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A Level H2 Biology Practice Paper 4
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TuitionGoWhere Practice Paper - Biology H2 A-Level
PRACTICE PAPER 4
Subject: Biology H2 (9477) Level: A-Level Paper: Practice Paper 4 (Cells & Biomolecules) Duration: 1 hour 15 minutes Total Marks: 60 Version: 4 of 5
Name: _________________________ Class: _________________________ Date: _________________________
Instructions to Candidates
- This paper consists of 20 questions in three sections.
- 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, 20 minutes on Section B, and 10 minutes on Section C.
- Where appropriate, show your working and reasoning clearly.
- Use diagrams and labelled sketches where they help your explanation.
Section A: Structured Questions (30 marks)
Answer all questions in this section.
1. Fig. 1.1 shows a diagram of a eukaryotic cell with several organelles labelled A to E.
(a) Identify the organelles labelled A, C, and E in Fig. 1.1. [3]
A: _________________________
C: _________________________
E: _________________________
(b) With reference to Fig. 1.1, describe how the structure of organelle B is adapted for its function in protein synthesis and transport. [3]
2. A student prepared a suspension of mitochondria isolated from liver cells in a buffer containing ADP and inorganic phosphate (Pi). The oxygen concentration in the buffer was monitored over time. At point X, sodium azide (an inhibitor of cytochrome c oxidase) was added to the suspension. The results are shown in Fig. 2.1.
(a) Describe the change in oxygen concentration before the addition of sodium azide. [1]
(b) Explain why oxygen is consumed by the mitochondrial suspension under these conditions. [2]
(c) With reference to Fig. 2.1, explain the effect of sodium azide on oxygen consumption by the mitochondria. [2]
3. Gel electrophoresis was used to analyse haemoglobin samples from four individuals (P, Q, R, and S). The results are shown in Fig. 3.1. Individual P is known to be homozygous for the normal haemoglobin allele (HbA). Individual Q is heterozygous (HbA/HbS).
(a) Explain why the haemoglobin from individual Q produces two bands on the gel, while individual P produces only one band. [3]
(b) Individual R shows a single band at the same position as the lower band of individual Q. Individual S shows a single band at the same position as the upper band of individual P. Deduce the genotypes of individuals R and S. Explain your reasoning. [3]
R: _________________________
S: _________________________
Reasoning:
4. Fig. 4.1 shows the fluid mosaic model of the cell surface membrane.
(a) Name the molecules labelled X and Y in Fig. 4.1. [2]
X: _________________________
Y: _________________________
(b) With reference to Fig. 4.1, explain how the structure of the cell surface membrane allows it to be selectively permeable. [3]
5. The lac operon in E. coli is an example of an inducible operon. The trp operon is an example of a repressible operon.
(a) Distinguish between an inducible operon and a repressible operon. [2]
(b) Suggest and explain why it is advantageous for E. coli to have the lac operon as an inducible operon rather than a constitutive operon (one that is always expressed). [3]
6. Bovine spongiform encephalopathy (BSE) is a disease caused by misfolded prion proteins. Fig. 6.1 shows the structure of a normal prion protein (PrP^C) and the misfolded form (PrP^Sc).
(a) With reference to Fig. 6.1, describe one structural difference between PrP^C and PrP^Sc. [1]
(b) Explain why misfolded prion proteins tend to aggregate in nerve cells. [3]
Section B: Data Interpretation and Analysis (20 marks)
Answer all questions in this section.
7. Fig. 7.1 shows the effect of substrate concentration on the rate of an enzyme-catalysed reaction, with and without a competitive inhibitor.
(a) Describe the effect of the competitive inhibitor on the rate of reaction at low substrate concentrations. [2]
(b) Explain why the maximum rate of reaction (Vmax) is the same with and without the competitive inhibitor. [2]
8. A student investigated the effect of temperature on the rate of diffusion of potassium permanganate crystals in agar jelly. The results are shown in Table 8.1.
