Cell Biology — Diagnostic Tests

Unit Tests

UT-1: Cell Structure and Ultrastructure

Question:

(a) Describe the structure and function of the following organelles: mitochondria, rough endoplasmic reticulum, Golgi apparatus, and lysosomes.

(b) Explain the difference between a prokaryotic cell and a eukaryotic cell, giving at least four structural differences.

(c) Describe the fluid mosaic model of cell membrane structure. Name the two main components of the membrane and explain how the term “fluid mosaic” reflects its structure.

(d) A cell has a high demand for protein synthesis. Explain how the endomembrane system (rough ER, Golgi apparatus, and vesicles) enables the production and secretion of proteins.

Solution:

(a)

• Mitochondria: Double membrane organelle; the inner membrane is folded into cristae to increase surface area for the reactions of aerobic respiration. The matrix contains enzymes for the Krebs cycle and its own DNA and ribosomes. Function: site of ATP production through aerobic respiration.
• Rough endoplasmic reticulum (RER): Network of membrane-bound flattened sacs (cisternae) studded with ribosomes on the outer surface. Function: synthesis and folding of proteins, particularly those destined for secretion or insertion into membranes.
• Golgi apparatus: Stack of flattened membrane-bound sacs. Receives proteins from the RER in transport vesicles, modifies them (e.g., adding carbohydrate groups to form glycoproteins), sorts them, and packages them into secretory vesicles for export from the cell.
• Lysosomes: Small membrane-bound vesicles containing hydrolytic (digestive) enzymes. Function: breakdown of waste materials, cellular debris, and engulfed pathogens (in phagocytosis). The membrane prevents the enzymes from damaging the rest of the cell.

(b) Four structural differences between prokaryotic and eukaryotic cells:

1. Nucleus: Eukaryotic cells have a membrane-bound nucleus containing DNA; prokaryotic cells have no nucleus — their DNA is free in the cytoplasm in a region called the nucleoid.
2. Mitochondria: Eukaryotic cells contain mitochondria; prokaryotic cells do not.
3. Ribosomes: Eukaryotic ribosomes are larger (80S); prokaryotic ribosomes are smaller (70S).
4. DNA structure: Eukaryotic DNA is linear and associated with histone proteins as chromosomes; prokaryotic DNA is circular and not associated with histones.

Other valid differences: cell size (eukaryotes are generally larger), presence of membrane-bound organelles (only in eukaryotes), cell wall composition (peptidoglycan in prokaryotes vs. cellulose in plant cells or chitin in fungi).

(c) The fluid mosaic model describes the cell membrane as a bilayer of phospholipids in which proteins are embedded. The phospholipids form a double layer with their hydrophilic (water-attracting) phosphate heads facing outward and their hydrophobic (water-repelling) lipid tails facing inward. Proteins are embedded within or attached to this bilayer and perform functions such as transport, enzyme activity, and cell signalling. The term “fluid” refers to the fact that the phospholipid molecules can move laterally within the bilayer, giving the membrane flexibility. “Mosaic” refers to the scattered, varied arrangement of different proteins within the phospholipid sea, like tiles in a mosaic.

(d) Proteins destined for secretion are synthesised on ribosomes attached to the rough ER. The newly formed polypeptide chain is fed into the RER lumen, where it folds and may be modified. Transport vesicles bud from the RER and carry the proteins to the Golgi apparatus. The Golgi further modifies the proteins (e.g., glycosylation), sorts them, and packages them into secretory vesicles. These vesicles move to the cell surface membrane, fuse with it, and release the proteins outside the cell by exocytosis. This coordinated system ensures proteins are properly folded, modified, and delivered to the correct destination.

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UT-2: Membrane Transport

Question:

(a) Explain the difference between passive transport and active transport across cell membranes. Give one example of each.

(b) Describe the process of osmosis. Explain what would happen to an animal cell placed in a hypotonic solution and why.

(c) Facilitated diffusion requires transport proteins. Explain why large or charged molecules cannot simply diffuse through the phospholipid bilayer.

