ROUTERA

Chapter 14 Biomolecules

Class 12th Chemistry Chapter Hots

1. Explain the primary, secondary, tertiary, and quaternary structures of proteins in detail, using diagrams where necessary. Discuss the forces stabilizing each structure.

Answer:
Proteins exhibit hierarchical levels of structure:

(a) Primary Structure:

  • Refers to the sequence of amino acids in a polypeptide chain linked by peptide bonds.
  • Determines the protein's overall structure and function.
  • Example: Insulin has two chains with a specific sequence.

(b) Secondary Structure:

  • Regular patterns of hydrogen bonding between the C=O and N-H groups of the backbone.
  • Alpha-Helix: A helical structure stabilized by intra-chain hydrogen bonds. Found in keratin (e.g., hair).
  • Beta-Pleated Sheets: Extended strands connected laterally by hydrogen bonds. Found in silk fibroin.

(c) Tertiary Structure:

  • The 3D arrangement of the entire polypeptide.
  • Stabilized by hydrophobic interactions, disulfide bonds, ionic interactions, and hydrogen bonds.
  • Example: Myoglobin (oxygen storage protein in muscles).

(d) Quaternary Structure:

  • Assembly of multiple polypeptide subunits.
  • Example: Hemoglobin consists of four subunits.

Stabilizing Forces:

  • Primary: Peptide bonds.
  • Secondary: Hydrogen bonds.
  • Tertiary: Ionic bonds, van der Waals forces, disulfide linkages.
  • Quaternary: Inter-subunit interactions.

2. Discuss the mechanism of enzyme action in detail, including the steps involved. Explain how enzymes lower activation energy with the help of diagrams.

Answer:
Enzymes catalyze biochemical reactions by lowering the activation energy required for the reaction to proceed.

Mechanism of Enzyme Action:

  1. Substrate Binding:

    • The substrate binds to the enzyme's active site, forming an enzyme-substrate complex.
    • This interaction may follow the lock and key model (rigid specificity) or induced fit model (flexible active site).

  2. Formation of Enzyme-Substrate Complex:

    • Temporary bonds form between the enzyme and substrate, such as hydrogen bonds or ionic interactions.

  3. Catalysis:

    • The enzyme stabilizes the transition state, reducing the energy barrier for product formation.

  4. Product Formation and Release:

    • The reaction proceeds to form the product, which is released, leaving the enzyme unchanged and ready for another cycle.

How Enzymes Lower Activation Energy:

  • Enzymes bring substrates closer together in the correct orientation.
  • They stabilize the transition state through binding energy.
  • They provide alternative pathways for the reaction.

Diagram: Include energy profiles comparing uncatalyzed and enzyme-catalyzed reactions.

3. Write a detailed note on DNA replication. Describe the role of enzymes such as helicase, DNA polymerase, primase, and ligase. Include a stepwise explanation with diagrams.

Answer:
DNA replication is a semiconservative process where each daughter DNA molecule consists of one parent and one newly synthesized strand.

Steps of DNA Replication:

  1. Initiation:

    • Helicase unwinds the double helix at the replication origin, creating a replication fork.
    • Single-strand binding proteins stabilize the unwound strands.

  2. Priming:

    • Primase synthesizes RNA primers complementary to the DNA template, providing a free 3'-OH group for DNA polymerase.

  3. Elongation:

    • DNA polymerase adds nucleotides in the 5' to 3' direction.
    • The leading strand is synthesized continuously, while the lagging strand is synthesized in fragments (Okazaki fragments).

  4. Joining:

    • DNA ligase seals the gaps between Okazaki fragments, forming a continuous strand.

  5. Proofreading:

    • DNA polymerase has exonuclease activity to correct errors.

Role of Enzymes:

  • Helicase: Unwinds DNA.
  • Primase: Synthesizes RNA primer.
  • DNA Polymerase: Adds nucleotides and proofreads.
  • Ligase: Joins DNA fragments.

