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A Level H2 Biology Practice Paper 4
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Questions
TuitionGoWhere Practice Paper - Biology H2 A-Level
TuitionGoWhere Practice Paper (AI)
Subject: Biology H2 Level: A-Level Paper: Practice Paper (Version 4 of 5) Duration: 2 hours Total Marks: 75
Name: _________________________ Class: _________________________ Date: _________________________
Instructions to Candidates
- This paper consists of three sections: Section A, Section B, and Section C.
- Answer all questions in the spaces provided.
- The number of marks is given in brackets [ ] at the end of each question or part question.
- You are advised to spend no more than 45 minutes on Section A, 45 minutes on Section B, and 30 minutes on Section C.
- You may use a scientific calculator.
- Where appropriate, show your working.
Section A: Structured Questions (30 marks)
Answer all questions in this section.
1. Figure 1.1 shows a diagram of a typical prokaryotic cell.
(a) Identify the structures labelled P, Q, and R. [3]
P: _________________________ Q: _________________________ R: _________________________
(b) State two structural differences between the ribosomes found in this cell and those found in the cytoplasm of a eukaryotic cell. [2]
(c) Explain why the absence of membrane-bound organelles in prokaryotic cells does not prevent them from carrying out metabolic processes such as respiration and photosynthesis. [2]
[Total: 7 marks]
2. Amylase is an enzyme that catalyses the hydrolysis of starch into maltose. A student investigated the effect of pH on the rate of this reaction. The results are shown in Table 2.1.
Table 2.1
| pH | Rate of reaction (mg starch hydrolysed min⁻¹) |
|---|---|
| 3 | 2.1 |
| 4 | 5.8 |
| 5 | 12.4 |
| 6 | 18.9 |
| 7 | 22.5 |
| 8 | 21.8 |
| 9 | 10.3 |
| 10 | 3.2 |
(a) Describe the relationship between pH and the rate of reaction shown in Table 2.1. [2]
(b) Explain why the rate of reaction is low at pH 3 and pH 10. [3]
(c) The student concluded that the optimum pH for amylase is 7. Suggest one limitation of this conclusion based on the data provided. [1]
[Total: 6 marks]
3. Figure 3.1 shows the fluid mosaic model of a cell surface membrane.
(a) Name the components labelled X and Y. [2]
X: _________________________ Y: _________________________
(b) Explain how the structure of component X allows it to perform its function in the membrane. [2]
(c) Aquaporins are channel proteins that facilitate the rapid movement of water across membranes. Explain why water molecules require aquaporins to cross the membrane efficiently despite being small enough to diffuse through the phospholipid bilayer. [2]
[Total: 6 marks]
4. Collagen is a fibrous protein found in connective tissues such as tendons and skin.
(a) Describe the quaternary structure of collagen. [2]
(b) Explain how the structure of collagen makes it suitable for its role in tendons. [3]
(c) Scurvy is a disease caused by vitamin C deficiency, which results in weakened collagen fibres. Suggest why a person with scurvy may experience poor wound healing. [2]
[Total: 7 marks]
5. Figure 5.1 shows the molecular structure of a triglyceride molecule.
(a) Name the type of chemical reaction that joins the components of a triglyceride together. [1]
(b) Explain why triglycerides are efficient energy storage molecules. [3]
[Total: 4 marks]
Section B: Data Interpretation and Application (30 marks)
Answer all questions in this section.
6. A group of researchers investigated the effect of temperature on the fluidity of membranes extracted from two different organisms: a cold-water fish (Species A) and a desert plant (Species B). Membrane fluidity was measured using fluorescence polarisation, where higher values indicate lower fluidity. The results are shown in Figure 6.1.
(a) Compare the effect of increasing temperature on the membrane fluidity of Species A and Species B. [3]
(b) With reference to the structure of phospholipids, suggest an explanation for the difference in membrane fluidity between the two species at 10°C. [3]
(c) Explain why it is essential for organisms to maintain membrane fluidity within a narrow range. [2]
[Total: 8 marks]
7. Figure 7.1 shows the oxygen dissociation curves for haemoglobin from two different mammals: a small, active mammal (Mammal X) and a large, sedentary mammal (Mammal Y).
