Chapter 14: Biomolecules
Exhaustive Revision Guide - 50 Subjective Questions with Detailed Solutions
Click on any question to reveal its answer.
1. Define Biomolecules.
Biomolecules are complex, lifeless organic organic molecules which build up living organisms and are required for their growth, maintenance, and ability to reproduce. Examples include carbohydrates, proteins, nucleic acids, and lipids.
2. What are Carbohydrates chemically?
Chemically, carbohydrates are defined as optically active polyhydroxy aldehydes or polyhydroxy ketones, or compounds which produce such units upon hydrolysis. Their general formula was traditionally written as $C_x(H_2O)_y$, though this isn't strictly true for all carbohydrates (e.g., rhamnose, $C_6H_{12}O_5$).
3. Classify carbohydrates on the basis of their behavior towards hydrolysis.
- Monosaccharides: Carbohydrates that cannot be hydrolyzed further into simpler polyhydroxy aldehydes or ketones. (e.g., Glucose, Fructose).
- Oligosaccharides: Carbohydrates that yield 2 to 10 monosaccharide units on hydrolysis. (e.g., Sucrose yields 2, Raffinose yields 3).
- Polysaccharides: Carbohydrates that yield a large number (hundreds to thousands) of monosaccharide units on hydrolysis. (e.g., Starch, Cellulose).
4. Distinguish between reducing and non-reducing sugars with examples.
- Reducing sugars: Carbohydrates that contain a free aldehyde or ketonic group and can reduce Fehling's solution or Tollens' reagent. All monosaccharides (e.g., Glucose, Fructose) and most disaccharides (e.g., Maltose, Lactose) are reducing sugars.
- Non-reducing sugars: Carbohydrates that do not contain a free aldehyde or ketonic group (the groups are involved in forming the glycosidic bond) and cannot reduce Fehling's or Tollens' reagent. Example: Sucrose.
5. Explain the preparation of glucose from sucrose (cane sugar).
Glucose is prepared in the laboratory by boiling sucrose (cane sugar) with dilute $HCl$ or $H_2SO_4$ in an alcoholic solution. It undergoes hydrolysis to yield an equimolar mixture of glucose and fructose.
$$C_{12}H_{22}O_{11} + H_2O \xrightarrow{H^+, \ \Delta} C_6H_{12}O_6 \text{ (Glucose)} + C_6H_{12}O_6 \text{ (Fructose)}$$
Glucose being less soluble in alcohol crystallizes out first upon cooling, allowing its separation.
6. Explain the commercial preparation of glucose from starch.
Commercially, glucose is obtained by the hydrolysis of starch by boiling it with dilute $H_2SO_4$ at 393 K under a pressure of 2-3 atm.
$$(C_6H_{10}O_5)_n \text{ (Starch)} + nH_2O \xrightarrow{H^+, \ 393K, \ 2-3 \text{ atm}} nC_6H_{12}O_6 \text{ (Glucose)}$$
7. What happens when glucose is heated with prolonged Hydrogen Iodide (HI)? What does it prove?
When glucose is heated with prolonged Hydrogen Iodide (HI) and red phosphorus, it is completely reduced to form n-hexane.
$$C_6H_{12}O_6 \xrightarrow{HI, \ \Delta} CH_3-CH_2-CH_2-CH_2-CH_2-CH_3 \text{ (n-hexane)}$$
Proof: This reaction proves that all six carbon atoms in the glucose molecule are linked in a straight, unbranched continuous chain.
8. How does glucose react with Hydroxylamine ($NH_2OH$) and Hydrogen Cyanide (HCN)? What do these reactions indicate?
- With $NH_2OH$: Glucose reacts to form an oxime (glucose oxime).
- With HCN: Glucose undergoes nucleophilic addition to form glucose cyanohydrin.
Indication: Both these reactions confirm the presence of a carbonyl group ($>C=O$) in the glucose molecule.
9. Write the reaction of glucose with Bromine water. What does this reaction prove?
Glucose is oxidized by a mild oxidizing agent like Bromine water to form a six-carbon carboxylic acid called Gluconic acid.
$$CHO-(CHOH)_4-CH_2OH \xrightarrow{Br_2 \text{ water}} COOH-(CHOH)_4-CH_2OH$$
Proof: Since bromine water is a mild oxidizing agent that specifically oxidizes aldehydes (and not ketones), this proves that the carbonyl group in glucose is an aldehyde group ($-CHO$).
