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A Level H2 Biology Practice Paper 1

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

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

Name: __________________________
Class: __________________________
Date: __________________________

Instructions to Candidates

Write your name and class on 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 10 minutes reading the paper and 65 minutes answering.

Section A: Structured Questions (40 Marks)

Answer all questions in this section.

1. Fig. 1.1 shows a transmission electron micrograph of a pancreatic acinar cell, which is specialised for the secretion of digestive enzymes.

(Note: In a real exam, Fig 1.1 would show a cell with prominent RER, Golgi apparatus, secretory vesicles, and mitochondria.)

(a) Identify the organelles labelled A, B, and C in Fig. 1.1. [3]

A: _______________________________________________________
B: _______________________________________________________
C: _______________________________________________________

(b) With reference to Fig. 1.1, describe the pathway taken by a newly synthesised enzyme from its site of production to its secretion from the cell. [4]

2. Haemoglobin is a globular protein found in red blood cells. Fig. 2.1 shows the structure of a haemoglobin molecule.

(a) State the level of protein structure represented by the entire haemoglobin molecule shown in Fig. 2.1. [1]

(b) Haemoglobin contains both $\alpha$ -helices and $\beta$ -pleated sheets. Name the type of bond that stabilises these secondary structures. [1]

(c) Sickle cell anaemia is caused by a mutation in the gene coding for the $\beta$ -globin chain, resulting in the substitution of valine for glutamic acid at position 6. (i) Explain how this substitution affects the primary structure of the protein. [1]

_________________________________________________________________________

(ii) With reference to the properties of amino acids, explain why this substitution causes haemoglobin molecules to aggregate under low oxygen conditions. [3]

_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________

3. Gel electrophoresis is used to analyse DNA fragments. Fig. 3.1 shows the results of gel electrophoresis for a specific gene locus in four individuals (P, Q, R, and S). The gene has two alleles, $A_1$ and $A_2$ . Allele $A_1$ produces a fragment of 200 base pairs (bp), and allele $A_2$ produces a fragment of 150 bp.

(Note: Fig 3.1 would show lanes. P has one band at 200bp. Q has two bands at 200bp and 150bp. R has one band at 150bp. S has one band at 200bp.)

(a) Determine the genotype of individual Q. [1]

(b) Explain why individual Q shows two bands while individual P shows only one band. [2]

4. The fluid mosaic model describes the structure of the plasma membrane.

(a) Define the term 'fluid mosaic' in the context of membrane structure. [2]

(b) Cholesterol is a component of animal cell membranes. Explain two roles of cholesterol in maintaining membrane stability. [4]

5. Enzymes are biological catalysts. Fig. 5.1 shows the effect of substrate concentration on the rate of an enzyme-catalysed reaction in the presence and absence of a competitive inhibitor.

(Note: Fig 5.1 would show two curves. Curve X reaches $V_{max}$ quickly. Curve Y rises more slowly but eventually reaches the same $V_{max}$ as X at high substrate concentrations.)

(a) Identify which curve represents the reaction in the presence of the competitive inhibitor. [1]

(b) Explain why the maximum rate of reaction ( $V_{max}$ ) is the same for both curves. [3]

Section B: Data Interpretation and Application (20 Marks)

Answer all questions in this section.

6. Mitochondria are the sites of aerobic respiration. An experiment was conducted to investigate the effect of ADP concentration on oxygen consumption by isolated mitochondria.

Mitochondria were suspended in a buffer containing excess substrate (pyruvate) and inorganic phosphate ( $P_i$ ). ADP was added at time $t=0$ . The oxygen concentration in the buffer was monitored over time. The results are shown in Table 6.1.

