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A Level H2 Biology Practice Paper 1

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TuitionGoWhere Practice Paper – Biology H2 A‑Level (Cells & Biomolecules)

PRACTICE · Version 1 of 5

Subject: Biology H2
Level: A‑Level
Paper: Topic Practice – Cells & Biomolecules
Duration: 60 minutes
Total Marks: 42

Name: _______________________________
Class: ________ Date: ___________

Instructions to Candidates

This paper consists of 20 questions based on the topic Cells & Biomolecules.
Write your answers in the spaces provided.
The marks for each question or part question are shown in square brackets.
You may use an approved calculator.
Where diagrams are provided, refer to them clearly in your answers.

SECTION A – Structured Response
Answer all questions in this section.

Question 1 (2 marks)
Fig. 1.1 shows a section of a prokaryotic cell.

Fig. 1.1

┌───────────────────────────────┐
│          Cell wall            │
│ ┌───────────────────────────┐ │
│ │     Plasma membrane       │ │
│ │ ┌─────────────────────┐   │ │
│ │ │   Cytoplasm          │   │ │
│ │ │   ·Ribosomes         │   │ │
│ │ │   ·Nucleoid (DNA)    │   │ │
│ │ └─────────────────────┘   │ │
│ └───────────────────────────┘ │
│   Flagellum                   │
└───────────────────────────────┘

With reference to Fig. 1.1, explain how the plasma membrane (labelled P) contributes to the maintenance of a chemical gradient across the membrane.

………………………………………………………………………………………………………………
………………………………………………………………………………………………………………
………………………………………………………………………………………………………………
………………………………………………………………………………………………………………
………………………………………………………………………………………………………………

[2]

Question 2 (3 marks)
Describe the role of cholesterol in regulating the fluidity of an animal cell membrane.

[3]

Question 3 (2 marks)
With reference to Fig. 3.1, explain how the presence of saturated fatty acid tails in phospholipids affects the permeability of the lipid bilayer to small polar molecules.

Fig. 3.1 (structure of phospholipid with straight saturated tails)

[2]

Question 4 (3 marks)
Explain how a sodium-potassium pump (Na⁺/K⁺-ATPase) can transport sodium ions against their concentration gradient using energy from ATP.

[3]

Question 5 (1 mark)
Name one organelle, other than a ribosome, that is not membrane-bound and state its function.

………………………………………………………………………………………………………………

[1]

SECTION B – Diagram / Data Interpretation
Refer to the diagrams and data provided. Answer all questions.

Question 6 (3 marks)
Fig. 6.1 shows the results of gel electrophoresis of haemoglobin from four individuals. A, B, C, and D represent the positions of haemoglobin variants.

Fig. 6.1

← Cathode (–)           Anode (+) →

Lane 1:  ────── A ────── B ──────  
Lane 2:  ────── A ──────  
Lane 3:  ────── B ────── C ──────  
Lane 4:  ────── A ────── B ────── C ──────

(a) Explain why the haemoglobin variants migrate to different positions in the gel. [2]

………………………………………………………………………………………………………………
………………………………………………………………………………………………………………

(b) Using the results in Fig. 6.1, determine the genotype of the individual in Lane 3 with respect to the gene encoding haemoglobin. Explain your reasoning. [1]

[3]

Question 7 (3 marks)
A suspension of isolated mitochondria was prepared in a buffer containing ADP and inorganic phosphate. The oxygen concentration of the suspension was monitored over time. At time X, sodium azide, a metabolic inhibitor, was added. Fig. 7.1 shows the oxygen concentration changes.

Fig. 7.1

Oxygen conc.
│
│  ────╲                         
│         ╲                      
│           ╲──────  (after addition)
│
└───────────────────────── Time
                    ↑
                 Time X

(a) Describe the effect of sodium azide on the rate of oxygen consumption by the mitochondria. [1]

………………………………………………………………………………………………………………

(b) Explain the effect you described in (a) in terms of the electron transport chain and chemiosmosis. [2]

………………………………………………………………………………………………………………
………………………………………………………………………………………………………………
………………………………………………………………………………………………………………

[3]

Question 8 (3 marks)
Fig. 8.1 shows a simplified diagram of the lac operon in the absence of lactose.

