A Level H1 Biology Practice Paper 2

Biology H1 A‑Level Practice Paper – Answer Key and Marking Scheme

Version 2
Total Marks: 53

Section A: Multiple Choice

1. B – Smooth endoplasmic reticulum is the site of lipid synthesis. [1]
2. C – The hydrophilic heads face both the cytoplasm and the exterior; hydrophobic tails face inwards. [1]
3. B – Palisade mesophyll cells contain numerous chloroplasts and a large central vacuole. [1]
4. B – Radioactive uracil is incorporated into RNA; transcription occurs throughout interphase but is high during G₁ when the cell is actively growing and synthesising proteins. (Accept G₁ or G₂; G₁ is more typical for initial detection after short exposure.) [1]
5. D – Increasing enzyme concentration increases rate until substrate becomes limiting. [1]

Section B: Short Structured Response

6(a) Mitochondrion. [1]
6(b) Produces ATP through aerobic respiration to fuel metabolic processes such as protein synthesis, detoxification or active transport. (Any valid liver‑related function.) [1]

7. Glucose is a large, polar molecule; it moves by facilitated diffusion. It binds to a specific carrier protein on the surface of the membrane. The carrier protein undergoes a conformational change, transporting glucose down its concentration gradient into the cell. (Award 2 marks for describing the mechanism, with reference to the figure showing carrier proteins.) [2]

8. Phospholipids form a bilayer. The hydrophilic (phosphate) heads face the aqueous environment on both sides (cytoplasm and exterior), while the hydrophobic (fatty acid) tails face inwards, away from water. This arrangement creates a selectively permeable barrier. [2]

9. Protease Y shows maximum activity at pH 2 (45 au). The stomach normally has a very low pH (~2), which provides the optimal pH for this enzyme. At this pH, the enzyme’s active site maintains its correct three‑dimensional shape, allowing the formation of enzyme‑substrate complexes at a high rate. Hence the stomach is suitable for protein digestion by this protease. (Reference to Table 1 data and enzyme‑pH relationship required.) [2]

10. Phase: S phase (DNA synthesis). Explanation: Radioactive thymine is a base specific to DNA; during S phase, DNA replication occurs, and thymine is incorporated into the newly synthesised DNA strands. Thus the radioactivity in the nucleus increases significantly during this phase. [2]

Section C: Data‑Based and Diagram Interpretation

11(a)

• CO₂ is produced during the Krebs cycle (citric acid cycle) in the mitochondrial matrix.
• Pyruvate can enter the mitochondrion directly, where it is converted to acetyl‑CoA and enters the Krebs cycle, releasing CO₂.
• Glucose cannot cross the mitochondrial membrane; it must first undergo glycolysis in the cytoplasm. Glycolysis does not produce CO₂. Since the incubation contains only isolated mitochondria (without cytoplasmic enzymes), glucose cannot be metabolised and no CO₂ is produced. [3]

12. Transport mechanism: Facilitated diffusion.
Explanation: The curve shows that the rate of uptake increases with concentration up to a maximum, then plateaus (becomes constant) at high concentrations. This indicates that uptake involves membrane proteins (carriers or channels) that become saturated at high substrate concentration; there is no further increase in rate because all transport proteins are occupied. Simple diffusion would produce a straight line through the origin; active transport would usually require energy. [3]

13(a) Rough endoplasmic reticulum (RER) is studded with ribosomes that synthesise (translate) polypeptide chains of the digestive enzymes. [1]
13(b) The Golgi apparatus (structure B) receives vesicles containing enzyme proteins from the RER. It modifies the proteins (e.g. glycosylation), packages them, and sorts them into secretory vesicles (structure C). The secretory vesicles transport the enzymes to the cell membrane for exocytosis. These organelles are abundant because the cell’s primary function is synthesis, modification and secretion of proteins; a large volume of membrane trafficking is required. [3]

14(a) At substrate concentrations above 4.0 mmol dm⁻³, the active sites of all enzyme molecules are saturated with substrate. The rate is limited by the rate at which the enzyme‑substrate complexes break down to release product. Increasing substrate concentration does not increase the rate because no free active sites are available. [2]
14(b) Vmax ≈ 33 μmol min⁻¹ (accept 32–33). [1]