Diagram: Include replication fork showing helicase, polymerase, and Okazaki fragments.

4. What are reducing and non-reducing sugars? Explain the differences with examples and chemical tests. Discuss why sucrose is non-reducing and glucose is reducing.

Answer:

Reducing Sugars:

  • Contain a free aldehyde (-CHO) or ketone (-CO-) group capable of reducing Tollens’ reagent (Ag⁺) or Fehling’s solution (Cu²⁺).
  • Example: Glucose, fructose, maltose.

Non-Reducing Sugars:

  • Do not have free aldehyde/ketone groups due to glycosidic bonds.
  • Example: Sucrose.

Tests to Identify Sugars:

  1. Fehling’s Test: Red precipitate indicates reducing sugar.
  2. Tollens’ Test: Silver mirror indicates reducing sugar.

Why Sucrose is Non-Reducing:

  • In sucrose, the anomeric carbons of glucose and fructose are involved in the glycosidic bond, preventing free aldehyde or ketone groups from forming.

Why Glucose is Reducing:

  • Glucose has a free aldehyde group in its open-chain form, which can reduce reagents.

5. Explain the role of carbohydrates, proteins, and fats in the human body. How does the deficiency of each affect human health? Provide examples.

Answer:

Carbohydrates:

  • Role: Primary energy source; stored as glycogen in the liver and muscles.
  • Deficiency Effects: Fatigue, ketosis (due to fat metabolism).
  • Example: Rice, bread.

Proteins:

  • Role: Structural components (keratin, collagen), enzymes, hormones, immunity (antibodies).
  • Deficiency Effects: Kwashiorkor, muscle wasting.
  • Example: Meat, eggs.

Fats:

  • Role: Energy storage, insulation, cell membranes, hormone synthesis.
  • Deficiency Effects: Dry skin, hormonal imbalance, poor brain function.
  • Example: Butter, olive oil.

6. Describe the biochemical role of vitamins and classify them into water-soluble and fat-soluble vitamins. Provide examples, functions, and deficiency disorders for each type.

Answer:

Classification:

  1. Water-Soluble Vitamins:

    • Vitamin C: Antioxidant; deficiency causes scurvy.
    • Vitamin B12: Coenzyme in DNA synthesis; deficiency causes anemia.

  2. Fat-Soluble Vitamins:

    • Vitamin A: Vision and immunity; deficiency causes night blindness.
    • Vitamin D: Calcium absorption; deficiency causes rickets.

Biochemical Role:

  • Serve as coenzymes or precursors for coenzymes.
  • Regulate metabolic pathways.

7. Calculate the osmotic pressure of a protein solution. Given: 2g of protein dissolved in 100mL solution exerts an osmotic pressure of 0.05 atm at 25°C. Calculate the molecular weight of the protein. (Use R = 0.0821 L atm K⁻¹ mol⁻¹)

Answer:
Use the formula for osmotic pressure:
Π=CRT\Pi = CRT
Where:

  • Π=0.05 atm\Pi = 0.05 \, \text{atm}
  • C=massmolar mass×volumeC = \frac{\text{mass}}{\text{molar mass} \times \text{volume}}
  • T=298 KT = 298 \, \text{K}

Substitute:
0.05=(2M×0.1)×0.0821×2980.05 = \left(\frac{2}{M \times 0.1}\right) \times 0.0821 \times 298
M=2×0.0821×2980.05×0.1=97628 g/molM = \frac{2 \times 0.0821 \times 298}{0.05 \times 0.1} = 97628 \, \text{g/mol}

Molecular weight = 97628 g/mol.

8. What is the role of nucleic acids in the cell? Describe the structure and function of DNA and RNA. How do they differ in terms of their chemical composition and function?

Answer:
Nucleic acids (DNA and RNA) are essential for storing and transmitting genetic information.

