Secondary 4 Pure Biology Preliminary Examination Paper 4

TuitionGoWhere Practice Paper - Pure Biology Secondary 4

Answer Key — Version 4 of 5

PRELIMINARY EXAMINATION — Paper 2

Section A: Multiple Choice Questions [10 marks]

1. B — Rough endoplasmic reticulum [1]

• Marking note: Free ribosomes (A) synthesise proteins for use within the cell, not for secretion. The RER has ribosomes attached and synthesises secretory proteins. The Golgi body (D) modifies and packages but does not synthesise proteins.

2. C — Mitochondrion [1]

• Marking note: The double membrane, cristae, and matrix are distinctive features of mitochondria. Chloroplasts (A) have thylakoids/grana. The nucleus (B) has nuclear pores and chromatin.

3. B — Provides shape and prevents the cell from bursting in a hypotonic solution [1]

• Marking note: The cell wall is rigid and fully permeable. It does not control entry/exit (that is the cell membrane, A). It does not store genetic information (C) or carry out photosynthesis (D).

4. B — It increases the surface area to volume ratio, allowing faster diffusion of oxygen. [1]

• Marking note: The biconcave shape increases SA:V ratio, which is the key adaptation for gas exchange. It does not increase volume (A).

5. B — The starch was broken down into reducing sugars. [1]

• Marking note: Iodine remaining brown (not blue-black) indicates starch is no longer present. This means the enzyme broke down the starch. The enzyme is therefore an amylase, not a protease (C).

6. B — Activity increases up to an optimum temperature, then decreases sharply. [1]

• Marking note: Beyond the optimum, the enzyme denatures — the active site changes shape and the enzyme can no longer bind substrate.

7. B — Osmosis [1]

• Marking note: Water moved from a region of higher water potential (distilled water) to lower water potential (potato cell contents) across a partially permeable membrane. This is osmosis.

8. D — Nucleus [1]

• Marking note: The nucleus contains DNA/chromosomes and controls all cellular activities including protein synthesis and cell division.

9. C — Synthesising proteins [1]

• Marking note: Protein synthesis occurs at ribosomes (on the RER or free in cytoplasm). The Golgi body modifies, packages, sorts, and transports proteins but does not synthesise them.

10. B — A large vacuole and long extension to increase surface area. [1]

• Marking note: The long root hair extension greatly increases surface area for absorption. The large vacuole maintains a water potential gradient. Root hair cells do not have chloroplasts (C) or cilia (D).

Section B: Structured Questions [40 marks]

11.

(a)

• B: Nucleus [1]
• C: Mitochondrion [1]
• E: Golgi body (or Golgi apparatus) [1]
• Marking note: Accept "Golgi apparatus". Do not accept "ER" for E — the Golgi body is a distinct stack of flattened sacs.

(b) Structure D = Rough endoplasmic reticulum

• Function: Synthesis of proteins (for secretion/transport) [1]
• Marking note: Accept "transports proteins" or "protein synthesis". Do not accept "stores proteins".

(c) Mitochondrion is called the "powerhouse of the cell" because:

• It is the site of aerobic respiration / cellular respiration [1]
• It produces ATP (adenosine triphosphate), which is the energy currency used for cellular activities [1]
• Marking note: Both points needed for 2 marks. Simply stating "it produces energy" without mentioning ATP or respiration is insufficient for full marks.

(d) Any two of the following differences [1 mark each, total 2]:

1. Plant cells have a cell wall; animal cells do not.
2. Plant cells have chloroplasts; animal cells do not.
3. Plant cells have a large permanent (central) vacuole; animal cells have small temporary vacuoles or no vacuoles.

• Marking note: Must state the feature in BOTH cell types for each mark. Stating only "plant cells have a cell wall" without mentioning animal cells is acceptable at this level, but full comparison is preferred.

[Total: 8 marks]

12.

(a) pH [1]

• Marking note: The independent variable is what the student deliberately changes. Here, the pH differs between test tubes.

