Class 11 Bio Botany · Chapter 8

Samacheer Class 11 Bio Botany - Biomolecules

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Chapter-wise textbook exercise answers for Biomolecules with validation-aware solutions.

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Sections in this chapter
II. a. Extracellular enzymes 1II. Choose the wrong answer. 2III. But A : T, need not be equal to G : C 16III. Co- Enzymes: 4III. Match The Following And Find The Correct Answer. 5III. Non-reversible/ Irreversible Inhibitors 2IV. Find Out The True And False Statements From The Following And That Basis Find Out The Right Answer. 4IV. Nucleoside 3Part I 5Part II 33VI. Assertion & Reason – Find Out The Correct Answer. 12
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1II. a. Extracellular enzymes1 question
Q.5Explain the three types of Co-Factors.v
Answer:

Co-factors are non-protein organic or inorganic substances that assist enzymes in catalyzing biochemical reactions. There are three main types of co-factors. First, metal ions such as Mg2+, Ca2+, Zn2+, and Fe2+ act as co-factors by facilitating enzyme-substrate interactions and stabilizing enzyme structure. Second, prosthetic groups are organic co-factors that remain permanently or tightly bound to the enzyme protein. A notable example is vitamin B2, also known as riboflavin, which forms flavin adenine dinucleotide (FAD), a crucial prosthetic group in oxidation-reduction reactions within the Krebs cycle and electron transport chain. Third, co-enzymes are organic co-factors that are loosely bound to enzymes and can dissociate from them after the reaction is complete. Important examples of co-enzymes include NAD (nicotinamide adenine dinucleotide), NADP (nicotinamide adenine dinucleotide phosphate), coenzyme A, and ATP (adenosine triphosphate). These co-enzymes function as carriers of chemical groups or electrons between different enzymes and are essential for numerous metabolic pathways. The distinction between prosthetic groups and co-enzymes lies in the strength of their association with the enzyme and whether they remain attached or dissociate after catalysis.

2II. Choose the wrong answer.2 questions
Q.2This has nothing to do with the structure of a. Cytosine b. Pyrimidine c. Adenine d. Thyaminev
Answer:

c. Adenine

Q.3Choose the right answer a) Amylose – linear unbranched polymer of with 20% starch b) Amylopectin – a polymer with some 1,6 linkages that give it a linear structure c) Inulin – Polymer of galactose d) Amino acid – Here a basic structure of carbon linked to a basic amino groupv
Answer:

The correct answer is d) Amino acid – Here a basic structure of carbon linked to a basic amino group. An amino acid is an organic compound with a characteristic structure consisting of a central carbon atom bonded to four different groups: an amino group (NH2), a carboxyl group (COOH), a hydrogen atom, and a variable side chain (R group). This basic structure is fundamental to all amino acids and allows them to link together through peptide bonds to form proteins. The other options are incorrect because amylose is a linear unbranched polymer of glucose that comprises approximately 20-25% of starch, not a polymer with a different composition. Amylopectin is a branched polymer of glucose with 1,4 glycosidic linkages in the linear chains and 1,6 glycosidic linkages at the branch points, giving it a branched structure rather than a linear one. Inulin is a polymer of fructose, not galactose. Understanding the correct structures of these biomolecules is essential for comprehending carbohydrate and protein chemistry.

3III. But A : T, need not be equal to G : C16 questions
Q.1Classify Polysaccharides.v
Answer:

Polysaccharides are complex carbohydrates composed of more than ten monosaccharide units linked together by glycosidic bonds. They serve various functions in living organisms, including energy storage and structural support. Polysaccharides can be broadly classified into two main categories: homopolysaccharides and heteropolysaccharides. Homopolysaccharides are composed of only one type of monosaccharide unit repeated many times. Examples of homopolysaccharides include starch, which is the primary energy storage polysaccharide in plants, glycogen, the main energy storage polysaccharide in animals and fungi, and cellulose, a major structural component of plant cell walls. Chitin, found in the exoskeletons of arthropods and cell walls of fungi, and inulin, a storage polysaccharide in some plants, are also homopolysaccharides. Heteropolysaccharides, on the other hand, are composed of two or more different types of monosaccharide units or their derivatives. Examples of heteropolysaccharides include peptidoglycan, a crucial component of bacterial cell walls, hyaluronic acid, a major component of connective tissues, chondroitin sulfate, found in cartilage, and keratan sulfate, present in cornea and cartilage. Agar-agar, a complex mixture of polysaccharides obtained from red algae, is also a heteropolysaccharide used extensively in microbiology and food industries.

Q.1How will you identify the presence of glucose in a given food sample?v
Answer:

Aldoses and ketoses are reducing sugars. This means that, when heated with an alkaline solution of copper (II) sulphate (a blue solution called Benedict’s solution), the aldehyde or ketone group reduces Cu 2+ ions to Cu + ions forming brick red precipitate of copper (I) oxide. In the process, the aldehyde or ketone group is oxidised to a carboxyl group (-COOH).
This reaction is used as test for reducing sugar and is known as Benedict’s test. The results of Benedict’s test depends on concentration of the sugar. If there is no reducing sugar it remains blue. Sucrose is not a reducing sugar The greater the concentration of reducing sugar, the more is the precipitate formed and greater is the colour change.

Q.2Why do we call Glucose and Fructose Isomers -Discuss.v
Answer:

Glucose and fructose are classified as isomers because they possess the same molecular formula, C6H12O6, but differ in their structural arrangements. Although both contain six carbon atoms, twelve hydrogen atoms, and six oxygen atoms, the atoms are arranged differently in space. Glucose is an aldohexose with an aldehyde group at the first carbon, forming a six-membered pyranose ring in its cyclic form. Fructose is a ketohexose with a ketone group at the second carbon, forming a five-membered furanose ring in its cyclic form. This difference in structure results in different chemical and biological properties despite their identical molecular formulas. Glucose serves as the primary energy source for cells and is the main product of photosynthesis, while fructose is sweeter than glucose and is commonly found in fruits and honey. These structural differences explain why the two sugars behave differently in metabolic pathways and have distinct physiological effects.

