O Level Biology Practice Paper 4

TuitionGoWhere Exam Practice (AI) - Answer Key

Subject: Biology (6093)
Level: O-Level
Paper: Practice Paper - Version 4 of 5
Topic: Cells and Biomolecules

Section A: Multiple Choice Answers

1. C
Reasoning: Animal cells do not have a large permanent vacuole (they may have small temporary ones), while plant cells typically have a large central vacuole. A is incorrect because animal cells lack cell walls. B is incorrect because animal cells lack chloroplasts. D is incorrect because both have nuclei.

2. C
Reasoning: The presence of a nucleus and mitochondria indicates a eukaryotic cell (ruling out bacteria). The absence of a cell wall and chloroplasts rules out plant cells (leaf palisade and root hair cells have cell walls; leaf cells have chloroplasts). A human liver cell fits this description.

3. C
Reasoning: Channel proteins (Z) facilitate the transport of specific ions or molecules, often via facilitated diffusion or active transport. X (heads) are hydrophilic. Y (tails) are hydrophobic and form the barrier. The membrane is selectively permeable, not fully permeable.

4. B
Reasoning: At isotonic concentration, there is no net change in mass. The graph crosses zero (no change) between +5% (0.2 mol/dm³) and -5% (0.4 mol/dm³). Therefore, the isotonic point is between 0.2 and 0.4 mol/dm³.

5. C
Reasoning: Active transport requires energy (ATP) to move substances against a concentration gradient. Diffusion, osmosis, and facilitated diffusion are passive processes.

6. A
Reasoning: Iodine turning blue-black indicates starch. Biuret turning purple indicates protein. Benedict’s remaining blue indicates no reducing sugar.

7. C
Reasoning: High temperatures (above optimum) cause the enzyme (protein) to denature. This changes the shape of the active site, preventing the substrate from binding.

8. C
Reasoning: Proteins contain Carbon, Hydrogen, Oxygen, and Nitrogen (CHON). Carbohydrates and Fats contain only Carbon, Hydrogen, and Oxygen (CHO).

9. D
Reasoning: Glucose is the monomer for starch, glycogen, and cellulose. However, the general polymer formed from alpha-glucose for storage in plants is starch. Glycogen is animal storage. Cellulose is structural. Given the generic context, Starch is the primary plant storage polymer example. Note: If the context implied animal storage, it would be Glycogen. Without context, Starch is the standard plant example often paired with this diagram style. However, looking at options, both C and D are polymers of glucose. Usually, "Molecule P" in this generic context refers to Starch (plant) or Glycogen (animal). Let's look at Q9 again. It just says "Molecule P". In O-Level, Starch is the most common example. Let's assume Starch. If the question specified animal, it would be Glycogen. Correction: The question doesn't specify plant or animal. However, Starch is the most common answer for "Glucose polymer" in basic contexts unless "animal" is specified. Let's stick with D as the primary example, but C is also chemically correct. In exam keys, usually, the context (e.g., leaf vs liver) dictates. Without context, Starch is the standard textbook example for glucose polymerization.

10. B
Reasoning: Catalysts speed up reactions without being consumed or permanently changed.

Section B: Structured Answers

11. (a) Any two from: [2]

1. Bacterial cell has a cell wall (made of peptidoglycan); Animal cell has no cell wall.
2. Bacterial cell has no nucleus (has nucleoid/DNA loop); Animal cell has a distinct nucleus.
3. Bacterial cell has plasmids; Animal cell does not.
4. Bacterial cell has 70S ribosomes (smaller); Animal cell has 80S ribosomes. (Note: Do not accept "Bacteria are smaller" as a structural feature visible in a standard diagram unless scale is given.)

(b) Binary fission [1] (Note: Mitosis is for eukaryotes. Bacteria undergo binary fission.)

(c) [2]

1. Human cells do not have cell walls. [1]
2. Therefore, antibiotics targeting cell wall synthesis have no target in human cells and do not harm them. [1]

12. (a) [2]

1. Enzymes are biological catalysts. [1]
2. They speed up chemical reactions without being used up. [1]

(b) [3]

1. pH 7 is the optimum pH for amylase. [1]
2. At optimum pH, the enzyme and substrate fit together best (active site shape is ideal). [1]
3. This results in the maximum number of enzyme-substrate complexes forming per unit time, leading to the fastest rate of reaction (shortest time). [1]

(c) [3]

1. The enzyme activity will decrease significantly or stop. [1]
2. pH 2 is very acidic and far from the optimum. [1]
3. The H+ ions disrupt the bonds holding the enzyme structure, causing the active site to change shape (denaturation), so the substrate can no longer bind. [1]