DNA (Deoxyribonucleic Acid):

  • Structure: Composed of two polynucleotide strands coiled into a double helix. The strands are held together by hydrogen bonds between complementary nitrogenous bases (A-T and C-G).
  • Function: DNA stores genetic information in the form of genes, which are inherited and provide instructions for protein synthesis.

RNA (Ribonucleic Acid):

  • Structure: A single-stranded molecule containing ribose sugar and nitrogenous bases (A, U, C, G).
  • Function: RNA transcribes genetic information from DNA and helps in protein synthesis (mRNA, tRNA, rRNA).

Differences:

  • Sugar: DNA has deoxyribose, while RNA has ribose.
  • Bases: DNA contains thymine (T), whereas RNA contains uracil (U).
  • Strands: DNA is double-stranded; RNA is single-stranded.
  • Function: DNA is for long-term storage of genetic information, while RNA is involved in protein synthesis.

9. Explain the importance of enzymes in metabolic pathways. How do enzyme inhibitors affect enzyme activity? Discuss the different types of enzyme inhibitors with examples.

Answer:
Enzymes play a crucial role in metabolic pathways by speeding up biochemical reactions and maintaining the body's homeostasis.

Types of Enzyme Inhibitors:

  1. Competitive Inhibitors:

    • Compete with the substrate for the active site.
    • Example: Methotrexate inhibits dihydrofolate reductase.

  2. Non-Competitive Inhibitors:

    • Bind to an allosteric site, altering the enzyme’s shape and preventing substrate binding.
    • Example: Cyanide inhibits cytochrome c oxidase.

  3. Uncompetitive Inhibitors:

    • Bind to the enzyme-substrate complex, preventing the reaction from proceeding.
    • Example: Lithium inhibits inositol monophosphatase.

Effect on Enzyme Activity:

  • Competitive inhibitors can be overcome by increasing substrate concentration.
  • Non-competitive inhibitors cannot be overcome by substrate concentration.
  • Uncompetitive inhibitors decrease both the enzyme's affinity for the substrate and the rate of the reaction.

10. Describe the concept of isomerism in carbohydrates. Explain the difference between D and L isomers with examples.

Answer:
Isomerism in carbohydrates refers to the existence of compounds with the same molecular formula but different structural arrangements.

Types of Isomerism:

  • Structural Isomerism: Different bonding patterns.
  • Stereoisomerism: Same bonding, but different spatial arrangements (including D/L and α/β isomerism).

D and L Isomers:

  • D and L refer to the configuration of the chiral center furthest from the carbonyl group in aldoses or the second chiral center in ketoses.
  • D-isomer: Has the same configuration as D-glyceraldehyde, the reference compound.
  • L-isomer: Has the opposite configuration.

Example:

  • D-glucose is a common D-isomer, while L-glucose is its mirror image and not commonly found in nature.

11. Discuss the types of biomolecules that make up the cell membrane. Explain the structure and function of phospholipids in the membrane.

Answer:
The cell membrane is primarily composed of lipids, proteins, and carbohydrates.

Phospholipids:

  • Structure: Phospholipids have a hydrophilic (water-attracting) "head" and hydrophobic (water-repelling) "tails" made of fatty acids.
  • Function: Phospholipids form a bilayer in the membrane. The hydrophilic heads face outward toward the aqueous environment, while the hydrophobic tails are oriented inward, creating a semi-permeable membrane. This structure allows selective transport of molecules.

12. Explain the concept of activation energy in biochemical reactions. How do enzymes alter the activation energy of a reaction? Use the Arrhenius equation to explain the relationship between activation energy and reaction rate.

Answer:
Activation energy is the minimum energy required for a chemical reaction to occur.

Effect of Enzymes on Activation Energy:

  • Enzymes lower the activation energy by stabilizing the transition state, making it easier for reactants to form products.
  • This results in an increased reaction rate at physiological temperatures.

