- Basic Math and Science Cellular Processes and Organelles Questions and Answers Flashcards

A cell biologist observes that a particular organelle has a double membrane, contains its own ribosomes (70S), and has circular DNA. However, inhibiting this organelle's function with oligomycin (an ATP synthase blocker) has no effect on the cell's survival when glucose is abundant. Which organelle is most likely being described, and why does the cell survive?

Answer: Mitochondrion; the cell switches to anaerobic glycolysis to generate ATP without oxidative phosphorylation

The organelle described — double membrane, 70S ribosomes, circular DNA — is the mitochondrion, consistent with the endosymbiotic theory. Oligomycin blocks ATP synthase (Complex V), halting oxidative phosphorylation. However, when glucose is abundant, the cell can generate ATP through glycolysis alone (producing a net 2 ATP per glucose via substrate-level phosphorylation), bypassing the need for the electron transport chain entirely. This is why cancer cells often favor glycolysis even with oxygen present (Warburg effect).

During receptor-mediated endocytosis, a ligand-receptor complex is internalized in a clathrin-coated vesicle. The vesicle fuses with an early endosome, which then matures into a late endosome. What critical pH-dependent event occurs in the late endosome that allows the cell to recycle its receptors?

Answer: The acidic pH (~5.5) causes conformational changes that dissociate the ligand from the receptor, freeing the receptor for recycling back to the plasma membrane

Proton pumps (V-type H⁺-ATPases) acidify the endosomal lumen progressively from early (~6.5) to late endosome (~5.5). This low pH induces conformational changes in the receptor that dramatically reduce its affinity for the ligand. The liberated receptor is then sorted into tubular extensions that bud off and fuse back with the plasma membrane for reuse. The ligand, now free, remains in the lumen and is eventually delivered to the lysosome (pH ~4.5–5.0) for enzymatic degradation. This separation of receptor fate from ligand fate is fundamental to efficient receptor recycling.

A mutation eliminates the function of the smooth endoplasmic reticulum (SER) in liver hepatocytes but leaves the rough ER intact. Which of the following consequences would be MOST directly attributable to the loss of SER function specifically?

Answer: Impaired detoxification of lipid-soluble drugs and accumulation of triglycerides due to blocked lipid metabolism

The smooth ER (lacking ribosomes) specializes in lipid metabolism (including phospholipid and cholesterol synthesis), drug/toxin detoxification via cytochrome P450 enzymes, and glycogen metabolism — all critical hepatocyte functions. Triglyceride synthesis and packaging into VLDL particles also depend heavily on SER. The other options describe rough ER functions: albumin/clotting factor synthesis involves RER-bound ribosomes; glycosylation of secretory proteins begins in the RER lumen; membrane protein folding relies on RER chaperones like BiP/GRP78. Ribosomal assembly itself occurs in the nucleolus, not in either ER compartment.

In a cell undergoing apoptosis, cytochrome c is released from the mitochondrial intermembrane space into the cytosol. Mathematically, if the probability that a single cytochrome c molecule activates one Apaf-1 monomer is 0.4, and 7 Apaf-1 monomers must all be activated to form one functional apoptosome complex, what is the approximate probability that a given apoptosome forms from exactly 7 independent cytochrome c activation events?

Answer: Approximately 0.016 (≈ 0.4⁷)

Since each of the 7 Apaf-1 activation events is stated to be independent with probability 0.4, the probability that all 7 are activated is the product of individual probabilities: 0.4⁷ = 0.4 × 0.4 × 0.4 × 0.4 × 0.4 × 0.4 × 0.4 = 0.0016384, which rounds to approximately 0.016. This is a direct application of the multiplication rule for independent events in probability. Biologically, this low probability per complex explains why massive cytochrome c release is needed to reliably trigger apoptosis — stochastic threshold behavior ensures the cell doesn't accidentally activate programmed death from minor mitochondrial stress.

A researcher treats cells with brefeldin A (BFA), a fungal toxin that blocks ARF1-GTPase activation, causing the Golgi apparatus to collapse into the endoplasmic reticulum within minutes. Which downstream cellular process would be LEAST affected by BFA treatment?

Answer: Cytoplasmic translation of mRNAs encoding cytosolic enzymes such as hexokinase

BFA disrupts COPI vesicle formation and causes Golgi collapse, blocking all anterograde (ER→Golgi→plasma membrane) and retrograde trafficking. This devastates: (B) insulin processing, which requires Golgi-derived secretory vesicles and convertase enzymes; (C) complex N-glycosylation, which occurs in Golgi compartments (medial/trans); and (D) mannose-6-phosphate tagging of lysosomal enzymes, which happens in the cis/medial Golgi. However, cytoplasmic translation of cytosolic proteins like hexokinase occurs entirely on free ribosomes in the cytoplasm — no ER entry, no vesicle trafficking required. BFA has no effect on the translation machinery or the cytosol itself.

Two cells are placed in solutions: Cell A is placed in a solution with a solute concentration of 0.15 M, and Cell B is placed in a solution with a solute concentration of 0.35 M. The cytoplasm of both cells has an internal solute concentration of 0.25 M. After equilibration, which statement correctly describes the RELATIVE volume changes and the MATHEMATICAL relationship between osmotic pressure differences?

Answer: Cell A swells and Cell B shrinks; the osmotic pressure driving water into Cell A is proportionally greater than the pressure driving water out of Cell B because the concentration gradient is larger for Cell A (ΔC = 0.10 M) than for Cell B (ΔC = 0.10 M) — they are equal, so volume changes are symmetric

Cell A's external solution (0.15 M) is hypotonic relative to its cytoplasm (0.25 M): water moves in by osmosis, causing swelling. Cell B's external solution (0.35 M) is hypertonic: water moves out, causing shrinkage (crenation in animal cells, plasmolysis in plant cells). The concentration gradient for Cell A = 0.25 − 0.15 = 0.10 M; for Cell B = 0.35 − 0.25 = 0.10 M. By van't Hoff's law (π = iMRT), with equal ΔC and assuming ideal solutions at the same temperature, the osmotic pressure driving force is equal in magnitude for both cells. Therefore, the volume changes (swelling vs. shrinkage) are symmetric in magnitude — a subtlety often missed. Option D is a mathematical error in reasoning: averaging external concentrations has no physical meaning for osmosis.