CBSE Class 11 Biology Biomolecules Assignments

QUESTIONS

Very Short Answer Questions (1 mark each)

Question 1. Why do oils generally remain in liquid state even in winters?
Answer: Oils are unsaturated lipids containing one or more double bonds in their fatty acid chains, which lower their melting point and keep them liquid at low temperatures.
In simple words: Unsaturated fats have kinks in their molecular structure that prevent tight packing, keeping them liquid in winter.

Exam Tip: Remember that saturated fats pack closely and are solid at room temperature, whereas unsaturated oils remain liquid.

Question 2. Name an element found in proteins but not in lipids and carbohydrates.
Answer: Nitrogen (N) is a characteristic component of amino acids and proteins, which is typically absent in simple carbohydrates and lipids.
In simple words: Proteins contain nitrogen in their amino groups, unlike typical fats and sugars which only contain carbon, hydrogen, and oxygen.

Exam Tip: Always associate nitrogen with proteins and nucleic acids when distinguishing them from carbohydrates and lipids.

Question 3. What is the difference between RNA and DNA in terms of nitrogenous base?
Answer: RNA contains the pyrimidine base uracil, whereas DNA contains thymine instead of uracil. Both share adenine, guanine, and cytosine.
In simple words: DNA has thymine while RNA has uracil as its pairing base for adenine.

Exam Tip: Keep this key distinction clear: T in DNA, U in RNA.

Question 4. What does an enzyme do in terms of energy requirement of a reaction?
Answer: An enzyme lowers the activation energy of a biochemical reaction, thereby increasing the rate at which the reaction proceeds.
In simple words: Enzymes act as biological catalysts that reduce the energy barrier needed to start a chemical reaction.

Exam Tip: Use the exact scientific term "activation energy" to secure full marks.

Question 5. What is the function of ATP in cell metabolism?
Answer: ATP (Adenosine triphosphate) serves as the primary energy currency of the cell, storing and transporting chemical energy for various metabolic processes.
In simple words: ATP provides the required energy to drive endergonic cellular reactions.

Exam Tip: Referring to ATP as the "energy currency of the cell" is a standard expected keyword.

Question 6. Name the protein which form the intercellular ground substance.
Answer: Collagen is the most abundant protein in animal world and forms the intercellular ground substance in connective tissues.
In simple words: Collagen provides structural support and forms the primary matrix between cells.

Exam Tip: Collagen is also noted as the most abundant protein in mammals.

Short Answer Questions-II (2 marks each)

Question 7. Why are amino acids also known as substituted methane?
Answer: Amino acids can be structurally derived from methane (\(\text{CH}_4\)), where the four hydrogen atoms are replaced by four different groups: an amino group (\(-\text{NH}_2\)), a carboxyl group (\(-\text{COOH}\)), a hydrogen atom (\(-\text{H}\)), and a variable R group.
In simple words: They share the central tetrahedral carbon of methane but have four distinct chemical groups attached to it instead of hydrogens.

Exam Tip: Mention all four substituting groups (\(-\text{H}, -\text{COOH}, -\text{NH}_2, -\text{R}\)) to get full credit.

Question 8. Amino acids exist as zwitter ions. Give its structure. Why is it formed?
Answer: Structure of a zwitterion:
\(\text{R}-\text{CH}(\text{NH}_3^+)-\text{COO}^-\)
It is formed due to the ionizable nature of amino and carboxyl groups, where the proton from the carboxyl group shifts to the amino group, creating a dipolar ion with both positive and negative charges.
In simple words: In a zwitterionic form, an amino acid carries both a positive and a negative charge simultaneously, making it overall neutral at a specific pH.

Exam Tip: Clearly draw or write the ionized groups (\(-\text{NH}_3^+\) and \(-\text{COO}^-\)) in your answer.