Q.2What is protein denaturation.v
Answer:

Protein denaturation is the process by which a protein loses its three-dimensional structure and biological function. Exposure to heat causes atoms within the protein to vibrate violently, which disrupts the weak chemical bonds that maintain the protein's structure, particularly hydrogen bonds and ionic bonds. As these bonds break, the protein unfolds from its compact, organized conformation into elongated, disorganized strands, losing its native shape and becoming non-functional. The primary structure of the protein, consisting of peptide bonds between amino acids, remains intact during denaturation, but the secondary, tertiary, and quaternary structures are disrupted. Beyond heat, various other agents can cause protein denaturation. Soaps and detergents disrupt the hydrophobic interactions that stabilize protein structure. Acids and bases alter the pH, affecting ionic bonds and the charge on amino acid side chains. Alcohol denatures proteins by disrupting hydrogen bonding and hydrophobic interactions. Certain disinfectants chemically modify amino acid residues, breaking interchain bonds and destroying the protein's functional architecture. Once denatured, most proteins cannot refold into their original structure, making denaturation generally irreversible under physiological conditions.

Q.4Classify enzyme reactions:v
Answer:

Enzymes can be classified based on their location and site of action into two main categories. Extracellular enzymes are secreted outside the cell and function in external environments. These enzymes work in the extracellular space, such as in the digestive tract, where they catalyze the breakdown of complex food molecules. Examples include digestive enzymes such as amylase, which breaks down starch; protease, which breaks down proteins; and lipase, which breaks down fats. These enzymes are produced by cells but released into the lumen of the digestive system to perform their catalytic functions. Intracellular enzymes, in contrast, remain within cells and catalyze reactions in the intracellular environment. These enzymes function within the cytoplasm, mitochondria, chloroplasts, and other cellular compartments to regulate metabolic pathways essential for cell survival and function. Examples include glycolytic enzymes involved in glucose metabolism, enzymes of the Krebs cycle, and enzymes involved in protein synthesis. It should be noted that insulin is actually a hormone, not an enzyme, though some enzymes do function intracellularly. This classification is important for understanding enzyme regulation and the compartmentalization of metabolic processes within cells.

Q.5Distinguish between Waxes & Steroids.v
Answer:

Waxes and steroids are both lipids but differ significantly in their structure and function. Waxes are esters formed between long-chain alcohols and saturated fatty acids. They are found as protective coatings on the fur, feathers, fruits, leaves, skin, and insect exoskeletons, providing waterproofing and reducing water loss. Steroids, in contrast, are complex compounds with a characteristic four-ring carbon structure. They are found in cell membranes and serve as animal hormones. Cholesterol is a well-known example of a steroid that reinforces the structure of the cell membrane in animal cells and also in Mycoplasma. While waxes function primarily as protective barriers due to their hydrophobic nature, steroids have diverse roles including structural support and hormonal regulation. The key distinction lies in their chemical composition, with waxes being fatty acid esters and steroids being polycyclic compounds with different biological functions.

Q.6Draw the structural formula of 3 simple amino acids – Glycine, Alanine & Valine.v
Answer:

Amino acids with non-polar aliphatic R groups include glycine, alanine, valine, leucine, methionine, and isoleucine. Glycine has the simplest structure with a hydrogen atom as its R group, making it the only amino acid without a chiral center. Its structural formula shows the central carbon bonded to an amino group (NH2), a carboxyl group (COOH), a hydrogen atom, and another hydrogen atom as the R group. Alanine has a methyl group (CH3) as its R group attached to the central carbon, giving it a small non-polar side chain. Valine has a branched isopropyl group as its R group, consisting of a carbon atom bonded to two methyl groups, making it larger and more hydrophobic than alanine. All three amino acids share the common backbone structure with the amino group at one end and the carboxyl group at the other end, differing only in their R group composition and size.

Q.7Distinguish between Macronutrients & Micronutrients.v
Answer:

Macronutrients and micronutrients are essential elements required by plants for their growth and development, but they differ significantly in the quantities needed. Macronutrients are those nutrients that plants require in relatively larger quantities for various physiological processes. These elements are crucial for building plant structures, carrying out metabolic reactions, and maintaining overall plant health. Examples of macronutrients include Potassium (K), which is vital for enzyme activation, water balance, and photosynthesis, and Calcium (Ca), essential for cell wall formation, cell division, and membrane permeability. Other macronutrients include Nitrogen, Phosphorus, Magnesium, and Sulfur. In contrast, micronutrients are trace elements that plants require in very small or minute amounts. Despite being needed in smaller quantities, they are equally vital for plant growth and play critical roles as cofactors for enzymes and in various metabolic pathways. Examples of micronutrients include Zinc (Zn), which is involved in enzyme activation and hormone synthesis, and Boron (B), essential for cell wall formation, sugar transport, and reproductive development. Other micronutrients include Iron, Manganese, Copper, Molybdenum, and Chlorine. Both categories of nutrients are indispensable for healthy plant life, and a deficiency in either can lead to significant growth abnormalities.

Q.8Define Activation energy?v
Answer:
  • The minimum quantity of energy the reactants must possess in order to undergo a specified reaction is known as Activation energy.
  • Energy being the biocatalysts reduce the activation energy, thereby help the reaction occurs.
  • The rate of reaction increases if activation energy decreases.
Q.9Differentiate Between DNA & RNA.v
Answer:

DNA and RNA are both nucleic acids but differ in several important ways. DNA is typically double-stranded while RNA is single-stranded. DNA serves as the genetic material in almost all living organisms except for RNA viruses, whereas RNA is not the primary genetic material except in RNA viruses. DNA exists in two forms depending on the organism: prokaryotic DNA is circular, while eukaryotic DNA is linear. RNA exists in three main types: messenger RNA (mRNA), transfer RNA (tRNA), and ribosomal RNA (rRNA), each with distinct functions. DNA contains the sugar deoxyribose while RNA contains ribose. DNA has thymine as one of its bases while RNA has uracil instead. DNA controls all aspects of cellular function and heredity by storing genetic information, while RNA plays important roles in protein synthesis by carrying genetic information from DNA and facilitating the translation process. The structural differences between DNA and RNA reflect their different biological roles in cells.

Q.9Distinguish between Primary metabolite & Secondary metabolite.v
Answer:

Primary metabolites and secondary metabolites are two categories of organic compounds produced by living organisms, differing in their direct roles in fundamental life processes. Primary metabolites are those compounds that are directly involved in the normal growth, development, and reproduction of an organism. They are essential for the basic metabolic processes that sustain life, such as photosynthesis, respiration, and nutrient assimilation. These metabolites are typically found in all cells and are crucial for the survival of the organism. Examples include amino acids, nucleotides, sugars, and fatty acids. For instance, lipase, a protein, is a primary metabolite as it plays a direct role in the digestion of fats, a fundamental metabolic process. Secondary metabolites, on the other hand, are compounds that are not directly involved in the normal growth, development, or reproduction of an organism. They do not show any direct function in the primary metabolic pathways but often play important roles in ecological interactions, such as defense against predators or pathogens, attraction of pollinators, or competition with other organisms. These compounds are often specific to certain species or groups of organisms and are produced in smaller quantities. Examples include ricin, a toxin, and various gums, which are polymeric substances. Other examples include alkaloids, terpenes, flavonoids, and antibiotics. While not essential for immediate survival, secondary metabolites contribute significantly to an organism's fitness in its environment.

