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CSIR UGC NET Life Sciences Unit 2 Mock Test 2026: Interactive Practice Paper

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CSIR UGC NET Life Sciences Mock Test 2026: Interactive Unit 2 Practice Paper
Square Google Blog thumbnail for CSIR UGC NET Life Sciences Unit 2 Mock Test 2026 featuring a professional green biology-themed design with DNA, molecular graphics, and exam icons. Includes the branding @indiabiologyneet with the DNA-leaf logo and the tagline "Fueling Biology Dreams". Highlights a free interactive practice paper with instant score analysis, detailed solutions, and high-quality MCQs.

Boost Your CSIR NET JRF Rank with High-Yield Cell Biology Questions | Fully Explained Interactive Practice Set

Are you looking to crack the CSIR UGC NET Life Sciences exam with a top rank? Mastering Unit 2: Cellular Organization (Cell Biology) is one of the most reliable strategies to guarantee maximum scores in both Part B and Part C. To assist your preparation, our expert panel has designed this exclusive, interactive full-length unit-specific mock test strictly matching the latest examination standards.

Exam Pattern Followed: This test contains exactly 50 questions broken down into Part A (General Aptitude), Part B (Core Cell Biology), and Part C (Analytical Cell Biology) with instant feedback & comprehensive explanations upon selecting your choice!

📋 Mock Test Scheme & Guidelines

  • Total Questions: 50 (Part A: 10 Questions | Part B: 25 Questions | Part C: 15 Questions)
  • Marking System: Part A (+2 | -0.5), Part B (+2 | -0.5), Part C (+5 | -1.0)
  • Syllabus Scope: General Aptitude & Unit 2 (Cellular Organization)

PART A: General Aptitude & General Science

(2 Marks each. Negative marking: 0.5)

Part A

Q1. If 4 men or 6 women can build a wall in 20 days, how long will it take 6 men and 3 women to build the same wall working together at the same pace?

  • A) 12 days
  • B) 10 days
  • C) 8 days
  • D) 15 days
Correct Answer: B
Explanation: 4 men = 6 women ⇒ 1 man = 1.5 women. Thus, 6 men + 3 women = 6(1.5) + 3 = 12 women. Since 6 women take 20 days, 12 women will take (6 × 20) / 12 = 10 days.
Part A

Q2. A group of cells doubles in count every 30 minutes. If a single petri dish is completely full of these cells at exactly 4:00 PM, at what time was the petri dish exactly 25% full?

  • A) 2:00 PM
  • B) 3:30 PM
  • C) 3:00 PM
  • D) 2:30 PM
Correct Answer: C
Explanation: If the dish is 100% full at 4:00 PM, it was 50% full 30 minutes earlier (3:30 PM), and 25% full another 30 minutes earlier, which is 3:00 PM.
Part A

Q3. Find the missing number in the following mathematical series: 2, 6, 12, 20, 30, ?, 56.

  • A) 38
  • B) 42
  • C) 40
  • D) 44
Correct Answer: B
Explanation: The difference between consecutive terms increases by 2 each time (+4, +6, +8, +10, +12...). So, 30 + 12 = 42. Next is 42 + 14 = 56.
Part A

Q4. In a cross-country sprint race, if runner A is faster than B, runner B runs at the same pace as C, and runner D is slower than C, who among them is definitely the slowest runner?

  • A) A
  • B) B
  • C) C
  • D) D
Correct Answer: D
Explanation: The speed relationships are: A > B, B = C, and C > D. Combining these gives us A > B = C > D, identifying D unambiguously as the slowest.
Part A

Q5. What is the angle between the hour hand and the minute hand of a standard clock at exactly 3:40?

  • A) 130 degrees
  • B) 140 degrees
  • C) 120 degrees
  • D) 150 degrees
Correct Answer: A
Explanation: Using the formula: Angle = |(30H - 5.5M)|, where H = 3 and M = 40. Angle = |30(3) - 5.5(40)| = |90 - 220| = 130 degrees.
Part A

Q6. The average weight of 5 distinct protein samples is 45 kDa. When a 6th protein sample is added to the batch, the new mean weight drops to 43 kDa. What is the molecular weight of the 6th protein?

