Secondary 3 Biology Cells Biomolecules Quiz

Secondary 3 Biology Quiz - Cells Biomolecules (Answer Key)

Total Marks: 40

Section A: Multiple Choice Questions (10 marks)

1. B — Rough endoplasmic reticulum [1]
Explanation: Radioactive amino acids are incorporated into proteins during translation. In eukaryotic cells, protein synthesis for secretion or membrane insertion begins on ribosomes attached to the rough endoplasmic reticulum (RER). The RER is the first organelle in the secretory pathway to receive newly synthesised polypeptides. The Golgi body receives proteins later via transport vesicles from the RER.
Common mistake: Choosing Golgi body (A) — this is the next organelle in the pathway, not the first.

2. B — Mitochondria [1]
Explanation: Glucose is broken down during cellular respiration. Aerobic respiration occurs mainly in the mitochondria (Krebs cycle and oxidative phosphorylation), where glucose-derived pyruvate is fully oxidised to CO₂, producing ATP. Radioactive glucose would therefore label mitochondrial components most rapidly during active respiration. Chloroplasts (A) are for photosynthesis, not respiration.

3. C — Glycogen → Glucose [1]
Explanation: Glycogen is a polysaccharide made of α-glucose monomers. Starch (A) is also made of glucose, not amino acids. Proteins (B) are made of amino acids, not nucleotides. DNA (D) is made of nucleotides, not fatty acids.

4. B — Reducing sugar [1]
Explanation: Benedict's test detects reducing sugars (e.g., glucose, fructose, maltose). A blue solution (Cu²⁺ ions) changes to green, yellow, orange, or brick-red precipitate (Cu₂O) upon heating with a reducing sugar. Starch (A) gives blue-black with iodine test. Protein (C) gives purple with biuret test. Fat (D) gives a cloudy white emulsion with ethanol emulsion test.

5. A A — Brown to blue-black [1]
Explanation: Iodine solution is brown/yellow. When it contacts starch, it forms a blue-black starch-iodine complex. A leaf exposed to sunlight photosynthesises and produces starch. Adding iodine to the decolourised leaf shows blue-black where starch is present. The colour change observed is from the brown iodine solution to blue-black on the leaf.

6. B — Rough ER: Synthesis of proteins; Golgi body: Modification and packaging of proteins [1]
Explanation: Ribosomes on the rough ER synthesise proteins destined for secretion or membranes. The Golgi body modifies (e.g., glycosylation), sorts, and packages these proteins into vesicles for transport. Option A swaps the functions. Option C reverses them. Option D is incorrect for both.

7. C — Liver cell [1]
Explanation: The cell has a nucleus, numerous mitochondria, rough ER, and Golgi bodies — typical of an active animal cell like a hepatocyte (liver cell). It lacks a cell wall and chloroplasts, ruling out plant cells (A: palisade mesophyll, B: root hair). Red blood cells (D) lack a nucleus and most organelles in mammals.

8. C — Enzymes are used up during the reaction they catalyse. [1]
Explanation: Enzymes are catalysts — they are not consumed in the reaction and can be reused. Statements A, B, and D are all correct: enzymes are proteins (mostly), lower activation energy, and are sensitive to temperature and pH.

9. C — The enzyme is denatured and loses its active site shape. [1]
Explanation: Above the optimum temperature (37°C), increased thermal energy breaks the weak bonds (hydrogen bonds, ionic bonds) maintaining the enzyme's tertiary structure. The active site loses its specific shape, so substrates can no longer bind effectively. The enzyme is denatured. Options A and B are incorrect — kinetic energy increases with temperature. Option D is unrelated to temperature effects.

10. A — Osmosis — water moves out of the potato cells [1]
Explanation: The concentrated sugar solution has a lower water potential (more negative) than the potato cell cytoplasm. Water moves by osmosis from higher water potential (inside cell) to lower water potential (outside solution). The cell loses water, becomes flaccid/plasmolysed, and the tissue softens and loses mass. Option D says "diffusion" — water movement across a partially permeable membrane is specifically osmosis.

Section B: Structured Questions (20 marks)

11. (a) A: Cell wall [1]
B: Cell membrane [1]
F: Chloroplast [1]
Marking note: Accept "cellulose cell wall" for A, "plasma membrane" for B.

