Cells — Diagnostic Tests

Unit Tests

UT-1: Prokaryotic vs Eukaryotic Cell Comparison

Question:

A student examines two cell samples under a microscope. Cell A is approximately 5 micrometres in diameter, lacks a nucleus, and has a cell wall. Cell B is approximately 20 micrometres in diameter, contains a nucleus, and has no cell wall.

(a) Compare and contrast the structure of Cell A and Cell B, making reference to at least five distinct structural features.

(b) Cell A contains small (70S) ribosomes, whereas Cell B contains larger (80S) ribosomes. Explain the significance of this difference for antibiotic therapy.

(c) Cell A can divide every 20 minutes under optimal conditions. Explain the process by which Cell A divides, naming the stage at which DNA replication occurs.

(d) Some prokaryotic cells contain plasmids. Describe the structure of a plasmid and explain its significance in genetic engineering.

Solution:

(a) Five structural comparisons:

Feature	Cell A (Prokaryotic)	Cell B (Eukaryotic)
Nucleus	No true nucleus; DNA free in cytoplasm (nucleoid)	True nucleus bounded by a nuclear envelope with nuclear pores
DNA form	Circular, naked DNA (no histones); may also have plasmids	Linear DNA associated with histone proteins to form chromosomes
Ribosomes	70S (smaller: 50S + 30S subunits)	80S (larger: 60S + 40S subunits)
Membrane-bound organelles	None (no mitochondria, ER, Golgi)	Present (mitochondria, ER, Golgi, lysosomes, etc.)
Cell wall	Present (made of peptidoglycan, not cellulose or chitin)	Absent (animal cell) or present (plant cell — cellulose, fungal — chitin)

(b) Many antibiotics (e.g., tetracycline, chloramphenicol) target bacterial 70S ribosomes, binding to them and inhibiting protein synthesis. Because eukaryotic cells have 80S ribosomes with a different structure, these antibiotics do not bind to human ribosomes at therapeutic concentrations, so human protein synthesis is unaffected. This selective toxicity is possible precisely because of the structural difference between 70S and 80S ribosomes. (Note: some antibiotics, e.g., streptomycin, target the 30S subunit specifically.)

(c) Cell A divides by binary fission. The process is: (1) the circular DNA replicates (DNA replication occurs in this stage), producing two identical copies; (2) the cell enlarges and the two DNA molecules move to opposite poles of the cell; (3) the cell membrane begins to grow inwards at the midpoint (septum formation); (4) the septum is completed, dividing the cytoplasm into two daughter cells, each with a copy of the DNA. Plasmids may also replicate independently.

(d) A plasmid is a small, circular, double-stranded DNA molecule found in the cytoplasm of many prokaryotic cells, separate from the main circular chromosome. Plasmids typically carry genes that confer advantageous traits (e.g., antibiotic resistance). In genetic engineering, plasmids are used as vectors — a gene of interest is inserted into the plasmid using restriction enzymes (which cut at specific recognition sequences, producing sticky ends) and DNA ligase (which joins the gene to the plasmid). The recombinant plasmid is then transferred into a host cell (e.g., using bacterial transformation), where the gene is expressed. Plasmids are ideal vectors because they replicate independently, can carry foreign DNA, and often contain antibiotic resistance genes that serve as selectable markers.

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UT-2: Organelle Functions and the Fluid Mosaic Model

Question:

The transmission electron microscope (TEM) reveals the ultrastructure of cells at magnifications far exceeding those of light microscopes.

(a) Describe the structure and function of the rough endoplasmic reticulum and explain its relationship with the Golgi apparatus in the production and secretion of proteins.

(b) Mitochondria are described as the "powerhouse of the cell." Describe the adaptations of mitochondrial structure that make them efficient at producing ATP.

(c) Describe the fluid mosaic model of cell membrane structure, naming all the molecules present and explaining the roles of cholesterol and glycoproteins.

(d) Explain how the phospholipid bilayer provides a barrier to the movement of water-soluble and lipid-soluble substances.

Solution:

(a) The rough endoplasmic reticulum (RER) is a network of flattened membrane-bound sacs (cisternae) studded with ribosomes on the outer surface. Ribosomes on the RER synthesise proteins (particularly those destined for secretion or for insertion into membranes). The newly synthesised polypeptide chain is fed into the cisternal space of the RER, where it folds and may undergo post-translational modification (e.g., glycosylation). Transport vesicles bud off from the RER and carry the proteins to the Golgi apparatus. The Golgi apparatus is a stack of membrane-bound, flattened sacs. It receives transport vesicles from the RER at its cis face, modifies the proteins further (e.g., adding carbohydrate groups, phosphorylation, folding), sorts them, and packages them into secretory vesicles that bud off from the trans face. These secretory vesicles move to the cell surface membrane, fuse with it (exocytosis), and release the proteins outside the cell.