  • A) 35 kDa
  • B) 33 kDa
  • C) 38 kDa
  • D) 40 kDa
Correct Answer: B
Explanation: Total weight of 5 samples = 5 × 45 = 225 kDa. Total weight of 6 samples = 6 × 43 = 258 kDa. Weight of the 6th sample = 258 - 225 = 33 kDa.
Part A

Q7. A bag contains 4 red labels, 5 blue labels, and 6 green labels. If two labels are drawn randomly from the bag one after another without replacement, what is the probability that both labels are blue?

  • A) 2/21
  • B) 2/21 (equivalent to 20/210)
  • C) 1/7
  • D) 5/21
Correct Answer: B
Explanation: Probability of first blue label = 5/15 = 1/3. Probability of second blue label = 4/14 = 2/7. Combined probability = (1/3) × (2/7) = 2/21.
Part A

Q8. In a certain scientific code language, 'RIBOSOME' is written as 'VMFPTOSJ'. How would 'NUCLEOPH' be written in that exact same code variant?

  • A) RVFMIPQS
  • B) RVFMJPQT
  • C) SVGNKPQT
  • D) RUELDOPH
Correct Answer: A
Explanation: The letters are shifted forward according to specific pattern variations. For NUCLEOPH, matching the same reverse-position alphanumeric increment yields RVFMIPQS.
Part A

Q9. A rectangular laboratory storage zone measures 12 meters in length, 9 meters in width, and 8 meters in height. What is the length of the longest structural diagnostic rod that can be fit completely inside this room corner-to-corner?

  • A) 15 meters
  • B) 17 meters
  • C) 19 meters
  • D) 21 meters
Correct Answer: B
Explanation: Longest diagonal of a cuboid = √(l² + w² + h²) = √(12² + 9² + 8²) = √(144 + 81 + 64) = √289 = 17 meters.
Part A

Q10. A student walks 6 km North, takes a precise left turn and walks 8 km, then takes a final left turn and walks 6 km. How far and in which direction is the student from the original starting point?

  • A) 8 km West
  • B) 8 km East
  • C) 14 km West
  • D) 6 km South
Correct Answer: A
Explanation: Moving North 6 km and then South 6 km cancels out the vertical displacement. The horizontal displacement remains exactly 8 km to the West.

PART B: Subject Core (Unit 2 - Cellular Organization)

(2 Marks each. Negative marking: 0.5)

Part B

Q11. Which of the following lipid types is explicitly restricted to the outer (exoplasmic) leaflet of the plasma membrane in healthy, non-apoptotic mammalian cells?

  • A) Phosphatidylserine
  • B) Phosphatidylethanolamine
  • C) Glycolipids like Gangliosides
  • D) Phosphatidylinositol
Correct Answer: C
Explanation: Gangliosides and other glycolipids are synthesized in the lumen of the Golgi apparatus and are exclusively displayed on the exoplasmic leaflet. Phosphatidylserine is flipped to the outside only during apoptosis.
Part B

Q12. During FRAP (Fluorescence Recovery After Photobleaching) kinetics analysis, which parameter measures the overall percentage of membrane protein molecules capable of free lateral diffusion?

  • A) Diffusion Coefficient (D)
  • B) Mobile Fraction
  • C) Bleaching Intensity Index
  • D) Immobile Threshold
Correct Answer: B
Explanation: The mobile fraction quantifies the extent of fluorescence recovery at the plateau phase relative to initial levels, defining the proportion of molecules free to diffuse.
Part B

Q13. Which transport protein catalyzes the active ATP-driven trans-bilayer movement of aminophospholipids from the exoplasmic leaflet to the cytosolic leaflet?

  • A) Flippase (P-type ATPase)
  • B) Floppase (ABC Transporter)
  • C) Scramblase
  • D) Phospholipid Transfer Protein
Correct Answer: A
Explanation: Flippases use ATP to transport lipids inwardly (outer to inner leaflet), while floppases move them outward, and scramblases move them bidirectionally independent of ATP.
Part B

Q14. Gram-negative bacteria contain a distinct structural layer between their inner plasma membrane and outer membrane known as:

  • A) Capsule
  • B) Periplasmic space containing a thin peptidoglycan layer
  • C) Teichoic acid meshwork
  • D) S-layer
Correct Answer: B
Explanation: Gram-negative bacteria possess a periplasmic space containing a thin peptidoglycan layer, bounded by the inner and outer membranes.
Part B

Q15. What is the core biochemical composition of the bacterial cell wall principal network?

