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A Level H2 Biology Practice Paper 5

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

Subject: Biology
Level: H2
Paper: Practice Paper (Version 5 of 5)
Topic: Cells & Biomolecules
Duration: 1 hour 15 minutes
Total Marks: 60

Name: __________________________
Class: __________________________
Date: __________________________

Instructions to Candidates

Write your name, class, and date in the spaces provided.
Answer all questions.
Write your answers in the spaces provided in this booklet.
The number of marks is given in brackets [ ] at the end of each question or part question.
You are advised to spend approximately 75 minutes on this paper.

Section A: Structured Questions

Answer all questions in this section.

1. Fig. 1.1 shows a simplified diagram of the fluid mosaic model of a cell surface membrane.

(Note: Imagine Fig 1.1 showing a phospholipid bilayer with embedded proteins, cholesterol, and glycoproteins. Label A points to a phospholipid head, Label B to a channel protein, Label C to cholesterol.)

(a) With reference to Fig. 1.1, describe the arrangement of the molecules labelled A and explain how this arrangement contributes to the barrier function of the membrane. [3]

(b) Explain the role of the molecule labelled C in maintaining membrane stability at varying temperatures. [2]

(c) A drug is designed to inhibit the transport of glucose into a specific cell type by binding to the protein labelled B. With reference to Fig. 1.1, suggest how this drug might prevent glucose entry without destroying the membrane structure. [2]

2. Mitochondria are often described as the "powerhouses" of the cell. Fig. 2.1 shows the structure of a mitochondrion.

(a) Identify the structures labelled X (inner membrane folds) and Y (fluid-filled space). [2]

(b) Explain why the inner mitochondrial membrane is highly folded, linking your answer to the efficiency of ATP production. [3]

(c) A suspension of isolated mitochondria was placed in a buffer containing ADP, inorganic phosphate (Pi), and succinate (a respiratory substrate). Oxygen concentration was monitored over time. (i) Explain why the oxygen concentration in the buffer decreases. [2]

(ii) If an inhibitor of the electron transport chain is added, predict and explain the effect on the pH of the intermembrane space. [3]

3. Proteins are essential biological molecules with diverse functions.

(a) Describe the formation of a peptide bond between two amino acids. Include the names of the functional groups involved. [3]

(b) Haemoglobin is a globular protein. Explain how the secondary and tertiary structures of haemoglobin allow it to function effectively in oxygen transport. [4]

(c) Suggest why fibrous proteins, such as collagen, are suitable for providing structural support in connective tissues, whereas globular proteins are not. [2]

4. Enzymes are biological catalysts. Fig. 4.1 shows the effect of substrate concentration on the rate of an enzyme-catalysed reaction in the presence and absence of a competitive inhibitor.

(a) Define the term active site. [1]

(b) With reference to Fig. 4.1, explain the difference in the reaction rates at low substrate concentrations between the uninhibited reaction and the reaction with the competitive inhibitor. [3]

(c) Explain why the maximum rate of reaction ( $V_{max}$ ) is the same for both curves at high substrate concentrations. [2]

5. DNA and RNA are nucleic acids involved in genetic information storage and expression.

(a) Compare the structural differences between DNA and RNA. Give three distinct differences. [3]

(b) Describe the process of DNA replication, focusing on the role of DNA helicase and DNA polymerase. [4]

Section B: Data Interpretation and Application

Answer all questions in this section.

6. Gel electrophoresis is a technique used to separate DNA fragments. Fig. 6.1 shows the results of a gel electrophoresis experiment performed on DNA samples from three individuals (P, Q, and R) and a DNA ladder (L).

(Note: Imagine Fig 6.1 showing bands. Individual P has one band at 500bp. Individual Q has two bands, one at 500bp and one at 300bp. Individual R has one band at 300bp.)

(a) Explain the principle behind the separation of DNA fragments in gel electrophoresis. [3]

(b) With reference to Fig. 6.1, deduce the genotype of individual Q if the gene being tested has two alleles, $A_1$ (500bp fragment) and $A_2$ (300bp fragment). Explain your answer. [3]

7. The lac operon in E. coli controls the production of enzymes involved in lactose metabolism.

(a) Distinguish between an inducible operon and a repressible operon. [2]

(b) Explain the role of the lac repressor protein in the absence of lactose. [3]

8. Fig. 8.1 shows the changes in membrane potential of a neuron during the transmission of an action potential.

(Note: Imagine Fig 8.1 showing a graph of Membrane Potential (mV) vs Time (ms). Phases: Resting (-70mV), Depolarisation (up to +40mV), Repolarisation (down to -70mV), Hyperpolarisation.)

(a) With reference to Fig. 8.1, describe the changes in membrane permeability to sodium ions ( $Na^+$ ) during the depolarisation phase. [2]

(b) Explain the role of the sodium-potassium pump in restoring the resting potential after hyperpolarisation. [3]

(c) Multiple sclerosis is a disease where the myelin sheath is damaged. Explain how this damage affects the speed of nerve impulse transmission. [2]

9. Cell division is essential for growth and repair.

(a) Outline the main events that occur during prophase of mitosis. [3]

(b) Explain the importance of the spindle fibres attaching to the centromeres of chromosomes during metaphase. [2]

10. Water is a vital molecule for life.

(a) Explain how the polarity of water molecules contributes to its effectiveness as a solvent for biological reactions. [3]

(b) Describe two other properties of water that are important for living organisms and explain the biological significance of each. [4]

Section C: Extended Response

Answer all questions in this section.

11. Discuss the relationship between the structure and function of the phospholipid bilayer in cell surface membranes. In your answer, include references to:

The amphipathic nature of phospholipids.
The role of membrane proteins in transport.
The importance of membrane fluidity.

[10]

...... [End of Paper] ......

Answers

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)

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)