A Level H2 Biology Practice Paper 2

TuitionGoWhere Practice Paper - Biology H2 A-Level

Answer Key & Marking Scheme (Version 2)

Topic: Cells & Biomolecules
Total Marks: 60

Section A: Structured Questions

1. Membrane Structure (a) Structure X: Cholesterol. [1]
Property: It is rigid/ring-structured and restricts the movement of phospholipid fatty acid tails, reducing fluidity at high temperatures / prevents membrane from becoming too fluid. [1]
(Note: If X is identified as a protein, max 1 mark for property if relevant to function, but X is typically cholesterol in this standard diagram context. Accept Phospholipid if labelled correctly, but property must be amphipathic nature.)

(b) Phospholipids are amphipathic molecules, having a hydrophilic phosphate head and hydrophobic fatty acid tails. [1]
In an aqueous environment, the hydrophilic heads face outward towards the water (cytoplasm and extracellular fluid). [1]
The hydrophobic tails face inward, away from water, forming a stable bilayer core. [1]

(c) At high temperatures, phospholipids have high kinetic energy and move more, increasing fluidity. [1]
Cholesterol restricts the movement of the fatty acid tails, thereby stabilising the membrane and preventing it from becoming too fluid/leaky. [1]

2. Co-transport (a) The Na⁺/K⁺ ATPase pump actively transports 3 Na⁺ ions out of the cell and 2 K⁺ ions into the cell using ATP. [1]
This creates a low concentration of Na⁺ inside the cell compared to the intestinal lumen. [1]
This establishes a steep electrochemical gradient for Na⁺ to move back into the cell. [1]

(b) Glucose moves against its concentration gradient into the cell, coupled with Na⁺ moving down its gradient. [1]
The energy for glucose transport is derived from the Na⁺ gradient (established by primary active transport), not directly from ATP hydrolysis at the co-transporter. Hence, it is secondary active transport. [1]
Glucose exits the cell into the blood via facilitated diffusion because the concentration of glucose is higher in the epithelial cell than in the blood, allowing it to move down its concentration gradient through a channel/carrier protein without energy input. [1]

3. Enzyme Kinetics (a) A competitive inhibitor binds to the active site, competing with the substrate. [1]
At high substrate concentrations, the substrate molecules outnumber the inhibitor molecules. [1]
The substrate successfully outcompetes the inhibitor for the active sites, allowing all enzyme molecules to form enzyme-substrate complexes, thus achieving the same $V_{max}$ as the uninhibited reaction. [1]

(b) The $K_m$ increases. [1]
Because the inhibitor occupies some active sites, a higher concentration of substrate is required to achieve half the maximum velocity ($V_{max}/2$) compared to the uninhibited reaction. [1]

4. Haemoglobin Structure (a) Primary: Sequence of amino acids in the polypeptide chains. [1]
Secondary: Hydrogen bonds between the backbone atoms form alpha-helices. [1]
Tertiary: 3D folding of each globin chain stabilised by hydrogen bonds, ionic bonds, disulfide bridges, and hydrophobic interactions between R-groups. [1]
Quaternary: Association of four polypeptide subunits (2 alpha, 2 beta) held together by similar interactions (hydrophobic, ionic, etc.) to form the functional haemoglobin molecule. [1]

(b) The substitution replaces a hydrophilic amino acid (glutamic acid) with a hydrophobic one (valine). [1]
Under low oxygen conditions, the haemoglobin molecule changes shape, exposing the hydrophobic valine residue. [1]
This hydrophobic patch interacts with hydrophobic regions on other haemoglobin molecules, causing them to polymerise/aggregate into long fibres, distorting the red blood cell. [1]

5. DNA Replication (a) Each new DNA molecule consists of one original (parental) strand and one newly synthesised strand. [1]

(b) DNA strands are antiparallel (one 5'→3', the other 3'→5'). [1]
DNA polymerase can only add nucleotides to the 3' end of the growing strand (synthesises 5'→3'). [1]
Therefore, one strand (leading) is synthesised continuously towards the replication fork, while the other (lagging) is synthesised discontinuously away from the fork in Okazaki fragments. [1]

(c) Joins the Okazaki fragments on the lagging strand by forming phosphodiester bonds. [1]

Section B: Data Interpretation and Application

6. Temperature and Enzymes (a) The rate of reaction increases from 10°C to 40°C (optimum). [1]
Above 40°C, the rate decreases rapidly, reaching zero at 70°C. [1]

(b) High temperatures cause the kinetic energy of the enzyme molecules to increase. [1]
This breaks the hydrogen bonds and ionic bonds maintaining the tertiary structure. [1]
The active site changes shape (denaturation), so the substrate can no longer bind to form an enzyme-substrate complex. [1]

(c) pH 3.0 is highly acidic and far from the optimum pH for amylase (approx pH 7). [1]
The excess H⁺ ions disrupt the ionic and hydrogen bonds in the enzyme's tertiary structure, causing denaturation and loss of activity. [1]

