Maharashtra Board Class 11 Biology Chapter 6 Biomolecules Solutions

Biomolecules Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Biology Chapter 6 Exercise Solutions Maharashtra Board

Biology Class 11 Chapter 6 Exercise Solutions

1. Choose Correct Option

Question (A)
Sugar, amino acids and nucleotides unite to their respective subunits to form -
(a) bioelements
(b) micromolecules
(c) macromolecules
(d) all of these
Answer: (c) macromolecules
In simple words: Sugars, amino acids, and nucleotides are small units (monomers) that link together to form larger complex molecules called macromolecules, such as polysaccharides, proteins, and nucleic acids, respectively.

🎯 Exam Tip: Understanding the basic building blocks and the larger structures they form is crucial for identifying biomolecules. Remember monomers combine to form polymers.

Question (B)
Glycosidic bond is found in -
(a) Disaccharide
(b) Nucleosides
(c) Polysaccharide
(d) all of these
Answer: (d) all of these
In simple words: A glycosidic bond is a covalent bond that links a carbohydrate (sugar) molecule to another group, which can be another sugar in disaccharides and polysaccharides, or a non-carbohydrate in nucleosides.

🎯 Exam Tip: Glycosidic bonds are fundamental to carbohydrate structures and nucleosides. Knowing their presence in various biological molecules helps in understanding their classification and function.

Question (C)
Amino acids in a polypeptide are joined by - bond.
(a) Disulphide
(b) glycosidic
(c) hydrogen bond
(d) none of these
Answer: (d) none of these
In simple words: Amino acids within a polypeptide chain are linked by a peptide bond, which is formed between the carboxyl group of one amino acid and the amino group of another, through a dehydration reaction.

🎯 Exam Tip: The peptide bond is the defining linkage in proteins. Distinguish it from other bonds like disulfide (stabilizes protein structure) or hydrogen bonds (secondary structure) or glycosidic (carbohydrates/nucleosides).

Question (D)
Lipids associated with cell membrane are -
(a) Sphingomyelin
(b) Isoprenoids
(c) Phospholipids
(d) Cholesterol
Answer: (c) Phospholipids
In simple words: Phospholipids are the primary components of cell membranes, forming a lipid bilayer that defines the cell's boundaries and regulates molecular passage.

🎯 Exam Tip: Phospholipids are crucial for cell membrane structure due to their amphiphilic nature. Cholesterol also plays a role in membrane fluidity, but phospholipids form the basic bilayer.

Question (E)
Linoleic, Linolenic and - acids are referred as essential fatty acids since they cannot be synthesized by the body and hence must be included in daily diet.
(a) Arachidonic
(b) Oleic
(c) Steric
(d) Palmitic
Answer: (a) Arachidonic
In simple words: Linoleic, Linolenic, and Arachidonic acids are essential fatty acids because the human body cannot synthesize them and thus they must be obtained through diet.

🎯 Exam Tip: Knowing essential nutrients like essential fatty acids (and amino acids) is important. They are vital for various bodily functions but must be consumed as they cannot be produced internally.

Question (F)
Hemoglobin is a type of - protein, which plays indispensable part in respiration.
(a) simple
(b) derived
(c) conjugated
(d) complex
Answer: (c) conjugated
In simple words: Hemoglobin is a conjugated protein, meaning it consists of a protein part (globin) linked to a non-protein part (heme containing iron), which is essential for oxygen transport in respiration.

🎯 Exam Tip: Differentiate between simple, conjugated, and derived proteins. Conjugated proteins have a non-protein component (prosthetic group) vital for their function, as seen in hemoglobin.

Question (G)
When inorganic ions or metallo-organic molecules bind to apoenzyme, they together form -
(a) isoenzyme
(b) holoenzyme
(c) denatured enzyme
(d) none of these
Answer: (b) holoenzyme
In simple words: An apoenzyme is the inactive protein part of an enzyme, and when it combines with a non-protein cofactor (like inorganic ions or coenzymes), it forms a complete, active enzyme complex called a holoenzyme.

🎯 Exam Tip: The terms apoenzyme, cofactor/coenzyme, and holoenzyme are key in understanding enzyme structure and function. Remember that the apoenzyme needs its cofactor to become fully functional.

Question (H)
In enzyme kinetics, Km = Vmax/2. If Km value is lower, it indicates -
(a) Enzyme has less affinity for substrate
(b) Enzyme has higher affinity towards substrate
(c) There will be no product formation
(d) All active sites of enzyme are saturated
Answer: (b) Enzyme has higher affinity towards substrate
In simple words: A lower Km value signifies that an enzyme can achieve half of its maximum reaction rate at a lower substrate concentration, indicating a strong binding affinity between the enzyme and its substrate.

🎯 Exam Tip: Km (Michaelis constant) is an important indicator in enzyme kinetics. A low Km means high affinity, and a high Km means low affinity. This concept is frequently tested.

2. Solve The Following Questions

Question (A)
Observe the following figures and write the differences between them.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक फॉस्फोलिपिड द्वितरफा परत (लिपिड बाइलेयर) को दर्शाता है जो पानी के माध्यम में मौजूद है। इसमें फॉस्फोलिपिड अणु होते हैं जिनके ध्रुवीय सिर (पोलर हेड ग्रुप) पानी की ओर बाहर की तरफ होते हैं और हाइड्रोफोबिक पूंछें अंदर की ओर होती हैं, जो केंद्र में 'X' क्षेत्र बनाती हैं, जिससे झिल्ली की संरचना बनती है।
Answer:

🎯 Exam Tip: Differentiating between saturated and unsaturated fats based on their chemical structure, physical state, and health implications is a common question. Focus on the presence/absence of double bonds.

3. Answer The Following Questions

Question (A)
What are building blocks of life?
Answer:Life is composed of four main building blocks: Carbohydrates, proteins, lipids and nucleic acids.
In simple words: The fundamental components that make up all living organisms and enable their functions are carbohydrates, proteins, lipids, and nucleic acids.

🎯 Exam Tip: This is a foundational concept. Remember the four main classes of biomolecules that constitute the basic structure and machinery of life.

Question (B)
Explain the peptide bond.
Answer:1. The covalent bond that links the two amino acids is called a peptide bond.
2. Peptide bond is formed by condensation reaction.
In simple words: A peptide bond is a strong covalent link that joins two amino acids together, forming through a reaction where a water molecule is removed.

🎯 Exam Tip: Understand that peptide bonds are crucial for protein formation. Mentioning "covalent bond" and "condensation reaction" are key points for a complete answer.

Question (C)
How many types of polysaccharides you know?
Answer:There are two types of polysaccharides:
1. Homopolysaccharides: It contains same type of monosaccharides. E.g. Starch, glycogen, cellulose.
2. Heteropolysaccharides: It contains two or more different monosaccharides. E.g. Hyaluronic acid, heparin, hemicellulose.
In simple words: Polysaccharides are large carbohydrate molecules categorized into homopolysaccharides (made of one type of sugar unit) and heteropolysaccharides (made of multiple types of sugar units).

🎯 Exam Tip: Classification of polysaccharides into homo- and hetero- forms, along with relevant examples for each, is important for demonstrating your understanding of carbohydrate diversity.

Question (D)
Enlist the significance of carbohydrates.
Answer:Significances of carbohydrates are as follows:
1. Carbohydrates provide energy for metabolism.
2. Glucose is the main substrate for ATP synthesis.
3. Lactose, a disaccharide present in the milk provides energy to babies.
4. Polysaccharide serves as a structural component of cell membrane, cell wall and reserved food as starch and glycogen.
In simple words: Carbohydrates are vital for living organisms, serving as primary energy sources, essential for ATP production, providing energy for infants, and forming structural components in cells.

🎯 Exam Tip: When discussing significance, ensure you cover both energy roles (immediate and stored) and structural roles, providing examples where possible.

Question (E)
What is reducing sugar?
Answer:1. A sugar that serves as a reducing agent due to presence of free aldehyde or ketone group is called a reducing sugar.
2. These sugars reduce the Benedict's reagent (\(Cu^{2+}\) to \(Cu^{+}\)) since they are capable of transferring hydrogens (electrons) to other compounds, a process called reduction.
3. All monosaccharides are reducing sugars.
In simple words: Reducing sugars are carbohydrates with a free aldehyde or ketone group that can donate electrons to reduce other compounds, such as in the Benedict's test, and all monosaccharides fall into this category.

