NCERT Solutions for Class 12 Chemistry Biomolecules

NCERT Solutions for Class 12 Chemistry Chapter 14 Biomolecules helps students to understand the structure and function of life’s chemical compounds. Biomolecules are the organic compounds which form the base of all living organism like carbohydrates, proteins, nucleic acids and lipids. In this chapter of class 12 chemistry biomolecules ncert solutions and MCQs on class 12 biomolecule, students learn about their classification, chemical nature, and biological importance. It also includes topics like enzymes and vitamins which are essential for body’s metabolic process.

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NCERT Solutions for Class 12 Chemistry Biomolecules PDF

NCERT Solutions for Class 12 Chemistry Biomolecules PDF cover topics like carbohydrates, proteins, nucleic acids and lipids, so students can learn simple concepts easily. These class 12 chemistry biomolecules ncert solutions help to understand the CBSE class 12 chemistry syllabus and score better. Biomolecules class 12 NCERT solutions chemistry chapter 14 makes tough questions simple, but sometimes student gets confused in PDF downloading.

Class 12 Chemistry Chapter 14 Biomolecules Questions with Solutions

Ques: Write the equation for hydrolysis of lactose. Name the products formed and explain the reaction mechanism.

Detailed Solution:

Lactose is a disaccharide with the molecular formula C₁₂H₂₂O₁₁. When lactose undergoes hydrolysis in the presence of dilute acids or the enzyme lactase, it breaks down into its constituent monosaccharides.

• Chemical Equation:

C₁₂H₂₂O₁₁ (Lactose) + H₂O → C₆H₁₂O₆ (β-D-Glucose) + C₆H₁₂O₆ (β-D-Galactose)

• Mechanism:

Lactose contains a glycosidic linkage between C₁ of β-D-galactose and C₄ of β-D-glucose. During hydrolysis, the glycosidic linkage is cleaved by addition of water molecule. The oxygen bridge between the two monosaccharide units is broken, resulting in formation of two separate sugar molecules.

The reaction requires either acidic conditions (H⁺ catalyst) or enzymatic action (lactase enzyme). This is an example of an acetal hydrolysis reaction where the hemiacetal carbon atom of one sugar is released from the glycosidic bond.

Ques: Write the equation when D-glucose reacts with hydroxylamine. What product is formed and what does this reaction indicate about glucose structure?

Detailed Solution:

When D-glucose reacts with hydroxylamine (NH₂OH), it forms glucose oxime.

• Chemical Equation:

C₆H₁₂O₆ (Glucose) + NH₂OH (Hydroxylamine) → C₆H₁₃NO₆ (Glucose oxime) + H₂O

• Structural Representation:

CHO-CHOH-CHOH-CHOH-CHOH-CH₂OH + NH₂OH → CH=NOH-CHOH-CHOH-CHOH-CHOH-CH₂OH + H₂O

• Significance:

This reaction confirms the presence of a carbonyl group (specifically an aldehyde group -CHO) in the open-chain structure of glucose. The aldehyde group reacts with hydroxylamine to form an oxime through a condensation reaction.

The formation of glucose oxime proves that glucose contains a free or potentially free aldehyde functional group. However, pentaacetate of glucose does not give this reaction because the cyclic hemiacetal form prevents the free aldehyde from being available.

Ques: Write the equation for oxidation of D-glucose with bromine water. Name the product and its significance.

Detailed Solution:

When D-glucose is treated with a mild oxidizing agent like bromine water, it gets oxidized to gluconic acid.

• Chemical Equation:

C₆H₁₂O₆ (D-Glucose) + Br₂ + H₂O → C₆H₁₂O₇ (D-Gluconic acid) + 2HBr

• Structural Equation:

CHO-(CHOH)₄-CH₂OH + [O] → COOH-(CHOH)₄-CH₂OH

• Detailed Explanation:

In this reaction, only the aldehyde group (-CHO) at C₁ position is oxidized to a carboxylic acid group (-COOH), while all other hydroxyl groups remain unchanged. Bromine water acts as a mild oxidizing agent that selectively oxidizes the aldehyde functional group without affecting the alcohol groups.