(d) Explain the role of ATP in active transport and sodium-potassium pumps in maintaining the resting potential of a neuron.

Solution:

(a) Passive transport is the movement of molecules across a membrane from a region of higher concentration to a region of lower concentration (down the concentration gradient), without requiring energy (ATP). Examples include simple diffusion (e.g., oxygen entering cells) and osmosis (water movement). Active transport is the movement of molecules against the concentration gradient (from lower to higher concentration), which requires energy in the form of ATP. An example is the sodium-potassium pump, which moves $\text{Na}^+$ out of the cell and $\text{K}^+$ into the cell against their concentration gradients.

(b) Osmosis is the net movement of water molecules across a selectively permeable membrane from a region of higher water potential (lower solute concentration) to a region of lower water potential (higher solute concentration). A hypotonic solution has a higher water potential than the cell cytoplasm. When an animal cell is placed in a hypotonic solution, water enters the cell by osmosis, causing it to swell. Because animal cells lack a rigid cell wall, the cell may burst (lyse) if too much water enters, as the membrane cannot withstand the expanding volume.

(c) The phospholipid bilayer is hydrophobic in its interior (the lipid tails face inward). Large molecules cannot pass through because they are too big to fit between the phospholipid molecules. Charged molecules (ions) and polar molecules cannot pass through because they are hydrophilic and are repelled by the hydrophobic core of the bilayer. Transport proteins (channel proteins and carrier proteins) provide a hydrophilic pathway through the membrane, allowing these molecules to cross.

(d) ATP provides the energy required for the sodium-potassium pump to move ions against their concentration gradients. The pump uses the energy from hydrolysing one ATP molecule to transport three $\text{Na}^+$ ions out of the cell and two $\text{K}^+$ ions into the cell. This active transport maintains the concentration gradients of sodium and potassium across the neuron membrane. The result is a higher concentration of $\text{Na}^+$ outside and $\text{K}^+$ inside, creating an electrochemical gradient (the resting potential of approximately $-70\,\text{mV}$). This gradient is essential for generating action potentials when the neuron is stimulated.

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UT-3: Enzymes

Question:

(a) Define the term “enzyme” and explain the “lock and key” model of enzyme action.

(b) Describe how temperature affects enzyme activity, referring to the concepts of kinetic energy, denaturation, and the optimum temperature.

(c) A student investigates the effect of pH on the enzyme catalase (which breaks down hydrogen peroxide). The results show maximum activity at pH 7. Sketch a graph of rate of reaction against pH and explain the shape.

(d) Explain the difference between competitive and non-competitive inhibition, giving an example of each.

Solution:

(a) An enzyme is a biological catalyst (typically a protein) that speeds up the rate of a biochemical reaction by lowering the activation energy, without being consumed in the reaction. The lock and key model proposes that the enzyme has a specific active site with a shape complementary to its substrate, like a key fitting into a lock. The substrate binds to the active site, forming an enzyme-substrate complex, and the products are released, leaving the enzyme unchanged and free to catalyse further reactions.

(b) As temperature increases from low levels, enzyme activity increases because molecules have more kinetic energy, leading to more frequent successful collisions between enzyme and substrate. The rate continues to rise until the optimum temperature (approximately $37\,^\circ\text{C}$ for human enzymes), at which the rate is highest. Above the optimum temperature, the enzyme begins to denature: the bonds holding the tertiary structure together (hydrogen bonds, ionic bonds) break, causing the active site to change shape. The substrate can no longer fit, and the rate of reaction drops sharply. At very high temperatures, the enzyme is permanently denatured and loses all catalytic function.

(c) The graph of rate of reaction against pH is a bell-shaped curve:

• Rate increases from low pH to a peak at pH 7 (the optimum pH for catalase).
• Rate decreases symmetrically on either side of the optimum.

This shape occurs because pH affects the ionisation of amino acid side chains in the enzyme’s active site. At the optimum pH, the ionisation of these residues gives the active site its correct shape for substrate binding. At pH values above or below the optimum, changes in ionisation alter the charge distribution in the active site, disrupting the tertiary structure and reducing the enzyme’s ability to bind the substrate. Extreme pH values can cause permanent denaturation.