Q.10Give examples for Secondary metabolites.v
Answer:

Secondary metabolites are organic compounds produced by bacteria, fungi, or plants that are not directly involved in the normal growth, development, or reproduction of the organism but often play significant ecological roles. These compounds are diverse in their chemical structures and biological activities. Examples of secondary metabolites include pigments such as carotenoids, which give plants and fruits their characteristic yellow, orange, and red colors, and anthocyanins, responsible for red, purple, and blue hues. Alkaloids are a large group of nitrogen-containing compounds, many of which have potent pharmacological effects, such as morphine, a powerful analgesic, and codeine, used as a cough suppressant. Essential oils, which are volatile aromatic compounds, include lemongrass oil and rose oil, used in perfumes and flavorings. Toxins, which are harmful substances, include abrin and ricin, both highly poisonous proteins. Lectins are carbohydrate-binding proteins, such as Concanavalin A, used in cell biology research. Drugs derived from plants include vinblastine, an anti-cancer agent, and curcumin, an anti-inflammatory compound from turmeric. Polymeric substances like rubber, gums, and cellulose also fall under secondary metabolites, with rubber being an elastic polymer and gums serving as protective or storage substances in plants.

Q.11How will you test reducing sugar?v
Answer:

To test for the presence of reducing sugars, the Benedict's test is commonly employed. Reducing sugars are carbohydrates that possess a free aldehyde or ketone group, which allows them to act as reducing agents. The procedure involves taking a sample, such as glucose solution, in a test tube. Glucose is an aldehyde sugar. An alkaline solution of copper(II) sulfate, known as Benedict's reagent, is then added to the test tube. The mixture is subsequently heated, typically in a boiling water bath. In the presence of a reducing sugar, the copper(II) ions (Cu²⁺), which are blue in Benedict's reagent, are reduced to copper(I) ions (Cu⁺). These copper(I) ions then form a brick-red precipitate of copper(I) oxide. This color change from blue to green, yellow, orange, or brick-red, depending on the concentration of the reducing sugar, indicates a positive test. Simultaneously, the aldehyde group of the reducing sugar is oxidized to a carboxylic acid group. This reaction is a classic example of a redox reaction used to detect the presence of reducing sugars in various biological samples.

Q.12Draw the structure of a basic amino acid.v
Answer:

A basic amino acid contains an amino group (NH2), a carboxyl group (COOH), a hydrogen atom, and an R group that is positively charged or contains an additional amino group. Examples of basic amino acids include lysine, arginine, and histidine. The structural formula shows a central carbon atom bonded to these four groups. In lysine, the R group is a long hydrocarbon chain ending with an amino group, making it basic in nature. In arginine, the R group contains a guanidinium group which is highly basic and positively charged at physiological pH. In histidine, the R group contains an imidazole ring that can accept a proton. These basic amino acids are important for protein structure and function, particularly in forming ionic bonds and interacting with negatively charged molecules.

Q.13Why do some people have curly hair?v
Answer:

Human hair is made primarily of protein, particularly a fibrous protein called keratin. The structure and texture of hair, including whether it is curly or straight, depends on the arrangement and bonding of these protein molecules. The distance between sulfur atoms in the protein chains is crucial in determining hair curliness. Sulfur atoms form disulfide bonds (also called disulfide bridges) between adjacent protein chains, which act as cross-links. When sulfur atoms are closer together, the protein chains are pulled tighter, causing the hair to curl more. Conversely, when sulfur atoms are farther apart, the protein chains are more relaxed, resulting in straighter hair. The number and positioning of these disulfide bonds determine the degree of curliness. This is why chemical treatments like perming or straightening work by breaking and reforming these disulfide bonds to alter the natural curl pattern of hair.

Q.14Deoxyribose (C 5 H 10 O 4 ) is not a carbohydrate – Discuss.v
Answer:

Deoxyribose with the molecular formula C5H10O4 is indeed a carbohydrate, but its formula does not follow the general formula for carbohydrates, which is typically represented as Cn(H2O)m or (CH2O)n. Carbohydrates are defined as hydrates of carbon, meaning they contain carbon, hydrogen, and oxygen in specific ratios. The general formula suggests that for every carbon atom, there should be two hydrogen atoms and one oxygen atom. If we apply this formula to deoxyribose, a pentose sugar with five carbon atoms, we would expect the formula to be C5H10O5. However, deoxyribose has the formula C5H10O4 because it is a deoxy sugar, meaning it has one less oxygen atom than the corresponding regular pentose sugar ribose (C5H10O5). The prefix 'deoxy' literally means lacking one oxygen. Despite not fitting the general carbohydrate formula perfectly, deoxyribose is still classified as a carbohydrate because it is a monosaccharide with the characteristic properties of sugars. It is a five-carbon sugar that serves as the sugar component of DNA, distinguishing DNA from RNA which contains ribose.

4III. Co- Enzymes:4 questions
Q.6Tabulate the uses of enzymesv
Answer:

Enzymes are biological catalysts that play crucial roles in various industrial and biotechnological applications due to their high specificity and efficiency. Bacterial proteases, often sourced from Bacillus species, are widely used in biological detergents to break down protein stains like blood and grass, enhancing cleaning efficiency. Bacterial glucose isomerase, also from Bacillus, is essential in the manufacture of high-fructose corn syrup, converting glucose into fructose, which is sweeter and more soluble. Fungal lactase, obtained from organisms like Kluyveromyces, is used to break down lactose into glucose and galactose, making dairy products suitable for lactose-intolerant individuals. Amylases, commonly sourced from Aspergillus fungi, are employed in the textile industry for the removal of starch sizing from woven cloth, a process known as desizing, to prepare the fabric for dyeing and finishing. These are just a few examples illustrating the diverse and significant uses of enzymes across different sectors, leveraging their catalytic power for specific biochemical transformations.