  • A) N-acetylglucosamine and N-acetylmuramic acid crosslinked by peptides
  • B) Beta-1,4-linked D-glucoronic polymers
  • C) Lipopolysaccharide-anchored glucans
  • D) Chitinous oligosaccharide matrix
Correct Answer: A
Explanation: The bacterial cell wall peptidoglycan (murein) consists of alternating NAG and NAM residues linked by beta-1,4-glycosidic bonds and crosslinked by amino acid side chains.
Part B

Q16. The nuclear localization signal (NLS) required for the importing of proteins into the nucleus is primarily enriched in which amino acids?

  • A) Hydrophobic residues (Leucine, Isoleucine)
  • B) Basic residues (Lysine, Arginine)
  • C) Acidic residues (Aspartate, Glutamate)
  • D) Aromatic residues (Tyrosine, Tryptophan)
Correct Answer: B
Explanation: Classical NLS sequences (like that of SV40 T-antigen) are rich in basic, positively charged amino acids like lysine and arginine.
Part B

Q17. Which GTPase is directly responsible for powering the nucleocytoplasmic cargo transport cycle through its differential compartmental concentration gradient?

  • A) Ran
  • B) Ras
  • C) Rab
  • D) Rho
Correct Answer: A
Explanation: Ran GTPase establishes a gradient (RanGTP high in nucleus, RanGDP high in cytoplasm) that controls cargo binding and release by importins and exportins.
Part B

Q18. Core N-linked glycosylation of nascent polypeptide chains initiates inside which organelle or compartment?

  • A) Golgi cis-cisternae
  • B) Lumen of the Rough Endoplasmic Reticulum
  • C) Cytosolic ribosomes
  • D) Trans-Golgi Network
Correct Answer: B
Explanation: N-linked glycosylation begins co-translationally in the ER lumen, where a pre-assembled 14-sugar oligosaccharide is transferred to an asparagine residue.
Part B

Q19. Proteins destined for degradation via the lysosomal pathway are tagged or modified in the Golgi apparatus with which specific chemical moiety?

  • A) Ubiquitin chains
  • B) Mannose-6-phosphate (M6P)
  • C) Glucose-1-phosphate
  • D) KDEL sequence
Correct Answer: B
Explanation: Soluble lysosomal enzymes are specifically modified with mannose-6-phosphate residues in the cis-Golgi to be recognized by M6P receptors in the TGN for sorting to lysosomes.
Part B

Q20. Which specific coat protein complex is responsible for mediating retrograde transport from the Golgi apparatus back to the Endoplasmic Reticulum?

  • A) COPII
  • B) COPI
  • C) Clathrin
  • D) Caveolin
Correct Answer: B
Explanation: COPI vesicles mediate retrograde transport (Golgi to ER), while COPII coats anterograde transport vesicles (ER to Golgi).
Part B

Q21. During cellular respiration, protons are actively pumped out of the mitochondrial matrix into which specific compartment to establish the proton motive force?

  • A) Cytoplasm
  • B) Intermembrane space
  • C) Outer membrane exterior pores
  • D) Thylakoid lumen
Correct Answer: B
Explanation: Complexes I, III, and IV pump protons from the matrix into the intermembrane space, creating the electrochemical gradient utilized by ATP synthase.
Part B

Q22. Plant vacuoles maintain an internal acidic pH primarily through the action of which specific ion pumps?

  • A) V-type H+ ATPase and Pyrophosphatases
  • B) P-type Ca2+ pumps
  • C) Na+/K+ exchangers
  • D) F-type ATP synthases
Correct Answer: A
Explanation: Vacuolar membranes (tonoplasts) feature active V-type ATPases and proton-pumping pyrophosphatases that move H+ into the vacuole against its gradient.
Part B

Q23. Which cytoskeletal filament system lacks intrinsic structural polarity and does not utilize motor proteins for direction-specific cargo translocation?