(b) Mitochondrial adaptations for ATP production:

• Double membrane: the outer membrane is permeable; the inner membrane is folded into cristae, greatly increasing the surface area for the electron transport chain and ATP synthase.
• Matrix: contains enzymes for the Krebs cycle and the link reaction, as well as mitochondrial DNA (circular, like prokaryotic DNA), 70S ribosomes, and its own pool of metabolites — allowing some degree of independent protein synthesis.
• Cristae: the folds of the inner membrane increase the surface area available for the electron carriers and ATP synthase, maximising the rate of oxidative phosphorylation.
• Small intermembrane space: allows the accumulation of H$^+$ ions to create the proton gradient (proton motive force) needed for chemiosmosis.

(c) The fluid mosaic model describes the cell membrane as a fluid bilayer of phospholipids with proteins embedded in it (forming a mosaic pattern). Components:

• Phospholipids: form the bilayer; hydrophilic heads face outwards (towards aqueous environments), hydrophobic tails face inwards. They can move laterally within their own monolayer (fluidity).
• Intrinsic (integral) proteins: span the entire bilayer; function as channels, carriers, receptors, or enzymes.
• Extrinsic (peripheral) proteins: attached to the surface; involved in cell signalling and maintaining the cytoskeleton.
• Cholesterol: a lipid molecule present between phospholipids in animal cell membranes. Its hydroxyl group interacts with phospholipid heads, while its ring structure interacts with fatty acid tails. Cholesterol regulates membrane fluidity — at high temperatures it restricts movement of phospholipids (making the membrane less fluid), and at low temperatures it prevents tight packing (preventing freezing).
• Glycoproteins: proteins with carbohydrate chains attached, projecting from the cell surface. They act as cell surface markers for cell recognition (e.g., ABO blood group antigens, MHC proteins for immune recognition), and as receptor molecules for signalling molecules such as hormones.

(d) The phospholipid bilayer is a barrier because the centre of the bilayer is hydrophobic (composed of fatty acid tails), whereas the surfaces are hydrophilic (phosphate heads). Lipid-soluble (non-polar) substances (e.g., oxygen, carbon dioxide, steroid hormones) can dissolve in the hydrophobic core and diffuse directly across the membrane by simple diffusion. Water-soluble (polar and ionic) substances (e.g., glucose, amino acids, sodium ions, chloride ions) cannot pass through the hydrophobic centre because they are not soluble in lipid; they repel the hydrophobic environment. These substances require the assistance of transport proteins (channel proteins or carrier proteins) to cross the membrane via facilitated diffusion or active transport.

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UT-3: Microscopy — Resolution, Magnification, and Cell Fractionation

Question:

A biologist wants to observe the detailed internal structure of a mitochondrion and determine whether a cell is prokaryotic or eukaryotic based on the presence of membrane-bound organelles.

(a) Define magnification and resolution, and explain why resolution is more important than magnification when choosing a microscope for viewing cell ultrastructure.

(b) Explain the advantages and limitations of using a transmission electron microscope (TEM) compared with a light microscope for viewing cell ultrastructure. Give two limitations of TEM.

(c) The biologist uses a TEM image where the scale bar is 50 nm long and measures 20 mm on the printed image. Calculate the magnification of the image.

(d) Describe the process of cell fractionation and ultracentrifugation, explaining how different organelles are separated based on their density.

Solution:

(a) Magnification is the number of times larger the image appears compared to the actual object: $\text{magnification} = \frac◆LB◆\text{image size}◆RB◆◆LB◆\text{actual size}◆RB◆$. Resolution (or resolving power) is the minimum distance between two points that can be distinguished as separate objects. Resolution is more important than magnification because increasing magnification without improving resolution simply produces a larger, blurry image — you cannot see more detail. The resolution of a light microscope is limited to approximately 200 nm (due to the wavelength of visible light), whereas electron microscopes have a resolution of approximately 0.2 nm (due to the much shorter wavelength of electrons). This means electron microscopes can reveal sub-cellular structures (organelles, ribosomes, membranes) that are invisible under a light microscope.