| Temperature (°C) | Distance diffused after 30 minutes (mm) |
|---|---|
| 10 | 3.2 |
| 20 | 5.1 |
| 30 | 7.8 |
| 40 | 10.4 |
| 50 | 12.9 |
(a) Calculate the percentage increase in distance diffused between 20°C and 40°C. Show your working. [2]
(b) Explain the effect of temperature on the rate of diffusion shown in Table 8.1. [3]
9. Fig. 9.1 shows the molecular structure of a phospholipid and its arrangement in water.
(a) With reference to Fig. 9.1, explain why phospholipids form a bilayer when placed in water. [3]
(b) State one function of phospholipids in cells other than forming membranes. [1]
10. A polypeptide has the following amino acid sequence:
Met – Ala – Gly – Ser – Lys – Phe – Trp – Cys – Leu – Arg
(a) State the number of peptide bonds in this polypeptide. [1]
(b) The polypeptide is treated with trypsin, which hydrolyses peptide bonds on the carboxyl side of lysine (Lys) and arginine (Arg). State the number of peptide fragments produced. Explain your answer. [2]
11. Fig. 11.1 shows the structure of ATP (adenosine triphosphate).
(a) Name the components labelled P, Q, and R in Fig. 11.1. [3]
P: _________________________
Q: _________________________
R: _________________________
(b) Explain why ATP is described as the universal energy currency of cells. [2]
12. A student carried out the Benedict's test on solutions of glucose, sucrose, and starch. The results are shown in Table 12.1.
| Solution | Colour after Benedict's test (with heating) |
|---|---|
| Glucose | Brick-red precipitate |
| Sucrose | Blue (no change) |
| Starch | Blue (no change) |
(a) Explain why glucose gives a positive result with Benedict's test but sucrose does not. [2]
(b) Describe an additional test that could be carried out to confirm the identity of the starch solution. State the expected result. [2]
Section C: Extended Response (10 marks)
Answer all questions in this section.
13. Discuss the importance of hydrogen bonding in determining the structure and properties of water, and explain how these properties make water essential for living organisms. [10]
In your answer, you should include:
- the molecular basis of hydrogen bonding in water
- at least three properties of water that arise from hydrogen bonding
- specific examples of how each property is important in biological systems.
END OF PAPER
TuitionGoWhere Secondary School (AI) – Practice Paper 4 – Biology H2 A-Level
Answers
TuitionGoWhere Practice Paper - Biology H2 A-Level
PRACTICE PAPER 4 – ANSWER KEY AND MARKING SCHEME
Subject: Biology H2 (9477) Level: A-Level Paper: Practice Paper 4 (Cells & Biomolecules) Version: 4 of 5
Section A: Structured Questions (30 marks)
1. (a) Identify organelles A, C, and E. [3]
| Label | Organelle | Mark |
|---|---|---|
| A | Nucleus / nuclear envelope | [1] |
| C | Mitochondrion | [1] |
| E | Golgi apparatus / Golgi body | [1] |
Accept: rough endoplasmic reticulum for A if clearly indicated in figure; accept "Golgi complex" for E.
1. (b) Describe how the structure of organelle B is adapted for its function in protein synthesis and transport. [3]
Organelle B = rough endoplasmic reticulum (RER)
| Mark | Point |
|---|---|
| [1] | RER has flattened membrane-bound sacs (cisternae) providing a large surface area for protein synthesis; |
| [1] | Ribosomes attached to the cytoplasmic surface of the membrane synthesise proteins and insert them into the RER lumen; |
| [1] | The RER lumen transports synthesised proteins, and vesicles bud off from the RER to transport proteins to the Golgi apparatus for further processing. |
Accept any three valid points. Award marks for linking structure to function explicitly.
2. (a) Describe the change in oxygen concentration before the addition of sodium azide. [1]
| Mark | Point |
|---|---|
| [1] | Oxygen concentration decreases (steadily/linearly) over time. |
Accept: "oxygen is consumed" or "oxygen concentration falls."
2. (b) Explain why oxygen is consumed by the mitochondrial suspension under these conditions. [2]
| Mark | Point |
|---|---|
| [1] | Oxygen acts as the final electron acceptor in the electron transport chain (ETC); |
| [1] | ADP and Pi are present, so ATP synthase is active and protons flow through ATP synthase, maintaining the proton gradient and allowing continued electron flow, which requires oxygen to accept electrons. |
Accept: "oxygen is reduced to water at Complex IV (cytochrome c oxidase)" as part of the explanation.
2. (c) With reference to Fig. 2.1, explain the effect of sodium azide on oxygen consumption by the mitochondria. [2]
| Mark | Point |
|---|---|
| [1] | After addition of sodium azide (at point X), oxygen consumption stops / oxygen concentration remains constant; |
| [1] | Sodium azide inhibits cytochrome c oxidase (Complex IV), preventing the transfer of electrons to oxygen, so the ETC is blocked and oxygen cannot be reduced to water. |
Must reference the figure (e.g., "after point X") for full marks.
