A Level H2 Biology Practice Paper 4

TuitionGoWhere Practice Paper - Biology H2 A-Level

Answer Key and Marking Scheme (Version 4)

Section A: Structured Questions

1. (a)

• P: Cell wall / peptidoglycan cell wall [1]
• Q: Circular DNA / nucleoid / bacterial chromosome [1]
• R: Plasmid [1]

(b) Any two from:

• Prokaryotic ribosomes are 70S; eukaryotic cytoplasmic ribosomes are 80S. [1]
• Prokaryotic ribosomes are smaller in size/mass than eukaryotic ribosomes. [1]
• Prokaryotic ribosomes have different subunit composition (50S + 30S vs 60S + 40S). [1] [Max 2]

(c)

• Prokaryotic cells have infoldings of the cell membrane (mesosomes) that provide a surface for respiratory enzymes / electron transport chain. [1]
• Photosynthetic prokaryotes have thylakoid membranes / chromatophores in the cytoplasm that contain photosynthetic pigments. [1]
• Therefore, the absence of membrane-bound organelles does not prevent metabolic processes because the cell membrane and internal membrane systems perform equivalent functions. [1] [Max 2]

2. (a)

• As pH increases from 3 to 7, the rate of reaction increases. [1]
• As pH increases from 7 to 10, the rate of reaction decreases. [1]
• The optimum pH is 7, where the rate is highest (22.5 mg min⁻¹). [1] [Max 2]

(b)

• At pH 3 and pH 10, the pH is far from the optimum. [1]
• The high concentration of H⁺ (low pH) or OH⁻ (high pH) ions disrupts the ionic and hydrogen bonds that maintain the tertiary structure of the enzyme. [1]
• This causes the active site to change shape / the enzyme to denature, so the substrate can no longer bind / enzyme-substrate complexes cannot form. [1] [3]

(c)

• The student only tested at whole pH unit intervals; the true optimum could be between pH 6 and 8 (e.g., pH 6.5 or 7.5) / more data points are needed around pH 7 to determine the precise optimum. [1]

3. (a)

• X: Phospholipid (bilayer) [1]
• Y: Glycoprotein / glycolipid (accept either) [1]

(b)

• Phospholipids have a hydrophilic (polar) phosphate head and two hydrophobic (non-polar) fatty acid tails. [1]
• This amphipathic nature allows them to form a bilayer in an aqueous environment, with the hydrophilic heads facing outwards towards the water and the hydrophobic tails facing inwards, creating a selectively permeable barrier. [1] [2]

(c)

• Although water molecules are small, the hydrophobic core of the phospholipid bilayer repels polar/charged molecules, including water. [1]
• Aquaporins provide hydrophilic channels/pores that allow water molecules to pass through rapidly by facilitated diffusion, bypassing the hydrophobic barrier. [1] [2]

4. (a)

• Collagen consists of three polypeptide chains (alpha chains) wound around each other in a triple helix. [1]
• The triple helix is held together by hydrogen bonds between the chains, and covalent cross-links between lysine residues of adjacent tropocollagen molecules. [1] [2]

(b)

• Collagen has a high tensile strength due to the triple helix structure and covalent cross-links between tropocollagen molecules, allowing it to withstand pulling forces without breaking. [1]
• The staggered arrangement of tropocollagen molecules forms fibrils and fibres, which provides flexibility while maintaining strength. [1]
• The fibrous, insoluble nature of collagen makes it suitable for structural support in tendons, which connect muscle to bone and must resist tension during muscle contraction. [1] [3]

(c)

• Vitamin C is required as a cofactor for the enzyme that hydroxylates proline and lysine residues in collagen synthesis. [1]
• Without vitamin C, collagen fibres are not properly cross-linked and are weaker; therefore, new collagen cannot be effectively synthesised to repair damaged tissue, leading to poor wound healing. [1] [2]

5. (a)

• Condensation (reaction) / esterification [1]

(b)

• Triglycerides are highly reduced molecules with many C-H bonds; when oxidised during respiration, they release a large amount of energy per gram (more than carbohydrates). [1]
• They are hydrophobic and can be stored without associated water (unlike glycogen), making them more compact / lighter for the same energy content. [1]
• They are insoluble and do not affect the osmotic balance of cells. [1] [3]

Section B: Data Interpretation and Application

6. (a)

• For both species, membrane fluidity increases (fluorescence polarisation decreases) as temperature increases. [1]
• At all temperatures, Species A (cold-water fish) has higher membrane fluidity (lower fluorescence polarisation) than Species B (desert plant). [1]
• The difference in fluidity between the two species is greater at lower temperatures and decreases as temperature increases. [1] [3]

(b)