(b) A test tube set up with starch solution and boiled/denatured enzyme X (at the same volume), or a test tube with starch solution and distilled water instead of enzyme [1]

• Marking note: The control must show that the enzyme is responsible for the breakdown. Accept any valid control that replaces the enzyme with an inactive equivalent.

(c) Optimum pH is 7 [1]

• This is because the starch was broken down in the shortest time (4 minutes) at pH 7, meaning the enzyme was most active at this pH [1]
• Marking note: Both the identification of pH 7 AND the explanation (shortest time = highest activity) are needed for 2 marks.

(d) At pH 3, the enzyme is denatured [1]

• The highly acidic conditions cause the active site of the enzyme to change shape, so the substrate can no longer fit into it, and the reaction cannot occur [1]
• Marking note: "Denatured" must be stated. Simply saying "the enzyme doesn't work" is insufficient for the first mark.

(e) Any two of the following [1 mark each]:

1. Temperature
2. Volume/concentration of starch solution
3. Volume/concentration of enzyme X

• Marking note: "Amount" is acceptable for volume. Do not accept "time" — time is the dependent variable being measured.

[Total: 8 marks]

13.

(a) Plasmolysis [1]

• Marking note: Accept "flaccid" but "plasmolysis" is the precise term for the condition shown.

(b) The distilled water has a higher water potential than the cell contents [1]

• Water molecules move into the cell by osmosis [1]
• The cell swells and the cell membrane is pushed against the rigid cell wall, making the cell turgid [1]
• Marking note: All three points needed. Must mention water potential gradient AND osmosis.

(c) The cell would become turgid (or return to its normal turgid state) [1]

• Distilled water has a higher water potential than the cell contents, so water moves into the cell by osmosis [1]
• The cell membrane pushes against the cell wall, restoring turgidity [1]
• Marking note: The cell was only slightly plasmolysed, so it can recover. Must explain the direction of water movement.

(d) Animal cells do not have a cell wall [1]

• When water enters an animal cell by osmosis, the cell swells and eventually bursts (lysis) because there is no rigid wall to resist the pressure [1]
• Plant cells have a rigid cell wall that prevents bursting; the cell becomes turgid instead [1]
• Marking note: Must contrast animal and plant cells. The key difference is the presence/absence of the cell wall.

[Total: 7 marks]

14.

(a) To make more space for haemoglobin, allowing the cell to carry more oxygen [1]

• Marking note: Accept "to increase surface area to volume ratio" or "to carry more oxygen". The absence of a nucleus is an adaptation for oxygen transport.

(b) Palisade mesophyll cells carry out photosynthesis [1]

• Chloroplasts contain chlorophyll, which absorbs light energy needed for photosynthesis; having many chloroplasts maximises light absorption [1]
• Marking note: Must link chloroplasts to photosynthesis. Simply stating "chloroplasts carry out photosynthesis" without mentioning light absorption is insufficient for the second mark.

(c) Any two of the following adaptations [1 mark each]:

1. Has a tail (flagellum) for swimming/motility to reach the egg cell
2. Contains many mitochondria in the midpiece to provide energy (ATP) for movement
3. Has an acrosome (enzyme-containing cap) at the head to digest the egg's outer layer

• Marking note: Must state the adaptation AND link it to function for each mark.

(d) Root hair cells have a long, thin extension that greatly increases the surface area for absorption of water [1]

• They have a large vacuole containing cell sap (a concentrated solution), which maintains a water potential gradient so water enters by osmosis [1]
• They have a thin cell membrane that allows water to pass through easily [1]
• Marking note: Award 2 marks for any two valid points with explanation. Surface area and water potential gradient are the key concepts.

[Total: 7 marks]

15.

(a) Sample Y [1]

• Benedict's test produced a brick-red precipitate, which is a positive result for reducing sugar [1]
• Note: This is 1 mark total — the identification with explanation.

(b) Sample Z [1]

• Iodine test turned blue-black, which is a positive result for starch [1]

(c) Samples X and Z [1]

• Biuret test turned purple, which is a positive result for protein [1]
• Marking note: Both X and Z must be named for the mark.

