Secondary 3 Biology Semestral Assessment 2 (End of Year) Paper 5

TuitionGoWhere Practice Paper - Biology Secondary 3

SA2 Practice Paper — Version 5 of 5

Answer Key and Marking Scheme

Section A: Multiple Choice Questions [10 marks]

1. B – Rough endoplasmic reticulum [1]
Note: The RER has ribosomes attached to its surface and is the site of protein synthesis, particularly for proteins destined for secretion. The Golgi body modifies and packages these proteins but is not the site of initial synthesis.

2. C – Golgi body [1]
Note: The Golgi body consists of stacked, flattened membrane cisternae (sacs) with vesicles budding off. This is a distinctive feature visible under electron microscopy.

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

4. A – Cell wall: Present (Plant Cell) / Absent (Animal Cell) [1]
Note: Plant cells have a cellulose cell wall outside the cell membrane; animal cells do not. Chloroplasts are present in plant cells, not animal cells. Both cell types have a nucleus and mitochondria.

5. B – It will swell and may burst due to water entering the cell by osmosis [1]
Note: Distilled water has a higher water potential than the cell sap. Water moves by osmosis from a region of higher water potential (distilled water) to a region of lower water potential (inside the cell). The cell swells and may undergo lysis.

6. C – Fat [1]
Note: Fats yield approximately 37 kJ per gram, compared to approximately 17 kJ per gram for carbohydrates and 16 kJ per gram for proteins.

7. B – Water moved out of the potato cells by osmosis into the sucrose solution [1]
Note: The 20% sucrose solution has a lower water potential than the cell sap of the potato cells. Water moves by osmosis from the potato cells (higher water potential) into the sucrose solution (lower water potential), causing a decrease in mass.

8. C – Enzymes speed up reactions by lowering the activation energy [1]
Note: Enzymes are biological catalysts. They are not used up in reactions. They lower the activation energy, allowing reactions to proceed faster. They have optimal temperature and pH ranges.

9. C – Reducing sugar [1]
Note: Benedict's test for reducing sugars produces a colour change from blue through green, yellow, orange to brick-red/orange-red precipitate in the presence of increasing concentrations of reducing sugar.

10. C – Mitochondrion [1]
Note: Mitochondria are present in both plant and animal cells and are the sites of aerobic respiration (Krebs cycle and oxidative phosphorylation). Chloroplasts, cell walls, and large central vacuoles are found only in plant cells.

Section B: Structured Questions [25 marks]

11. Cell Structure and Organisation [4 marks]

(a) Identify structures B and D. [2]

• B: Nucleus [1]
• D: Rough endoplasmic reticulum [1]
  Note: Accept "rough ER" or "RER".

(b) Function of structure C (mitochondrion). [1]

• Aerobic respiration / site of ATP production / releases energy from glucose [1]
  Note: Accept any one correct function. "Respiration" alone is acceptable.

(c) Why structure D appears "rough". [1]

• Ribosomes are attached to its surface [1]
  Note: The "rough" appearance is due to the presence of ribosomes bound to the cytoplasmic face of the endoplasmic reticulum membrane.

12. Biomolecules – Food Tests [4 marks]

(a) Which sample contains reducing sugar? [2]

• Sample Q [1] – because it turned orange-red with Benedict's solution after heating, indicating the presence of reducing sugar [1]
  Note: Benedict's solution changes from blue to green/yellow/orange/brick-red in the presence of reducing sugars. Orange-red indicates a moderate to high concentration.

(b) Which sample contains both protein and starch? [2]

• Sample R [1] – because it turned blue-black with iodine solution (indicating starch) and turned purple with Biuret test (indicating protein) [1]
  Note: Iodine turns blue-black in the presence of starch. Biuret test turns purple/violet in the presence of protein.