3. (a) Explain why the haemoglobin from individual Q produces two bands on the gel, while individual P produces only one band. [3]
| Mark | Point |
|---|---|
| [1] | Individual P is homozygous (HbA/HbA) and produces only normal haemoglobin (HbA), which migrates as a single band; |
| [1] | Individual Q is heterozygous (HbA/HbS) and produces two different haemoglobin variants – normal (HbA) and sickle (HbS); |
| [1] | HbA and HbS have different charges (due to the amino acid substitution: glutamic acid → valine in HbS), so they migrate at different rates in the electric field, producing two distinct bands. |
Must explain the charge difference between HbA and HbS for full marks.
3. (b) Deduce the genotypes of individuals R and S. Explain your reasoning. [3]
| Individual | Genotype | Mark |
|---|---|---|
| R | HbS/HbS (homozygous for sickle cell allele) | [1] |
| S | HbA/HbA (homozygous for normal allele) | [1] |
Reasoning [1]:
- R shows a single band at the same position as the lower band of Q (HbS band), indicating R produces only HbS → homozygous HbS/HbS.
- S shows a single band at the same position as the upper band of P (HbA band), indicating S produces only HbA → homozygous HbA/HbA.
4. (a) Name the molecules labelled X and Y in Fig. 4.1. [2]
| Label | Molecule | Mark |
|---|---|---|
| X | Glycoprotein / carbohydrate chain (attached to protein) | [1] |
| Y | Cholesterol | [1] |
Accept: "glycolipid" if X is attached to a lipid in the figure; accept "integral protein" or "transmembrane protein" if X is a protein spanning the membrane – depends on figure labelling.
4. (b) With reference to Fig. 4.1, explain how the structure of the cell surface membrane allows it to be selectively permeable. [3]
| Mark | Point |
|---|---|
| [1] | The phospholipid bilayer has a hydrophobic core, which prevents the free passage of ions and large polar molecules; |
| [1] | Small, non-polar molecules (e.g., O₂, CO₂) can diffuse through the phospholipid bilayer directly; |
| [1] | Transport proteins (channel and carrier proteins) embedded in the membrane allow specific ions and polar molecules to cross the membrane by facilitated diffusion or active transport. |
Must reference the figure (e.g., "as shown by the phospholipid bilayer / transport proteins in Fig. 4.1") for full marks.
5. (a) Distinguish between an inducible operon and a repressible operon. [2]
| Mark | Point |
|---|---|
| [1] | An inducible operon is normally turned OFF and is switched ON in the presence of an inducer (e.g., substrate); |
| [1] | A repressible operon is normally turned ON and is switched OFF in the presence of a co-repressor (e.g., end-product). |
Accept: "inducible operon: transcription is induced by substrate; repressible operon: transcription is repressed by product."
5. (b) Suggest and explain why it is advantageous for E. coli to have the lac operon as an inducible operon rather than a constitutive operon. [3]
| Mark | Point |
|---|---|
| [1] | The lac operon codes for enzymes involved in lactose metabolism; |
| [1] | If the operon were constitutive, E. coli would continuously produce these enzymes even when lactose is absent, which wastes energy and resources (ATP, amino acids); |
| [1] | As an inducible operon, the enzymes are produced only when lactose is present, allowing E. coli to conserve energy and resources when lactose is not available, which is advantageous in competitive environments. |
Must link to metabolic efficiency and resource conservation for full marks.
6. (a) With reference to Fig. 6.1, describe one structural difference between PrP^C and PrP^Sc. [1]
| Mark | Point |
|---|---|
| [1] | PrP^C has a higher proportion of alpha-helical structure, while PrP^Sc has a higher proportion of beta-pleated sheet structure. |
Accept: "PrP^C is predominantly alpha-helical; PrP^Sc is predominantly beta-sheet" or "different secondary structure."
6. (b) Explain why misfolded prion proteins tend to aggregate in nerve cells. [3]
| Mark | Point |
|---|---|
| [1] | Misfolded prion proteins (PrP^Sc) expose hydrophobic amino acid residues that are normally buried in the core of the correctly folded protein (PrP^C); |
| [1] | The exposed hydrophobic regions on different PrP^Sc molecules interact with each other via hydrophobic interactions; |
| [1] | These hydrophobic interactions cause PrP^Sc molecules to clump together, forming insoluble aggregates that accumulate in nerve cells, leading to cell damage and death. |
Must explain the role of hydrophobic interactions in aggregation for full marks.
