A Level H2 Biology Practice Paper 5

TuitionGoWhere Exam Practice (AI) - Biology H2 A-Level

Subject: Biology
Level: H2
Paper: Practice Paper (Version 5 of 5) - Answer Key
Topic: Cells & Biomolecules

Section A: Structured Questions

1. (a)

• Description: Molecules labelled A are phospholipids arranged in a bilayer. The hydrophilic phosphate heads face outwards towards the aqueous environment (extracellular fluid and cytoplasm), while the hydrophobic fatty acid tails face inwards, away from water. [1]
• Explanation: This arrangement creates a hydrophobic core that prevents the passage of water-soluble (polar/charged) molecules and ions, acting as a barrier to free diffusion. [1] It allows the membrane to be selectively permeable. [1] (Max 3 marks)

(b)

• Molecule C is cholesterol. [1]
• At high temperatures, cholesterol restricts the movement of phospholipid fatty acid tails, reducing membrane fluidity and preventing it from becoming too fluid/leaky. At low temperatures, it prevents tight packing of tails, maintaining fluidity and preventing the membrane from becoming too rigid. [1] (Max 2 marks)

(c)

• The drug acts as a competitive inhibitor. [1]
• It has a similar shape to glucose and binds to the active site of the channel/carrier protein (B), blocking glucose from binding. [1]
• It does not disrupt the phospholipid bilayer structure itself. (Max 2 marks)

2. (a)

• X: Cristae (inner membrane folds). [1]
• Y: Matrix. [1] (Max 2 marks)

(b)

• The folding increases the surface area of the inner membrane. [1]
• This allows for a greater number of electron transport chain components and ATP synthase enzymes to be embedded. [1]
• This maximizes the rate of oxidative phosphorylation and ATP production. [1] (Max 3 marks)

(c) (i)

• Oxygen acts as the final electron acceptor in the electron transport chain. [1]
• It accepts electrons and protons to form water, thus being consumed from the buffer. [1] (Max 2 marks)

(ii)

• The inhibitor stops the flow of electrons through the ETC. [1]
• Protons ($H^+$) are no longer pumped from the matrix into the intermembrane space. [1]
• Protons may leak back or diffuse, causing the pH of the intermembrane space to increase (become less acidic/more alkaline) as $[H^+]$ decreases. [1] (Max 3 marks)

3. (a)

• A condensation reaction occurs between the carboxyl group (-COOH) of one amino acid and the amino group (-NH2) of another. [1]
• A molecule of water is removed. [1]
• A covalent peptide bond (-CO-NH-) is formed. [1] (Max 3 marks)

(b)

• Secondary structure: The polypeptide chain folds into alpha-helices due to hydrogen bonding between backbone atoms. This provides a compact structure. [1]
• Tertiary structure: The helices fold further into a specific 3D globular shape due to interactions between R-groups (hydrogen bonds, ionic bonds, disulfide bridges, hydrophobic interactions). [1]
• This specific shape creates a hydrophobic pocket for the haem group and exposes hydrophilic residues on the surface, making haemoglobin soluble in blood plasma. [1]
• The quaternary structure (4 subunits) allows for cooperative binding of oxygen. [1] (Max 4 marks)

(c)

• Fibrous proteins like collagen have long, parallel polypeptide chains cross-linked by covalent bonds, forming strong fibres. [1]
• This provides high tensile strength suitable for structural support. Globular proteins are compact and soluble, lacking this structural strength. [1] (Max 2 marks)

4. (a)

• The active site is a specific region on the enzyme surface with a unique shape and chemical environment that binds to the substrate. [1] (Max 1 mark)

(b)

• At low substrate concentrations, there are many available active sites. [1]
• The competitive inhibitor competes with the substrate for the active sites. [1]
• This reduces the frequency of successful enzyme-substrate collisions, lowering the initial rate of reaction compared to the uninhibited reaction. [1] (Max 3 marks)

(c)

• At high substrate concentrations, the substrate molecules outnumber the inhibitor molecules. [1]
• The substrate successfully out-competes the inhibitor for the active sites, allowing all enzyme molecules to be saturated with substrate. [1]
• Thus, $V_{max}$ is reached. (Max 2 marks)

5. (a)

• DNA contains deoxyribose sugar; RNA contains ribose sugar. [1]
• DNA contains thymine; RNA contains uracil. [1]
• DNA is typically double-stranded (helix); RNA is typically single-stranded. [1] (Max 3 marks)

(b)

• DNA helicase breaks hydrogen bonds between complementary base pairs, unwinding the double helix to form two template strands. [1]
• Free nucleotides align opposite their complementary bases on the template strands. [1]
• DNA polymerase joins adjacent nucleotides by forming phosphodiester bonds between the sugar and phosphate groups. [1]
• DNA polymerase also proofreads the new strand to correct errors. [1] (Max 4 marks)

(c)

• Each new DNA molecule consists of one original (parental) strand and one newly synthesized strand. [1]
• This ensures genetic continuity and accurate transmission of genetic information to daughter cells. [1] (Max 2 marks)

Section B: Data Interpretation and Application

6. (a)