13. Cell Specialisation [3 marks]

(a) Define cell differentiation. [1]

• The process by which unspecialised cells become specialised to carry out specific functions [1]
  Note: Accept equivalent wording, e.g., "the process where cells develop specific structures to perform specific roles/functions."

(b) Two adaptations of a sperm cell for its function. [2]

Any two of the following, [1] each:

• Many mitochondria in the middle piece – to provide energy (ATP) for swimming/movement towards the egg
• Acrosome (at the tip of the head) – contains enzymes to digest the outer membrane of the ovum
• Long, whip-like flagellum (tail) – for motility/swimming towards the egg
• Streamlined/compact head – to reduce resistance during swimming
• Haploid nucleus – contains half the chromosome number so that upon fertilisation, the diploid number is restored

Note: Each adaptation must be linked to its function for the mark. Stating the adaptation alone without the function explanation is acceptable at this level.

14. Osmosis – Data Interpretation [4 marks]

(a) Why did the potato cylinder in solution X increase in mass? [2]

• Solution X has a concentration of 0.0 mol/dm³ (pure water), which has a higher water potential than the cell sap of the potato cells [1]
• Water moved into the potato cells by osmosis, from a region of higher water potential to a region of lower water potential, causing the cells to gain mass [1]
  Note: Must mention osmosis and water potential gradient for full marks.

(b) Estimate the concentration of the cell sap. [2]

• Approximately 0.1–0.2 mol/dm³ [1]
• This is the concentration at which there would be no net movement of water (no change in mass). Since solution X (0.0) caused a mass increase and solution Y (0.3) caused a mass decrease, the isotonic point lies between 0.0 and 0.3 mol/dm³. A reasonable estimate is around 0.1–0.2 mol/dm³ [1]
  Note: Accept any value between 0.05 and 0.25 mol/dm³ with appropriate reasoning. The key is that the student explains the estimate is between the concentration that caused gain and the concentration that caused loss.

15. Enzymes [4 marks]

(a) Describe the relationship between temperature and amylase activity. [2]

• As temperature increases from 10 °C to 37 °C, the time taken for starch to disappear decreases (from 25 to 4 minutes), meaning amylase activity increases [1]
• Above 37 °C, as temperature increases further to 70 °C, the time taken increases (from 4 to 25 minutes), meaning amylase activity decreases [1]
  Note: Accept equivalent descriptions. Students may state that 37 °C is the optimum temperature for amylase.

(b) Why is the time longest at 70 °C? [2]

• At 70 °C, the amylase enzyme is denatured [1]
• The high temperature causes the enzyme's active site to lose its specific shape, so substrate molecules can no longer fit into it, and the reaction slows down or stops [1]
  Note: Must mention denaturation and change in active site shape for full marks. Simply stating "the enzyme is destroyed" is not sufficient.

16. Radioactive Tracer Pathway [3 marks]

(a) Sequence of organelles where radioactivity is detected. [2]

• Rough endoplasmic reticulum → Golgi body → Secretory vesicles (or cell membrane) [2]
  Note: Award [2] for all three in correct order. Award [1] if two are in correct order or if the sequence is partially correct.

(b) Why does radioactivity appear first in the rough endoplasmic reticulum? [1]

• Because the rough endoplasmic reticulum is the site of protein synthesis (ribosomes on its surface assemble amino acids into proteins), so radioactive amino acids are first incorporated into proteins here [1]
  Note: Must link the function (protein synthesis) to the appearance of radioactivity.

17. Biomolecules – Structure and Function [3 marks]

(a) Two differences between a fat molecule and a starch molecule. [2]

Any two of the following, [1] each:

• Fats are made of glycerol and fatty acids; starch is made of glucose monomers
• Fats do not dissolve in water; starch is insoluble in water (both are insoluble, but fats are hydrophobic while starch is a large polymer)
• Fats yield more energy per gram than starch when oxidised
• Fats are not a polymer of repeating subunits in the same way starch is (starch is a polysaccharide made of many glucose units)
• Fats contain the elements C, H, and O with a much higher proportion of C–H bonds; starch also contains C, H, and O but with a lower proportion of C–H bonds

Note: Accept any two valid, distinct differences.