(d) Sample X [1]

• Ethanol emulsion test produced a cloudy white emulsion, which is a positive result for fat [1]

(e) Steps for Benedict's test [2]:

1. Add Benedict's reagent (blue solution) to the food sample in a test tube [1]
2. Heat the test tube in a water bath at 60–80 °C (or boil) for 2–3 minutes [1]
3. Observe the colour change: blue → green → yellow → orange → brick-red precipitate indicates increasing amounts of reducing sugar

• Marking note: Must mention both adding the reagent AND heating. Award 1 mark for each step.

[Total: 6 marks]

16.

(a) 40 °C [1]

• Marking note: Read from the peak of the curve (point P).

(b) As temperature increases from 0 °C to 40 °C [1]:

• The kinetic energy of both enzyme and substrate molecules increases, leading to more frequent successful collisions between enzyme and substrate [1]
• The rate of the enzyme-catalysed reaction therefore increases [1]
• Note: 2 marks available — award for explaining the increase in kinetic energy/collision frequency and linking it to increased rate.

(c) Above 40 °C, the enzyme begins to denature [1]

• The high temperature disrupts the hydrogen bonds (and other bonds) that maintain the three-dimensional shape of the enzyme's active site [1]
• The active site changes shape so the substrate can no longer bind to it, and the rate of reaction decreases sharply [1]
• Note: 2 marks available — award for stating denaturation and explaining the change in active site shape.

(d) Curve N should be drawn below the original curve, rising to a lower peak at the same optimum temperature (40 °C) [1]

• Marking note: A non-competitive inhibitor binds to a site other than the active site, reducing the maximum rate of reaction but not changing the optimum temperature. The curve should be lower but peak at the same temperature.

[Total: 6 marks]

17.

(a) Transport of oxygen (from lungs to body tissues) [1]

• Marking note: Accept "carries oxygen" or "transports oxygen". Also accept "contains haemoglobin".

(b) The biconcave shape increases the surface area to volume ratio of the cell [1]

• This allows faster diffusion of oxygen into and out of the cell [1]
• The thin centre of the biconcave disc also means oxygen has a shorter distance to diffuse [1]
• Note: 2 marks available — award for SA:V ratio and faster/shorter diffusion.

(c) Any two of the following [1 mark each]:

1. Cell wall
2. Chloroplasts
3. Large permanent/central vacuole

• Marking note: These are structures present in plant cells (palisade mesophyll) but absent in animal cells (red blood cell). Nucleus is NOT acceptable as red blood cells lack a nucleus but the question asks for structures visible in the palisade cell — however, the nucleus is present in both cell types shown (sperm cell has a nucleus, but red blood cell does not). Accept cell wall, chloroplast, and large vacuole as the clearest answers.

(d) Palisade mesophyll cells are the main site of photosynthesis in the leaf [1]

• Chloroplasts contain chlorophyll, which absorbs light energy needed for photosynthesis [1]
• Having many chloroplasts maximises the amount of light energy captured [1]
• Note: 2 marks available — award for linking chloroplasts to photosynthesis and light absorption.

[Total: 7 marks]

Section C: Free Response Questions [20 marks]

18.

(a) Graph plotting [3 marks]:

• Axes correctly labelled with units: x-axis = "Substrate concentration (mmol/dm³)", y-axis = "Rate of reaction (arbitrary units)" [1]
• Appropriate scale used on both axes (evenly spaced, using at least half the grid) [1]
• All points correctly plotted and a smooth curve drawn through the points [1]
• Marking note: Deduct 1 mark if no units on axes. Deduct 1 mark if scale is inappropriate or points are misplotted. The curve should rise steeply at first and then level off (plateau) from 50 mmol/dm³ onwards.

(b) As substrate concentration increases, the rate of reaction increases [1]

• The rate of reaction levels off / reaches a maximum / plateaus at higher substrate concentrations (from 50 mmol/dm³ onwards) [1]
• Marking note: Must describe both the initial increase AND the levelling off.

