Cell Biology — Diagnostic Tests

Unit Tests

UT-1: Prokaryotic vs Eukaryotic Cell Comparison

Question: Create a comparison table for prokaryotic and eukaryotic cells covering: (a) cell wall composition, (b) DNA organisation, (c) ribosome size, (d) membrane-bound organelles, (e) cell division mechanism, (f) size range. Explain why prokaryotes can reproduce more rapidly than eukaryotes.

Solution:

Feature	Prokaryotic	Eukaryotic
Cell wall	Peptidoglycan (murein)	None in animals; cellulose in plants; chitin in fungi
DNA	Single circular chromosome; no histones; naked in nucleoid	Multiple linear chromosomes; wrapped around histones; in nucleus
Ribosomes	70S (30S + 50S subunits)	80S (40S + 60S subunits); 70S in mitochondria/chloroplasts
Organelles	None (no nucleus, mitochondria, ER, etc.)	Membrane-bound organelles present (nucleus, mitochondria, ER, Golgi)
Cell division	Binary fission (no spindle)	Mitosis (and meiosis); spindle apparatus involved
Size	0.5--5 $\mu$m	10--100 $\mu$m

Prokaryotes reproduce faster because: (1) binary fission is simpler than mitosis -- no spindle formation or nuclear envelope breakdown; (2) they have a smaller genome to replicate (circular DNA, no histones); (3) they have no membrane-bound organelles to distribute; (4) their small size means a high surface-area-to-volume ratio for rapid nutrient uptake; (5) generation times can be as short as 20 minutes (E. coli) compared to hours or days for eukaryotic cells.

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UT-2: Organelle Structure-Function Relationships

Question: Explain how the structure of each organelle relates to its function: (a) mitochondrion (inner membrane cristae), (b) rough endoplasmic reticulum (ribosome-studded membrane), (c) Golgi apparatus (stacked cisternae with vesicles).

Solution:

(a) Mitochondrion: The inner membrane is highly folded into cristae, increasing the surface area available for the electron transport chain (ETC) and ATP synthase. The ETC proteins and ATP synthase are embedded in this membrane. The large surface area allows more copies of these proteins, maximising the rate of oxidative phosphorylation and ATP production. The intermembrane space is a small, confined compartment that allows efficient proton gradient accumulation for chemiosmosis.

(b) Rough ER: The ribosomes studding the outer surface translate mRNA into polypeptide chains that are fed directly into the ER lumen. The membrane provides a large surface area for ribosome attachment and protein processing. Inside the lumen, proteins undergo folding (assisted by chaperone proteins) and post-translational modifications such as disulfide bond formation and glycosylation. The continuity between the rough ER and the nuclear envelope allows efficient mRNA transport from the nucleus to the ribosomes.

(c) Golgi apparatus: The stacked cisternae provide separate processing compartments. Proteins arrive at the cis face in vesicles from the ER, and are progressively modified (glycosylation, phosphorylation, proteolytic cleavage) as they pass through the medial and trans cisternae. The vesicles budding from the trans face (trans Golgi network) sort and package modified proteins for transport to their final destinations (lysosomes, plasma membrane, or secretion). The modular stack structure allows sequential processing steps in an assembly-line fashion.

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UT-3: Fluid Mosaic Model and Membrane Transport

Question: Describe the fluid mosaic model of membrane structure. Explain why small non-polar molecules (e.g., $\text{O}_2$, $\text{CO}_2$) can diffuse across the membrane easily, while ions (e.g., $\text{Na}^+$) require transport proteins. Include the roles of cholesterol and integral proteins.

Solution:

The fluid mosaic model (Singer and Nicolson, 1972) describes the membrane as a fluid bilayer of phospholipids with embedded proteins. The phospholipids have hydrophilic phosphate heads facing outward and hydrophobic fatty acid tails facing inward. The membrane is "fluid" because phospholipids can move laterally within their layer (but rarely flip-flop between layers), and proteins can drift within the bilayer. "Mosaic" refers to the scattered pattern of different proteins.

Small non-polar molecules diffuse easily because the hydrophobic core of the bilayer (fatty acid tails) presents no energetic barrier to them -- they are soluble in the lipid environment. They move down their concentration gradient by simple diffusion.

Ions are charged and hydrophilic. The hydrophobic core is highly unfavourable for charged species -- the dielectric constant is very different from water. Ions cannot shed their hydration shell to enter the non-polar interior. Transport proteins (channel proteins or carrier proteins) provide a hydrophilic pathway through the membrane. Channel proteins form pores with polar amino acid linings; carrier proteins undergo conformational changes to shuttle ions across.

Cholesterol is embedded between phospholipids in animal cell membranes. It acts as a bidirectional fluidity buffer: at high temperatures, it restricts phospholipid movement (reducing excessive fluidity), and at low temperatures, it prevents tight packing (maintaining fluidity). This ensures membrane integrity across a range of temperatures.

Integration Tests

IT-1: Cell Division and Cancer (with Molecular Biology)

Question: Explain how a mutation in a proto-oncogene can lead to uncontrolled cell division. Include the roles of: (a) the normal proto-oncogene product, (b) how the mutation changes its function, (c) the cell cycle checkpoints that should prevent this, and (d) why multiple mutations are typically required for cancer development.

