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IGNOU BBCCT-101 Previous Year Solved Question Paper in Hindi

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Q1. (a) “Biochemistry is an interdisciplinary subject.” Discuss. (b) Which properties of water make it suitable for aquatic life ? (c) Calculate the concentration of OH⁻ in a solution of 0.07 M HCl.

Ans. (a) The statement “Biochemistry is an interdisciplinary subject” is entirely accurate. Biochemistry, the chemistry of life, serves as a bridge between biology and chemistry, studying the chemical processes and molecular basis of living organisms. Its interdisciplinary nature is evident from its connections to various fields:

• Connection with Biology: Biochemistry is the foundation for molecular biology, cell biology, genetics, and physiology. For instance, understanding the structure and replication of DNA is fundamental to genetics, while enzyme kinetics and metabolic pathways are essential for understanding physiology.
• Connection with Chemistry: It extensively uses principles of organic chemistry to understand the structure and properties of biomolecules (carbohydrates, proteins, lipids, etc.). Principles of physical chemistry (like thermodynamics and kinetics) are vital for studying enzyme kinetics and bioenergetics.
• Connection with Medicine: Biochemistry plays a crucial role in understanding the molecular basis of diseases. The diagnosis and treatment of conditions like diabetes (insulin metabolism), cancer (abnormal cell growth), and many genetic disorders are based on biochemical principles. Pharmacology also relies heavily on biochemistry to understand the mechanisms of drug action.
• Connection with Agriculture and Nutrition: Biochemistry is important in crop improvement, development of pesticides, and analyzing the nutritional value of food.

Thus, biochemistry integrates knowledge and techniques from diverse scientific disciplines, making it a truly interdisciplinary field.

(b) The following properties of water make it uniquely suitable for aquatic life:

1. High Specific Heat Capacity: Water has a high specific heat capacity, meaning it resists changes in its temperature. This property helps to stabilize the temperature of aquatic environments (like lakes and oceans), protecting aquatic organisms from extreme temperature fluctuations.
2. Anomalous Expansion of Water: Water reaches its maximum density at 4°C. As water cools further and freezes, it expands, making ice less dense than liquid water. Consequently, ice floats and forms an insulating layer on the surface, which prevents the water below from freezing solid and enables aquatic life to survive.
3. Universal Solvent: Water is an excellent polar solvent, capable of dissolving many nutrients, minerals, and gases (like oxygen) that aquatic organisms require for survival.

(c) Hydrochloric acid (HCl) is a strong acid, which means it dissociates completely in water. HCl → H⁺ + Cl⁻ Therefore, in a 0.07 M solution of HCl, the concentration of hydrogen ions [H⁺] will be 0.07 M. [H⁺] = 0.07 M

We know the ion product of water (Kₑ) is a constant: Kₑ = [H⁺] [OH⁻] = 1.0 x 10⁻¹⁴ M² (at 25°C)

To calculate the concentration of hydroxide ions [OH⁻], we rearrange the equation: [OH⁻] = Kₑ / [H⁺]

Substituting the values: [OH⁻] = (1.0 x 10⁻¹⁴ M²) / (0.07 M) [OH⁻] = (1.0 x 10⁻¹⁴) / (7 x 10⁻²) M [OH⁻] = (1/7) x 10⁻¹² M [OH⁻] = 0.1428 x 10⁻¹² M [OH⁻] = 1.43 x 10⁻¹³ M (to three significant figures)

Thus, the concentration of OH⁻ in a 0.07 M HCl solution is 1.43 x 10⁻¹³ M .

Q2. (a) Name the secondary structures found in a protein. Describe four features of any one of these. (b) Calculate pI of glycine; given pK₁ =2.34 and pK₂ = 9.6. Draw the titration curve for glycine and indicate the ionization states of glycine at given pK values.

Ans. (a) The main secondary structures found in a protein are:

1. The Alpha-helix (α-helix)
2. The Beta-pleated sheet (β-pleated sheet)

Other structures include β-turns and loops, which connect these main structural elements.

