401. Examine the following cellular records.
| Record | Genetic organisation | Ribosomes | Internal organisation |
|---|
| P | Circular DNA in a nucleoid | \(70S\) | No typical membrane-bound organelles |
| Q | Chromosomes inside a nucleus | \(80S\) in cytoplasm | Extensive compartmentalisation |
| R | Circular DNA inside a double-membrane organelle | \(70S\) within the organelle | Occurs inside a eukaryotic cell |
| S | Chromosomes inside a nucleus | \(80S\) | Filament-hook-basal-body flagellum |
The record requiring correction is:
ⓐ. P
ⓑ. Q
ⓒ. R
ⓓ. S
Correct Answer: S
Explanation: Records P, Q and R are internally consistent. P describes a prokaryotic cell through its nucleoid, \(70S\) ribosomes and absence of typical membrane-bound organelles. Q represents a eukaryotic cell with nuclear chromosomes, cytoplasmic \(80S\) ribosomes and extensive compartmentalisation. R can describe a mitochondrion or chloroplast inside a eukaryotic cell, since these double-membrane organelles contain circular DNA and \(70S\) ribosomes. Record S mixes a eukaryotic nucleus and \(80S\) cytoplasmic ribosomes with the bacterial filament-hook-basal-body flagellar structure. A eukaryotic flagellum instead possesses a membrane-covered microtubular axoneme. S must be corrected to make the appendage consistent with the rest of the cellular organisation. Replacing the bacterial flagellar description with a \(9+2\) axoneme would make record S internally consistent. Record S is inconsistent because a eukaryotic nucleus and cytosolic \(80S\) ribosomes cannot be combined with the filament-hook-basal-body flagellar plan characteristic of bacteria.
402. An unknown cell is disrupted and analysed. No nuclear envelope, ER or Golgi apparatus is detected. The genetic fraction contains one principal circular DNA molecule and several small circular DNA molecules. The cytoplasmic fraction contains \(70S\) ribosomes, and the locomotory structure separates into a filament, hook and basal body. The strongest inference is:
ⓐ. The sample is a prokaryotic cell containing genomic DNA and plasmids
ⓑ. The sample is a eukaryotic cell whose nucleus was lost during disruption
ⓒ. The sample is a plant cell containing damaged plastids
ⓓ. The sample is an animal cell containing a centrosome
Correct Answer: The sample is a prokaryotic cell containing genomic DNA and plasmids
Explanation: The evidence converges on a prokaryotic organisation. Absence of a nuclear envelope and endomembrane organelles is accompanied by a principal circular chromosome located outside a nucleus. The additional small circular DNA molecules fit plasmids, which occur separately from the main bacterial genome. Cytoplasmic \(70S\) ribosomes provide another prokaryotic marker, while the filament-hook-basal-body arrangement identifies a bacterial flagellum. Interpreting the observations as accidental loss of a eukaryotic nucleus would not explain the plasmid-like DNA, ribosome class and bacterial appendage structure. The experiment supports identification of a prokaryotic cell rather than a damaged plant or animal cell. Several independent observations converge on the same cellular plan, making accidental preparation damage an inadequate explanation. Cell classification and prediction require several compatible features rather than one isolated resemblance. The combined profile is characteristically bacterial and prokaryotic.
403. Use the two cellular arrangements described below. Cell P has an outer glycocalyx, a wall, a plasma membrane, a nucleoid, plasmids, \(70S\) ribosomes and pili. Cell Q has only a plasma membrane at its outer living boundary, a membrane-bound nucleus, mitochondria, ER and a membrane-covered flagellum with a \(9+2\) axoneme. Which interpretation is most accurate?
ⓐ. P and Q possess the same organisational plan but differ only in their surface coverings
ⓑ. P is eukaryotic because it possesses both hereditary DNA and protein-synthesising ribosomes
ⓒ. P is prokaryotic and Q eukaryotic; their flagella have different structural plans
ⓓ. Q is prokaryotic because its membrane-covered flagellum is used for locomotion
Correct Answer: P is prokaryotic and Q eukaryotic; their flagella have different structural plans
Explanation: Cell P displays a bacterial envelope, nucleoid, plasmids, \(70S\) ribosomes and pili, forming a coherent prokaryotic profile. Possession of DNA and ribosomes does not make a cell eukaryotic, since both cellular organisations contain these basic components. Cell Q has a nuclear envelope, mitochondria and endoplasmic reticulum, which establish eukaryotic compartmentalisation. Its flagellum contains a membrane-covered \(9+2\) axoneme, unlike the filament-hook-basal-body structure of bacterial flagella. Shared locomotory function does not imply structural identity. The figure must be interpreted through several internal and boundary features rather than through the general presence of DNA, ribosomes or a flagellum. The complete organisation of each cell, not locomotory function alone, fixes both the cell type and the structural distinction between its flagellum. The conclusion integrates envelope, genetic region, ribosome class, organelles and appendage ultrastructure, so no single shared feature controls the identification.