  • A) Microtubules
  • B) Actin filaments (Microfilaments)
  • C) Intermediate Filaments
  • D) Mitotic spindle fibers
Correct Answer: C
Explanation: Intermediate filaments are symmetric, non-polar polymers assembled from staggered tetramers, unlike polar microtubules or actin tracks.
Part B

Q24. Which motor protein complex moves progressively along microtubule tracks exclusively toward the minus (-) end (retrograde transport)?

  • A) Kinesin-1
  • B) Cytoplasmic Dynein
  • C) Myosin V
  • D) Kinesin-5
Correct Answer: B
Explanation: Cytoplasmic dyneins are minus-end directed microtubule motors, while most kinesins migrate toward the plus end.
Part B

Q25. The critical concentration (Cc) threshold for actin filament assembly is lowest at which structural end?

  • A) Plus (+) barbed end
  • B) Minus (-) pointed end
  • C) Both ends have identical Cc
  • D) Mid-filament core zone
Correct Answer: A
Explanation: The barbed (+) end has a much lower critical concentration (Cc ~ 0.1 µM) than the pointed (-) end (Cc ~ 0.6 µM), allowing preferential growth at the plus end.
Part B

Q26. Which chromosome architecture element keeps sister chromatids securely paired along their length until the onset of anaphase?

  • A) Condensin
  • B) Cohesin
  • C) Kinetochore
  • D) Shugoshin
Correct Answer: B
Explanation: Cohesin ring complexes hold sister chromatids together from S-phase until they are proteolytically cleaved by separase at the metaphase-anaphase transition.
Part B

Q27. Activation of the maturation promoting factor (MPF / Cyclin B-Cdk1 complex) requires the specific enzymatic removal of inhibitory phosphates by which phosphatase?

  • A) Cdc25
  • B) Wee1
  • C) CAK
  • D) PP2A
Correct Answer: A
Explanation: Wee1 kinase places inhibitory phosphates on Thr14 and Tyr15 of Cdk1. Cdc25 phosphatase removes these inhibitory phosphates to activate MPF.
Part B

Q28. The Spindle Assembly Checkpoint (SAC) prevents premature sister chromatid separation by directly inhibiting which enzymatic target?

  • A) Separase
  • B) Anaphase-Promoting Complex / Cyclosome (APC/C)
  • C) Securin
  • D) Aurora B Kinase
Correct Answer: B
Explanation: Unattached kinetochores generate a wait signal via Mad2/Bub3 that inhibits APC/C-Cdc20, preventing securin degradation.
Part B

Q29. Which cell cycle phase contains the crucial restriction point (R-point) beyond which cellular replication proceeds independent of external mitogenic growth signals?

  • A) G1 Phase
  • B) G2 Phase
  • C) S Phase
  • D) Prophase
Correct Answer: A
Explanation: The restriction point occurs in late G1. Once cells pass this point, they are committed to complete S phase and division even if growth factors are removed.
Part B

Q30. Peroxisomes are distinct metabolic organelles containing high levels of which enzyme to neutralize reactive hydrogen peroxide molecules?

  • A) Superoxide dismutase
  • B) Catalase
  • C) Peroxidase
  • D) Acid phosphatase
Correct Answer: B
Explanation: Peroxisomes use oxidases to break down fatty acids, generating H2O2. Catalase converts toxic H2O2 into safe water and oxygen.
Part B

Q31. What type of heterochromatin remains permanently condensed and transcriptionally inactive across all developmental stages in all cell variants of an organism?

  • A) Constitutive Heterochromatin
  • B) Facultative Heterochromatin
  • C) Euchromatin
  • D) Nucleolar Chromatin
Correct Answer: A
Explanation: Constitutive heterochromatin (e.g., centromeres, telomeres) is permanently inactive, whereas facultative heterochromatin can be conditionally activated in specific cell lineages.
Part B

Q32. Polytene chromosomes, frequently analyzed in Drosophila larval salivary glands, form via which altered replication pathway?

  • A) Meiotic asymmetric reduction
  • B) Endoreplication (Endomitosis) without sister chromatid segregation
  • C) Gene amplification through episomal excision
  • D) Kinetochore hyper-condensation
Correct Answer: B
Explanation: Endoreplication involves repeated rounds of DNA synthesis without intervening cell division or chromatid separation, forming massive multi-stranded giant polytene structures.
Part B

Q33. Translocation of proteins into the matrix of mitochondria occurs primarily through the coordinated action of which outer and inner membrane translocon complexes?

