RBSE Solutions Class 12 Chemistry Chapter 14 Bio-Molecules

RBSE Class 12 Chemistry Chapter 14 Text Book Questions

RBSE Class 12 Chemistry Chapter 14 Multiple Choice Questions

 

Question 1. Which is known as the powerhouse of the cell?
(a) Golgi body
(b) Mitochondria
(c) Cytosome
(d) Ribosome
Answer: (b) Mitochondria
In simple words: Mitochondria make most of the energy needed for the cell to work. That is why they are called the cell's "powerhouses".

🎯 Exam Tip: Remember that the mitochondria are responsible for cellular respiration, which generates ATP, the cell's energy currency.

 

Question 2. Which of the following is a disaccharide?
(a) Starch
(b) Fructose
(c) Sucrose
(d) Glucose
Answer: (c) Sucrose
In simple words: A disaccharide is a sugar made from two smaller sugar units joined together. Sucrose, which is common table sugar, is an example because it is made of glucose and fructose.

🎯 Exam Tip: Disaccharides are formed when two monosaccharides are linked by a glycosidic bond. Common disaccharides include sucrose (glucose + fructose), lactose (glucose + galactose), and maltose (glucose + glucose).

 

Question 4. The molecular formula of most common disaccharide is:
(a) \( (\text{C}_6\text{H}_{12}\text{O}_6)_6 \)
(b) \( (\text{C}_{12}\text{H}_{22}\text{O}_{11}) \)
(c) \( (\text{C}_{10}\text{H}_{22}\text{O}_{11}) \)
(d) \( (\text{C}_{18}\text{H}_{22}\text{O}_{11}) \)
Answer: (b) \( (\text{C}_{12}\text{H}_{22}\text{O}_{11}) \)
In simple words: Most common disaccharides, like sucrose, have the chemical formula \( \text{C}_{12}\text{H}_{22}\text{O}_{11} \). This formula shows how many carbon, hydrogen, and oxygen atoms are in one molecule.

🎯 Exam Tip: The general formula for disaccharides is \( \text{C}_{12}\text{H}_{22}\text{O}_{11} \), which is formed when two monosaccharides (e.g., \( \text{C}_6\text{H}_{12}\text{O}_6 \)) join, losing one molecule of water (\( \text{H}_2\text{O} \)).

 

Question 5. Which of the following is not a reducing sugar?
(a) Glucose
(b) Fructose
(c) Sucrose
(d) Maltose
Answer: (c) Sucrose
In simple words: A reducing sugar can react with certain chemicals because it has a special aldehyde or ketone group. Sucrose cannot do this reaction, so it is not a reducing sugar.

🎯 Exam Tip: Reducing sugars have a free aldehyde or ketone group that can be oxidized. Sucrose lacks this free group because the anomeric carbons of both glucose and fructose are involved in the glycosidic bond, preventing it from acting as a reducing agent.

 

Question 6. The product obtained after hydrolysis of protein in the presence of an enzyme is
(a) Amino acids
(b) Hydroxy acid
(c) Aromatic acids
(d) Dicarboxylic acid
Answer: (a) Amino acids
In simple words: When proteins are broken down by water with the help of enzymes, they turn into their basic building blocks, which are called amino acids.

🎯 Exam Tip: Proteins are long chains of amino acids. Hydrolysis (breaking with water) reverses the process of protein synthesis, yielding the individual amino acid units.

 

Question 7. The example of globular protein is
(a) Collagen
(b) Insulin
(c) Myosin
(d) Keratin
Answer: (b) Insulin
In simple words: Globular proteins are usually spherical and soluble in water. Insulin, which helps control blood sugar, is an example of this type of protein.

🎯 Exam Tip: Globular proteins have compact, roughly spherical shapes and are often involved in metabolic functions (e.g., enzymes, hormones like insulin). Fibrous proteins (like collagen, myosin, keratin) are elongated and provide structural support.

 

Question 9. Which of the following is a basic amino acid?
(a) Glycine
(b) Aspartic acid
(c) Lysine
(d) Glutamine
Answer: (c) Lysine
In simple words: Basic amino acids have an extra amino group in their side chain, which makes them basic. Lysine is one such amino acid.

🎯 Exam Tip: Amino acids are classified based on the nature of their side chain (R-group). Basic amino acids have an extra amine group (like lysine, arginine, histidine), while acidic amino acids have an extra carboxyl group (like aspartic acid, glutamic acid).

 

Question 10. Enzymes are
(a) Carbohydrates
(b) Proteins
(c) Fats
(d) Salts
Answer: (b) Proteins
In simple words: Enzymes are special kinds of proteins that help speed up chemical reactions in living things. They act as biological catalysts.

🎯 Exam Tip: Almost all enzymes are proteins, except for a few catalytic RNA molecules called ribozymes. They are highly specific in their action and sensitive to temperature and pH.

 

Question 11. The conversion of protein to amino acids takes place by which of the following enzyme?
(a) Lipase
(b) Maltase
(c) Trypsin
(d) Renin
Answer: (c) Trypsin
In simple words: Trypsin is an enzyme that helps break down proteins into smaller amino acids in the body. It works in the digestive system.

🎯 Exam Tip: Trypsin is a protease enzyme found in the digestive system, where it helps in the breakdown of dietary proteins. Lipase breaks down fats, maltase breaks down maltose, and renin is involved in milk coagulation.

 

Question 12. Chemical messengers are
(a) Hormones
(b) Enzymes
(c) Vitamins
(d) Nucleic acids
Answer: (a) Hormones
In simple words: Hormones are substances that act like messengers in the body, traveling through the blood to tell different parts of the body what to do.

🎯 Exam Tip: Hormones are signaling molecules produced by glands that regulate various physiological processes, often acting on target cells far from their site of production.

 

Question 13. Number of the thyroid gland in humans is
(a) One
(b) Two
(c) Three
(d) Four
Answer: (a) One
In simple words: Humans have one thyroid gland, which is shaped like a butterfly and located in the neck. It controls many important body functions.

🎯 Exam Tip: The thyroid gland is a single, bilobed endocrine gland located in the neck, responsible for producing thyroid hormones that regulate metabolism.

 

Question 15. Vitamin A deficiency causes
(a) Night blindness
(b) Scurvy
(c) Beriberi
(d) Anaemia
Answer: (a) Night blindness
In simple words: Not having enough Vitamin A can make it hard to see in dim light, a condition known as night blindness. Vitamin A is important for good vision.

🎯 Exam Tip: Vitamin A (retinol) is crucial for vision, immune function, and cell growth. Scurvy is caused by Vitamin C deficiency, beriberi by Vitamin B1 deficiency, and anaemia can be caused by various deficiencies, including iron and Vitamin B12.

