Secondary 3 Biology Semestral Assessment 2 (End of Year) Paper 3

TuitionGoWhere Practice Paper - Biology Secondary 3

SA2 (Version 3 of 5) — Answer Key & Marking Scheme

Total Marks: 50

Section A: Multiple Choice Questions [10 marks]

1. B — Rough endoplasmic reticulum [1]
Note: RER has ribosomes attached and synthesises proteins destined for secretion. Students commonly select Golgi body (C), but proteins are first synthesised on RER before being transported to the Golgi body for modification and packaging.

2. C — Golgi body [1]
Note: The Golgi body consists of stacked, flattened membrane sacs (cisternae) with vesicles budding off from the trans face.

3. C — Nucleus [1]
Note: Thymine is a nitrogenous base found in DNA. During cell division, DNA replication occurs in the nucleus, so radioactive thymine would first be incorporated there.

4. B — Chloroplast: Present (Plant Cell) / Absent (Animal Cell) [1]
Note: Option A is incorrect because animal cells lack a cell wall. Option C is incorrect because both cell types have a nucleus. Option D is incorrect because both have mitochondria.

5. B — It will swell and may burst due to water entering the cell by osmosis [1]
Note: Distilled water is hypotonic relative to the cell contents. Water moves into the cell by osmosis, causing it to swell and potentially lyse (haemolysis in red blood cells).

6. D — Carbohydrates [1]
Note: Glucose (a carbohydrate) is the primary immediate energy source. Lipids are for long-term energy storage. Proteins and nucleic acids are not primary energy sources.

7. C — 37 °C [1]
Note: Human enzymes have an optimal temperature around 37 °C (body temperature). At 60 °C, the enzyme would be denatured.

8. B — It controls the movement of substances into and out of the cell [1]
Note: The cell membrane is partially permeable and regulates transport. Rigid support is provided by the cell wall (plants only). Photosynthesis occurs in chloroplasts. Genetic material is stored in the nucleus.

9. B — Water moved out of the potato cells by osmosis, causing them to become plasmolysed [1]
Note: The concentrated salt solution is hypertonic relative to the cell sap. Water moves out by osmosis, causing the cytoplasm to shrink away from the cell wall (plasmolysis), making the tissue feel flaccid.

10. B — Rough endoplasmic reticulum and Golgi body [1]
Note: Hormones that are proteins (e.g., insulin) are synthesised on the RER, modified and packaged by the Golgi body, and secreted via vesicles. A cell secreting large amounts of hormones would have abundant RER and Golgi body.

Section B: Structured Questions [25 marks]

11.

(a) [2]
B: Mitochondrion [1]
D: Golgi body [1]
Note: Award 1 mark for each correct identification. Accept "Golgi apparatus" for D.

(b) [1]
The rough endoplasmic reticulum is the site where proteins are synthesised (by ribosomes attached to its surface). [1]
Note: Accept any equivalent phrasing, e.g., "protein synthesis" or "synthesis of proteins for secretion".

(c) [2]
Structure A is the nucleus, which contains DNA / genetic material [1]. The DNA carries the genetic instructions / codes for the synthesis of proteins, which are essential for controlling cell activities and cell division [1].
Note: Award 1 mark for identifying that the nucleus contains DNA/genetic material. Award 1 mark for linking DNA to protein synthesis or control of cell activities. Simply stating "controls cell activities" without reference to DNA/proteins scores only 1 mark.

12.

(a) [2]
Independent variable: Temperature [1]
Dependent variable: Time taken for starch to break down [1]
Note: Award 1 mark for each correct variable. "Rate of reaction" is not acceptable as the dependent variable because the student measured time, not rate.

(b) [1]
As the temperature increases from 10 °C to 37 °C, the time taken for starch to break down decreases [1], meaning the rate of reaction increases.
Note: Award 1 mark for describing the inverse relationship between temperature and time (or direct relationship between temperature and rate).

(c) [2]
At 70 °C, the enzyme amylase was denatured [1]. The high temperature caused the enzyme's active site to change shape, so the substrate (starch) could no longer fit into the active site, and the reaction could not occur [1].
Note: Award 1 mark for stating denaturation. Award 1 mark for explaining that the active site shape is altered. Simply stating "the enzyme is destroyed" without reference to active site shape scores only 1 mark.

(d) [1]
Using five test tubes (one at each temperature) ensures that all tests are conducted simultaneously / under the same conditions, which makes the experiment a fair test [1].
Note: Accept equivalent answers such as "to ensure a fair test" or "to control variables". Simply stating "to get accurate results" without explanation is insufficient.

