GSEB Class 12 Chemistry Solutions Chapter 14 Biomolecules

GSEB Class 12 Chemistry Biomolecules InText Questions and Answers

Question 1. Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six-membered ring compounds) are insoluble in water. Explain.
Answer: Benzene and cyclohexane lack polarity and hydroxyl (-OH) groups, preventing them from forming hydrogen bonds with water molecules, which makes them insoluble. In contrast, glucose and sucrose are polar and contain numerous -OH groups (five in glucose, eight in sucrose). These groups enable them to form extensive hydrogen bonds with water, thereby rendering them soluble.
In simple words: Glucose and sucrose dissolve in water because their many -OH groups allow them to form strong bonds with water. Benzene and cyclohexane don't have these groups and aren't polar, so they can't bond with water and remain insoluble.

🎯 Exam Tip: Focus on the presence or absence of polar groups (like -OH) and their ability to form hydrogen bonds with water for solubility explanations.

 

Question 2. What are the expected products of hydrolysis of lactose?
Answer: When lactose undergoes hydrolysis, it yields an equimolar combination of D-glucose and D-galactose.
In simple words: Lactose breaks down into equal parts of D-glucose and D-galactose when water is added.

🎯 Exam Tip: Remember that hydrolysis of disaccharides like lactose breaks them into their constituent monosaccharides.

 

Question 3. How do you explain the absence of aldehyde group in the pentaacetate of D-glucose?
Answer: The pentaacetate of D-glucose does not exhibit an aldehyde group because when alpha or beta glucose is treated with acetic anhydride, it produces a pentaacetyl derivative. This derivative lacks a free hydroxyl (-OH) group at the C-1 position, preventing its hydrolysis in an aqueous solution to an open-chain aldehyde form. Consequently, glucose penta-acetate cannot react with NH₂OH to form glucose oxime.
In simple words: The aldehyde group is absent in glucose pentaacetate because the C-1 hydroxyl group, which is needed for the aldehyde form, is blocked by acetylation, preventing it from showing aldehyde reactions.

🎯 Exam Tip: Understand that the C-1 hydroxyl group in the cyclic form of glucose is crucial for its open-chain aldehyde reactivity. Acetylation blocks this.

 

Question 4. The melting points and solubility in water of amino acids are generally higher than that of the corresponding halo acids. Explain.
Answer: Amino acids typically have higher melting points and greater water solubility compared to their corresponding halo acids. This is because the amino (-NH₂) group within amino acids can form extensive intermolecular hydrogen bonds, significantly contributing to these elevated physical properties.
In simple words: Amino acids melt at higher temperatures and dissolve better in water than halo acids because the -NH₂ group in amino acids allows them to form strong hydrogen bonds with each other and with water.

🎯 Exam Tip: The ability to form intermolecular hydrogen bonds (due to -NH₂ and -COOH groups) is a key factor in explaining the higher melting points and solubility of amino acids.

 

Question 5. Where does the water present in the egg go after boiling the egg?
Answer: Upon boiling an egg, the proteins undergo denaturation and subsequent coagulation, primarily facilitated by hydrogen bonding. During this process, the water originally present within the egg is absorbed or adsorbed as the globular proteins transform into an insoluble fibrous structure, effectively disappearing from its free state.
In simple words: When an egg is boiled, proteins change shape and clump together (denaturation and coagulation). The water gets trapped or bound within this new protein structure, making it seem to disappear.

🎯 Exam Tip: Relate the "disappearance" of water to the denaturation and coagulation of egg proteins, which traps or binds the water within the new protein structure.

 

Question 6. Why cannot vitamin C be stored in our body?
Answer: Vitamin C cannot be retained by the human body because it is water-soluble. Consequently, it is easily expelled through urine, necessitating a consistent intake from the diet.
In simple words: Vitamin C can't be stored because it dissolves in water and leaves the body through urine, so we need to eat it regularly.

🎯 Exam Tip: Water-solubility is the key reason for the non-storage and regular dietary requirement of Vitamin C.

 

Question 7. What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?
Answer: When a DNA nucleotide containing thymine undergoes hydrolysis, the resulting products are 2-deoxy-D-ribose (the sugar component), thymine (the nitrogenous base), and phosphoric acid.
In simple words: Breaking down a DNA nucleotide with thymine gives you deoxyribose sugar, thymine, and phosphoric acid.

🎯 Exam Tip: Remember the three components of a DNA nucleotide: a deoxyribose sugar, a nitrogenous base (thymine in this case), and phosphoric acid. Hydrolysis separates these components.

 

Question 8. When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA?
Answer: The observation that hydrolysis of RNA yields varying quantities of different bases, without any specific stoichiometric relationship, indicates that RNA lacks complementary base pairing and possesses a single-stranded structural configuration.
In simple words: Since the amounts of bases in RNA aren't paired up, it means RNA doesn't have complementary base pairs and is a single strand, unlike DNA's double helix.

🎯 Exam Tip: The lack of fixed ratios between bases in RNA is a strong indicator of its single-stranded nature, contrasting with the double-stranded, complementary base-paired structure of DNA.

GSEB Class 12 Chemistry Biomolecules Text Book Questions and Answers

Question 1. What are monosaccharides?
Answer: Monosaccharides are defined as polyhydroxy aldehydes or polyhydroxy ketones that cannot be broken down into simpler carbohydrate units through hydrolysis. Examples include glucose and fructose.
In simple words: Monosaccharides are the simplest sugars; they are either aldehydes or ketones with many -OH groups and can't be split into smaller sugar molecules.

🎯 Exam Tip: The key characteristics of monosaccharides are their polyhydroxy aldehyde/ketone structure and their inability to be hydrolyzed further.

 

Question 2. What are reducing sugars?
Answer: Reducing sugars are carbohydrates capable of reducing Fehling's solution, leading to the formation of a red precipitate of cuprous oxide (Cu₂O). This category includes all monosaccharides and the majority of disaccharides, with sucrose being a notable exception.
In simple words: Reducing sugars are carbohydrates that can react with and reduce Fehling's solution, like all monosaccharides and most disaccharides (but not sucrose).

🎯 Exam Tip: The ability to reduce Fehling's solution (or Tollen's reagent) is the defining test for reducing sugars. Remember sucrose is a non-reducing sugar.

 

Question 3. Write two main functions of carbohydrates in plants.
Answer:
(i) Cellulose functions as the primary structural component of plant cell walls.
(ii) Starch serves as the main stored food reserve within plants.
In simple words: In plants, carbohydrates like cellulose build cell walls, and starch stores energy.

🎯 Exam Tip: When discussing plant carbohydrates, always link cellulose to structure and starch to energy storage.

 

Question 4. Classify the following into monosaccharides and disaccharides. Ribose, 2 – deoxyribose, maltose, galactose, fructose and lactose.
Answer: Monosaccharides: Ribose, 2-deoxyribose, galactose, fructose
Disaccharides: Maltose, lactose
In simple words: Ribose, deoxyribose, galactose, and fructose are single sugar units (monosaccharides), while maltose and lactose are made of two sugar units (disaccharides).

🎯 Exam Tip: To classify, remember monosaccharides are single sugar units, and disaccharides are composed of two monosaccharide units linked together.

 

Question 5. What do you understand by the term glycosidic linkage?
Answer: A glycosidic linkage refers to the C-O-C ether bond that connects two monosaccharide units, forming a disaccharide or polysaccharide.
In simple words: A glycosidic linkage is the oxygen-containing bond that joins two simple sugar molecules together, like in a disaccharide.

🎯 Exam Tip: Clearly define it as an "ether linkage" and specify that it connects monosaccharide units.

 

Question 6. What is glycogen? How is it different from starch?
Answer: Glycogen is the primary storage carbohydrate in animals, often called "animal starch." Its structure resembles amylopectin but features even more extensive branching. Upon hydrolysis with dilute acids, it yields D-glucose. Starch, in contrast, is a plant storage carbohydrate, comprising a mixture of amylose (a linear α-D-glucose polymer) and amylopectin (a branched α-D-glucose polymer). While amylopectin chains contain 20-25 glucose units, glycogen chains are shorter, typically 10-14 units. Hydrolysis of starch produces dextrins of various complexities, maltose, and ultimately D-glucose.
In simple words: Glycogen is how animals store sugar, highly branched like amylopectin but with shorter branches. Starch is how plants store sugar, made of a linear part (amylose) and a branched part (amylopectin). Both break down into glucose.

🎯 Exam Tip: Highlight glycogen's role as animal storage, its highly branched structure, and compare it to starch's components (amylose and amylopectin) and their respective degrees of branching and glucose unit lengths.