Section B: Data Interpretation and Analysis (20 marks)
7. (a) Describe the effect of the competitive inhibitor on the rate of reaction at low substrate concentrations. [2]
| Mark | Point |
|---|---|
| [1] | At low substrate concentrations, the rate of reaction is lower in the presence of the competitive inhibitor compared to the uninhibited reaction; |
| [1] | The competitive inhibitor competes with the substrate for the active site of the enzyme, so fewer enzyme-substrate complexes form at low substrate concentrations. |
7. (b) Explain why the maximum rate of reaction (Vmax) is the same with and without the competitive inhibitor. [2]
| Mark | Point |
|---|---|
| [1] | At high substrate concentrations, the substrate greatly outnumbers the inhibitor molecules; |
| [1] | The substrate outcompetes the inhibitor for the active site, so all enzyme active sites become occupied by substrate, and the maximum rate (Vmax) can still be achieved. |
Accept: "competitive inhibition can be overcome by increasing substrate concentration."
8. (a) Calculate the percentage increase in distance diffused between 20°C and 40°C. Show your working. [2]
| Mark | Point |
|---|---|
| [1] | Correct working: [(10.4 − 5.1) / 5.1] × 100 = (5.3 / 5.1) × 100; |
| [1] | Correct answer: 103.9% (accept 104% or 103.9). |
Allow ecf (error carried forward) if working is correct but final answer has minor rounding error.
8. (b) Explain the effect of temperature on the rate of diffusion shown in Table 8.1. [3]
| Mark | Point |
|---|---|
| [1] | As temperature increases, the distance diffused increases (rate of diffusion increases); |
| [1] | At higher temperatures, molecules have greater kinetic energy, so they move faster; |
| [1] | Faster-moving molecules diffuse more rapidly through the agar jelly, covering a greater distance in the same time. |
Accept reference to increased molecular motion / increased kinetic energy of both potassium permanganate and water molecules.
9. (a) With reference to Fig. 9.1, explain why phospholipids form a bilayer when placed in water. [3]
| Mark | Point |
|---|---|
| [1] | Phospholipids are amphipathic – they have a hydrophilic (polar) phosphate head and two hydrophobic (non-polar) fatty acid tails; |
| [1] | In water, the hydrophilic heads orient towards the aqueous environment (both outside and inside the cell), while the hydrophobic tails orient away from water, facing each other; |
| [1] | This arrangement forms a stable bilayer, with the hydrophobic tails shielded from water in the interior of the membrane. |
Must reference the figure and use the term "amphipathic" or equivalent description for full marks.
9. (b) State one function of phospholipids in cells other than forming membranes. [1]
| Mark | Point |
|---|---|
| [1] | Any one of: cell signalling (e.g., phospholipase C pathway producing IP₃ and DAG) / source of second messengers / component of lung surfactant / involved in blood clotting (platelet-activating factor) / emulsification of fats in digestion (as part of bile). |
Accept any valid function with brief context.
10. (a) State the number of peptide bonds in this polypeptide. [1]
| Mark | Point |
|---|---|
| [1] | 9 peptide bonds. |
Reasoning: 10 amino acids → 9 peptide bonds (n amino acids → n−1 peptide bonds).
10. (b) State the number of peptide fragments produced. Explain your answer. [2]
| Mark | Point |
|---|---|
| [1] | 3 fragments; |
| [1] | Trypsin cleaves on the carboxyl side of Lys (position 5) and Arg (position 10). Cleavage after Lys produces two fragments (Met–Ala–Gly–Ser–Lys and Phe–Trp–Cys–Leu–Arg). Cleavage after Arg produces a further fragment (Phe–Trp–Cys–Leu and Arg). Total: 3 fragments (Met–Ala–Gly–Ser–Lys, Phe–Trp–Cys–Leu, Arg). |
Accept clear explanation with correct fragment count. Allow 4 fragments if student counts the terminal Arg separately after cleavage.
11. (a) Name the components labelled P, Q, and R in Fig. 11.1. [3]
| Label | Component | Mark |
|---|---|---|
| P | Adenine | [1] |
| Q | Ribose | [1] |
| R | Three phosphate groups / triphosphate | [1] |
Accept: "adenosine" for P+Q combined if figure labels them together; adjust based on figure.
11. (b) Explain why ATP is described as the universal energy currency of cells. [2]
| Mark | Point |
|---|---|
| [1] | ATP releases a small, manageable amount of energy (approximately 30.5 kJ mol⁻¹) when the terminal phosphate bond is hydrolysed; |
| [1] | This energy is used to drive energy-requiring processes (e.g., active transport, muscle contraction, biosynthesis) in all living cells, and ATP is continuously regenerated from ADP and Pi during cellular respiration. |
Accept: "ATP couples energy-releasing reactions (catabolism) to energy-requiring reactions (anabolism)."