Arrhenius Equation:

k=Ae−EaRTk = Ae^{-\frac{E_a}{RT}}

Where:

  • kk is the reaction rate constant.
  • AA is the pre-exponential factor (frequency of collisions).
  • EaE_a is the activation energy.
  • RR is the gas constant.
  • TT is the temperature in Kelvin.

As activation energy decreases (EaE_a ), the reaction rate kk increases. Enzymes reduce EaE_a , speeding up the reaction.

13. Discuss the concept of the double helix model of DNA. What are the forces responsible for the stability of this structure?

Answer:
The double helix model of DNA, proposed by Watson and Crick, consists of two strands of nucleotides coiled around each other.

Structure:

  • The two strands run in opposite directions (antiparallel).
  • The strands are connected by hydrogen bonds between complementary nitrogenous bases (A-T, C-G).
  • The sugar-phosphate backbone is on the outside, and the bases face inward.

Forces Stabilizing the Structure:

  1. Hydrogen Bonds: Between complementary bases (A-T, C-G).
  2. Hydrophobic Interactions: Between stacked base pairs.
  3. Van der Waals Forces: Between base pairs.
  4. Ionic Interactions: Between the negatively charged phosphate backbone and cations in the solution.

14. Explain the structure of amino acids and their classification. Discuss how the sequence of amino acids determines the structure and function of proteins.

Answer:
Amino acids are the building blocks of proteins. Each amino acid consists of a central carbon (alpha-carbon) bonded to an amino group (-NH2), a carboxyl group (-COOH), a hydrogen atom, and a variable side chain (R-group).

Classification of Amino Acids:

  1. Nonpolar (Hydrophobic): e.g., Alanine, Valine.
  2. Polar (Hydrophilic): e.g., Serine, Glutamine.
  3. Acidic: e.g., Aspartic acid.
  4. Basic: e.g., Lysine, Arginine.

Protein Structure:

The sequence of amino acids (primary structure) determines how the protein folds into its 3D structure (secondary, tertiary, and quaternary structures), which in turn determines its function.

15. Define the term ‘enzyme specificity’. Discuss the factors affecting enzyme specificity.

Answer:
Enzyme specificity refers to the ability of an enzyme to recognize and bind to a specific substrate or group of related substrates.

Factors Affecting Specificity:

  1. Shape of the Active Site: The enzyme's active site has a specific shape that fits the substrate (lock-and-key model).
  2. Electrostatic Interactions: Charge and polarity of the substrate and active site determine binding.
  3. Steric Compatibility: The size and shape of the substrate must match the enzyme's active site.

16. Discuss the structure of fatty acids and their classification. Explain the difference between saturated and unsaturated fatty acids with examples.

Answer:
Fatty acids are long chains of hydrocarbons with a carboxyl group (-COOH) at one end.

Classification:

  1. Saturated Fatty Acids:

    • Contain only single bonds between carbon atoms.
    • Example: Stearic acid.
    • Solid at room temperature.

  2. Unsaturated Fatty Acids:

    • Contain one or more double bonds between carbon atoms.
    • Example: Oleic acid (monounsaturated), Linoleic acid (polyunsaturated).
    • Liquid at room temperature.

17. Describe the structure and function of glycolipids. How do they contribute to cell-cell communication?

Answer:
Glycolipids are lipids with a carbohydrate attached to them. They are primarily found on the outer surface of cell membranes.

Structure:

  • Consist of a lipid tail (usually a fatty acid) and a carbohydrate chain.
  • The carbohydrate chain is often branched and varies in composition.

Function:

  • Glycolipids play a crucial role in cell recognition and communication.
  • They help in the formation of cell membranes and are involved in immune responses, serving as markers for cell identification.

18. What are the different types of RNA and their functions?

Answer:
There are three main types of RNA:

  1. mRNA (Messenger RNA):

    • Carries genetic information from DNA to the ribosome for protein synthesis.