• Species A (cold-water fish) likely has a higher proportion of unsaturated fatty acids in its phospholipids. [1]
• The double bonds in unsaturated fatty acids introduce kinks in the hydrocarbon tails, preventing tight packing of phospholipids. [1]
• This increases membrane fluidity at low temperatures, which is an adaptation to maintain membrane function in cold environments. Species B (desert plant) has more saturated fatty acids, resulting in tighter packing and lower fluidity at 10°C. [1] [3]

(c)

• Membrane fluidity affects the permeability of the membrane and the movement/function of membrane proteins (e.g., enzymes, transport proteins, receptors). [1]
• If the membrane is too rigid, transport processes and cell signalling are impaired; if too fluid, the membrane loses its integrity and becomes leaky. Maintaining fluidity within a narrow range ensures proper cellular function. [1] [2]

7. (a)

• Curve B represents Mammal X (small, active mammal). [1]
• Curve B is shifted to the right / has a lower oxygen affinity, meaning haemoglobin releases oxygen more readily to the tissues. [1] [2]

(b)

• Mammal X has a higher metabolic rate and therefore a higher oxygen demand in its tissues. [1]
• The lower oxygen affinity (right-shifted curve) means that for a given partial pressure of oxygen in the tissues, a greater proportion of oxygen is unloaded from haemoglobin, supplying the respiring tissues with more oxygen. [1] [2]

(c)

• Actively respiring tissues produce more CO₂, which dissolves to form carbonic acid, lowering the pH. [1]
• The lower pH causes the oxygen dissociation curve to shift to the right (Bohr effect), reducing haemoglobin's affinity for oxygen. [1]
• This promotes the unloading of more oxygen to the tissues that need it most, enhancing the efficiency of oxygen delivery during exercise/high metabolic activity. [1] [3]

8. (a)

• Competitive inhibition. [1]
• The maximum rate of reaction (V_max) is the same for both curves at high substrate concentrations, but a higher substrate concentration is needed to reach V_max in the presence of the inhibitor (Curve B has a higher K_m). [1] [2]

(b)

• The competitive inhibitor has a shape similar to the substrate and competes for the active site of the enzyme. [1]
• When the inhibitor is bound to the active site, the substrate cannot bind, preventing enzyme-substrate complex formation and reducing the rate of reaction. This can be overcome by increasing substrate concentration. [1] [2]

(c)

• Curve C should be drawn to the right of Curve B, with the same V_max but a further increased K_m (shallower initial slope, reaching plateau at even higher substrate concentration). [1]

(d)

• Methotrexate competitively inhibits dihydrofolate reductase, an enzyme required for the synthesis of thymine nucleotides / DNA precursors. [1]
• Cancer cells divide rapidly and require high rates of DNA synthesis; by inhibiting this enzyme, methotrexate selectively reduces DNA replication and slows cancer cell proliferation. [1] [2]

9. (a)

• Individuals P and S are homozygous for the normal allele. [1]
• They show only one band (the larger fragment), indicating that both alleles lack the additional restriction site and produce the same-sized fragment. [1] [2]

(b)

• Individual Q shows two bands: one corresponding to the larger (normal) fragment and one corresponding to the smaller (mutant) fragment. [1]
• This indicates that Q has one normal allele (producing the larger fragment) and one mutant allele (producing the smaller fragment due to the extra restriction site), confirming heterozygosity. [1] [2]

(c)

• An electric current is applied across the gel, and DNA fragments (negatively charged due to phosphate groups) migrate towards the positive electrode. [1]
• Smaller fragments move through the pores of the gel more easily and travel further than larger fragments, separating the fragments by size. [1] [2]

(d)

• The DNA ladder contains fragments of known sizes, allowing the sizes of the sample DNA fragments to be estimated by comparison. [1]

10. (a)

• Immediately after antibiotic addition, the bacterial population continues to increase briefly before declining. [1]
• The population then decreases sharply as most bacteria are killed, but after some time, the population begins to increase again as resistant bacteria multiply. [1] [2]

(b)

• A random mutation in a bacterial gene may alter the target site of the antibiotic (e.g., a ribosomal protein or cell wall synthesis enzyme), so the antibiotic can no longer bind and exert its effect. [1]
• Bacteria with this mutation survive the antibiotic treatment, while susceptible bacteria die. [1]
• The resistant bacteria reproduce, passing the resistance allele to their offspring, so the population becomes predominantly resistant over time (natural selection). [1] [3]

(c)

• Peptidoglycan cross-links provide strength and rigidity to the bacterial cell wall, preventing osmotic lysis. [1]
• Without cross-link formation, the cell wall is weakened; water enters the cell by osmosis, causing the cell to swell and burst (lyse). [1] [2]

Section C: Extended Response

11. Discuss how the structure of DNA relates to its functions in the storage of genetic information and in replication. [15]

Marking scheme:

Indicative content:

Storage of genetic information:

• DNA is a double-stranded polymer of nucleotides, each consisting of deoxyribose sugar, phosphate group, and nitrogenous base (A, T, C, G).
• The two strands are antiparallel and held together by hydrogen bonds between complementary base pairs (A-T: 2 H-bonds; C-G: 3 H-bonds).
• The sugar-phosphate backbone is on the outside, protecting the bases on the inside.
• The sequence of bases along the DNA molecule encodes genetic information in the form of the genetic code (triplet code).
• The double-helix structure provides stability and protects the genetic code from chemical damage.
• The complementary base pairing allows accurate copying of information.