13.

(a) [2]

1. Palisade mesophyll cells have a cell wall, which animal cells lack. [1]
2. Palisade mesophyll cells contain chloroplasts, which animal cells lack. [1]
   Note: Also accept: large central vacuole (present in plant cells, small/absent in animal cells). Award 1 mark for each correct difference. The difference must be stated as a comparison (i.e., what the plant cell has that the animal cell does not).

(b) [3]

• Palisade mesophyll cells contain many chloroplasts [1], which contain the pigment chlorophyll that absorbs light energy for photosynthesis.
• These cells are located near the upper surface of the leaf [1], where they receive the most sunlight.
• The cells are elongated and tightly packed [1], which maximises the number of chloroplasts exposed to light.
  Note: Award 1 mark each for any three valid adaptation-function links. Each point must link a structural feature to its function in photosynthesis. Simply listing adaptations without explaining their function scores a maximum of 1 mark.

14.

(a) [1]
Carotenoid [1]
Note: Accept "carotene" or "xanthophyll". The orange colour in citrus peel cells is due to carotenoid pigments stored in the vacuole or chromoplasts.

(b) [2]
The large central vacuole is filled with cell sap (a solution of sugars, salts, and pigments in water) [1]. When the vacuole is full of water, it exerts turgor pressure against the cell wall, keeping the cell turgid / firm and providing structural support to the plant [1].
Note: Award 1 mark for identifying cell sap or water in the vacuole. Award 1 mark for linking turgor pressure to structural support/rigidity. Simply stating "stores water" without linking to cell structure scores only 1 mark.

15.

(a) [2]
The raisins would increase in size in Setup X (distilled water) [1]. The distilled water is hypotonic compared to the contents of the raisin cells, so water moves into the raisin cells by osmosis, causing them to swell [1].
Note: Award 1 mark for identifying Setup X. Award 1 mark for explaining osmosis (water moving into the cells due to the water potential gradient).

(b) [1]
Osmosis [1]
Note: Accept "osmosis" only. Do not accept "diffusion" — osmosis is the specific term for the movement of water across a partially permeable membrane.

16.

(a) [2]
Cell Type P is likely a secretory cell (e.g., a cell in the pancreas that secretes digestive enzymes, or a plasma cell that secretes antibodies) [1]. This is because it has a high abundance of rough endoplasmic reticulum and Golgi body, which are involved in the synthesis, modification, and packaging of proteins for secretion [1].
Note: Award 1 mark for identifying a secretory cell type. Award 1 mark for linking the organelle abundance to the function of protein secretion. Accept any valid example of a secretory cell.

(b) [2]
Cell Type Q is likely a muscle cell (or sperm cell / actively contracting cell) [1]. This is because it has a high abundance of mitochondria, which carry out aerobic respiration to produce ATP needed for muscle contraction / high energy demand [1].
Note: Award 1 mark for identifying a cell with high energy demands. Award 1 mark for linking mitochondria abundance to ATP production for energy. Accept any valid example (e.g., muscle cell, sperm cell, liver cell).

17.

(a) [1]
Water (H₂O) [1]
Note: The reaction is: 2H₂O₂ → 2H₂O + O₂. Award 1 mark for "water" or "H₂O".

(b) [2]
Catalase would show the highest activity at pH 7 [1]. Catalase is an enzyme found in living cells (e.g., potato, liver), and enzymes have an optimum pH at which they work best. At pH 3 and pH 11, the enzyme would be denatured because the extreme pH changes the shape of the active site [1].
Note: Award 1 mark for predicting pH 7. Award 1 mark for explaining that extreme pH denatures the enzyme / alters the active site shape. Accept "the enzyme works best at neutral pH" as a valid explanation. Note: catalase from potato has an optimum around pH 7; students should not assume pH 2 (which is the optimum for pepsin, a stomach enzyme).

Section C: Data-Based and Extended Response Questions [15 marks]

18.

(a) [2]
The iodine solution remained amber/yellow, indicating that starch is not present in the water outside the Visking tubing [1]. This is because starch molecules are too large to pass through the pores of the partially permeable Visking tubing membrane [1].
Note: Award 1 mark for stating that starch is absent. Award 1 mark for explaining that starch molecules are too large to pass through the membrane.

(b) [2]
The Benedict's test showed an orange-red precipitate, indicating that glucose is present in the water outside the Visking tubing [1]. This is because glucose molecules are small enough to pass through the pores of the partially permeable Visking tubing membrane by diffusion [1].
Note: Award 1 mark for stating that glucose is present. Award 1 mark for explaining that glucose molecules are small enough to pass through the membrane. Accept "small molecules can pass through" as a valid explanation.