• DNA fragments are negatively charged due to phosphate groups. [1]
• When an electric field is applied, they migrate towards the positive electrode (anode). [1]
• Smaller fragments move faster through the gel matrix pores than larger fragments, resulting in separation by size. [1] (Max 3 marks)

(b)

• Individual Q is heterozygous ($A_1A_2$). [1]
• Q shows two bands: one at 500bp (allele $A_1$) and one at 300bp (allele $A_2$). [1]
• This indicates the presence of both alleles in the genotype. [1] (Max 3 marks)

(c)

• The DNA ladder contains fragments of known sizes, allowing the estimation of the size of the unknown DNA fragments in the samples. [1] (Max 1 mark)

7. (a)

• Inducible operon: Normally switched off (repressed) and is turned on (induced) in the presence of a specific substrate (e.g., lactose). [1]
• Repressible operon: Normally switched on and is turned off (repressed) in the presence of a specific end-product (e.g., tryptophan). [1] (Max 2 marks)

(b)

• In the absence of lactose, the lac repressor protein is active. [1]
• It 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] (Max 3 marks)

(c)

• It prevents the wasteful synthesis of enzymes (beta-galactosidase, permease) when lactose is not available. [1]
• This conserves energy and resources (amino acids, ATP) for the cell. [1] (Max 2 marks)

8. (a)

• Voltage-gated sodium channels open. [1]
• This causes a rapid influx of $Na^+$ ions into the axon, making the inside more positive (depolarisation). [1] (Max 2 marks)

(b)

• The sodium-potassium pump actively transports 3 $Na^+$ ions out of the axon and 2 $K^+$ ions into the axon. [1]
• This uses ATP. [1]
• It restores the ion concentration gradients and the resting potential (-70mV) after the action potential. [1] (Max 3 marks)

(c)

• Myelin acts as an electrical insulator. [1]
• Damage to myelin prevents saltatory conduction (jumping of impulses between Nodes of Ranvier), forcing the impulse to travel along the entire membrane, which is slower. [1] (Max 2 marks)

9. (a)

• Chromosomes condense and become visible (shorten and thicken). [1]
• The nucleolus disappears and the nuclear envelope breaks down. [1]
• Centrioles move to opposite poles and spindle fibres begin to form. [1] (Max 3 marks)

(b)

• Attachment ensures that chromosomes are aligned at the equator. [1]
• This allows for the equal separation of sister chromatids to opposite poles during anaphase, ensuring each daughter cell receives the correct number of chromosomes. [1] (Max 2 marks)

(c)

• Mitosis: Produces two genetically identical daughter cells with the same chromosome number (diploid) as the parent. No genetic variation introduced (except mutation). [2]
• Meiosis: Produces four genetically different daughter cells with half the chromosome number (haploid) of the parent. Genetic variation is introduced via crossing over and independent assortment. [2] (Max 4 marks)

10. (a)

• Water molecules are polar (dipole), with a partial negative charge on oxygen and partial positive charge on hydrogen. [1]
• They form hydrogen bonds with other polar molecules and ions. [1]
• This surrounds solute particles, keeping them in solution and allowing metabolic reactions to occur in aqueous media. [1] (Max 3 marks)

(b)

• High specific heat capacity: Water can absorb/release large amounts of heat with little temperature change. Significance: Helps organisms maintain stable internal body temperature (homeostasis). [2]
• High latent heat of vaporisation: Large amount of energy required to evaporate water. Significance: Effective cooling mechanism via sweating/transpiration without excessive water loss. [2]
• (Alternative: Cohesion/Tension - supports transport in xylem. Ice density - ice floats, insulating aquatic life.) (Max 4 marks)

Section C: Extended Response

11. Marking Scheme Guidelines:

Introduction:

• Define the cell surface membrane as a partially permeable barrier.
• State the Fluid Mosaic Model.

Amphipathic Nature of Phospholipids (3-4 marks):

• Phospholipids have hydrophilic heads and hydrophobic tails.
• Form a bilayer in aqueous environments.
• Hydrophobic core prevents passage of water-soluble substances (ions, polar molecules).
• Allows passage of lipid-soluble substances (steroids, $O_2$, $CO_2$) via simple diffusion.
• Provides structural integrity and flexibility.

Role of Membrane Proteins in Transport (3-4 marks):

• Channel proteins: Form hydrophilic pores for facilitated diffusion of ions/small polar molecules (e.g., $Na^+$, $K^+$). Specificity based on size/charge.
• Carrier proteins: Bind specific molecules (e.g., glucose), change shape to transport them across. Can be facilitated diffusion or active transport.
• Pump proteins: Use ATP (active transport) to move substances against concentration gradient (e.g., Na+/K+ pump).
• Mention specificity and saturation kinetics.

Importance of Membrane Fluidity (2-3 marks):

• Phospholipids and proteins can move laterally (fluidity).
• Allows for membrane self-sealing, endocytosis/exocytosis, and cell division.
• Allows proteins to diffuse and interact (e.g., enzyme-substrate, receptor-ligand).
• Role of cholesterol and unsaturated fatty acids in regulating fluidity at different temperatures.

Conclusion:

• Summarize how structure (bilayer + proteins + fluidity) enables function (selective permeability, transport, signalling).

Quality of Communication (1 mark):

• Clear, logical structure.
• Correct use of biological terminology.

(Total 10 marks)