heating with HI converts glucose to n-hexane, proving all six carbons are linked in a single continuous chain.

    Why does glucose form a pentaacetate product, and what does this tell us?

    Five —OH groups are present in glucose attached to different carbon atoms; the stable pentaacetate confirms five acetylable hydroxyl groups.

    Distinguish between the Fischer projection of glucose and gluconic acid.

    Glucose has —CHO at C1 and —CH₂OH at C6; gluconic acid has —COOH at C1 and —CH₂OH at C6 (C1 oxidised to carboxylic acid).

    What do the prefixes D and L mean in carbohydrate nomenclature?

    D and L denote stereochemical configuration at the highest-numbered chiral carbon, not optical rotation; they are unrelated to (+) or (−) activity.

    How is glucose prepared commercially from starch?

    Starch is hydrolysed by boiling with dilute H₂SO₄ at 393 K under 2–3 atm pressure to yield glucose as the major product.

    Important Board Questions

    Define monosaccharides and classify them based on the number of carbon atoms and the type of carbonyl group present. Give one example of an aldopentose. [2 marks]

    State: monosaccharides cannot hydrolyse further; classify as triose, tetrose, pentose, hexose, etc., and as aldose (—CHO) or ketose (C=O); example: ribose or arabinose (aldopentose with 5 carbons and —CHO).

    Glucose is oxidised with bromine water to give gluconic acid. Further oxidation of gluconic acid with nitric acid gives saccharic acid (a dicarboxylic acid). Explain what these oxidation reactions reveal about the structure of glucose, and write the Fischer projection of glucose. [5 marks]

    Br₂ oxidises —CHO to —COOH (proving aldehyde at C1); HNO₃ oxidises both —CHO and —CH₂OH to —COOH (proving primary —OH at C6); Fischer projection shows C1(—CHO) at top, —CH₂OH at C6 at bottom, with chiral carbons C2–C5 in between with —OH groups; show all three structures (glucose, gluconic acid, saccharic acid).

    Glucose pentaacetate is formed when glucose is treated with acetic anhydride. The molecular formula of glucose pentaacetate is C₁₆H₂₂O₁₁. Show that glucose must contain exactly five hydroxyl groups, and explain how this evidence combined with oxidation reactions proves the complete structural formula of glucose as C₆H₁₂O₆ with the structure given in the Fischer projection. [6 marks]

    Glucose C₆H₁₂O₆ + 5 (CH₃CO)₂O → pentaacetate + 5 CH₃COOH; calculate: glucose (180 g/mol) + 5 acetyl groups (5 × 42 = 210) − 5 water molecules (5 × 18 = 90) = 300 g/mol, matching C₁₆H₂₂O₁₁; five separate —OH groups on different carbons; combined with Br₂ water (—CHO at C1), HNO₃ (—CH₂OH at C6), HI degradation (straight chain), oxime + HCN (carbonyl present), derive that glucose = open-chain aldohexose with structure CHO-CHOH-CHOH-CHOH-CHOH-CH₂OH; show Fischer projection with correct stereochemistry.

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