 

Question 16. In nucleic acids, nucleotides are attached to one another by
(a) Hydrogen bond
(b) Peptide bond
(c) Phosphorous group
(d) Glycosidic bond
Answer: (c) Phosphorous group
In simple words: In DNA and RNA, the building blocks called nucleotides are linked together by special bonds involving a phosphorus group. These are called phosphodiester bonds.

🎯 Exam Tip: Nucleotides within a single strand of a nucleic acid are linked by phosphodiester bonds between the sugar of one nucleotide and the phosphate group of the next. Hydrogen bonds link complementary base pairs between two strands (e.g., in DNA).

 

Question 17. How many nucleotides sequence forms one codon in mRNA for an amino acid?
(a) One
(b) Two
(c) Three
(d) Four
Answer: (c) Three
In simple words: A codon is a set of three chemical units in mRNA that tells the cell which specific amino acid to add when building a protein.

🎯 Exam Tip: The genetic code is a triplet code, meaning three consecutive nucleotides (a codon) specify one amino acid. This triplet nature ensures enough combinations to code for all 20 standard amino acids.

 

Question 18. DNA and RNA have chiral asymmetric carbon atom. The reason for their chirality is
(a) Asymmetric base
(b) D-sugar component
(c) L-sugar component
(d) Asymmetric phosphate ester units
Answer: (d) Asymmetric phosphate ester units
In simple words: DNA and RNA are chiral because of their D-sugar component, which contains asymmetric carbon atoms. This structural feature makes them optically active. The phosphate ester linkages also contribute to the overall chirality.

🎯 Exam Tip: The D-ribose (in RNA) and D-deoxyribose (in DNA) sugars are chiral molecules with asymmetric carbon atoms, which is the primary reason for the overall chirality of nucleic acids. The asymmetric phosphate ester units are a consequence of this chiral sugar component.

 

Question 20. The sequence in nucleic acids is
(a) Base - sugar - phosphate
(b) Sugar - base - phosphate
(c) Phosphate - base - sugar
(d) Base - phosphate-sugar
Answer: (a) Base - sugar - phosphate
In simple words: Each building block of DNA and RNA, called a nucleotide, is made up of three main parts: a nitrogenous base, a sugar molecule, and a phosphate group, linked in that order.

🎯 Exam Tip: A nucleotide is the monomer unit of nucleic acids, consisting of a nitrogenous base attached to the 1' carbon of a pentose sugar, which is in turn attached to one or more phosphate groups at the 5' carbon.

RBSE Class 12 Chemistry Chapter 14 Very Short Answer Type Questions

 

What are monosaccharides?
Answer: Monosaccharides are a type of carbohydrate that cannot be broken down further into simpler sugar units by hydrolysis. They are the simplest forms of sugar. Common examples are glucose and fructose, which are basic energy sources for cells.
In simple words: Monosaccharides are simple sugars that cannot be split into smaller sugar units. Glucose and fructose are examples.

🎯 Exam Tip: When defining monosaccharides, clearly state that they are the simplest carbohydrates and cannot be hydrolyzed. Providing examples like glucose and fructose strengthens your answer.

 

Question 3. What are non- sugars?
Answer: Non-sugars are carbohydrates that do not taste sweet. These are usually complex carbohydrates that do not dissolve easily in water. For instance, starch and cellulose are common examples of non-sugars found in plants.
In simple words: Non-sugars are carbohydrates that do not taste sweet. Starch and cellulose are examples.

🎯 Exam Tip: Differentiate non-sugars from sugars based on taste and often solubility. Non-sugars are typically polysaccharides, while sugars are monosaccharides or disaccharides.

 

Question 4. What is the main structural difference between starch and cellulose?
Answer: The primary structural difference between starch and cellulose lies in how their glucose molecules are linked. Starch consists of \( \alpha \)-D-(+) glucose molecules linked in a \( \text{C}_1 - \text{C}_4 \) manner (amylose) and sometimes \( \text{C}_1 - \text{C}_6 \) manner (amylopectin for branches). Cellulose, however, is made of \( \beta \)-D-(+) glucose molecules linked in a \( \text{C}_1 - \text{C}_4 \) manner. This seemingly small difference in linkage ( \( \alpha \) vs. \( \beta \) ) leads to vastly different overall structures and properties, with starch being a branched energy storage molecule and cellulose being a linear structural component of plants. Thus, the arrangement of glucose units is key.
In simple words: Starch and cellulose are both made of glucose, but the way these glucose units are connected is different. Starch has \( \alpha \)-linkages, making it easily digestible, while cellulose has \( \beta \)-linkages, making it a strong fiber.

🎯 Exam Tip: Focus on the \( \alpha \) and \( \beta \) glycosidic linkages between glucose units as the key distinguishing feature. Also, briefly mention the resulting structural implications (branched vs. linear, energy storage vs. structural).

 

Question 5. Define essential and non-essential amino acids with examples.
Answer:
(i) Essential Amino acids: These are amino acids that the human body cannot produce on its own. Therefore, they must be obtained through a person's diet. It is important to consume these through food, as the body needs them for various functions. Examples include arginine, valine, and histidine.
(ii) Non-essential Amino Acids: These are amino acids that the human body can synthesize (make) from other molecules. This means they do not strictly need to be included in the diet. The body can create them as needed. Examples include glycine, alanine, serine, and cysteine.
In simple words: Essential amino acids must come from food because the body cannot make them. Non-essential amino acids can be made by the body itself.

🎯 Exam Tip: Clearly state whether the body can or cannot synthesize each type of amino acid and provide at least two correct examples for both essential and non-essential categories to score full marks.

 

Question 7. Why are hormones known as 'glandular juice'?
Answer: Hormones are called 'Glandular Juices' because they are chemical substances produced and secreted by specialized glands in the body, known as endocrine glands. These secretions act as messengers to regulate various physiological processes. The term "juice" here simply refers to the fluid they are released in, similar to digestive juices.
In simple words: Hormones are called 'glandular juices' because they are made and released by special glands in the body, just like juices are squeezed from fruits.

🎯 Exam Tip: The key points are that hormones are (1) chemical messengers (2) secreted by endocrine glands. Emphasize these aspects in your answer.

 

Question 8. Which vitamin, is soluble in water?
Answer: Vitamins B and C are soluble in water. This means they dissolve in water and are easily carried to the body's tissues. The body does not store these vitamins, so they need to be consumed regularly. Other vitamins like A, D, E, and K are fat-soluble.
In simple words: Vitamins B and C can dissolve in water.

🎯 Exam Tip: Remember the two main categories: water-soluble (B and C) and fat-soluble (A, D, E, K). Knowing examples for each group is important.

 

Question 9. Write the name of nitrogenous bases found in DNA.
Answer: The nitrogenous bases found in DNA are grouped into two types: Purines and Pyrimidines. The Purines are Adenine (A) and Guanine (G). The Pyrimidines are Cytosine (C) and Thymine (T). These four bases form the genetic alphabet.
In simple words: The nitrogen bases in DNA are Adenine (A), Guanine (G), Cytosine (C), and Thymine (T).

