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A Level H2 Biology Practice Paper 4

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

Subject: Biology H2
Level: A-Level
Paper: Practice Paper (Version 4 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 above.
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 simplified diagram of the fluid mosaic model of a cell surface membrane.

(Note: In a real exam, Fig 1.1 would show a phospholipid bilayer with embedded proteins, cholesterol, and glycoproteins. Assume standard labelling: A=Phospholipid head, B=Fatty acid tail, C=Channel protein, D=Glycoprotein, E=Cholesterol).

(a) Identify the structures labelled C and E. [2] C: _______________________________________________________ E: _______________________________________________________

(b) With reference to Fig. 1.1, explain how the structure of the phospholipid molecule contributes to the formation of the bilayer in an aqueous environment. [3]

2. A student investigated the effect of temperature on the activity of the enzyme amylase. The student measured the time taken for starch to be completely hydrolysed at different temperatures. The results are shown in Table 2.1.

Table 2.1

Temperature / °C	Time for starch hydrolysis / s	Rate of reaction / s⁻¹
20	120	0.0083
30	60	0.0167
40	30	0.0333
50	45	0.0222
60	> 300	< 0.0033

(a) Calculate the rate of reaction at 50°C. Show your working. [1] Working:

Answer: ____________________ s⁻¹

(b) Explain the difference in the rate of reaction between 40°C and 60°C with reference to enzyme structure. [4]

3. Fig. 3.1 shows the results of a gel electrophoresis analysis of haemoglobin from four individuals. The alleles for haemoglobin are $Hb^A$ (normal) and $Hb^S$ (sickle cell).

(Note: Fig 3.1 shows 4 lanes. Lane 1: One band at top (slow). Lane 2: One band at bottom (fast). Lane 3: Two bands (top and bottom). Lane 4: One band at bottom.)

(a) Identify which lane represents an individual who is heterozygous for sickle cell anaemia. [1] Lane: ______

(b) Explain why the heterozygous individual shows two bands on the gel. [2]

4. Mitochondria are often described as the "powerhouses" of the cell.

(a) Name the process that occurs in the mitochondrial matrix that produces reduced NAD and reduced FAD. [1]

(b) Explain the role of oxygen in aerobic respiration. [2]

(c) Cyanide is a poison that inhibits the enzyme cytochrome c oxidase in the electron transport chain. Explain why cyanide poisoning leads to a rapid decrease in ATP production. [3]

5. Fig. 5.1 shows the structure of a triglyceride and a phospholipid.

(a) State one structural difference between a triglyceride and a phospholipid. [1]

(b) Explain why phospholipids are suitable for forming cell membranes, whereas triglycerides are not. [3]

(c) Triglycerides are used as energy storage molecules in animals. Explain two properties of triglycerides that make them suitable for this function. [2]

Section B: Data Interpretation and Application

Answer all questions in this section.

6. The lac operon in E. coli controls the production of enzymes required for lactose metabolism.

(a) Define the term 'operon'. [1]

(b) In the absence of lactose, the lac operon is switched off. Explain the molecular mechanism that prevents transcription of the structural genes in the absence of lactose. [3]

(c) Suggest why it is advantageous for E. coli to have an inducible operon for lactose metabolism rather than constitutively expressing the enzymes. [2]

7. A researcher studied the uptake of glucose into red blood cells. The rate of uptake was measured at varying external glucose concentrations. The results are plotted in Fig. 7.1.

(Note: Fig 7.1 shows a curve that rises steeply at low concentrations and then plateaus at high concentrations.)

