A Level H2 Biology Practice Paper 2

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TuitionGoWhere Exam Practice (AI)

Subject: Biology H2
Level: A-Level
Paper: Practice Paper 2 (Version 2 of 5)
Duration: 2 hours
Total Marks: 75
Name: __________________________ Class: __________ Date: __________

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Instructions to Candidates

1. Answer all questions.
2. Write your answers in the spaces provided.
3. Use a black or dark blue pen.
4. For any diagrams, use a sharp pencil.
5. Show all working for calculations.

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Section A: Structured Questions

Question 1 Fig 1.1 shows a schematic diagram of the lac operon in Escherichia coli.

(Imagine Fig 1.1: A diagram showing the promoter, operator, and structural genes lacZ, lacY, and lacA, with a repressor protein bound to the operator in the absence of lactose).

(a) The lac operon is described as an inducible system. Explain why it is metabolically advantageous for a prokaryote to utilize an inducible operon for the breakdown of lactose. [3]



(b) Describe the molecular changes that occur when lactose is present in the medium, leading to the expression of the structural genes. [4]



(c) A mutation occurs in the operator region such that the repressor protein can no longer bind. Predict and explain the effect of this mutation on the production of $\beta$-galactosidase in the absence of lactose. [3]


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Question 2 A researcher investigated the effect of different inhibitors on the rate of oxygen consumption in a suspension of isolated mitochondria provided with ADP and inorganic phosphate ($\text{P}_i$). The results are shown in Table 2.1.

Table 2.1: Oxygen Consumption Rate of Mitochondria

Treatment	Rate of $\text{O}_2$ Consumption ($\text{nmol min}^{-1} \text{mg}^{-1}$)
Control (ADP + $\text{P}_i$)	120
+ Sodium Azide (Complex IV inhibitor)	12
+ Oligomycin (ATP synthase inhibitor)	25
+ 2,4-Dinitrophenol (Uncoupler)	155

(a) Explain why the rate of oxygen consumption is significantly lower in the "Sodium Azide" treatment compared to the control. [3]



(b) With reference to the chemiosmotic theory, explain why the inhibition of ATP synthase by Oligomycin also leads to a decrease in the rate of oxygen consumption. [4]



(c) Explain the observation that 2,4-Dinitrophenol (DNP) increases the rate of oxygen consumption beyond the control level. [4]


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Question 3 Fig 3.1 shows the primary structure of a segment of a normal protein and a mutated version of the same protein associated with a neurodegenerative disease.

(Imagine Fig 3.1: Normal protein showing a folded globular structure with hydrophobic residues in the core; Mutated protein showing a misfolded state with exposed hydrophobic patches).

(a) With reference to Fig 3.1, suggest why the mutated proteins tend to aggregate within the cell. [3]



(b) Describe how the formation of these protein aggregates can lead to cellular dysfunction and eventual cell death. [3]



(c) Explain the role of chaperone proteins in preventing the process described in part (a). [2]


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Question 4 Gel electrophoresis was used to analyze the haemoglobin of four individuals (A, B, C, and D) to screen for sickle cell anaemia.

(Imagine Fig 4.1: A gel image showing: Individual A: one band at HbA position; Individual B: one band at HbS position; Individual C: two bands (HbA and HbS); Individual D: one band at HbA position).

(a) Describe the principle of gel electrophoresis in separating different variants of haemoglobin. [4]



(b) Identify the genotype of Individual C. Explain your answer with reference to the banding pattern shown in Fig 4.1. [3]



(c) Explain why the HbS variant migrates to a different position on the gel compared to the HbA variant, referring to the molecular change in the $\beta$-globin chain. [3]


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Question 5 (a) Compare and contrast the structure and function of the plasma membrane of a prokaryotic cell with that of a eukaryotic cell. [5]




(b) Describe the process of facilitated diffusion and explain how it differs from active transport in terms of energy requirements and the use of concentration gradients. [5]



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Question 6 (a) Describe the structure of a phospholipid molecule and explain how this structure contributes to the formation of the lipid bilayer. [4]




(b) Explain the "Fluid Mosaic Model" of the cell membrane, describing the roles of cholesterol and integral proteins in maintaining membrane stability and function. [6]