10. What happens when glucose is oxidized with concentrated Nitric acid ($HNO_3$)? What does it indicate?
Upon oxidation with a strong oxidizing agent like dilute or concentrated $HNO_3$, glucose yields a dicarboxylic acid called Saccharic acid (Glucaric acid).
$$CHO-(CHOH)_4-CH_2OH \xrightarrow{HNO_3, \ [O]} COOH-(CHOH)_4-COOH$$
Indication: This reaction indicates the presence of a primary alcoholic group ($-CH_2OH$) at the terminal end of the glucose chain, as $HNO_3$ oxidizes both the aldehyde and the primary alcohol groups.
11. Write the reaction of glucose with Acetic anhydride. What structural feature does this confirm?
Glucose reacts with acetic anhydride to form Glucose pentaacetate.
$$CHO-(CHOH)_4-CH_2OH + 5(CH_3CO)_2O \rightarrow CHO-(CHOCOCH_3)_4-CH_2OCOCH_3 + 5CH_3COOH$$
Confirmation: The formation of a pentaacetate derivative confirms the presence of exactly five hydroxyl ($-OH$) groups in the glucose molecule, attached to different carbon atoms (making it a stable compound).
12. What are the limitations of the open-chain structure of glucose?
The open-chain structure cannot explain the following facts:
- Despite having an aldehyde group, glucose does not give the 2,4-DNP test, Schiff's test, and it does not form the hydrogen sulfite addition product with $NaHSO_3$.
- Glucose pentaacetate does not react with hydroxylamine, indicating the absence of a free $-CHO$ group.
- Glucose exists in two different crystalline forms: $\alpha$-glucose and $\beta$-glucose. The open chain structure cannot account for these two anomeric forms.
13. Describe the cyclic structure of Glucose. What is a pyranose ring?
To overcome the limitations of the open-chain structure, glucose is proposed to exist as an intramolecular hemiacetal. The $-OH$ group at C-5 adds to the $-CHO$ group at C-1.
This forms a six-membered ring containing 5 carbon atoms and 1 oxygen atom. Because this ring resembles the cyclic ether Pyran, the cyclic structures of glucose are called Glucopyranose.
14. What are Anomers? Give an example.
Anomers: These are a pair of optical isomers of cyclic carbohydrates which differ in their spatial configuration only at the hemiacetal (or hemiketal) carbon, also known as the anomeric carbon (C-1 in glucose, C-2 in fructose).
Example: $\alpha$-D-glucopyranose and $\beta$-D-glucopyranose are anomers of each other. In the $\alpha$-form, the $-OH$ group at C-1 is on the right (or 'down' in Haworth), while in the $\beta$-form, the $-OH$ at C-1 is on the left (or 'up' in Haworth).
15. What is Mutarotation?
Mutarotation: It is the spontaneous change in the specific optical rotation of an optically active compound in solution, until an equilibrium value is reached.
When either pure $\alpha$-D-glucose (specific rotation $+112^\circ$) or $\beta$-D-glucose (specific rotation $+18.7^\circ$) is dissolved in water, their specific rotation gradually changes until it settles at an equilibrium value of $+52.7^\circ$. This happens because the two anomers interconvert via the open-chain form in solution.
16. Describe the Haworth projection formula for $\alpha$-D-glucopyranose and $\beta$-D-glucopyranose.
Haworth projection is a standard way to draw cyclic sugars as flat hexagons.
- The oxygen atom is at the top-right corner. C-1 is at the far right.
- For D-glucose, the terminal $-CH_2OH$ group attached to C-5 points upwards.
- $\alpha$-D-glucopyranose: At the anomeric carbon (C-1), the $-OH$ group points downwards. (Positions for C1 to C4: Down, Down, Up, Down).
- $\beta$-D-glucopyranose: At the anomeric carbon (C-1), the $-OH$ group points upwards. (Positions for C1 to C4: Up, Down, Up, Down).
17. Define Glycosidic linkage.
Glycosidic Linkage: It is an oxide linkage (an ether bond, $C-O-C$) formed by the loss of a water molecule between two monosaccharide units. It is the chemical bond that holds monosaccharides together to form disaccharides, oligosaccharides, and polysaccharides.
18. Describe the composition and glycosidic linkage in Sucrose. Why is it a non-reducing sugar?
Composition: Sucrose is composed of one unit of $\alpha$-D-glucose and one unit of $\beta$-D-fructose.
Linkage: It is held together by an $\alpha-1, \beta-2$ glycosidic linkage between C-1 of $\alpha$-glucose and C-2 of $\beta$-fructose.
Non-reducing: Because the glycosidic bond involves the reducing groups of both monosaccharides (the anomeric C-1 of glucose and the anomeric C-2 of fructose), there is no free aldehyde or ketone group available. Thus, it cannot reduce Fehling's solution or Tollens' reagent.