 

Question 7. What are the hydrolysis products of i. sucrose and ii. lactose?
Answer:
(i) Hydrolysis of sucrose yields one molecule of glucose and one molecule of fructose.
(ii) Hydrolysis of lactose produces one molecule of glucose and one molecule of galactose.
In simple words: When sucrose is broken down, it gives glucose and fructose. When lactose is broken down, it gives glucose and galactose.

🎯 Exam Tip: Know the specific monosaccharides that make up common disaccharides (sucrose: glucose + fructose; lactose: glucose + galactose; maltose: glucose + glucose).

 

Question 8. What is the basic structural difference between starch and cellulose?
Answer: The fundamental structural distinction between starch and cellulose lies in their constituent glucose units and the type of glycosidic linkages. Starch is a polysaccharide composed of α-D-glucose units, primarily linked by 1-4 glycosidic bonds, and includes both linear (amylose) and branched (amylopectin) components. In contrast, cellulose is a linear polysaccharide made entirely of β-D-glucose units, also joined by 1-4 glycosidic linkages.
In simple words: Starch is made of alpha-glucose units, giving it a coiled structure for energy storage. Cellulose is made of beta-glucose units, forming straight chains that provide structural support.

🎯 Exam Tip: The critical difference is the type of anomeric carbon in the glucose unit (alpha vs. beta) which dictates the glycosidic linkage and the overall macromolecular structure (coiled vs. linear).

 

Question 9. What happens when D-glucose is treated with the following reagents?
i. HI
ii. Bromine water
iii. HNO3
Answer:
(i) When D-glucose reacts with HI, it undergoes reduction, yielding n-hexane and 2-iodohexane.
\( C_6H_{12}O_6 \xrightarrow{\text{HI, Reduction}} CH_3-(CH_2)_4-CH_3 + CH_3-(CH_2)_3-CHI-CH_3 \)
(n-hexane and 2-iodohexane)
(ii) Treating D-glucose with bromine water results in its oxidation to D-gluconic acid.
\( C_6H_{12}O_6 \xrightarrow{Br_2/H_2O} \)
\[ \begin{array}{c} COOH \\ | \\ H-OH \\ | \\ HO-H \\ | \\ H-OH \\ | \\ H-OH \\ | \\ CH_2OH \end{array} \]
(D-Gluconic acid)
(iii) Reaction of D-glucose with HNO₃ causes oxidation at both ends, forming D-saccharic acid.
\( C_6H_{12}O_6 \xrightarrow{HNO_3} \)
\[ \begin{array}{c} COOH \\ | \\ H-OH \\ | \\ HO-H \\ | \\ H-OH \\ | \\ H-OH \\ | \\ COOH \end{array} \]
(D-Saccharic acid)
In simple words: With strong reducing agent HI, glucose forms hexane derivatives. With mild oxidizing agent (bromine water), glucose forms gluconic acid. With strong oxidizing agent (HNO₃), glucose forms saccharic acid, oxidized at both ends.

🎯 Exam Tip: Memorize these three key reactions of glucose with HI (reduction to alkane), bromine water (oxidation to gluconic acid), and HNO₃ (oxidation to saccharic acid). They demonstrate different functional groups present in glucose.

 

Question 10. Enumerate the reactions of D – glucose which cannot be explained by its open chain structure.
Answer: The open-chain structure of D-glucose fails to account for several observed reactions:
(i) The aldehydic group in glucose does not give a positive Schiff's test, nor does it form a bisulphite addition product with NaHCO₃.
(ii) Glucose penta-acetate does not react with hydroxylamine, which suggests the absence of a free aldehyde (-CHO) group in this derivative.
In simple words: Glucose's open-chain model can't explain why it doesn't always act like an aldehyde (no Schiff's test, no bisulphite product) or why its penta-acetate form has no free aldehyde group.

🎯 Exam Tip: These "anomalous" reactions led to the proposal of cyclic structures for glucose. Focus on Schiff's test, bisulphite addition, and the penta-acetate reaction with hydroxylamine.

 

Question 11. What are essential and non-essential amino acids? Give two examples of each type.
Answer:
(i) Essential amino acids are those that the human body cannot synthesize internally and, therefore, must be obtained through dietary intake. Examples include valine and leucine.
(ii) Non-essential amino acids are those that the body's tissues can produce from other precursors, eliminating the need for their direct dietary supply. Examples include glycine and alanine.
In simple words: Essential amino acids must come from food because the body can't make them (like valine). Non-essential amino acids can be made by the body (like glycine).

🎯 Exam Tip: The distinction hinges on whether the body can synthesize them or if they must be obtained from the diet. Provide examples for each.

 

Question 12. Define the following as related to proteins.
(i) Peptide linkage
(ii) Primary structure
(iii) Denaturation.
Answer:
(i) **Peptide linkage:** This is an amide bond (—CO—NH—) formed when the carboxyl group of one alpha-amino acid condenses with the amino group of another alpha-amino acid, with the elimination of a water molecule. It links amino acids together to form polypeptide chains.
(ii) **Primary structure:** This refers to the specific, linear sequence of alpha-amino acids linked by peptide bonds within one or more polypeptide chains that constitute a protein. It's the unique order of amino acids from the N-terminus to the C-terminus.
(iii) **Denaturation:** This is the process by which a protein's natural, biologically active three-dimensional structure is disrupted and unfolded, typically by changes in pH, temperature, or the addition of certain chemicals. This disruption leads to a loss of the protein's biological function.
In simple words: Peptide linkage: The bond that connects amino acids together in a protein. Primary structure: The exact order of amino acids in a protein chain. Denaturation: When a protein loses its normal shape and function due to heat, pH changes, or other factors.

🎯 Exam Tip: For peptide linkage, specify it as an amide bond and the groups involved. For primary structure, emphasize the "linear sequence." For denaturation, focus on the disruption of 3D structure and loss of function.

 

Question 13. What are the common types of secondary structure of proteins?
Answer: The prevalent types of secondary structures found in proteins are the alpha-helix and the beta-pleated sheet.
In simple words: The two main secondary structures of proteins are the alpha-helix and the beta-pleated sheet.

🎯 Exam Tip: Just listing the names is sufficient here, but understanding what stabilizes them (hydrogen bonding) is also important.

 

Question 14. What type of bonding helps in stabilising the a helix structure of proteins?
Answer: The alpha-helix conformation in proteins is primarily stabilized by intramolecular hydrogen bonding. Specifically, these bonds form between the carbonyl oxygen (C=O) of an amino acid residue and the amide hydrogen (N-H) of an amino acid residue located four positions further along the polypeptide chain.
In simple words: The alpha-helix is held stable by hydrogen bonds forming within the same protein chain, specifically between the oxygen of one amino acid and the hydrogen of an amino acid four spots away.

🎯 Exam Tip: Clearly state "intramolecular hydrogen bonding" and specify the donor/acceptor groups (C=O and N-H) and their relative positions (n and n+4 residues).

 

Question 15. Differentiate between globular and fibrous proteins.
Answer: Proteins are categorized into two main types based on their shape and function: fibrous and globular proteins.
(i) **Fibrous proteins:** These proteins have an elongated, thread-like structure, with polypeptide chains arranged in parallel or intertwined forms. They serve primarily as structural components in animal tissues, are typically water-insoluble, and chemically inert. Examples include keratin (in skin, hair, nails) and myosin (in muscles).
(ii) **Globular proteins:** In contrast, globular proteins have polypeptide chains that are tightly folded, looped, and twisted into compact, spherical shapes. They are generally water-soluble and biologically active, performing various metabolic functions. Key examples include egg albumin, hemoglobin, enzymes, and many hormones (e.g., insulin).
In simple words: Fibrous proteins are long, tough, water-insoluble, and provide structure (like hair). Globular proteins are folded into compact, spherical shapes, are water-soluble, and carry out biological functions (like enzymes).

🎯 Exam Tip: Create a mental table of differences: shape (thread-like vs. spherical), solubility (insoluble vs. soluble), function (structural vs. functional/catalytic), and give examples for each.

 

Question 16. How do you explain the amphoteric behaviour of amino acids?
Answer: Amino acids exhibit amphoteric behavior because, in an aqueous solution, their carboxyl (-COOH) group can donate a proton while their amino (-NH₂) group can accept one. This simultaneous proton donation and acceptance leads to the formation of a dipolar ion, known as a zwitterion, which contains both a positive and a negative charge, although the molecule itself is electrically neutral. In this zwitterionic state, amino acids can react with both acids (acting as a base) and bases (acting as an acid).
In simple words: Amino acids are amphoteric because they can act as both an acid and a base. In water, they form a zwitterion, where the -COOH group loses a proton and the -NH₂ group gains one, resulting in a molecule with both positive and negative charges.