13. (a) [2] Tissue: Upper Epidermis [1] Function: Protection / Secretes cuticle to prevent water loss / Transparent to allow light through. [1] (Accept "Protection" or "Prevents water loss")

(b) [2]

1. Contains many chloroplasts to maximize light absorption for photosynthesis. [1]
2. Arranged in a columnar/palisade layer near the upper surface to receive maximum light. [1]

(c) [2]

1. Stomata open to allow carbon dioxide to diffuse into the leaf for photosynthesis. [1]
2. They also allow oxygen (a waste product of photosynthesis) to diffuse out. [1]

14. (a) Protein and Fat (Lipid) [2] (Biuret +ve = Protein; Ethanol emulsion +ve = Fat)

(b) [2]

1. Starch is insoluble in water, so it does not affect the water potential (osmotic balance) of the cell. [1]
2. Glucose is soluble and would lower water potential, causing water to enter the cell by osmosis, potentially causing it to burst or swell. / Starch is a compact storage form. [1]

(c) [3]

1. Crush the food sample and mix with ethanol. [1]
2. Shake well to dissolve any fats. [1]
3. Pour the solution into water. A cloudy white emulsion indicates the presence of fat. [1]

15. (a) [3]

1. The level of liquid inside the tubing will rise. [1]
2. Water molecules move from the beaker (distilled water, high water potential) into the tubing (sucrose solution, lower water potential) by osmosis. [1]
3. This occurs through the partially permeable visking tubing membrane. [1]

(b) [2]

1. The mass of the tubing will decrease. [1]
2. Water moves out of the tubing (distilled water, high water potential) into the beaker (sucrose solution, lower water potential) by osmosis. [1]

Section C: Free Response Answers

16. (a) [3]

1. The enzyme has an active site with a specific shape. [1]
2. The substrate has a complementary shape that fits into the active site (like a key in a lock). [1]
3. This forms an enzyme-substrate complex, allowing the reaction to occur. [1]

(b) [5]

1. As temperature increases, kinetic energy of enzyme and substrate molecules increases. [1]
2. This leads to more frequent collisions between enzyme and substrate. [1]
3. Therefore, the rate of reaction increases up to the optimum temperature. [1]
4. Above the optimum temperature, the heat energy breaks the bonds holding the enzyme's structure. [1]
5. The active site changes shape (denaturation), the substrate no longer fits, and the reaction stops. [1]

17. (a) [6]

(1 mark for each correct feature, 1 mark for each correct functional link. Max 6 marks.)

(b) [3]

1. Mineral ions need to be absorbed against the concentration gradient (from low concentration in soil to high concentration in root). [1]
2. Diffusion only moves substances down a concentration gradient. [1]
3. Active transport uses energy (ATP) to pump ions into the root hair cell against the gradient. [1]

18. (a) [4]

(1 mark per correct cell.)

(b) [3]

1. Add Biuret solution (or Sodium Hydroxide and Copper Sulphate) to the food sample. [1]
2. Shake/mix gently. [1]
3. A purple/violet colour indicates the presence of protein. [1]

19. (a) [2]

1. Diffusion is the passive movement of particles from high to low concentration; Active transport is the movement against the concentration gradient. [1]
2. Diffusion does not require energy; Active transport requires energy (ATP). [1]

(b) [3]

1. Urea moves by diffusion from the blood (high concentration) to the dialysis fluid (zero concentration) down its concentration gradient. [1]
2. Glucose concentration is the same in the blood and dialysis fluid. [1]
3. Therefore, there is no concentration gradient for glucose, so there is no net movement of glucose. [1]

20. (a) [2]

1. At low substrate concentrations, there are many empty active sites on the enzyme molecules. [1]
2. Increasing substrate concentration increases the frequency of collisions and formation of enzyme-substrate complexes, increasing the rate. [1]

(b) [2]

1. At high substrate concentrations, all enzyme active sites are occupied (saturated). [1]
2. The enzyme is working at its maximum rate (Vmax), so adding more substrate cannot increase the rate further. [1]

(c) [1] Sketch: The line I should start at the same origin (0,0) but rise more slowly (lower gradient) and reach a lower plateau (or the same plateau if non-competitive, but competitive usually reaches same Vmax at very high substrate, just slower initial rate. Correction for O-Level: Competitive inhibitors compete for the active site. At very high substrate concentrations, the substrate outcompetes the inhibitor, so Vmax is eventually reached. The line should be below the original line initially but converge to the same plateau at high substrate concentrations. Marking: Line starts at origin, stays below the original curve, and eventually approaches the same maximum rate.