🎯 Exam Tip: The presence of a free aldehyde or ketone group is the defining characteristic of a reducing sugar. Mentioning its ability to reduce substances like Benedict's reagent and providing examples like monosaccharides will earn full marks.

Question (F)
Enlist the examples of simple proteins and their significance.
Answer:Examples of simple proteins are: E.g.: Albumins and histones.
Significance:
1. Albumin:
a. It is the main protein in the blood.
b. It maintains the pressure in the blood vessels.
c. It helps in transportation of substances like hormone and drugs in the body.
2. Histones:
a. It is the chief protein of chromatin.
b. They are involved in packaging of DNA into structural units called nucleosomes.
In simple words: Simple proteins like albumin and histones are essential; albumin maintains blood pressure and transports substances, while histones package DNA into compact structures within the cell nucleus.

🎯 Exam Tip: When listing examples, always provide their key functions or significance to show a comprehensive understanding of their biological roles.

Question (G)
Describe the secondary structure of protein with examples.
Answer:1. There are two types of secondary structure of protein: a-helix and P-pleated sheets.
2. The polypeptide chain is arranged in a spiral helix. These spiral helices are of two types: a-helix (right handed) and P-helix (left handed).
3. This spiral configuration is held together by hydrogen bonds.
4. The sequence of amino acids in the polypeptide chain determines the location of its bend or fold and the position of formation of hydrogen bonds between different portions of the chain or between different chains. Thus, peptide chains form an a-helix structure.
5. Example of a-helix structure is keratin.
6. In some proteins two or more peptide chains are linked together by intermolecular hydrogen bonds. Such structures are called P-pleated sheets.
7. Example of P-pleated sheet is silk fibres.
8. Due to formation of hydrogen bonds peptide chains assume a secondary structure.
In simple words: Protein secondary structure refers to local folded patterns like alpha-helices (e.g., keratin) and beta-pleated sheets (e.g., silk fibers), formed by hydrogen bonds between the polypeptide backbone.

🎯 Exam Tip: Focus on the two main types (alpha-helix and beta-pleated sheet), the type of bond responsible (hydrogen bonds between backbone atoms), and a relevant example for each to score well.

Question (H)
Explain the induced fit model for mode of enzyme action.
Answer:1. The induced fit model shows that enzymes are flexible structures in which the active site continually reshapes by its interactions with the substrate until the time the substrate is completely bound to it. It is also the point at which the final form and shape of the enzyme is determined.
2. Three-Dimensional conformation:
a. All enzymes have specific 3-dimensional conformation.
b. They have one or more active sites to which substrate (reactant) combines.
c. The points of active site where the substrate joins with the enzyme is called substrate binding site.
In simple words: The induced fit model proposes that an enzyme's active site is flexible and changes shape to precisely fit the substrate upon binding, optimizing the enzyme-substrate interaction for catalysis.

🎯 Exam Tip: Contrast the induced fit model with the lock-and-key model by emphasizing the flexibility of the enzyme's active site and its dynamic interaction with the substrate for optimal binding and reaction.

Question (I)
What is RNA? Enlist types of RNA.
Answer:1. RNA stands for Ribonucleic Acid. It is a long single stranded polynucleotide chain which helps in protein synthesis, functions as a messenger and translates messages coded in DNA into protein.
2. There are three types of RNA:
mRNA (messenger RNA), rRNA (ribosomal RNA) and tRNA (transfer RNA)
In simple words: RNA, or Ribonucleic Acid, is a single-stranded nucleic acid crucial for protein synthesis, existing in three main types: messenger RNA (mRNA), ribosomal RNA (rRNA), and transfer RNA (tRNA).

🎯 Exam Tip: Clearly define RNA and list its three primary types. A concise explanation of each type's general function (messenger, ribosomal, transfer) is often expected.

Question (J)
Describe the concept of metabolic pool.
Answer:1. Metabolic pool is the reservoir of biomolecules in the cell on which enzymes can act to produce useful products as per the need of the cell.
2. The concept of metabolic pool is significant in cell biology because it allows one type of molecule to change into another type E.g. Carbohydrates can be converted to fats and vice-versa.
In simple words: The metabolic pool is the collection of readily available biomolecules within a cell that can be interconverted and used as building blocks or energy sources for various cellular processes as needed.

🎯 Exam Tip: Emphasize that the metabolic pool is a dynamic reservoir where different biomolecules can be interconverted, highlighting its flexibility in meeting cellular demands.

Question (K)
How do secondary metabolites useful for mankind?
Answer:1. Drugs developed from secondary metabolites have been used to treat infectiou diseases, cancer, hypertension and inflammation.
2. Morphine, the first alkaloid isolated from Papaver somniferum is used as pain reliver and cough suppressant.
3. Secondary metabolites like alkaloids, nicotine, cocaine and the terpenes, cannabinol are widely used for recreation and stimulation.
4. Flavours of secondary metabolites improve our food preferences.
5. Tannins are added to wines and chocolate for improving astringency.
6. Since most secondary metabolites have antibiotic property, they are also used as food preservatives.
7. Glucosinolates is a secondary metabolite which is naturally present in cabbage imparts a characteristic flavour and aroma because of nitrogen and sulphur-containing chemicals. It also offers protection to these plants from many pests.
In simple words: Secondary metabolites from plants offer numerous benefits to humans, including serving as sources for drugs, pain relievers, stimulants, flavor enhancers, food preservatives, and even pest protection.

🎯 Exam Tip: When listing uses of secondary metabolites, provide specific examples for medical, recreational, and food-related applications, as this showcases a strong grasp of their diverse utility.

4. Solve The Following Questions

Question (A)
Complete the following chart.

🎯 Exam Tip: For protein functions, it's crucial to link specific proteins to their primary biological roles. A clear and accurate mapping demonstrates a good understanding of protein diversity and importance.

Question (B)
Answer the following with reference

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो पॉलीपेप्टाइड श्रृंखलाओं के बीच एक डाइसल्फाइड बंधन के निर्माण और कमी को दर्शाता है। यह ऑक्सीकरण प्रतिक्रिया के माध्यम से दो थायोल (SH) समूहों से एक डाइसल्फाइड (S-S) बंधन के निर्माण और कमी प्रतिक्रिया के माध्यम से डाइसल्फाइड बंधन के टूटने को दिखाता है।
i. Name the type of bond formed between two polypeptides.
ii. Which amino acid is involved in the formation of such bond?
iii. Amongst I, II, III and IV structural level of protein, which level of structure includes such bond? Answer:
i. Disulfide bond.
ii. Cysteine
iii. Tertiary structure.
[Note: Quaternary structure of protein also have disulfide bond, for stabilization of protein structure.!
In simple words: The diagram illustrates the formation of a disulfide bond, which links two polypeptides or parts of a single polypeptide, involves the amino acid cysteine, and is a key feature stabilizing tertiary and sometimes quaternary protein structures.

🎯 Exam Tip: Disulfide bonds are critical covalent linkages in protein structure. Remember that cysteine is the only amino acid capable of forming these bonds, and they are important for stabilizing higher-order protein structures (tertiary and quaternary).

Question (C)
Match the following items given in column I and II.

🎯 Exam Tip: For matching questions, identify key terms and their most direct associations. For instance, Uracil is exclusive to RNA, Koshland proposed the induced fit model, and Sucrase is an enzyme that performs hydrolysis (hydrolase).