🎯 Exam Tip: Explain the formation of the zwitterion (internal proton transfer from -COOH to -NH₂) as the basis for amphoteric behavior, allowing reaction with both acids and bases.

 

Question 17. What are enzymes?
Answer: Enzymes are specialized biological catalysts synthesized by living cells. Their primary role is to accelerate the rate of biochemical reactions occurring within living organisms.
In simple words: Enzymes are natural catalysts made by living things that speed up chemical reactions in the body.

🎯 Exam Tip: Key terms: "biological catalysts," "produced by living cells," "accelerate biochemical reactions."

 

Question 18. What is the effect of denaturation on the structure of proteins?
Answer: Denaturation of a protein leads to the disruption and loss of its secondary (2°) and tertiary (3°) structures. However, the primary (1°) structure, which is the sequence of amino acids, remains unaffected during this process.
In simple words: Denaturation causes a protein to lose its folded 3D shape (secondary and tertiary structures), but the basic chain of amino acids (primary structure) stays the same.

🎯 Exam Tip: Emphasize that primary structure is preserved while secondary and tertiary (and quaternary, if applicable) structures are lost.

 

Question 19. How are vitamins classified? Name the vitamin responsible the coagulation of blood.
Answer: Vitamins are broadly categorized into two main groups: water-soluble vitamins (e.g., B-complex vitamins, vitamin C) and fat-soluble vitamins (e.g., vitamins A, D, E, and K). Fat-soluble vitamins are typically stored in the liver and adipose tissue. Among these, Vitamin K plays a crucial role in the coagulation of blood.
In simple words: Vitamins are grouped into water-soluble (like B and C) and fat-soluble (like A, D, E, K). Fat-soluble vitamins are stored in the body. Vitamin K is essential for blood clotting.

🎯 Exam Tip: Remember the two classifications and examples for each, and specifically identify Vitamin K for blood coagulation.

 

Question 20. Why are vitamin A and vitamin C essential to us? Give their important sources.
Answer: Both Vitamin A and Vitamin C are vital for human health.
**Vitamin A:** It is crucial for growth and development, especially in childhood. Even a mild deficiency can cause night blindness (nyctalopia), while prolonged lack leads to xerophthalmia, a serious eye condition.
**Vitamin C:** Its deficiency results in scurvy, characterized by symptoms such as swollen, bleeding gums and subcutaneous hemorrhages.
**Important Sources:**
(i) Vitamin A: Rich sources include fish liver oil, carrots, butter, and milk.
(ii) Vitamin C: Found abundantly in citrus fruits, amla, and green leafy vegetables.
In simple words: Vitamin A is vital for growth and good vision, preventing night blindness. Vitamin C prevents scurvy, which causes bleeding gums. Vitamin A comes from things like carrots and milk, and Vitamin C from citrus fruits.

🎯 Exam Tip: For each vitamin, link it to specific functions (growth, vision, collagen formation) and a deficiency disease, then provide a few common dietary sources.

 

Question 21. What are nucleic acids? Mention their two important functions.
Answer: Nucleic acids are fundamental biomolecules located within the nuclei of all living cells, typically associated with proteins as nucleoproteins. The two primary types are deoxyribonucleic acid (DNA) and ribonucleic acid (RNA).
Their two crucial functions include:
(i) DNA's role in the transmission of hereditary characteristics from one generation to the next.
(ii) Both DNA and RNA collectively direct the synthesis of all proteins essential for the body's growth and maintenance.
In simple words: Nucleic acids (DNA and RNA) are key molecules in cell nuclei. DNA carries genetic information from parents to offspring, and both DNA and RNA work together to build all the proteins our bodies need.

🎯 Exam Tip: Define nucleic acids as biomolecules in cell nuclei (DNA and RNA). The two main functions are heredity (DNA) and protein synthesis (DNA and RNA).

 

Question 22. What is the difference between a nucleoside and a nucleotide?
Answer: A nucleotide is a complex molecule composed of three distinct parts: a sugar molecule (either ribose or deoxyribose), a heterocyclic nitrogenous base, and a phosphoric acid group. Examples include adenosine monophosphate (AMP) and adenosine diphosphate (ADP). In contrast, a nucleoside is formed solely by the combination of a nitrogenous base with a sugar molecule, lacking the phosphate group. Adenosine and deoxyadenosine are examples of nucleosides.
In simple words: A nucleotide has three parts: a sugar, a base, and a phosphate. A nucleoside is just a sugar and a base, without the phosphate.

🎯 Exam Tip: The key difference is the presence or absence of the phosphate group. Nucleotide = Base + Sugar + Phosphate. Nucleoside = Base + Sugar.

 

Question 23. The two strands in DNA are not identical but are complementary. Explain.
Answer: The explanation for DNA's complementary, rather than identical, strands lies in its double-helix structure, as proposed by Watson and Crick in 1953. This structure features two polynucleotide strands coiled around each other, with each strand's nucleotides linked by phosphodiester bonds, forming the sugar-phosphate backbone on the exterior. The nitrogenous bases project inward, pairing specifically between the two strands through hydrogen bonds. Adenine (A) always pairs with thymine (T) via two hydrogen bonds, while guanine (G) consistently pairs with cytosine (C) through three hydrogen bonds. This highly specific base pairing ensures that the sequence of bases on one strand dictates the exact, corresponding sequence on the other strand, making them complementary rather than identical. This principle is fundamental to DNA's structure and function.
In simple words: DNA strands are complementary, not identical. This means A always pairs with T, and G always pairs with C through hydrogen bonds. So, if one strand has 'A', the other must have 'T' at that spot, making them mirror images, not duplicates.

🎯 Exam Tip: Focus on the concept of specific base pairing (A-T with two H-bonds, G-C with three H-bonds) and how this dictates the sequence on the opposite strand, leading to complementarity.

 

Question 24. Write the important structural and functional differences between DNA and RNA.
Answer: DNA serves as the genetic material within the cell nucleus, orchestrating RNA synthesis, which in turn directs protein synthesis. DNA's unique ability for self-replication ensures the transmission of hereditary information across generations. The key structural and functional distinctions between DNA and RNA are as follows:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र डीएनए की दोहरी हेलिक्स संरचना को दर्शाता है। इसमें दो पॉली न्यूक्लियोटाइड स्ट्रैंड्स एक-दूसरे के चारों ओर कुंडलित होते हैं। सीढ़ियों की तरह दिखने वाले ये स्ट्रैंड्स चीनी और फॉस्फेट से बने बैकबोन बनाते हैं, जबकि उनके बीच के 'रंग' नाइट्रोजनस बेस पेयर (A-T और G-C) द्वारा बनते हैं जो हाइड्रोजन बॉन्ड से जुड़े होते हैं। यह डीएनए की आनुवंशिक जानकारी को संग्रहीत करने की विशेषता को दर्शाता है।
**DNA:**
(i) The sugar component in DNA is deoxyribose.
(ii) Its nitrogenous bases are adenine (A), thymine (T), guanine (G), and cytosine (C).
(iii) DNA exists as a double-helix nucleotide chain.
**RNA:**
(i) The sugar component in RNA is ribose.
(ii) Its nitrogenous bases are adenine (A), uracil (U), guanine (G), and cytosine (C).
(iii) RNA typically forms a single-helix nucleotide chain.
In simple words: DNA, found in the nucleus, carries genetic info and makes copies of itself. RNA helps make proteins. DNA uses deoxyribose sugar and thymine, forming a double helix. RNA uses ribose sugar and uracil (instead of thymine), forming a single strand.

🎯 Exam Tip: Create a comparative table for DNA vs. RNA, listing differences in sugar, bases, and strandedness. Also, mention their primary functions (heredity for DNA, protein synthesis for RNA).

 

Question 25. What are the different types of RNA found in the cell?
Answer: Cells contain three main types of RNA, each with a specialized function:
(i) Ribosomal RNA (rRNA)
(ii) Messenger RNA (mRNA)
(iii) Transfer RNA (tRNA)
In simple words: There are three main kinds of RNA in cells: rRNA (ribosomal), mRNA (messenger), and tRNA (transfer).

🎯 Exam Tip: Just list the three types of RNA and their abbreviations.

GSEB Class 12 Chemistry Biomolecules Additional Important Questions and Answers

Question 1. Write the chemical name formula source and disease caused by Vitamin 'C'
Answer:
**Chemical Name:** Ascorbic acid.
**Formula:** \(C_6H_8O_6\).
**Sources:** Abundant in green vegetables, salads, and citrus fruits.
**Deficiency Disease:** Scurvy, characterized by symptoms such as spongy, bleeding gums.
In simple words: Vitamin C is called Ascorbic acid (\(C_6H_8O_6\)), found in greens and citrus fruits. Not enough Vitamin C causes scurvy, leading to bleeding gums.