Solution:

(a) Normal proto-oncogenes encode proteins that promote cell division, such as growth factor receptors, signal transduction proteins (e.g., Ras), and transcription factors (e.g., Myc). In a healthy cell, these proteins are activated only when appropriate growth signals are received, and their activity is tightly regulated. For example, Ras is active only when bound to GTP and is inactivated by its intrinsic GTPase activity.

(b) A mutation can convert a proto-oncogene into an oncogene through several mechanisms: point mutation producing a constitutively active protein (e.g., Ras locked in GTP-bound state), gene amplification (multiple copies producing excess protein), or chromosomal translocation placing the gene under a stronger promoter. The result is continuous, unregulated promotion of cell division regardless of external signals.

(c) Cell cycle checkpoints that should prevent uncontrolled division:

• G1 checkpoint: checks cell size, nutrient availability, DNA damage (mediated by p53). If damage is detected, p53 activates p21 which inhibits cyclin-dependent kinases (CDKs).
• G2 checkpoint: verifies DNA replication is complete and error-free before mitosis.
• M checkpoint (spindle assembly checkpoint): ensures all chromosomes are properly attached to spindle fibres before anaphase.

(d) Multiple mutations are required because cells have redundant tumour suppressor mechanisms. Typically, cancer requires: activation of at least one oncogene, inactivation of at least one tumour suppressor gene (e.g., p53, Rb), and mutations enabling angiogenesis and metastasis. A single mutation is usually insufficient because the remaining checkpoints can still halt division or trigger apoptosis.

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IT-2: Endosymbiosis and Evolution (with Evolution and Genetics)

Question: Outline three pieces of evidence supporting the endosymbiotic theory for the origin of mitochondria and chloroplasts. Explain how this theory accounts for the observation that mitochondrial DNA is inherited maternally in most animals.

Solution:

Three pieces of evidence:

1. Double membrane structure: Both mitochondria and chloroplasts have two membranes. The outer membrane is thought to derive from the host cell's phagocytic vesicle, while the inner membrane derives from the original prokaryote's plasma membrane.

2. Own genome and ribosomes: Mitochondria and chloroplasts contain circular DNA (like prokaryotes, not linear like eukaryotic nuclear DNA), and 70S ribosomes (prokaryotic size, not 80S). They can transcribe and translate their own proteins independently of the nucleus.

3. Binary fission reproduction: Both organelles divide by binary fission, the same process used by prokaryotes, rather than by mitosis. They cannot be synthesised de novo -- new organelles arise only from pre-existing ones.

Maternal inheritance explanation: During fertilisation, the sperm contributes primarily nuclear DNA to the zygote. The sperm's midpiece contains mitochondria for powering flagellar movement, but these mitochondria (along with the rest of the tail) are typically excluded from the zygote upon fusion. The egg cell (ovum) is much larger and contains thousands of mitochondria in its cytoplasm. Therefore, the zygote inherits its mitochondria almost exclusively from the mother. Additionally, in many species, there is an active mechanism to selectively degrade sperm-derived mitochondria in the zygote through ubiquitin-mediated autophagy.

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IT-3: Membrane Transport and Neurone Function (with Human Physiology)

Question: The sodium-potassium pump ($\text{Na}^+/\text{K}^+$-ATPase) moves $3\ \text{Na}^+$ out and $2\ \text{K}^+$ in per ATP hydrolysed. Calculate the percentage of a typical neuron's resting energy budget consumed by this pump, given that a neuron at rest has a membrane potential of $-70\ \text{mV}$ and the pump maintains concentration gradients of $[\text{Na}^+]_{\text{out}}/[\text{Na}^+]_{\text{in}} = 10$ and $[\text{K}^+]_{\text{in}}/[\text{K}^+]_{\text{out}} = 30$. One ATP hydrolysis releases approximately $50\ \text{kJ mol}^{-1}$.

Solution:

The work done by the pump per cycle (moving 3 Na$^+$ out and 2 K$^+$ in) against both concentration and electrical gradients:

For $\text{Na}^+$ (moving out against both gradient and electrical gradient, since inside is negative): $\Delta G_{\text{Na}} = 3RT\ln\frac{[\text{Na}^+]_{\text{out}}}{[\text{Na}^+]_{\text{in}}} + 3F \times V_m$

$= 3(8.314)(310)\ln(10) + 3(96485)(0.070)$

$= 3(2577)(2.303) + 3(6754) = 17800 + 20262 = 38062\ \text{J mol}^{-1}$

For $\text{K}^+$ (moving in against the concentration gradient but with the electrical gradient): $\Delta G_{\text{K}} = 2RT\ln\frac{[\text{K}^+]_{\text{in}}}{[\text{K}^+]_{\text{out}}} - 2F \times V_m$

$= 2(2577)\ln(30) - 2(6754) = 2(2577)(3.401) - 13508 = 17534 - 13508 = 4026\ \text{J mol}^{-1}$

Total work per cycle: $38062 + 4026 = 42088\ \text{J mol}^{-1} \approx 42\ \text{kJ mol}^{-1}$

Efficiency: $\frac{42}{50} \times 100 = 84\%$

The $\text{Na}^+/\text{K}^+$ pump is estimated to consume approximately 60--70% of a neuron's total ATP at rest. This is because the pump must continuously work against the leak channels that allow $\text{Na}^+$ to diffuse back in and $\text{K}^+$ to diffuse back out, maintaining the resting potential that is essential for action potential generation.