(a) Describe the relationship between glucose concentration and rate of uptake shown in Fig. 7.1. [2]

(b) Explain the shape of the curve at high glucose concentrations. [3]

(c) The researcher repeated the experiment in the presence of a competitive inhibitor. Sketch the expected curve on Fig. 7.1 and label it 'I'. [2] (Space for sketch/description)

8. DNA replication is a semi-conservative process.

(a) Define 'semi-conservative replication'. [1]

(b) Describe the role of DNA helicase and DNA polymerase in replication. [4] DNA Helicase:

DNA Polymerase:

9. Fig. 9.1 shows a dipeptide formed from two amino acids.

(a) Name the type of bond linking the two amino acids. [1]

(b) Describe the reaction that forms this bond, including the molecules produced. [3]

10. Water is essential for life.

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

(b) State one other biological importance of water and explain it. [2] Importance: _____________________________________________________________ Explanation: ____________________________________________________________

Section C: Extended Response

Answer all questions in this section.

11. Compare and contrast the structures of prokaryotic and eukaryotic cells. In your answer, refer to specific organelles and genetic material. [8]

12. Discuss the importance of enzymes in metabolic pathways. Include references to activation energy, specificity, and the effect of environmental factors. [10]

Answers

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

Subject: Biology H2
Paper: Practice Paper (Version 4 of 5) - Answer Key
Topic: Cells & Biomolecules

Section A: Structured Questions

1. (a) C: Channel protein / Transport protein [1] E: Cholesterol [1]

(b)

Phospholipids are amphipathic / have a hydrophilic head and hydrophobic tails. [1]
In an aqueous environment, the hydrophilic heads face outward towards the water. [1]
The hydrophobic tails face inward, away from water, forming a bilayer. [1]

(c)

Acts as a receptor / antigen for cell recognition / cell signalling. [1]

2. (a)

Rate = 1 / Time
Rate = 1 / 45
Answer: 0.0222 s⁻¹ [1]

(b)

At 40°C, the enzyme is near its optimum temperature; kinetic energy is high, leading to frequent successful collisions. [1]
At 60°C, the high temperature breaks hydrogen bonds / ionic bonds holding the tertiary structure. [1]
The active site changes shape / denatures. [1]
Substrate can no longer bind to the active site / enzyme-substrate complexes cannot form. [1]

(c)

To maintain a constant pH / prevent pH from changing, which could affect enzyme activity. [1]

3. (a)

Lane 3 [1]

(b)

Heterozygous individuals have both the $Hb^A$ and $Hb^S$ alleles. [1]
These alleles code for haemoglobin proteins with different charges / masses, causing them to migrate to different positions on the gel. [1]

(c)

Proteins are placed in a gel matrix. [1]
An electric field / potential difference is applied. [1]
Proteins migrate based on their charge and size / molecular mass. [1]

4. (a)

Krebs cycle / Link reaction (Accept either, though Krebs produces more reduced coenzymes). [1]

(b)

Oxygen acts as the final electron acceptor. [1]
It combines with electrons and protons to form water. [1]

(c)

Cyanide blocks the electron transport chain. [1]
Electrons cannot flow, so protons are not pumped into the intermembrane space. [1]
No proton gradient is established, so ATP synthase cannot produce ATP via chemiosmosis. [1]

5. (a)

Phospholipids have a phosphate group / two fatty acid tails, whereas triglycerides have three fatty acid tails and no phosphate group. [1]

(b)

Phospholipids are amphipathic (hydrophilic head, hydrophobic tail), allowing them to form a bilayer in water. [1]
Triglycerides are non-polar / hydrophobic. [1]
Triglycerides would form droplets rather than a membrane barrier. [1]

(c)

High energy content per gram (due to many C-H bonds). [1]
Insoluble in water, so they do not affect water potential / osmotic balance of cells. [1]

Section B: Data Interpretation and Application

6. (a)

A group of genes controlled by a single promoter / operator region. [1]

(b)

A repressor protein is bound to the operator region. [1]
This prevents RNA polymerase from binding to the promoter. [1]
Therefore, transcription of the structural genes cannot occur. [1]

(c)

It saves energy / resources. [1]
Enzymes are only produced when the substrate (lactose) is present, preventing waste. [1]

7. (a)

As glucose concentration increases, the rate of uptake increases. [1]
Up to a certain point, after which the rate levels off / reaches a maximum. [1]

(b)

Glucose enters via facilitated diffusion using carrier proteins. [1]
At high concentrations, all carrier proteins are saturated / occupied. [1]
The rate is limited by the number of carrier proteins available. [1]

(c)

Sketch: Curve starts at origin, rises less steeply than original, and plateaus at the same maximum rate (Vmax) but at a higher concentration. [2]
- Marking note: 1 mark for lower initial slope, 1 mark for same Vmax.

