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A Level H1 Biology Cells Biomolecules Quiz
Free A Level H1 Biology Cells Biomolecules quiz, HY3 Exam version, with questions, answers, and A Level-style practice for Singapore students.
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A-Level Biology H1 Quiz - Cells Biomolecules: Answer Key
Total Marks: 40
Topic: Cells & Biomolecules (Core Idea 1)
Section A Answers (1–10)
1. [1 mark]
Answer: Any one of: cells are the smallest unit of life; all cells come from pre-existing cells; living organisms are composed of cells.
Teaching note: Cell theory is a foundational concept. Mark awarded for one correct statement. Common mistake: confusing with "cells can arise spontaneously" (false).
2. [2 marks]
Answer: Organelle: bacterial cell (or prokaryote) has no membrane-bound nucleus; difference: lacks membrane-bound organelles such as mitochondria / has peptidoglycan wall (vs cellulose in plant).
Mark breakdown: 1 for naming bacterium/prokaryote context, 1 for structural difference.
Note: Syllabus specifies bacterial cell has circular DNA, 70S ribosomes, peptidoglycan wall, no membrane-bound organelles.
3. [2 marks]
Answer: Phospholipids form a bilayer; hydrophilic heads face outward to aqueous environment, hydrophobic tails face inward away from water.
Mark breakdown: 1 for bilayer, 1 for orientation of heads/tails.
Teaching: This arrangement creates a selective barrier.
4. [2 marks]
Answer: Glycosidic bond; enzyme class: carbohydrase (or glycosidase / amylase).
Mark breakdown: 1 bond name, 1 enzyme class.
Note: α-glucose polymers (starch) linked by α-1,4 glycosidic bonds.
5. [1 mark]
Answer: Cholesterol provides stability / reduces fluidity at high temp and prevents crystallisation at low temp.
Teaching: It fits between phospholipid tails.
6. [2 marks]
Answer: Osmosis; passive transport (or simple diffusion of water).
Mark breakdown: 1 term, 1 type.
Note: Water moves out to region of higher solute concentration (lower water potential).
7. [2 marks]
Answer: Glucose moves down its concentration gradient via a carrier protein; no ATP needed because no energy input required for passive movement.
Mark breakdown: 1 gradient/carrier, 1 no ATP.
Teaching: Facilitated diffusion is passive.
8. [2 marks]
Answer: Tertiary (and quaternary); other bond: hydrogen / ionic / hydrophobic interaction.
Mark breakdown: 1 level, 1 bond type.
Note: Disulfide bonds stabilise tertiary structure.
9. [1 mark]
Answer: Totipotent.
Teaching: Zygotic stem cells can form all cell types including extra-embryonic.
10. [2 marks]
Answer: S phase; thymine is incorporated into DNA during replication.
Mark breakdown: 1 phase, 1 reason.
Teaching: Radioactive thymine labels new DNA strands.
Section B Answers (11–15)
11. [3 marks]
(a) A: Golgi body (or Golgi apparatus); C: mitochondrion (with cristae). [2]
(b) B (smooth ER): lipid synthesis / detoxification. [1]
Teaching: Smooth ER lacks ribosomes; identified by tubular network.
12. [3 marks]
Answer: Uptake plateaus because carriers saturate (all transporter proteins occupied) OR internal concentration equals external (equilibrium). [2] External X constant but finite number of carriers / gradient reduced. [1]
Mark breakdown: 2 for saturation/equilibrium, 1 for constant external not sufficient.
Teaching: Shows facilitated diffusion, not simple diffusion.
13. [2 marks]
Answer: G (glycoprotein): cell recognition / receptor; Integral protein: channel / carrier / pump. [1+1]
Teaching: Glycoproteins have carbohydrate chains external.
14. [3 marks]
(a) 40 °C [1]
(b) At 60 °C enzyme denatured: tertiary structure disrupted, active site lost. [2]
Mark breakdown: 1 opt, 2 explanation.
Teaching: Beyond optimum, H-bonds/ionic bonds break.
15. [2 marks]
Answer: Any two: circular DNA (no nucleus); 70S ribosomes; peptidoglycan wall; no mitochondria. [1 each]
Teaching: Diagram must show nucleoid not enclosed.
Section C Answers (16–20)
16. [4 marks]
Answer: Cellulose is polymer of β-glucose with β-1,4 glycosidic bonds [1]; straight chains form H-bonds with adjacent chains → microfibrils [1]; provides tensile strength [1]; rigid wall supports plant cells against turgor [1].
Teaching: Contrast with starch (α, coiled, storage).
17. [3 marks]
Answer: Pyruvate enters mitochondrial matrix directly and is decarboxylated in Krebs cycle → CO₂ [1+1]; glucose requires glycolysis in cytoplasm first (not in mitochondrion) so isolated mitochondria cannot metabolise it [1].
Mark breakdown: 2 pyruvate reason, 1 glucose location.
18. [4 marks]
Answer: Enzyme active site initially not complementary [1]; substrate binding induces conformational change improving fit [1]; stresses substrate bonds / lowers Ea [1]; enzyme–substrate complex forms, product released [1].
Teaching: Induced-fit vs lock-and-key.
19. [3 marks]
Answer: Multipotent blood stem cells [1]; can differentiate into limited range of blood cell types (RBC, WBC, platelets) [1]; self-renew to maintain population [1].
Teaching: Multipotent < pluripotent < totipotent.
20. [4 marks]
Answer: Haemoglobin has 4 subunits (quaternary) [1]; cooperative binding: binding of O₂ to one subunit increases affinity of others [1]; graph sigmoid shows steep load at lung pO₂ [1]; plateau at tissue ensures release only at low pO₂ [1].
Teaching: Quaternary structure enables allostery.