5. Long Answer Questions

Question (A)
What are biomolecules? Explain the building blocks of life.
Answer:Biomolecules are essential substances produced by our body which are necessary for life.
The building blocks of life are carbohydrates, lipids, proteins and nucleic acids.
1. Carbohydrates:
a. Carbohydrates are biomolecules made from carbon, hydrogen and oxygen.
b. The general formula of carbohydrates is (CH20)n.
c. They contain hydrogen and oxygen in the same ratio as in water (2:1).
d. Carbohydrates can be broken down to release energy.
e. Based on sugar units, carbohydrates are classified into three types: Monosaccharides, disaccharides and polysaccharides.
2. Lipids:
a. These are group of substances with greasy consistency with long hydrocarbon chain containing carbon, hydrogen and oxygen.
b. In lipids hydrogen to oxygen ration is greater than 2:1.
c. Lipid is a broader term used for fatty acids and their derivatives.
d. They are soluble in organic solvents (non-polar solvents).
e. Fatty acids are organic acids which are composed of hydrocarbon chain ending in carboxyl group (COOH) -.
f. These are divided into: Saturated fatty acids and unsaturated fatty acids.
g. Fatty acids are basic molecules which form different kinds of lipids.
h. Lipids are classified into three types: Simple lipids, Compound lipids, Derived lipids.
3. Proteins:
a. Proteins are large molecules containing amino acid units ranging from 100 to 3000.
b. They have higher molecular weight.
c. In proteins, amino acids are linked together by peptide bonds which join the carboxyl group of one amino acid residue to the amino group of another residue.
d. A protein molecule consists of one or more polypeptide chains.
e. Proteins contain any or all twenty naturally occurring amino acid types.
f. Proteins have different structures like primary structure, secondary structure, tertiary structure and quaternary structure.
g. Proteins are classified into three types: Simple proteins: Simple proteins on hydrolysis yield only amino acids. E.g. Histones and albumins. Conjugated proteins: It consists of a simple protein united with some non-protein substance. E.g. Haemoglobin. Derived proteins: These proteins are not found in nature as such but are derived from native protein molecules on hydrolysis. E.g. Metaproteins, peptones.
4. Nucleic Acids:
a. Nucleic acids are macromolecules composed of many small units or monomers called nucleotides.
b. Each nucleotide is formed of three components i.e. pentose sugar, a nitrogen base and a phosphate (phosphoric acid).
c. When sugar combine with nitrogenous base it forms nucleoside. Nucleotides can be called as nucleoside phosphate.
d. There are two types of nucleic acids, i.e. DNA and RNA.
DNA (Deoxyribonucleic acid) is a genetic material of a cell. It is double stranded helix. Each strand of helix is made up of deoxyribose nucleotides.
RNA (Ribonucleic Acid) is a single stranded structure having fewer nucleotides as compared to DNA. The strands may be straight or variously folded upon itself. It is made up of nucleotides.
In simple words: Biomolecules are the organic compounds essential for life, and the four main building blocks are carbohydrates (for energy), lipids (for storage and structure), proteins (for diverse functions), and nucleic acids (for genetic information), each composed of specific smaller units.

🎯 Exam Tip: For this comprehensive question, ensure you define biomolecules, then detail each of the four building blocks (carbohydrates, lipids, proteins, nucleic acids) including their general composition, key characteristics, and primary biological roles. Examples for each type are highly recommended.

Question (B)
Explain the classes of carbohydrates with examples.
Answer:Based on number of sugar units, carbohydrates are classified into three types namely, monosaccharides, disaccharides and polysaccharides.
1. Monosaccharides:
a. Monosaccharides are the simplest sugars having crystalline structure, sweet taste and soluble in water.
b. They cannot be further hydrolyzed into smaller molecules.
c. They are the building blocks or monomers of complex carbohydrates.
d. They have the general molecular formula (CH20)n, where n can be 3, 4, 5, 6 and 7.
e. They can be classified as triose, tetrose, pentose, etc.
f. Monosaccharides containing the aldehyde (-CHO) group are classified as aldoses e.g. glucose, xylose, and those with a ketone(-C=0) group are classified as ketoses. E.g. ribulose, fructose.
2. Disaccharides:
a. Disaccharide is formed when two monosaccharide react by condensation reaction releasing a water molecule. This process requires energy.
b. A glycosidic bond forms and holds the two monosaccharide units together.
c. Sucrose, lactose and maltose are examples of disaccharides.
d. Sucrose is a nonreducing sugar since it lacks free aldehyde or ketone group.
e. Lactose and maltose are reducing sugars.
f. Lactose also exists in beta form, which is made from P-galactose and p-glucose.
g. Disaccharides are soluble in water, but they are too big to pass through the cell membrane by diffusion.
3. Polysaccharides:
a. Monosaccharides can undergo a series of condensation reactions, adding one unit after the other to the chain till a very large molecule (polysaccharide) is formed. This is called polymerization.
b. Polysaccharides are broken down by hydrolysis into monosaccharides.
c. The properties of a polysaccharide molecule depends on its length, branching, folding and coiling.
d. Examples: Starch, glycogen, cellulose.
In simple words: Carbohydrates are classified into monosaccharides (simple sugars like glucose), disaccharides (two sugar units like sucrose), and polysaccharides (many sugar units like starch), differing in complexity, solubility, and biological roles, all formed through condensation or broken down by hydrolysis reactions.

🎯 Exam Tip: Clearly define and provide examples for each class of carbohydrate: monosaccharides, disaccharides, and polysaccharides. Highlight key characteristics like hydrolyzability, solubility, and their roles (e.g., energy, structure).

Question (C)
Describe the types of lipids and mention their biological significance.
Answer:Lipids are classified into three main types:
1. Simple lipids:
a. These are esters of fatty acids with various alcohols. Fats and waxes are simple lipids.
b. Fats are esters of fatty acids with glycerol (CH2OH-CHOH-CH2OH).
c. Triglycerides are three molecules of fatty acids and one molecule of glycerol.
d. Unsaturated fats are liquid at room temperature and are called oils. Unsaturated fatty acids are hydrogenated to produce fats e.g. Vanaspati ghee.
Biological significance:
a. Fats are a nutritional source with high calorific value and they act as reserved food materials.
b. In plants, fat is stored in seeds to nourish embryo during germination.
c. In animals, fat is stored in the adipocytes of the adipose tissue.
d. Fats deposited in subcutaneous tissue act as an insulator and minimize loss of body heat.
e. Fats deposited around the internal organs act as cushions to absorb mechanical shocks.
f. Wax is another example of simple lipid. They are esters of long chain fatty acids with long chain alcohols.
g. They are found in the blood, gonads and sebaceous glands of the skin.
h. Waxes are not as readily hydrolyzed as fats.
i. They are solid at ordinary temperature.
j. Waxes form water insoluble coating on hair and skin in animals, waxes form an outer coating of stems, leaves and fruits.
2. Compound lipids:
a. These are ester of fatty acids containing other groups like phosphate (Phospholipids), sugar (glycolipids), etc.
b. They contain a molecule of glycerol, two molecules of fatty acids and a phosphate group or simple sugar.
c. Some phospholipids such as lecithin also have a nitrogenous compound attached to the phosphate group.
d. Phospholipids have both hydrophilic polar groups (phosphate and nitrogenous group) and hydrophobic non-polar groups (hydrocarbon chains of fatty acids).
e. Glycolipids contain glycerol, fatty acids, simple sugars such as galactose. They are also called cerebrosides.
Biological significance:
a. Phospholipids contribute in the formation of cell membrane.
b. Large amounts of glycolipids are found in the brain white matter and myelin sheath.
3. Derived Lipids:
a. They are composed of fused hydrocarbon rings (steroid nucleus) and a long hydrocarbon side chain.
b. One of the most common sterols is cholesterol.
Biological significance:
a. It is widely distributed in all cells of the animal body, but particularly in nervous tissue.
b. Cholesterol exists either free or as cholesterol ester.
c. Adrenocorticoids, sex hormones (progesterone, testosterone) and vitamin D are synthesized from cholesterol.
d. Cholesterol is not found in plants.
e. Sterols exist as phytosterols in plants.
f. Yam Plant (Dioscorea) produces a steroid compound called diosgenin. It is used in the manufacture of antifertility pills, i.e. birth control pills.
In simple words: Lipids are diverse molecules classified into simple (fats, waxes for energy storage, insulation, protection), compound (phospholipids, glycolipids forming cell membranes and myelin sheath), and derived lipids (steroids like cholesterol, hormones, vitamin D, with structural and signaling roles).

🎯 Exam Tip: For lipids, classify them into simple, compound, and derived, giving examples for each. Crucially, detail their varied biological significances, covering energy storage, structural roles (membranes), insulation, protection, and hormonal functions.