  2. tRNA (Transfer RNA):

    • Transports amino acids to the ribosome, matching them with the appropriate codons on the mRNA.

  3. rRNA (Ribosomal RNA):

    • Makes up the structure of ribosomes, where protein synthesis occurs.

19. Explain the importance of the central dogma of molecular biology. How does transcription and translation contribute to protein synthesis?

Answer:
The central dogma of molecular biology states that genetic information flows from DNA to RNA to proteins.

Transcription:

  • In the nucleus, DNA is transcribed into mRNA, which carries the genetic code for protein synthesis.

Translation:

  • The mRNA moves to the ribosome in the cytoplasm, where it is translated into a sequence of amino acids, forming a protein.

The central dogma highlights the fundamental process by which genetic information is used to produce proteins, which carry out cellular functions.

20. What is the significance of proteins in the body? How are proteins classified based on their structure? Discuss the primary, secondary, tertiary, and quaternary structures of proteins with examples.

Answer:

Proteins are fundamental biomolecules that perform various vital functions in the body such as enzyme catalysis, immune response, transport of molecules, and structural support.

Classification of Proteins based on Structure:

  1. Primary Structure:

    • The primary structure is the sequence of amino acids in the polypeptide chain.
    • It determines the protein's identity and function.
    • Example: Insulin's primary structure consists of two chains of amino acids linked by disulfide bonds.

  2. Secondary Structure:

    • The folding or coiling of the polypeptide chain into regular patterns such as α-helix and β-pleated sheets.
    • Stabilized by hydrogen bonds between the backbone atoms.
    • Example: The α-helix in keratin (hair, skin, nails).

  3. Tertiary Structure:

    • The three-dimensional shape of a protein formed by the folding of the secondary structure into a functional unit.
    • Stabilized by various interactions like hydrogen bonds, hydrophobic interactions, ionic bonds, and disulfide bridges.
    • Example: Myoglobin, a protein that carries oxygen in muscles, has a globular tertiary structure.

  4. Quaternary Structure:

    • Some proteins consist of more than one polypeptide chain that are held together by non-covalent interactions.
    • Example: Hemoglobin, composed of four polypeptide subunits, carries oxygen in the blood.

21. Explain the concept of an enzyme-substrate complex. How does the Michaelis-Menten theory describe enzyme kinetics? Provide the equation and explain its significance.

Answer:

The enzyme-substrate complex is formed when an enzyme binds to its substrate at the active site. The enzyme catalyzes the conversion of the substrate into products.

Michaelis-Menten Theory:

  • This theory explains the relationship between enzyme concentration, substrate concentration, and the rate of reaction.
  • It is based on the assumption that the enzyme binds reversibly to the substrate, forming an enzyme-substrate complex (ES), which then converts into the enzyme-product complex (EP).

Michaelis-Menten Equation:

V0=Vmax[S]Km+[S]V_0 = \frac{V_{max}[S]}{K_m + [S]}

Where:

  • V0V_0 = Initial reaction rate.
  • VmaxV_{max} = Maximum rate achieved by the enzyme.
  • [S][S] = Concentration of substrate.
  • KmK_m = Michaelis constant, which represents the substrate concentration at which the reaction rate is half of VmaxV_{max} .

Significance:

  • The equation describes how the reaction rate varies with the concentration of substrate.
  • VmaxV_{max} indicates the maximum rate of reaction when all enzyme active sites are saturated with substrate.
  • KmK_m provides information about the affinity of the enzyme for the substrate—lower KmK_m means higher affinity.

22. What is the role of vitamins in the body? Discuss the classification of vitamins into water-soluble and fat-soluble vitamins with examples.

Answer:

Vitamins are organic compounds required in small amounts for normal growth and metabolic functions. They act as coenzymes or cofactors in enzymatic reactions, and deficiency in vitamins can lead to various disorders.

