A Level H2 Biology Cells Biomolecules Quiz

A-Level Biology H2 Quiz - Cells Biomolecules (Answer Key)

Total Marks: 40
Topic: Cells & Biomolecules (Syllabus 9477 Core Idea 1)

Section A Answers (1–8)

Q1 [2 marks]

• Both have ribosomes for protein synthesis.
• Both are surrounded by a cell membrane (cell surface membrane) and contain genetic material (DNA).
  Marking: 1 mark each for any two correct shared features (e.g., ribosomes, plasma membrane, DNA, cytoplasm). Common mistake: stating "nucleus" – prokaryotes lack a membrane-bound nucleus.

Q2 [1 mark]

• Glycosidic bond.
  Teaching note: Formed by condensation reaction between two glucose monomers; O-glycosidic in starch (α-1,4 and α-1,6).

Q3 [2 marks]

• Bacterial cell lacks membrane-bound organelles (e.g., no mitochondrion); plant cell has mitochondria and other membrane-bound organelles.
  OR Bacterial cell wall is made of peptidoglycan; plant cell wall is cellulose.
  Marking: 1 for correct structure named, 1 for accurate comparison. Electron micrograph would show absence of nuclear envelope in bacteria.

Q4 [2 marks]

• Cholesterol restricts movement of phospholipids, reducing membrane fluidity at high temperature and preventing tight packing at low temperature, thus maintaining stability.
  Marking: 1 for role in fluidity regulation, 1 for temperature buffering effect.

Q5 [1 mark]

• Influenza virus (or HIV / any enveloped virus from syllabus).
  Teaching: Enveloped viruses have a lipid envelope derived from host membrane.

Q6 [2 marks]

• The enzyme's active site changes shape slightly to fit the substrate upon binding, forming an enzyme-substrate complex that lowers activation energy.
  Marking: 1 for conformational change, 1 for induced fit concept vs rigid lock-and-key.

Q7 [2 marks]

• Totipotent. Means it can differentiate into all cell types including extra-embryonic tissues (full organism).
  Marking: 1 for term, 1 for explanation.

Q8 [2 marks]

• Phagocytosis; lysosome (fuses with phagosome to digest).
  Marking: 1 each.

Section B Answers (9–14)

Q9 [3 marks]

• Structure C = Golgi body (Golgi apparatus). [1]
• Structure B (mitochondrion) function: site of aerobic respiration / ATP production via oxidative phosphorylation. [2]
  Marking descriptors: ID 1 mark; function 2 marks for detail (double membrane, cristae, ETC). Image must show labelled C as stacked membranes.

Q10 [3 marks]

• At high [substrate], all enzyme active sites are saturated with substrate. [1]
• Adding more substrate cannot increase rate because enzyme concentration is fixed. [1]
• Rate is limited by enzyme availability / turnover number. [1]
  Teaching: Hyperbolic curve from graph shows zero-order kinetics at plateau.

Q11 [2 marks]

• The two fatty acid tails (hydrocarbon chains) are hydrophobic.
  Marking: 1 for fatty acids, 1 for hydrophobic tail concept.

Q12 [4 marks]

• Treatment 1 (ADP+Pi): oxygen consumed at 12.0 because ADP accepts electrons via ETC, driving proton gradient and ATP synthase; O₂ is final electron acceptor. [2]
• Treatment 3 (no ADP): low O₂ use (1.5) because lack of ADP means proton gradient not dissipated, ETC backs up (respiratory control). [2]
  Marking: 2+2 for linking ADP to ETC activity and explaining control.

Q13 [2 marks]

• Ionic bond (salt bridge) / disulfide bridge / hydrophobic interaction. (Any one other than H-bond.)
  Marking: 1 mark.

Q14 [3 marks]

• Genetic material = DNA (double-stranded linear). [1]
• Challenges cell theory because virus is not cellular, cannot reproduce independently without host cell, and is not composed of cells. [2]
  Marking: challenges cell theory by lacking cellular organisation.

Section C Answers (15–20)

Q15 [2 marks]

• Glycogen: short-term energy store in animals (liver/muscle). [1]
• Cellulose: structural component of plant cell walls, not used for energy directly. [1]

Q16 [3 marks]

• Triglyceride has 3 fatty acids → high reduced C-H bonds → high energy yield per gram. [1]
• Insoluble in water → no osmotic effect. [1]
• Compact storage in adipose tissue. [1]

Q17 [4 marks]

• Quaternary = 4 polypeptide subunits (2α, 2β). [1]
• Each subunit has haem group binding O₂ cooperatively. [1]
• Globular soluble in RBC cytoplasm. [1]
• Conformational change on O₂ binding increases affinity (sigmoid curve). [1]

Q18 [3 marks]

• Active transport: requires ATP, against gradient, specific carrier. [1.5]
• Facilitated diffusion: down gradient, no ATP, channel/carrier. [1.5]
  Table acceptable.

Q19 [3 marks]

• Multipotent: can form multiple specialised cell types of a tissue lineage but not all. [2]
• Use: treat leukaemia via bone marrow transplant (lymphoid/myeloid restore). [1]

Q20 [4 marks]

• Competitive inhibitor binds active site, blocking substrate → lower rate at given [S]. [2]
• Induced-fit: inhibitor mimics substrate shape; increasing [S] outcompetes inhibitor → Vmax restored. [2]
  Marking: 2 for mechanism, 2 for substrate reversal.