Q.7Enumerate the properties of Enzyme.v
Answer:

Enzymes possess several distinct properties that make them highly efficient and specific biological catalysts. Firstly, enzymes are typically globular proteins, meaning they have a complex three-dimensional structure that is crucial for their function. Secondly, they act as catalysts, significantly increasing the rate of biochemical reactions without being consumed in the process, and are effective even in very small quantities. Consequently, they remain unchanged at the end of the reaction and can be reused. Thirdly, enzymes are highly specific; each enzyme usually catalyzes only one or a very limited number of reactions, acting on specific substrates. This specificity arises from the unique shape of their active site. Fourthly, they have an active site, which is a specific region on the enzyme molecule where the substrate binds and the catalytic reaction takes place. Finally, a fundamental property of enzymes is their ability to lower the activation energy of the reactions they catalyze. By reducing the energy barrier, enzymes allow reactions to proceed much faster at physiological temperatures, which is vital for life processes.

Q.8Explain lock and key mechanism of enzymes.v
Answer:

The lock and key mechanism is a model proposed by Emil Fischer in 1894 to explain the specificity of enzyme action. According to this model, the enzyme has a specific three-dimensional structure with a unique region called the active site, which is analogous to a 'lock'. The substrate, which is the molecule upon which the enzyme acts, has a complementary shape that fits precisely into the active site, much like a 'key' fits into a specific lock. When the substrate binds to the active site of the enzyme, they form an enzyme-substrate (ES) complex. This binding is highly specific due to the complementary shapes and chemical interactions between the enzyme and the substrate. As the enzyme and substrate form the ES-complex, the substrate is often raised in energy, reaching a transition state where the chemical bonds are strained or altered, making them more susceptible to reaction. The enzyme then facilitates the chemical transformation of the substrate into products. After the reaction is complete, the products are released from the active site, and the enzyme remains unchanged and is free to bind to another substrate molecule and catalyze the reaction again. This mechanism highlights the precise fit and specificity that characterize enzyme catalysis.

Q.9What are the various types of inhibitors of enzymes.v
Answer:

Definition:
Substances present in the cells may react with enzyme and lower the rate of reactions Inhibitors
I. Competitive Inhibitors:
Substances resemble the shape of substrate & compete to occupy active sites
Eg. 1. Enzyme RUBISCO – is competitively inhibited by oxygen/carbon dioxide in the chloroplast
2. Succinic dehydrogenase – Inhibited by malonate.
II. Non-Competitive Inhibitors
Unlike substrates, blocks by binding on active sites, change its shape so enzyme unable to accept substrate.
Enzyme – pyruvate kinase- inhibited by amino acid Alanine.
III. Non-reversible/ Irreversible Inhibitors
They bind to an enzyme tightly & permanently destroying their catalytic nature Enzyme cytochrome oxidase inhibited by cyanide ions Neurotransmitter – blocked by nerve gas sarin.

5III. Match The Following And Find The Correct Answer.5 questions
Q.1(I) Morphine – A. Hectins (II) Concanavalin A – B. Drug (III) Vinblastin – C. Pigment (IV) Anthocyanin – D. Toxinv
Answer:

(I) Morphine – B. Drug (II) Concanavalin A – A. Lectins (III) Vinblastin – D. Toxin (IV) Anthocyanin – C. Pigment. Morphine is an alkaloid drug derived from the opium poppy plant used for pain relief. Concanavalin A is a lectin, a protein that binds specifically to carbohydrates. Vinblastin is an alkaloid toxin from the periwinkle plant used in cancer chemotherapy. Anthocyanin is a water-soluble pigment responsible for red, purple, and blue colors in plants.

Q.2(I) Lactose – A. Penta saccharide (II) Ramnose – B. Tetra saccharide (III) Stachyose – C. Disaccharide (IV) Verbascose – D. Tri saccharidev
Answer:

(b) C-D-B-A

Q.3(I) Fred Sanger 1st sequenced (II) Di sulphide bridges formed between sulphur & amino acids (III) non-protein enzyme (IV) homo polysaccharide with amino acidv
Answer:

(I) Fred Sanger – A. 1st sequenced insulin (II) Disulfide bridges – B. formed between sulfur atoms in amino acids (III) Non-protein enzyme – D. ribozyme (IV) Homopolysaccharide with amino acid – C. peptidoglycan. Fred Sanger was the first scientist to determine the complete amino acid sequence of a protein, specifically insulin, earning him the Nobel Prize. Disulfide bridges are covalent bonds formed between sulfur atoms in cysteine amino acids, providing structural stability to proteins. Ribozymes are non-protein enzymes made of RNA that can catalyze biochemical reactions. Peptidoglycan is a homopolysaccharide that contains amino acids and forms the structural component of bacterial cell walls.

Q.4(I) Amino acid chain is twisted into coiled configuration call a helix – A. Tertiary Protein (II) Protein fold into a globular structure called domains – B. Quaternary protein (III) Linear arrangement of amino acids in a Polypeptide chain – C. Secondary protein (IV) more than one polypeptide forms a large multiunit multimer – D. Primary Proteinv
Answer:

(I) Amino acid chain is twisted into coiled configuration called a helix – C. Secondary protein (II) Protein fold into a globular structure called domains – A. Tertiary Protein (III) Linear arrangement of amino acids in a Polypeptide chain – D. Primary Protein (IV) More than one polypeptide forms a large multiunit multimer – B. Quaternary protein. Primary structure refers to the linear sequence of amino acids connected by peptide bonds. Secondary structure involves the coiling of the polypeptide chain into an alpha helix or beta sheet configuration, stabilized by hydrogen bonds. Tertiary structure is the three-dimensional folding of the protein into a globular shape with specific domains, stabilized by various interactions including disulfide bonds and hydrophobic interactions. Quaternary structure occurs when multiple polypeptide chains associate together to form a larger protein complex.

Q.5(I)) Esters formed between long-chain alcohol another negative. – A. a molecule with two or more & saturated fatty acid function group one +ve and (II) lipids have both hydrophobic & hydrophilic end known for permeability – B. fluid nature & selective (III) The amino acids are both acidic & basic exoskeleton of insects – C. waxy substance coating (IV) Zwitter is also called dipolar – D. amophoteric in naturev
Answer:

(I) Esters formed between long-chain alcohol and saturated fatty acid – C. waxy substance coating exoskeleton of insects (II) Lipids have both hydrophobic and hydrophilic end known for permeability – B. fluid nature and selective (III) The amino acids are both acidic and basic – D. amphoteric in nature (IV) Zwitterion is also called dipolar – A. a molecule with two or more function groups one positive and one negative. Waxes are esters that provide protective coatings. Amphipathic lipids have both water-repelling and water-attracting regions, enabling selective permeability. Amino acids contain both carboxyl (acidic) and amino (basic) groups, making them amphoteric. Zwitterions are dipolar molecules with both positive and negative charges.