The product gluconic acid is a sugar acid with both carboxylic acid and multiple hydroxyl groups. This reaction confirms that glucose has a reducing property and contains an aldehyde group. This is a characteristic reaction of all aldoses (aldehyde sugars).

Ques: What happens when glucose is oxidized with concentrated nitric acid? Write the equation and name the product.

Detailed Solution:

When D-glucose is oxidized with concentrated nitric acid (a strong oxidizing agent), both the aldehyde group and the primary alcoholic group are oxidized.

• Chemical Equation:

C₆H₁₂O₆ (D-Glucose) + 3[O] (from HNO₃) → C₆H₁₀O₈ (Saccharic acid/Glucaric acid) + H₂O

• Structural Equation:

CHO-(CHOH)₄-CH₂OH + 3[O] → COOH-(CHOH)₄-COOH

• Detailed Explanation:

In this reaction, concentrated nitric acid acts as a powerful oxidizing agent. It oxidizes both terminal groups of glucose:

• The aldehyde group (-CHO) at C₁ position is oxidized to carboxylic acid (-COOH)
• The primary alcoholic group (-CH₂OH) at C₆ position is also oxidized to carboxylic acid (-COOH)

The product formed is saccharic acid (also called glucaric acid), which is a dicarboxylic acid with the molecular formula C₆H₁₀O₈. The four secondary hydroxyl groups in the middle remain unchanged.

This reaction demonstrates that glucose has both aldehyde and primary alcohol functional groups at its terminal positions.

Ques: Write the reaction showing that all carbon atoms in glucose are linked in a straight chain. What product is formed?

Detailed Solution:

When glucose is heated with hydroiodic acid (HI) for a prolonged period, it gets reduced to n-hexane.

• Chemical Equation:

C₆H₁₂O₆ (Glucose) + 6HI → C₆H₁₄ (n-Hexane) + 3I₂ + 3H₂O

• Detailed Explanation:

In this reduction reaction, hydroiodic acid (HI) acts as a strong reducing agent. All the oxygen-containing functional groups (aldehyde group and hydroxyl groups) are removed and replaced by hydrogen atoms.

The product formed is n-hexane (CH₃-CH₂-CH₂-CH₂-CH₂-CH₃), which is a straight-chain saturated hydrocarbon with six carbon atoms.

• Significance: The formation of n-hexane (not any branched isomer) proves conclusively that all six carbon atoms in glucose are arranged in a continuous straight chain without any branching. If glucose had a branched structure, the product would be a branched-chain alkane like isohexane or neohexane.

This reaction is crucial evidence for establishing the linear structure of the carbon skeleton in glucose.

Ques: Write the equation when D-glucose reacts with acetic anhydride. What does this reaction reveal about glucose structure?

Detailed Solution:

When D-glucose is treated with acetic anhydride in the presence of pyridine or zinc chloride, it forms glucose pentaacetate.

• Chemical Equation:

C₆H₁₂O₆ + 5(CH₃CO)₂O → C₆H₇O(OCOCH₃)₅ + 5CH₃COOH

• Structural Representation:

CHO-(CHOH)₄-CH₂OH + 5(CH₃CO)₂O → [Acetylated product with 5 acetate groups]

• Detailed Explanation:

Glucose contains five hydroxyl (-OH) groups - one at each of the C₂, C₃, C₄, C₅ positions and one primary -OH at C₆ position. When these hydroxyl groups react with acetic anhydride, acetylation occurs and all five -OH groups are converted to acetate groups (-OCOCH₃).

The product is called glucose pentaacetate because it contains five acetate groups.

• Important Observation:

The fact that glucose forms pentaacetate (not hexaacetate) indicates that the aldehyde group is not free in the stable form of glucose. Instead, it exists in a cyclic hemiacetal form where the aldehyde group reacts internally with one of the -OH groups.

Glucose pentaacetate does not react with hydroxylamine (NH₂OH), confirming the absence of a free aldehyde group. This observation led to the proposal of the cyclic structure of glucose.

Ques: Write the equation for hydrolysis of sucrose. Name the products and explain why sucrose is called invert sugar.