Four features of the Alpha-helix (α-helix):

1. Helical Structure: The alpha-helix is a right-handed, rod-like coiled structure. The polypeptide backbone is tightly coiled, and the side chains (R-groups) project outwards from the axis of the helix.
2. Stabilization by Hydrogen Bonds: The structure is stabilized by extensive intramolecular hydrogen bonds . These bonds form between the carbonyl oxygen (C=O) of one amino acid residue (n) and the amide hydrogen (N-H) of the fourth amino acid residue down the chain (n+4). This regular bonding pattern provides stability to the helix.
3. Structural Parameters: There are 3.6 amino acid residues per turn of the alpha-helix. The pitch of the helix (the height of one complete turn) is 5.4 Å (0.54 nm).
4. Absence of Proline: The amino acid proline is considered a “helix breaker” because its rigid cyclic structure prevents the N-H group from participating in hydrogen bonding and introduces a kink in the polypeptide backbone. Therefore, it is rarely found within an alpha-helix.

(b) Calculation of pI for Glycine: The isoelectric point (pI) is the pH at which a molecule has no net electrical charge (i.e., it exists as a zwitterion). For a simple amino acid like glycine, which has no ionizable side chain, the pI is calculated as the average of the pK values for the carboxyl group (pK₁) and the amino group (pK₂).

Given: pK₁ = 2.34 (for the carboxyl group) and pK₂ = 9.6 (for the amino group). pI = (pK₁ + pK₂) / 2 pI = (2.34 + 9.6) / 2 pI = 11.94 / 2 pI = 5.97

Titration Curve and Ionization States of Glycine: The titration curve shows the relationship between pH and the amount of base (e.g., NaOH) added.

• At low pH (pH < 2.34): Both the carboxyl and amino groups are protonated. Glycine exists in its cationic form (⁺H₃N-CH₂-COOH) with a net charge of +1.
• At pH = pK₁ (2.34): The solution contains an equimolar mixture of the cationic form (⁺H₃N-CH₂-COOH) and the zwitterion (⁺H₃N-CH₂-COO⁻). This is the first buffering region.
• At pH = pI (5.97): Glycine exists predominantly as the zwitterion (⁺H₃N-CH₂-COO⁻) , which has a net charge of zero.
• At pH = pK₂ (9.6): The solution contains an equimolar mixture of the zwitterion (⁺H₃N-CH₂-COO⁻) and the anionic form (H₂N-CH₂-COO⁻). This is the second buffering region.
• At high pH (pH > 9.6): The amino group is deprotonated. Glycine exists in its anionic form (H₂N-CH₂-COO⁻) with a net charge of -1.

(A diagram of the titration curve for glycine would be drawn here, with pH on the Y-axis and equivalents of OH⁻ on the X-axis. The points pK₁, pK₂, and pI would be clearly marked, and the structures of the ionization states of glycine at the respective pH values would be shown on the curve.)

Q3. Write short notes on the following: (i) Glycosidic bond and its types (ii) Cellulose

Ans. (i) Glycosidic bond and its types A glycosidic bond is a type of covalent bond that joins a carbohydrate (sugar) molecule to another group, which may be another carbohydrate or a non-carbohydrate moiety. The bond is formed through a condensation (dehydration) reaction between the hemiacetal or hemiketal group of a monosaccharide and a hydroxyl (-OH) group of a second molecule, resulting in the elimination of a water molecule.

The main types of glycosidic bonds are:

• O-glycosidic bond: This is the most common type, where an anomeric carbon is linked to another molecule through an oxygen atom. This bond links monosaccharides together to form disaccharides, oligosaccharides, and polysaccharides.
  • α-linkage: When the bond on the anomeric carbon is axial (pointing down in a Haworth projection), it is an α-linkage. Example: α(1→4) linkage in maltose.
  • β-linkage: When the bond on the anomeric carbon is equatorial (pointing up), it is a β-linkage. Example: β(1→4) linkage in lactose.

  The linkages are also identified by the numbers of the carbon atoms involved in the bond, such as α(1→4), β(1→4), α(1→6), etc.

• N-glycosidic bond: In this type of bond, the anomeric carbon of a sugar links to a nitrogen atom. This bond is crucial in nucleosides and nucleotides, where a ribose or deoxyribose sugar is linked to a nitrogenous base. For example, in adenosine, ribose is linked to N-9 of adenine.

The nature of the glycosidic bond (α or β) greatly influences the structure and function of the resulting polysaccharide.

(ii) Cellulose Cellulose is the most abundant organic polymer on Earth and is the primary structural component of plant cell walls. It is a structural polysaccharide .