Table 6.1: Oxygen concentration in mitochondrial suspension over time

Time (min)	Oxygen Concentration (arbitrary units)
0	100
2	92
4	84
6	76
8	68
10	60

(a) Calculate the rate of oxygen consumption between 0 and 10 minutes. Show your working. [2]

Answer space

(b) Explain the relationship between ADP concentration and the rate of oxygen consumption in mitochondria. [4]

(c) Sodium azide is a poison that inhibits cytochrome c oxidase (Complex IV) in the electron transport chain. Predict and explain the effect of adding sodium azide to the mitochondrial suspension on: (i) The rate of oxygen consumption. [2]

_________________________________________________________________________
_________________________________________________________________________

(ii) The production of ATP. [2]

_________________________________________________________________________
_________________________________________________________________________

7. The lac operon in E. coli controls the expression of genes involved in lactose metabolism.

(a) State the function of the following components of the lac operon: (i) Promoter [1]

_________________________________________________________________________

(ii) Operator [1]

_________________________________________________________________________

(b) Explain why the lac operon is described as an inducible operon. [3]

(c) Suggest why it is metabolically advantageous for E. coli to regulate the lac operon in this way, rather than expressing the genes constitutively (continuously). [2]

8. Water is essential for life due to its unique physical and chemical properties.

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

(b) Describe how hydrogen bonding between water molecules contributes to temperature regulation in organisms. [2]

End of Paper

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Answers

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

Answer Key & Marking Scheme

Topic: Cells & Biomolecules
Paper: Practice Paper 1 (Version 1 of 5)

Section A: Structured Questions

1. Pancreatic Acinar Cell

(a) Identification [3]

A: Rough Endoplasmic Reticulum (RER) [1]
B: Golgi Apparatus [1]
C: Mitochondria [1]
- Note: Accept "Mitochondrion". Do not accept "Ribosomes" for A unless clearly pointing to dots on membrane, but RER is the organelle.

(b) Pathway of Enzyme Secretion [4]

Protein/enzyme synthesised by ribosomes on the RER [1].
Transported in vesicles to the Golgi apparatus [1].
Modified/processed/packaged in the Golgi apparatus [1].
Transported in secretory vesicles to the plasma membrane and released via exocytosis [1].

(c) Role of Mitochondria [2]

Mitochondria are the site of aerobic respiration / ATP production [1].
ATP is required for protein synthesis, vesicle transport, and exocytosis (active processes) [1].

2. Haemoglobin Structure

(a) Level of Structure [1]

Quaternary structure [1].

(b) Bond in Secondary Structure [1]

Hydrogen bonds [1].
- Note: Must specify hydrogen bonds between peptide backbone / amide and carbonyl groups if asked for detail, but "Hydrogen bonds" is sufficient for 1 mark here.

(c) Sickle Cell Mutation (i) Effect on Primary Structure [1]

Change in the sequence/order of amino acids [1].

(ii) Aggregation Explanation [3]

Glutamic acid is hydrophilic/polar/charged, while valine is hydrophobic/non-polar [1].
The substitution exposes a hydrophobic region on the surface of the haemoglobin molecule [1].
Hydrophobic interactions cause haemoglobin molecules to stick together/aggregate to minimise contact with water [1].

3. Gel Electrophoresis

(a) Genotype of Q [1]

Heterozygous ( $A_1A_2$ ) [1].

(b) Explanation of Bands [2]

Individual Q has two different alleles ( $A_1$ and $A_2$ ), which produce DNA fragments of different sizes/masses [1].
Individual P is homozygous ( $A_1A_1$ ), so both alleles produce fragments of the same size, appearing as a single band [1].

(c) Principle of Separation [3]

DNA is negatively charged (due to phosphate groups) [1].
An electric field/potential difference is applied across the gel [1].
DNA fragments migrate towards the positive anode; smaller fragments move faster/further through the gel matrix than larger fragments [1].

4. Fluid Mosaic Model

(a) Definition [2]

Fluid: Phospholipids and proteins can move laterally within the layer [1].
Mosaic: Proteins are embedded in the phospholipid bilayer in a scattered/patterned arrangement [1].