404. Cell P possesses a cellulose-containing wall, plasmodesmata, chloroplasts and a large central vacuole. Cell Q lacks a wall and plastids but possesses centrioles and numerous microvilli. Both cells contain mitochondria, ER, Golgi apparatus and ribosomes. The most accurate conclusion is:
ⓐ. P is prokaryotic, while Q is eukaryotic
ⓑ. P is a typical plant cell, while Q is a typical animal cell
ⓒ. P and Q are both plant cells with different surface modifications
ⓓ. P and Q are both animal cells because they contain mitochondria
Correct Answer: P is a typical plant cell, while Q is a typical animal cell
Explanation: Cell P combines a cellulose-containing wall, plasmodesmata, chloroplasts and a large central vacuole, a characteristic set of plant-cell features. Cell Q lacks the wall and plastids but possesses centrioles and numerous microvilli, supporting a typical animal-cell pattern. Mitochondria, endoplasmic reticulum, Golgi apparatus and ribosomes occur in both cells because they belong to the shared eukaryotic organisation and cannot serve as exclusive identifiers. The classification therefore depends on the contrasting feature combinations rather than on one organelle considered alone. P is not prokaryotic because it contains the organelles and cellular organisation of a plant eukaryote. Q is not a plant cell merely because it shares mitochondria and ribosomes with P. The complete evidence identifies P as a typical plant cell and Q as a typical animal cell.
405. A cell diagram shows an outer rigid boundary, an inner plasma membrane, chloroplasts, mitochondria, a nucleus and a large vacuole enclosed by a tonoplast. Channels cross the outer boundary to connect the cell with its neighbours. The cell is best identified as:
ⓐ. A typical plant cell with plasmodesmata
ⓑ. A typical animal cell with microvilli
ⓒ. A bacterium with pili and gas vacuoles
ⓓ. A fungal cell containing chloroplasts
Correct Answer: A typical plant cell with plasmodesmata
Explanation: The outer rigid boundary is a cell wall, while the large tonoplast-bound vacuole and chloroplasts strongly support a plant-cell identity. Mitochondria and a nucleus confirm eukaryotic organisation but are shared with animal and fungal cells. The channels traversing the wall and connecting neighbouring cytoplasms are plasmodesmata, another characteristic feature of plant tissues. Microvilli project from animal-cell surfaces and do not cross a wall to establish cytoplasmic continuity. Bacteria lack chloroplasts, mitochondria and a membrane-bound nucleus, while fungi possess walls but do not contain chloroplasts. The combined boundary, organelle and intercellular-channel evidence identifies a typical plant cell.
406. Examine the following plant-animal comparison records.
| Record | Feature | Proposed distribution |
|---|
| P | Plasma membrane | Shared by plant and animal cells |
| Q | Large central vacuole | Characteristic of a typical mature plant cell |
| R | Centrioles | Commonly associated with a typical animal cell |
| S | Mitochondria | Exclusive to animal cells |
The record requiring correction is:
ⓐ. P
ⓑ. Q
ⓒ. R
ⓓ. S
Correct Answer: S
Explanation: The plasma membrane is shared by plant and animal cells, so P is valid. A prominent central vacuole is characteristic of the typical mature plant-cell plan, supporting Q. Centrioles commonly provide a distinguishing cue for typical animal cells, making R acceptable. Record S is incorrect because mitochondria occur in both plant and animal cells and support aerobic ATP production in each. Treating a shared organelle as an exclusive animal feature would misclassify plant cells. The comparison must separate common eukaryotic structures from features that are especially useful for distinguishing the two typical cellular patterns. Mitochondria belong to the shared category rather than the animal-exclusive category. The correction is specifically to classify mitochondria as shared organelles, not to alter the valid plant-animal distinctions in the other rows. Row S is the mismatch: mitochondria occur in both typical plant and animal cells, so they cannot serve as an animal-cell-exclusive feature.
407. A typical plant cell is treated with enzymes that remove its wall. Its plasma membrane, nucleus, mitochondria and large vacuole remain intact, but the particular tissue examined contains no developed chloroplasts. Which conclusion is most justified?