(b) Mitochondrion (G) is the site of aerobic respiration, producing ATP for cellular activities. [1]
Accept: "Releases energy from glucose", "Produces ATP", "Site of cellular respiration".

(c) The large central vacuole (E) maintains turgor pressure against the rigid cell wall, providing structural support to the plant. It also stores water, nutrients, and waste. [2]
Mark breakdown: 1 mark for turgor pressure/support; 1 mark for storage function. Animal cells lack a cell wall, so large vacuoles would cause bursting; they have small temporary vacuoles instead.

12. (a) Solution Y [1]
Explanation: Biuret test: blue → purple/violet indicates protein (peptide bonds). Blue (no change) means no protein.

(b) Biuret reagent contains copper(II) sulfate in alkaline solution. The Cu²⁺ ions form a violet/purple complex with peptide bonds (–CO–NH–) in proteins. [2]
Mark breakdown: 1 mark for alkaline copper(II) sulfate / Cu²⁺ ions; 1 mark for peptide bond complex formation. Do not accept "nitrogen atoms" alone — must specify peptide bonds.

(c) Reducing sugar (e.g., glucose, maltose) [1]
Explanation: Benedict's test positive (brick-red precipitate) indicates reducing sugar.

13. (a) Rough endoplasmic reticulum (RER) [1]
Explanation: Secretory proteins are synthesised by ribosomes on the RER. The nucleus (DNA → mRNA) directs synthesis but the polypeptide chain is assembled at the RER.

(b) The Golgi body modifies the protein (e.g., adds carbohydrate chains to form glycoproteins), sorts it, and packages it into secretory vesicles for transport to the cell membrane. [2]
Mark breakdown: 1 mark for modification (glycosylation/folding); 1 mark for sorting/packaging into vesicles.

(c) Exocytosis [1]
Explanation: Secretory vesicles fuse with the cell membrane and release contents outside the cell.

14. (a) Both are polymers of α-glucose / both are polysaccharides / both have glycosidic bonds. [1]
Accept any valid structural similarity.

(b) Glycogen is more highly branched (α-1,6 branches every 8–12 glucose units) than starch (amylopectin branches every 24–30 units; amylose is unbranched). [1]
Accept: "Glycogen has more branches" or "Starch has two forms (amylose and amylopectin) while glycogen is a single highly branched molecule".

(c) Glycogen's highly branched structure allows rapid hydrolysis at many ends simultaneously, releasing glucose quickly for energy. Its compact granular form allows dense storage in liver and muscle cells. [2]
Mark breakdown: 1 mark for many ends → rapid glucose release; 1 mark for compact storage. Link structure to function.

15. (a) The active site of the enzyme has a specific 3D shape complementary to the substrate. Only substrates with the matching shape can fit into the active site, forming an enzyme-substrate complex. This ensures each enzyme catalyses only a specific reaction. [2]
Mark breakdown: 1 mark for complementary/specific shape; 1 mark for only matching substrates fit → specificity.

(b) A competitive inhibitor has a similar shape to the substrate and competes for the active site. It blocks the active site, preventing substrate binding. This reduces the rate of reaction because fewer enzyme-substrate complexes form. The inhibition can be overcome by increasing substrate concentration. [2]
Mark breakdown: 1 mark for similar shape/competes for active site; 1 mark for reduced rate / fewer complexes / overcome by high substrate.

Section C: Extended Response Questions (10 marks)

16. (a) Graph plotting [3]
Marking points:

• Axes correctly labelled with units: x-axis "pH", y-axis "Time taken for starch digestion (s)" [1]
• Appropriate linear scales covering all data points (pH 3–11, time 0–200 s) [1]
• All 5 points plotted accurately (± half a small square) and connected with a smooth curve or ruled lines [1]
  Note: Time is inversely related to rate. Minimum time = maximum rate at pH 7.

(b) pH 7 [1]
Explanation: Shortest time (30 s) = fastest digestion = highest enzyme activity.