(b) Advantages of TEM over light microscopy:

• Much higher resolution (approximately 0.2 nm vs 200 nm), revealing fine ultrastructural detail such as mitochondrial cristae, ribosomes, and nuclear pores.
• Much higher useful magnification (up to approximately $\times 500,000$ vs approximately $\times 1,500$).

Limitations of TEM:

1. The specimen must be placed in a vacuum, so only dead, fixed, dehydrated specimens can be viewed — no living cells can be observed.
2. The preparation process is complex and may introduce artefacts (structural features that are not present in the living cell, caused by chemical fixation and staining).
3. The image is 2D (a thin section), so 3D reconstruction requires multiple sections.
4. The image appears in black and white (density-dependent contrast from heavy metal staining), not in natural colour.

(c) Magnification $= \frac◆LB◆\text{image size}◆RB◆◆LB◆\text{actual size}◆RB◆$.

First, convert units consistently:

• Image size of scale bar $= 20\text{ mm} = 20 \times 10^6\text{ nm}$
• Actual size of scale bar $= 50\text{ nm}$

$\text{Magnification} = \frac◆LB◆20 \times 10^6◆RB◆◆LB◆50◆RB◆ = \frac{20,000,000}{50} = 400,000$

The magnification is $\times 400,000$.

(d) Cell fractionation is the process of breaking open cells and separating their organelles. The steps are:

1. Homogenisation: the tissue is placed in a cold, isotonic, buffered solution and homogenised (blended) to break the cell membranes, releasing the organelles. The solution is cold to reduce enzyme activity that could damage organelles; isotonic to prevent osmotic lysis or shrinkage of organelles; and buffered to maintain a constant pH.
2. Filtration: the homogenate is filtered through a gauze to remove debris (unbroken cells and connective tissue).
3. Ultracentrifugation: the filtrate is placed in centrifuge tubes and spun at increasing speeds. At each speed, the heaviest (densest) organelles sediment first to form a pellet at the bottom of the tube, while lighter organelles remain in the supernatant (which is poured off and re-centrifuged). Typical order of sedimentation (low to high speed): nuclei (heaviest, pellet first), mitochondria and chloroplasts, lysosomes, endoplasmic reticulum, ribosomes (lightest, require highest speed).

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Integration Tests

IT-1: Prokaryotic Cell Structure and Antibiotic Resistance (with Genetics and DNA)

Question:

Bacterial resistance to the antibiotic ampicillin is commonly conferred by the bla gene, which encodes the enzyme beta-lactamase. This gene is often carried on a plasmid.

(a) Describe how a bacterium can acquire a plasmid carrying the bla gene from another bacterium through the process of conjugation.

(b) Beta-lactamase catalyses the hydrolysis of the beta-lactam ring of ampicillin, rendering the antibiotic inactive. Explain why the production of beta-lactamase does not harm the bacterial cell itself, despite ampicillin's target being the bacterial cell wall.

(c) Explain how the widespread use of ampicillin in medicine and agriculture has led to the increase in ampicillin-resistant bacteria, with reference to natural selection.

(d) A scientist extracts the bla gene from a resistant bacterium and inserts it into a plasmid vector. Describe the role of restriction enzymes and DNA ligase in this process, and explain why the plasmid must also contain a promoter sequence.

Solution:

(a) During conjugation, two bacterial cells form a physical connection via a sex pilus (a thin protein tube). The donor cell (containing the plasmid with the bla gene) and the recipient cell are drawn together. A copy of the plasmid DNA is made (by rolling circle replication) and transferred through the pilus to the recipient cell. The recipient cell now contains the plasmid and expresses the bla gene, gaining ampicillin resistance. Both cells now carry copies of the plasmid.

(b) Ampicillin targets bacterial cell wall synthesis by inhibiting the transpeptidase enzymes (penicillin-binding proteins) that cross-link peptidoglycan chains. Beta-lactamase is secreted into the periplasmic space (between the cell membrane and the cell wall) or the surrounding medium, where it hydrolyses the beta-lactam ring of ampicillin before the antibiotic can reach its target. Because the enzyme acts on the antibiotic molecule itself (destroying it), and because the bacterial cell wall has already been synthesised (beta-lactamase is produced by an existing, intact cell), the enzyme does not interfere with normal cell wall synthesis. The enzyme is specific to the beta-lactam ring and does not hydrolyse peptidoglycan or other cell components.