• Structure: Cellulose is a long, unbranched, linear polymer of β-D-glucose units. These glucose units are linked to each other by β(1→4) glycosidic bonds .
• Linear Chains and Fibrils: The β(1→4) linkage causes each glucose unit to be rotated 180° relative to the preceding unit, resulting in a straight, extended chain. This is in contrast to the helical structure of starch (which has α-linkages).
• Hydrogen Bonding: Parallel cellulose chains can form numerous inter-chain and intra-chain hydrogen bonds with each other. This extensive hydrogen bonding allows the cellulose chains to pack tightly into strong, water-insoluble microfibrils . These microfibrils, in turn, aggregate to form larger fibers.
• Insolubility and Indigestibility: Due to these strong fibrils, cellulose is insoluble in water and has high tensile strength, which provides structural support to plants. Most animals, including humans, cannot digest cellulose because they lack the enzyme cellulase , which can break the β(1→4) bonds. It is, however, an important source of dietary fiber.
• Biological Function: Its main function is to provide structural support in plants, making the cell walls rigid and strong.

Q4. Draw structures of any two of the following and write their biological function(s) : (i) Hyaluronic acid (ii) Raffinose (iii) Maltose

Ans. (i) Hyaluronic acid Hyaluronic acid is a type of glycosaminoglycan (GAG), which is an unbranched polysaccharide.

Structure: It is composed of a repeating disaccharide unit of D-glucuronic acid and N-acetyl-D-glucosamine . These units are linked by alternating β(1→3) and β(1→4) glycosidic bonds. The repeating unit is: [β(1→4)-D-glucuronic acid-β(1→3)-N-acetyl-D-glucosamine]ₙ. (A diagram of the repeating disaccharide unit of hyaluronic acid would be drawn here.)

Biological Function(s):

• Tissue Lubrication and Hydration: Hyaluronic acid can bind a large amount of water, making it an excellent lubricant and shock absorber. It is a major component of the synovial fluid in joints, providing lubrication.
• Component of Extracellular Matrix (ECM): It is a major component of the ECM in connective, epithelial, and neural tissues, providing structural support to cells.
• Role in the Eye: It is abundant in the vitreous humor of the eye, helping to maintain the eye’s shape.
• Skin Health: It helps retain moisture in the skin, keeping it supple and hydrated.

(iii) Maltose Maltose, also known as malt sugar, is a disaccharide.

Structure: Maltose is formed from two units of α-D-glucose joined by an α(1→4) glycosidic bond . It is a reducing sugar because the anomeric carbon (C1) of one of the glucose units is not involved in the glycosidic bond and can exist in the hemiacetal form. (A diagram of the structure of maltose would be drawn here, showing two glucose units and the α-1,4 bond.)

Biological Function(s):

• Intermediate in Starch Digestion: Maltose is formed as an intermediate product during the digestion of starch in animals. The enzyme amylase , found in saliva and the pancreas, breaks down starch into maltose.
• Energy Source: In the intestines, the enzyme maltase hydrolyzes maltose into two glucose molecules, which are then absorbed and used for energy.
• Use in Food Industry: It is used in brewing, where yeast ferments maltose to produce ethanol. It is also used in food products for its sweet taste.

Q5. (a) What are lectins ? Illustrate their functions with the help of suitable examples. (b) Identify the fatty acid from the given shorthand notation and draw its structure : (i) 16:1(Δ⁹) (ii) 20:0 (iii) 18:2(Δ⁹,¹²)

Ans. (a) Lectins Lectins are carbohydrate-binding proteins that exhibit high specificity and affinity for recognizing and reversibly binding to specific sugar structures (carbohydrate moieties). They are not enzymes or antibodies. These proteins are found in nearly all living organisms, including plants, animals, and microorganisms. Their specificity allows them to play crucial roles in many biological processes.

Functions and Examples of Lectins:

1. Cell-Cell Recognition and Adhesion: Lectins mediate cell-cell contact by binding to glycoproteins and glycolipids on cell surfaces.
   • Example: A family of lectins called selectins plays a critical role during inflammation. Selectins on the surface of endothelial cells lining blood vessels bind to specific carbohydrates on white blood cells (leukocytes), causing them to slow down and roll along the vessel wall before migrating into the tissue at the site of infection.
2. Binding of Pathogens: Many viruses and bacteria use lectins to attach to host cells to initiate infection.
   • Example: The influenza virus has a lectin on its surface called hemagglutinin , which binds to sialic acid residues (a type of sugar) present on the surface of host respiratory tract cells, allowing the virus to enter the cell.
3. Defense Mechanism in Plants: Many plants produce toxic lectins that protect them from herbivores and insects.
   • Example: Ricin , a lectin found in castor beans, is an extremely potent toxin. It inactivates ribosomes, thereby halting protein synthesis and causing cell death.

