Chapter 14: Biomolecules
Highly Detailed Mark-wise Solutions for Board Exam Preparation
1 Mark Questions (Very Short Answer)
Answer: A peptide bond is an amide linkage ($-CO-NH-$) formed between the carboxyl group ($-COOH$) of one $\alpha$-amino acid and the amino group ($-NH_2$) of another $\alpha$-amino acid, with the elimination of a water molecule.
Answer: Uracil (U) is the pyrimidine base present in RNA but absent in DNA (DNA contains Thymine instead).
Answer: Glucose (or Maltose, Lactose, Fructose).
Answer: Monosaccharides are the simplest carbohydrates that cannot be hydrolyzed further into smaller polyhydroxy aldehyde or ketone units. (e.g., Glucose, Fructose).
2 Mark Questions (Short Answer-I)
Answer:
In aqueous solution, the acidic carboxyl group ($-COOH$) of an $\alpha$-amino acid can lose a proton ($H^+$), while the basic amino group ($-NH_2$) can accept that proton. This internal acid-base reaction results in the formation of a dipolar ion, which is electrically neutral overall but contains both a positive and a negative charge. This is known as a Zwitterion.
Example (Glycine):
$H_2N-CH_2-COOH \rightleftharpoons H_3N^+-CH_2-COO^-$
(Glycine) (Zwitterion of Glycine)
Answer:
Proteins have a unique 3D structure necessary for their biological activity (native state). When a native protein is subjected to a physical change (like heating) or a chemical change (like change in pH or addition of heavy metal salts), the hydrogen bonds stabilizing the structure are disturbed.
Due to this, the protein globules unfold and the helices uncoil, causing the protein to lose its biological activity. This phenomenon is called denaturation. During denaturation, the secondary and tertiary structures are destroyed, but the primary structure (amino acid sequence) remains intact.
Example: Coagulation of egg white on boiling, or curdling of milk.
Answer:
Aqueous solution of sucrose is dextrorotatory ($+66.5^\circ$). Upon hydrolysis with dilute acid or the enzyme invertase, it yields an equimolar mixture of D-(+)-glucose ($+52.5^\circ$) and D-(-)-fructose ($-92.4^\circ$).
Because the laevorotation of fructose ($-92.4^\circ$) is greater in magnitude than the dextrorotation of glucose ($+52.5^\circ$), the resulting mixture becomes laevorotatory. This change in the sign of specific rotation (from positive/dextro to negative/laevo) is called inversion, and the equimolar mixture of glucose and fructose formed is known as invert sugar.
3 Mark Questions (Short Answer-II / Differentiation)
Answer:
| Property | DNA (Deoxyribonucleic Acid) | RNA (Ribonucleic Acid) |
|---|---|---|
| 1. Sugar Unit | The pentose sugar is $\beta$-D-2-deoxyribose. | The pentose sugar is $\beta$-D-ribose. |
| 2. Pyrimidine Bases | Contains Cytosine and Thymine. | Contains Cytosine and Uracil. |
| 3. Structure | It has a double-stranded $\alpha$-helix structure. | It is generally single-stranded (which may fold back on itself). |
| 4. Function | Controls heredity; transfers genetic information from generation to generation. | Primarily controls protein synthesis in the cell. |
| 5. Replication | It has the unique property of self-replication. | It usually does not replicate (synthesized from DNA). |
| 6. Location in cell | Found mainly in the nucleus (chromosomes). | Found mainly in the cytoplasm and nucleolus. |
Answer:
Based on their behavior towards hydrolysis, carbohydrates are classified into three main types:
- Monosaccharides: These are the simplest carbohydrates that cannot be hydrolyzed further to yield simpler polyhydroxy aldehyde or ketone units. They usually contain 3 to 7 carbon atoms.
Example: Glucose ($C_6H_{12}O_6$), Fructose. - Oligosaccharides: These carbohydrates yield 2 to 10 monosaccharide units upon hydrolysis. Depending on the number of units yielded, they are further classified as disaccharides, trisaccharides, etc.
Example: Sucrose ($C_{12}H_{22}O_{11}$) is a disaccharide yielding one molecule of glucose and one of fructose on hydrolysis. - Polysaccharides: These are high molecular mass carbohydrates that yield a very large number (hundreds or thousands) of monosaccharide units on complete hydrolysis. They are generally tasteless and non-reducing.
Example: Starch, Cellulose, Glycogen.
4 Mark Questions (Long Answer / Combined)
Note: In the HSC Board Exam, 4-mark questions in Chemistry are frequently split into sub-questions (e.g., a 3-mark structural elucidation combined with a 1-mark definition).
(b) What are essential amino acids? [1 Mark] March 2017, Oct 2020, March 2023
Answer (a): Structural Elucidation of Glucose
(i) Reaction with Hydroxylamine ($NH_2OH$):
When D-glucose reacts with hydroxylamine, it forms an oxime.
Conclusion: This reaction proves the presence of a carbonyl group ($>C=O$) in the glucose molecule.
(ii) Reaction with Bromine water:
When D-glucose is treated with mild oxidizing agents like bromine water, it gets oxidized to a six-carbon carboxylic acid called Gluconic acid.
Conclusion: This reaction proves that the carbonyl group present in glucose is an aldehyde group ($-CHO$) located at one end of the carbon chain.
(iii) Reaction with HI and Red Phosphorus:
Prolonged heating of D-glucose with Hydrogen Iodide and Red Phosphorus yields n-hexane.
Conclusion: This reaction conclusively proves that all six carbon atoms in the glucose molecule are linked in a straight, unbranched continuous chain.
Answer (b): Essential Amino Acids
Essential Amino Acids: Amino acids which cannot be synthesized within the human body and must necessarily be supplied through our daily diet are called essential amino acids. (e.g., Valine, Leucine, Lysine).
(b) Distinguish between Globular and Fibrous proteins. [2 Marks] Oct 2013, March 2021
Answer (a): Glycosidic Linkage
Definition: The oxide linkage (ether linkage) formed between two monosaccharide units by the elimination of a water molecule is called a glycosidic linkage.
Formation in Maltose: Maltose is a disaccharide composed of two $\alpha$-D-glucose units. The glycosidic linkage is formed between the $C_1$ hydroxyl group of one $\alpha$-D-glucose unit and the $C_4$ hydroxyl group of the second $\alpha$-D-glucose unit. Hence, it is called an $\alpha$-(1$\rightarrow$4) glycosidic linkage.
Answer (b): Globular vs Fibrous Proteins
| Globular Proteins | Fibrous Proteins |
|---|---|
| 1. Polypeptide chains coil around to give a spherical (3D) shape. | 1. Polypeptide chains run parallel and are held together to form fiber-like structures. |
| 2. They are generally soluble in water and aqueous solutions. | 2. They are generally insoluble in water. |
| 3. Sensitive to small changes in temperature and pH. | 3. They are stable to moderate changes in temperature and pH. |
| 4. Examples: Insulin, Albumin, Enzymes. | 4. Examples: Keratin (hair, wool), Myosin (muscles). |
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