Detailed Solution:

Sucrose is a disaccharide that undergoes hydrolysis in the presence of dilute acids or the enzyme invertase (sucrase).

• Chemical Equation:

C₁₂H₂₂O₁₁ (Sucrose) + H₂O → C₆H₁₂O₆ (D-Glucose) + C₆H₁₂O₆ (D-Fructose)

• Detailed Explanation:

Sucrose, commonly known as cane sugar or table sugar, is a non-reducing disaccharide. Upon hydrolysis, it yields an equimolar mixture of D-(+)-glucose and D-(-)-fructose. This mixture is called "invert sugar."

Why is it called Invert Sugar?

Sucrose has a specific rotation of +66.5°. However, after hydrolysis, the resulting mixture has a negative specific rotation of -19.8° because:

• D-Glucose has specific rotation = +52.5°
• D-Fructose has specific rotation = -92.4°
• Net rotation = (+52.5° + (-92.4°))/2 = -19.95°

Since the sign of rotation changes (inverts) from positive (+) to negative (-) upon hydrolysis, the resulting mixture is called "invert sugar." The enzyme that catalyzes this reaction is called invertase.

Note: Honey contains invert sugar, which is why it is sweeter than sucrose.

Ques: Write the equation for complete hydrolysis of maltose and explain the type of glycosidic linkage present.

Detailed Solution:

Maltose is a disaccharide that undergoes hydrolysis to yield glucose molecules.

• Chemical Equation:

C₁₂H₂₂O₁₁ (Maltose) + H₂O → 2C₆H₁₂O₆ (D-Glucose)

• Detailed Explanation:

Maltose, also called malt sugar, is a reducing disaccharide. It is composed of two α-D-glucose units. Upon complete hydrolysis with dilute acids or the enzyme maltase, maltose yields two molecules of D-(+)-glucose.

• Glycosidic Linkage:

In maltose, the two glucose units are connected through α(1→4) glycosidic linkage. This means that the C₁ of one α-D-glucose unit (in its hemiacetal form) is connected to the C₄ hydroxyl group of the second α-D-glucose unit through an oxygen atom.

Since the second glucose unit has a free anomeric carbon (C₁ is free to open), maltose is a reducing sugar and shows mutarotation.

Maltose is obtained by the partial hydrolysis of starch by the enzyme diastase present in malt. It is also present in germinating seeds.

Ques: Write the equation when glucose reacts with phenylhydrazine. Name the product and its significance.

Detailed Solution:

When glucose reacts with excess phenylhydrazine, it forms glucose osazone.

• Chemical Equation:

C₆H₁₂O₆ (Glucose) + 3C₆H₅NHNH₂ (Phenylhydrazine) →

C₆H₁₀O₄(N₂HC₆H₅)₂ (Glucose osazone) + C₆H₅NH₂ (Aniline) + 2H₂O + NH₃

• Detailed Explanation:

In this reaction, glucose reacts with three molecules of phenylhydrazine:

• First molecule: Reacts with the aldehyde group at C₁ to form phenylhydrazone
• Second molecule: Oxidizes the -CHOH group at C₂ to C=O (carbonyl), simultaneously getting reduced to aniline and ammonia
• Third molecule: Reacts with the newly formed carbonyl at C₂ to form another phenylhydrazone

The final product, glucose osazone, contains two phenylhydrazone groups at C₁ and C₂ positions. Osazones are crystalline compounds with characteristic melting points and crystal shapes.

• Significance:

Both glucose and fructose give the same osazone because they differ only in the structure at C₁ and C₂, which both get converted to the same osazone structure. This reaction was historically important for identifying and characterizing sugars.

Ques: Write the equation showing the formation of a peptide bond between two glycine molecules. Explain the peptide linkage.

Detailed Solution:

A peptide bond is formed when two amino acids condense together with the elimination of a water molecule.