🎯 Exam Tip: List all four bases and optionally group them into purines and pyrimidines. Accuracy in names and symbols is important.

 

Question 10. What are the main functions of nucleic acids?
Answer: Nucleic acids, specifically DNA and RNA, have two main functions:
1. They play a crucial role in passing on hereditary traits from parents to offspring. DNA contains the genetic instructions for the development and function of all known living organisms.
2. They are involved in the creation of proteins, a process known as protein biosynthesis. RNA molecules help carry out these instructions to build proteins, which perform most of the work in cells.
In simple words: Nucleic acids help pass on traits from parents to children and help make proteins.

🎯 Exam Tip: Key functions to mention are heredity (DNA) and protein synthesis (RNA). Briefly explain how each contributes to these roles.

RBSE Class 12 Chemistry Chapter 14 Short Answer Type Questions

 

Question 1. Write the functions of carbohydrates.
Answer: Carbohydrates perform several vital functions in living organisms:
1. They serve as the primary source of energy for all living beings, providing fuel for cellular activities.
2. They are a major component of plant structures, forming the bulk of plant bodies (e.g., cellulose in cell walls).
3. Sugars like deoxyribose and ribose are fundamental building blocks of genetic material, specifically DNA and RNA, respectively.
4. Starch functions as an important food storage molecule, particularly in plants. Cellulose provides structural support in plants and is also used by humans for shelter and clothing (e.g., wood, cotton).
In simple words: Carbohydrates give us energy, build plant parts, are part of DNA and RNA, and store food.

🎯 Exam Tip: Mention at least three distinct functions, such as energy source, structural component, and genetic material constituent, to provide a comprehensive answer.

 

Question 3. What is Sickle Cell Anaemia?
Answer: Sickle Cell Anaemia is a genetic blood disorder where the red blood cells, which normally carry oxygen, become crescent-shaped (like a sickle) instead of their usual round shape. This happens because of a change in just one specific amino acid in the hemoglobin molecule, which is responsible for carrying oxygen. This change causes the hemoglobin to clump together, making the red blood cells stiff and misshapen. These sickle cells can block blood flow and burst easily, leading to symptoms like pain, fatigue, and other health issues. In normal hemoglobin, glutamic acid is present, but in sickle cell hemoglobin, valine replaces it.
In simple words: Sickle Cell Anaemia is a disease where red blood cells become C-shaped and block blood flow because of a small change in hemoglobin.

🎯 Exam Tip: Key aspects to mention are the abnormal "sickle" shape of red blood cells, the cause being a single amino acid change in hemoglobin, and the resulting issues like blocked blood vessels and cell bursting.

 

Question 4. Write the reactions of glucose with Fehling's solution and Tollen's reagent.
Answer:
(i) Reaction with Fehling's Solution:
Fehling's solution is prepared by mixing Fehling A (which is an aqueous \( \text{CuSO}_4 \) solution) and Fehling B (which contains aqueous \( \text{NaOH} \) and Rochelle's salt) in equal amounts. When glucose is heated with Fehling's solution, it reduces the \( \text{Cu}^{2+} \) ions in the solution to \( \text{Cu}_2\text{O} \) (cuprous oxide), forming a characteristic red precipitate. This reaction demonstrates the presence of an aldehyde group in glucose. Glucose itself gets oxidized to gluconic acid.
\[ \text{CHO}(\text{CHOH})_4\text{CHO} + 2\text{Cu}^{2+} + 4\text{OH}^- \longrightarrow \text{CH}_2\text{OH}(\text{CHOH})_4\text{COOH} + \text{Cu}_2\text{O}\downarrow + 2\text{H}_2\text{O} \]Glucose, which contains an aldehyde group, is oxidized to gluconic acid, while the blue copper(II) ions are reduced to form a brick-red precipitate of copper(I) oxide.
(ii) Reaction with Tollen's Reagent:
Tollen's reagent is an ammoniacal silver nitrate solution. When glucose is heated with Tollen's reagent in a clean test tube, the aldehyde group in glucose reduces the silver ions (\( \text{Ag}^+ \)) in the reagent to metallic silver (\( \text{Ag} \)). This metallic silver then deposits on the inner surface of the test tube, creating a shining silver mirror. This reaction is a classic test for aldehydes.
\[ \text{CHO}(\text{CHOH})_4\text{CHO} + 2\text{Ag}(\text{NH}_3)_2\text{OH} \longrightarrow \text{CH}_2\text{OH}(\text{CHOH})_4\text{COOH} + 2\text{Ag}\downarrow + 4\text{NH}_3 + \text{H}_2\text{O} \]In this reaction, glucose is oxidized to gluconic acid, and silver ions are reduced to metallic silver, forming a mirror. This shows that glucose has an aldehyde functional group.
In simple words: Glucose reacts with Fehling's solution to make a red solid, and with Tollen's reagent, it makes a silver mirror inside the test tube. Both reactions show that glucose has an aldehyde part.

🎯 Exam Tip: For each reaction, clearly state the reagent used, the observable change (red precipitate for Fehling's, silver mirror for Tollen's), and the product glucose is oxidized to. Make sure to write the balanced chemical equations accurately.

 

Question 5. Why are hormones known as “Chemical messengers"?
Answer: Hormones are called "chemical messengers" because they are substances secreted by specific glands into the bloodstream, and they travel to target cells or tissues located far from their origin. Once they reach these target sites, they bind to specific receptors and trigger particular responses, thereby regulating various bodily functions like growth, metabolism, and reproduction. They transmit information and instructions between different cells and organs, coordinating complex processes in the body.
In simple words: Hormones are called chemical messengers because they carry signals from one part of the body to another, telling cells what to do.

🎯 Exam Tip: Emphasize that hormones are chemicals, travel through the blood, and act on distant target cells to regulate physiological processes. This highlights their "messenger" role.

 

Question 6. What is the isoelectric point of amino acids? Define.
Answer: The isoelectric point (pI) of an amino acid is the specific pH at which the amino acid exists predominantly as a zwitterion and has a net charge of zero. At this pH, the amino acid does not migrate towards either the anode (positive electrode) or the cathode (negative electrode) when an electric current is passed through its solution. Amino acids exist as zwitterions (dipolar ions) where a proton (H+) from the carboxylic acid group transfers to the amino group within the same molecule. This creates a molecule with both positive and negative charges, but with a neutral overall charge. The isoelectric points for most \( \alpha \)-amino acids typically range from 5.5 to 6.3.
\[ \text{R-CH-COOH} \rightleftharpoons \text{R-CH-COO}^- \]\[ \qquad\quad |\qquad\qquad\quad |\ \]\[ \qquad\quad \text{NH}_2\qquad\qquad\text{NH}_3^+ \]This diagram shows the conversion to a zwitterion (dipolar ion) form, which has both a positive and negative charge, but a net zero charge.
In simple words: The isoelectric point is a special pH level where an amino acid has no overall electrical charge and will not move in an electric field.

