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A Level H2 Biology Practice Paper 5

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TuitionGoWhere Practice Paper - Biology H2 A-Level

TuitionGoWhere Practice Paper (AI)

Subject: Biology H2
Level: A-Level
Paper: Practice Paper - Version 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 15 minutes reading the paper and 60 minutes answering.

Section A: Structured Questions

Answer all questions in this section.

1. Fig. 1.1 shows a diagram of a phospholipid bilayer with embedded proteins.

(Note: In a real exam, Fig 1.1 would show a membrane with a channel protein, a carrier protein, and cholesterol molecules.)

(a) Identify the component labelled X in Fig. 1.1 and state its role in membrane fluidity at low temperatures. [2]

(b) Explain why potassium ions ( $K^+$ ) cannot diffuse directly through the phospholipid bilayer, despite being small ions. [2]

(c) Describe the mechanism by which glucose enters a red blood cell from the plasma. In your answer, refer to the specific type of transport protein involved. [3]

2. Enzymes are biological catalysts that speed up metabolic reactions.

(a) Define the term activation energy. [1]

(b) Fig. 2.1 shows the effect of temperature on the rate of an enzyme-controlled reaction. (Note: Fig 2.1 shows a bell-shaped curve peaking at 40°C and dropping sharply to zero at 60°C.)

Explain the shape of the curve between: (i) 10°C and 40°C. [2]

(ii) 45°C and 60°C. [3]

(c) A student investigates the effect of pH on the activity of amylase. Suggest two variables, other than temperature and pH, that must be controlled in this experiment. [2]

3. Mitochondria are known as the "powerhouses" of the cell.

(a) State the specific location within the mitochondrion where the Krebs cycle occurs. [1]

(b) Explain the role of the inner mitochondrial membrane in ATP production. Refer to the electron transport chain and chemiosmosis in your answer. [4]

(c) Cyanide is a poison that inhibits cytochrome c oxidase, the final enzyme in the electron transport chain. Explain why cyanide poisoning leads to a rapid decrease in ATP production. [2]

4. DNA and RNA are nucleic acids essential for life.

(a) Compare the structural differences between DNA and RNA. Complete Table 4.1. [3]

Feature	DNA	RNA
Sugar
Number of strands
Bases present

(b) Describe the process of DNA replication. Include the roles of helicase and DNA polymerase in your answer. [4]

5. Proteins have complex structures determined by their amino acid sequence.

(a) Name the type of bond that links amino acids together in a polypeptide chain. [1]

(b) Explain how the primary structure of a protein determines its tertiary structure. [3]

(c) Haemoglobin is a globular protein, while collagen is a fibrous protein. Distinguish between globular and fibrous proteins in terms of their solubility and function. [2]

Section B: Data Interpretation and Application

Answer all questions in this section.

6. Fig. 6.1 shows the results of an experiment investigating the uptake of substance X by cells. The cells were placed in solutions with varying concentrations of substance X. Curve A represents uptake in the presence of oxygen, and Curve B represents uptake in the absence of oxygen (anaerobic conditions).

(Note: Fig 6.1 shows Curve A rising linearly then plateauing. Curve B rises linearly then plateaus at a much lower level than A.)

(a) Identify the mechanism of transport for substance X. Justify your answer with reference to the effect of oxygen. [2]

(b) Explain why the rate of uptake plateaus at high concentrations of substance X in Curve A. [2]

(c) Predict how the rate of uptake would change if a metabolic inhibitor, such as cyanide, was added to the solution in Curve A. Explain your prediction. [2]

7. Gel electrophoresis is used to separate DNA fragments.

(a) Explain the principle behind the separation of DNA fragments during gel electrophoresis. [2]

(b) A DNA sample is cut with a restriction enzyme. The resulting fragments are separated by gel electrophoresis. (i) State why DNA fragments move towards the anode (positive electrode). [1]

(ii) Explain why smaller DNA fragments travel further through the gel than larger fragments. [2]

8. Fig. 8.1 shows a simplified diagram of the lac operon in E. coli.

(Note: Fig 8.1 shows the Promoter, Operator, and Structural Genes (lacZ, lacY, lacA), along with the Regulator Gene producing a Repressor.)