19. What is Invert Sugar? Why is it called so?
Sucrose is dextrorotatory ($+66.5^\circ$). Upon hydrolysis, it gives an equimolar mixture of D-(+)-glucose ($+52.5^\circ$) and D-(-)-fructose ($-92.4^\circ$).
Because the levorotation of fructose is greater than the dextrorotation of glucose, the resulting mixture is levorotatory. Since the sign of specific rotation changes from dextro (+) to laevo (-) during hydrolysis, the process is called inversion of cane sugar, and the resulting mixture of glucose and fructose is called Invert Sugar.
20. Describe the composition and glycosidic linkage in Maltose. Is it a reducing sugar?
Composition: Maltose is composed of two units of $\alpha$-D-glucose.
Linkage: The units are linked by an $\alpha-1,4$ glycosidic linkage between C-1 of one glucose unit and C-4 of the other glucose unit.
Reducing nature: Yes, it is a reducing sugar because the anomeric carbon (C-1) of the second glucose unit is free (unlinked) and can open to form a free aldehyde group in solution.
21. Describe the composition and glycosidic linkage in Lactose. Where is it found?
Composition: Lactose is composed of one unit of $\beta$-D-galactose and one unit of $\beta$-D-glucose.
Linkage: The linkage is between C-1 of galactose and C-4 of glucose, known as a $\beta-1,4$ glycosidic linkage.
Source: It is naturally found in the milk of mammals, hence it is known as milk sugar. (It is a reducing sugar).
22. What is Starch? Name its two components and state their main differences.
Starch is the main storage polysaccharide of plants. It is a polymer of $\alpha$-D-glucose. It consists of two components: Amylose and Amylopectin.
- Amylose (15-20%): It is a water-soluble, unbranched, straight-chain polymer of $\alpha$-D-glucose units held together by $C_1-C_4$ glycosidic linkages.
- Amylopectin (80-85%): It is a water-insoluble, highly branched polymer. The main chain has $C_1-C_4$ linkages, while the branching occurs via $C_1-C_6$ glycosidic linkages.
23. Write a short note on Cellulose.
Cellulose is an exclusively plant-based polysaccharide and is the most abundant organic substance in the plant kingdom (the main constituent of plant cell walls).
It is a straight-chain, unbranched polymer composed only of $\beta$-D-glucose units. The glucose units are joined by $\beta-1,4$ glycosidic linkages. Because human digestive enzymes (like amylase) cannot break $\beta$-linkages, humans cannot digest cellulose.
24. What is Glycogen? Where is it found?
Glycogen is the storage polysaccharide found in animals, hence it is also called animal starch. It is a highly branched polymer of $\alpha$-D-glucose.
Its structure is similar to amylopectin but it is much more highly branched. It is mainly present in the liver, muscles, and brain. When the body needs glucose, enzymes break glycogen down to glucose.
25. What are Amino Acids? Write the general formula for an $\alpha$-amino acid.
Amino acids are organic bifunctional molecules containing both an amino group ($-NH_2$) and a carboxylic acid group ($-COOH$). They are the fundamental building blocks (monomers) of proteins.
In $\alpha$-amino acids, both groups are attached to the same carbon ($\alpha$-carbon). General formula: $H_2N-CH(R)-COOH$, where R is a side chain that determines the specific amino acid.
26. Distinguish between Essential and Non-essential amino acids. Give examples.
- Essential amino acids: These cannot be synthesized by the human body and must be obtained through the diet. Example: Valine, Leucine, Lysine.
- Non-essential amino acids: These can be synthesized by the human body and do not necessarily need to be included in the diet. Example: Glycine, Alanine, Serine.
27. What is a Zwitterion? Explain its formation in amino acids.
In aqueous solution, the carboxyl group of an amino acid can lose a proton ($H^+$) and the amino group can accept that proton. This internal proton transfer forms a dipolar ion known as a Zwitterion.
$$H_2N-CH(R)-COOH \rightleftharpoons H_3N^+-CH(R)-COO^- \text{ (Zwitterion)}$$
The zwitterion is electrically neutral overall but contains both positive and negative charges. It explains why amino acids behave like salts (high melting points, water solubility).
28. What is the isoelectric point of an amino acid?
The isoelectric point (pI) is the specific pH at which an amino acid exists almost entirely in its zwitterionic form (net charge is zero). At this pH, the amino acid does not migrate towards either the anode or the cathode under the influence of an applied electric field.