Question (D)

Explain the chemical nature, structure and role of phospholipids in biological membrane.
Answer:Chemical nature: Phospholipids are amphiphilic in nature. As they have hydrophilic head and hydrophobic tail. Structure: It contains an alcohol, two fatty acid chains and a phosphate group. Role: Phospholipids forms the membranes around the cells and cellular organelles. They form a lipid bilayer membrane. The phospholipids are arranged tail to tail. It serves as a barrier against movement of any ions or polar compounds into and out of the cell.
In simple words: Phospholipids are crucial for cell membranes, forming a double layer where their water-attracting heads face outwards and water-repelling tails face inwards, creating a barrier that controls what enters and leaves the cell.

🎯 Exam Tip: Focus on explaining the amphiphilic nature and bilayer formation as key components for high scores.

Question (E)

Describe classes of proteins with their importance.
Answer:On the basis of structure, proteins are classified into three categories: 1. Simple proteins: a. Simple proteins on hydrolysis yield only amino acids. b. These are soluble in one or more solvents. c. Simple proteins may be soluble in water. d. Histones of nucleoproteins are soluble in water. e. Globular molecules of histones are not coagulated by heat. f. Albumins are also soluble in water but they get coagulated on heating. g. Albumins are widely distributed e.g. egg albumin, serum albumin and legumelin of pulses are albumins. Importance: They are involved in structural components; they also act as a storage kind of protein. Some are associated with nucleic acids in nucleoproteins of cell. 2. Conjugated proteins: a. Conjugated proteins consist of a simple protein united with some non-protein substance. b. The non-protein group is called prosthetic group e.g. haemoglobin. c. Globin is the protein and the iron containing pigment haem is the prosthetic group. d. Similarly, nucleoproteins have nucleic acids. e. Proteins are classified as glycoproteins and mucoproteins. f. Mucoproteins are carbohydrate-protein complexes e.g. mucin of saliva and heparin of blood. g. Lipoproteins are lipid-protein complexes e.g. conjugate protein found in brain, plasma membrane, milk etc. Importance: They are involved in structural components of cell membranes and organelles. They also act as a transporter. Some conjugated proteins are important in electron transport chain in respiration. 3. Derived proteins: a. These proteins are not found in nature as such. b. These proteins are derived from native protein molecules on hydrolysis. c. Metaproteins, peptones are derived proteins. Importance: They act as a precursor for many molecules which are essential for life
In simple words: Proteins are essential macromolecules classified into simple, conjugated, and derived types based on their composition. Simple proteins are made only of amino acids, conjugated proteins have an additional non-protein part, and derived proteins result from the breakdown of native proteins, all serving diverse structural, catalytic, and transport roles in the body.

🎯 Exam Tip: Remember to provide examples for each class of protein and explain their specific functions to secure full marks.

Question (F)

What are enzymes? How are they classified? Mention example of each class.
Answer:1. Enzymes are biological macromolecules which act as a catalyst and accelerates the reaction in the body. 2. Enzymes are classified into six classes: a. Oxidoreductases: These enzymes catalyze oxidation and reduction reactions by the transfer of hydrogen and/or oxygen, e.g. alcohol dehydrogenase Alcohol + NAD+
\( \xrightarrow{\text{Alcohol dehydroenase}} \) Aldehyde + NADH\(_{2}\) b. Transferases: These enzymes catalyse the transfer of certain groups between two molecules, e.g. glucokinase Glucose + ATP
\( \xrightarrow{\text{Glucokinase}} \) Glucose-6-Phosphate + ADP c. Hydrolases: These enzymes catalyse hydrolytic reactions. This class includes amylases, proteases, lipases etc. e.g. Sucrase Sucrose + water
\( \xrightarrow{\text{Sucrase}} \) Glucose + Fructose d. Lyases: These enzymes are involved in elimination reactions resulting in the removal of a group of atoms from substrate molecule to leave a double bond. It includes aldolases, decarboxylases, and dehydratases, e.g. fumarate hydratase. Histidine
\( \xrightarrow{\text{Histidine decarboxylase}} \) Histamine + CO\(_{2}\) e. Isomerases: These enzymes catalyze structural rearrangements within a molecule. Their nomenclature is based on the type of isomerism. Thus, these enzymes are identified as racemases, epimerases, isomerases, mutases, e.g. xylose isomerase. Glucose-6-Phosphate
\( \xrightarrow{\text{Isomerase}} \) Fructose-6 - Phosphate f. Ligases or Synthetases: These are the enzymes which catalyze the covalent linkage of the molecules utilizing the energy obtained from hydrolysis of an energy-rich compound like ATP, GTP e.g. glutathione synthetase, Pyruvate carboxylase. Pyruvate + CO\(_{2}\) + ATP
\( \xrightarrow{\text{Pyruvate carboxylase}} \) Oxaloacetate + ADP + Pi
In simple words: Enzymes are biological catalysts that speed up chemical reactions in the body without being used up. They are categorized into six main classes—oxidoreductases, transferases, hydrolases, lyases, isomerases, and ligases—each performing a specific type of chemical transformation.

🎯 Exam Tip: When classifying enzymes, ensure to mention an example reaction and the specific enzyme for each class to illustrate your understanding clearly.

Question (G)

Explain the properties of enzyme? Describe the models for enzyme actions.
Answer:1. Proteinaceous Nature: All enzymes are basically made up of protein. 2. Three-Dimensional conformation: a. All enzymes have specific 3-dimensional conformation. b. They have one or more active sites to which substrate (reactant) combines. c. The points of active site where the substrate joins with the enzyme is called substrate binding site. 3. Catalytic property: a. Enzymes are like inorganic catalysts and influence the speed of biochemical reactions but themselves remain unchanged. b. After completion of the reaction and release of the product they remain active to catalyze again. c. A small quantity of enzymes can catalyze the transformation of a very large quantity of the substrate into an end product. d. For example, sucrase can hydrolyze 100000 times of sucrose as compared with its own weight. 4. Specificity of action: a. The ability of an enzyme to catalyze one specific reaction and essentially no other is perhaps its most significant property. Each enzyme acts upon a specific substrate or a specific group of substrates. b. Enzymes are very sensitive to temperature and pH. c. Each enzyme exhibits its highest activity at a specific pH i.e. optimum pH. d. Any increase or decrease in pH causes decline in enzyme activity e.g. enzyme pepsin (secreted in stomach)shows highest activity at an optimum pH of 2 (acidic) 5. Temperature: a. Enzymes are destroyed at higher temperature of 60-70°C or below, they are not destroyed but become inactive. b. This inactive state is temporary and the enzyme can become active at suitable temperature. c. Most of the enzymes work at an optimum temperature between 20°C and 35°C. There are two types of models: 1. Lock and Key model: a. Lock and Key model was first postulated in 1894 by Emil Fischer. b. This model explains the specific action of an enzyme with a single substrate. c. In this model, lock is the enzyme and key is the substrate. d. The correctly sized key (substrate) fits into the key hole (active site) of the lock (enzyme).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख लॉक और कुंजी मॉडल (Lock and Key model) को दर्शाता है, जिसमें एंजाइम (लॉक) और सब्सट्रेट (कुंजी) को एक विशिष्ट आकार के रूप में दिखाया गया है जो सक्रिय स्थल पर एक-दूसरे में पूरी तरह से फिट होते हैं। यह मॉडल दर्शाता है कि एंजाइम की सक्रिय साइट सब्सट्रेट के लिए एक निश्चित, कठोर आकार रखती है। 2. Induced Fit model (Flexible Model): a. Induced Fit model was first proposed in 1959 by Koshland. b. This model states that approach of a substrate induces a conformational change in the enzyme. c. It is the more accepted model to understand mode of action of enzyme. d. The induced fit model shows that enzymes are rather flexible structures in which the active site continually reshapes by its interactions with the substrate until the time the substrate is completely bound to it. e. It is also the point at which the final form and shape of the enzyme is determined. [Note: Temperature is a factor affecting enzyme activity and not a property of enzyme.]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक जटिल लचीले मॉडल (Complex Flexible model) को दिखाता है, जहाँ एंजाइम (E) और सब्सट्रेट (S) शुरू में पूरी तरह से फिट नहीं होते हैं। जैसे ही सब्सट्रेट एंजाइम से संपर्क करता है, एंजाइम की सक्रिय साइट अपने आकार को समायोजित करती है ताकि सब्सट्रेट पूरी तरह से फिट हो सके, एक एंजाइम-सब्सट्रेट कॉम्प्लेक्स (E-S) बनाती है, जिसके बाद उत्पाद (P) जारी होते हैं (E+P)।
In simple words: Enzymes are proteins with specific 3D shapes and active sites that act as catalysts, speeding up reactions without being consumed, and are highly sensitive to temperature and pH. Two main models, Lock and Key and Induced Fit, explain how enzymes bind to substrates, with the Induced Fit model suggesting a flexible active site that reshapes upon substrate binding.