(a) State which curve (A or B) is likely to represent Mammal X. Give a reason for your answer. [2]
(b) Explain the advantage of the oxygen dissociation curve you selected in (a) for Mammal X. [2]
(c) The Bohr effect describes how increased carbon dioxide concentration affects the oxygen dissociation curve of haemoglobin. Explain the physiological importance of the Bohr effect in actively respiring tissues. [3]
[Total: 7 marks]
8. A student carried out an investigation into the effect of substrate concentration on the rate of an enzyme-catalysed reaction. The student measured the initial rate of reaction at different substrate concentrations, first in the absence of an inhibitor (Curve A) and then in the presence of a fixed concentration of an inhibitor (Curve B). The results are shown in Figure 8.1.
(a) Identify the type of inhibition shown by Curve B. Explain your answer with reference to Figure 8.1. [2]
(b) Explain, at the molecular level, how this type of inhibitor reduces the rate of reaction. [2]
(c) The student repeated the experiment with a higher concentration of the inhibitor. On Figure 8.1, sketch the curve you would expect to see. Label this curve C. [1]
(d) Methotrexate is a competitive inhibitor of the enzyme dihydrofolate reductase, which is involved in DNA synthesis. Explain why methotrexate is used as a chemotherapy drug to treat cancer. [2]
[Total: 7 marks]
9. Figure 9.1 shows the results of gel electrophoresis of DNA fragments obtained from four individuals (P, Q, R, and S) at a locus associated with a genetic disorder. The disorder is caused by a mutation that creates an additional restriction site, resulting in smaller DNA fragments.
(a) Which individual(s) is/are homozygous for the normal allele? Explain your answer. [2]
(b) Individual Q is heterozygous for this locus. Explain how the banding pattern for Q supports this conclusion. [2]
(c) Explain how gel electrophoresis separates DNA fragments of different sizes. [2]
(d) Suggest why a DNA ladder (marker) is included in the gel. [1]
[Total: 7 marks]
10. Figure 10.1 shows the effect of an antibiotic on the growth of a bacterial population over time. The antibiotic was added at the time indicated by the arrow.
(a) Describe the changes in the bacterial population after the addition of the antibiotic. [2]
(b) Explain how a mutation could enable some bacteria to survive and reproduce in the presence of the antibiotic. [3]
(c) The antibiotic in Figure 10.1 inhibits the formation of cross-links in the peptidoglycan cell wall of bacteria. Explain why this leads to bacterial cell death. [2]
[Total: 7 marks]
Section C: Extended Response (15 marks)
Answer one question from this section. Write your answer in the space provided.
EITHER
11. Discuss how the structure of DNA relates to its functions in the storage of genetic information and in replication. In your answer, you should include:
- the structural features of DNA that allow stable storage of genetic information;
- the mechanism of semi-conservative replication;
- the roles of the key enzymes involved in replication. [15]
OR
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]
END OF PAPER
Answers
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:
| Level | Descriptor | Marks |
|---|---|---|
| 3 | Comprehensive answer demonstrating detailed knowledge of DNA structure and replication. Clear links between structure and function. All key enzymes and their roles described. Well-structured and coherent. | 11–15 |
| 2 | Good knowledge of DNA structure and replication, but some details missing or links not fully explained. Most key enzymes mentioned. Generally well-organised. | 6–10 |
| 1 | Basic knowledge with significant omissions or errors. Limited links between structure and function. Poorly structured. | 1–5 |
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:
| Level | Descriptor | Marks |
|---|---|---|
| 3 | Comprehensive description of eukaryotic cell structure with detailed explanation of structure-function relationships for at least four organelles. Accurate terminology and clear organisation. | 11–15 |
| 2 | Good description with some structure-function links for at least three organelles. Some details missing or minor errors. | 6–10 |
| 1 | Basic description with limited or inaccurate structure-function links. Fewer than three organelles discussed in detail. | 1–5 |
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