29. Define a Peptide bond. How is it formed?
A peptide bond (or peptide linkage) is an amide linkage ($-CO-NH-$) formed between two amino acid molecules.
It is formed by a condensation reaction between the carboxyl group ($-COOH$) of one amino acid and the amino group ($-NH_2$) of the next amino acid, with the elimination of a water molecule ($H_2O$).
30. Distinguish between Fibrous and Globular proteins with examples.
| Fibrous Proteins | Globular Proteins |
|---|---|
| Polypeptide chains run parallel to form fiber-like structures. | Polypeptide chains fold upon themselves into a spherical/globular shape. |
| Usually insoluble in water. | Usually soluble in water. |
| Provide structural support. Example: Keratin (hair, nails), Myosin (muscles). | Perform metabolic functions. Example: Insulin, Enzymes, Hemoglobin. |
31. Explain the primary structure of proteins.
The primary structure refers to the exact, linear sequence in which various $\alpha$-amino acids are linked to one another by peptide bonds in a polypeptide chain. Any change in this specific sequence results in a completely different protein with a different biological function (e.g., normal hemoglobin vs sickle-cell hemoglobin).
32. Discuss the secondary structure of proteins ($\alpha$-helix and $\beta$-pleated sheet).
The secondary structure refers to the local spatial conformation of the polypeptide backbone, stabilized by hydrogen bonds between the $>C=O$ and $-NH$ groups of peptide bonds.
- $\alpha$-Helix: The chain coils into a right-handed spiral. Hydrogen bonds form between the $C=O$ of one amino acid and the $N-H$ of the fourth amino acid down the chain.
- $\beta$-Pleated Sheet: Polypeptide chains lie side-by-side in a zigzag, pleated pattern, held together by intermolecular hydrogen bonds between adjacent chains.
33. Explain the tertiary structure of proteins. What bonds stabilize it?
The tertiary structure refers to the overall, 3-dimensional folding of the entire polypeptide chain into a specific functional shape (globular or fibrous). It represents further folding of the secondary structure.
It is stabilized by interactions between the side chains (R-groups) of amino acids. The stabilizing forces include: hydrogen bonds, disulfide linkages ($-S-S-$), Van der Waals forces, and electrostatic (ionic) interactions.
34. What is the Denaturation of proteins? Explain with an example.
Denaturation: When a protein in its native form is subjected to physical change (like heating) or chemical change (like change in pH), the stabilizing hydrogen bonds and secondary/tertiary structures are disrupted. The protein unfolds, uncoils, and loses its specific 3D shape and its biological activity. This is denaturation.
During denaturation, secondary and tertiary structures are destroyed, but the primary structure (peptide bonds) remains intact.
Example: Coagulation of egg white on boiling; curdling of milk by lactic acid bacteria.
35. What are Enzymes? State their chemical nature.
Enzymes are biological catalysts that speed up chemical reactions occurring in living organisms. They are highly specific and efficient.
Chemically, almost all enzymes are globular proteins. (Some contain a non-protein part called a co-factor or co-enzyme essential for activity).
36. Explain the 'Lock and Key' mechanism of enzyme action.
Enzymes have specific regions on their surface called active sites. These active sites have a specific size, shape, and functional groups.
According to this mechanism, the substrate molecule (the key) has a complementary shape that perfectly fits into the active site of the enzyme (the lock). They bind to form a short-lived Enzyme-Substrate (E-S) complex. The reaction occurs, and the products are released, freeing the enzyme active site for another cycle.
$E + S \rightarrow [E-S] \rightarrow E + P$
37. What are Nucleic Acids? Name the two main types.
Nucleic Acids are biologically important polymers present in the nuclei of all living cells. They are responsible for the transmission of inherent characters from parents to offspring (heredity) and for the synthesis of proteins. Chemically, they are polynucleotides.
Two main types are: DNA (Deoxyribonucleic acid) and RNA (Ribonucleic acid).
38. Name the three chemical components of a nucleotide.
Complete hydrolysis of DNA or RNA yields three fundamental components which make up a nucleotide:
- A pentose sugar (ribose or deoxyribose).
- A nitrogenous base (purine or pyrimidine).
- A phosphate group ($PO_4^{3-}$).
39. Which pentose sugars are present in DNA and RNA?
- In RNA: The pentose sugar is $\beta$-D-ribose.
- In DNA: The pentose sugar is $\beta$-D-2-deoxyribose. (It lacks an oxygen atom at the C-2 position compared to ribose).
40. Name the nitrogenous bases present in DNA and RNA. Which base is unique to each?
Nitrogenous bases are derivatives of Purines (Adenine, Guanine) and Pyrimidines (Cytosine, Thymine, Uracil).