• Chemical Equation:

NH₂-CH₂-COOH (Glycine) + NH₂-CH₂-COOH (Glycine) →

NH₂-CH₂-CO-NH-CH₂-COOH (Glycylglycine/Dipeptide) + H₂O

• Detailed Explanation:

The peptide bond (also called peptide linkage) is an amide linkage (-CO-NH-) formed between the carboxyl group (-COOH) of one α-amino acid and the amino group (-NH₂) of another α-amino acid through the loss of a water molecule.

• Mechanism:

The -OH from the carboxyl group of one glycine and -H from the amino group of the second glycine are eliminated as H₂O. The resulting bond -CO-NH- is called a peptide bond.

When two amino acids are joined, the product is called a dipeptide. Similarly, three amino acids form a tripeptide, and many amino acids (more than 10) form a polypeptide.

Characteristics of Peptide Bond:

• It is a rigid, planar bond due to partial double bond character (resonance)
• The C-N bond has restricted rotation
• The peptide bond is the fundamental linkage in all protein structures
• Primary structure of proteins is determined by the sequence of amino acids linked by peptide bonds

Ques: Explain the amphoteric behavior of amino acids with the help of equations. What is a zwitter ion?

Detailed Solution:

Amino acids exhibit amphoteric behavior because they contain both acidic (carboxyl) and basic (amino) groups.

• Zwitter Ion Formation:

NH₂-CHR-COOH ⇌ ⁺NH₃-CHR-COO^- (Zwitter ion)

• In Acidic Medium:

⁺NH₃-CHR-COO^- (Zwitter ion) + H⁺ (from acid) → ⁺NH₃-CHR-COOH (Cation)

• In Basic Medium:

⁺NH₃-CHR-COO^- (Zwitter ion) + OH^- (from base) → NH₂-CHR-COO^- (Anion) + H₂O

• Detailed Explanation:

In aqueous solution, amino acids exist predominantly as zwitter ions (also called dipolar ions). The carboxyl group loses a proton (H⁺) and becomes negatively charged (-COO^-), while the amino group accepts this proton and becomes positively charged (-NH₃⁺). This results in a molecule with both positive and negative charges but overall neutral.

Amphoteric Nature:

• In acidic solution: The zwitter ion accepts H⁺ on the -COO^- group, behaving as a base
• In basic solution: The zwitter ion donates H⁺ from the -NH₃⁺ group, behaving as an acid

This amphoteric behavior is responsible for the high melting points and water solubility of amino acids compared to carboxylic acids of similar molecular mass.

Ques: What happens when an egg is boiled? Write the changes occurring during denaturation of egg protein and explain where the water goes.

Detailed Solution:

When an egg is boiled, the protein present in the egg white (albumin) undergoes denaturation.

• Process of Denaturation:

Native Protein (Globular, soluble) + Heat → Denatured Protein (Fibrous, insoluble)

• Detailed Explanation:

Egg white contains a globular protein called albumin which is soluble in water and has a specific three-dimensional structure stabilized by hydrogen bonds, disulfide bridges, and other weak interactions. When heat is applied during boiling:

1. Hydrogen bonds break: The weak hydrogen bonds maintaining the secondary and tertiary structures of the protein are disrupted by thermal energy
2. Protein unfolds: The globular protein loses its specific 3D structure and unfolds
3. Coagulation: The unfolded protein chains aggregate and coagulate to form an insoluble, rubber-like mass
4. Water absorption: The water present in the egg gets absorbed by the coagulated protein mass through hydrogen bonding with peptide groups

Where does water go?

The water doesn't disappear; it becomes bound to the denatured protein through extensive hydrogen bonding. The coagulated protein network traps water molecules within its structure, making the boiled egg appear solid but still retaining moisture internally.

Important Note:

Denaturation destroys the secondary and tertiary structures of proteins but does not break peptide bonds, so the primary structure (amino acid sequence) remains intact.

Ques: Explain the complementary base pairing in DNA with equations. Why are the two strands complementary and not identical?

Detailed Solution:

In DNA, the two strands are held together by hydrogen bonds between complementary base pairs.

