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

During oxidative phosphorylation, the F₀F₁-ATP synthase uses a proton gradient to synthesize ATP. If a poison specifically blocks the proton channel in the F₀ subunit without affecting the electron transport chain, what is the MOST LIKELY immediate consequence?

Answer: ATP synthesis halts but oxygen consumption initially continues, increasing heat production

Blocking F₀ prevents proton re-entry through ATP synthase, causing the proton gradient (Δp) to build up in the intermembrane space. The electron transport chain continues pumping protons (consuming NADH and O₂) until back-pressure from the excessive gradient eventually slows it — but the immediate effect is uncoupling of ATP synthesis from electron transport, dissipating energy as heat. Glycolysis is a cytosolic process not directly regulated by membrane potential in this acute scenario.

A cell biologist treats cultured cells with brefeldin A (BFA), a drug that causes the Golgi apparatus to merge with the endoplasmic reticulum. Which of the following proteins would be MOST disrupted in its normal processing pathway?

Answer: A lysosomal hydrolase that requires mannose-6-phosphate tagging

Lysosomal hydrolases are translated on rough ER ribosomes, enter the ER lumen, are glycosylated, and then receive a mannose-6-phosphate (M6P) tag in the Golgi — specifically in the cis-Golgi. BFA disrupts Golgi structure by blocking COPI vesicle formation, merging Golgi with ER. Without a functional Golgi, M6P tagging cannot occur, and lysosomal enzymes are secreted extracellularly instead of reaching lysosomes. Cytosolic, nuclear, and mitochondrially-encoded proteins bypass this ER–Golgi trafficking route entirely.

In meiosis I, homologous chromosomes separate. A cell undergoes meiosis where one pair of homologs fails to separate (nondisjunction) during meiosis I, but all other pairs separate normally. Meiosis II then proceeds normally for all cells. How many of the four resulting gametes will have an abnormal chromosome number?

Answer: All four gametes

When nondisjunction occurs in meiosis I, BOTH homologs of the affected pair migrate to the same secondary oocyte or secondary spermatocyte. That cell then undergoes meiosis II, splitting its two copies of the same homolog into two cells — each with one extra chromosome (n+1). The other secondary cell that received neither homolog also undergoes meiosis II, producing two cells each missing that chromosome (n−1). Result: 2 cells with n+1 and 2 cells with n−1 — all four gametes are aneuploid.

The sodium-potassium ATPase (Na⁺/K⁺-ATPase) is classified as a P-type ion pump. During its catalytic cycle, it undergoes phosphorylation of a conserved aspartate residue. If a mutation prevents dephosphorylation of this residue, which outcome is MOST accurate?

Answer: The pump would be locked in the E2 conformation, unable to release K⁺ on the intracellular side

In the Na⁺/K⁺-ATPase cycle: E1 binds 3 Na⁺ intracellularly → phosphorylation → E2-P releases Na⁺ extracellularly → K⁺ binds extracellularly → dephosphorylation converts E2-P to E2 → E2 releases K⁺ intracellularly → conformational change returns to E1. If dephosphorylation is blocked, the pump cannot transition from E2-P to E2. In the E2-P state, K⁺ is already bound on the extracellular side but cannot be translocated and released intracellularly because that requires dephosphorylation. The pump stalls with K⁺ trapped, unable to complete the cycle.

Peroxisomes oxidize very-long-chain fatty acids (VLCFAs) via beta-oxidation. Unlike mitochondrial beta-oxidation, the peroxisomal version uses FAD-linked oxidases that transfer electrons directly to O₂, producing H₂O₂. Catalase then degrades H₂O₂. If catalase activity in peroxisomes is completely abolished, what would be the DIRECT biochemical consequence?

Answer: VLCFAs would accumulate because peroxisomal beta-oxidation would be inhibited by H₂O₂ buildup

Peroxisomal beta-oxidation generates H₂O₂ as a byproduct of the FAD-linked acyl-CoA oxidase step. Catalase normally converts this toxic H₂O₂ to H₂O and O₂. Without catalase, H₂O₂ accumulates in the peroxisome. H₂O₂ is a reactive oxygen species (ROS) that oxidatively inactivates peroxisomal enzymes — including those of beta-oxidation itself — leading to impaired VLCFA breakdown and their accumulation. This mirrors the pathology seen in Zellweger syndrome-related disorders where peroxisomal function is compromised.

The signal recognition particle (SRP) cycle directs proteins destined for the secretory pathway to the rough ER. SRP is a ribonucleoprotein that recognizes the signal peptide as it emerges from the ribosome. In which scenario would a protein with a functional N-terminal signal peptide FAIL to be inserted into the ER lumen?

Answer: Loss of the SRP receptor's GTPase activity that is required for ribosome docking to the ER membrane

The SRP cycle requires GTP hydrolysis at two points: SRP54 (on SRP) and the α-subunit of the SRP receptor (SR). When SRP-bound ribosome meets the SR at the ER membrane, both GTPases stimulate each other's activity. GTP hydrolysis by the SRP receptor is required for SRP to release the ribosome and signal peptide, allowing the translocon (Sec61) to accept the ribosome and initiate co-translational translocation. If the SRP receptor's GTPase is non-functional, it cannot release SRP, the ribosome remains stalled, and the protein cannot be threaded into the ER lumen. Option A would also prevent targeting (SRP54 recognizes hydrophobic signal peptides), but option D specifically disrupts the docking/release step after SRP has already recognized the signal, making it a distinct mechanistic failure point for a protein with a 'functional' signal peptide.