(c) When ampicillin is used, susceptible bacteria (lacking the bla gene) are killed. Any bacteria that happen to carry a plasmid with the bla gene survive and reproduce. These resistant bacteria pass the bla gene to their offspring (vertical gene transfer) and to other bacteria via conjugation (horizontal gene transfer). Over time, the proportion of resistant bacteria in the population increases because: (1) resistant bacteria have a selective advantage in the presence of the antibiotic; (2) horizontal gene transfer spreads the resistance gene rapidly through the population. The continued use of ampicillin maintains the selective pressure, favouring resistant strains. This is a classic example of natural selection acting on genetic variation in a population.

(d) Restriction enzymes (restriction endonucleases) cut DNA at specific recognition sequences (palindromic sequences). The scientist uses a restriction enzyme to cut both the plasmid vector and the DNA containing the bla gene at complementary sites, producing sticky ends (single-stranded overhangs with complementary base sequences). The bla gene and the cut plasmid are mixed; the sticky ends of the bla gene anneal (bind by complementary base pairing) with the sticky ends of the plasmid. DNA ligase then catalyses the formation of phosphodiester bonds between adjacent nucleotides, sealing the bla gene into the plasmid to form a recombinant plasmid. A promoter sequence must also be present on the plasmid upstream of the bla gene because RNA polymerase binds to the promoter to initiate transcription. Without a promoter, the bla gene cannot be transcribed and the host bacterium will not produce beta-lactamase, even though it carries the gene.

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IT-2: Cell Fractionation and Mitochondrial Respiration (with Biological Molecules)

Question:

A student isolates mitochondria from liver tissue using cell fractionation and ultracentrifugation. The isolated mitochondria are placed in a solution containing pyruvate, ADP, and inorganic phosphate. Oxygen consumption is measured.

(a) Describe why the isolation buffer must be ice-cold, isotonic, and buffered.

(b) Explain the role of the electron transport chain and ATP synthase in the production of ATP during oxidative phosphorylation in the mitochondria. Include the term chemiosmosis.

(c) The student adds a substance that makes the inner mitochondrial membrane permeable to H$^+$ ions. Predict and explain the effect on ATP production and oxygen consumption.

(d) Pyruvate is the end product of glycolysis. Describe where glycolysis occurs, explain why it does not require oxygen, and state the net products of glycolysis per molecule of glucose.

Solution:

(a) The buffer must be ice-cold to reduce the activity of digestive enzymes (lysosomal enzymes and proteases) that could damage or degrade the mitochondria once cells are broken open. It must be isotonic (same water potential as the cytoplasm of the cells) to prevent osmotic gain or loss of water by the organelles — if the solution were hypotonic, mitochondria would absorb water and burst; if hypertonic, they would shrink and be damaged. It must be buffered to maintain a constant pH, because enzyme activity (including mitochondrial enzymes for the Krebs cycle and oxidative phosphorylation) is pH-dependent, and the metabolic activity of broken cells could alter the pH of the solution.

(b) Reduced coenzymes (NADH and FADH$_2$) produced in the Krebs cycle donate electrons to the electron transport chain on the inner mitochondrial membrane. The electrons pass through a series of electron carriers (protein complexes I, III, and IV), losing energy at each transfer. This energy is used to pump H$^+$ ions from the mitochondrial matrix into the intermembrane space, creating an electrochemical gradient (proton gradient) across the inner membrane. The H$^+$ ions accumulate in the intermembrane space, creating a high concentration and a more positive charge. H$^+$ ions diffuse back into the matrix through the enzyme ATP synthase (a transmembrane protein). The flow of H$^+$ through ATP synthase drives the rotation of its stalk, catalysing the phosphorylation of ADP to ATP. This process — the movement of H$^+$ ions down their electrochemical gradient driving ATP synthesis — is called chemiosmosis. The electrons are ultimately transferred to oxygen (the terminal electron acceptor), which combines with H$^+$ to form water.

(c) If the inner membrane is made permeable to H$^+$, the proton gradient is dissipated — H$^+$ ions leak back into the matrix without passing through ATP synthase. ATP production will decrease or stop because there is no proton gradient to drive ATP synthase. Oxygen consumption will increase (or continue at a high rate) because the electron transport chain will continue to operate (electrons still flow from NADH/FADH$_2$ to oxygen) and will pump H$^+$ ions, but the gradient cannot be maintained. The uncoupled mitochondria will consume oxygen and oxidise substrates without producing ATP — the energy is released as heat instead.