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Question 7 (a) Explain the importance of the surface area to volume ratio in determining the maximum size of a cell. [3]




(b) Describe how the internal membrane systems of a eukaryotic cell (e.g., endoplasmic reticulum, Golgi apparatus) allow the cell to overcome the limitations of a low surface area to volume ratio. [4]



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Question 8 (a) Describe the structure and function of the nucleolus within the nucleus. [3]




(b) Explain the role of the nuclear pore complex in regulating the transport of materials between the nucleus and the cytoplasm. [4]



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Question 9 (a) Distinguish between the roles of the smooth endoplasmic reticulum (SER) and the rough endoplasmic reticulum (RER) in protein and lipid synthesis. [4]




(b) Describe the pathway a secreted protein takes from its site of synthesis to its exit from the cell. [6]



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Question 10 (a) Explain how the structure of a lysosome is adapted to its function of intracellular digestion. [4]




(b) Describe the process of autophagy and explain its significance in cellular homeostasis. [3]



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Answer Key - Biology H2 Practice Paper 2 (Version 2)

Question 1 (a)

• Inducible operons are only expressed when the substrate (lactose) is present. [1]
• This prevents the cell from wasting energy and amino acids/resources [1] synthesizing enzymes ($\beta$-galactosidase, permease) when they are not needed. [1]

(b)

• Lactose (or allolactose) acts as an inducer. [1]
• It binds to the repressor protein. [1]
• This causes a conformational change in the repressor, preventing it from binding to the operator. [1]
• RNA polymerase can then bind to the promoter and transcribe the structural genes. [1]

(c)

• The repressor cannot bind to the operator regardless of lactose presence. [1]
• The operon is constitutively expressed (always "on"). [1]
• $\beta$-galactosidase will be produced continuously even in the absence of lactose. [1]

Question 2 (a)

• Sodium azide inhibits Complex IV (cytochrome c oxidase). [1]
• This blocks the final transfer of electrons to oxygen. [1]
• Consequently, the electron transport chain (ETC) stops, and oxygen consumption ceases/drops. [1]

(b)

• Oligomycin inhibits ATP synthase. [1]
• This prevents protons ($\text{H}^+$) from flowing back into the matrix. [1]
• This leads to an accumulation of protons in the intermembrane space, creating an excessively high proton gradient. [1]
• The high gradient makes it energetically unfavorable for the ETC to pump more protons, slowing down electron flow and $\text{O}_2$ consumption. [1]

(c)

• DNP is an uncoupler that makes the inner mitochondrial membrane permeable to protons. [1]
• Protons leak back into the matrix bypassing ATP synthase. [1]
• The proton gradient is dissipated, removing the "brake" on the ETC. [1]
• The ETC operates at maximum speed to try and restore the gradient, leading to increased $\text{O}_2$ consumption. [1]

Question 3 (a)

• Misfolding exposes hydrophobic amino acid residues. [1]
• These residues are normally buried in the core of the globular protein. [1]
• Exposed hydrophobic regions interact with each other to avoid the aqueous environment of the cytoplasm, leading to aggregation. [1]

(b)

• Aggregates form insoluble plaques/fibrils. [1]
• These can physically disrupt cellular organelles or interfere with intracellular transport. [1]
• This triggers stress responses (e.g., ER stress) or apoptosis, leading to cell death. [1]

(c)

• Chaperones bind to unfolded or misfolded polypeptide chains. [1]
• They provide an isolated environment or use ATP to help the protein fold into its correct native conformation. [1]

Question 4 (a)

• An electric field/potential difference is applied across the gel. [1]
• Proteins migrate through the gel matrix based on their net charge and size/shape. [1]
• The gel acts as a molecular sieve, slowing down larger or less charged molecules. [1]
• Different haemoglobin variants have different charges/sizes and thus migrate different distances. [1]

(b)

• Heterozygous (HbAS). [1]
• The presence of two distinct bands indicates two different types of haemoglobin are present. [1]
• One band corresponds to the normal HbA and the other to the mutant HbS. [1]

(c)