(a) State the function of the operator region in the lac operon. [1]

(b) Explain what happens to the lac operon when lactose is absent from the environment. [3]

9. Water is essential for life due to its unique properties.

(a) Explain how hydrogen bonding between water molecules contributes to its high specific heat capacity. [2]

(b) Describe the importance of water’s high latent heat of vaporization to living organisms. [2]

10. Fig. 10.1 shows a diagram of a chloroplast.

(a) Identify the structure where the light-dependent reactions of photosynthesis occur. [1]

(b) Describe the role of water in the light-dependent reactions. [2]

(c) Explain how the products of the light-dependent reactions are used in the Calvin cycle (light-independent reactions). [3]

Section C: Extended Response

Answer all questions in this section.

11. Discuss the importance of membrane transport mechanisms in maintaining homeostasis in multicellular organisms. In your answer, compare and contrast simple diffusion, facilitated diffusion, and active transport, providing specific biological examples for each. [10]

12. "Enzymes are highly specific biological catalysts." With reference to the lock-and-key hypothesis and the induced-fit model, explain enzyme specificity. Discuss how environmental factors such as pH and substrate concentration affect enzyme activity. [10]

End of Paper

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Answers

TuitionGoWhere Practice Paper - Biology H2 A-Level

Answer Key and Marking Scheme Topic: Cells & Biomolecules (Version 5)

Section A: Structured Questions

1. (a) Component: Cholesterol. [1] Role: Prevents phospholipid tails from packing too closely together/crystallising at low temperatures, maintaining fluidity. [1]

(b) Potassium ions are charged/polar. [1] The interior of the phospholipid bilayer is hydrophobic/non-polar. [1] Therefore, ions are repelled by the hydrophobic core and cannot pass through.

(c) Mechanism: Facilitated diffusion. [1] Protein: Carrier protein (or channel protein, specifically GLUT1 for RBCs). [1] Description: Glucose binds to the carrier protein, causing a conformational change that releases glucose into the cell down its concentration gradient. No energy/ATP is required. [1]

2. (a) The minimum amount of energy required for reactant molecules to collide successfully and form the transition state. [1]

(b) (i) As temperature increases, kinetic energy of enzyme and substrate molecules increases. [1] This leads to more frequent successful collisions per unit time, increasing the rate of reaction. [1]

(ii) High temperature breaks the hydrogen bonds/ionic bonds holding the tertiary structure. [1] The enzyme denatures/loses its specific 3D shape. [1] The active site changes shape and is no longer complementary to the substrate, so enzyme-substrate complexes cannot form. [1]

Substrate concentration. [1]
Enzyme concentration. [1]
Volume of solution. [1]

3. (a) Mitochondrial matrix. [1]

(b) The inner membrane contains the electron transport chain (ETC) proteins. [1] Electrons pass along the ETC, releasing energy. [1] This energy is used to pump protons ( $H^+$ ) from the matrix into the intermembrane space, creating an electrochemical gradient. [1] Protons flow back into the matrix through ATP synthase, driving the synthesis of ATP from ADP and Pi (chemiosmosis). [1]

(c) Cyanide blocks electron transport. [1] This prevents the pumping of protons, so the proton gradient is not maintained, and ATP synthase cannot function. [1]

4. (a)

Feature	DNA	RNA
Sugar	Deoxyribose	Ribose
Number of strands	Double-stranded (helix)	Single-stranded
Bases present	A, T, C, G	A, U, C, G
(1 mark for each correct row) [3]

(b) Helicase: Breaks hydrogen bonds between base pairs, unzipping the DNA double helix. [1] DNA Polymerase: Adds free nucleotides to the exposed template strands. [1] It joins nucleotides via phosphodiester bonds. [1] Replication is semi-conservative (each new molecule has one old and one new strand). [1]

5. (a) Peptide bond. [1]

(b) The primary structure is the specific sequence of amino acids. [1] The R-groups (side chains) of the amino acids interact (via hydrogen bonds, ionic bonds, disulfide bridges, hydrophobic interactions). [1] These interactions fold the polypeptide into a specific 3D tertiary shape. [1]