Fig. 8.1 (operator bound by repressor; RNA polymerase blocked)

(a) State the role of the repressor protein under these conditions. [1]

………………………………………………………………………………………………………………

(b) Explain why it is advantageous for E. coli to have an inducible operon such as the lac operon. [2]

[3]

Question 9 (2 marks)
Fig. 9.1 illustrates a misfolded prion protein (PrP^Sc) interacting with a normal cellular prion protein (PrP^C), causing the latter to adopt the misfolded conformation.

Fig. 9.1 (diagram showing conversion of α-helical PrP^C to β-sheet PrP^Sc)

With reference to Fig. 9.1, suggest why misfolded prion proteins are resistant to normal cellular degradation and tend to accumulate in nervous tissue.

[2]

Question 10 (3 marks)
The table below shows the amino acid sequence of a small part of the active site of an enzyme obtained from two different species.

Species	Active site sequence (one-letter code)
P	Gly–His–Ser–Glu–Gly
Q	Gly–His–Ser–Asp–Gly

(a) State one type of bond or interaction that stabilises the three‑dimensional shape of the active site. [1]

………………………………………………………………………………………………………………

(b) Explain why the replacement of glutamate (Glu) with aspartate (Asp) at position 4 might not abolish enzyme activity, even though the amino acid has changed. [2]

[3]

SECTION C – Calculation and Application
Show your working clearly where necessary.

Question 11 (2 marks)
A student measures the rate of oxygen consumption by a mitochondrial suspension in the presence of different substrates. The results are shown below.

Substrate	Rate of O₂ consumption (nmol O₂ min⁻¹ mg⁻¹ protein)
Succinate	120
Pyruvate	80
No substrate	10

Calculate the percentage increase in oxygen consumption when succinate is added compared with the no‑substrate control. Show your working.

[2]

Question 12 (2 marks)
A red blood cell has a diameter of 7.5 µm. A mitochondrion has a diameter of approximately 0.5 µm. Calculate the ratio of the red blood cell diameter to the mitochondrion diameter.

[2]

Question 13 (2 marks)
A phospholipid bilayer has a thickness of 6 nm. A transmembrane protein extends 2 nm beyond the bilayer into the cytoplasm. What is the total length of the protein embedded in the membrane?

[2]

Question 14 (2 marks)
In a cell, the concentration of K⁺ ions inside the cell is 140 mmol dm⁻³ and outside is 5 mmol dm⁻³. Calculate the concentration gradient ratio (inside : outside) for K⁺.

[2]

Question 15 (2 marks)
The enzyme catalase has a turnover number of 4 × 10⁷ molecules of substrate per second. If a catalase solution produces 1.2 × 10⁸ molecules of product per second under saturating substrate conditions, calculate the minimum number of catalase molecules present.

[2]

SECTION D – Extended Application
Answer the following questions, which involve synthesising concepts from the topic.

Question 16 (3 marks)
Explain how the structure of a channel protein determines its ability to transport specific ions, such as K⁺, across the cell membrane.

[3]

Question 17 (3 marks)
A student proposes that the rate of facilitated diffusion of glucose into a cell will increase linearly with increasing extracellular glucose concentration. Discuss whether this prediction is correct by referring to the properties of GLUT transporter proteins found in the plasma membrane.

[3]

Question 18 (3 marks)
Using your knowledge of protein folding, explain why a single amino acid substitution (e.g., glutamate → valine) in the β‑globin chain can cause the symptoms of sickle cell anaemia.

[3]

Question 19 (2 marks)
Suggest why glycoproteins on the cell surface membrane are important for cell‑cell recognition, but glycolipids are also necessary for maintaining membrane stability.

[2]

Question 20 (1 mark)
State one function of the nucleolus that involves the synthesis of cellular components.

………………………………………………………………………………………………………………

[1]

END OF PAPER
Check your answers if time permits.

Answers

TuitionGoWhere Practice Paper – Biology H2 A‑Level (Cells & Biomolecules)

ANSWER KEY & MARKING SCHEME

Question 1 (2 marks)

Answer:
The plasma membrane is selectively permeable due to the phospholipid bilayer (hydrophobic core). It allows non‑polar, small molecules to diffuse but restricts larger/polar/charged molecules, maintaining different concentrations of ions and solutes on each side. The presence of transport proteins enables controlled movement, further sustaining the gradient.

1 mark: reference to selective permeability / hydrophobic core restricts movement of charged/polar solutes.
1 mark: transport proteins maintain gradient by active/passive transport.