🎯 Exam Tip: Clearly differentiate between the Lock and Key and Induced Fit models, including their proponents and the flexibility of the active site, to earn maximum points.

Question (H)

Describe the factors affecting enzyme action.
Answer:The factors affecting the enzyme activity are as follows: 1. Concentration of substrate: a. Increase in the substrate concentration gradually increases the velocity of enzyme activity within the limited range of substrate levels. b. A rectangular hyperbola is obtained when velocity is plotted against the substrate concentration. c. Three distinct phases (A, B and C) of the reaction are observed in the graph. Where V - Measured velocity, V\(_{\text{max}}\) - Maximum velocity, S - Substrate concentration, K\(_{\text{m}}\) - Michaelis-Menten constant. d. K\(_{\text{m}}\) or the Michaelis-Menten constant is defined as the substrate concentration (expressed in moles/lit) to produce half of maximum velocity in an enzyme catalyzed reaction. e. It indicates that half of the enzyme molecules (i.e. 50%) are bound with the substrate molecules when the substrate concentration equals the K\(_{\text{m}}\) value. f. K\(_{\text{m}}\) value is a constant and a characteristic feature of a given enzyme. g. It is a representative for measuring the strength of ES complex. h. A low K\(_{\text{m}}\) value indicates a strong affinity between enzyme and substrate, whereas a high K\(_{\text{m}}\) value reflects a weak affinity between them. 1. For majority of enzymes, the K\(_{\text{m}}\) values are in the range of 10\(^{-5}\) to 10\(^{-2}\) moles.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एंजाइम गतिविधि पर सब्सट्रेट सांद्रता के प्रभाव को दर्शाता है। वक्र से पता चलता है कि सब्सट्रेट सांद्रता बढ़ने के साथ प्रतिक्रिया की गति भी बढ़ती है, जब तक कि वह Vmax (अधिकतम वेग) तक नहीं पहुंच जाती, जिसके बाद यह स्थिर हो जाती है। K\(_{\text{m}}\) मान सब्सट्रेट सांद्रता को इंगित करता है जहाँ वेग Vmax का आधा होता है। 2. Enzyme Concentration:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एंजाइम सांद्रता के प्रभाव को दर्शाता है। इसमें दिखाया गया है कि एंजाइम की सांद्रता (1X Enzyme बनाम 2X Enzyme) बढ़ने पर उत्पाद की सांद्रता (Product concentration) समय के साथ तेज़ी से बढ़ती है, जो यह दर्शाता है कि अधिक एंजाइम अधिक तेजी से प्रतिक्रिया को उत्प्रेरित करते हैं, जबकि "No Enzyme" रेखा कोई उत्पाद नहीं दिखाती है। a. The rate of an enzymatic reaction is directly proportional to the concentration of the substrate. b. The rate of reaction is also directly proportional to the square root of the concentration of enzymes. c. It means that the rate of reaction also increases with the increasing concentration of enzyme and the rate of reaction can also decrease by decreasing the concentration of enzyme. 3. Temperature:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एंजाइम गतिविधि पर तापमान के प्रभाव को दर्शाता है, जिसमें एक इष्टतम तापमान (Optimum temperature) होता है जहाँ एंजाइम की गतिविधि अधिकतम होती है। इस इष्टतम तापमान से कम या अधिक तापमान पर एंजाइम की गतिविधि घट जाती है। a. The temperature at which the enzymes show maximum activity is called Optimum temperature. b. The rate of chemical reaction is increased by a rise in temperature but this is true only over a limited range of temperature. c. Enzymes rapidly denature at temperature above 40°C. d. The activity of enzymes is reduced at low temperature. e. The enzymatic reaction occurs best at or around 37°C which is the average normal body temperature in homeotherms. 4. Effect of pH:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एंजाइम गतिविधि पर pH के प्रभाव को दर्शाता है, जिसमें एक इष्टतम pH (Optimum pH) होता है जहाँ एंजाइम की गतिविधि अधिकतम होती है। इस इष्टतम pH से कम या अधिक pH पर एंजाइम की गतिविधि घट जाती है, जिससे यह पता चलता है कि एंजाइम केवल एक संकीर्ण pH सीमा के भीतर प्रभावी ढंग से कार्य करते हैं। a. The pH at which an enzyme catalyzes the reaction at the maximum rate is known as optimum pH. b. The enzyme cannot perform its function beyond the range of its pH value. 5. Other substances: a. The enzyme action is also increased or decreased in the presence of some other substances such as co-enzymes, activators and inhibitors. b. Most of the enzymes are combination of a co-enzyme and an apo-enzyme. c. Activators are the inorganic substances which increase the enzyme activity. d. Inhibitor is the substance which reduces the enzyme activity.
In simple words: Enzyme activity is influenced by several factors, including substrate concentration (initial increase, then saturation), enzyme concentration (directly proportional to reaction rate), temperature (optimal range, denatures at high temperatures), pH (optimal range, activity declines outside it), and other substances like activators and inhibitors.

🎯 Exam Tip: When discussing factors affecting enzyme action, always mention the optimal conditions (temperature, pH) and explain how deviations from these optima impact enzyme efficiency, often by affecting their 3D structure.

Question (I)

What are the types of RNA? Mention the role of each class of RNA.
Answer:There are three types of cellular RNAs: 1. messenger RNA (mRNA), 2. ribosomal RNA (rRNA), 3. transfer RNA (tRNA). ' 1. Messenger RNA (mRNA): a. It is a linear polynucleotide. b. It accounts 3% of cellular RNA. c. Its molecular weight is several million. d. mRNA molecule carrying information to form a complete polypeptide chain is called cistron. e. Size of mRNA is related to the size of message it contains. f. Synthesis of mRNA begins at 5' end of DNA strand and terminates at 3' end. Role of messenger RNA: It carries genetic information from DNA to ribosomes, which are the sites of protein synthesis.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख mRNA की संरचना को दर्शाता है, जिसमें 5' और 3' सिरे, एक प्रारंभिक कोडॉन (AUG), एक सिस्ट्रॉन क्षेत्र और एक समाप्ति कोडॉन (UAA) शामिल हैं। यह एक रैखिक पॉली न्यूक्लियोटाइड है जो प्रोटीन संश्लेषण के लिए आनुवंशिक जानकारी ले जाता है। 2. Ribosomal RNA (rRNA): a. rRNA was discovered by Kurland in 1960. b. It forms 50-60% part of ribosomes. c. It accounts 80-90% of the cellular RNA. d. It is synthesized in nucleus. e. It gets coiled at various places due to intrachain complementary base pairing. Role of ribosomal RNA: It provides proper binding site for m-RNA during protein synthesis.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख rRNA की एक सामान्यीकृत संरचना को दर्शाता है, जिसमें हाइड्रोजन बंधन द्वारा जुड़े हुए युग्मित और अयुग्मित क्षार होते हैं, तथा एक चीनी-फॉस्फेट बैकबोन भी दिखाई गई है। यह संरचना राइबोसोम का एक मुख्य घटक है और प्रोटीन संश्लेषण के दौरान mRNA के लिए एक उचित बंधन स्थल प्रदान करती है। 3. Transfer RNA (tRNA): a. These molecules are much smaller consisting of 70-80 nucleotides. b. Due to presence of complementary base pairing at various places, it is shaped like clover-leaf. c. Each tRNA can pick up particular amino acid. d. Following four parts can be recognized on tRNA 1. DHU arm (Dihydroxyuracil loop/ amino acid recognition site 2. Amino acid binding site 3. Anticodon loop / codon recognition site 4. Ribosome recognition site. e. In the anticodon loop of tRNA, three unpaired nucleotides are present called as anticodon which pair with codon present on mRNA. f. The specific amino acids are attached at the 3' end in acceptor stem of clover leaf of tRNA. Role of transfer RNA: It helps in elongation of polypeptide chain during the process called translation.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख tRNA की क्लोवर-लीफ मॉडल संरचना को दर्शाता है, जिसमें एक 3' अमीनो एसिड अटैचमेंट साइट, 5' G लूप, D लूप, T लूप, परिवर्तनीय लूप (Variable loop), और एक एंटीकोडॉन लूप शामिल है, जिसमें mRNA कोडॉन के साथ युग्मित होने वाला एंटीकोडॉन होता है। यह संरचना प्रोटीन संश्लेषण के दौरान विशिष्ट अमीनो एसिड को राइबोसोम तक ले जाने के लिए महत्वपूर्ण है।
In simple words: RNA comes in three main types: messenger RNA (mRNA) carries genetic instructions from DNA to ribosomes, ribosomal RNA (rRNA) forms the structural and catalytic core of ribosomes for protein synthesis, and transfer RNA (tRNA) brings specific amino acids to the ribosome during protein production based on the mRNA code.

