DSE Biology Diagnostic: Cell Biology and Biochemistry

Unit Test 1: Prokaryotic vs Eukaryotic Cell Features

Question

A student examines two cell types under an electron microscope and records the following observations:

• Cell A: No membrane-bound nucleus visible; ribosomes approximately 20 nm in diameter; cell wall present; circular DNA molecule visible in the cytoplasm; mesosomes present; capsule surrounding the cell.
• Cell B: Membrane-bound nucleus with nucleolus; ribosomes approximately 25 nm in diameter; mitochondria present; rough endoplasmic reticulum with ribosomes attached; Golgi apparatus visible.

(a) Identify Cell A and Cell B as prokaryotic or eukaryotic. Justify your answer with three structural differences from the observations. [3 marks]

(b) Cell A is placed in a hypertonic solution and undergoes plasmolysis. Explain why plasmolysis occurs, referring to the water potential of the cell and the external solution. [3 marks]

(c) A student claims that Cell B must be an animal cell because it has no cell wall. Evaluate this claim. [2 marks]

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Worked Solution

(a) Cell A is prokaryotic and Cell B is eukaryotic.

Three structural differences:

1. Cell A has no membrane-bound nucleus; Cell B has a membrane-bound nucleus with a nucleolus.
2. Cell A has 70S ribosomes (approximately 20 nm); Cell B has 80S ribosomes (approximately 25 nm).
3. Cell A has no membrane-bound organelles (no mitochondria, ER, or Golgi); Cell B possesses mitochondria, rough ER, and Golgi apparatus.

(Alternative acceptable difference: Cell A has circular DNA; Cell B has linear chromosomes within the nucleus.)

(b) In a hypertonic solution, the external water potential is lower (more negative) than the water potential inside Cell A. Water moves out of the cell by osmosis down the water potential gradient, through the selectively permeable cell surface membrane. The cytoplasm shrinks and pulls away from the cell wall, causing plasmolysis.

(c) The claim is not necessarily correct. While animal cells lack cell walls, some eukaryotic cells such as protozoa (e.g. Amoeba, Paramecium) also lack cell walls. The absence of a cell wall and presence of organelles (mitochondria, ER, Golgi) could indicate either an animal cell or a protozoan. The observation of these organelles alone confirms only that the cell is eukaryotic, not specifically animal.

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Unit Test 2: Phospholipid Bilayer and Fluid Mosaic Model

Question

(a) Describe the arrangement of phospholipids in the cell surface membrane and explain how this arrangement makes the membrane partially permeable. [4 marks]

(b) Cholesterol molecules are embedded in the phospholipid bilayer of animal cell membranes. Explain two roles of cholesterol in the cell surface membrane. [4 marks]

(c) Glycolipids and glycoproteins are found on the outer surface of the cell surface membrane. State two functions of these molecules. [2 marks]

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Worked Solution

(a) Phospholipids form a bilayer in which the hydrophilic phosphate heads face outwards (towards the aqueous environment on both sides) and the hydrophobic fatty acid tails face inwards, away from water. This arrangement creates a hydrophobic core in the centre of the bilayer.

The membrane is partially permeable because:

• Small non-polar molecules (e.g. $O_{2}$, $CO_{2}$, lipid-soluble steroid hormones) can diffuse directly through the hydrophobic core.
• Large polar molecules (e.g. glucose), ions (e.g. $Na^{+}$, $Cl^{-}$), and charged molecules cannot pass through the hydrophobic region and require transport proteins or vesicles.

(b) Two roles of cholesterol:

1. Cholesterol molecules fit between phospholipids and bind to the hydrophobic tails, restricting their movement. This increases membrane stability and reduces fluidity at high temperatures, preventing the membrane from becoming too permeable.
2. At low temperatures, cholesterol prevents tight packing of phospholipid fatty acid tails, maintaining some membrane fluidity and preventing crystallisation.

(c) Two functions of glycolipids/glycoproteins:

1. Cell recognition and cell signalling: They act as chemical markers (antigens) that allow cells to recognise one another, important in immune response and tissue organisation.
2. Receptor sites for hormones and neurotransmitters: glycoproteins can bind specific signalling molecules, triggering intracellular responses.

(Alternative: forming mucus for protection; attachment sites for pathogens.)