(b) Identification and Structure of Fatty Acids from Shorthand Notation:

(i) 16:1(Δ⁹)

• Identification: This is Palmitoleic Acid . The notation means:
  • 16: The fatty acid chain has 16 carbon atoms.
  • 1: It has one double bond.
  • Δ⁹: The double bond is located between carbon-9 and carbon-10 (counting from the carboxyl end). It is typically a cis double bond.
• Structure: (A drawing of the 16-carbon chain of palmitoleic acid would be shown here, with a cis-double bond at C9.)

(ii) 20:0

• Identification: This is Arachidic Acid . The notation means:
  • 20: The fatty acid chain has 20 carbon atoms.
  • 0: It has zero double bonds; it is a saturated fatty acid.
• Structure: (A drawing of the straight 20-carbon saturated chain of arachidic acid would be shown here.)

(iii) 18:2(Δ⁹,¹²)

• Identification: This is Linoleic Acid , an omega-6 essential fatty acid. The notation means:
  • 18: The fatty acid chain has 18 carbon atoms.
  • 2: It has two double bonds.
  • Δ⁹,¹²: The double bonds are located between C9-C10 and C12-C13. Both are typically cis double bonds.
• Structure: (A drawing of the 18-carbon chain of linoleic acid would be shown here, with two cis-double bonds at C9 and C12.)

Q6. (a) What are the key features of rRNA ? Mention its biological functions. (b) Arrange the following in terms of decreasing UV absorption values and give reason for your answer : ssDNA, dsDNA, Free bases

Ans. (a) Key Features and Biological Functions of rRNA (ribosomal RNA)

Key Features:

1. Abundance: rRNA is the most abundant type of RNA in the cell, constituting about 80% of the total cellular RNA.
2. Location: It is an integral component of ribosomes and is found in the cytoplasm within ribosomes. In eukaryotes, its synthesis takes place in the nucleolus.
3. Structure: rRNA molecules have complex and specific three-dimensional (3D) structures, containing both single-stranded and double-stranded regions (formed by base-pairing). This structure is essential for binding with ribosomal proteins.
4. Conservation: The genes for rRNA and their sequences are highly conserved throughout evolution, meaning their structure and function are very similar across different species. This property makes them useful for studying evolutionary relationships.

Biological Functions:

1. Structural Role: rRNA acts as the scaffold for the ribosome. Ribosomal proteins assemble onto the rRNA framework to form the small and large subunits of the ribosome.
2. Catalytic Role (Ribozyme): rRNA is not just a structural molecule; it also functions as a catalyst. The rRNA in the large ribosomal subunit (e.g., 23S rRNA in prokaryotes) possesses peptidyl transferase activity. This enzymatic activity catalyzes the formation of peptide bonds between amino acids during protein synthesis. Such a catalytic RNA is called a ribozyme .
3. Role in Translation: rRNA plays a crucial role in the process of protein synthesis (translation). In prokaryotes, a part of the 16S rRNA recognizes and binds to the Shine-Dalgarno sequence on the mRNA, ensuring that translation starts at the correct initiation site. It also provides the binding sites for tRNAs.

(b) Arrangement by Decreasing UV Absorption and Reason

The correct arrangement in order of decreasing UV absorption at ~260 nm is: Free bases > ssDNA (single-stranded DNA) > dsDNA (double-stranded DNA)

Reason: This phenomenon is related to the hyperchromic effect and hypochromicity . The purine and pyrimidine bases in nucleic acids absorb UV light maximally around 260 nm due to their aromatic ring structures.

1. Free bases: When the bases are free and not part of a polymer chain, they are completely unconstrained and their aromatic rings are fully exposed to photons. Therefore, they absorb the maximum amount of UV light.
2. ssDNA (single-stranded DNA): In ssDNA, the bases are linked by a phosphodiester backbone but are still relatively unstacked and disordered. Because they are largely exposed, they absorb a high amount of UV light, although slightly less than free bases.
3. dsDNA (double-stranded DNA): In dsDNA, the two complementary strands form a double helix. In this structure, the bases are stacked on top of each other in the interior of the helix. This base stacking causes electronic interactions between the π-electron clouds of adjacent bases. These interactions constrain the resonance of the bases and reduce their ability to absorb UV light. This phenomenon is called hypochromicity . As a result, the UV absorbance of dsDNA per nucleotide is the lowest. When dsDNA is denatured (melted) into ssDNA by heating, the absorbance increases, an effect known as the hyperchromic shift.