6III. Non-reversible/ Irreversible Inhibitors2 questions
Q.10Distinguish between feedback allosteric inhibition negative feedback (end product) inhibition.v
Answer:

Allosteric inhibition and negative feedback inhibition are two distinct regulatory mechanisms for controlling enzyme activity. In allosteric inhibition, an inhibitor molecule binds to a site on the enzyme other than the active site, called the allosteric site. This binding causes a conformational change in the enzyme's three-dimensional structure, modifying the active site and reducing its ability to bind substrate. Allosteric inhibition is reversible, meaning the inhibitor can dissociate and the enzyme can regain its original activity. A classic example is the inhibition of hexokinase by glucose-6-phosphate, where the product of the reaction binds to an allosteric site and reduces enzyme activity. In negative feedback or end-product inhibition, the accumulation of the final product of a metabolic pathway inhibits an enzyme earlier in the pathway, typically the first committed enzyme. This prevents overproduction of the product and maintains metabolic balance. Once the end products are consumed and their concentration decreases, the inhibition is relieved and the enzyme reaction is switched back on. Both mechanisms are reversible and serve to regulate metabolic pathways, but allosteric inhibition involves binding at a different site on the enzyme, while negative feedback inhibition typically involves the product of the pathway itself.

Q.11Tabulate other sugar compounds.v
Answer:

Other polysaccharides
Structure
Functions
Inulin
Polymer of fructose
It is not metabolized in the human body and is readily filtered through the kidney
Hyaluronic acid
Heteropolymer of d glucuronic acid and D-N acetyl glucosamine
It accounts for the toughness and flexibility of cartilage and tendon
Agar
Mucopolysaccharide from red algae
Used as a solidifying agent in culture medium in a laboratory
Heparin
Glucosamine glycan contains variably sulphated disaccharide unit present in liver
Used as anticoagulant
Used as anticoagulant
Sulphated glycosaminoglycan composed of altering sugars (N-acetylglucosamine and glucuronic acid)
Dietary supplement for treatment of osteoarthritis
Keratan sulphate
Sulphated glycosaminoglycan and is a structural carbohydrate
Acts as cushion to absorb mechanical shock

7IV. Find Out The True And False Statements From The Following And That Basis Find Out The Right Answer.4 questions
Q.1(I) Esters are formed between long-chain alcohol & saturated fatty acid. (II) Lecithin is a food additive & dietary supplement. ‘ (III) Lipids in their structure have two hydrophilic ends (IV) Solid fats are usually unsaturatedv
Answer:

(I) Esters are formed between long-chain alcohol and saturated fatty acid – True. This is the correct definition of waxes, which are esters composed of a long-chain alcohol and a saturated fatty acid. (II) Lecithin is a food additive and dietary supplement – True. Lecithin is a phospholipid commonly used as an emulsifier in food products and is also available as a dietary supplement. (III) Lipids in their structure have two hydrophilic ends – False. Lipids, particularly amphipathic lipids like phospholipids, have one hydrophilic (water-loving) head region and one or more hydrophobic (water-repelling) tail regions, not two hydrophilic ends. (IV) Solid fats are usually unsaturated – False. Solid fats at room temperature are typically saturated fats, which have no double bonds between carbon atoms and pack tightly together. Unsaturated fats, which contain double bonds, are usually liquid at room temperature.

Q.1Label the diagram parts correctly by choosing the right option. A B C D a Deoxyribose sugar Nitrogen base Nucleotide Phosphate b Deoxyribose sugar Phosphate Nucleotide Nitrogen base c Deoxyribose sugar Nitrogen base Nucleotide Phosphate d Deoxyribose sugar Phosphate Nitrogen base Nucleotidev
Answer:

(d) Deoxyribose sugar – Phosphate Nitrogen base – Nucleotide

Q.2(I) In saturated fatty acids, the hydrocarbon chain is single-bonded (II) Triglycerides are composed of a single molecule of glycerol bound to 2 fatty acids (III) Palmitic acid is an example of saturated fatty acid. (IV) Oleic acid is an example of unsaturated fatty acid.v
Answer:

Statement (I) is correct because in saturated fatty acids, the hydrocarbon chain contains only single bonds between carbon atoms, with no double bonds present. Statement (II) is incorrect because triglycerides are composed of one molecule of glycerol bound to three fatty acids, not two. Statement (III) is correct as palmitic acid is indeed a saturated fatty acid with a 16-carbon chain and no double bonds. Statement (IV) is correct because oleic acid is an unsaturated fatty acid containing one double bond in its hydrocarbon chain. Therefore, statements (I), (III), and (IV) are correct while statement (II) is incorrect.

Q.2A B C D a Q arm Centromere Sister Chromatids Q arm b P arm Centromere Sister Chromatids Q arm c Sister Chromatids Centromere Q arm P arm d Q arm Centromere P arm Sister Chromatidsv
Answer:

The correct arrangement of chromosome parts from top to bottom is: Q arm, Centromere, Sister Chromatids, and P arm. The centromere is the constricted region that divides the chromosome into two arms. The shorter arm is called the P arm and the longer arm is called the Q arm. Sister chromatids are the two identical copies of a chromosome held together at the centromere after DNA replication. The correct answer is option (b).

8IV. Nucleoside3 questions
Q.7What are the factors affecting the rate of enzyme reactions?v
Answer:

The rate of enzyme-catalyzed reactions is influenced by several environmental and intrinsic factors, as enzymes are biomolecules highly sensitive to their surroundings. Firstly, temperature significantly affects enzyme activity. Generally, heating increases molecular motion, which quickens enzyme reactions up to a certain point. However, there is an optimum temperature at which an enzyme exhibits maximum activity. Beyond this optimum, higher temperatures can cause denaturation, where the enzyme's three-dimensional structure is irreversibly altered, leading to a loss of activity. Secondly, pH is another critical factor. Changes in pH can alter the ionization state of amino acid residues in the enzyme's active site, leading to an alteration of the enzyme's shape and affecting its ability to bind the substrate. Extremes of pH can denature enzymes. Similar to temperature, there is an optimum pH at which the maximum rate of reaction occurs. Thirdly, substrate concentration plays a crucial role. For a given enzyme concentration, the rate of reaction increases with increasing substrate concentration until all active sites are saturated. At this point, the reaction rate reaches its maximum (Vmax) and further increases in substrate concentration will not increase the rate. Finally, enzyme concentration is directly proportional to the rate of reaction. If the substrate is in excess, increasing the enzyme concentration will increase the number of available active sites, thereby increasing the overall rate of product formation.