🎯 Exam Tip: Define the isoelectric point as the pH at which the amino acid has a net zero charge and exists as a zwitterion. Explain that at this point, it does not migrate in an electric field. Mentioning the proton transfer to form a dipolar ion is also crucial.

 

Question 7. Write one difference and one similarity between enzymes and hormones.
Answer:
Similarity: Both hormones and enzymes act as biocatalysts, meaning they help speed up or regulate biological processes within living systems. They are both crucial for maintaining life functions.
Difference: Hormones are chemical messengers that are produced in glands and then transported through the blood to act on distant target organs or tissues. They are typically used up or inactivated after performing their function. In contrast, enzymes are usually proteins that function as catalysts within the cells where they are produced or in specific extracellular locations, directly facilitating biochemical reactions without being consumed in the process. Enzymes are not generally released into the blood before use; instead, they act locally where they are needed.
In simple words: Both enzymes and hormones help control body processes. Hormones travel far to send signals, while enzymes speed up reactions without being used up.

🎯 Exam Tip: For similarity, mention their role in regulating biological processes. For difference, highlight hormones as messengers acting remotely via the bloodstream, and enzymes as catalysts acting locally, often without being consumed.

 

Question 9. What is the genetic code?
Answer: The genetic code is a set of rules that living cells use to translate information encoded within DNA or messenger RNA (mRNA) sequences into proteins. This code specifies how sequences of three nucleotide bases, called codons, correspond to specific amino acids. Each protein in a cell is determined by a corresponding sequence of DNA called a gene. The mRNA, which is copied from DNA, then guides the assembly of amino acids in a specific order to create a protein. This relationship between the nucleotide triplets and the amino acids they specify is known as the genetic code. It is a fundamental mechanism that underlies all life.
In simple words: The genetic code is like a secret language that tells cells how to build proteins using sets of three DNA or RNA letters called codons.

🎯 Exam Tip: Define the genetic code as the set of rules relating nucleotide sequences to amino acids. Crucially mention codons (triplets of nucleotides) and their role in specifying amino acids during protein synthesis.

 

Question 10. Differentiate between primary and secondary structure of a protein.
Answer:

In simple words: The primary structure is just the order of amino acids in a protein chain. The secondary structure is about the simple folding patterns (like coils or zigzags) formed by hydrogen bonds in that chain.

🎯 Exam Tip: Clearly state that primary structure is the linear sequence of amino acids, while secondary structure refers to local folding patterns (like alpha-helix, beta-sheet) formed by hydrogen bonding between backbone atoms.

 

Question 11. Explain mutarotation.
Answer: Mutarotation is the change in the optical rotation of a solution containing a chiral molecule, such as a sugar, as it equilibrates between different anomeric forms. When a cyclic sugar, like D-(+)glucose, dissolves in water, it exists in two main cyclic forms: \( \alpha \)-D-(+)glucose and \( \beta \)-D-(+)glucose. These two forms have different specific optical rotations. In an aqueous solution, these cyclic forms slowly convert into each other, passing through a short-lived open-chain aldehyde form. This interconversion continues until an equilibrium is reached, at which point the optical rotation of the solution stabilizes at a constant value (e.g., around +52.7\(^\circ\) for D-glucose). At equilibrium, the mixture typically contains about 36% of the \( \alpha \)-form, 63.5% of the \( \beta \)-form, and a very small amount (about 0.5%) of the open-chain structure. This dynamic process of interconversion and changing optical rotation is called mutarotation.
\[ \alpha\text{-D-(+)Glucose} \rightleftharpoons \text{Open chain structure} \rightleftharpoons \beta\text{-D-(+)Glucose} \]\[ [\alpha] = +112^\circ \qquad \quad (0.5\%) \qquad \quad [\alpha] = +19^\circ \]\[ \quad (36\%) \qquad \qquad \qquad \quad (63.5\%) \]This shows the equilibrium between the alpha, open-chain, and beta forms, each with their specific rotation and percentage at equilibrium.
In simple words: Mutarotation is when a sugar solution's ability to rotate light slowly changes until it settles at a steady value. This happens because the sugar molecules are switching between different forms.

🎯 Exam Tip: Define mutarotation as the change in optical rotation due to the interconversion of anomers in solution. Mention the \( \alpha \), \( \beta \), and open-chain forms and their eventual equilibrium mixture with a stable optical rotation.

 

Question 12. Write the name of the disease caused by the deficiency of Vitamin-B12 and Vitamin A. Also, write the source of these vitamins.
Answer:

In simple words: Lack of Vitamin B12 causes serious anemia, found in animal foods. Lack of Vitamin A causes night blindness and poor growth, found in dairy, eggs, and colorful vegetables.

🎯 Exam Tip: For each vitamin, clearly state one or two key deficiency diseases and at least two common food sources. Ensure accuracy in disease names and sources.

 

Question 13. Write four differences between RNA and DNA.
Answer:

In simple words: RNA has ribose sugar and uracil base, is usually a single strand, and helps make proteins. DNA has deoxyribose sugar and thymine base, is a double strand, and carries genetic information for heredity.

🎯 Exam Tip: When differentiating, clearly state corresponding differences for each point (e.g., sugar type, specific base, strand structure, and primary function). This helps highlight the contrast.

 