Question 2 (3 marks)

Answer:
Cholesterol is interspersed among phospholipids in animal membranes. At high temperatures it reduces fluidity by restraining phospholipid movement; at low temperatures it prevents tight packing, maintaining fluidity. It thus acts as a fluidity buffer and also reduces permeability to small water‑soluble molecules.

1 mark: cholesterol restricts movement of fatty acid tails at high temperatures.
1 mark: prevents crystallisation / maintains fluidity at low temperatures.
1 mark: overall buffer effect / reduces permeability.

Question 3 (2 marks)

Answer:
Saturated fatty acid tails have straight chains that pack closely together, increasing the number of van der Waals interactions. This dense packing reduces the movement of lipid molecules and decreases the passage of small polar molecules, making the bilayer less permeable.

1 mark: straight tails pack closely, increasing van der Waals forces.
1 mark: reduces permeability to small polar molecules.

Question 4 (3 marks)

Answer:
The Na⁺/K⁺ pump uses energy from ATP hydrolysis to phosphorylate itself, causing a conformational change. This change expels three Na⁺ ions out of the cell and imports two K⁺ ions, both against their gradients. The cycle restores the pump’s original shape, enabling repeated active transport.

1 mark: ATP hydrolysis and phosphorylation of pump.
1 mark: conformational change moves ions against concentration gradient.
1 mark: 3 Na⁺ out, 2 K⁺ in per ATP, maintaining gradient.

Question 5 (1 mark)

Answer:
Nucleoid (region containing DNA) – function: stores genetic information.

1 mark: any correct non‑membrane‑bound organelle (e.g., nucleoid, cytoskeleton filament) with function.
(Accept: nucleolus – ribosome assembly, but nucleolus not in prokaryotes; in eukaryotes, nucleolus is not membrane‑bound.)
Ideal answer: Centriole – organisation of spindle fibres; or Ribosome is membrane‑bound? Ribosomes are not membrane‑bound but the question says “other than a ribosome”. So answer: Cytoskeleton – provides mechanical support.

Correct: Nucleolus – site of ribosomal RNA synthesis and ribosome assembly.

Accept any one valid.

Question 6 (3 marks)

(a) 2 marks
Answer: A potential difference (electric field) is applied. Proteins migrate according to their net charge and molecular mass; smaller or more highly charged proteins move faster/farther. Variation in amino acid composition (e.g., HbA vs HbS with different charge) causes different electrophoretic mobility.

1 mark: electric field / potential difference applied.
1 mark: migration depends on charge and size/mass, yielding different positions.

(b) 1 mark
Answer: Lane 3 shows two bands (B and C), indicating heterozygosity – the individual has two different alleles, e.g., Hba/Hbs.

1 mark: two bands = heterozygous genotype (any correct notation).

Question 7 (3 marks)

(a) 1 mark
Answer: Sodium azide reduced/stopped oxygen consumption.

(b) 2 marks
Answer: Sodium azide inhibits cytochrome c oxidase (Complex IV), blocking electron transfer to oxygen. This halts the electron transport chain, so no proton gradient is generated, and chemiosmosis cannot produce ATP. Consequently, oxygen is no longer consumed as the final electron acceptor.

1 mark: inhibition of electron transport chain / complex IV.
1 mark: prevents proton gradient formation, stops ATP synthesis, and oxygen consumption ceases.

Question 8 (3 marks)

(a) 1 mark
Answer: The repressor binds to the operator, blocking RNA polymerase from transcribing the structural genes.

(b) 2 marks
Answer: An inducible operon saves energy and resources because the enzymes are synthesised only when the substrate (lactose) is present. In the absence of lactose, no unnecessary proteins are made, allowing the bacterium to allocate resources to other metabolic needs.

1 mark: prevents wasteful enzyme synthesis.
1 mark: energy/resource conservation, advantage in changing environments.

Question 9 (2 marks)

Answer: Misfolded prion proteins have a high proportion of β‑sheet structures that aggregate via exposed hydrophobic regions, forming stable, insoluble fibrils. These aggregates resist proteolytic degradation and accumulate, especially in nervous tissue, because nervous cells have limited protein turnover capacity.

1 mark: exposure of hydrophobic regions / aggregation into stable fibrils.
1 mark: resistance to degradation leads to accumulation.

Question 10 (3 marks)

(a) 1 mark
Answer: Hydrogen bonds, ionic bonds, hydrophobic interactions, or disulfide bonds – any one.

