A Level H1 Biology Practice Paper 1

TuitionGoWhere Exam Practice (AI) — Biology H1 A-Level

Practice Paper: Cells & Biomolecules (Version 1 of 5) — Answer Key

Total Marks: 60
Duration: 60 minutes

Section A: Short Structured Questions (20 marks)

1. [1 mark]
Answer: The nucleolus is the site of ribosome subunit assembly (or rRNA synthesis).
Teaching note: The nucleolus is a region within the nucleus where ribosomal RNA is transcribed and ribosomal subunits are assembled before export to the cytoplasm.

2. [1 mark]
Answer: Glycosidic bond.
Teaching note: α-glucose monomers link via condensation to form glycosidic bonds in starch (amylose/amylopectin).

3. [1 mark]
Answer: Rough ER has ribosomes attached to its surface; smooth ER does not.
Teaching note: The presence of ribosomes gives rough ER a granular appearance and links it to protein synthesis.

4. [2 marks]
Answer: Phospholipids form a bilayer; hydrophilic heads face outward to aqueous environments, hydrophobic tails face inward away from water. [1+1]
Teaching note: This arrangement creates a stable barrier; the bilayer is central to the fluid mosaic model.

5. [1 mark]
Answer: Totipotent (or zygotic stem cells are totipotent).
Teaching note: Totipotency = can form all cell types including extra-embryonic tissues.

6. [1 mark]
Answer: Endocytosis (or phagocytosis if particle is solid).
Teaching note: Endocytosis invaginates membrane to internalise large materials.

7. [1 mark]
Answer: β-glucose has a straight chain conformation allowing H-bonds between adjacent chains.
Teaching note: The β(1→4) linkage produces linear chains for cellulose microfibrils.

8. [2 marks]
Answer: Facilitated diffusion uses channel/carrier proteins down a concentration gradient; no ATP needed because no energy input against gradient. [1 for gradient, 1 for no ATP]
Teaching note: It is passive, unlike active transport which needs ATP.

9. [1 mark]
Answer: The cytoplasm (or freely in cytosol; bacterial ribosomes are 70S, not in organelle).
Teaching note: Bacteria lack membrane-bound organelles; ribosomes are free in cytoplasm.

10. [1 mark]
Answer: High temperature breaks hydrogen bonds, causing denaturation/unfolding.
Teaching note: Secondary structure (α-helix, β-sheet) relies on H-bonds; heat disrupts them.

Section B: Diagram and Data Interpretation (20 marks)

11. [2 marks]
Answer: Structure C = rough endoplasmic reticulum; role = synthesis and transport of proteins for secretion. [1 name, 1 role]
Teaching note: Rough ER is identified by ribosomes; in secretory cells it makes proteins for export.

12. [3 marks]
Answer: Molecule X enters via facilitated diffusion (or active transport if against gradient, but plateau suggests carrier saturation) [1]; curve rises as gradient drives entry [1]; plateau occurs when all carriers saturated [1].
Teaching note: Saturation indicates protein-mediated transport, not simple diffusion.

13. [2 marks]
Answer: (i) Lack of membrane-bound organelles; (ii) circular DNA / peptidoglycan wall / 70S ribosomes. [1 each]
Teaching note: Eukaryotic cells have nucleus and organelles; bacteria do not.

14. [3 marks]
Answer: Optimum pH = 7 (highest rate 45). [1] Percentage decrease = (45−14)/45 × 100 = 68.9% (or 69%). [2 for calculation]
Working: Decrease = 45 − 14 = 31; 31/45 × 100 = 68.9%.
Teaching note: Optimum is peak rate; % decrease from pH 7 to 9 uses pH 7 as base.

15. [2 marks]
Answer: Cholesterol restricts phospholipid movement, reducing fluidity at high temp and preventing solidification at low temp. [1 role, 1 stability]
Teaching note: It modulates membrane fluidity and mechanical stability.

Section C: Extended Structured Responses (20 marks)

16. [4 marks]
Answer:

• Formation: amino acids join by condensation; carboxyl group of one + amino group of next release water, forming peptide bond. [2]
• Breakage: hydrolysis adds water to split peptide bond. [1]
• Reactants: amino acids; product: dipeptide/polypeptide + H₂O. [1]
  Teaching note: Peptide bond is –CO–NH–; condensation vs hydrolysis are reverse reactions.

17. [3 marks]
Answer: Pyruvate enters mitochondrial matrix and Krebs cycle → CO₂ released. [1] Glucose needs glycolysis in cytoplasm first; mitochondria alone lack glycolytic enzymes. [1] No glycolysis → no pyruvate → no CO₂ in mitochondria. [1]
Teaching note: Glycolysis occurs in cytosol; mitochondria only process pyruvate onward.

18. [5 marks]
Answer:

• Primary: sequence of amino acids; peptide bonds. [1]
• Secondary: α-helix/β-sheet; H-bonds. [1]
• Tertiary: 3D folding; H-bonds, ionic, disulfide, hydrophobic. [1+1]
• Quaternary: multiple polypeptides; same bonds as tertiary. [1]
  Teaching note: Each level adds stabilising interactions; quaternary only in multi-subunit proteins (e.g. haemoglobin).

19. [3 marks]
Answer: Blood stem cells (haematopoietic). [1] They are multipotent [1] – can differentiate into limited range (e.g. RBCs, WBCs) but not all types. [1]
Teaching note: Multipotent = tissue-specific restricted potency; replaces worn cells continuously.

20. [5 marks]
Answer:

• Plateau: all active sites occupied at saturating substrate; rate independent of [S]. [2]
• More enzyme: more active sites → higher max rate, curve shifts up/plateau higher. [2]
• Reference to graph shape required. [1]
  Teaching note: Michaelis-Menten style saturation; enzyme concentration changes Vmax not Km.