Question 14. Write the Haworth structure of glucose and fructose.
Answer:
(i) Haworth Structure of Glucose:
Glucose primarily exists in two cyclic anomeric forms, \( \alpha \)-D-(+)glucopyranose and \( \beta \)-D-(+)glucopyranose. These pyranose ring structures are six-membered rings containing five carbon atoms and one oxygen atom. This ring formation is a result of the aldehyde group reacting with a hydroxyl group within the same molecule. The main difference between the \( \alpha \) and \( \beta \) forms is the position of the hydroxyl group at the anomeric carbon (C-1).
\[ \text{ }^{6}\text{CH}_2\text{OH} \]\[ \quad |\ \]\[ \text{ }^{5}\text{O} \text{-----}\text{H} \]\[ \quad |\qquad\quad |\qquad\quad |\ \]\[ \text{H---}^{4}\text{C---}\text{OH}\quad \text{H---}^{2}\text{C---}\text{OH} \]\[ \quad |\qquad\quad |\qquad\quad |\ \]\[ \text{HO---}^{3}\text{C---}\text{H}\quad \text{HO---}^{1}\text{C---}\text{H} \]\[ \qquad\quad |\ \]\[ \qquad\text{Pyran} \]\[ \alpha\text{-D-(+)Glucopyranose} \]This is the general form showing the pyranose ring structure with its carbon and oxygen numbering. The \( \alpha \) and \( \beta \) forms differ in the orientation of the \( \text{OH} \) group at C-1.
\[ \text{ }^{6}\text{CH}_2\text{OH} \]\[ \quad |\ \]\[ \text{ }^{5}\text{O} \text{-----}\text{H} \]\[ \quad |\qquad\quad |\qquad\quad |\ \]\[ \text{H---}^{4}\text{C---}\text{OH}\quad \text{H---}^{2}\text{C---}\text{OH} \]\[ \quad |\qquad\quad |\qquad\quad |\ \]\[ \text{HO---}^{3}\text{C---}\text{H}\quad \text{H---}^{1}\text{C---}\text{OH} \]\[ \qquad\quad |\ \]\[ \qquad\text{Pyran} \]\[ \beta\text{-D-(+)Glucopyranose} \]The \( \beta \) anomer has the C1-OH in an equatorial position relative to the \( \text{CH}_2\text{OH} \) group.
(ii) Haworth Structure of Fructose:
Fructose also exists in two anomeric forms, \( \alpha \)-D-(-)fructofuranose and \( \beta \)-D-(-)fructofuranose. These furanose ring structures are five-membered rings containing four carbon atoms and one oxygen atom. This cyclic structure is formed by the reaction of the ketone group with an internal hydroxyl group. Similar to glucose, the \( \alpha \) and \( \beta \) forms of fructose differ in the orientation of the hydroxyl group at the anomeric carbon (C-2). Furan itself is a five-membered heterocyclic ring with an oxygen atom, which forms the basis for the cyclic structures of D-(-)fructose.
\[ \qquad\qquad\text{OH} \]\[ \qquad\qquad |\ \]\[ \qquad\text{O}\text{----}\text{CH}_2 \]\[ \quad /\quad |\qquad\quad |\ \]\[ \text{HO---}\text{C}\quad\text{H---}\text{OH} \]\[ \quad |\quad |\qquad\quad |\ \]\[ \text{H---}\text{C---}\text{OH}\quad\text{H---}\text{C---}\text{OH} \]\[ \qquad\qquad |\ \]\[ \qquad\text{O} \]\[ \alpha\text{-D-(-)Fructofuranose} \]This image depicts the Haworth projection for the alpha form of D-fructofuranose.
\[ \qquad\qquad\text{H} \]\[ \qquad\qquad |\ \]\[ \qquad\text{O}\text{----}\text{CH}_2 \]\[ \quad /\quad |\qquad\quad |\ \]\[ \text{HO---}\text{C}\quad\text{HO---}\text{C} \]\[ \quad |\quad |\qquad\quad |\ \]\[ \text{H---}\text{C---}\text{OH}\quad\text{H---}\text{C---}\text{OH} \]\[ \qquad\qquad |\ \]\[ \qquad\text{O} \]\[ \beta\text{-D-(-)Fructofuranose} \]This image depicts the Haworth projection for the beta form of D-fructofuranose.
In simple words: The Haworth structures show glucose as a six-sided ring (pyranose) and fructose as a five-sided ring (furanose). Both have alpha and beta forms depending on where a certain \( \text{OH} \) group points.

🎯 Exam Tip: For Haworth structures, ensure correct ring size (pyranose for glucose, furanose for fructose), proper placement of all atoms and hydroxyl groups, and distinct representation of \( \alpha \) and \( \beta \) anomers (especially at C-1 for glucose and C-2 for fructose).

 

Question 15. Define proteins and give their classification.
Answer:
Proteins: Proteins are complex macromolecules essential for all life processes. They are polypeptides, meaning they are long chains made up of more than a hundred amino acid residues, linked together by peptide bonds. These chains typically have a molecular mass greater than 10,000 unified atomic mass units (u). Proteins are fundamental for the growth, repair, and development of life, found in all living cells, and form the basis of structures like hair, skin, and muscles, and act as enzymes and hormones.
Classification of Proteins:
Proteins can be classified in several ways, including based on their composition, shape, or products of hydrolysis. A common classification is based on their structure and solubility:
1. Fibrous Proteins: These are linear, elongated, and typically insoluble in water. They provide structural support. Examples include Keratin (in hair and nails), Collagen (in connective tissues), and Myosin (in muscles).
2. Globular Proteins: These are compact, spherical, and generally water-soluble. They often have active roles in metabolism. Examples include Hemoglobin (oxygen transport), Insulin (hormone), and Albumins and Globulins (blood proteins).
3. Simple Proteins: These proteins yield only amino acids upon hydrolysis. Examples include Albumins and Globulins.
4. Conjugated Proteins: These proteins consist of an amino acid part and a non-protein part (prosthetic group). Examples include Nucleoproteins (protein + nucleic acid), Glycoproteins (protein + carbohydrate), Chromoproteins (protein + pigment like hemoglobin, chlorophyll), and Phosphoproteins (protein + phosphate group).
5. Derived Proteins: These are products obtained by the partial hydrolysis of native proteins. They include Peptones and Proteoses.
In simple words: Proteins are large molecules made of many amino acids linked together, important for life. They are classified by their shape and what they are made of, like fibrous (long) or globular (round) proteins.

🎯 Exam Tip: Provide a clear definition of proteins, highlighting their composition (amino acids, peptide bonds) and size. For classification, list at least two categories with a brief description and an example for each.

RBSE Class 12 Chemistry Chapter 14 Long Answer Type Questions

 