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Unit Test 3: Enzyme Kinetics and Inhibition

Question

An experiment was conducted to investigate the effect of substrate concentration on the rate of an enzyme-catalysed reaction. The results are shown below:

Substrate concentration (mmol/dm$^{3}$)	Rate of reaction (arbitrary units)
2	8
4	15
6	20
8	23
10	24
12	24.5
14	24.8

(a) Explain why the rate of reaction increases at low substrate concentrations but plateaus at high substrate concentrations. [3 marks]

(b) A competitive inhibitor is added to the reaction mixture at a substrate concentration of 4 mmol/dm$^{3}$. State the effect on the reaction rate and explain the mechanism of competitive inhibition. [3 marks]

(c) Explain how a non-competitive inhibitor differs from a competitive inhibitor in terms of (i) where it binds to the enzyme, and (ii) its effect on Vmax and Km. [4 marks]

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Worked Solution

(a) At low substrate concentrations, not all active sites are occupied. Increasing substrate concentration increases the frequency of successful enzyme-substrate collisions, so the rate increases proportionally (first-order kinetics with respect to substrate).

At high substrate concentrations, all active sites are saturated with substrate at any given time. The rate is now limited by the turnover number of the enzyme (the rate at which the enzyme can convert substrate to product and release it). Adding more substrate has no effect because there are no free active sites available. The maximum rate is Vmax.

(b) Adding a competitive inhibitor decreases the reaction rate (from 15 to a lower value at 4 mmol/dm$^{3}$).

Mechanism: The competitive inhibitor has a similar shape to the substrate and binds to the active site of the enzyme, forming an enzyme-inhibitor complex. This blocks the substrate from binding. Since the inhibitor and substrate compete for the same active site, increasing substrate concentration can overcome competitive inhibition.

(c) Differences between non-competitive and competitive inhibition:

(i) Binding site: A non-competitive inhibitor binds to the allosteric site (a site other than the active site), whereas a competitive inhibitor binds to the active site.

(ii) Effect on Vmax and Km:

• Non-competitive inhibition decreases Vmax (because some enzyme molecules are permanently inactivated regardless of substrate concentration) but does not change Km (the remaining active enzymes still have the same affinity for the substrate).
• Competitive inhibition does not change Vmax (at very high substrate concentration, the substrate outcompetes the inhibitor) but increases Km (a higher substrate concentration is needed to reach half Vmax, indicating reduced apparent affinity).

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Integration Test 1: Biological Molecules and Biochemical Tests

Question

A student is given four unknown solutions (W, X, Y, Z) and asked to identify the biological molecules they contain. The student performs the following tests:

Test	Result
Benedict's test on W	Brick-red precipitate after heating
Biuret test on X	Violet/purple colour
Emulsion test on Y	White emulsion forms
Iodine test on Z	Blue-black colour

(a) Identify the biological molecule present in each solution. For W, explain the chemical basis of the colour change. [4 marks]

(b) The student then hydrolyses solution W using dilute hydrochloric acid and neutralises the mixture. A second Benedict's test on the hydrolysed product gives a much more intense brick-red precipitate than the first test. Explain these observations. [4 marks]

(c) Solution X contains a polypeptide. Describe the structure of a polypeptide, naming the type of bond that joins amino acids and the type of bond responsible for the secondary structure. [4 marks]

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Worked Solution

(a) - W: Reducing sugar (e.g. glucose). The Benedict's reagent contains copper(II) sulphate. Reducing sugars reduce blue $Cu^{2+}$ ions to red insoluble copper(I) oxide ($Cu_{2}O$), forming a brick-red precipitate.

• X: Protein (or polypeptide). The biuret reagent (containing copper(II) sulphate in alkaline solution) reacts with peptide bonds ($-CO-NH-$), producing a violet/purple colour.
• Y: Lipid. When ethanol is added and the mixture is poured into water, a cloudy white emulsion forms, indicating the presence of lipids.
• Z: Starch. Iodine solution turns blue-black in the presence of starch due to iodine molecules fitting into the helical structure of amylose.

(b) W gives a brick-red precipitate in the first test, confirming it contains a reducing sugar (e.g. glucose or maltose). After acid hydrolysis and neutralisation, the more intense brick-red precipitate indicates that W also contained a polysaccharide (e.g. starch). Hydrolysis breaks the polysaccharide into many more reducing sugar units (glucose), producing more reducing ends and hence a more intense Benedict's result. For example, W might contain a mixture of maltose (a reducing disaccharide) and starch.

(c) A polypeptide is a chain of amino acids joined by peptide bonds (formed by condensation reactions between the carboxyl group of one amino acid and the amino group of the next, with the elimination of water).

The secondary structure (e.g. alpha-helix, beta-pleated sheet) is stabilised by hydrogen bonds between the $-C=O$ group of one amino acid and the $-NH$ group of another amino acid further along the chain.

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Integration Test 2: Cell Membrane Transport and Enzyme Activity

Question

Glucose enters intestinal epithelial cells from the lumen via secondary active transport. The process involves the sodium-potassium pump ($Na^{+}/K^{+}$ pump) on the basolateral membrane and a sodium-glucose cotransporter (SGLT1) on the apical membrane.

