Secondary 4 Pure Biology Preliminary Examination Paper 3

TuitionGoWhere Practice Paper - Pure Biology Secondary 4

PRELIM Paper 2 (Version 3 of 5) — Answer Key

Section A: Multiple Choice Questions [10 marks]

1. C — Smooth endoplasmic reticulum
Explanation: The smooth ER synthesises lipids and detoxifies harmful substances. The rough ER has ribosomes and is involved in protein synthesis.

2. C — Plant cell
Explanation: The presence of a cell wall, chloroplasts, and a large central vacuole are characteristic features of plant cells.

3. B — It controls the movement of substances into and out of the cell.
Explanation: The cell membrane is partially permeable and regulates what enters and leaves the cell.

4. C — Glucose
Explanation: Starch is a polysaccharide made up of many glucose monomers joined together by glycosidic bonds.

5. B — Modifying, sorting, and packaging proteins for secretion
Explanation: The Golgi body receives proteins from the rough ER, modifies them (e.g., adding carbohydrate groups), sorts them, and packages them into vesicles for transport.

6. C — The enzyme has an optimum pH of 7.
Explanation: The highest activity at pH 7 indicates this is the optimum pH. Very low activity at extreme pH values suggests denaturation.

7. C — Ribosomes and rough endoplasmic reticulum
Explanation: Digestive enzymes are proteins. Cells actively secreting proteins require many ribosomes (for protein synthesis) and extensive rough ER (for protein processing and transport).

8. B — The cell will swell and may burst due to water entering by osmosis.
Explanation: Distilled water has a higher water potential than the cytoplasm of the red blood cell. Water moves into the cell by osmosis, causing it to swell and potentially burst (haemolysis).

9. C — Lipid
Explanation: Lipids are insoluble in water and serve as long-term energy stores. They also provide insulation and form cell membranes.

10. B — How substrates bind to the active site of an enzyme.
Explanation: The lock-and-key hypothesis states that the substrate fits precisely into the active site of the enzyme, like a key fits into a lock.

Section B: Structured Questions [35 marks]

11.

(a)
A: Nucleus [1]
B: Mitochondrion [1]
Marking note: Accept "mitochondria" (plural). Spelling must be reasonably accurate.

(b) One function of rough endoplasmic reticulum (C):

• Synthesis/transport of proteins [1]
  Accept any one valid function: protein synthesis, transport of proteins, provides a large surface area for chemical reactions.

(c) Structure D (Golgi body) is important for cells that secrete hormones because:

• Hormones (protein-based) are synthesised in the rough ER and then transported to the Golgi body [1]
• The Golgi body modifies, sorts, and packages the hormones into vesicles for secretion from the cell [1]
  Marking note: Students must link the Golgi body's function to the secretion process. Simply naming the Golgi body is not sufficient.

(d) Structure E (cell membrane) is adapted by:

• Being partially/permeable, allowing it to control which substances enter and leave the cell [1]
• Having a phospholipid bilayer structure with embedded proteins that act as channels or carriers for specific molecules [1]
  Marking note: Accept references to fluid mosaic structure, receptor proteins, or cholesterol for stability.

12.

(a)
Independent variable: Temperature [1]
Dependent variable: Time taken for starch to be broken down [1]
Marking note: "Rate of reaction" is not acceptable as the dependent variable here because the student measured time, not rate directly.

(b) Highest activity at 40°C [1]
Explanation: The shortest time (3 minutes) was recorded at 40°C, meaning starch was broken down the fastest, indicating the highest enzyme activity [1]
Marking note: Students must link shortest time to highest activity.

(c) At temperatures above 40°C:

• The enzyme molecules gain excessive kinetic energy [1]
• This causes the bonds maintaining the enzyme's tertiary structure to break [1]
• The active site changes shape (denaturation), so the substrate can no longer fit, and the rate of reaction decreases [1]
  Marking note: All three points required for full marks. Simply stating "the enzyme is denatured" without explaining why is insufficient for 3 marks.

(d) One way to improve reliability:

• Repeat the experiment at each temperature and calculate the mean/average [1]
  Accept: use a water bath for more accurate temperature control, use a colorimeter for more objective measurements, ensure the same volume/concentration of enzyme and starch.

13.

(a)

• Hydrophilic head labelled on the phosphate-containing end (water-loving, polar end) [1]
• Hydrophobic tail labelled on the fatty acid chains (water-fearing, non-polar end) [1]
  Marking note: Only 1 mark available — accept either label correctly placed, or both for 1 mark total.

(b) The phospholipid arrangement acts as a barrier because:

• Phospholipids are arranged in a bilayer with hydrophobic tails facing inwards and hydrophilic heads facing outwards [1]
• Water-soluble (polar) substances cannot pass through the hydrophobic core of the bilayer [1]
• Only small, non-polar molecules can diffuse through the lipid bilayer freely [1]
  Marking note: Students must explain the barrier property, not just describe the structure.

(c)
Component 1: Proteins
Function: Act as channel/carrier proteins for facilitated diffusion or active transport of specific molecules; or act as receptor sites for cell signalling [1]

Component 2: Cholesterol
Function: Stabilises the membrane by fitting between phospholipids, preventing the membrane from becoming too fluid at high temperatures or too rigid at low temperatures [1]

Alternative acceptable answers:

• Glycoproteins: cell recognition, cell adhesion
• Glycolipids: cell recognition, maintaining membrane stability

Marking note: Each component-function pair is worth 2 marks (1 for component, 1 for function).