(c) Solubility: Globular proteins are generally soluble in water; fibrous proteins are insoluble. [1] Function: Globular proteins are often metabolic (e.g., enzymes, transport); fibrous proteins are structural (e.g., collagen, keratin). [1]

Section B: Data Interpretation and Application

6. (a) Mechanism: Active transport. [1] Justification: Uptake is higher in the presence of oxygen (Curve A) because aerobic respiration produces ATP, which is required for active transport. Curve B (anaerobic) has lower uptake due to less ATP production. [1]

(b) The carrier proteins/transport proteins become saturated. [1] All available binding sites are occupied, so the rate cannot increase further despite higher concentration. [1]

(c) The rate of uptake would decrease significantly/stop. [1] Cyanide inhibits aerobic respiration, stopping ATP production. Without ATP, active transport cannot occur. [1]

7. (a) DNA fragments are separated based on their size/molecular mass. [1] An electric field is applied, and fragments move through a gel matrix; smaller fragments move faster/further. [1]

(b) (i) DNA is negatively charged (due to phosphate groups). [1] (ii) Smaller fragments encounter less resistance/friction from the gel matrix. [1] Therefore, they can migrate through the pores of the gel more easily/quickly than larger fragments. [1]

8. (a) It is the binding site for the repressor protein. [1]

(b) The regulator gene produces an active repressor protein. [1] The repressor binds to the operator region. [1] This blocks RNA polymerase from binding to the promoter, preventing transcription of the structural genes. [1]

(c) It prevents the waste of energy and resources. [1] The enzymes for lactose digestion are only produced when lactose is present (the substrate is available). [1]

9. (a) Hydrogen bonds between water molecules require a large amount of energy to break. [1] Therefore, water can absorb a lot of heat energy with only a small rise in temperature. [1]

(b) Evaporation of water (e.g., sweating) removes a large amount of heat energy from the body. [1] This helps to cool the organism and maintain a stable body temperature. [1]

10. (a) Thylakoid membrane (or grana). [1]

(b) Water is photolysed (split) by light energy. [1] It provides electrons to replace those lost by chlorophyll in Photosystem II, and releases protons ( $H^+$ ) and oxygen. [1]

(c) ATP provides energy for the reduction of GP to TP. [1] Reduced NADP (NADPH) provides hydrogen/electrons for the reduction of GP to TP. [1] These products allow the Calvin cycle to continue fixing carbon dioxide. [1]

Section C: Extended Response

11. Marking Guidance:

Simple Diffusion: Movement of small, non-polar molecules (e.g., $O_2$ , $CO_2$ ) down a concentration gradient directly through the bilayer. No energy/protein required. Example: Gas exchange in alveoli. [3 marks]
Facilitated Diffusion: Movement of polar/charged molecules (e.g., glucose, ions) down a concentration gradient via channel or carrier proteins. No energy required. Example: Glucose uptake in liver cells. [3 marks]
Active Transport: Movement of molecules/ions against a concentration gradient via carrier pumps. Requires ATP. Example: Na+/K+ pump in neurons to maintain resting potential. [3 marks]
Homeostasis Link: Explanation of how these mechanisms regulate internal environment (e.g., ion balance, nutrient uptake, waste removal). [1 mark]
Quality of Communication: Clear, logical structure, correct terminology. [1 mark] Total: 10 marks

12. Marking Guidance:

Lock-and-Key: Active site is rigid and complementary to substrate. [1 mark]
Induced-Fit: Active site is flexible; changes shape upon substrate binding to fit more tightly. [1 mark]
Specificity: Only specific substrates fit the active site due to shape/charge complementarity. [1 mark]
pH Effect: Changes in pH alter ionization of R-groups, disrupting bonds holding tertiary structure. Denaturation changes active site shape. Optimum pH exists. [3 marks]
Substrate Concentration: Rate increases as [S] increases due to more collisions. Plateaus when enzymes are saturated (Vmax). [3 marks]
Quality of Communication: Clear explanation, correct terminology. [1 mark] Total: 10 marks