Question 1. Write the general chemical reactions of glucose.
Answer: Glucose, being an aldopentose, exhibits various chemical reactions due to its aldehyde and alcoholic hydroxyl groups. These reactions can be broadly categorized as follows:
(a) Reactions due to alcoholic groups:
Glucose contains five hydroxyl (\( \text{OH} \)) groups: one primary alcoholic group (\( \text{CH}_2\text{OH} \)) and four secondary alcoholic groups (\( \text{CHOH} \)).
(i) Acetylation: Glucose reacts with acetic anhydride to form glucose pentaacetate, indicating the presence of five hydroxyl groups.
\[ \text{C}_6\text{H}_7\text{O}(\text{OH})_5 + 5(\text{CH}_3\text{CO})_2\text{O} \longrightarrow \text{C}_6\text{H}_7\text{O}(\text{OCOCH}_3)_5 + 5\text{CH}_3\text{COOH} \]Glucose pentaacetate is formed, confirming five hydroxyl groups.
(ii) Methylation: Glucose reacts with methyl alcohol in the presence of dry \( \text{HCl} \) gas to form methyl glucoside (an ether), showing that it can form acetals/ketals from its hemiacetal form. This forms an acetal, indicating the cyclic structure of glucose.
\[ \text{C}_6\text{H}_{11}\text{O}_5\text{OH} + \text{HOCH}_3 \xrightarrow{\text{HCl gas}} \text{C}_6\text{H}_{11}\text{O}_5\text{OCH}_3 + \text{H}_2\text{O} \]
(iii) Reaction with \( \text{PCl}_5 \): Glucose reacts with phosphorus pentachloride (\( \text{PCl}_5 \)) to form glucose pentachloride, also confirming the presence of five hydroxyl groups.
(b) Reactions due to aldehydic (-\( \text{CHO} \)) group:
The aldehyde group in glucose is responsible for its reducing properties and several characteristic reactions.
(i) Reduction: When glucose is reduced, the aldehyde group can be converted to an alcohol.
- With \( \text{Na-Hg} \) and water, the aldehyde group is reduced to a primary alcohol, yielding sorbitol (a hexahydric alcohol).
\[ \text{CH}_2\text{OH}(\text{CHOH})_4\text{CHO} + 2[\text{H}] \longrightarrow \text{CH}_2\text{OH}(\text{CHOH})_4\text{CH}_2\text{OH} \]This converts glucose to sorbitol.
- With \( \text{HI} \) and red phosphorus, glucose is completely reduced to n-hexane, suggesting a straight-chain arrangement of all six carbon atoms.
\[ \text{CHO}(\text{CHOH})_4\text{CH}_2\text{OH} \xrightarrow{\text{HI/Red P}} \text{CH}_3(\text{CH}_2)_4\text{CH}_3 \]
(ii) Oxidation: The aldehyde group can be oxidized to a carboxylic acid group.
- With mild oxidizing agents like bromine water (\( \text{Br}_2\text{/H}_2\text{O} \)), glucose is oxidized to gluconic acid (a carboxylic acid). This reaction is specific for aldehydes and confirms the presence of an aldehydic group.
\[ \text{CH}_2\text{OH}(\text{CHOH})_4\text{CHO} \xrightarrow{\text{Br}_2\text{/H}_2\text{O}} \text{CH}_2\text{OH}(\text{CHOH})_4\text{COOH} \]This produces gluconic acid.
- With strong oxidizing agents like concentrated nitric acid (\( \text{HNO}_3 \)), both the aldehyde group and the primary alcoholic group at the other end are oxidized to carboxylic acid groups, forming saccharic acid (a dicarboxylic acid).
\[ \text{CH}_2\text{OH}(\text{CHOH})_4\text{CHO} \xrightarrow{\text{HNO}_3} \text{HOOC}(\text{CHOH})_4\text{COOH} \]This forms saccharic acid.
(iii) Reaction with hydrogen cyanide: Glucose reacts with hydrogen cyanide (\( \text{HCN} \)) to form glucose cyanohydrin. This addition reaction is characteristic of carbonyl compounds and confirms the presence of a carbonyl group.
\[ \text{CHO}(\text{CHOH})_4\text{CH}_2\text{OH} + \text{HCN} \longrightarrow \text{CN}(\text{CHOH})_5\text{H}_2 \]This forms glucose cyanohydrin.
(iv) Reaction with hydroxylamine: Glucose reacts with hydroxylamine (\( \text{NH}_2\text{OH} \)) to form glucose oxime. This condensation reaction is also characteristic of carbonyl compounds.
\[ \text{CH}_2\text{OH}(\text{CHOH})_4\text{CHO} + \text{NH}_2\text{OH} \longrightarrow \text{CH}_2\text{OH}(\text{CHOH})_4\text{CH}=\text{NOH} + \text{H}_2\text{O} \]This forms glucose oxime.
(v) Action with Tollen's Reagent: As discussed earlier, glucose reduces Tollen's reagent (ammoniacal silver nitrate) to metallic silver, forming a silver mirror. This is a crucial test for the presence of an aldehyde group.
\[ \text{CHO}(\text{CHOH})_4\text{CH}_2\text{OH} + 2\text{Ag}(\text{NH}_3)_2\text{OH} \longrightarrow \text{CH}_2\text{OH}(\text{CHOH})_4\text{COOH} + 2\text{Ag}\downarrow + 4\text{NH}_3 + \text{H}_2\text{O} \]
(vi) Action with Fehling's Solution: Glucose reduces Fehling's solution (a complex of \( \text{Cu}^{2+} \) ions) to red cuprous oxide (\( \text{Cu}_2\text{O} \)) precipitate. This is another important test for reducing sugars.
\[ \text{CHO}(\text{CHOH})_4\text{CH}_2\text{OH} + 2\text{Cu}^{2+} + 4\text{OH}^- \longrightarrow \text{CH}_2\text{OH}(\text{CHOH})_4\text{COOH} + \text{Cu}_2\text{O}\downarrow + 2\text{H}_2\text{O} \]
(vii) Reaction with Phenyl Hydrazine: Glucose reacts with phenylhydrazine to form glucose phenylhydrazone. If an excess of phenylhydrazine is used, it forms an osazone (glucosazone), which is a crystalline compound used for identification.
\[ \text{CH}_2\text{OH}(\text{CHOH})_3\text{CHO} + \text{C}_6\text{H}_5\text{NHNH}_2 \longrightarrow \text{CH}_2\text{OH}(\text{CHOH})_3\text{CH}=\text{NNHC}_6\text{H}_5 + \text{H}_2\text{O} \]This initial reaction forms phenylhydrazone. With excess phenylhydrazine, further reactions occur to form glucosazone.
In simple words: Glucose undergoes many reactions because it has both alcohol and aldehyde parts. It can be reduced to sorbitol or n-hexane, oxidized to gluconic or saccharic acid, and reacts with substances like HCN, hydroxylamine, Tollen's reagent, Fehling's solution, and phenylhydrazine.

🎯 Exam Tip: When writing reactions, ensure you categorize them based on the functional group (alcoholic or aldehydic). For each reaction, state the reactant, product, and balanced equation if possible. Pay attention to specific conditions (e.g., mild vs. strong oxidizing agents) and the implications for glucose's structure.

 

Question 2. What are the main sources of cellulose and starch? Explain their structure in brief.
Answer:
(i) Cellulose \( (\text{C}_6\text{H}_{10}\text{O}_5)_n \)
Sources: Cellulose is a natural material found only in plants and is the most common organic substance on Earth. It is the main part of plant cell walls. We find cellulose in wood, cotton, and jute. For example, wood has about 50% cellulose, dry grasses have 40-45%, and cotton clothes are 90-95% cellulose. The rest consists of fats and waxes.
Structure: Cellulose is made of many \( \beta \)-D-glucose units linked together by (1-4) glycosidic bonds. X-ray studies show that these glucose units form long, straight chains that lie next to each other in bundles. These bundles are held together by hydrogen bonding between the hydroxyl groups of neighboring chains. Each D(+) glucose unit is connected to the next by C1 to C4 bonds through \( \beta \)-glycosidic linkages.
(ii) Starch \( (\text{C}_6\text{H}_{10}\text{O}_5)_n \)
Sources: Starch is a key source of carbohydrates, which are very important for human bodies because they provide energy. It is found as small grains mainly in seeds, fruits, underground stems (tubers), and roots of plants like wheat, maize, rice, and potatoes. Starch is a stored form of energy in plants.
In simple words: Cellulose comes from plants and makes up their cell walls, found in things like wood and cotton. Starch also comes from plants, found in seeds and roots, and gives us energy. Both are made of many glucose units linked together.