- In DNA: Adenine (A), Guanine (G), Cytosine (C), and Thymine (T).
- In RNA: Adenine (A), Guanine (G), Cytosine (C), and Uracil (U).
Thymine is unique to DNA, and Uracil is unique to RNA.
41. Distinguish between a Nucleoside and a Nucleotide.
- Nucleoside: A unit formed by the attachment of a nitrogenous base to the 1'-position of a pentose sugar. It lacks a phosphate group.
(Nucleoside = Sugar + Base). - Nucleotide: A unit formed when a phosphate group is attached to the 5'-OH (or 3'-OH) of a nucleoside. It is the basic monomer of nucleic acids.
(Nucleotide = Sugar + Base + Phosphate).
42. How are nucleotides linked together to form a nucleic acid chain (polynucleotide)?
Nucleotides are linked together by phosphodiester linkages. Specifically, the phosphate group at the 5'-carbon of the sugar of one nucleotide forms an ester bond with the 3'-hydroxyl group of the sugar of the adjacent nucleotide. This creates a continuous sugar-phosphate backbone (5' $\rightarrow$ 3' direction).
43. Describe the Watson-Crick double helix model of DNA.
According to James Watson and Francis Crick, DNA exists as a double helix.
- Two right-handed polynucleotide chains are coiled around a common axis.
- The chains are anti-parallel (one runs 5'$\rightarrow$3', the other 3'$\rightarrow$5').
- The sugar-phosphate backbone lies on the outside, while the nitrogenous bases point inwards.
- The two strands are held together by hydrogen bonds formed between specific, complementary base pairs.
44. Explain the base-pairing rule in DNA.
Due to size and geometry constraints, a purine must always pair with a specific pyrimidine via hydrogen bonding. This is strict complementarity:
- Adenine (A) always pairs with Thymine (T) by forming exactly two hydrogen bonds ($A=T$).
- Guanine (G) always pairs with Cytosine (C) by forming exactly three hydrogen bonds ($G \equiv C$).
45. What are the different types of RNA? Mention their functions briefly.
RNA molecules exist primarily as single strands and are of three types, all involved in protein synthesis:
- Messenger RNA (mRNA): Carries the genetic code (message) from DNA in the nucleus to the ribosomes.
- Ribosomal RNA (rRNA): A structural component of ribosomes, which act as the site of protein synthesis.
- Transfer RNA (tRNA): Transfers specific amino acids from the cytoplasm to the ribosomes for incorporation into the growing protein chain.
46. Differentiate between DNA and RNA. (Any 4 points)
| Feature | DNA | RNA |
|---|---|---|
| Sugar | Contains $\beta$-D-2-deoxyribose. | Contains $\beta$-D-ribose. |
| Bases | A, G, C, and Thymine (T). | A, G, C, and Uracil (U). |
| Structure | Double-stranded $\alpha$-helix. | Usually single-stranded. |
| Function | Responsible for heredity (genetic material). | Responsible for protein synthesis. |
47. What is DNA replication? Why is it important?
DNA Replication: It is the biological process by which a unique DNA molecule produces two identical exact copies of itself. During cell division, the two strands of the parent DNA unzip, and each acts as a template for synthesizing a new complementary strand.
Importance: It ensures that the exact genetic information is inherited by the daughter cells when a cell divides, maintaining genetic continuity.
48. What are vitamins? Classify them based on their solubility.
Vitamins are specific organic compounds required in the diet in small amounts to perform specific biological functions for normal maintenance of optimum growth and health of the organism.
- Fat-soluble vitamins: Soluble in fat/lipids, insoluble in water. They are stored in the liver and adipose tissues. (Vitamins A, D, E, K).
- Water-soluble vitamins: Soluble in water. They must be supplied regularly in diet as they are readily excreted in urine (except B12). (B group vitamins and Vitamin C).
49. Name the deficiency diseases caused by the lack of Vitamin A, Vitamin C, and Vitamin D.
- Vitamin A (Retinol): Night blindness, Xerophthalmia (hardening of cornea).
- Vitamin C (Ascorbic acid): Scurvy (bleeding gums).
- Vitamin D (Calciferol): Rickets in children (bone deformities) and Osteomalacia in adults.
50. Why cannot Vitamin C be stored in our body?
Vitamin C (Ascorbic acid) is a water-soluble vitamin. Because it easily dissolves in the body's water, it is readily excreted in the urine and cannot be stored in our body tissues (unlike fat-soluble vitamins). Therefore, it must be supplied continuously through our daily diet (like citrus fruits).
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