Q.9Write down the characteristic features of DNA?v
Answer:

The characteristic feature of DNA.
* If one strand runs in the 5′ – 3′ direction, the other runs in 3′ – 5′ direction and thus are antiparallel (they run in the opposite direction). The 5′ end has the phosphate group and 3’end has the OH group.
* The angle at which the two sugars protrude from the base pairs is about 120°, for the narrow-angle and 240° for the wide-angle. The narrow-angle between the sugars generates a minor groove and the large angle on the other edge generates major groove.
* Each base is 0.34 nm apart and a complete turn of the helix comprises 3.4 nm or 10 base pairs per turn in the predominant B form of DNA.
* DNA helical structure has a diameter of 20 Å and a pitch of about 3 Å. X-ray crystal study of DNA takes a stack of about 10 bp to go completely around the helix (360°).
* Thermodynamic stability of the helix and specificity of base pairing includes
* The hydrogen bonds between the complementary bases of the double helix
* stacking interaction between bases tend to stack about each other perpendicular to the direction of the helical axis.
* Electron cloud interactions (\({ \Pi -{ \Pi } }\)) between the bases in the helical stacks contribute to the stability of the double helix.
* The phosphodiester linkages give an inherent polarity to the DNA helix. They form strong covalent bonds, gives strength and stability to the polynucleotide chain.
* Plectonemic coiling – the two strands of the DNA are wrapped around each other in a helix, making it impossible to simply move them apart without breaking the entire structure. Whereas in paranemic coiling the two strands simply lie alongside one another, making them easier to pull apart.
* Based on the helix and the distance between each turn, the DNA is of three forms – A DNA, B DNA and Z DNA.

Q.10Explain the structure and function of different types of RNA?v
Answer:

I. mRNA (messenger RNA)
* single-stranded
* carries a copy of instructions to carry out amino acid assembling &protein synthesis
* unstable
* 5% of total RNA
* In Prokaryotes – it is (polycistronic carrying coding sequence for many polypeptides
* Eukaryotes – (monocistronic) contain information for only one polypeptide
II. tRNA (transfer RNA)
* single-stranded clover-shaped with 4 arms highly folded -3 D structure
* translates the code from mRNA and transfers amino acid to ribosomes (to built proteins)
* unstable (also known as soluble RNA)
* 15% of total RNA
III. rRNA (ribosomal RNA)
* single-stranded
* make up the 2 subunits of ribosomes
* metabolically stable
* 80% total RNA
* A polymer with varied length from 120 – 3000 nucleotides & give ribosomes their shape
* Genes of rRNA employed for phylogenetic studies
Part II
11th Bio Botany Guide Biomolecules Additional Important Questions and Answers
I Choose the right answer.

9Part I5 questions
Q.1The most basic amino acid is a. Arginine b. Histidene c. Glycine d. Glutaminev
Answer:

c. GIycine

Q.2An example of feed back inhibition is a. cyanide action on cytochrome b. Sulpha drug on folic acid c. Allosteric inhibition of hexokinase by glucose- 6- phosphate d. The inhibition of succinic dehydrogenase by malonatev
Answer:

c. Allosteric inhibition of hexokinase by glucose-6-phosphate

Q.3Enzymes the catalyse interconversion of optical, geometrical or positional isomers are a. Ligases b. Lyases c. Hydrolases d. Isomerasesv
Answer:

d. Isomerases

Q.4Proteins perform many physiological functions, for example, some functions as enzymes one of the following represents an additional function that some proteins discharge a. Antibiotics b. Pigment conferring colour to skin c. Pigments making colours of flowers d. Hormonesv
Answer:

d.Hormones

Q.5Given below is the diagrammatic representation of one of the categories of small molecular weight organic compounds in the living tissues. Identify the category shown &one Blank component ‘X’ in it. Category Compound I. Cholesterol A. Guanine II. Amino acid B. IVH 2 III. Nucleotide C. Adenine IV. Nucleoside D. Uracilv
Answer:

IV

10Part II33 questions
Q.1Who invented the electron microscope? (2010 AIIMS, 2008 JIPMER)v
  1. (a) Janssen
  2. (b) Edison
  3. (c) Knoll and Ruska
  4. (d) Landsteiner
Answer:

(c) Knoll and Ruska

Q.2Polysaccharides also called a. Polymers b. Glycans c. Glycosidic compounds d. Glyconesv
Answer:

b. Glycans

Q.3Omnis – cellula – e – cellula was given by ……………. (2007 AIIMS)v
  1. (a) Virchow
  2. (b) Hooke
  3. (c) Leeuwenhoek
  4. (d) Robert Brown
Answer:

(a) Virchow

Q.4Nitrocellulose is used in making a. cellophane b. drapers c. explosives d. pain balmsv
Answer:

c. explosives

Q.5Genes present in the cytoplasm of eukaryotic cells are found in ……………. (2006 AIIMS)v
  1. (a) mitochondria and inherited via egg cytoplasm
  2. (b) lysosomes and peroxisomes
  3. (c) Golgi bodies and smooth endoplasmic reticulum
  4. (d) Plastids inherited via male gametes
Answer:

(a) mitochondria and inherited via egg cytoplasm

Q.6Chitin when added with amino acid becomes a.myeopolysaccharide b. amylopolysaceharide c.mucopolysaccharide d. peptidopolysaccharidev
Answer:

c. mucopolysaccharide

Q.7A quantosome is present in ……………. (JIPMER 2012)v
  1. (a) Mitochondria
  2. (b) Chloroplast
  3. (c) Golgi bodies
  4. (d) ER
Answer:

(b) Chloroplast

Q.8Among the following one is not a non-polar solvent a. benzene b. sulphuric acid c. ether d. chloroformv
Answer:

b. sulphuric acid

Q.9One of the given below is a complex found in the cell membrane of animal cell a. cholesterol b. myelin c. proline d. Ieeithinv
Answer:

a. cholesterol

Q.10A major site for the synthesis of lipids ……………. (2013 NEET)v
  1. (a) Rough ER
  2. (b) smooth ER
  3. (c) Centriole
  4. (d) Lysosome
Answer:

(b) smooth ER

Q.11Principle information molecules of the cell are known as a. Nucleus b. DNA c. RNA d. Nucleic acidsv
Answer:

d. Nucleic acids

Q.12(I) Cellulose – A most abundant organic compound (II) Morphine – Pain relieving alkaloid (III) Aldose – reducing sugar & Ketose (IV) Glycogen – mucopolysaccharidev
Answer:

The incorrect statement is (IV) Glycogen – mucopolysaccharide. Glycogen is a polysaccharide, not a mucopolysaccharide. It is a branched polymer of glucose that serves as the storage form of carbohydrates in animals, primarily found in liver and muscle cells. Statement (I) is correct as cellulose is the most abundant organic compound on Earth. Statement (II) is correct as morphine is an alkaloid with pain-relieving properties. Statement (III) is correct as aldose sugars are reducing sugars, and ketose sugars are also reducing sugars. Therefore, the incorrect statement is (IV).