🎯 Exam Tip: For vitamins, know the chemical name, formula, common sources, and deficiency disease/symptoms.

 

Question 2. The Haworth structure of a monosaccharide sugar is drawn on a chart as follows.
(i) Identify the compound and draw the Haworth structure of its anomer.
(ii) Explain the phenomenon of mutarotation.
Answer:
(i) The given Haworth structure (on page 11) represents \( \alpha -D-(+)-glucopyranose \). Its anomer, \( \beta -D-(+)-glucopyranose \), has the hydroxyl group at the anomeric carbon (C-1) pointing upwards.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र β-D-ग्लूकोपाइरोनोज (बीटा-डी-ग्लूकोज) की हॉवर्थ संरचना को दर्शाता है। इसमें C-1 पर हाइड्रॉक्सिल समूह रिंग के तल से ऊपर की ओर इंगित करता है, जो इसे α-एनोमर से अलग करता है। अन्य हाइड्रॉक्सिल समूह और CH₂OH समूह भी अपनी विशिष्ट स्थिति में दिखाए गए हैं, जो ग्लूकोज के इस एनोमेरिक रूप की चक्रीय संरचना को पूरा करते हैं।
(ii) Mutarotation is the phenomenon observed when the specific optical rotation of an aqueous solution of a pure anomer (either α-D-glucose or β-D-glucose) gradually changes over time until it reaches a constant equilibrium value. This occurs because the α and β forms interconvert through an open-chain aldehyde intermediate, establishing an equilibrium mixture of both anomers and the open-chain form.
In simple words: (i) The drawing shows alpha-D-glucose. Its anomer, beta-D-glucose, has the -OH group on the first carbon pointing up. (ii) Mutarotation is when the rotation of polarized light by a sugar solution slowly changes until it stabilizes. This happens as the alpha and beta forms of the sugar convert back and forth through an open-chain form until a balance is reached.

🎯 Exam Tip: For mutarotation, define it as the change in optical rotation, and explain that it's due to the interconversion of anomers through the open-chain form. For structures, pay attention to the orientation of the C-1 hydroxyl group for anomers.

 

Question 3. What is Zwitter ion? Give the Zwitter ion structure of a – amino acid.
Answer: A zwitterion is a dipolar ion formed through an intramolecular proton transfer within the same molecule. In amino acids, which possess both an acidic carboxyl (-COOH) group and a basic amino (-NH₂) group, a proton from the carboxyl group migrates to the amino group. This results in a molecule carrying both a positive charge (on the amino group, forming -NH₃⁺) and a negative charge (on the carboxyl group, forming -COO⁻), rendering the overall molecule electrically neutral despite having separate positive and negative centers.
\[ \overset{+}{H}_3N - CH(R) - COO^- \]
(Zwitter ion)
In simple words: A zwitterion is an amino acid form where a hydrogen ion moves from its acid part to its base part, creating a molecule with both positive and negative charges, but overall neutral.

🎯 Exam Tip: Define zwitterion, explain the intramolecular proton transfer, and show the structure with the charged amino and carboxyl groups.

 

Question 4. Give the names and functions of any four proteins.
Answer: Here are four proteins along with their respective functions:
(i) **Hemoglobin:** Responsible for transporting oxygen from the lungs to various tissues throughout the body.
(ii) **Myosin:** A key protein involved in muscle contraction and movement.
(iii) **Pepsin:** Acts as a catalyst (enzyme) in various biochemical digestive reactions.
(iv) **Keratin:** A structural protein found in hair, nails, and teeth, providing strength and protection.
In simple words: Hemoglobin carries oxygen. Myosin helps muscles move. Pepsin speeds up digestion. Keratin makes up hair, nails, and teeth.

🎯 Exam Tip: Choose proteins with distinct and clear functions (transport, movement, catalysis, structural) and briefly describe their role.

 

Question 5. What are vitamins? Name the vitamin whose deficiency causes scurvy.
Answer: Vitamins are organic micronutrients that are indispensable for the body's proper growth and overall physiological functioning. They are primarily categorized into two classes: water-soluble vitamins (e.g., B-complex vitamins, vitamin C) and fat-soluble vitamins (e.g., vitamins A, D, E). The deficiency of Vitamin C specifically leads to the disease known as scurvy.
In simple words: Vitamins are essential organic substances for normal body growth. They are either water-soluble (like Vitamin B and C) or fat-soluble (like Vitamin A, D, E). Lack of Vitamin C causes scurvy.

🎯 Exam Tip: Define vitamins, mention the two solubility-based classifications with examples, and identify Vitamin C for scurvy.

 

Question 6. Some vitamins and their deficiency diseases are given in the wrong order. Match them in the correct order.
Answer: The correct matching of vitamins with their respective deficiency diseases is as follows:

Vitamin	Deficiency diseases
i. Vitamin A	c. Night blindness
ii. Vitamin B	e. Beriberi
iii. Vitamin B6	a. Anaemia
iv. Vitamin D	f. Rickets
v. Vitamin K	b. Impaired clotting of blood
vi. Vitamin C	d. Scurvy

In simple words: This table correctly links each vitamin to the disease caused by its deficiency, such as Vitamin A with night blindness and Vitamin C with scurvy.

🎯 Exam Tip: A clear understanding of vitamin names and their associated deficiency diseases is crucial for matching questions.

 

Question 7. What are essential and non-essential amino acids? Give two examples of each type.
Answer:
**Essential Amino Acids:** These are amino acids that the human body cannot produce on its own and, therefore, must be acquired through dietary sources. Examples include valine and leucine.
**Non-essential Amino Acids:** These are amino acids that the body is capable of synthesizing from other precursor molecules, meaning they do not necessarily need to be obtained from the diet. Examples include glycine and glutamic acid.
In simple words: Essential amino acids must come from your diet (like valine), while non-essential amino acids are made by your body (like glycine).

🎯 Exam Tip: The main criterion for essential vs. non-essential amino acids is the body's ability to synthesize them. Provide two distinct examples for each category.

GSEB Class 12 Chemistry Biomolecules InText Questions and Answers

 

Question 1. Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six-membered ring compounds) are insoluble in water. Explain.
Answer: Benzene and cyclohexane lack polarity and do not possess hydroxyl (-OH) groups. Consequently, they are unable to form hydrogen bonds with water molecules, rendering them insoluble in water. In contrast, glucose and sucrose are polar molecules and contain numerous -OH groups (glucose has five, sucrose has eight). This allows them to establish extensive hydrogen bonding with water molecules, making them highly soluble.
In simple words: Glucose and sucrose dissolve in water because they have many -OH groups that can form hydrogen bonds with water. Benzene and cyclohexane don't have these groups and are not polar, so they can't form hydrogen bonds and thus don't dissolve.

🎯 Exam Tip: When explaining solubility, always refer to the molecule's polarity and its ability to form hydrogen bonds with the solvent, especially for organic compounds in water.

 

Question 2. What are the expected products of hydrolysis of lactose?
Answer: The hydrolysis of lactose yields an equimolar mixture comprising D-glucose and D-galactose.
In simple words: When lactose is broken down with water, it produces equal amounts of D-glucose and D-galactose.

🎯 Exam Tip: Remember the specific monosaccharide units that constitute common disaccharides like lactose, sucrose, and maltose, as their hydrolysis products are frequently tested.

 

Question 3. How do you explain the absence of aldehyde group in the pentaacetate of D-glucose?
Answer: When either α- or β-glucose is treated with acetic anhydride, it forms a pentaacetyl derivative. This derivative lacks a free -OH group at carbon-1 and cannot undergo hydrolysis in an aqueous solution to yield the open-chain aldehyde form. Therefore, glucose penta-acetate does not react with NH2OH to produce glucose oxime, confirming the absence of a free aldehyde group.
In simple words: Glucose penta-acetate doesn't show an aldehyde group because the -OH group at carbon-1, which is necessary for the open-chain aldehyde form, is acetylated, preventing it from reacting like a typical aldehyde.

🎯 Exam Tip: Understanding the cyclic versus open-chain forms of glucose and how derivatives like pentaacetate lock specific functional groups is key to explaining its reactions.

 

Question 4. The melting points and solubility in water of amino acids are generally higher than that of the corresponding halo acids. Explain.
Answer: Amino acids possess an -NH2 group, which facilitates the formation of intermolecular hydrogen bonds. Additionally, in aqueous solutions, amino acids exist as zwitterions (dipolar ions) with strong electrostatic forces between them. These factors lead to significantly higher melting points and increased solubility in water for amino acids compared to their corresponding halo acids.
In simple words: Amino acids have higher melting points and are more water-soluble than similar halo acids because their amino and carboxyl groups allow them to form strong hydrogen bonds and exist as charged zwitterions, leading to stronger intermolecular attractions.