8. (a)

Each new DNA molecule consists of one original (parental) strand and one newly synthesized strand. [1]

(b)

DNA Helicase: Breaks hydrogen bonds between base pairs; unzips the double helix. [2]
DNA Polymerase: Joins nucleotides together; forms phosphodiester bonds; adds nucleotides in the 5' to 3' direction. [2]

(c)

To ensure that each daughter cell receives a complete set of genetic information. [1]
To maintain the chromosome number / genetic identity. [1]

9. (a)

Peptide bond. [1]

(b)

Condensation reaction. [1]
Between the amine group of one amino acid and the carboxyl group of another. [1]
A molecule of water is released. [1]

(c)

The sequence of amino acids (primary structure) determines the interactions between R-groups. [1]
These interactions (hydrogen bonds, ionic bonds, disulfide bridges) cause folding into secondary and tertiary structures. [1]
The specific 3D shape determines the protein's function. [1]

10. (a)

Water molecules are polar (dipole). [1]
Positive ends attract negative ions/parts of molecules; negative ends attract positive ions/parts. [1]
This surrounds solute molecules, keeping them in solution. [1]

(b)

Importance: High specific heat capacity. [1]
Explanation: Buffers temperature changes, providing a stable environment for enzymes. [1]
- Alternative: High latent heat of vaporisation -> Cooling effect via sweating.

Section C: Extended Response

11. Compare and contrast prokaryotic and eukaryotic cells. [8]

Genetic Material:
- Prokaryotes: DNA is circular, naked (no histones), located in the nucleoid region. [1]
- Eukaryotes: DNA is linear, associated with histone proteins, enclosed within a nuclear envelope. [1]
Organelles:
- Prokaryotes: Lack membrane-bound organelles (e.g., no mitochondria, Golgi, ER). [1]
- Eukaryotes: Possess membrane-bound organelles (mitochondria, ER, Golgi, lysosomes). [1]
Ribosomes:
- Prokaryotes: Have 70S ribosomes. [1]
- Eukaryotes: Have 80S ribosomes (and 70S in mitochondria/chloroplasts). [1]
Cell Wall:
- Prokaryotes: Cell wall made of peptidoglycan (murein). [1]
- Eukaryotes: Plant cell walls made of cellulose; animal cells lack cell walls. [1]
Size:
- Prokaryotes are generally smaller (1-10 µm). [1]
- Eukaryotes are generally larger (10-100 µm). [1]
- (Award max 8 marks. Points must be comparative.)

12. Discuss the importance of enzymes in metabolic pathways. [10]

Activation Energy:
- Enzymes lower the activation energy required for reactions. [1]
- This allows metabolic reactions to occur rapidly at body temperature. [1]
- Without enzymes, reactions would be too slow to sustain life. [1]
Specificity:
- Enzymes are specific due to the complementary shape of the active site (lock and key / induced fit). [1]
- This ensures that metabolic pathways are controlled and specific products are formed. [1]
- Prevents unwanted side reactions. [1]
Control/Regulation:
- Enzyme activity can be regulated (e.g., by inhibitors, pH, temperature, cofactors). [1]
- This allows cells to respond to changes in environment or metabolic demand. [1]
- Example: End-product inhibition prevents waste of resources. [1]
Environmental Factors:
- Temperature: Increases kinetic energy up to optimum; denaturation above optimum. [1]
- pH: Affects ionization of R-groups and active site shape; each enzyme has an optimum pH. [1]
- Substrate concentration: Rate increases until saturation (Vmax). [1]
- (Award marks for clear explanation and linkage to metabolic importance. Max 10 marks.)