Semi-conservative replication:

• DNA replication is semi-conservative: each new DNA molecule consists of one original (parental) strand and one newly synthesised strand.
• The double helix unwinds and the hydrogen bonds between base pairs are broken by DNA helicase, forming a replication fork.
• Single-strand binding proteins stabilise the separated strands.
• DNA polymerase synthesises the new strand in the 5' to 3' direction, using the parental strand as a template and adding complementary nucleotides.
• The leading strand is synthesised continuously; the lagging strand is synthesised discontinuously in Okazaki fragments.
• DNA ligase joins the Okazaki fragments together.
• Primase synthesises short RNA primers to provide a free 3'-OH group for DNA polymerase to start synthesis.
• The complementary base pairing ensures accurate replication; proofreading by DNA polymerase corrects errors.

Links between structure and function:

• The double-stranded, helical structure with hydrogen bonds provides stability for long-term information storage.
• Complementary base pairing is essential for accurate replication.
• The antiparallel nature explains the different modes of leading and lagging strand synthesis.
• The large size of DNA allows storage of vast amounts of information.

12. Describe the structure of a typical eukaryotic cell and explain how the ultrastructure of at least four different organelles is related to their functions. [15]

Marking scheme:

Indicative content:

General eukaryotic cell structure:

• Membrane-bound nucleus containing genetic material.
• Cytoplasm containing membrane-bound organelles.
• 80S ribosomes.
• Cell surface membrane (phospholipid bilayer with proteins).

Organelle structure-function relationships (any four):

Nucleus:

• Surrounded by a double membrane (nuclear envelope) with nuclear pores.
• Nuclear pores allow selective transport of molecules (e.g., mRNA, ribosomes) between nucleus and cytoplasm.
• Contains chromatin (DNA associated with histones) which condenses into chromosomes during cell division.
• Nucleolus is the site of ribosomal RNA synthesis and ribosome assembly.
• Function: stores genetic information and controls cellular activities through gene expression.

Mitochondrion:

• Double membrane: outer membrane is smooth; inner membrane is highly folded into cristae.
• Cristae increase surface area for electron transport chain and ATP synthase enzymes.
• Matrix contains enzymes for the Krebs cycle and mitochondrial DNA/ribosomes.
• Intermembrane space allows proton accumulation for chemiosmosis.
• Function: site of aerobic respiration and ATP synthesis.

Rough Endoplasmic Reticulum (RER):

• Network of flattened membrane-bound sacs (cisternae) studded with ribosomes.
• Ribosomes synthesise proteins that enter the RER lumen for folding and modification.
• Vesicles bud off and transport proteins to the Golgi apparatus.
• Function: synthesis, folding, and transport of proteins.

Golgi Apparatus:

• Stack of flattened membrane-bound sacs (cisternae).
• Receives vesicles from RER at the cis face; modifies proteins (e.g., glycosylation).
• Sorts and packages proteins into vesicles at the trans face for secretion or delivery to other organelles.
• Function: modification, sorting, and packaging of proteins for transport.

Chloroplast (plant cells):

• Double membrane envelope.
• Internal thylakoid membranes stacked into grana, containing chlorophyll and other photosynthetic pigments.
• Thylakoid membranes provide a large surface area for light-dependent reactions and ATP synthase.
• Stroma contains enzymes for the Calvin cycle, starch grains, and chloroplast DNA/ribosomes.
• Function: site of photosynthesis.

Lysosome:

• Membrane-bound vesicle containing hydrolytic enzymes (e.g., proteases, lipases, nucleases).
• Enzymes are active at acidic pH maintained by proton pumps in the lysosomal membrane.
• Function: intracellular digestion of worn-out organelles (autophagy) and engulfed pathogens (phagocytosis).

Ribosome:

• Composed of two subunits (60S and 40S in eukaryotes) made of rRNA and proteins.
• Site of protein synthesis (translation): mRNA binds to the small subunit, and tRNA brings amino acids to the ribosome.
• Function: translation of mRNA into polypeptide chains.

END OF ANSWER KEY