🎯 Exam Tip: The zwitterionic nature and strong hydrogen bonding capacity are fundamental properties of amino acids that explain many of their physical characteristics, including melting point and solubility.

 

Question 5. Where does the water present in the egg go after boiling the egg?
Answer: When an egg is boiled, the proteins undergo denaturation, which is followed by coagulation. This process likely involves the rearrangement of hydrogen bonds. The water molecules originally present within the egg become absorbed or adsorbed during this denaturation, effectively disappearing as the globular proteins transform into insoluble fibrous proteins, leading to a solidified structure.
In simple words: When an egg boils, its proteins denature and coagulate, trapping the water molecules within the newly formed insoluble protein network, making the egg solidify.

🎯 Exam Tip: Protein denaturation is a key concept; remember that it involves a change in protein structure (2°, 3°, 4°), often affecting solubility and function, while the primary structure remains intact.

 

Question 6. Why cannot vitamin C be stored in our body?
Answer: Vitamin C is a water-soluble vitamin, meaning it readily dissolves in water. As a result, it is easily excreted from the body via urine and cannot be stored for extended periods. This is why it must be consumed regularly through the diet to maintain adequate levels.
In simple words: Vitamin C cannot be stored in the body because it dissolves in water and is quickly flushed out through urine, requiring constant dietary replenishment.

🎯 Exam Tip: Differentiate between water-soluble and fat-soluble vitamins; water-soluble vitamins (like C and B complex) need regular intake, while fat-soluble vitamins (A, D, E, K) can be stored in the body.

 

Question 7. What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?
Answer: The hydrolysis products of a DNA nucleotide containing thymine would be: 2-deoxy-D-ribose (sugar), Thiamine (the nitrogenous base), and Phosphoric acid.
In simple words: When a DNA nucleotide with thymine breaks down, it yields deoxyribose sugar, thymine base, and phosphoric acid.

🎯 Exam Tip: Know the three components of a nucleotide: a pentose sugar (deoxyribose in DNA, ribose in RNA), a nitrogenous base, and a phosphate group. Also, remember the specific bases for DNA (A, T, C, G) and RNA (A, U, C, G).

 

Question 8. When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA?
Answer: This observation implies that complementary base pairing, which is characteristic of DNA's double helix, is absent in RNA. Therefore, this fact strongly suggests that RNA typically possesses a single-strand structure, unlike the double-stranded DNA where base quantities follow Chargaff's rules (A=T, C=G).
In simple words: The unequal amounts of bases after RNA hydrolysis indicate that RNA is a single-stranded molecule and does not have complementary base pairing like DNA.

🎯 Exam Tip: The absence of Chargaff's base pairing rules (A=U, C=G in equal amounts) in RNA is a key indicator of its predominantly single-stranded nature, a critical structural difference from DNA.

 

GSEB Class 12 Chemistry Biomolecules Text Book Questions and Answers

 

Question 1. What are monosaccharides?
Answer: Monosaccharides are the simplest forms of carbohydrates. They are polyhydroxy aldehydes or polyhydroxy ketones that cannot be further hydrolyzed into smaller carbohydrate units. Examples include glucose and fructose.
In simple words: Monosaccharides are simple sugars that cannot be broken down further, like glucose or fructose.

🎯 Exam Tip: Always remember that monosaccharides are the basic building blocks of carbohydrates, and they are defined by their inability to be hydrolyzed further.

 

Question 2. What are reducing sugars?
Answer: Reducing sugars are carbohydrates capable of reducing Fehling's solution to form a red precipitate of copper (I) oxide (\( \text{Cu}_2\text{O} \)). All monosaccharides exhibit this property, and most disaccharides, with the notable exception of sucrose, are also reducing sugars.
In simple words: Reducing sugars are carbohydrates that can reduce Fehling's solution, like all monosaccharides and most disaccharides (except sucrose).

🎯 Exam Tip: The presence of a free aldehyde or ketone group (or one that can isomerize to such) is characteristic of reducing sugars. Sucrose is a crucial non-reducing sugar example.

 

Question 3. Write two main functions of carbohydrates in plants.
Answer:
• Cellulose functions as the primary structural material, forming the cell walls of plants.
• Starch serves as the major reserved food material, storing energy for the plants.
In simple words: In plants, carbohydrates like cellulose provide structure for cell walls, and starch stores energy.

🎯 Exam Tip: Understand the dual role of carbohydrates in biology: structural support (e.g., cellulose) and energy storage (e.g., starch, glycogen).

 

Question 4. Classify the following into monosaccharides and disaccharides. Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose.
Answer:
Monosaccharides: Ribose, 2-deoxyribose, galactose, fructose
Disaccharides: Maltose, lactose
In simple words: Ribose, deoxyribose, galactose, and fructose are simple sugars (monosaccharides), while maltose and lactose are made of two simple sugars (disaccharides).

🎯 Exam Tip: To classify carbohydrates, recall that monosaccharides are single sugar units, and disaccharides are formed from two monosaccharide units joined together.

 

Question 5. What do you understand by the term glycosidic linkage?
Answer: A glycosidic linkage refers to the C-O-C (ether) bond that connects two monosaccharide units, allowing them to join and form a disaccharide or polysaccharide. This bond is typically formed between the anomeric carbon of one monosaccharide and a hydroxyl group of another.
In simple words: A glycosidic linkage is the ether bond that connects two sugar molecules, forming larger carbohydrates like disaccharides.

🎯 Exam Tip: Recognize the glycosidic linkage as the defining covalent bond that links monosaccharides in complex carbohydrates. It's an ether linkage formed by dehydration.

 

Question 6. What is glycogen? How is it different from starch?
Answer: Glycogen is the primary form of carbohydrate storage in animal bodies, often referred to as "animal starch." Structurally, glycogen is similar to amylopectin (a component of starch) but exhibits a higher degree of branching. Upon hydrolysis with dilute acids, both glycogen and starch yield D-glucose.
Starch is a mixture of two polymers: amylose and amylopectin. Amylose is a linear polymer of α-D-glucose, whereas amylopectin is a branched polymer. Amylopectin chains typically consist of 20-25 glucose units, while glycogen chains are more extensively branched, containing 10-14 glucose units between branches. Hydrolysis of starch yields dextrins of varying complexity, followed by maltose, and finally D-glucose.
In simple words: Glycogen is animal starch, more branched than amylopectin. Starch is a mix of linear amylose and branched amylopectin, both provide energy, but glycogen is stored in animals and starch in plants.

🎯 Exam Tip: Focus on the structural differences (branching) and physiological roles (animal vs. plant storage) to distinguish between glycogen and starch.

 

Question 7. What are the hydrolysis products of i. sucrose and ii. lactose?
Answer:
i. Sucrose on hydrolysis yields one molecule each of glucose and fructose.
ii. Lactose on hydrolysis yields one molecule each of glucose and galactose.
In simple words: Sucrose breaks down into glucose and fructose, while lactose breaks down into glucose and galactose.

🎯 Exam Tip: Memorize the monosaccharide components of common disaccharides (sucrose: glucose + fructose; lactose: glucose + galactose; maltose: glucose + glucose).

 

Question 8. What is the basic structural difference between starch and cellulose?
Answer: Starch is a polysaccharide composed of α-D-glucose units, linked by 1-4 glycosidic bonds. It consists of two fractions: water-soluble amylose (a linear polymer) and water-insoluble amylopectin (a branched-chain polymer). Both amylose and amylopectin are polymers of α-D-glucose.
Cellulose, conversely, is a polysaccharide made up of β-D-glucose units, linked by 1-4 glycosidic bonds. It is a linear polymer of β-D-glucose, characterized by its straight, unbranched chains. The difference in the anomeric form (α vs. β) of glucose units fundamentally distinguishes their structures and properties.
In simple words: Starch is made of alpha-glucose and can be branched or linear, while cellulose is a linear polymer of beta-glucose. This small difference in how the glucose units link makes them have very different properties.

🎯 Exam Tip: The key structural difference lies in the anomeric carbon linkage: α-glycosidic bonds in starch make it digestible for humans, while β-glycosidic bonds in cellulose make it fibrous and indigestible for most.