🎯 Exam Tip: When explaining structures of complex molecules, remember to mention the monomer units (like glucose for cellulose and starch) and the type of bonds linking them, along with their general shape (e.g., linear chains).

 

Question 3. What is the final product obtained by hydrolysis of the following:
(1) Fructose
(2) Lactose
(3) Sucrose
(4) Starch
(5) Maltose
(6) Cellulose
Answer:
(1) Fructose: Fructose is a monosaccharide, meaning it is a simple sugar. Because of this, it does not break down further into smaller units when hydrolyzed under normal conditions.
(2) Lactose: When lactose is hydrolyzed with a dilute mineral acid or the enzyme lactase, it produces equal amounts of \( \beta \)-D-glucose and \( \beta \)-D-galactose, which have similar chemical formulas but different structures.
\[ \text{C}_{12}\text{H}_{22}\text{O}_{11} \text{ (Lactose)} + \text{H}_2\text{O} \xrightarrow{\text{Hydrolysis}} \text{C}_6\text{H}_{12}\text{O}_6 \text{ (}\beta\text{-D-Glucose)} + \text{C}_6\text{H}_{12}\text{O}_6 \text{ (}\beta\text{-D-Galactose)} \]
(3) Sucrose: Hydrolysis of sucrose using a dilute mineral acid or the enzyme sucrase forms a mixture containing equal molar amounts of D-(+) glucose and D-(-) fructose. This mixture is often called invert sugar.
\[ \text{C}_{12}\text{H}_{22}\text{O}_{11} \text{ (Sucrose)} + \text{H}_2\text{O} \xrightarrow{\text{Hydrolysis}} \text{C}_6\text{H}_{12}\text{O}_6 \text{ (D-(+) Glucose)} + \text{C}_6\text{H}_{12}\text{O}_6 \text{ (D-(-) Fructose)} \]
(4) Starch: When starch is hydrolyzed by the enzyme diastase, it yields maltose. Starch is a polysaccharide that breaks down into disaccharides.
\[ 2(\text{C}_6\text{H}_{10}\text{O}_5)_n \text{ (Starch)} + n\text{H}_2\text{O} \xrightarrow{\text{Diastase}} n\text{C}_{12}\text{H}_{22}\text{O}_{11} \text{ (Maltose)} \]
(5) Maltose: Hydrolysis of maltose in the presence of an acid produces two molecules of D-glucose. Maltose is a disaccharide made of two glucose units.
\[ \text{C}_{12}\text{H}_{22}\text{O}_{11} \text{ (Maltose)} + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{C}_6\text{H}_{12}\text{O}_6 \text{ (D-Glucose)} + \text{C}_6\text{H}_{12}\text{O}_6 \text{ (D-Glucose)} \]
(6) Cellulose: Hydrolysis of cellulose, a polysaccharide, yields D-glucose units. This process often requires strong acids or specific enzymes due to cellulose's robust structure.
In simple words: When bigger sugar molecules are broken down with water (hydrolysis), they turn into smaller sugars. Fructose is already small, so it doesn't break down. Lactose breaks into glucose and galactose. Sucrose breaks into glucose and fructose. Starch breaks into maltose, and maltose breaks into two glucose units. Cellulose also breaks down into glucose.

🎯 Exam Tip: Remember the specific products of hydrolysis for common carbohydrates (disaccharides and polysaccharides), as this is a fundamental concept in biochemistry. Pay attention to the types of glucose/galactose units (alpha or beta, D or L forms).

 

Question 4. Define proteins. Explain its hydrolysis. Explain the primary and secondary structures of proteins.
Answer: The term 'Protein' was introduced by Berzelius in 1838. The word comes from the Greek word 'proteios', meaning 'primary' or 'of prime importance', highlighting their crucial role. Proteins are essential chemicals for growth, repair, and overall life development. They are found in all living cells and are the basic building blocks for life's structures and functions. In humans, proteins make up hair, skin, hemoglobin, nails, enzymes, and cells. Main sources include milk, cheese, pulses, peanuts, fish, and meat.
Composition of Proteins: Proteins are large and complex molecules (biopolymers) made of amino acids and other organic compounds, vital for life. They are carbon compounds that contain nitrogen. All proteins contain elements like carbon, hydrogen, oxygen, nitrogen, and sulfur. Some may also contain phosphorus, iodine, and small amounts of metals such as iron, copper, zinc, and manganese. Proteins are long chains (polymers) of alpha-amino acids.
Hydrolysis of Proteins: When proteins are broken down by hydrolysis, they first yield peptides, which are shorter chains of amino acids.
\( \text{Proteins} \xrightarrow{\text{Hydrolysis}} \text{Peptides} \xrightarrow{\text{Hydrolysis}} \text{Amino acids} \)
Complete hydrolysis breaks proteins down entirely into their individual alpha-amino acid units. This shows that proteins are polypeptides, meaning long chains of alpha-amino acids. Their molecular weight is typically over 10,000, making them consistent natural polymers.
Primary Structure of Proteins: This refers to the specific, unique sequence in which amino acids are linked together in a polypeptide chain. This sequence determines the protein's overall shape and function.
Secondary Structure of Proteins: This describes the local folding or coiling patterns within a polypeptide chain. Common secondary structures include alpha-helices and beta-pleated sheets, which are formed and stabilized by hydrogen bonds between amino acid residues along the backbone of the protein. These regular patterns add to the protein's stability.
In simple words: Proteins are very important building blocks for life, made from smaller units called amino acids. When proteins break down (hydrolysis), they turn back into these amino acids. Proteins have a specific order of amino acids (primary structure) and also fold into certain shapes like spirals or zig-zags (secondary structure), held together by weak bonds.

🎯 Exam Tip: When defining proteins, always mention their role in growth and repair, and clearly explain hydrolysis as the process of breaking them down into amino acids. For structures, remember to differentiate primary (sequence) from secondary (local folding) and the bonds involved.

 

Question 5. Write the functions of enzymes. Also, give their classification.
Answer:
Functions of Enzymes:
1. Enzymes act as biological catalysts, significantly speeding up biochemical reactions within living organisms without being consumed in the process. They help make reactions happen millions of times faster than they would naturally.
2. They are highly specific, meaning each enzyme usually acts on only one type of substrate or a very small group of related substrates. This ensures that only specific reactions occur in the body.
3. Enzymes operate under mild conditions of temperature and pH, which are typical for living cells.
4. They help regulate metabolic pathways by controlling the rate of various reactions, ensuring proper cell function and maintaining balance in the body.

Classification of Enzymes:
The IUB (International Union of Biochemistry) classified enzymes into six main groups in 1965 based on the type of reaction they catalyze:

In simple words: Enzymes make body reactions faster and are very specific. They are classified into six groups based on what kind of chemical change they help with, like moving parts, breaking things with water, or changing shapes.