Q7. (a) Discuss the role of ATP as a source of energy. (b) Write names of nitrogenous bases present in DNA and draw their structures (any two).

Ans. (a) The Role of ATP as a Source of Energy Adenosine Triphosphate (ATP) is universally known as the “energy currency” of the cell. It is a nucleotide that plays a central role in the transfer and storage of chemical energy within living cells.

Structure and High-Energy Bonds: An ATP molecule is composed of three main components:

1. A nitrogenous base: Adenine
2. A five-carbon sugar: Ribose
3. Three phosphate groups linked in a series (alpha, beta, and gamma)

The bonds linking the last two phosphate groups (between beta and gamma, and alpha and beta) are known as

high-energy phosphoanhydride bonds

. The hydrolysis of these bonds releases a large amount of free energy.

Energy Release and Utilization: The cell harnesses energy for its needs by hydrolyzing ATP. When the terminal phosphate group of ATP is cleaved, it is converted to Adenosine Diphosphate (ADP) and an inorganic phosphate (Pi). ATP + H₂O → ADP + Pi + Energy This reaction releases approximately -30.5 kJ/mol of free energy under standard conditions. The cell uses this released energy to drive various energy-requiring (endergonic) processes, such as:

• Mechanical Work: Muscle contraction and cellular movement.
• Active Transport: Pumping ions and molecules across membranes against their concentration gradients (e.g., the Na⁺/K⁺ pump).
• Biosynthesis: The synthesis of large macromolecules (proteins, nucleic acids, etc.).
• Signal Transduction: Activating or deactivating enzymes and other proteins via phosphorylation.

This is achieved through

reaction coupling

, where the exergonic hydrolysis of ATP is paired with an endergonic reaction, making the overall process energetically favorable.

ATP Regeneration: The cell’s supply of ATP is limited, so it must be constantly regenerated from ADP and Pi. This process occurs primarily through cellular respiration (glycolysis, Krebs cycle, and oxidative phosphorylation) and, in plants, photosynthesis .

(b) Nitrogenous Bases in DNA and their Structures There are four main nitrogenous bases present in DNA. They are classified into two categories:

1. Purines: These are double-ring structures.
   • Adenine (A)
   • Guanine (G)
2. Pyrimidines: These are single-ring structures.
   • Cytosine (C)
   • Thymine (T)

In DNA, adenine always pairs with thymine (via two hydrogen bonds), and guanine always pairs with cytosine (via three hydrogen bonds).

Structures of two bases:

1. Adenine (A) – A Purine (A diagram of the double-ring structure of adenine would be drawn here, showing an amino group (-NH₂) at the C6 position.)

2. Cytosine (C) – A Pyrimidine (A diagram of the single-ring structure of cytosine would be drawn here, showing an amino group (-NH₂) at the C4 position and a keto group (=O) at the C2 position.)

Q8. Write short notes on the following: (i) Endocrine hormones (ii) Functions of a biological membrane

Ans. (i) Endocrine hormones Endocrine hormones are chemical messengers synthesized and secreted by specialized glands known as endocrine glands (e.g., pituitary, thyroid, adrenal glands). These glands are ductless, meaning they release their hormones directly into the bloodstream . The blood then transports these hormones throughout the body, but they only affect cells that possess specific receptors for them. These cells are called target cells .

Classification based on Chemical Nature:

1. Peptide/Protein Hormones: Composed of chains of amino acids (e.g., insulin, glucagon ). They are water-soluble and typically bind to receptors on the surface of the target cell.
2. Steroid Hormones: Derived from cholesterol (e.g., cortisol, testosterone, estrogen ). They are lipid-soluble and can pass through the cell membrane to bind to intracellular receptors (in the cytoplasm or nucleus).
3. Amino Acid-derived Hormones: Formed from a single amino acid, usually tyrosine or tryptophan (e.g., thyroxine, epinephrine ).

Function:

Hormones regulate and coordinate a wide variety of bodily processes, including metabolism, growth and development, water and electrolyte balance, reproduction, and response to stress. They are essential for maintaining the body’s internal balance, or

homeostasis

.