🎯 Exam Tip: For each RNA type, focus on its distinct role and structural features (e.g., mRNA as a linear message, rRNA as ribosomal component, tRNA as clover-leaf carrying amino acids) for a comprehensive answer.

Question (J)

What is metabolism? How metabolic pool is formed in the cell.
Answer:1. Metabolism is the sum of the chemical reactions that take place within each cell of a living organism and provide energy for vital processes and for synthesizing new' organic material. 2. Metabolic pool in the cell is formed due to glycolysis and Krebs cycle. 3. The catabolic chemical reaction of glycolysis and Krebs cycle provides ATP and biomolecules. These biomolecules form the metabolic pool of the cell. 4. These biomolecules can be utilized for synthesis of many important cellular components. 5. The metabolites can be added or withdrawn from the pool according to the need of the cell.
In simple words: Metabolism refers to all chemical reactions in a living cell that generate energy and synthesize new materials. The metabolic pool is a collection of intermediate biomolecules, primarily generated by processes like glycolysis and the Krebs cycle, which can be drawn upon or added to as needed to produce cellular components or energy.

🎯 Exam Tip: When defining metabolism, emphasize both catabolic and anabolic aspects. For the metabolic pool, linking it to central pathways like glycolysis and the Krebs cycle is key.

Question 6.

If double stranded DNA has 14% C (cytosine) what percent A (adenine), T (thymine) and G (guanine) would you expect?
Answer:A purine always pairs with pyrimidine. Adenine pairs with thymine and cytosine pairs with guanine. Therefore, as per the given data If cytosine = 14% then guanine = 14%. According to Chargaff's rule, (C+G) = 14+ 14 = 28% Therefore, (A+T) = 72% So, A= 36%, T= 36%, G = 14%.
In simple words: According to Chargaff's rules, in double-stranded DNA, the amount of cytosine always equals guanine, and adenine always equals thymine. If cytosine is 14%, then guanine is also 14%, making (C+G) 28%. The remaining 72% must be (A+T), so both adenine and thymine would be 36% each.

🎯 Exam Tip: Remember Chargaff's rules (A=T and C=G) as the fundamental principle for solving DNA base composition problems.

Question 7.

Name 1. The reagent used for testing for reducing sugar. 2. The form in which carbohydrate is transported in a plant. 3. The term that describes all the chemical reactions taking place in an organism.
Answer:1. Benedict's reagent 2. Sucrose 3. Metabolism
In simple words: Benedict's reagent tests for reducing sugars, sucrose is the transport form of carbohydrates in plants, and metabolism encompasses all chemical reactions within an organism.

🎯 Exam Tip: For "Name" questions, a direct and accurate one-word or short phrase answer is usually sufficient.

Practical / Project:

Question 1.

Perform an experiment to study starch granules isolated from potato.
Answer:Isolation of starch granules from potato: 1. Peal the potato with a clean knife. 2. Grind the potato till the homogenous mixture is formed. 3. Then strain the mixture through a cheese cloth into a beaker. 4. Keep it standing for some time. 5. Throw the supernatant and fill the beaker containing starch with water. 6. Stir it well and again allow the starch to settle. 7. After sometime, again through the supernatant. 8. Repeat this for 2-3 times. 9. Collect the white starch in the watch glass and keep it in the oven for drying. To study the isolated starch granules: 1. Examination under microscope: Examine starch granules under microscope by using a mixture of equal volumes of glycerol and distilled water. Result: The potato starch granules appears transparent granules. They are irregularly shaped. 2. Using Iodine solution: Boil a little amount of starch with water. Cool it. Add iodine solution to it. Result: The solution changes colour to blue. This indicates the presence of starch.
In simple words: To isolate starch from a potato, peel and grind it, then repeatedly strain and wash the mixture, allowing starch to settle and collecting the white sediment. Under a microscope, the granules appear transparent and irregularly shaped, and they turn blue with iodine solution, confirming the presence of starch.

🎯 Exam Tip: Detail the steps for isolation and testing carefully, focusing on observations and conclusions for both microscopic examination and iodine test.

Question 2.

Study the action of enzyme urease on urea.
Answer:Urease is an enzyme which exists in a dimer form. It has two active sites which are highly specific and only bind to urea or hydroxy urea. The active sites of urease contain nickel atoms. Urease catalyzes the hydrolysis of substrate urea into carbon dioxide and ammonia. It attacks the nitrogen and carbon bond in amide compounds and forms alkaline product like ammonia. (NH\(_{2}\))\(_{2}\)CO + 2H\(_{2}\)O
\( \xrightarrow{\text{urease}} \) CO\(_{2}\) + H\(_{2}\)O + 2NH\(_{3}\)
In simple words: Urease is an enzyme with nickel-containing active sites that specifically breaks down urea, converting it into carbon dioxide and ammonia, thereby producing an alkaline product.

🎯 Exam Tip: Highlight the specificity of urease for urea and the end products of its catalytic action (CO2 and NH3) in your explanation.

11th Biology Digest Chapter 6 Biomolecules Intext Questions And Answers

Can You Recall? (Textbook Page No. 59)

Question (i)

Which are different cell components?
Answer:a. The three main components of any cell are: Cell membrane, Cytoplasm, Nucleus. b. The components present in both plant and animal cells are: Endoplasmic reticulum, ribosomes, golgi apparatus, lysosomes, mitochondria, vacuoles. c. The components present in plant cell and not in animal cell: Cell wall and plastids. d. The components present in animal cell and not in plant cell: Cilia and flagella.
In simple words: Cell components include the cell membrane, cytoplasm, and nucleus. Both plant and animal cells share organelles like endoplasmic reticulum and mitochondria, but plant cells uniquely have cell walls and plastids, while animal cells have cilia and flagella.

🎯 Exam Tip: Clearly distinguish between components common to both plant and animal cells, and those specific to each, for a well-rounded answer.

Question (ii)

What is the role of each component of cell?
Answer:The role of each component of a cell is as follows: a. Cell membrane: Cell membrane separates the cytoplasmic contents from external environment. b. Cytoplasm: Site for metabolic activities and organelles. c. Nucleus: It is the control center of the cell. Genetic material is present in the nucleus. d. Endoplasmic reticulum: It produces, processes and transports proteins and lipids. e. Ribosomes: Ribosome is the site for protein synthesis. f. Golgi apparatus: It is involved in modifying, sorting and packing of proteins for secretion. It also transports lipids around the cell. g. Lysosomes: It is involved in digestion of worn out organelles and waste removal. h. Mitochondria: It is responsible for production of energy. i. Vacuoles: It has various functions like storage, waste disposal, protection and growth. j. Cell wall: It provides strength and support to the cell. k. Plastids: They are responsible for production and storage of food. It also contains photosynthetic pigments (Chloroplasts). l. Cilia and flagella: Help in motility.
In simple words: Each cell component has a vital role: the cell membrane controls entry and exit, cytoplasm hosts reactions, the nucleus manages cell functions, ER processes proteins and lipids, ribosomes synthesize proteins, Golgi modifies and packages, lysosomes digest waste, mitochondria generate energy, vacuoles store substances, cell walls provide support, plastids handle food, and cilia/flagella aid movement.