(b) Why are fats a more efficient energy store than carbohydrates? [1]

• Fats contain a higher proportion of C–H bonds, which release more energy when broken during oxidation/respiration [1]
  Note: Accept "fats yield more energy per gram" or equivalent.

Section C: Free Response Questions [15 marks]

18. Cell Structure and Function [5 marks]

(a) Three cell structures in a palisade mesophyll cell but NOT in a human cheek cell, with functions. [3]

Any three of the following, [1] each (structure + function):

• Cell wall – provides structural support and maintains cell shape; prevents the cell from bursting when it gains water by osmosis
• Chloroplast – site of photosynthesis; contains chlorophyll which absorbs light energy
• Large central vacuole (permanent vacuole) – stores cell sap; maintains turgidity of the cell; pushes cytoplasm and chloroplasts to the edge of the cell

Note: Award [1] for each correct structure with its correct function. Structure alone without function = 0 marks for that part.

(b) How the structure and position of palisade mesophyll cells are adapted for photosynthesis. [2]

• Position: Located near the upper surface of the leaf, where they receive the most sunlight for photosynthesis [1]
• Structure: They contain many chloroplasts (more than other leaf cells), which can move within the cell to absorb maximum light; they are elongated and closely packed, providing a large surface area for light absorption [1]
  Note: Award [1] for position adaptation and [1] for structural adaptation. Accept any valid structural adaptation.

19. Osmosis – Experimental Design [5 marks]

(a) Independent and dependent variables. [2]

• Independent variable: Concentration of sucrose solution [1]
• Dependent variable: Mass of potato cylinder (or percentage change in mass) [1]
  Note: Must be correctly identified. Swapping the variables = 0 marks.

(b) Two variables to be kept constant. [2]

Any two of the following, [1] each:

• Length/diameter/size of potato cylinders
• Type/variety of potato (same source)
• Duration of time in the solution (30 minutes)
• Temperature of the experiment
• Volume of sucrose solution used
• Same method of blotting dry before reweighing

Note: "Mass of potato cylinder" is not acceptable as it is the dependent variable. "Concentration of sucrose" is the independent variable and is not acceptable.

(c) One safety precaution. [1]

Any one of the following:

• Be careful when using the cork borer to cut potato cylinders (risk of cutting fingers)
• Handle the scalpel/cutter with care
• Wear safety goggles to protect eyes from splashes
• Clean up any spills immediately to prevent slipping
• Wash hands after handling chemicals/potato

Note: Accept any reasonable safety precaution relevant to the experiment.

20. Enzymes and Biomolecules – Extended Response [5 marks]

(a) Enzyme that digests proteins in the stomach and its conditions. [2]

• Enzyme: Pepsin [1]
• Conditions: Acidic pH (pH 1.5–2) / works best in acidic conditions / at body temperature (~37 °C) [1]
  Note: Accept "protease" as a general term, but "pepsin" is the specific enzyme. The stomach produces hydrochloric acid which creates the acidic environment.

(b) How products of protein digestion are absorbed in the small intestine. [3]

• Proteins are digested into amino acids by proteases (e.g., pepsin, trypsin) [1]
• The villi in the small intestine provide a large surface area for absorption; they have a thin epithelial lining (one cell thick) for short diffusion distance [1]
• Each villus contains a network of blood capillaries into which amino acids are absorbed by diffusion and active transport; the capillaries transport the amino acids away in the blood, maintaining a concentration gradient [1]
  Note: Award [1] for mentioning amino acids as the product, [1] for villus structure adaptations (large surface area, thin wall, rich blood supply), and [1] for the process of absorption (diffusion/active transport into capillaries). Accept reference to microvilli for additional surface area.

Mark Summary

End of Answer Key