(c) At higher substrate concentrations, all active sites of the enzyme molecules are occupied/saturated [1]

• Adding more substrate does not increase the rate because there are no free active sites available [1]
• The enzyme is working at its maximum rate (Vmax) [1]
• Marking note: The key concept is enzyme saturation. All three points needed for 3 marks.

(d) 20 mmol/dm³ [1]

• Working: Maximum rate = 38 arbitrary units. Half of maximum = 19. From the table, at 20 mmol/dm³ the rate is 22, which is approximately half. Alternatively, read from the graph where rate = 19.
• Marking note: Accept 18–22 mmol/dm³ depending on graph reading.

(e) The rate of reaction would decrease [1]

• With fewer enzyme molecules, there are fewer active sites available to bind with substrate at any given time [1]
• The maximum rate (Vmax) would be lower [1]
• Note: 2 marks available — award for predicting a decrease and explaining fewer active sites.

[Total: 11 marks]

19.

(a) Labels [3 marks]:

• A: Upper epidermis [1]
• B: Palisade mesophyll [1]
• C: Spongy mesophyll [1]
• D: Stoma (plural: stomata) [1]
• E: Vascular bundle [1]
• Note: 3 marks available — award 1 mark for each correct label, up to 3. If all 5 are correct, award 3 marks.

(b) Palisade mesophyll adaptations for photosynthesis [3]:

• Palisade cells are located near the upper surface of the leaf, where they receive the most sunlight [1]
• They contain many chloroplasts (more than spongy mesophyll cells), which contain chlorophyll to absorb light energy [1]
• The cells are elongated and closely packed, maximising the number of chloroplasts per unit area exposed to light [1]
• Marking note: Must refer to BOTH cell structure (chloroplasts) AND position (near top of leaf) for full marks.

(c) The spongy mesophyll layer has large air spaces between the cells [1]

• These air spaces allow gases (carbon dioxide and oxygen) to diffuse freely through the leaf [1]
• The irregular shape of spongy mesophyll cells also increases the surface area for gas exchange [1]
• Note: 2 marks available — award for air spaces and free diffusion of gases.

(d) The upper epidermis is transparent to allow light to pass through to the palisade mesophyll layer below [1]

• Marking note: Must mention light transmission. Accept "so light can reach the chloroplasts in palisade cells".

[Total: 9 marks]

20.

(a) Enzyme (or protein) [1]

• Marking note: Accept "enzyme" or "protein". Lactase is an enzyme, which is a type of protein.

(b)

• Substrate: Lactose [1]
• Products: Glucose and Galactose [1]
• Marking note: Both products must be named for 1 mark. Spelling must be reasonably correct.

(c) The stomach has a very low pH (approximately pH 1–2), which is highly acidic [1]

• At this pH, lactase would be denatured because the active site changes shape [1]
• Additionally, the stomach contains pepsin (a protease), which would digest/break down the lactase enzyme (since lactase is a protein) [1]
• Therefore, the enzyme would no longer function to break down lactose [1]
• Note: 3 marks available — award for low pH/denaturation and protease digestion.

(d) Predictions [3]:

• At pH 2: Very low/no rate of reaction — the enzyme is denatured by the highly acidic conditions [1]
• At pH 6.5: Highest rate of reaction — this is the optimum pH for lactase [1]
• At pH 10: Very low/no rate of reaction — the enzyme is denatured by the highly alkaline conditions [1]
• Marking note: Must predict AND explain for each pH. Award 1 mark per pH condition.

(e) 37 °C is the normal body temperature of humans [1]

• At this temperature, the enzyme and substrate molecules have sufficient kinetic energy for optimal collision frequency without the enzyme being denatured [1]
• Marking note: Must link 37 °C to body temperature. The second mark is for explaining that this temperature provides optimal conditions for enzyme activity.

[Total: 11 marks]

Mark Summary

END OF ANSWER KEY