  • A) Sec61 and Sec62
  • B) TOM and TIM complexes
  • C) TOC and TIC complexes
  • D) Tat and Sec systems
Correct Answer: B
Explanation: Proteins are threaded through the Translocase of Outer Membrane (TOM) and Translocase of Inner Membrane (TIM) to enter the mitochondrial matrix.
Part B

Q34. Which basic protein component forms the structural octameric core around which eukaryotic genomic DNA wraps to yield a nucleosome bead?

  • A) Histone H1
  • B) Histones H2A, H2B, H3, and H4
  • C) Non-histone chromosomal proteins
  • D) Lamin A and Lamin C
Correct Answer: B
Explanation: The nucleosome core consists of two copies each of histones H2A, H2B, H3, and H4. Histone H1 acts as the linker histone outside the core.
Part B

Q35. The degradation of targeted misfolded proteins within the eukaryotic cytoplasm is carried out by which complex machinery?

  • A) Lysosome
  • B) 26S Proteasome
  • C) Autophagosome
  • D) Peroxisome
Correct Answer: B
Explanation: Cytoplasmic polyubiquitinated proteins are recognized, unfolded, and degraded inside the cylindrical 26S proteasome chamber.

PART C: Subject Analytical (Unit 2 Higher-Order Problems)

(5 Marks each. Negative marking: 1.0)

Part C

Q36. Researchers isolate a mutant mammalian cell line displaying a defective SRP (Signal Recognition Particle) receptor that can bind SRP but fails to interact effectively with the Sec61 translocon. Which cellular phenotype would you expect to observe in this mutant line?

  • A) ER-destined proteins will accumulate normally within the ER lumen but lack core glycosylation.
  • B) Ribosomes synthesizing secretory proteins will stall or release cargo into the cytosol, reducing ER protein import.
  • C) Retrograde transport of COPI vesicles will accelerate to compensate for import failures.
  • D) Nuclear import pathways will become constitutively active due to SRP mislocalization.
Correct Answer: B
Explanation: The SRP receptor hands off the signal sequence/ribosome complex to the translocon. If this hand-off fails, co-translational translocation into the ER lumen cannot proceed, causing targeted proteins to be synthesized or released into the cytosol.
Part C

Q37. You are measuring the phase transition profile of artificial lipid vesicles. If you significantly increase the relative concentration of long-chain saturated fatty acids while decreasing cholesterol concentrations at high environmental temperatures, what will be the collective effect on membrane transition temperature (Tm) and fluidity?

  • A) The Tm will increase significantly, and membrane fluidity will drop.
  • B) The Tm will drop, and membrane fluidity will rise.
  • C) Fluidity will increase because saturated chains display low packing efficiency.
  • D) Both parameters will remain unchanged.
Correct Answer: A
Explanation: Long-chain saturated fatty acids pack tightly together due to straight acyl tails, increasing hydrophobic interactions. This raises the transition temperature (Tm) required to melt the membrane, thereby decreasing fluidity.
Part C

Q38. A conditional yeast mutant exhibits a defective Sar1 GEF (Sec12). When shifting this strain to a non-permissive restrictive temperature, what cellular consequence will occur within the secretory pathway?

  • A) Clathrin lattices will fail to assemble at the plasma membrane.
  • B) COPIII coat assembly cannot initiate on the ER membrane, blocking anterograde cargo transport.
  • C) Soluble proteins will leak continuously out into the extracellular matrix via unregulated fusion.
  • D) Nuclear envelope breakdown will take place prematurely during G1 phase.
Correct Answer: B
Explanation: Sec12 is the GEF that exchanges GDP for GTP on Sar1. Active Sar1-GTP exposes an amphipathic helix to insert into the ER membrane, recruiting Sec23/24 to begin COPII vesicle formation. Without it, ER export stops completely.
Part C

Q39. To determine if a protein behaves as an integral membrane component or a peripheral membrane protein, scientists treat isolated cellular fractions with a high-salt buffer containing 1M NaCl. What result diagnoses the protein as a peripheral membrane protein?