• Base Pairing Rules:

Adenine (A) ≡ Thymine (T) [Two hydrogen bonds]

Guanine (G) ≡ Cytosine (C) [Three hydrogen bonds]

• Structural Representation:

A = T (via 2 H-bonds)

G ≡ C (via 3 H-bonds)

• Detailed Explanation:

The two strands of DNA are antiparallel (one runs 5'→3' and the other runs 3'→5') and are held together by hydrogen bonds between nitrogenous bases. Due to the specific geometries and sizes of purine and pyrimidine bases, only certain base pairings are possible:

Why A pairs with T:

Adenine (purine) and thymine (pyrimidine) form two hydrogen bonds. Their sizes and shapes are complementary, allowing stable pairing.

Why G pairs with C:

Guanine (purine) and cytosine (pyrimidine) form three hydrogen bonds, creating a stronger interaction than A-T pairing.

Why Complementary, Not Identical?

If one strand has the sequence 5'-ATGC-3', the other strand must have 3'-TACG-5'. The sequences are different but complementary. This base-pairing principle means that if you know the sequence of one strand, you can determine the sequence of the other strand.

This complementarity is essential for:

• DNA replication (each strand serves as a template)
• DNA repair mechanisms
• Transcription processes

The complementary nature ensures faithful transmission of genetic information.

Ques: Differentiate between a nucleoside and nucleotide. Write equations showing their formation.

Detailed Solution:

• Nucleoside Formation:

Pentose sugar (Ribose/Deoxyribose) + Nitrogenous base → Nucleoside + H₂O

Example: Ribose + Adenine → Adenosine

• Nucleotide Formation:

Nucleoside + Phosphoric acid → Nucleotide

OR

Pentose sugar + Nitrogenous base + Phosphoric acid → Nucleotide + 2H₂O

Example: Adenosine + H₃PO₄ → Adenosine monophosphate (AMP)

• Detailed Comparison:

• Significance:

Nucleotides are the monomeric units of nucleic acids (DNA and RNA). They are joined together through phosphodiester linkages to form polynucleotide chains. Nucleosides, on the other hand, are intermediates in nucleotide biosynthesis and can also have independent biological roles.

Ques: Write the structural and functional differences between DNA and RNA. Include equations showing their hydrolysis products.

Detailed Solution:

• DNA Hydrolysis:

DNA → 2-Deoxyribose + Phosphoric acid + Adenine + Guanine + Cytosine + Thymine

• RNA Hydrolysis:

RNA → Ribose + Phosphoric acid + Adenine + Guanine + Cytosine + Uracil

Detailed Structural Differences:

Functional Differences:

DNA Functions:

• Stores genetic information
• Responsible for transmission of hereditary characters
• Controls protein synthesis indirectly
• Capable of self-replication

RNA Functions:

• Transfers genetic information from DNA
• Directly involved in protein synthesis
• Three types: mRNA (carries genetic message), tRNA (transfers amino acids), rRNA (provides site for protein synthesis)
• Cannot self-replicate

Chemical Significance:

The presence of 2'-OH group in RNA makes it chemically more reactive and less stable than DNA. This is why DNA is the storage molecule for genetic information while RNA is used for temporary functions like protein synthesis.

Ques: Differentiate between α-helix and β-pleated sheet structures of proteins. What type of bonding stabilizes these structures?

Detailed Solution:

• α-Helix Structure:

Polypeptide chain coils into right-handed helix with intramolecular H-bonding

C=O (of nth amino acid) ----H-N (of (n+4)th amino acid)

• β-Pleated Sheet Structure:

Extended polypeptide chains arranged side by side with intermolecular H-bonding

C=O (of one chain) ----H-N (of adjacent chain)

• Detailed Comparison:

Stabilizing Forces:

Both structures are stabilized by hydrogen bonding:

• In α-helix: H-bonds form between C=O of amino acid at position n and N-H of amino acid at position n+4
• In β-sheet: H-bonds form between C=O of one strand and N-H of adjacent strand

Mechanical Properties:

• α-helix proteins are elastic because stretching breaks weak H-bonds which can reform
• β-sheet proteins like silk are strong but not elastic because stretching would require breaking covalent bonds

These secondary structures are fundamental to protein architecture and determine many physical properties of proteins.

