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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.
These static practice materials are generated from the site's syllabus and paper-generation workflow, with source and model context shown so students and parents can evaluate the material before use.
Questions
A-Level Biology H1 Quiz - Cells Biomolecules
Name: ___________________________
Class: ___________________________
Date: ___________________________
Score: ___________ / 40
Duration: 60 minutes
Total Marks: 40
Instructions:
- Answer all 20 questions.
- Section A: Short structured questions (1–10).
- Section B: Diagram and data interpretation (11–15).
- Section C: Extended application (16–20).
- Write your answers in the spaces provided. Use pen. Show working where requested.
Section A: Short Structured Questions (1–10)
1. State one feature of the cell theory. [1]
2. Name the organelle in a eukaryotic cell that contains circular DNA and 70S ribosomes, and state one structural difference from a typical plant cell. [2]
3. Describe the arrangement of phospholipids in the cell surface membrane. [2]
4. α-glucose and β-glucose differ in the position of the hydroxyl group on carbon 1. Name the bond formed when two α-glucose molecules join, and state the enzyme class that breaks it. [2]
5. With reference to the fluid mosaic model, state the role of cholesterol in the membrane. [1]
6. A bacterial cell is exposed to a solution with a higher solute concentration than its cytoplasm. State the term for the movement of water out of the cell and the type of transport involved. [2]
7. Explain why facilitated diffusion of glucose does not require ATP. [2]
8. State the level of protein structure that is stabilised by disulfide bonds and give one other bond type involved at that level. [2]
9. Describe the unique property of zygotic stem cells using the correct term from the syllabus. [1]
10. A cell is incubated with radioactive thymine. State which phase of the cell cycle would first show increased radioactivity in the nucleus and why. [2]
Section B: Diagram and Data Interpretation (11–15)
11. The electron micrograph below shows part of a eukaryotic cell.
Image pending generation: diagram for Q11.
(a) Name structures A and C. [2]
(b) State one function of structure B in this cell. [1]
12. The graph shows the uptake of molecule X by a cell over 10 minutes at 25 °C.
Image pending generation: graph for Q12.
Explain why the uptake of X plateaus despite a constant external concentration. [3]
13. The figure shows a section of a phospholipid bilayer with proteins.
Image pending generation: diagram for Q13.
With reference to Fig., state the role of glycoprotein G and one role of an integral protein. [2]
14. Table shows results of an experiment on enzyme activity.
| Temperature (°C) | Rate of reaction (μmol min⁻¹) |
|---|---|
| 20 | 12 |
| 30 | 25 |
| 40 | 38 |
| 50 | 30 |
| 60 | 8 |
(a) State the optimum temperature for this enzyme. [1]
(b) Explain the decrease in rate from 50 °C to 60 °C. [2]
15. The diagram shows a bacterial cell.
Image pending generation: diagram for Q15.
State two features visible in the diagram that show this is a bacterial cell, not a eukaryotic cell. [2]
Section C: Extended Application (16–20)
16. Explain how the structure of cellulose relates to its function in plant cell walls. [4]
17. A mitochondrion was incubated with pyruvate and separately with glucose. CO₂ was produced only with pyruvate. Explain this observation using knowledge of cellular compartmentation. [3]
18. Describe the induced-fit hypothesis of enzyme action and explain how it lowers activation energy. [4]
19. Blood stem cells replace worn-out red blood cells. State the type of stem cell involved and describe its unique properties using syllabus terms. [3]
20. The graph shows haemoglobin saturation with oxygen at different partial pressures.
Image pending generation: graph for Q20.
Explain how the quaternary structure of haemoglobin supports its oxygen transport role with reference to the graph shape. [4]
Answers
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.
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