7. Gel Electrophoresis (a) Heterozygous ($R_1 R_2$). [1]

(b) Individual B has two different alleles ($R_1$ and $R_2$). [1]
Each allele produces a DNA fragment of a different size (200bp and 300bp), which migrate to different positions on the gel. Individual A is homozygous ($R_1 R_1$), producing only one fragment size. [1]

(c) DNA phosphate groups are negatively charged. [1]
(Accept: DNA is negatively charged)

8. Mitochondrial Respiration (a) Oxygen acts as the final electron acceptor in the electron transport chain (ETC). [1]
As electrons pass down the ETC, they combine with oxygen and protons to form water, causing oxygen concentration in the buffer to decrease. [1]

(b) (i) Oxygen consumption will stop (or decrease to zero). [1]
Cyanide blocks the ETC at Complex IV, preventing electrons from being passed to oxygen. Without electron flow, oxygen cannot be reduced to water. [1]

(ii) ATP production will stop. [1]
Blocking the ETC prevents the pumping of protons into the intermembrane space. Without a proton gradient, chemiosmosis cannot occur, and ATP synthase cannot produce ATP. [1]

9. Lac Operon (a) In the absence of lactose, the regulator gene produces a repressor protein. [1]
The repressor protein binds to the operator region of the DNA. [1]
This blocks RNA polymerase from binding to the promoter, preventing transcription of the structural genes. [1]

(b) Lactose acts as an inducer. [1]
Lactose binds to the repressor protein, causing a conformational change that makes the repressor release from the operator. [1]
RNA polymerase can now bind to the promoter and transcribe the structural genes (lacZ, lacY, lacA). [1]

10. Water Properties (a) Water molecules are polar and form extensive hydrogen bonds with each other. [1]
A large amount of heat energy is required to break these hydrogen bonds before the molecules can move faster (increase in temperature), resulting in high specific heat capacity. [1]

(b) It helps buffer temperature changes in organisms/cells, maintaining a stable internal environment for enzyme activity. [1]
(Or: Large bodies of water act as thermal buffers for aquatic habitats.)

Section C: Extended Response

11. RNA Structures and Functions (10 Marks)

Marking Guidance:

• mRNA (3-4 marks):
  • Structure: Single-stranded, linear, contains codons (triplets), Uracil instead of Thymine.
  • Function: Carries genetic code from DNA in nucleus to ribosomes in cytoplasm.
  • Adaptation: Short-lived (allows rapid control of protein synthesis); sequence determines amino acid order.
• tRNA (3-4 marks):
  • Structure: Cloverleaf shape (2D) / L-shape (3D), contains anticodon loop, amino acid attachment site (3' end), extensive hydrogen bonding within molecule.
  • Function: Transports specific amino acids to ribosome; matches codon on mRNA via anticodon.
  • Adaptation: Specific shape ensures correct amino acid attachment; anticodon allows precise base pairing with mRNA.
• rRNA (2-3 marks):
  • Structure: Globular, complex folding, associated with proteins to form ribosomes (large and small subunits).
  • Function: Catalytic role (ribozyme) in peptide bond formation; structural component of ribosome.
  • Adaptation: Stable structure provides platform for mRNA and tRNA interaction; catalytic site facilitates translation.
• Comparison/Quality (1 mark): Clear distinction between roles (carrier vs translator vs catalyst/structure) and link to protein synthesis stages (transcription/translation).

12. Membrane Transport and Homeostasis (10 Marks)

Marking Guidance:

• Definition: Homeostasis is the maintenance of a stable internal environment despite external changes. Membrane transport is crucial for regulating solute and water balance.
• Passive Transport Examples (3-4 marks):
  • Oxygen/CO2 exchange: Simple diffusion in lungs/alveoli maintains blood gas levels for respiration.
  • Glucose uptake: Facilitated diffusion in liver/muscle (via GLUT4) lowers blood glucose after a meal, maintaining normoglycemia.
  • Osmosis: Water movement maintains cell turgor (plants) or volume (animal cells), preventing lysis or crenation.
• Active Transport Examples (3-4 marks):
  • Na+/K+ Pump: Maintains resting potential in neurons (essential for nerve impulse transmission). Establishes gradients for secondary active transport.
  • Proton Pumps: In stomach lining (H+/K+ ATPase) maintains low pH for digestion/enzyme activity. In plants/fungi, creates electrochemical gradient for nutrient uptake.
  • Reabsorption in Kidneys: Active transport of glucose/ions from filtrate back to blood prevents loss of essential nutrients and maintains blood composition.
• Synthesis/Conclusion (2 marks):
  • Link specific mechanisms to the concept of homeostasis (e.g., negative feedback loops involving transporters).
  • Emphasise that selective permeability allows cells to control internal composition distinct from the external environment.
  • Mention energy cost (ATP) for active transport as a trade-off for precise control.

(Note: Answers must be coherent, use biological terminology correctly, and directly address the prompt.)