• In HbS, glutamic acid (polar/negatively charged) is replaced by valine (non-polar/neutral) at position 6 of the $\beta$-globin chain. [1]
• This changes the overall net charge of the protein. [1]
• The difference in charge causes HbS to migrate at a different rate/distance in the electric field compared to HbA. [1]

Question 5 (a)

• Both have a phospholipid bilayer. [1]
• Prokaryotes lack membrane-bound organelles; eukaryotes have specialized internal membranes (e.g., nuclear envelope). [1]
• Prokaryotic membranes often perform functions like respiration/photosynthesis (mesosomes/thylakoids) which are delegated to organelles in eukaryotes. [1]
• Eukaryotic membranes contain cholesterol for stability; most prokaryotes do not (except Mycoplasma). [1]
• Both use transport proteins for selective permeability. [1]

(b)

• Facilitated diffusion uses channel or carrier proteins to move molecules across the membrane. [1]
• It moves substances down a concentration gradient (high to low). [1]
• It is a passive process requiring no metabolic energy (ATP). [1]
• Active transport moves substances against a concentration gradient (low to high). [1]
• Active transport requires ATP to power the pump/carrier protein. [1]

Question 6 (a)

• Phospholipid has a hydrophilic phosphate head and two hydrophobic fatty acid tails. [1]
• In water, they spontaneously arrange into a bilayer. [1]
• Hydrophilic heads face the aqueous environment (extracellular/cytoplasmic). [1]
• Hydrophobic tails face inward, away from water, creating a hydrophobic core. [1]

(b)

• "Fluid": Phospholipids and proteins can move laterally within the layer. [1]
• "Mosaic": Proteins are embedded in or attached to the bilayer in a random pattern. [1]
• Cholesterol: Wedges between phospholipids to regulate fluidity (prevents too fluid at high temp, too rigid at low temp). [2]
• Integral proteins: Act as channels/carriers for transport or as receptors for cell signaling. [2]

Question 7 (a)

• As a cell grows, volume increases faster (cubed) than surface area (squared). [1]
• The surface area (plasma membrane) must be large enough to supply nutrients and remove wastes for the entire volume. [1]
• If the ratio is too low, diffusion is too slow to sustain the cell's metabolic needs. [1]

(b)

• Internal membranes increase the total surface area available for reactions. [1]
• Specialized compartments (organelles) concentrate enzymes and substrates. [1]
• This increases efficiency of biochemical processes. [1]
• Allows for the separation of incompatible reactions (e.g., synthesis in RER, degradation in lysosomes). [1]

Question 8 (a)

• Structure: Dense region of RNA, proteins, and chromatin within the nucleus. [1]
• Function: Site of rRNA synthesis. [1]
• Function: Assembly of ribosomal subunits. [1]

(b)

• Nuclear pores are large protein complexes spanning the nuclear envelope. [1]
• They allow free diffusion of small molecules (water, ions). [1]
• They regulate the selective transport of large molecules (proteins, mRNA). [1]
• Use of nuclear localization signals (NLS) and transport proteins (importins/exportins) for gated transport. [1]

Question 9 (a)

• RER: Studded with ribosomes; synthesizes proteins for secretion or membrane insertion. [2]
• SER: Lacks ribosomes; synthesizes lipids (phospholipids, steroids) and detoxifies drugs/toxins. [2]

(b)

• Synthesized by ribosomes on the RER. [1]
• Transported into the RER lumen for folding/initial modification. [1]
• Transported via transport vesicles to the Golgi apparatus. [1]
• Modified (e.g., glycosylation) and sorted in the Golgi cisternae. [1]
• Packaged into secretory vesicles. [1]
• Vesicles fuse with the plasma membrane (exocytosis) to release the protein. [1]

Question 10 (a)

• Single membrane encloses hydrolytic enzymes. [1]
• Maintains an acidic internal pH (approx. pH 5) via proton pumps. [1]
• This acidic environment is optimal for the activity of acid hydrolases. [1]
• Membrane protects the rest of the cell from autodigestion. [1]

(b)

• Process: Cell engulfs its own damaged organelles or cytoplasm into a vesicle (autophagosome). [1]
• The autophagosome fuses with a lysosome for degradation. [1]
• Significance: Recycles nutrients (amino acids/lipids) and removes dysfunctional organelles to prevent cellular damage. [1]