🎯 Exam Tip: Be precise and concise when describing the function of each organelle; use keywords that highlight their primary role (e.g., "protein synthesis" for ribosomes, "energy production" for mitochondria).

Can You Tell? (Textbook Page No. 62)

Question.

What are carbohydrates?
Answer:1. The word carbohydrates mean 'hydrates of carbon'. 2. They are also called saccharides. 3. They are biomolecules made from just three elements: carbon, hydrogen and oxygen with the general formula C\(_{\text{x}}\)(H\(_{2}\)0)\(_{\text{y}}\). 4. They contain hydrogen and oxygen in the same ratio as in water (2:1). 5. Carbohydrates can be broken down (oxidized) to release energy.
In simple words: Carbohydrates, also known as saccharides, are biomolecules composed of carbon, hydrogen, and oxygen, typically in a 2:1 hydrogen-to-oxygen ratio like water, and serve as a primary source of energy when broken down.

🎯 Exam Tip: Include the general formula (C\(_{\text{x}}\)(H\(_{2}\)0)\(_{\text{y}}\)) and their primary function as energy sources for a complete definition.

Can You Tell? (Textbook Page No. 62)

Question (i)

Enlist the natural sources, structural units and functions of the following polysaccharides. a. Starch b. Cellulose c. Glycogen
Answer:a. Starch: 1. Natural Sources: Cereals (wheat, maize, rice), root vegetables (potato, cassava etc.) 2. Structural units: Starch consist of two types of molecules - Amylose and amylopectin. 3. Functions: It acts as a reserve food and supply energy. b. Cellulose: 1. Natural sources: Plant fibers (cotton, flax, hemp, jute, etc.), wood. 2. Structural units: It is made from p glucose molecules. 3. Functions: It in a major component of cell wall. It provides structural support. c. Glycogen: 1. Natural sources: Fruits, starchy vegetables, whole grain foods. 2. Structural units: It consists of linear chains of glucose residues. The glucose is linked linearly by a (1 → 4) glycosidic bonds and branches are linked to the linear chain by a (1 → 6) glycosidic bonds. 3. Functions: It is stored in liver and muscles and it readily provides energy when the blood glucose level decreases.
In simple words: Starch (from cereals) is a plant energy reserve made of amylose and amylopectin; cellulose (from plant fibers) provides structural support to plant cell walls using beta-glucose units; and glycogen (from fruits/starchy foods) is an animal energy reserve in liver and muscles, composed of branched glucose chains.

🎯 Exam Tip: For each polysaccharide, clearly list its natural sources, the specific monosaccharide units or forms, and its primary biological function to ensure a complete answer.

Question (ii)

The exoskeleton of insects is made up of chitin. This is a _______.

🎯 Exam Tip: Remember chitin's classification as a polysaccharide and its structural role in insects and fungi.

Question (iii)

List names of structural polysaccharides.
Answer:Arabinoxylans, cellulose, chitin, pectin.
In simple words: Structural polysaccharides, which provide support, include arabinoxylans, cellulose, chitin, and pectin.

🎯 Exam Tip: Be sure to list at least three examples of structural polysaccharides and associate them with their biological roles (e.g., cellulose in plant cell walls, chitin in exoskeletons).

Question (iv)

Write a note on oligosaccharide and glycosidic bond.
Answer:Oligosaccharides: a. A carbohydrate polymer comprising of two to six monosaccharide molecules is called oligosaccharide. b. They are linked together by glycosidic bond. c. They are classified on the basis of monosaccharide units: Disaccharides: These are the sugars containing two monosaccharide units and can be further hydrolysed into smaller components. E.g.: Sucrose, maltose, lactose, etc. Trisaccharides: These contain three monomers. E.g. Raffmose. Tetrasaccharides: These contain four monomers. E.g.: Stachyose. Glycosidic bond: a. Glycosidic bond is a covalent bond that forms a linkage between two monosaccharides by a dehydration reaction. b. It is formed when a hydroxyl group of one sugar reacts with the anomeric carbon of the other. c. Glycosidic bonds are readily hydrolyzed by acid but resist cleavage by base. d. There are two types of glycosidic bonds: a-glycosidic bond and P-glycosidic bond.
In simple words: Oligosaccharides are carbohydrates made of 2-6 monosaccharide units linked by glycosidic bonds. A glycosidic bond is a covalent link formed by a dehydration reaction between two sugar molecules, crucial for forming di-, tri-, and tetrasaccharides.

🎯 Exam Tip: Clearly define oligosaccharides by their monomer count and glycosidic bonds by their chemical nature and formation (dehydration reaction).

Can You Tell? (Textbook Page No. 63)

Question. What are lipids? Classify them and give at least one example of each.
Answer: Lipids are a group of heterogeneous compounds like fats, oils, steroids, waxes, etc. They are macro-biomolecules. These are group of substances with greasy consistency with long hydrocarbon chain containing carbon, hydrogen and oxygen. Lipids are classified into:
1. Saturated fatty acids: They contain single chain of carbon atoms with single bonds.
E.g. Palmitic acid, stearic acid
2. Unsaturated fatty acids: They contain one or more double bonds between the carbon atoms of the hydrocarbon chain.
a. Simple lipids: These are esters of fatty acids with various alcohols.
E.g. Fats, wax.
b. Compound lipids: These are ester of fatty acids containing other groups like phosphate (Phospholipids), sugar (glycolipids), etc.
E.g. Lecithin
c. Sterols: They are derived lipids. They are composed of fused hydrocarbon rings
In simple words: Lipids are organic molecules like fats and oils, essential for life. They are categorized into simple, compound, and derived lipids based on their chemical structure, such as saturated or unsaturated fatty acids.

🎯 Exam Tip: Pay attention to the classification of lipids and be able to provide at least one example for each type, as this demonstrates a fundamental understanding of biomolecules.

Find Out (Textbook Page No. 63)

Question. (i) Why do high cholesterol level in the blood cause heart diseases?
Answer: a. When there is high level of cholesterol in the blood, the cholesterol builds up on the walls of arteries causing a condition called atherosclerosis (a form of heart disease).
b. Because of this the arteries are narrowed and the blood flow to the heart is slowed down.
c. The blood carries oxygen to the heart, but because of this condition enough blood and oxygen does not reach to the heart and causes heart diseases.
d. If the condition increases, the supply of oxygen and blood is completely cut off to the heart and this can lead to heart attack.
In simple words: High cholesterol causes plaque buildup in arteries, narrowing them and restricting blood flow to the heart. This reduced oxygen supply can lead to various heart diseases, including heart attacks.

🎯 Exam Tip: Understand the mechanism of atherosclerosis and its direct link to restricted blood flow and oxygen supply to the heart, which is a common concept in cardiovascular health.

Question. (ii) Polyunsaturated fatty acids are believed to decrease blood cholesterol level. How?
Answer: a. The liver converts polyunsaturated fatty acids into ketones instead of cholesterol.
b. Therefore, polyunsaturated fatty acids are transported directly to tissues for oxidation without leaving behind any lipoprotein in the form of cholesterol as it is seen in the case of saturated fatty acids.
c. Thus, polyunsaturated fatty acids are believed to decrease blood cholesterol level.
In simple words: Polyunsaturated fatty acids help lower cholesterol by being converted into ketones in the liver instead of cholesterol. They are then directly used by tissues for energy, preventing cholesterol buildup.

🎯 Exam Tip: Focus on the metabolic pathways and the role of polyunsaturated fatty acids in diverting precursors away from cholesterol synthesis, leading to a reduction in blood cholesterol levels.