Ques: Define essential and non-essential amino acids with examples. Write equations showing how they are utilized in the body.

Detailed Solution:

• Essential Amino Acids:

These are amino acids that cannot be synthesized by the human body and must be obtained through diet.

Examples:

• Valine (Val): (CH₃)₂CH-CH(NH₂)-COOH
• Leucine (Leu): (CH₃)₂CH-CH₂-CH(NH₂)-COOH
• Phenylalanine (Phe): C₆H₅-CH₂-CH(NH₂)-COOH
• Methionine (Met): CH₃-S-CH₂-CH₂-CH(NH₂)-COOH
• Lysine (Lys): NH₂-(CH₂)₄-CH(NH₂)-COOH

Non-Essential Amino Acids:

These are amino acids that can be synthesized by the human body from other compounds.

Examples:

• Glycine (Gly): NH₂-CH₂-COOH
• Alanine (Ala): CH₃-CH(NH₂)-COOH
• Aspartic acid (Asp): HOOC-CH₂-CH(NH₂)-COOH
• Glutamic acid (Glu): HOOC-CH₂-CH₂-CH(NH₂)-COOH
• Serine (Ser): HO-CH₂-CH(NH₂)-COOH

Utilization in Body (Protein Synthesis):

Multiple amino acids + Energy (ATP) → Protein + nH₂O

General Equation:

nNH₂-CHR-COOH → [-NH-CHR-CO-]_n (Protein) + (n-1)H₂O

Both essential and non-essential amino acids are required for protein synthesis. They are joined together by peptide bonds through condensation reactions. The sequence of amino acids determines the primary structure of proteins.

Dietary Importance:

• Essential amino acids must be consumed in adequate amounts for normal growth and health
• Protein-rich foods like meat, eggs, dairy, soy contain all essential amino acids (complete proteins)
• Some plant proteins lack one or more essential amino acids (incomplete proteins)
• Deficiency of essential amino acids leads to protein malnutrition (kwashiorkor, marasmus)

The body pools all amino acids (from diet and biosynthesis) and uses them as needed for synthesizing various proteins, enzymes, hormones, antibodies, and other nitrogen-containing compounds.

Ques: How are vitamins classified? Write the deficiency diseases and one chemical reaction involving any water-soluble vitamin.

Detailed Solution:

Classification of Vitamins:

1. Water-Soluble Vitamins:

• Vitamin B-complex (B₁, B₂, B₆, B₁₂, Niacin, Pantothenic acid, Biotin, Folic acid)
• Vitamin C (Ascorbic acid)

2. Fat-Soluble Vitamins:

• Vitamin A (Retinol)
• Vitamin D (Calciferol)
• Vitamin E (Tocopherol)
• Vitamin K (Phylloquinone)

Vitamin C (Ascorbic Acid) Reaction:

Oxidation of Vitamin C:

C₆H₈O₆ (Ascorbic acid) + [O] → C₆H₆O₆ (Dehydroascorbic acid) + H₂O

This reversible oxidation-reduction is important for:

• Acting as antioxidant
• Electron transport in metabolism
• Protecting cells from oxidative damage

Deficiency Diseases Table:

Why Water-Soluble Vitamins Must Be Supplied Regularly:

Water-soluble vitamins (B-complex and C) cannot be stored in the body in significant amounts. They are readily excreted through urine and must be replenished daily through diet. In contrast, fat-soluble vitamins (A, D, E, K) are stored in the liver and adipose tissue and need not be consumed daily.

Ques: What are enzymes? Explain enzyme-substrate specificity. Write a general equation for enzyme-catalyzed reaction.

Detailed Solution:

Definition:

Enzymes are biological catalysts (biocatalysts) that are proteinaceous in nature and catalyze biochemical reactions in living organisms.

General Enzyme-Catalyzed Reaction Equation:

E + S ⇌ ES → EP → E + P

Where:

• E = Enzyme
• S = Substrate
• ES = Enzyme-Substrate Complex
• EP = Enzyme-Product Complex
• P = Product

Detailed Mechanism (Lock and Key Model):

Step 1: Substrate Binding

Enzyme (E) + Substrate (S) ⇌ Enzyme-Substrate Complex (ES)

The substrate binds to the active site of the enzyme, forming a temporary enzyme-substrate complex. The active site has a specific three-dimensional shape complementary to the substrate (like a lock and key).