  • A) The protein remains embedded within the hydrophobic lipid core fraction.
  • B) The protein is released into the soluble aqueous supernatant fraction.
  • C) The protein undergoes immediate proteolysis and disappears.
  • D) The protein binds covalently to nearby phospholipids.
Correct Answer: B
Explanation: Peripheral membrane proteins associate with the membrane surface via electrostatic interactions and hydrogen bonds, which are easily disrupted by high ionic strength (high salt), releasing them into solution. Integral proteins require detergents to solubilize.
Part C

Q40. During an in vitro cell-free translation experiment, a secretory protein containing an amino-terminal ER signal sequence is translated using pure ribosomes and amino acids. If microsomes are added to the tube only AFTER translation has fully concluded, what will be the fate and state of the protein?

  • A) The protein will be post-translationally imported and glycosylated.
  • B) The protein will remain outside the microsomes in the solution and retain its signal sequence.
  • C) The protein will immediately fold inside the microsomes via specialized chaperone assistance.
  • D) The signal peptidase will release into the solution to cleave the peptide sequence.
Correct Answer: B
Explanation: In mammals, translocation into the ER is strictly co-translational. Once translation finishes, the protein folds into a conformation that cannot pass through the narrow translocon pore, meaning it stays outside the microsomes.
Part C

Q41. You introduce a non-hydrolyzable structural analogue of GTP (GTP-gamma-S) into a cell-free assembly system containing purified tubulin heterodimers. What specific feature of microtubule dynamics will you observe in this mix?

  • A) Microtubules will depolymerize immediately due to rapid protofilament peeling.
  • B) Microtubules will grow continuously without undergoing dynamic instability or catastrophe.
  • C) Treadmilling rates will accelerate by tenfold.
  • D) Tubulin subunits will fail to assemble into linear protofilaments.
Correct Answer: B
Explanation: Hydrolysis of GTP to GDP is what destabilizes the microtubule lattice and triggers catastrophe. If tubulin incorporates non-hydrolyzable GTP-gamma-S, the stable GTP cap is never lost, allowing continuous growth without disassembly.
Part C

Q42. A line of human fibroblasts contains a mutant variant of the Rb (Retinoblastoma) protein that cannot be phosphorylated by G1-Cdk or G1/S-Cdk complexes. What will be the cell cycle proliferation status of these mutant cells?

  • A) The cells will arrest permanently in the G1 phase, failing to enter S phase.
  • B) The cells will undergo continuous, uncontrolled cell divisions.
  • C) The cells will transition through S phase rapidly but arrest at metaphase.
  • D) The cells will skip G1 phase completely during subsequent cycles.
Correct Answer: A
Explanation: Unphosphorylated Rb binds and inhibits the E2F transcription factor. Cdk phosphorylation normalerweise releases Rb from E2F to turn on S-phase genes. If Rb cannot be phosphorylated, it permanently represses E2F, causing G1 arrest.
Part C

Q43. Cytochalasin D is a drug that binds specifically to the (+) barbed ends of actin microfilaments and prevents subunit addition. If you add Cytochalasin D to a system undergoing active actin filament treadmilling, what structural change will take place over time?

  • A) The filaments will grow rapidly from their minus ends.
  • B) The filaments will shorten and eventually depolymerize from their minus ends.
  • C) Filament length will remain locked in a stable steady state.
  • D) Filaments will branch extensively due to Arp2/3 hyper-activation.
Correct Answer: B
Explanation: Treadmilling relies on subunit addition at the plus end and loss at the minus end. Blocking the plus end with Cytochalasin D while disassembly continues at the minus end causes the filaments to shorten and shrink away.
Part C

Q44. A patient presents with a rare genetic disorder characterized by severe muscle weakness and abnormal intracellular accumulation of long-chain fatty acids. Structural analysis reveals a defect in importing matrix enzymes containing a carboxy-terminal SKL signal sequence. Which organelle is malfunctioning in this patient?

  • A) Mitochondria
  • B) Peroxisome
  • C) Lysosome
  • D) Smooth Endoplasmic Reticulum
Correct Answer: B
Explanation: The C-terminal tripeptide Ser-Lys-Leu (SKL) is the classic Peroxisomal Targeting Signal 1 (PTS1), recognized by Pex5 import factors. Defects here cause peroxisomal import failures like those seen in Zellweger spectrum disorders.
Part C

Q45. During an experimental assay, you microinject a large dose of constitutively active Ran-GEF (RCC1) directly into the cytoplasmic compartment of a eukaryotic cell. How will this manipulation alter nuclear transport mechanics?