(c) [2]
The Benedict's test would show a blue solution (negative result) [1]. This is because starch is not a reducing sugar and would not react with Benedict's solution. Since there is no glucose in the Visking tubing, no glucose would diffuse out, and the test would remain negative [1].
Note: Award 1 mark for predicting a negative result (blue solution). Award 1 mark for explaining that starch is not a reducing sugar / no glucose is present to diffuse out. Simply stating "no glucose" without explanation scores only 1 mark.

19.

(a) [2]
As the substrate concentration increases, the rate of reaction increases [1]. This is because there are more substrate molecules available to bind to the active sites of enzymes, leading to more enzyme-substrate complexes being formed per unit time [1].
Note: Award 1 mark for describing the increase. Award 1 mark for explaining the increase in terms of more substrate molecules binding to active sites. At low enzyme concentration, the curve eventually plateaus — if the student describes the full trend (increase then plateau), this is also acceptable for 2 marks.

(b) [2]
At high substrate concentration, all the 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 for additional substrate molecules to bind to [1].
Note: Award 1 mark for stating that active sites are saturated/occupied. Award 1 mark for explaining that no free active sites are available. Simply stating "all enzyme molecules are used up" is incorrect — enzymes are not used up in reactions.

(c) [2]
At high enzyme concentration, there are more enzyme molecules present [1], which means there are more active sites available to bind with substrate molecules, so more enzyme-substrate complexes can form per unit time, resulting in a higher maximum rate [1].
Note: Award 1 mark for stating that there are more enzyme molecules. Award 1 mark for linking more enzyme molecules to more active sites and a higher maximum rate.

20.

(a) [3]
Solution P: Starch and protein [1]
Solution Q: Reducing sugar (glucose) [1]
Solution R: None of the three biomolecules tested (or no reducing sugar, no starch, no protein) [1]
Note: Award 1 mark for each correct identification. For Solution P, the blue-black iodine test indicates starch, and the purple Biuret test indicates protein. Benedict's was negative, so no reducing sugar. For Solution Q, the orange-red Benedict's test indicates reducing sugar; iodine and Biuret were negative. For Solution R, all three tests were negative.

(b) [1]
Solution P gave a positive Biuret test (purple colour) because it contains protein [1]. The Biuret reagent reacts with peptide bonds in proteins to produce a purple colour.
Note: Award 1 mark for stating that Solution P contains protein. Accept "peptide bonds are present" as an alternative explanation.

(c) [2]
The student could perform the emulsion test (ethanol test) [1]. Add ethanol to Solution Q, shake, and then pour the mixture into water. If a white emulsion / cloudy white suspension forms, this confirms the presence of lipid [1].
Note: Alternatively, accept: The student could use Benedict's test with dilution — if the student suspects the reducing sugar is glucose specifically, they could use glucose test strips or perform chromatography to identify the specific sugar. Award 1 mark for naming a valid test. Award 1 mark for describing the expected positive result. Note: Since Solution Q tested negative for starch and protein, the student may want to test for lipid as the remaining biomolecule.

Mark Summary

Common Mistakes to Watch For

1. Confusing RER and Golgi body in protein secretion pathways — Students often select Golgi body as the site of protein synthesis. Emphasise that proteins are synthesised on ribosomes on the RER, then transported to the Golgi body for modification and packaging.

2. Using "diffusion" when "osmosis" is required — Osmosis specifically refers to the movement of water across a partially permeable membrane. In questions involving water movement into or out of cells, "osmosis" is the expected term.

3. Stating enzymes are "killed" by high temperature — The correct term is "denatured". Enzymes are proteins, not living organisms. Students should explain that denaturation involves a change in the shape of the active site.

4. Not linking structure to function in adaptation questions — In questions about cell adaptations (e.g., palisade mesophyll cells), students must explain how each structural feature helps the cell carry out its function. Simply listing adaptations without explanation will not earn full marks.

5. Confusing the order of the radioactive tracer pathway — For amino acids: RER → Golgi body → secretory vesicles → cell membrane. Students should trace the pathway of protein synthesis and secretion carefully.

6. Incorrectly identifying the Benedict's test result — A positive Benedict's test produces a colour change from blue → green → yellow → orange → brick-red precipitate. Students should not describe the result as just "red" — "orange-red precipitate" or "brick-red precipitate" is more precise.