Q.13The following is a general formula of a. Amino acid b. Fatty acid c. Nucleotide d. Monosaccharidev
Answer:

a. Amino acid

Q.14Lactose is a disaccharide of a. Glucose – Glucose b. Fructose – Fructose c. Glucose – Galactose d. Fructose – Galactosev
Answer:

c. Glucose – Galactose

Q.15Number of fatty acids in triglyceride is …………….v
  1. (a) 1
  2. (b) 2
  3. (c) 3
  4. (d) 4
Answer:

(c) 3

Q.16Heparin the anti-coagulant is got from a. D – glucuronic acid b. Polymer of fructose c. Mucopolysaccharide from red algae d. Glucosaminoglycanv
Answer:

d. Glycosaminoglycan

Q.17The p H at which Zwitterion is formed is known as a. Iso ionic balance b. Isoelectric potential c. Isoelectric point d. Iso ionic pointv
Answer:

c. Isoelectric point

Q.18Aspartate and Glutamate are amino acids of a. Negatively charged ‘R’ groups b. Positively charged ‘R’ groups c. Non-polar aliphatic ‘W groups d. Non-polar aromatic ‘R’ groupsv
Answer:

a. Negatively charged ‘R’ groups

Q.19The test for protein ¡s a. iodine test b. Biuret test c. Benedict’s test d. Hydrolysis testv
Answer:

b. Biuret test

Q.20The competitive inhibitor is …………… for succinic dehydrogenase.v
  1. (a) malonate
  2. (b) succinate
  3. (c) oxalate
  4. (d) citrate
Answer:

(a) malonate

Q.21Formation of new chemical bonds using ATP as a source of energy a. Lyase b. Hydrolase c. Telomerase d. Ligasev
Answer:

d. Ligase

Q.22Uridylic acid is an a. Dinucleotide b. Nucleoside c. Nucleotide d. Ribo nucleotidev
Answer:

d. Ribonucleotide

Q.23Phosphate forming linkage with sugar is known as a. diester linkage b. peptide linkage c. phosphodiester linkage d. Ionic linkagev
Answer:

c. phosphodiester linkage

Q.24…………… is a catalytic RNA.v
  1. (a) mRNA
  2. (b) Ribozyme
  3. (c) Ribonuclease
  4. (d) rRNA
Answer:

(b) Ribozyme

Q.25A class of lipid that serves as a major component of the cell membrane is a. triglyceride b. glycerol c. phospho lipid d. lipoproteinv
Answer:

c. phospholipid

Q.26One molecule of sucrose on hydrolysis give a. 2 molecules of glucose b. 1 molecule glucose & 1 molecule fructose c. 2 molecules of glucose & 1 molecule of fructose d. 2 molecules of fructosev
Answer:

b. 1 molecule glucose & 1 molecule fructose

Q.27In fibrous proteins polypeptide chai are held together by a. Vander Waals forces b. disulphide linkage c. electrostatic forces d. hydrogen bondsv
Answer:

a. Vander Waals forces

Q.28According to Chargaff’s rule, the hydrogen bonding between Adenine and Thymine is …………….v
  1. (a) 2
  2. (b) 3
  3. (c) 4
  4. (d) Nil
Answer:

(a) 2

Q.29Which polymer is stored in liver a. Amylose b. Amylo pectin c. Cellulose d. Glycogenv
Answer:

d. Glycogen

Q.30The bond that is not needed for protein formation is a. Hydrogen bond b. Peptide bond c. Ionic bond d. glucosidic bondv
Answer:

d. glucosidic bond

Q.31A complete turn of the helix comprises …………….v
  1. (a) 34 nm
  2. (b) 3.4 nm
  3. (c) 20 nm
  4. (d) 2nm
Answer:

(b) 3.4 nm

Q.32The acid is also known as vitamin C a. Aspartic acid b. Tartaric acid c. Ascorbic acid d. Adipic acidv
Answer:

c. Ascorbic acid

Q.33Which is the left-handed DNA?v
  1. (a) B – DNA
  2. (b) A – DNA
  3. (c) Z – DNA
  4. (d) dsDNA
Answer:

(c) Z – DNA
II. Choose the wrong answer.

11VI. Assertion & Reason – Find Out The Correct Answer.12 questions
Q.1Assertion (A): Adhesion refers to the tendency of water molecules to cling together Reason (R): Because of hydrogen bonding, water molecules interact with one another continuous column of water is raised in xylem vessels. (a) Assertion & Reason correct Reason Explaining Assertion (b) Assertion & Reason correct- Reason not explaining Assertion (c) Assertion is true but Reason is wrong (d) Assertion is true but Reason is not explaining Assertion.v
Answer:

(a) Assertion & Reason correct Reason Explaining Assertion.

Q.1Define Micronutrients.v
Answer:

Micronutrients are nutrients that are required by plants in very small or trace amounts but are essential for normal growth, development, and metabolic functions. These include elements such as cobalt, zinc, boron, copper, molybdenum, and manganese. Micronutrients are primarily required as cofactors and prosthetic groups for various enzymes, enabling their catalytic activity. For example, molybdenum is an essential component of the enzyme nitrogenase, which is responsible for the fixation of atmospheric nitrogen into usable forms in nitrogen-fixing plants and microorganisms. Similarly, zinc is required for the functioning of multiple enzymes involved in carbohydrate metabolism and protein synthesis. Copper is necessary for photosynthesis and respiration, while boron plays a role in cell wall formation and carbohydrate transport. Although required in minute quantities, the deficiency of any micronutrient can lead to severe metabolic disorders and stunted plant growth.

Q.2Assertion (A): Glycine is a non-essential amino acid Reason (R): It must be taken through diet (a) Assertion & Reason correct Reason Explaining Assertion (b) Assertion & Reason correct- Reason not explaining Assertion (c) Assertion is true but Reason is wrong (d) Assertion is true but Reason is not explaining Assertion.v
Answer:

(c) Assertion is true but Reason is wrong.