(ii) Functions of a biological membrane A biological membrane (or cell membrane) is a semipermeable lipid bilayer that surrounds cells and their organelles. Its roles extend far beyond being just a physical barrier. Its major functions include:

1. Compartmentalization: Membranes separate the cell from the external environment and create distinct organelles within eukaryotic cells (e.g., nucleus, mitochondria, lysosomes). This creates specialized compartments for specific biochemical reactions to occur efficiently without interfering with each other.
2. Selective Permeability: The cell membrane regulates the passage of substances into and out of the cell. Small, nonpolar molecules (like O₂, CO₂) can pass directly through the lipid bilayer, while ions and large polar molecules require specific transport proteins , such as channels and pumps.
3. Signal Transduction: Receptor proteins embedded in the membrane bind to external signals (like hormones, neurotransmitters). This binding triggers a response inside the cell, leading to a change in the cell’s behavior.
4. Energy Transduction: Certain membranes are specialized in converting energy from one form to another. For example, the inner mitochondrial membrane houses the electron transport chain , which generates ATP through oxidative phosphorylation. Similarly, the thylakoid membrane in chloroplasts is the site of photosynthesis.
5. Cell-Cell Recognition and Communication: Glycoproteins and glycolipids on the outer surface of the membrane act as “identity markers.” These help in distinguishing the body’s own cells from foreign cells (immune response) and enable cells to adhere to one another to form tissues.
6. Anchoring the Cytoskeleton: Membrane proteins provide anchor points for filaments of the cytoskeleton, which helps maintain cell shape and is involved in cell movement.

Q9. Write the biochemical functions and dietary sources of the following : (i) Pyridoxine (Vitamin B₆) (ii) Vitamin B₁₂ (iii) Riboflavin (Vitamin B₂) (iv) Thiamine (Vitamin B₁)

Ans. (i) Pyridoxine (Vitamin B₆)

• Biochemical Functions: The active coenzyme form of Vitamin B₆ is Pyridoxal Phosphate (PLP) . PLP is an essential coenzyme in amino acid metabolism . It is involved in over 100 enzymatic reactions, including:
  • Transamination: Crucial for the synthesis and degradation of amino acids.
  • Decarboxylation: Necessary for the synthesis of neurotransmitters (e.g., serotonin, dopamine, GABA).
  • Glycogenolysis: As a component of the enzyme glycogen phosphorylase, which releases glucose from glycogen.
• Dietary Sources: Meat, fish, poultry, potatoes, chickpeas, bananas, and fortified cereals.

(ii) Vitamin B₁₂ (Cobalamin)

• Biochemical Functions: The coenzyme forms of Vitamin B₁₂ are methylcobalamin and adenosylcobalamin . It is required for two major enzymatic reactions in humans:
  1. Conversion of homocysteine to methionine: Methylcobalamin is a cofactor for the enzyme methionine synthase. This is vital for DNA synthesis and cell division.
  2. Conversion of methylmalonyl-CoA to succinyl-CoA: Adenosylcobalamin is a cofactor for methylmalonyl-CoA mutase. This is essential for the metabolism of some fatty acids and amino acids.

  Deficiency can lead to megaloblastic anemia and neurological damage.

• Dietary Sources: Vitamin B₁₂ is found almost exclusively in animal products , such as meat, fish, poultry, eggs, and dairy products. For vegetarians, fortified foods (like cereals, plant-based milks) and supplements are the main sources.

(iii) Riboflavin (Vitamin B₂)

• Biochemical Functions: Riboflavin is an integral component of two major coenzymes: Flavin Mononucleotide (FMN) and Flavin Adenine Dinucleotide (FAD) . These coenzymes (also called flavoproteins) play a critical role in numerous oxidation-reduction (redox) reactions in metabolism. They act as electron carriers in processes such as:
  • The Krebs cycle (e.g., succinate dehydrogenase uses FAD).
  • The electron transport chain (Complex I uses FMN and Complex II uses FAD).
  • β-oxidation of fatty acids.
• Dietary Sources: Dairy products (milk, yogurt, cheese), eggs, lean meats, leafy green vegetables (e.g., spinach), and fortified cereals and breads.

(iv) Thiamine (Vitamin B₁)

• Biochemical Functions: The active form of thiamine is Thiamine Pyrophosphate (TPP) . TPP is a critical coenzyme in carbohydrate metabolism , essential for energy production. It is primarily involved in decarboxylation reactions:
  • Pyruvate dehydrogenase complex: This enzyme links glycolysis to the Krebs cycle, converting pyruvate to acetyl-CoA.
  • α-ketoglutarate dehydrogenase complex: A key enzyme within the Krebs cycle.
  • Transketolase: An enzyme in the pentose phosphate pathway, which is necessary for nucleic acid synthesis.

  Deficiency of thiamine causes the disease beriberi.

• Dietary Sources: Whole grains, pork, legumes, nuts, seeds, and fortified breads and cereals.