(d) Glycolysis occurs in the cytoplasm of the cell (not in the mitochondria). It does not require oxygen because it is an anaerobic process — no electron transport chain or oxidative phosphorylation is involved. Glycolysis involves the phosphorylation of glucose (using 2 ATP), its splitting into two triose phosphate molecules, and the oxidation of these to pyruvate (producing 4 ATP and 2 NADH by substrate-level phosphorylation). The net products of glycolysis per molecule of glucose are: 2 pyruvate, 2 ATP (net), and 2 reduced NAD (NADH). (Note: 2 ATP are used and 4 ATP are produced, giving a net gain of 2 ATP.)

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IT-3: Microscopy and Cell Membrane Transport (with Exchange and Transport)

Question:

A researcher investigates the uptake of the amino acid leucine by epithelial cells from the small intestine. Leucine is taken up against its concentration gradient by active transport.

(a) Explain why light microscopy cannot be used to directly observe the uptake of leucine by epithelial cells. Suggest an alternative technique the researcher could use.

(b) Describe the structure of a channel protein and a carrier protein, and explain why leucine uptake requires a carrier protein rather than a channel protein.

(c) The uptake of leucine is coupled to the movement of sodium ions (Na$^+$) into the cell. Explain how this co-transport mechanism works, including the role of ATP.

(d) The researcher uses an inhibitor that stops ATP synthesis. Predict and explain the effect on leucine uptake. Would the effect be the same if a competitive inhibitor of the Na$^+$-leucine co-transporter were used instead? Explain.

Solution:

(a) Light microscopy cannot directly observe leucine uptake because: (1) leucine molecules are far smaller than the resolution limit of a light microscope (approximately 200 nm), so individual molecules cannot be seen; (2) the light microscope cannot track the movement of individual molecules across membranes. An alternative technique is to use radioactive labelling — leucine can be labelled with a radioactive isotope (e.g., $^{14}$C-leucine or $^{3}$H-leucine). The researcher measures the radioactivity taken up by the cells over time using a scintillation counter. Alternatively, fluorescent labelling (e.g., using a fluorescently tagged leucine analogue) with fluorescence microscopy or confocal microscopy could allow visualisation of uptake patterns.

(b) A channel protein is a transmembrane protein that forms a hydrophilic pore through the bilayer, allowing specific ions or small polar molecules to pass through by facilitated diffusion. It does not undergo a conformational change and is typically gated (opens or closes in response to a stimulus). A carrier protein is a transmembrane protein that binds to a specific molecule, undergoes a conformational change, and releases the molecule on the other side of the membrane. Leucine uptake requires a carrier protein because: (1) leucine is an amino acid (a relatively large polar molecule) that cannot pass through a channel protein, which typically accommodates only ions and very small molecules; (2) leucine is transported against its concentration gradient, which requires a carrier protein that can couple the movement of leucine to the movement of Na$^+$ down its gradient — channel proteins cannot perform this coupled transport.

(c) In co-transport, the $\text{Na}^+/\text{K}^+$ pump on the basolateral membrane of the epithelial cell uses ATP to actively transport $\text{Na}^+$ out of the cell and $\text{K}^+$ into the cell. This maintains a low intracellular concentration of $\text{Na}^+$, creating a concentration gradient of $\text{Na}^+$ across the apical membrane (high $\text{Na}^+$ in the intestinal lumen, low $\text{Na}^+$ in the cell). $\text{Na}^+$ diffuses down its concentration gradient into the cell through the $\text{Na}^+$-leucine co-transporter protein on the apical membrane. The co-transporter has binding sites for both $\text{Na}^+$ and leucine; $\text{Na}^+$ binding causes a conformational change that simultaneously transports leucine into the cell against its concentration gradient. Leucine then exits the cell via facilitated diffusion on the basolateral membrane.

(d) If ATP synthesis is inhibited: the $\text{Na}^+/\text{K}^+$ pump stops working, so $\text{Na}^+$ is no longer pumped out of the cell. The $\text{Na}^+$ concentration gradient across the apical membrane dissipates (intracellular $\text{Na}^+$ increases). Without the $\text{Na}^+$ gradient, there is no driving force for $\text{Na}^+$ to enter through the co-transporter, so leucine uptake stops. This would not be the case with a competitive inhibitor of the co-transporter: a competitive inhibitor binds to the $\text{Na}^+$-leucine co-transporter's active site, competing with $\text{Na}^+$ and/or leucine for binding. Increasing the concentration of $\text{Na}^+$ and leucine in the lumen could partially or fully overcome this inhibition (as substrate outcompetes the inhibitor for the binding site). Therefore, unlike ATP inhibition (which indirectly affects the gradient and cannot be overcome by increasing substrate), competitive inhibition of the co-transporter can be partially overcome by increasing substrate concentration.