15(a) Net movement from X to Y (down the concentration gradient). Process: simple diffusion. [1]
15(b) Oxygen is a small, non‑polar molecule. It can dissolve in the hydrophobic core of the phospholipid bilayer and diffuse through without requiring transport proteins. The concentration gradient drives the movement. [2]

Section D: Extended Response

16. Significance of membrane transport to photosynthesis (6 marks)

• Photosynthesis requires CO₂, water, and the export of carbohydrate products. All these substances must cross membranes.
• CO₂ uptake: CO₂ diffuses through stomata and across the plasma membrane of spongy mesophyll cells down a concentration gradient. It then diffuses into chloroplasts. Without CO₂ entry, the Calvin cycle cannot proceed.
• Water uptake: Water enters root hairs by osmosis across the plasma membrane; it travels from cell to cell, eventually reaching photosynthetic tissues. Water is the electron donor in the light‑dependent reactions.
• Ion transport: Active transport of mineral ions (e.g. Mg²⁺ for chlorophyll, K⁺ for stomatal opening) across cell membranes is essential for chloroplast function and stomatal control.
• Product export: Triose phosphates (G3P) are exported from the chloroplast via phosphate translocators in the inner membrane. Sucrose loaded into phloem requires co‑transport with H⁺ (active transport) across companion cell membranes.
• Thus membrane transport regulates the supply of raw materials, the internal concentrations, and the removal of photosynthetic products, directly influencing the rate and efficiency of photosynthesis.

(Award marks for covering at least three substances, linking transport mechanism to photosynthetic importance, and showing overall integration.) [6]

17. Endomembrane system role (4 marks)

• Ribosomes on rough ER synthesise the enzyme as an inactive precursor.
• The polypeptide enters the RER lumen, where it folds and is packaged into transport vesicles.
• Vesicles fuse with the Golgi apparatus; the Golgi modifies the protein (e.g. by adding carbohydrate groups) and packages it into secretory vesicles.
• Secretory vesicles move to the cell membrane, fuse with it, and release the enzyme by exocytosis.
  (Any four steps clearly linked to the secretion pathway.) [4]

18. Prokaryotic vs. eukaryotic animal cell (4 marks)
Three differences with functional significance:

• Nucleus: Eukaryotic cell has a true nucleus bounded by a double membrane; prokaryotic DNA is free in the nucleoid. This allows compartmentalisation of transcription and translation in eukaryotes, offering more control over gene expression.
• Membrane‑bound organelles: Eukaryotic cells contain mitochondria, ER, etc.; prokaryotes lack them. The presence of mitochondria enables efficient aerobic respiration and much higher ATP yield.
• Ribosomes: Prokaryotes have 70S ribosomes (smaller); eukaryotes, 80S. This difference is exploited by certain antibiotics that selectively inhibit prokaryotic protein synthesis.
  (Any three valid differences, with a clear explanation of functional significance, are acceptable.) [4]

19. Properties of water (4 marks)

• High specific heat capacity: Water absorbs a lot of heat with little temperature change, providing a stable environment for aquatic organisms and temperature regulation in cells.
• Cohesion/tension: Hydrogen bonds create cohesion between water molecules, enabling the transpiration pull in plants, which aids upward movement of water.
• Excellent solvent: Water’s polarity allows it to dissolve ions and polar molecules, making it an ideal medium for metabolic reactions and transport of solutes (e.g., glucose, amino acids).
  (Award up to 4 marks for clear explanation of three properties linked to specific biological roles.) [4]

20. Enzyme inhibition comparison (4 marks)

• Competitive inhibition: The inhibitor resembles the substrate and binds to the active site, blocking substrate binding. This is reversible; increasing substrate concentration can outcompete the inhibitor and restore maximum rate (Vmax remains unchanged; Km increases). Diagram: active site with inhibitor similar to substrate.
• Non‑competitive inhibition: The inhibitor binds to an allosteric site (distinct from the active site), altering the enzyme’s tertiary structure and shape of the active site. The substrate may still bind but the enzyme is less catalytically active. This cannot be overcome by increasing substrate concentration (Vmax decreases; Km unchanged). Diagram: enzyme with separate allosteric site and inhibitor binding, distorted active site.
  (Award 2 marks for mechanism explanation plus diagram elements, and 2 marks for overcoming each type.) [4]

End of Answer Key