Question 9. Why do starch give blue black colour with iodine?
Answer: Starch forms helical secondary structures which can hold iodine molecules in the helical cavity, forming a starch-iodine complex that absorbs light and produces a characteristic blue-black color.
In simple words: Iodine fits snugly inside the spiral coils of starch molecules, changing the light absorption properties to blue-black.

Exam Tip: Mention the term "helical secondary structure" as it explains the trapping of iodine molecules.

Question 10. Why are starch and glycogen more suitable than glucose as a storage product?
Answer: Starch and glycogen are large polymers that occupy lesser osmotic space, are less bulky inside the cell, and are insoluble, preventing them from diffusing out of the cell. They can be readily hydrolyzed to glucose when energy is required.
In simple words: Storing thousands of glucose units as a single large polymer prevents osmotic disruption and saves cellular space.

Exam Tip: Focus on osmotic pressure and insolubility as the core biological advantages.

Question 11. What would happed when salivary amylase which acts on starch in mouth and in stomach?
Answer: In the mouth, salivary amylase breaks down starch into maltose. When it reaches the stomach, the highly acidic environment (\(\text{pH}\) roughly 1.8 to 2.0) denatures the enzyme, halting its action completely.
In simple words: Salivary amylase digests starch in the neutral mouth cavity, but stomach acid destroys its active structure.

Exam Tip: Highlight that enzymes are sensitive to pH changes, which inactivates salivary amylase in the stomach.

Question 12. Differentiate between homopolysaccharides and heteropolysaccharides
Answer:

In simple words: Homopolysaccharides contain repeating units of only one sugar, while heteropolysaccharides contain a mix of different sugars.

Exam Tip: Use a clear comparative table for differences to ensure full marks.

Question 13. Why do physicians recommend vegetable oils rich in polyunsaturated fat for persons suffering from cardiovascular diseases?
Answer: Polyunsaturated oils contain fatty acids with multiple double bonds that help lower blood cholesterol levels and do not clog arteries compared to saturated fats, thereby reducing the risk of heart blockages and cardiovascular diseases.
In simple words: Unsaturated plant oils prevent cholesterol buildup in blood vessels, keeping the heart healthier.

Exam Tip: Connect polyunsaturated fats with lower cholesterol deposition in blood vessels.

Question 14. Why does the shelf life of fruits and vegetables increase in a refrigerator?
Answer: Low temperatures in a refrigerator inhibit the growth and reproduction of food-spoiling microorganisms and reduce metabolic activity. It also slows down endogenous enzymatic reactions that lead to ripening and decay.
In simple words: Cold temperatures slow down both bacterial growth and the natural chemical reactions that cause food to spoil.

Exam Tip: Mention both microbial inhibition and enzyme slowdown for a complete answer.

Short Answer Questions-I (3 marks each)

Question 15. Explain the structure of proteins.
Answer: Proteins are heteropolymers of amino acids linked by peptide bonds, organized into four structural levels:
1. Primary structure: Linear sequence of amino acids with an N-terminal and a C-terminal end.
2. Secondary structure: Polypeptide chains fold or coil into regular patterns like alpha-helices or beta-pleated sheets, stabilized by hydrogen bonds.
3. Tertiary structure: The protein chain folds further upon itself like a hollow woolen ball, providing a functional 3-dimensional conformation (e.g., myoglobin).
4. Quaternary structure: Association of two or more folded polypeptide subunits to form a functional protein complex (e.g., hemoglobin with two alpha and two beta chains).
In simple words: Protein structure ranges from a simple chain of amino acids to complex 3D folded shapes made of multiple subunits.

Exam Tip: Outline all four levels (Primary, Secondary, Tertiary, Quaternary) clearly with brief descriptions.