🎯 Exam Tip: When classifying enzymes, remember the six main classes and a key characteristic reaction type for each. Knowing one example enzyme for each class can also help secure full marks.

 

Question 6. Write the names and biological functions of hormones secreted by the pituitary gland and thyroid gland.
Answer: The pituitary and thyroid glands produce several hormones that are vital for many body functions. These hormones act as chemical messengers to control growth, reproduction, and metabolism.

In simple words: The pituitary gland, located under the brain, releases hormones like FSH, LH, STH, Prolactin, ACTH, and Oxytocin that control reproduction, growth, and other body functions. The thyroid gland, in the front of the throat, releases Thyroxine, Triiodothyronine, and Calcitonin, which manage metabolism, development, and mineral balance.

🎯 Exam Tip: When describing hormones and their functions, ensure you specify the gland that secretes them, their chemical nature (protein, peptide, steroid), and a clear, concise biological effect. For deficiency/excess, mention the resulting conditions.

 

Question 7. What is vitamin B complex? Write the name of the disease caused by a deficiency of these vitamins.
Answer: Vitamin B complex is a collection of at least 13 different vitamins, often called B1, B2, B3, and so on. To avoid confusion, people now often use their chemical names instead of just B numbers. Although these vitamins do not share similar chemical structures or biological actions, they are grouped together because they often occur together in foods and share some overlapping roles in the body. A lack of these vitamins can cause various health problems, as each B vitamin plays a unique role in cell metabolism.
Diseases caused by the deficiency of various vitamins in the B complex group are summarized below:

In simple words: Vitamin B complex is a group of many B vitamins, each with a different chemical name. Not having enough of these vitamins can lead to different sicknesses. For example, a lack of Vitamin B1 causes Beri-beri, B6 deficiency can cause general weakness, and B12 deficiency leads to Pernicious Anaemia, causing confusion and face swelling.

🎯 Exam Tip: When defining vitamin B complex, emphasize it as a group of distinct vitamins rather than a single entity. For deficiency diseases, learn specific names associated with each vitamin, as this is often tested, and recall at least one major symptom for each.

 

Question 8. Explain the molecular structure of DNA.
Answer:
Molecular structure of DNA:
The DNA molecule is built from repeating units called nucleotides. Each nucleotide has three main parts:
1. A pentose sugar, specifically deoxyribose.
2. A phosphate group.
3. A nitrogenous base.
DNA contains two main types of nitrogenous bases: purine bases and pyrimidine bases.
Purine Bases: These are double-ring structures and include Adenine (A) and Guanine (G).
Pyrimidine Bases: These are single-ring structures and include Thymine (T) and Cytosine (C).
[The specific chemical structures of deoxyribose sugar, phosphate group, and the nitrogenous bases (Adenine, Guanine, Thymine, Cytosine) are fundamental to understanding DNA.]

Arrangement of constituents in DNA molecule:
When a nitrogenous base combines with a deoxyribose sugar, it forms a nucleoside. Adding a phosphate group to a nucleoside creates a nucleotide. DNA molecules are long chains, or polymers, made of many such nucleotides linked together.
\( \text{Nitrogenous base} + \text{Sugar} = \text{Nucleoside} \)
\( \text{Nucleoside} + \text{Phosphoric acid} = \text{Nucleotide} \)
\( \text{Many Nucleotides} = \text{Polynucleotide (DNA Chain)} \)

DNA is a macromolecule consisting of two long chains of deoxyribonucleotides twisted around each other to form a double helix. Each polydeoxyribonucleotide chain is formed by many deoxyribonucleotides linked by strong phosphodiester bonds. These phosphodiester bonds connect the sugar of one nucleotide to the phosphate of the next, forming the backbone of each strand. The two strands are held together by hydrogen bonds between specific pairs of bases: Adenine (A) always pairs with Thymine (T), and Guanine (G) always pairs with Cytosine (C). This specific pairing ensures the stability and function of the DNA molecule.

In simple words: DNA looks like a twisted ladder. Each side of the ladder is made of sugar and phosphate. The rungs are made of pairs of bases (A with T, and G with C). This special structure helps DNA carry genetic information.

🎯 Exam Tip: For DNA structure, remember the double helix model, the components of a nucleotide, and the specific base pairing rules (A-T, G-C). A simple diagram showing the ladder-like structure and dimensions (20Å width, 34Å per turn) can boost your score.

 

Question 9. Explain the process of protein synthesis, including its main steps.
Answer: Protein synthesis is one of the most important functions of nucleic acids (DNA and RNA). It is a complex process where living cells create proteins. About 20 different amino acids combine in various ways to create proteins. Living cells use over 200 enzymes and more than 70 types of RNA molecules for protein synthesis. DNA holds the genetic instructions for making proteins.
The synthesis of protein takes place in two main steps:
1. Transcription: This is the copying of genetic information from DNA to RNA.
Transcription is the process of copying genetic information from one strand of the DNA into a messenger RNA (mRNA) molecule. Three main types of RNA are involved in protein synthesis: messenger RNA (mRNA), ribosomal RNA (rRNA), and transfer RNA (tRNA). All of these are made from DNA.
Transcription begins when the RNA polymerase enzyme attaches to a specific start region (promoter site) on the DNA. As the enzyme moves along the DNA's structural gene, the DNA unwinds, and one strand acts as a blueprint to create a new RNA molecule. This new RNA strand is built with matching bases: A (adenine), U (uracil, which replaces thymine in DNA), C (cytosine), and G (guanine). RNA production stops when the RNA polymerase enzyme reaches a special 'stop' sequence (terminator) on the DNA.

2. Translation: This process is very complex. The formation of protein from messenger RNA (mRNA) is known as translation. During translation, the mRNA sequence is read in groups of three bases, known as a codon. Each codon acts like a code word for a specific amino acid. Transfer RNA (tRNA) molecules have special ends: one end matches the codon on mRNA, and the other end carries a specific amino acid.
Inside ribosomes (which are made of ribosomal RNA or rRNA), tRNA molecules bring the correct amino acids to the mRNA. The amino acids are then linked together to form a polypeptide chain through peptide bonds, following the sequence dictated by the mRNA. Ultimately, the order of amino acids in a protein is determined by the mRNA sequence, which itself comes from the DNA sequence. Therefore, translation, the process of assembling amino acids, takes place on ribosomes where the mRNA instructions are read.
In simple words: Protein synthesis is how cells make proteins. First, DNA's instructions are copied into RNA (Transcription). Then, this RNA information is used to build a chain of amino acids, which makes a protein (Translation). It's like DNA giving a recipe, and RNA helping to cook the meal.

🎯 Exam Tip: Clearly distinguish between transcription (DNA to RNA) and translation (RNA to protein). Remember the key molecules involved in each step (RNA polymerase in transcription, mRNA, tRNA, rRNA in translation) and the locations within the cell (nucleus for transcription, ribosomes for translation).