(d) Competitive inhibition: the inhibitor molecule has a similar structure to the substrate and competes with the substrate for binding to the active site. It blocks the active site, preventing the substrate from binding. The effect can be overcome by increasing substrate concentration. Example: malonate competitively inhibits succinate dehydrogenase in the Krebs cycle.

Non-competitive inhibition: the inhibitor binds to the enzyme at a site other than the active site (the allosteric site), causing a conformational change in the enzyme’s tertiary structure that alters the shape of the active site. The substrate can no longer bind effectively. Increasing substrate concentration does not overcome non-competitive inhibition. Example: heavy metal ions such as lead ($\text{Pb}^{2+}$) can bind to enzymes and inhibit them non-competitively.

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Integration Tests

IT-1: Cell Biology in Context

Question:

(a) Insulin is a protein hormone produced by the beta cells of the pancreas and secreted into the bloodstream. Describe the pathway of insulin synthesis and secretion, referring to the relevant organelles and processes.

(b) Type 1 diabetes results from the destruction of beta cells in the pancreas. Explain how the absence of insulin affects glucose uptake by cells, referring to membrane transport mechanisms.

(c) A researcher wants to isolate mitochondria from liver cells. Explain which cell fractionation technique would be used, naming the organelles that would be separated first at each stage (based on size and density).

(d) Evaluate the importance of the fluid mosaic model in understanding how substances enter and leave cells. Use specific examples of transport mechanisms in your answer.

Solution:

(a) Insulin synthesis and secretion:

1. Insulin mRNA is translated on ribosomes attached to the rough ER, where the polypeptide chain is synthesised and enters the RER lumen.
2. In the RER lumen, the polypeptide folds and disulphide bonds form to create the correct tertiary structure.
3. Transport vesicles bud from the RER and carry insulin to the Golgi apparatus.
4. The Golgi modifies insulin (removing the signal peptide and packaging it into secretory vesicles).
5. Secretory vesicles carry insulin to the cell surface membrane.
6. Vesicles fuse with the membrane and release insulin into the bloodstream by exocytosis.

(b) Insulin normally stimulates the insertion of glucose transporter proteins (GLUT4) into the cell surface membrane, allowing glucose to enter cells by facilitated diffusion down its concentration gradient. In Type 1 diabetes, the absence of insulin means GLUT4 transporters remain in vesicles inside the cell and are not inserted into the membrane. As a result, glucose cannot enter cells efficiently by facilitated diffusion, leading to high blood glucose concentration (hyperglycaemia). Cells are deprived of glucose for respiration, and the body attempts to excrete excess glucose in urine.

(c) Cell fractionation involves homogenising the cells (breaking them open) in a cold, isotonic buffer to preserve organelle structure, then separating organelles by differential centrifugation:

1. Low-speed centrifugation (e.g., $1000\,g$ for 10 minutes): pellets the nucleus (largest organelle).
2. Medium-speed centrifugation (e.g., $10000\,g$ for 20 minutes): pellets mitochondria and lysosomes.
3. High-speed centrifugation (e.g., $100000\,g$ for 60 minutes): pellets ribosomes and fragments of the ER.
4. The supernatant after the final spin contains soluble cytoplasmic proteins and small molecules.

(d) The fluid mosaic model is essential for understanding membrane transport because it explains that the membrane is a dynamic, selectively permeable barrier rather than a rigid wall. The phospholipid bilayer allows small, non-polar molecules (e.g., $\text{O}_2$, $\text{CO}_2$) to diffuse through directly, while the embedded proteins provide pathways for ions and larger molecules. For example, channel proteins allow rapid passage of ions during nerve impulse transmission, carrier proteins enable facilitated diffusion of glucose, and the sodium-potassium pump actively transports ions using ATP. The fluidity of the membrane allows these proteins to move and assemble as needed. Without understanding the fluid mosaic model, it would be impossible to explain how the same membrane can simultaneously be a barrier, a gateway, and a signalling platform.