(c) At pH 3 (acidic) and pH 11 (alkaline), the H⁺ or OH⁻ ions disrupt the ionic and hydrogen bonds maintaining the enzyme's tertiary structure. The active site loses its specific shape (denaturation), so substrate (starch) cannot bind effectively. Fewer enzyme-substrate complexes form, reducing the reaction rate, so digestion takes longer. [3]
Mark breakdown: 1 mark for H⁺/OH⁻ disrupt bonds; 1 mark for active site shape lost / denaturation; 1 mark for fewer complexes → slower rate → longer time.

(d) Adding HCl lowers the pH from 7 (optimum) to acidic. The enzyme (amylase) will denature — its active site shape is permanently altered. The reaction will slow down and eventually stop. Starch digestion will be incomplete. [3]
Mark breakdown: 1 mark for pH drops / becomes acidic; 1 mark for enzyme denatures / active site shape lost; 1 mark for reaction slows/stops / incomplete digestion.

17. (a) P: Hypotonic [1]
Q: Isotonic [1]
R: Hypertonic [1]
Explanation: Hypotonic = higher water potential than cell → water enters → lysis. Isotonic = same water potential → no net water movement. Hypertonic = lower water potential → water leaves → crenation.

(b) Solution P has a higher water potential (less negative) than the red blood cell cytoplasm. Water enters the cell by osmosis down the water potential gradient. The cell swells and eventually bursts (lyses) because it lacks a cell wall to resist the pressure. [2]
Mark breakdown: 1 mark for water potential gradient / water enters by osmosis; 1 mark for swelling and bursting due to no cell wall.

(c) Plant cells have a rigid, fully permeable cell wall made of cellulose. As water enters by osmosis, the cell membrane pushes against the cell wall, generating turgor pressure. The cell wall prevents further expansion and bursting. The cell becomes turgid. [2]
Mark breakdown: 1 mark for cell wall presence; 1 mark for cell wall resists pressure / prevents bursting / turgidity.

18. Ethanol emulsion test: [4]

1. Add about 2 cm³ of the food sample (liquid) or crushed solid sample to a test tube. [1]
2. Add an equal volume of ethanol and shake vigorously to dissolve any lipids. [1]
3. Pour the mixture into a test tube containing an equal volume of distilled water. [1]
4. A cloudy white emulsion (milky suspension) indicates the presence of fat/lipid. [1]
   Marking notes: Must mention ethanol, water, and cloudy white emulsion. "Shake" or "mix" required. If solid sample, "crush/grind" first. Do not accept "add Sudan III" — not standard in Sec 3 syllabus.

19. (a) Condensation reaction (or esterification) [1]
Explanation: Three water molecules are removed as three ester bonds form between glycerol's –OH groups and fatty acids' –COOH groups.

(b) Saturated fatty acids have no C=C double bonds (all C–C single bonds); unsaturated fatty acids have one or more C=C double bonds. [1]
Accept: "Saturated = straight chains; unsaturated = kinked/bent chains due to double bonds".

(c) Fats yield more energy per gram (~37 kJ/g) than carbohydrates (~17 kJ/g) because they have a higher proportion of C–H bonds and fewer oxygen atoms, so they are more reduced. More energy is released when they are oxidised during respiration. Fats are also hydrophobic and stored without water, making them more compact. [2]
Mark breakdown: 1 mark for higher energy yield per gram / more reduced / more C–H bonds; 1 mark for hydrophobic/compact storage (no water of hydration).

20. (a) The cell membrane pulls away from the cell wall. The cytoplasm shrinks and becomes concentrated in the centre of the cell. The cell becomes plasmolysed. [2]
Mark breakdown: 1 mark for membrane pulls away from wall / cytoplasm shrinks; 1 mark for "plasmolysed" or description of plasmolysis.

(b) Plasmolysis (or exosmosis) [1]
Accept: "Osmosis — water moves out of the cell".

(c) Distilled water has a higher water potential than the cell cytoplasm. Water enters the cell by osmosis. The cytoplasm expands, pushing the cell membrane back against the cell wall. The cell becomes turgid (deplasmolysis). [2]
Mark breakdown: 1 mark for water enters by osmosis / higher water potential outside; 1 mark for membrane pushes against wall / cell becomes turgid / deplasmolysis.

End of Answer Key