Question 16. (a) What is an enzyme? (b) Give an example of co-enzyme. (c) Distinguish between apoenzyme and co-enzyme.
Answer:
(a) Enzymes are biological catalysts (mostly proteins, except ribozymes) that accelerate biochemical reactions without being consumed in the process.
(b) Examples of co-enzymes include NAD, NADP, and FAD.
(c) Distinction:
- Apoenzyme: The protein portion of an enzyme, which is catalytically inactive by itself.
- Co-enzyme: A non-protein, organic molecule that binds loosely to the apoenzyme to make it catalytically active.
In simple words: An enzyme needs a non-protein helper (co-enzyme) attached to its protein part (apoenzyme) to function properly.

Exam Tip: Clearly separate your answers into parts (a), (b), and (c) as presented in the question paper.

Question 17. Explain Watson-Crick model on DNA structure.
Answer: According to the Watson and Crick model:
1. DNA is a double helix consisting of two polynucleotide chains that run antiparallel to each other.
2. The backbone is formed by alternating sugar and phosphate groups, while the nitrogenous bases project inward.
3. The two chains are held together by hydrogen bonds between complementary bases: Adenine pairs with Thymine (A=T with two hydrogen bonds) and Cytosine pairs with Guanine (C\(\equiv\)G with three hydrogen bonds).
4. The helix is right-handed and exhibits major and minor grooves.
In simple words: DNA is a twisted ladder made of two antiparallel strands connected by hydrogen-bonded base pairs.

Exam Tip: Mention antiparallel strands, sugar-phosphate backbone, and base pairing rules (A with T, C with G).

Question 18. Explain peptide bond, glycosidic bond and phosphodiester bond.
Answer:
1. Peptide bond: Formed between the carboxyl group (\(-\text{COOH}\)) of one amino acid and the amino group (\(-\text{NH}_2\)) of the next amino acid with the release of a water molecule (dehydration synthesis).
2. Glycosidic bond: A bond formed between two carbon atoms of two adjacent monosaccharide units through an oxygen atom, formed by dehydration.
3. Phosphodiester bond: A covalent bond in nucleic acids where a phosphate group links the 3-carbon atom of one sugar molecule to the 5-carbon atom of the adjacent sugar molecule.
In simple words: These are chemical bonds that join amino acids in proteins, sugars in carbohydrates, and nucleotides in nucleic acids.

Exam Tip: Clearly state the participating functional groups for each type of bond.

Question 19. Explain competitive inhibition along with an example.
Answer: Competitive inhibition occurs when an inhibitor molecule closely resembles the substrate in its molecular structure and binds to the active site of the enzyme, blocking the actual substrate from entering.
Example: The inhibition of succinic dehydrogenase by malonate, which resembles succinate.
In simple words: An impostor molecule competes with the real substrate for the enzyme's active site, slowing down the reaction.

Exam Tip: Always include the classic example of succinic dehydrogenase and malonate.

Long Answer Questions (5 marks each)

Question 20. List the 6 classes of enzymes along with their functions.
Answer: Enzymes are grouped into 6 major classes based on the type of reactions they catalyze:
1. Oxidoreductases / Dehydrogenases: Catalyze oxidation-reduction reactions, transferring electrons or hydrogens between substrates.
2. Transferases: Catalyze the transfer of a specific functional group (other than hydrogen) from one pair of substrates to another.
3. Hydrolases: Catalyze the hydrolysis of ester, ether, peptide, glycosidic, carbon-carbon, or phosphorus-nitrogen bonds by adding water.
4. Lyases: Catalyze the removal of groups from substrates by mechanisms other than hydrolysis, leaving double bonds.
5. Isomerases: Catalyze the interconversion of optical, geometric, or positional isomers.
6. Ligases: Catalyze the linking together of two compounds utilizing energy (typically from ATP hydrolysis), such as joining C-O, C-S, C-N, or P-O bonds.
In simple words: Enzymes are classified into six groups depending on whether they oxidize, transfer, hydrolyze, cleave, rearrange, or join molecules.

Exam Tip: Learn the names of all 6 classes (Oxidoreductases, Transferases, Hydrolases, Lyases, Isomerases, Ligases) along with one-line functions for each.