Can You Tell? (Textbook Page No. 64)

Question. Which of the following is a simple protein?
(A) nucleoprotein
(B) mucoprotein
(C) chromoprotein
(D) globulin
Answer: (D) Globulin
In simple words: Globulin is a type of simple protein, meaning it yields only amino acids upon hydrolysis, unlike conjugated proteins which have non-protein components.

🎯 Exam Tip: Be familiar with the classification of proteins (simple vs. conjugated) and common examples for each type, as this helps differentiate their structural compositions.

Can You Tell? (Textbook Page No. 64)

Question. What are conjugated proteins? How do they differ from simple ones? Give one example of each.
Answer: 1. Conjugated proteins consist of a simple protein attached with some non-protein substance. The non-protein group is called prosthetic group.
2. The conjugated protein functions in interaction with other chemical group whereas simple proteins contain only amino acids and no other chemical group attached to it.
3. Example of conjugated protein is haemoglobin. Globin is the protein and iron containing pigment and haem is the prosthetic group.
In simple words: Conjugated proteins are proteins linked to a non-protein part (prosthetic group), like hemoglobin with its iron-containing heme. Simple proteins, in contrast, are made solely of amino acids.

🎯 Exam Tip: Clearly define conjugated proteins by their non-protein component (prosthetic group) and contrast them with simple proteins which are purely amino acid chains. Providing hemoglobin as an example is key.

Can You Tell? (Textbook Page No. 64)

Question. All Proteins are made up of the same amino acids; then how proteins found in human beings and animals may be different from those of other?
Answer: The proteins found in human beings and animals may be different from those of others because the ratio of amino acids present in the protein differs.
In simple words: Even though all proteins use the same basic amino acids, their specific sequence, number, and ratio of these amino acids create unique protein structures and functions, making proteins different across species.

🎯 Exam Tip: Understand that while the building blocks (amino acids) are universal, the specific arrangement and quantity of these amino acids dictate the unique structure and function of proteins, leading to species-specific variations.

Can You Tell? (Textbook Page No. 67)

Question. What is a nucleotide? How is it formed? Mention the names of all nucleotides.
Answer: 1. Nucleotide is a unit which consists of a sugar, phosphate and a base. Nucleotides are basic units of nucleic acids.
2. The nitrogen base and a sugar form a nucleoside. In a nucleoside, nitrogenous base is attached to the first carbon atom (C-1) of the sugar and when a phosphate group gets attached with that of the carbon (C-5) atom of the sugar molecule a nucleotide molecule is formed.
3. The names of all nucleotides are:

In simple words: Nucleotides are the fundamental units of DNA and RNA, composed of a sugar, a phosphate group, and a nitrogenous base. They are formed when a nucleoside (base + sugar) links with a phosphate.

🎯 Exam Tip: Know the three components of a nucleotide and how they assemble. Memorize the names of the common nucleotides and their respective bases for both RNA and DNA.

Can You Tell? (Textbook Page No. 67)

Question. Describe the structure of DNA molecule as proposed by Watson and Crick.
Answer: 1. According to Watson and Crick, DNA molecule consists of two strands twisted around each other in the form of a double helix.
2. The two strands i.e. polynucleotide chains are supposed to be in opposite direction so end of one chain having 3' lies beside the 5' end of the other.
3. One turn of the double helix of the DNA measures about 34A.
4. It consists paired nucleotides and the distance between two neighboring pair nucleotides is 3.4A.
5. The diameter of the DNA molecule has been found be 20A.
In simple words: Watson and Crick described DNA as a double helix with two polynucleotide strands twisted around each other, running in opposite directions. Key features include specific base pairing, a consistent diameter, and specific measurements for each turn.

🎯 Exam Tip: Focus on the key characteristics of the Watson and Crick DNA model: double helix, antiparallel strands, base pairing rules, and the dimensions of the helix (e.g., 34Å per turn, 3.4Å between base pairs, 20Å diameter).

Can You Tell (Textbook Page No. 70)

Question. Name the chemical found in the living cell which has necessary message for the production of all enzymes required by it.
Answer: DNA found in the nucleus of a living cell has necessary message for the production of all enzymes required by it. DNA forms mRNA through the process of transcription. This mRNA through the process of translation forms proteins.
In simple words: DNA, located in the cell's nucleus, carries the genetic instructions for producing all enzymes. It directs the synthesis of mRNA (transcription), which then guides protein formation (translation).

🎯 Exam Tip: Understand the central dogma of molecular biology: DNA stores genetic information, which is transcribed into mRNA, and then translated into proteins (enzymes).

Can You Tell? (Textbook Page No. 67)

Question. Difference between DNA and RNA is because of
(A) sugar and base
(B) sugar and phosphate
(C) phosphate and base
(D) sugar only
Answer: (A) sugar and base
In simple words: DNA and RNA primarily differ in their sugar (deoxyribose vs. ribose) and one of their nitrogenous bases (thymine in DNA vs. uracil in RNA).

🎯 Exam Tip: Key differences between DNA and RNA are the sugar component (deoxyribose vs. ribose) and one pyrimidine base (thymine vs. uracil). Be able to recall these fundamental distinctions.

Can You Tell? (Textbook Page No. 67)

Question. Differentiate between DNA and RNA.
Answer:

In simple words: DNA is typically a double-stranded genetic blueprint with deoxyribose sugar and thymine, primarily found in the nucleus. RNA is usually single-stranded, uses ribose sugar and uracil, and functions in protein synthesis, found in both nucleus and cytoplasm.

🎯 Exam Tip: Master the comparative differences between DNA and RNA across several criteria: structure (double vs. single strand), sugar type, nitrogenous bases, base pairing, purine/pyrimidine ratios, location, and primary biological roles.

Can You Tell? (Textbook Page No. 70)

Question. Co-enzyme is -
(A) often a metal
(B) often a vitamin
(C) always as organic molecule
(D) always an inorganic molecule
Answer: (B) often a vitamin
In simple words: Co-enzymes are non-protein organic molecules, often derived from vitamins, that assist enzymes in catalyzing biochemical reactions.

🎯 Exam Tip: Remember that co-enzymes are typically organic molecules, and a significant number are derived from vitamins, which is crucial for their function in enzymatic reactions.

Can You Tell? (Textbook Page No. 70)

Question. (i) Which enzyme is needed to digest food reserve in castor seed?
(A) Amylase
(B) Diastase
(C) Lipase
(D) Protease
Answer: (C) Lipase
In simple words: Castor seeds primarily store food reserves as lipids, and lipase is the enzyme responsible for breaking down lipids (fats) into smaller molecules for energy.

🎯 Exam Tip: Relate the type of food reserve (lipids in castor seeds) to the specific enzyme (lipase) required for its digestion, showcasing knowledge of enzyme specificity and plant physiology.

Question. (ii) List the important properties of enzymes.
Answer: a. Proteinaceous Nature
b. Three-Dimensional conformation
c. Catalytic property
d. Specificity of action
e. Temperature
In simple words: Enzymes are proteins with a specific 3D shape, acting as highly efficient catalysts for specific reactions, and are sensitive to temperature and pH.

🎯 Exam Tip: Focus on the five core properties of enzymes: their protein nature, specific 3D structure (active site), catalytic efficiency, specificity for substrates, and sensitivity to environmental factors like temperature and pH.

Try This: (Textbook Page No. 70)

Question. To demonstrate the effect of heat on the activities of inorganic catalysts and enzymes.
Answer: 1. Using MnO2 and Enzymes without any heat treatment:
Mn02 and cellular enzymes (catalase/peroxidase) causes breakdown of H202 and evolution of oxygen.
2. Using Mn02 and Enzymes after heat treatment:
Oxygen evolves in the H202 solution containing boiled and cooled Mn02. But oxygen does not evolve in the tube containing the enzyme.
3. This confirms that heat affects the enzyme and inactivates it whereas heat does not have any effect on inorganic catalyst.
In simple words: This experiment shows that heat inactivates enzymes, preventing them from breaking down hydrogen peroxide, while an inorganic catalyst like MnO2 remains active even after heating.

🎯 Exam Tip: Understand that enzymes, being proteins, are denatured by heat and lose their activity, unlike inorganic catalysts which are generally more heat-stable. This distinction highlights enzyme specificity and sensitivity.