 

Question 9. What happens when D-glucose is treated with the following reagents?
i. HI
ii. Bromine water
iii. HNO3
Answer:
i. When D-glucose is treated with HI (hydroiodic acid) under prolonged heating, it undergoes reduction and forms n-hexane and 2-iodohexane.
\( \text{C}_6\text{H}_{12}\text{O}_6 \xrightarrow{\text{HI, Reduction}} \text{CH}_3-(\text{CH}_2)_4-\text{CH}_3 + \text{CH}_3-(\text{CH}_2)_3-\text{CHI}-\text{CH}_3 \)
(Glucose) (n-hexane) (2-iodohexane)
ii. When D-glucose is treated with bromine water (a mild oxidizing agent), it oxidizes the aldehyde group to a carboxylic acid group, forming D-gluconic acid.
\( \text{C}_6\text{H}_{12}\text{O}_6 \xrightarrow{\text{Br}_2/\text{H}_2\text{O}} \text{COOH}-(\text{CHOH})_4-\text{CH}_2\text{OH} \)
(Glucose) (D-Gluconic acid)
iii. When D-glucose is treated with concentrated nitric acid (\( \text{HNO}_3 \)) (a strong oxidizing agent), both the aldehyde group and the primary alcohol group at C-6 are oxidized to carboxylic acid groups, forming D-saccharic acid (glucaric acid).
\( \text{C}_6\text{H}_{12}\text{O}_6 \xrightarrow{\text{HNO}_3} \text{COOH}-(\text{CHOH})_4-\text{COOH} \)
(Glucose) (D-Saccharic acid)
In simple words: HI reduces glucose to hexanes, bromine water oxidizes it to gluconic acid, and strong nitric acid oxidizes it further to saccharic acid.

🎯 Exam Tip: Remember these characteristic reactions of glucose as they reveal its open-chain structure and the presence of aldehyde and hydroxyl groups. HI confirms the straight chain, bromine water confirms the aldehyde, and nitric acid confirms the primary alcohol.

 

Question 10. Enumerate the reactions of D-glucose which cannot be explained by its open chain structure.
Answer: The following reactions of D-glucose cannot be fully explained by its simple open-chain aldehyde structure:
(i) Glucose does not give a positive Schiff's test, which is typically observed for aldehydes. Furthermore, it does not form the bisulphite addition product with \( \text{NaHCO}_3 \).
(ii) The penta-acetate derivative of glucose does not react with hydroxylamine, indicating the absence of a free -CHO (aldehyde) group in the penta-acetate, even though glucose itself possesses one.
(iii) Glucose exists in two crystalline forms, α- and β-glucose, which differ in their specific rotation and undergo mutarotation in aqueous solution, reaching an equilibrium. This behavior is inconsistent with a simple open-chain structure.
In simple words: Glucose doesn't give a Schiff's test, its penta-acetate form doesn't react with hydroxylamine, and it exists in two forms that interconvert (mutarotation). These observations contradict its simple open-chain aldehyde structure.

🎯 Exam Tip: These anomalies in glucose's reactions were crucial evidence that led to the postulation of its cyclic (hemiacetal) structure, which better explains its properties like mutarotation and non-reactivity with certain aldehyde tests.

 

Question 11. What are essential and non-essential amino acids? Give two examples of each type.
Answer:
(i) Essential amino acids are those that cannot be synthesized by the human body and, therefore, must be acquired through dietary intake. Examples include Valine and Leucine.
(ii) Non-essential amino acids are those that the body's tissues can synthesize from other compounds. Examples include Glycine and Alanine.
In simple words: Essential amino acids must come from food because the body can't make them (like Valine, Leucine), while non-essential amino acids can be made by the body (like Glycine, Alanine).

🎯 Exam Tip: A good way to remember this is "essential" means it's essential to get it from your diet. Knowing a couple of examples for each category is often sufficient.

 

Question 12. Define the following as related to proteins.
(i) Peptide linkage
(ii) Primary structure
(iii) Denaturation.
Answer:
(i) Peptide linkage: A peptide linkage is an amide bond that connects two α-amino acid units within a protein. Chemically, it is formed through a condensation reaction between the carboxyl (-COOH) group of one α-amino acid and the amino (-NH2) group of another α-amino acid, with the elimination of a water molecule.
(ii) Primary structure: The primary structure of a protein refers to the specific, linear sequence in which various α-amino acids are linked together by peptide bonds within one or more polypeptide chains. This precise order of amino acids defines the protein's fundamental identity.
(iii) Denaturation: Denaturation is the process by which the complex three-dimensional native structure of a protein is disrupted or destroyed. This typically occurs due to changes in environmental conditions such as pH, temperature, or the addition of certain salts, leading to the loss of the protein's biological activity without breaking the primary peptide bonds.
In simple words: A peptide linkage is the bond joining amino acids. The primary structure is the unique sequence of amino acids in a protein. Denaturation is when a protein loses its 3D shape and function due to external factors like heat or pH.

🎯 Exam Tip: For denaturation, remember that only the secondary, tertiary, and quaternary structures are affected, while the primary structure (amino acid sequence) remains intact. Peptide linkages are the backbone of proteins.

 

Question 13. What are the common types of secondary structure of proteins?
Answer: The common types of secondary structure observed in proteins are the α-helix and the β-pleated sheet.
In simple words: The two main types of protein secondary structures are the alpha-helix and the beta-pleated sheet.

🎯 Exam Tip: Secondary structures are stabilized primarily by hydrogen bonding between the backbone atoms of the polypeptide chain, forming repeating local conformations.

 

Question 14. What type of bonding helps in stabilising the α-helix structure of proteins?
Answer: The α-helix structure of proteins is primarily stabilized by intramolecular hydrogen bonding. These hydrogen bonds form specifically between the carbonyl oxygen (C=O) of one amino acid residue and the amide hydrogen (N-H) of the amino acid residue located four positions further along the polypeptide chain.
In simple words: The alpha-helix structure in proteins is held stable by hydrogen bonds forming within the same chain, specifically between the oxygen of one amino acid's carbonyl group and the hydrogen of an amide group four amino acids away.

🎯 Exam Tip: Understand that intramolecular hydrogen bonding within the polypeptide backbone, rather than side chains, is the defining characteristic for stabilizing α-helices and β-sheets.

 

Question 15. Differentiate between globular and fibrous proteins.
Answer: Proteins are broadly classified into two main types based on their shape and functional characteristics: fibrous proteins and globular proteins.
(i) Fibrous proteins: These molecules are thread-like and elongated, often arranging themselves side-by-side to form strong fibers. Their polypeptide chains are extended and intertwine or lie parallel to neighboring chains, resembling strands in a rope or fabric threads. Fibrous proteins typically serve as the primary structural material in animal tissues, are generally insoluble in water, and are chemically less reactive. Examples include keratin (found in skin, hair, nails) and myosin (in muscles).
(ii) Globular proteins: In contrast, globular proteins have polypeptide chains that are tightly folded, looped, and twisted, resulting in molecules that are more or less spherical or compact. These proteins are generally soluble in water and exhibit chemical reactivity. They perform various biological functions such as enzymatic catalysis, transport, and hormonal regulation. Examples include egg albumin, hemoglobin, enzymes, and certain hormones (e.g., insulin).
In simple words: Fibrous proteins are long, thread-like, water-insoluble, and provide structural support (like keratin). Globular proteins are compact, spherical, water-soluble, and perform active roles like enzymes or transport (like hemoglobin).

🎯 Exam Tip: Key distinguishing features are shape (thread-like vs. spherical), solubility in water (insoluble vs. soluble), and primary function (structural vs. dynamic/metabolic).

 

Question 16. How do you explain the amphoteric behaviour of amino acids?
Answer: In an aqueous solution, amino acids exhibit amphoteric behavior because they can act as both acids and bases. This is due to the presence of both an acidic carboxyl group (-COOH) and a basic amino group (-NH2) within the same molecule. The carboxyl group can donate a proton, and the amino group can accept a proton, leading to the formation of a dipolar ion known as a zwitterion. Although electrically neutral overall, a zwitterion carries both a positive and a negative charge, allowing it to react with acids (by accepting a proton) and with bases (by donating a proton).
In simple words: Amino acids are amphoteric because they contain both an acidic carboxyl group and a basic amino group, forming a zwitterion in water. This allows them to react with both acids and bases.

🎯 Exam Tip: The zwitterion form is critical for understanding the amphoteric nature of amino acids and their behavior at different pH levels, including the concept of the isoelectric point.

 

Question 17. What are enzymes?
Answer: Enzymes are biological catalysts produced by living cells. Their primary function is to accelerate or catalyze specific biochemical reactions that occur within living organisms, making them essential for virtually all metabolic processes.
In simple words: Enzymes are natural catalysts made by cells that speed up biochemical reactions in living things.