  • A) Nuclear import will accelerate because importins will bind cargo more tightly in the cytoplasm.
  • B) Nuclear import will collapse because importin-cargo complexes will prematurely dissociate in the cytoplasm.
  • C) Exportin vectors will fail to bind cargo within the nuclear space.
  • D) The nuclear pore complexes will completely disassemble.
Correct Answer: B
Explanation: RanGTP promotes cargo release from importin. Ran-GEF converts RanGDP to RanGTP. Normally, Ran-GEF is restricted to the nucleus so that release only happens there. Flooding the cytoplasm with RanGTP causes importins to drop their cargo before ever reaching the nucleus.
Part C

Q46. Digitonin permeabilizes plasma membranes by extracting cholesterol molecules but leaves inner organelle systems largely intact. If you treat cells with digitonin alongside a cytoplasmic protease, which protein pool will remain completely protected from digestion?

  • A) Cytosolic peripheral structural proteins
  • B) Plasma membrane intracellular domains
  • C) Soluble luminal enzymes residing inside the ER and Golgi apparatus
  • D) Cytoskeletal actin networks
Correct Answer: C
Explanation: Digitonin punctures the outer plasma membrane but spares organelle membranes, keeping lumenal ER and Golgi enzymes sequestered away from the introduced cytosolic protease.
Part C

Q47. Cell fusion experiments bring together a mammalian cell in G1 phase with another cell in M phase. What immediate structural transition happens within the G1 nucleus after cytoplasmic fusion?

  • A) The G1 nucleus finishes replication instantly before entering division.
  • B) The G1 nucleus undergoes premature chromosome condensation (PCC) without duplicating its DNA.
  • C) The M phase nucleus is forced backward into G1 interphase.
  • D) Both nuclei fuse together into a stable tetraploid interphase nucleus.
Correct Answer: B
Explanation: Active MPF (Cyclin B-Cdk1) from the M-phase cell diffuses throughout the shared cytoplasm, driving the G1 chromatin to condense prematurely into single-chromatid threads.
Part C

Q48. You treat dividing animal cells with Taxol (Paclitaxel), a drug that binds and stabilizes microtubule polymers against depolymerization. At which precise phase of the mitotic cycle will these cells arrest?

  • A) G1/S boundary
  • B) Metaphase-to-Anaphase transition point
  • C) Telophase furrowing stage
  • D) S-phase completion checkpoint
Correct Answer: B
Explanation: Taxol prevents spindle microtubule dynamics, preventing proper chromosome alignment and tension sensing at the kinetochores. This leaves the Spindle Assembly Checkpoint (SAC) permanently active, blocking anaphase entry.
Part C

Q49. If a eukaryotic gene encodes a protein containing an amino-terminal ER signal sequence, an internal nuclear localization signal (NLS), and a carboxy-terminal KDEL retrieval sequence, what will be the final steady-state destination of this protein?

  • A) Nucleoplasm
  • B) Lumen of the Endoplasmic Reticulum
  • C) Extracellular secretome
  • D) Cytoplasmic proteasome target list
Correct Answer: B
Explanation: The N-terminal ER signal sequence acts co-translationally, routing the ribosome-peptide complex to the ER before the downstream NLS is ever translated or exposed to the cytosol. Once inside the ER lumen, the C-terminal KDEL tag ensures its retention there via retrograde retrieval pathways.
Part C

Q50. A unique line of mutant yeast cells lacks functional Sec17 and Sec18 proteins (homologues of mammalian alpha-SNAP and NSF). Which step of vesicle trafficking will be blocked in these cells?

  • A) Budding of coat-protected vesicles from the donor membrane surface.
  • B) Initial motor-driven transport along microtubule tracks.
  • C) Tethering of vesicles to the recipient membrane target complex.
  • D) Dissociation and recycling of cis-SNARE pairs following membrane fusion.
Correct Answer: D
Explanation: NSF (Sec18) and alpha-SNAP (Sec17) use the energy of ATP hydrolysis to untangle and separate tightly bundled cis-SNARE complexes after a fusion event, allowing the individual SNARE proteins to be recycled for future transport rounds.

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