14.

(a) Plasmolysis [1]
Marking note: Accept "osmosis" but plasmolysis is the specific term for this observation.

(b) Explanation of plasmolysis:

• The concentrated salt solution has a lower water potential than the cell sap [1]
• Water moves out of the cell by osmosis, from a region of higher water potential to lower water potential [1]
• The cell membrane pulls away from the cell wall as the vacuole and cytoplasm shrink [1]
  Marking note: Students must use the term "water potential" or equivalent concept for full marks.

(c) If placed in distilled water:

• The cell would become turgid / regain its original shape (deplasmolysis) [1]
• Distilled water has a higher water potential than the cell sap [1]
• Water enters the cell by osmosis, causing the vacuole to expand and push the cell membrane against the cell wall [1]
  Marking note: The cell wall prevents bursting, unlike animal cells.

15.

(a) Sample X contains starch [1]
Explanation: The iodine test turned blue-black, which is the positive result for starch [1]

(b) Sample Y contains reducing sugar [1]
Explanation: Benedict's test turned orange-red, indicating the presence of reducing sugar [1]
Marking note: Blue = no reducing sugar; green/yellow/orange/red = increasing amounts of reducing sugar.

(c) Samples Y and Z contain protein [1]
Explanation: The Biuret test turned purple, which is the positive result for protein [1]

(d) Samples X and Z contain lipid [1]
Explanation: The ethanol emulsion test produced a cloudy white precipitate, indicating the presence of lipid [1]

Marking note: Students must correctly identify the sample AND explain using the test result. Each part is worth 2 marks.

16.

(a) Relationship:

• As substrate concentration increases, the rate of reaction increases [1]
• The rate eventually levels off / reaches a maximum at high substrate concentrations [1]
  Marking note: Both parts required for 2 marks.

(b) The rate levels off because:

• At high substrate concentrations, all active sites of the enzyme molecules are occupied [1]
• The enzyme is working at its maximum capacity [1]
• Adding more substrate has no further effect because there are no free active sites available [1]
  Marking note: Students must explain the concept of enzyme saturation.

(c) Curve X should be drawn:

• Starting from the origin with a similar initial gradient [1]
• Levelling off at a lower maximum rate than the original curve [1]
  Marking note: With half the enzyme concentration, there are fewer active sites available, so the maximum rate is lower. The initial gradient may also be less steep.

17.

(a) Two structural features:
Feature 1: Long, thin extension (hair-like projection) that increases surface area [1]
Feature 2: Thin cell wall / cell membrane that allows easy passage of water [1]
Alternative acceptable features: large vacuole with concentrated cell sap (lower water potential), numerous mitochondria (for active transport of mineral ions).

(b) How each feature helps:
Feature 1: The increased surface area allows more water to be absorbed per unit time [1]
Feature 2: The thin cell wall reduces the distance water must travel, speeding up absorption [1]

If vacuole mentioned: The concentrated cell sap creates a lower water potential inside the cell, maintaining a water potential gradient for osmosis [1]
If mitochondria mentioned: Mitochondria provide ATP for active transport of mineral ions, which lowers the water potential in the cell [1]

(c) Osmosis [1]
Marking note: Must specify osmosis, not just "diffusion" or "active transport."

Section C: Free Response Question [15 marks]

18.

(a) A biological catalyst is:

• A substance (enzyme) that speeds up the rate of a chemical reaction [1]
• Without being used up or permanently changed in the reaction [1]
  Marking note: Both points required. Simply saying "speeds up reactions" is insufficient.

(b) Lock-and-key hypothesis:

• Each enzyme has a uniquely shaped active site [1]
• The substrate has a complementary shape to the active site [1]
• The substrate fits into the active site like a key fits into a lock [1]
• An enzyme-substrate complex is formed, the reaction occurs, and products are released [1]

Diagram should show:

• Enzyme with active site (clearly labelled)
• Substrate with complementary shape
• Enzyme-substrate complex
• Products being released
• Enzyme unchanged after reaction

Marking note: 2 marks for description, 2 marks for diagram (if included). Diagram must be labelled.

(c) Effect of temperature:

Low temperatures:

• Enzyme and substrate molecules have low kinetic energy [1]
• Fewer successful collisions between enzyme and substrate, so the rate of reaction is slow [1]

Optimum temperature:

• Enzyme activity is at its maximum [1]
• Molecules have sufficient kinetic energy for frequent successful collisions without denaturation [1]

High temperatures (above optimum):

• Excessive kinetic energy causes bonds in the enzyme to break [1]
• The active site changes shape (denaturation) and the substrate can no longer fit [1]
• The enzyme is permanently inactivated and the rate of reaction drops sharply [1]

Marking note: 6 marks total — 2 for each temperature condition. Students must explain the molecular mechanism, not just state the effect.

(d) Pepsin's low optimum pH:

• Pepsin is adapted to work in the acidic environment of the stomach (pH ~2), where hydrochloric acid is secreted [1]
• Its active site is shaped to function optimally at this pH [1]

Effect in the small intestine (pH ~8):

• The alkaline pH would denature pepsin [1]
• The active site would change shape, and pepsin would lose its activity / become inactive [1]
• This is why trypsin (optimum pH ~8) is used in the small intestine instead [bonus knowledge, not required]

Marking note: 3 marks — 1 for explaining the stomach adaptation, 2 for explaining the effect of alkaline pH.

END OF ANSWER KEY

Total Marks: 60