12. (a) Explain why glucose gives a positive result with Benedict's test but sucrose does not. [2]
| Mark | Point |
|---|---|
| [1] | Glucose is a reducing sugar because it has a free aldehyde group (or can form one in solution) that can reduce Cu²⁺ ions to Cu⁺ ions (forming a brick-red precipitate of Cu₂O); |
| [1] | Sucrose is a non-reducing sugar because its anomeric carbons (C1 of glucose and C2 of fructose) are involved in the glycosidic bond, so no free aldehyde or ketone group is available to act as a reducing agent. |
12. (b) Describe an additional test that could be carried out to confirm the identity of the starch solution. State the expected result. [2]
| Mark | Point |
|---|---|
| [1] | Add iodine solution (iodine in potassium iodide) to the starch solution; |
| [1] | A blue-black colour develops (positive result for starch). |
Accept: "iodine test" with correct colour change.
Section C: Extended Response (10 marks)
13. Discuss the importance of hydrogen bonding in determining the structure and properties of water, and explain how these properties make water essential for living organisms. [10]
Marking Scheme
| Mark Band | Descriptor |
|---|---|
| 9–10 | Comprehensive discussion covering molecular basis of H-bonding, at least three properties with clear biological examples, well-structured, accurate scientific language. |
| 7–8 | Good discussion covering H-bonding and three properties with biological examples, mostly accurate, some minor omissions. |
| 5–6 | Adequate discussion covering H-bonding and two properties with some biological examples, some inaccuracies or lack of detail. |
| 3–4 | Limited discussion, one or two properties mentioned, weak biological examples, significant omissions. |
| 1–2 | Very limited, only basic points, little or no biological context. |
| 0 | No creditworthy content. |
Indicative Content
Molecular basis of hydrogen bonding in water:
- Water is a polar molecule – oxygen is more electronegative than hydrogen, creating a dipole (δ⁻ O, δ⁺ H).
- Hydrogen bonds form between the slightly positive hydrogen atom of one water molecule and the slightly negative oxygen atom of a neighbouring water molecule.
- Each water molecule can form up to four hydrogen bonds (two via its hydrogen atoms, two via its lone pairs on oxygen).
- Hydrogen bonds are individually weak but collectively strong, giving water its unique properties.
Property 1: High specific heat capacity
- Hydrogen bonds absorb a lot of heat energy before breaking, so water resists temperature changes.
- Biological importance: Provides a stable thermal environment for aquatic organisms; helps maintain constant body temperature in endotherms (e.g., mammals); prevents rapid temperature fluctuations in cells.
Property 2: High latent heat of vaporisation
- Many hydrogen bonds must be broken for water to evaporate, requiring substantial energy.
- Biological importance: Evaporation of sweat/perspiration is an effective cooling mechanism in mammals (e.g., humans); transpiration in plants cools leaves.
Property 3: Cohesion and surface tension
- Hydrogen bonds cause water molecules to stick together (cohesion).
- Biological importance: Cohesion-tension theory explains water transport in xylem vessels – continuous water columns are pulled up from roots to leaves under tension; surface tension allows small organisms (e.g., pond skaters) to walk on water surfaces.
Property 4: Water as a solvent
- Water's polarity and hydrogen bonding ability allow it to dissolve many ionic and polar substances (e.g., salts, sugars, amino acids).
- Biological importance: Acts as a transport medium in blood plasma (dissolved nutrients, gases, wastes); cytoplasm is an aqueous solution where metabolic reactions occur; excretion of nitrogenous wastes (e.g., urea dissolved in urine).
Property 5: Density and freezing behaviour
- Hydrogen bonds hold water molecules in a crystalline lattice when frozen, making ice less dense than liquid water.
- Biological importance: Ice floats, insulating the water below and preventing aquatic ecosystems from freezing solid, allowing organisms to survive winter.
Property 6: High specific heat of fusion
- Water releases significant heat when freezing.
- Biological importance: Provides a buffer against freezing in cells and tissues.
Credit any three properties with clear biological examples. Award marks for accurate scientific language, logical structure, and integration of molecular explanation with biological significance.
END OF ANSWER KEY
TuitionGoWhere Secondary School (AI) – Practice Paper 4 – Biology H2 A-Level – Marking Scheme