🎯 Exam Tip: Remember that enzymes are highly specific to their substrates and reactions, and their activity is sensitive to factors like temperature and pH. Most enzymes are proteins.

 

Question 18. What is the effect of denaturation on the structure of proteins?
Answer: During denaturation, the higher-order structures of proteins – specifically the secondary (2°) and tertiary (3°) structures – are disrupted and destroyed. However, the primary (1°) structure, which is the sequence of amino acids linked by peptide bonds, remains intact. This loss of complex folding often results in the loss of the protein's biological activity.
In simple words: Denaturation breaks down a protein's 3D shape and its secondary and tertiary structures, but it leaves the basic amino acid sequence (primary structure) untouched.

🎯 Exam Tip: Emphasize that denaturation involves loss of 3D conformation and biological function, but *not* the peptide bonds of the primary structure. This is a common point of confusion.

 

Question 19. How are vitamins classified? Name the vitamin responsible the coagulation of blood.
Answer: Vitamins are broadly classified into two main categories based on their solubility:
(i) Water-soluble vitamins: These include the vitamin B-complex group and vitamin C. They are not stored in the body and need to be consumed regularly.
(ii) Fat-soluble vitamins: These include vitamins A, D, E, and K. They can be stored in the body's liver and adipose (fat-storing) tissues.
Vitamin K is the specific vitamin responsible for initiating the coagulation (clotting) of blood.
In simple words: Vitamins are classified as either water-soluble (B-complex, C) or fat-soluble (A, D, E, K). Vitamin K is crucial for blood clotting.

🎯 Exam Tip: Knowing the solubility classification helps understand how vitamins are absorbed, stored, and excreted. Always link specific vitamins to their key physiological functions, like Vitamin K for clotting.

 

Question 20. Why are vitamin A and vitamin C essential to us? Give their important sources.
Answer: Vitamin A and Vitamin C are essential for several vital bodily functions:
Vitamin A: In childhood, a deficiency in Vitamin A can stunt growth and is thus considered a growth-promoting factor. Mild deficiency can lead to night blindness (nyctalopia), while prolonged deficiency can result in xerophthalmia (severe dry eye leading to blindness).
Vitamin C: A deficiency of Vitamin C causes scurvy, characterized by a tendency to hemorrhage, swollen bleeding gums, and bleeding into the skin and deeper tissues.
Sources of vitamin A: Rich sources include fish liver oil, carrots, butter, and milk.
Sources of vitamin C: Abundant in citrus fruits, amla (Indian gooseberry), and green leafy vegetables.
In simple words: Vitamin A is vital for growth and good vision, preventing night blindness. Vitamin C is crucial for healthy gums and preventing scurvy. Vitamin A sources are fish oil, carrots; Vitamin C sources are citrus fruits, amla.

🎯 Exam Tip: When discussing vitamins, always connect their importance to specific biological roles and the diseases caused by their deficiencies. Also, be prepared to list common dietary sources.

 

Question 21. What are nucleic acids? Mention their two important functions.
Answer: Nucleic acids are essential biomolecules found within the nuclei of all living cells, typically existing as nucleoproteins. The two primary types are deoxyribonucleic acid (DNA) and ribonucleic acid (RNA).
Two important functions of nucleic acids are:
• DNA is solely responsible for the transmission of hereditary characteristics from one generation to the next.
• Both DNA and RNA play crucial roles in synthesizing all the proteins required for the body's growth, repair, and maintenance.
In simple words: Nucleic acids (DNA and RNA) are biomolecules in cell nuclei. DNA carries genetic information for heredity, and both DNA and RNA work together to make proteins for growth and maintenance.

🎯 Exam Tip: Highlight the distinction in their primary functions: DNA for heredity, and RNA for protein synthesis (under DNA's direction). This is a foundational concept in molecular biology.

 

Question 22. What is the difference between a nucleoside and a nucleotide?
Answer: A nucleotide is a fundamental building block of nucleic acids, composed of three distinct parts: a sugar molecule (either ribose or deoxyribose), a heterocyclic nitrogenous base (adenine, guanine, cytosine, thymine, or uracil), and a phosphoric acid group. Examples include adenosine monophosphate (AMP) and adenosine diphosphate (ADP).
A nucleoside, on the other hand, is a simpler compound formed solely by the combination of a nitrogenous base and a sugar molecule, lacking the phosphoric acid group. Examples include adenosine and deoxyadenosine.
In simple words: A nucleotide has three parts: a sugar, a base, and a phosphate. A nucleoside has only two parts: a sugar and a base, without the phosphate.

🎯 Exam Tip: The key differentiator is the presence or absence of the phosphate group. Nucleotides are "nucleosides with a phosphate." This distinction is critical for understanding nucleic acid structure.

 

Question 23. The two strands in DNA are not identical but are complementary. Explain.
Answer:
Structure of DNA: The three-dimensional structure of DNA was elucidated by James Watson and Francis Crick in 1953. They proposed that a DNA chain consists of two polynucleotide strands coiled around each other, forming a double helix. The nucleotides within each strand are connected by phosphodiester bonds, which form the sugar-phosphate backbone.
The nitrogenous bases (purines: adenine, guanine; pyrimidines: cytosine, thymine) project inwards from this backbone. The two strands are held together by hydrogen bonding between these bases. This hydrogen bonding is highly specific due to the precise structures of the bases, allowing only one mode of pairing: adenine (A) always pairs with thymine (T) via two hydrogen bonds, and guanine (G) always pairs with cytosine (C) via three hydrogen bonds. Other combinations are energetically less favorable and do not occur in normal DNA.
Therefore, the two strands of DNA molecules are complementary because the sequence of bases in one strand dictates the sequence of bases in the other. For instance, if one strand has 'A', the opposite strand must have 'T'. This specific base pairing is the most important principle governing the structure of the DNA molecule. Watson, Crick, and Wilkins were awarded the Nobel Prize in 1962 for this achievement.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र डीएनए की दोहरी हेलिक्स संरचना को दर्शाता है। इसमें दो पॉलीपेप्टाइड स्ट्रैंड एक दूसरे के चारों ओर घुमावदार रूप में दिखाए गए हैं। प्रत्येक स्ट्रैंड में शर्करा-फॉस्फेट की एक रीढ़ होती है जिससे नाइट्रोजनस बेस (जैसे A, T, C, G) जुड़े होते हैं। ये बेस एक दूसरे से हाइड्रोजन बॉन्ड (A-T के बीच दो, G-C के बीच तीन) के माध्यम से जुड़कर दो स्ट्रैंड्स को एक साथ रखते हैं, जिससे एक सीढ़ी जैसी संरचना बनती है जो मुड़ी हुई होती है। चित्र में 34Å की कुल हेलिक्स पिच और प्रति मोड़ 10 बेस पेयर दर्शाए गए हैं।
In simple words: DNA has two strands that are complementary, not identical. This means that a specific base on one strand (like A) always pairs with a specific base on the other (like T), and C with G, held together by hydrogen bonds, making their sequences perfectly matching.

🎯 Exam Tip: Understanding base pairing rules (Chargaff's rules) and their role in DNA's double helix and complementary nature is fundamental. The concept of complementary strands is vital for DNA replication and transcription.

 

Question 24. Write the important structural and functional differences between DNA and RNA.
Answer:
**DNA:**
• DNA contains 2-deoxyribose as its sugar unit.
• The nitrogenous bases found in DNA are adenine (A), thymine (T), guanine (G), and cytosine (C).
• A DNA nucleotide chain typically forms a double helix structure.
• DNA is the primary constituent of the genetic material in the cell nucleus. It controls RNA synthesis, which in turn directs protein synthesis.
• DNA possesses the unique property of self-replication, enabling it to create identical copies of itself, which is crucial for maintaining heredity across generations.
**RNA:**
• RNA contains ribose as its sugar unit.
• The nitrogenous bases found in RNA are adenine (A), uracil (U), guanine (G), and cytosine (C).
• An RNA nucleotide chain is typically single-stranded, although it can fold into complex secondary and tertiary structures.
• RNA plays a direct role in protein synthesis, translating the genetic information from DNA into functional proteins.
In simple words: DNA uses deoxyribose sugar and thymine, is double-stranded, and stores genetic information. RNA uses ribose sugar and uracil, is typically single-stranded, and helps in making proteins.

🎯 Exam Tip: Crucial differences include the sugar (deoxyribose vs. ribose), a nitrogenous base (thymine vs. uracil), and the number of strands (double helix vs. single strand). These differences directly relate to their distinct functions.