Q.2Write down the properties of Water.v
Answer:
  • It has Adhesion & cohesion property
  • High latent heat of vaporisation
  • High melting and boiling point
  • Universal solvent
  • Has specific heat capacity.
Q.3Assertion (A): In the presence of enzyme substance molecules can be attached by the reagent. Reason (R): Active sites of enzymes hold the substance in a suitable position. (a) Assertion & Reason correct Reason Explaining Assertion (b) Assertion & Reason correct- Reason not explaining Assertion (c) Assertion is true but Reason is wrong (d) Assertion is true but Reason is not explaining Assertionv
Answer:

(a) Assertion & Reason correct Reason Explaining Assertion.

Q.3Differentiate between Primary and Secondary Metabolites.v
Answer:

Primary metabolites are organic compounds that are directly involved in the basic metabolic processes of organisms, including photosynthesis, respiration, protein synthesis, and lipid metabolism. They are essential for the growth, development, and survival of all living organisms and are produced by all plants. Examples include carbohydrates, proteins, lipids, and nucleic acids. Secondary metabolites, on the other hand, are organic compounds that are not directly involved in the primary metabolic processes and do not have an obvious role in the immediate growth and development of organisms. They are produced by only certain plants or specific plant tissues and are often involved in plant defense, pigmentation, and adaptation to environmental stress. Examples include alkaloids like morphine and quinine, phenolic compounds, terpenes, and essential oils. While primary metabolites are universal and fundamental, secondary metabolites are specialized compounds that provide plants with competitive advantages in their ecological niches.

Q.4Assertion (A): Aminoacids behave like salt rather than simple amines or carboxylic acid. Reason (R): In aqueous solution, the COOH group of amino acid loses a protein and the NH2 group accepts a proton to form zwitterion (salt). (a) Assertion & Reason correct Reason Explaining Assertion (b) Assertion & Reason correct- Reason not explaining Assertion (c) Assertion is true but Reason is wrong (d) Assertion is true but Reason is not explaining Assertionv
Answer:

(a) Assertion & Reason correct Reason Explaining Assertion.
2 Marks

Q.4Define Polymerisation.v
Answer:

Polymerization is a chemical process in which small organic molecules called monomers are chemically bonded together in long chains or networks to form larger molecules called polymers. During polymerization, monomers undergo condensation reactions where water molecules are released as the bonds form between adjacent units. This process can occur through various mechanisms and can produce polymers of different lengths and structures. Examples of polymerization include the formation of starch from glucose monomers, the formation of cellulose from glucose units, the formation of proteins from amino acids, and the formation of polynucleotides like DNA and RNA from nucleotide monomers. Polymerization is fundamental to the formation of all biological macromolecules essential for life.

Q.5Distinguish between Glycogen and Cellulose.v
Answer:

Glycogen and cellulose are both polysaccharides composed of glucose units, but they differ significantly in their structure, function, and occurrence. Glycogen is a storage polysaccharide found in animals, particularly in liver cells and skeletal muscle fibers throughout the human body. It is composed of glucose units linked by α-1,4 glucosidic bonds in the main chains with α-1,6 glucosidic bonds creating branch points, giving it a highly branched structure that allows for rapid mobilization of glucose during energy demands. Cellulose, in contrast, is a structural polysaccharide found in plants, serving as a major component of plant cell walls. It is composed of thousands of glucose units held together by β-1,4 glucosidic bonds in a linear, unbranched structure, which provides rigidity and strength to plant tissues. While glycogen is readily soluble and easily broken down for quick energy release, cellulose is insoluble and provides structural support. Cellulose is also used industrially in the production of nitrocellulose, which has applications as an explosive and in various manufacturing processes. The different linkages and structures of these two polysaccharides reflect their distinct biological roles.

Q.6Distinguish between Dinucleotide & Polynucleotide.v
Answer:

Dinucleotides and polynucleotides are both nucleic acid molecules composed of nucleotide units, but they differ in their size and complexity. A dinucleotide is formed when two nucleotides are joined together through a condensation reaction. The two nucleotides are linked by a 3′-5′ phosphodiester linkage, which forms between the phosphate group of one nucleotide and the sugar (ribose or deoxyribose) of the adjacent nucleotide. This creates a covalent bond that connects the nucleotides in a chain. Polynucleotides, on the other hand, are formed when many nucleotides, typically hundreds to millions, are joined together through the same 3′-5′ phosphodiester linkage mechanism. Like dinucleotides, each nucleotide in a polynucleotide is connected to the next through condensation reactions between the phosphate group of one nucleotide and the sugar of the following nucleotide. Examples of polynucleotides include DNA (deoxyribonucleic acid) and RNA (ribonucleic acid), which are the major nucleic acids found in all living organisms. The repeated formation of phosphodiester bonds creates the sugar-phosphate backbone of these polynucleotide chains, with nitrogenous bases extending from the sugar molecules.

Q.7Differentiate between Nucleoside & Nucleotide.v
Answer:

A nucleoside is formed by the condensation of a nitrogenous base (purine or pyrimidine) with a pentose sugar (ribose in RNA or deoxyribose in DNA). This bond is typically a N-glycosidic bond. For example, adenine combined with ribose forms adenosine, and guanine with deoxyribose forms deoxyguanosine. Nucleosides are important intermediates in the synthesis of nucleotides. A nucleotide, on the other hand, is a nucleoside with one or more phosphate groups attached to the sugar molecule, usually at the 5' carbon. This addition of a phosphate group converts the nucleoside into a nucleotide. For instance, adenosine combined with a phosphoric acid molecule forms adenylic acid (adenosine monophosphate, AMP). Nucleotides are the fundamental building blocks of nucleic acids like DNA and RNA, and they also play crucial roles in cellular energy transfer (e.g., ATP) and signaling pathways.

Q.8State Chargaff’s Law.v
Answer:

Chargaff's Law, proposed by Erwin Chargaff in 1949, describes specific quantitative relationships between the amounts of nitrogenous bases in DNA. The law states that in any double-stranded DNA molecule, the amount of adenine (A) is approximately equal to the amount of thymine (T), and the amount of guanine (G) is approximately equal to the amount of cytosine (C). This can be expressed as A = T and G = C. Furthermore, it implies that the total amount of purines (A + G) is equal to the total amount of pyrimidines (T + C). These equivalences are due to the specific base pairing rules in the DNA double helix, where adenine forms two hydrogen bonds with thymine, and guanine forms three hydrogen bonds with cytosine. However, the ratio of A:T or G:C to the total number of bases can vary significantly between different species, meaning that the ratio of (A+T) to (G+C) is not necessarily equal to 1, and the ratio of A:T need not be equal to G:C.