(b) 2 marks
Answer: Glutamate (Glu, E) and aspartate (Asp, D) are both negatively charged, acidic amino acids with similar side‑chain properties (carboxyl group). The substitution conserves the charge and shape, so the active site’s interaction with the substrate may be retained. Even if the side chain is slightly shorter, it may still stabilise the transition state or bind the substrate.

1 mark: both are acidic/negatively charged, similar chemical property.
1 mark: because side chain is similar, the change is conservative and likely maintains catalytic function.

Question 11 (2 marks)

Answer:
Percentage increase = ((120 – 10) / 10) × 100% = (110 / 10) × 100% = 1100%.

1 mark: correct subtraction.
1 mark: correct calculation and final answer.

Question 12 (2 marks)

Answer:
Ratio = 7.5 µm / 0.5 µm = 15 : 1.

1 mark: division set‑up.
1 mark: correct simplified ratio.

Question 13 (2 marks)

Answer:
Transmembrane segment embedded in bilayer = total protein length – protruding part.
Embedded length = (bilayer thickness 6 nm) because the protein spans the membrane; protruding 2 nm is outside, so the embedded span is 6 nm. Alternatively, if the protein extends 2 nm into cytoplasm beyond the bilayer, the total protein length is 6 + 2 = 8 nm, but the embedded portion is the 6 nm within the bilayer. The question asks “total length of the protein embedded in the membrane” – that’s the 6 nm.

1 mark: recognition that embedded part spans the bilayer thickness.
1 mark: answer 6 nm.

Question 14 (2 marks)

Answer:
Inside : outside = 140 : 5 = 28 : 1.

1 mark: correct ratio set‑up.
1 mark: simplification to 28:1.

Question 15 (2 marks)

Answer:
Number of enzyme molecules = total product molecules per second / turnover number = (1.2 × 10⁸) / (4 × 10⁷) = 3.

1 mark: division.
1 mark: answer 3.

Question 16 (3 marks)

Answer:
Channel proteins have a specific, hydrophilic pore lined with amino acid residues that interact selectively with ions. For K⁺ channels, a selectivity filter with carbonyl oxygen atoms mimics the hydration shell of K⁺ but not Na⁺, allowing only K⁺ to pass. The channel’s diameter and charge distribution determine ion selectivity, and gating mechanisms regulate opening.

1 mark: structure has a hydrophilic pore with specific amino acid residues.
1 mark: selectivity filter discriminates between ions (e.g., size, dehydration energy).
1 mark: reference to gating / conformational change for regulated transport.

Question 17 (3 marks)

Answer:
Facilitated diffusion via GLUT transporters is saturable because transporter proteins have a finite number of binding sites. At low concentrations, rate increases as more binding sites are occupied. At high concentrations, all transporters become occupied (Vmax is reached), and the rate plateaus, not increasing linearly. Therefore the prediction is incorrect beyond the saturation point.

1 mark: GLUT transporters have specific binding sites and number limited.
1 mark: rate increases initially but reaches maximum velocity (Vmax).
1 mark: correctly explains non‑linearity / plateau.

Question 18 (3 marks)

Answer:
Glutamate (polar, hydrophilic) is replaced by valine (non‑polar, hydrophobic) at position 6 of β‑globin. This substitution creates a hydrophobic “sticky” patch on the haemoglobin molecule. Under low‑oxygen conditions, deoxygenated HbS molecules aggregate via these hydrophobic patches, forming long fibres that distort red blood cells into a sickle shape. Sickled cells block capillaries and are fragile, causing symptoms.

1 mark: substitution introduces hydrophobic valine.
1 mark: hydrophobic interactions cause aggregation/polymerisation of deoxy‑HbS.
1 mark: fibres distort RBC shape → vaso‑occlusive crises / anaemia.

Question 19 (2 marks)

Answer:
Glycoproteins have carbohydrate chains extending extracellularly that act as recognition sites (antigens, receptors) for cell‑cell communication. Glycolipids also contribute carbohydrate chains for recognition but additionally stabilise the membrane through hydrogen bonding with water and neighbouring lipids, and by contributing to the glycocalyx that protects the cell surface.

1 mark: glycoproteins for recognition/signalling.
1 mark: glycolipids stabilise membrane structure / glycocalyx protection.

Question 20 (1 mark)

Answer:
The nucleolus synthesises ribosomal RNA (rRNA) and assembles ribosomal subunits from rRNA and ribosomal proteins.

1 mark: correct function involving synthesis (rRNA, ribosome assembly).

END OF ANSWER KEY