(b) Roles of Cholesterol [4]

At high temperatures, cholesterol restricts the movement of phospholipid fatty acid tails, reducing membrane fluidity and preventing it from becoming too fluid [2].
At low temperatures, cholesterol prevents fatty acid tails from packing closely together, maintaining fluidity and preventing the membrane from becoming too rigid/solidifying [2].

5. Enzyme Kinetics

(a) Identification [1]

Curve Y [1].

(b) $V_{max}$ Explanation [3]

Competitive inhibitors bind to the active site, competing with the substrate [1].
At high substrate concentrations, the substrate outcompetes the inhibitor for the active site [1].
Therefore, all enzyme active sites can eventually be occupied by substrate, allowing the reaction to reach the same maximum rate as without inhibitor [1].

(c) Non-competitive Inhibitor Effect [2]

$V_{max}$ decreases [1].
$K_m$ remains unchanged (or increases slightly depending on pure/mixed, but typically "unchanged" is accepted for pure non-competitive in H2 context unless specified otherwise; however, strictly, pure non-competitive affects $V_{max}$ only. Accept: $V_{max}$ lower, $K_m$ same) [1].

Section B: Data Interpretation and Application

6. Mitochondrial Respiration

(a) Calculation [2]

Change in oxygen = $100 - 60 = 40$ arbitrary units [1].
Time = 10 minutes.
Rate = $40 / 10 = 4$ $40/10 = 4$ arbitrary units per minute [1].
- Note: Accept correct working even if final answer is wrong due to calculation error.

(b) ADP and Oxygen Consumption [4]

ADP is required for ATP synthesis via ATP synthase [1].
Electron transport chain (ETC) pumps protons to create a gradient [1].
Protons flow back through ATP synthase, driving ATP production from ADP + Pi [1].
If ADP is available, ATP synthase operates, allowing proton flow, which allows the ETC to continue passing electrons to oxygen (final electron acceptor), thus consuming oxygen [1].
- Alternative phrasing: Coupling of oxidation and phosphorylation. High ADP stimulates respiration (acceptor control).

(c) Sodium Azide Effect (i) Oxygen Consumption [2]

Rate of oxygen consumption decreases/stops [1].
Because cytochrome c oxidase is inhibited, electrons cannot be passed to oxygen, so oxygen is not reduced/consumed [1].

(ii) ATP Production [2]

ATP production decreases/stops [1].
Because the electron transport chain stops, no proton gradient is generated, so chemiosmosis/ATP synthase cannot function [1].

7. Lac Operon

(a) Functions [2] (i) Promoter: Site where RNA polymerase binds to initiate transcription [1]. (ii) Operator: Site where the repressor protein binds to block transcription [1].

(b) Inducible Operon Explanation [3]

The operon is normally switched off (repressed) because the repressor protein is bound to the operator [1].
In the presence of lactose (inducer), lactose binds to the repressor [1].
This causes a conformational change in the repressor, causing it to detach from the operator, allowing transcription to proceed [1].

(c) Metabolic Advantage [2]

Prevents waste of energy and resources (amino acids/ATP) synthesising enzymes when lactose is not present [1].
Allows the bacterium to respond rapidly to changes in environmental nutrient availability [1].

8. Water Properties

(a) Solvent Property [3]

Water molecules are polar (dipole), with partial positive charge on H and partial negative charge on O [1].
Polar/ionic solutes are attracted to water molecules (hydration shells form) [1].
This allows solutes to dissolve and remain dispersed, facilitating metabolic reactions in aqueous solution [1].

(b) Temperature Regulation [2]

Water has a high specific heat capacity due to hydrogen bonding [1].
Large amounts of heat energy are required to raise the temperature of water, helping organisms maintain stable internal temperatures / buffer against temperature fluctuations [1].
- Alternative: High latent heat of vaporisation allows cooling via sweating/evaporation.