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

A cell biologist treats cells with oligomycin, a drug that blocks ATP synthase. Which cellular process would be MOST directly impaired as a result, and why?

Answer: The electron transport chain's ability to maintain the proton gradient, because ATP synthase blockage causes protons to accumulate in the intermembrane space

ATP synthase uses the proton gradient (proton-motive force) across the inner mitochondrial membrane to synthesize ATP. When oligomycin blocks ATP synthase, protons can no longer flow back into the matrix through the synthase, causing proton accumulation in the intermembrane space. This backs up the entire electron transport chain because the chain's function depends on continuously pumping protons out — without a way to dissipate the gradient, the ETC stalls. Glycolysis and fermentation do not involve ATP synthase, and the citric acid cycle uses substrate-level phosphorylation for its small ATP yield.

Which of the following best explains why the rough endoplasmic reticulum (RER) is morphologically distinct from the smooth ER, and what functional consequence follows from that distinction?

Answer: The RER is studded with ribosomes translating mRNAs that encode a signal peptide, co-translationally inserting nascent polypeptides into the ER lumen for processing

The 'rough' appearance of the RER comes from ribosomes docked on its cytoplasmic face. These ribosomes are translating mRNAs whose proteins carry a signal peptide recognized by the signal recognition particle (SRP), which docks the ribosome to the RER membrane. The growing polypeptide is threaded into the ER lumen, where it undergoes folding, N-linked glycosylation, and disulfide bond formation. This co-translational import is the key functional consequence of the rough morphology. Cristae are mitochondrial structures, not ER features, and lipid synthesis occurs in the smooth ER.

In a cell undergoing aerobic respiration, the net ATP yield per glucose molecule is commonly cited as approximately 30–32 ATP. However, the theoretical maximum from chemiosmosis alone would be higher. Which factor MOST accounts for this gap?

Answer: NADH generated in the cytoplasm during glycolysis cannot directly enter the mitochondrial matrix due to the impermeability of the inner membrane, and the shuttle mechanisms have an energetic cost

The inner mitochondrial membrane is impermeable to NADH itself. The 2 NADH produced in glycolysis (in the cytoplasm) must be shuttled indirectly into the mitochondria. The malate-aspartate shuttle preserves their full reducing equivalent (yielding ~2.5 ATP each), but the glycerol-3-phosphate shuttle sacrifices one proton-pumping step, yielding only ~1.5 ATP per cytoplasmic NADH. This energetic cost of shuttling, plus the ATP consumed by the mitochondrial ATP/ADP translocase (which costs one proton per ATP exported), accounts for the shortfall from theoretical maximum. Oxygen is not a competitive inhibitor of complex I; it is the final electron acceptor at complex IV.

A researcher discovers a mutation that causes lysosomes to fuse prematurely with early endosomes before cargo is fully sorted in the late endosome. What would be the MOST likely consequence at the cellular level?

Answer: Degradation of receptor-ligand complexes before receptors can be recycled to the plasma membrane, leading to reduced receptor availability on the cell surface

Normally, early endosomes serve as a sorting station: some receptors (e.g., LDL receptors) are recycled back to the plasma membrane while their cargo is sent to late endosomes for degradation. If lysosomes fuse prematurely with early endosomes, lysosomal hydrolases (active at pH ~4.5–5) would degrade both ligands AND receptors before recycling can occur, depleting the cell surface of functional receptors. Option A is backward — lysosomal enzymes would degrade receptors, not rescue them. Option C is wrong because lysosomal enzymes would still work (the early endosome is acidic, ~pH 6, though not optimal). Option D is incorrect; MHC II antigen loading occurs in specialized late endosome/lysosome compartments under tightly regulated conditions.

Which of the following scenarios would cause a cell to undergo apoptosis via the INTRINSIC (mitochondrial) pathway rather than the extrinsic pathway?

Answer: Irreparable double-strand DNA breaks activate p53, which upregulates pro-apoptotic Bcl-2 family proteins such as Bax, leading to cytochrome c release from mitochondria

The intrinsic pathway is triggered by internal cellular stress signals — such as DNA damage, oxidative stress, or oncogene activation — rather than external death signals. Irreparable DNA damage activates the tumor suppressor p53, which transcriptionally upregulates pro-apoptotic Bcl-2 family members (Bax, Bak). These proteins permeabilize the outer mitochondrial membrane, releasing cytochrome c into the cytoplasm. Cytochrome c then forms the apoptosome with Apaf-1, activating caspase-9 and subsequently caspase-3. Options A, B, and D all describe the extrinsic pathway, which is initiated by extracellular death ligands binding cell-surface receptors (Fas, TNFR) or by immune cell attack.

A student calculates that a particular membrane protein spans the lipid bilayer seven times (a 7-TM or GPCR-type topology). If each transmembrane helix is composed of approximately 20 hydrophobic amino acids and the average amino acid contributes 0.15 nm to the length of an α-helix, what is the CLOSEST estimate of the total length of transmembrane sequence this protein buries within a typical 3.5 nm hydrophobic core?

Answer: 21.0 nm

Each transmembrane helix spans the hydrophobic core with ~20 amino acids × 0.15 nm/residue = 3.0 nm per helix. With 7 transmembrane helices, the total length of transmembrane sequence is 7 × 3.0 nm = 21.0 nm. Note that each individual helix is ~3.0 nm, which matches the ~3.5 nm hydrophobic core of a bilayer (the small discrepancy reflects that real helices are slightly longer and may tilt). The question asks for TOTAL length of transmembrane sequence across all seven helices combined, so the answer is 21.0 nm. This calculation tests understanding of both α-helix geometry and membrane protein topology.