(a) Describe how the $Na^{+}/K^{+}$ pump establishes and maintains the sodium concentration gradient across the cell membrane. Include the role of ATP in your answer. [4 marks]

(b) Explain how this sodium gradient enables glucose to be absorbed against its concentration gradient via SGLT1. [3 marks]

(c) A toxin inhibits ATP production in these epithelial cells. Predict and explain the effect on glucose absorption. [3 marks]

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Worked Solution

(a) The $Na^{+}/K^{+}$ pump is a transmembrane carrier protein that actively transports $3Na^{+}$ ions out of the cell and $2K^{+}$ ions into the cell per ATP molecule hydrolysed.

ATP is hydrolysed to ADP and inorganic phosphate ($P_{i}$), and the phosphate group attaches to the pump protein, causing a conformational change. This change:

1. Reduces the pump's affinity for $Na^{+}$, releasing $3Na^{+}$ outside the cell.
2. Increases the pump's affinity for $K^{+}$, allowing $2K^{+}$ to bind from outside.
3. Dephosphorylation returns the pump to its original shape, releasing $K^{+}$ inside the cell.

This maintains a high $Na^{+}$ concentration outside the cell and a low $Na^{+}$ concentration inside, creating a steep sodium concentration gradient.

(b) SGLT1 is a cotransporter (symporter) on the apical membrane that binds both $Na^{+}$ and glucose. $Na^{+}$ moves down its concentration gradient (from high concentration in the lumen to low concentration in the cell) through SGLT1. The energy released from $Na^{+}$ moving down its gradient is used to transport glucose against its concentration gradient into the cell. This is secondary active transport because the energy source (the $Na^{+}$ gradient) was established by primary active transport (the $Na^{+}/K^{+}$ pump using ATP).

(c) Glucose absorption would stop or decrease significantly.

The toxin inhibits ATP production, so the $Na^{+}/K^{+}$ pump cannot function (it requires ATP hydrolysis). Without the pump, the $Na^{+}$ concentration gradient across the apical membrane will dissipate as $Na^{+}$ diffuses back into the cell. Without the $Na^{+}$ gradient, SGLT1 cannot cotransport glucose against its concentration gradient. Glucose absorption ceases or is greatly reduced.

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Integration Test 3: Enzyme Denaturation and Industrial Application

Question

An industrial process uses the enzyme catalase to break down hydrogen peroxide ($H_{2}O_{2}$) into water and oxygen. The reaction is carried out at pH 7 and $37^{\circ}C$.

(a) Explain why catalase has an optimum temperature of $37^{\circ}C$ in terms of molecular kinetics and enzyme structure. [3 marks]

(b) The temperature is raised to $80^{\circ}C$ and the rate of reaction drops to near zero. A student claims the enzyme has been "killed." Evaluate this claim using the terms denaturation and coagulation. [3 marks]

(c) To improve the efficiency of the process, immobilised catalase beads are used instead of free catalase in solution. Explain two advantages of using immobilised enzymes in industrial processes. [4 marks]

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Worked Solution

(a) At $37^{\circ}C$ (below the optimum), increasing temperature increases the kinetic energy of both enzyme and substrate molecules, leading to more frequent and more energetic collisions between them, increasing the rate of successful enzyme-substrate complex formation.

However, as temperature approaches $37^{\circ}C$ and beyond, the vibrational energy of the enzyme molecule increases to the point where it begins to disrupt the weak bonds (hydrogen bonds, ionic bonds, hydrophobic interactions) maintaining the enzyme's tertiary structure. At the optimum temperature, the rate of productive collisions is maximised before denaturation becomes significant.

(b) The claim is imprecise. Enzymes are not living organisms and cannot be "killed." The correct term is denaturation: at $80^{\circ}C$, the excessive thermal energy breaks the hydrogen bonds, ionic bonds, and other weak interactions maintaining the enzyme's active site shape. The tertiary structure unfolds irreversibly, and the active site can no longer bind the substrate.

Coagulation is a related but distinct process: it refers specifically to the aggregation of denatured protein molecules, often resulting in a visible solid or semi-solid mass (e.g. cooking egg white). Coagulation is a consequence of denaturation, not the same process.

(c) Two advantages of immobilised enzymes:

1. Reusability: Immobilised enzymes can be easily separated from the reaction mixture (e.g. by filtration) and reused multiple times, reducing costs.
2. Stability: Immobilisation can protect the enzyme from denaturation by restricting molecular movement, allowing the enzyme to function at higher temperatures and a wider pH range compared to free enzymes.

(Alternative: ease of product purification; continuous flow processing; prevention of enzyme contamination of the product.)