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IT-2: Experimental Design in Cell Biology

Question:

(a) Design an experiment to investigate the effect of temperature on the rate of catalase activity. Include a hypothesis, variables (independent, dependent, controlled), method, and how results would be analysed.

(b) In the experiment described in part (a), a student measures the volume of oxygen gas produced in 60 seconds. At $20\,^\circ\text{C}$, the volume is $12\,\text{cm}^3$. At $40\,^\circ\text{C}$, the volume is $28\,\text{cm}^3$. At $60\,^\circ\text{C}$, the volume is $5\,\text{cm}^3$. Explain these results with reference to enzyme structure and function.

(c) A second student repeats the experiment but uses boiled catalase. Predict the result and explain why.

(d) Evaluate the reliability and validity of this experimental method. Suggest one improvement.

Solution:

(a) Hypothesis: As temperature increases towards the optimum, the rate of catalase activity will increase. Above the optimum, the rate will decrease due to denaturation.

Variables:

• Independent: temperature of the reaction mixture ($10, 20, 30, 40, 50, 60\,^\circ\text{C}$).
• Dependent: volume of oxygen gas produced in a fixed time period (measured using a gas syringe).
• Controlled: concentration and volume of hydrogen peroxide solution; concentration and volume of catalase extract; pH of the solution; time allowed for the reaction.

Method:

1. Prepare a water bath at the target temperature.
2. Add a fixed volume of catalase extract to a test tube and place it in the water bath for 5 minutes to equilibrate.
3. Add a fixed volume of hydrogen peroxide to a second test tube and equilibrate in the same bath.
4. Connect a gas syringe to the test tube containing catalase.
5. Pour the hydrogen peroxide into the catalase and start a timer.
6. Record the volume of oxygen collected in 60 seconds.
7. Repeat at each temperature, performing three trials per temperature for reliability.

Analysis: Plot a graph of volume of oxygen (y-axis) against temperature (x-axis). The graph should show a peak at the optimum temperature.

(b) At $20\,^\circ\text{C}$, the moderate rate ($12\,\text{cm}^3$) indicates that molecules have sufficient kinetic energy for reactions but are below the optimum. At $40\,^\circ\text{C}$, the maximum rate ($28\,\text{cm}^3$) indicates this is close to the optimum temperature — molecules have high kinetic energy and the enzyme’s active site is fully functional. At $60\,^\circ\text{C}$, the sharp drop to $5\,\text{cm}^3$ indicates that the enzyme has been significantly denatured — the high temperature has broken bonds maintaining the tertiary structure, altering the active site so substrate binding is greatly reduced.

(c) Boiled catalase would produce approximately $0\,\text{cm}^3$ of oxygen. Boiling denatures the catalase completely, permanently destroying the active site’s shape so that hydrogen peroxide can no longer bind and be broken down. This serves as a useful control to confirm that the observed gas production is enzymatic and not due to spontaneous decomposition of hydrogen peroxide.

(d) Reliability: Performing three trials at each temperature and calculating a mean improves reliability by reducing the effect of anomalies. Validity: The experiment measures the dependent variable (oxygen volume) in a way that directly reflects enzyme activity, making it a valid measure. However, some hydrogen peroxide decomposes spontaneously without catalase, which could introduce a small systematic error. Improvement: include a control tube without catalase at each temperature to measure the background rate of spontaneous decomposition, and subtract this from the experimental values.

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Common Pitfalls

• Confusing osmosis with diffusion: osmosis refers specifically to the movement of water, not solutes.
• Describing active transport without mentioning ATP or the requirement to move against the concentration gradient.
• Stating that enzymes are “used up” in a reaction — enzymes are biological catalysts and are not consumed.
• Confusing competitive and non-competitive inhibition: competitive inhibitors compete for the active site; non-competitive inhibitors bind to the allosteric site.
• Describing the fluid mosaic model without explaining both terms: “fluid” (lateral movement of phospholipids) and “mosaic” (scattered arrangement of proteins).