 

Question 25. What are the different types of RNA found in the cell?
Answer: There are three primary types of RNA found within a cell, each playing a specific role in protein synthesis:
• Ribosomal RNA (rRNA): A structural component of ribosomes, where protein synthesis occurs.
• Messenger RNA (mRNA): Carries genetic information from DNA in the nucleus to the ribosomes in the cytoplasm, serving as a template for protein synthesis.
• Transfer RNA (tRNA): Transports specific amino acids to the ribosome during protein synthesis, matching them to the codons on mRNA.
In simple words: Cells have three main types of RNA: ribosomal RNA (rRNA) for ribosome structure, messenger RNA (mRNA) for carrying genetic code, and transfer RNA (tRNA) for delivering amino acids during protein making.

🎯 Exam Tip: Understand the distinct roles of rRNA, mRNA, and tRNA in the central dogma of molecular biology (DNA -> RNA -> Protein). Their functions are interdependent.

 

GSEB Class 12 Chemistry Biomolecules Additional Important Questions and Answers

 

Question 1. Write the chemical name, formula, source, and disease caused by Vitamin 'C'.
Answer:
• The chemical name of Vitamin 'C' is Ascorbic acid.
• The chemical formula for Vitamin 'C' is \( \text{C}_6\text{H}_8\text{O}_6 \).
• Important sources of Vitamin 'C' include green vegetables, various salads, and citrus fruits.
• A deficiency of Vitamin 'C' causes the disease Scurvy, which manifests as spongy and bleeding gums, along with bleeding into the skin and deeper tissues.
In simple words: Vitamin C, chemically called Ascorbic acid (\( \text{C}_6\text{H}_8\text{O}_6 \)), comes from green vegetables and citrus fruits. Lack of it causes Scurvy, leading to bleeding gums and skin issues.

🎯 Exam Tip: For vitamins, always know their chemical name, formula (if specified), key dietary sources, and the specific deficiency disease they prevent. This is standard biological chemistry knowledge.

 

Question 2. The Haworth structure of a monosaccharide sugar is drawn on a chart as follows. (i) Identify the compound and draw the Haworth structure of its anomer. (ii) Explain the phenomenon of mutarotation.
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): दिया गया चित्र α-D-ग्लूकोज के हॉवर्थ संरचना को दर्शाता है, जिसमें एक छह-सदस्यीय रिंग होती है। इस संरचना में, कार्बन-1 पर हाइड्रॉक्सिल (-OH) समूह रिंग के नीचे की ओर (यानी α स्थिति में) स्थित है। कार्बन-6 पर -CH2OH समूह रिंग के ऊपर की ओर होता है। यह संरचना ग्लूकोज के एक विशेष साइक्लिक आइसोमर को प्रस्तुत करती है।
(i) The compound identified from the given Haworth structure is α-D-glucose (specifically, α-D-(+)-glucopyranose).
The Haworth structure of its anomer, β-D-glucose, would have the -OH group at C-1 positioned above the plane of the ring (or opposite to the -CH2OH group at C-6), whereas in α-D-glucose, the -OH group at C-1 is below the plane.

(ii) Mutarotation is the phenomenon observed when the specific optical rotation of an aqueous solution of either pure α-D-glucose or pure β-D-glucose gradually changes over time until it reaches a constant equilibrium value. This change occurs because the cyclic α and β forms interconvert through a transient open-chain aldehyde form in solution. Eventually, an equilibrium mixture of α-D-glucose, β-D-glucose, and a small amount of the open-chain form is established, leading to a stable specific rotation for the solution.
In simple words: The given structure is alpha-D-glucose. Its anomer, beta-D-glucose, has the C-1 -OH group positioned differently. Mutarotation is the process where the optical rotation of glucose solutions changes until equilibrium is reached between its alpha, beta, and open-chain forms.

🎯 Exam Tip: Be able to draw and differentiate between the α and β anomers of glucose using Haworth projections. Mutarotation is a key characteristic of cyclic monosaccharides and indicates their dynamic equilibrium in solution.

 

Question 3. What is Zwitter ion? Give the Zwitter ion structure of a – amino acid.
Answer: A zwitterion is a dipolar ion formed when a molecule contains both a positively charged group and a negatively charged group, making the molecule electrically neutral overall. In the context of amino acids, this occurs due to an intramolecular proton transfer: the acidic carboxyl (-COOH) group donates a proton to the basic amino (-NH2) group within the same molecule. This results in the formation of a positively charged ammonium group (-\(\text{NH}_3^+\)) and a negatively charged carboxylate group (-\(\text{COO}^-\)).
The zwitterion structure of an α-amino acid can be represented as:
\( \text{H}_3\text{N}^+ - \text{CH}(\text{R}) - \text{COO}^- \)
Where 'R' represents the side chain.
In simple words: A zwitterion is a molecule, like an amino acid, that has both a positive and negative charge on different parts, making it neutral overall. For amino acids, the amino group takes a proton and the carboxyl group loses one.

🎯 Exam Tip: The zwitterionic form is the predominant structure of amino acids in neutral aqueous solutions. Understanding its formation is crucial for explaining amino acid properties like amphoteric behavior and isoelectric point.

 

Question 4. Give the names and functions of any four proteins.
Answer: Here are four proteins and their key functions:
• Hemoglobin: Responsible for the transport of oxygen from the lungs to various tissues throughout the body.
• Myosin: A motor protein essential for the contraction and motion of muscles.
• Pepsin: An enzyme that acts as a catalyst in biochemical reactions, specifically breaking down proteins in the stomach.
• Keratin: A structural protein found in hairs, nails, and teeth, providing strength and protection.
In simple words: Hemoglobin carries oxygen, Myosin helps muscles move, Pepsin digests proteins, and Keratin provides structure to hair and nails.

🎯 Exam Tip: Be prepared to list various protein functions (e.g., transport, structural, enzymatic, hormonal) and provide specific examples for each. This demonstrates a broad understanding of protein diversity.

 

Question 5. What are vitamins? Name the vitamin whose deficiency causes scurvy.
Answer: Vitamins are organic compounds that are essential for the normal growth, development, and metabolic functions of the body. They are required in small quantities and cannot typically be synthesized by the body in sufficient amounts, hence they must be obtained through the diet.
Vitamins are generally categorized into two types:
• Water-soluble vitamins: For instance, Vitamin B complex and Vitamin C.
• Fat-soluble vitamins: Such as Vitamin A and Vitamin E.
The deficiency of Vitamin C is the cause of scurvy.
In simple words: Vitamins are vital organic compounds needed for normal body functions. They're split into water-soluble (like B, C) and fat-soluble (like A, E). A lack of Vitamin C causes scurvy.

🎯 Exam Tip: Define vitamins accurately and remember their classification by solubility. Crucially, always recall that Vitamin C deficiency leads to scurvy, a frequently asked fact.

 

Question 6. Some vitamins and their deficiency diseases are given in the wrong order. Match them in the correct order.
Answer:

Vitamin	Deficiency diseases
i. Vitamin A	a. Anaemia
ii. Vitamin B	b. Impaired clotting of blood
iii. Vitamin B6	c. Night blindness
iv. Vitamin D	d. Scurvy
v. Vitamin K	e. Beriberi
vi. Vitamin C	f. Rickets

**Correct Match:**
i. c (Vitamin A - Night blindness)
ii. e (Vitamin B - Beriberi)
iii. a (Vitamin B6 - Anaemia)
iv. f (Vitamin D - Rickets)
v. b (Vitamin K - Impaired clotting of blood)
vi. d (Vitamin C - Scurvy)
In simple words: Matching vitamins to their deficiency diseases reveals Vitamin A causes night blindness, Vitamin B causes beriberi, B6 causes anaemia, Vitamin D causes rickets, Vitamin K causes impaired blood clotting, and Vitamin C causes scurvy.

🎯 Exam Tip: Creating a concise table or flashcards linking specific vitamins to their deficiency diseases and functions is an effective way to memorize this information, which is a common exam topic.

 

Question 7. What are essential and non-essential amino acids? Give two examples of each type.
Answer: Essential Amino Acids: These are amino acids that the human body cannot synthesize on its own and, therefore, must be supplied through dietary intake. They are indispensable for various physiological processes.
Example: Valine, Leucine, Isoleucine.
Non-essential Amino Acids: These are amino acids that can be synthesized by the human body from other compounds, meaning they do not necessarily need to be obtained directly from the diet.
Example: Glycine, Glutamic acid, Alanine.
In simple words: Essential amino acids are those we must get from food (like Valine, Leucine) because our body can't make them, while non-essential amino acids are those our body can produce itself (like Glycine, Glutamic acid).

🎯 Exam Tip: The primary distinction lies in whether the body can synthesize the amino acid. While "non-essential" implies they aren't crucial, they are still vital for protein synthesis and function.