Classification of Vitamins:

  1. Water-Soluble Vitamins:

    • These vitamins dissolve in water and are not stored in the body. They need to be consumed regularly in the diet.
    • Examples:
      • Vitamin C (Ascorbic Acid): Essential for collagen formation and immune function.
      • Vitamin B Complex: Includes B1 (thiamine), B2 (riboflavin), B3 (niacin), B6 (pyridoxine), and B12 (cobalamin), important for energy metabolism and red blood cell production.

  2. Fat-Soluble Vitamins:

    • These vitamins are absorbed along with dietary fat and can be stored in the liver and adipose tissues.
    • Examples:
      • Vitamin A: Important for vision, growth, and immune function.
      • Vitamin D: Essential for calcium absorption and bone health.
      • Vitamin E: An antioxidant that protects cell membranes.
      • Vitamin K: Important for blood clotting.

23. What are lipids? Discuss their structure and different types, including triglycerides, phospholipids, and steroids.

Answer:

Lipids are a diverse group of hydrophobic or amphipathic molecules involved in energy storage, insulation, and structural components of cell membranes.

Types of Lipids:

  1. Triglycerides:

    • Consist of three fatty acids esterified to a glycerol molecule.
    • Main energy storage form in the body.
    • Example: Fats and oils.

  2. Phospholipids:

    • Composed of two fatty acid tails, a glycerol backbone, and a phosphate group attached to the third carbon of glycerol.
    • Key components of cell membranes, forming a bilayer with hydrophilic heads facing the outside and hydrophobic tails facing the inside.
    • Example: Phosphatidylcholine.

  3. Steroids:

    • Lipids with a structure based on four fused carbon rings.
    • Involved in regulating metabolic functions and acting as hormones.
    • Example: Cholesterol, estrogen, and testosterone.

24. What is the significance of the structure of carbohydrates in biological systems? Describe the classification of carbohydrates with examples.

Answer:

Carbohydrates are vital energy sources and structural components of living organisms. They consist of carbon, hydrogen, and oxygen atoms, typically in a 1:2:1 ratio.

Classification of Carbohydrates:

  1. Monosaccharides:

    • Simplest form of carbohydrates, consisting of one sugar unit.
    • Example: Glucose, Fructose, Galactose.

  2. Disaccharides:

    • Composed of two monosaccharide units joined by a glycosidic bond.
    • Example: Sucrose (glucose + fructose), Lactose (glucose + galactose).

  3. Polysaccharides:

    • Complex carbohydrates made of many monosaccharides linked together.
    • Example:
      • Starch: Energy storage in plants.
      • Glycogen: Energy storage in animals.
      • Cellulose: Structural component in plant cell walls.

25. Explain the role of DNA replication in cell division. How do enzymes like DNA polymerase and helicase contribute to the process?

Answer:

DNA replication is essential for cell division, allowing genetic material to be accurately copied and passed on to daughter cells. This process occurs during the S-phase of the cell cycle.

Steps in DNA Replication:

  1. Initiation:

    • The DNA double helix is unwound by helicase, which breaks the hydrogen bonds between complementary bases.

  2. Elongation:

    • DNA polymerase adds complementary nucleotides to the growing DNA strand, synthesizing a new strand in the 5' to 3' direction.
    • Primase synthesizes a short RNA primer to provide a starting point for DNA polymerase.

  3. Proofreading and Termination:

    • DNA polymerase also has proofreading ability to ensure accuracy during replication.
    • Once the entire DNA molecule has been replicated, the newly formed strands rewind into a double helix.

Key Enzymes:

  • DNA Helicase: Unwinds the DNA helix to separate the strands.
  • DNA Polymerase: Synthesizes the new DNA strand by adding nucleotides to the 3' end of the template strand.
  • Primase: Synthesizes RNA primers required for initiating DNA synthesis.
  • Ligase: Joins the Okazaki fragments on the lagging strand.