Step 2: Catalysis

ES → EP (Enzyme-Product Complex)

The enzyme facilitates the conversion of substrate to product by lowering the activation energy of the reaction.

Step 3: Product Release

EP → E + P

The product is released, and the enzyme is regenerated in its original form to catalyze another reaction cycle.

Enzyme Specificity:

Enzymes are highly specific in their action:

1. Substrate Specificity: Each enzyme catalyzes only a particular substrate or class of substrates
   • Example: Urease catalyzes only hydrolysis of urea, not other amides
2. Reaction Specificity: Each enzyme catalyzes only one type of reaction
   • Example: Maltase catalyzes only hydrolysis of maltose, not synthesis

Factors Affecting Enzyme Activity:

• Temperature (optimum around 37°C for human enzymes)
• pH (each enzyme has optimum pH)
• Substrate concentration
• Enzyme concentration
• Presence of inhibitors/activators

Example Reactions:

Maltase:

Maltose + H₂O --Maltase--> 2 Glucose

Invertase (Sucrase):

Sucrose + H₂O --Invertase--> Glucose + Fructose

Urease:

Urea + H₂O --Urease--> 2NH₃ + CO₂

Enzymes are crucial for life as they allow complex biochemical reactions to occur at body temperature and physiological conditions.

Ques: Define primary, secondary, and tertiary structures of proteins. Write equations showing the forces involved in stabilizing each structure.

Detailed Solution:

1. Primary Structure:

The specific linear sequence of amino acids linked by peptide bonds in a polypeptide chain.

Equation:

Amino acid₁ + Amino acid₂ + ... + Amino acid_n →

[-NH-CHR₁-CO-NH-CHR₂-CO-...-NH-CHR_n-CO-] + (n-1)H₂O

Representation:

NH₂-Gly-Ala-Val-Leu-...-COOH

Stabilizing Force: Covalent peptide bonds (-CO-NH-)

Characteristics:

• Determines the unique identity of each protein
• Any change in sequence creates a different protein
• Sequence determines how protein will fold into higher structures
• Mutations that change primary structure can cause diseases (e.g., sickle cell anemia)

2. Secondary Structure:

The regular, repeating spatial arrangement of the polypeptide backbone, stabilized by hydrogen bonding.

Two Types:

α-Helix:

...NH-CHR-CO... (hydrogen bond) ...NH-CHR-CO...

C=O···H-N (intramolecular H-bonding pattern)

β-Pleated Sheet:

Chain 1: ...NH-CHR-CO...

||||| (H-bonds)

Chain 2: ...OC-RHC-NH...

Stabilizing Force: Hydrogen bonds between C=O and N-H groups

Characteristics:

• α-helix: right-handed coil, H-bonds between 1st and 4th amino acids
• β-sheet: extended chains lying side by side with intermolecular H-bonds
• Proteins can have regions of both α-helix and β-sheet

3. Tertiary Structure:

The overall three-dimensional folding of the entire polypeptide chain, creating a globular or fibrous shape.

Stabilizing Forces (Multiple):

1. Disulfide bonds (covalent):

-CH₂-S-S-CH₂- (between cysteine residues)

Formation:

2 Cysteine residues: 2(-CH₂-SH) → -CH₂-S-S-CH₂- + 2H⁺ + 2e^-

2. Hydrogen bonds:

-OH···O=C- or -NH···O=C-

3. Ionic/Electrostatic interactions:

-COO^-···⁺NH₃- (salt bridges)

4. Hydrophobic interactions:

Nonpolar R-groups cluster in protein interior away from water

5. Van der Waals forces:

Weak attractions between all atoms in close proximity

Characteristics:

• Determines biological activity of protein
• Hydrophobic residues fold inward, hydrophilic outward (in globular proteins)
• Most stable conformation under physiological conditions
• Denaturation disrupts tertiary structure, causing loss of function