ⓐ. the treated cell should be reclassified as prokaryotic because the wall is absent
ⓑ. the treated cell should be reclassified as animal because developed chloroplasts are absent
ⓒ. Removing the wall does not alter the cell's plant origin or eukaryotic organisation
ⓓ. retention of a large vacuole is sufficient to classify the treated cell as fungal
Correct Answer: Removing the wall does not alter the cell's plant origin or eukaryotic organisation
Explanation: Cell classification cannot be changed merely by experimentally removing one structure. The treated cell retains a membrane-bound nucleus, mitochondria and a large vacuolar compartment, so it remains eukaryotic. Absence of developed chloroplasts is not sufficient to identify an animal cell. Many plant tissues contain non-green cells or plastids specialised for functions other than photosynthesis. The wall was removed by treatment rather than being naturally absent from the original cellular plan, and a large vacuole alone does not establish fungal identity. The evidence supports a wall-less plant-derived cell whose experimental alteration has not erased its origin or its remaining eukaryotic organisation. The decisive reasoning combines experimental history with retained structures, rather than treating absence of one visible feature as proof of a different natural cell type.
408. Arrange the following events in the pathway of a newly synthesised secretory protein.
P. Translation on a ribosome attached to rough ER
Q. Entry into an ER-derived transport vesicle
R. Arrival at the Golgi cis face
S. Modification and movement toward the Golgi trans face
T. Delivery in a vesicle to the plasma membrane
ⓐ. P → Q → R → S → T
ⓑ. R → P → Q → T → S
ⓒ. P → S → R → Q → T
ⓓ. Q → P → R → T → S
Correct Answer: P → Q → R → S → T
Explanation: The pathway begins when an RER-associated ribosome synthesises the secretory polypeptide. The product enters the ER pathway and is packaged into a transport vesicle that moves toward the Golgi apparatus. Incoming ER-derived vesicles fuse with the convex cis face, the receiving side of the Golgi stack. The protein is modified and sorted as it progresses through the cisternae toward the concave trans face. Vesicles leaving the trans side then carry the processed product toward the plasma membrane for secretion. Each stage depends on completion of the previous one, so the order reflects the directional organisation of the endomembrane system rather than a memorised list of organelles. The order follows the route of the secretory product from rough endoplasmic reticulum through Golgi cisternae to the plasma membrane. This directional sequence also explains why a block at any earlier stage reduces delivery to every later compartment.
409. Assertion: Blocking vesicle release from the Golgi trans face can cause processed secretory products to accumulate within the Golgi apparatus.
Reason: The trans face is the principal site at which ER-derived vesicles first enter the Golgi stack.
ⓐ. Both Assertion and Reason are true, and Reason correctly explains the Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain the Assertion
ⓒ. Assertion is true, but Reason is false; the Reason cannot explain the Assertion
ⓓ. Assertion is false, but Reason is true; the Reason cannot explain the Assertion
Correct Answer: Assertion is true, but Reason is false; the Reason cannot explain the Assertion
Explanation: Golgi polarity supports the Assertion: blocking trans-face dispatch can retain processed products within the pathway. The trans face is the releasing side of the Golgi apparatus, so a block in vesicle formation or departure there can prevent processed products from leaving and lead to their accumulation within the Golgi pathway. The Reason fails its truth condition because ER-derived vesicles normally arrive at the convex cis face, not the trans face. Material then progresses toward the concave trans side for sorting and dispatch. Golgi polarity separates two steps: entry occurs at the cis face, while exit occurs at the trans face. Failure of the exit step explains accumulation, but the Reason assigns the receiving function to the wrong face. The Assertion concerns failure of dispatch, whereas the false Reason incorrectly assigns the cis-face receiving role to the trans face.
410. A graph plots radioactive label intensity against time for four locations after a short pulse labels a newly synthesised secretory protein. The RER curve peaks first, the cis-Golgi curve peaks next, the trans-Golgi curve peaks later, and the extracellular-medium curve rises last. The trend most strongly supports:
ⓐ. independent synthesis of the labelled protein at each recorded cellular location
ⓑ. sequential transfer from RER through cis- and trans-Golgi before secretion
ⓒ. reverse uptake from extracellular medium through Golgi cisternae into RER
ⓓ. direct RER release while Golgi labelling reflects separate protein synthesis
Correct Answer: sequential transfer from RER through cis- and trans-Golgi before secretion
Explanation: Time is plotted on the horizontal axis, while radioactive label intensity records the relative presence of the newly synthesised protein at each location. The first peak occurs in the rough ER, where ribosomes associated with the membrane synthesise proteins entering the secretory pathway. The later cis-Golgi peak indicates arrival at the receiving face of the Golgi apparatus. Label subsequently peaks at the trans face, showing movement through the Golgi stack toward its releasing side. The extracellular signal appears last, placing secretion after Golgi processing and transit. This ordered displacement of the same radioactive label supports directional transfer rather than independent synthesis at every location. It also excludes reverse uptake and direct release from RER without Golgi participation. The peak sequence therefore traces the secretory route from RER to cis-Golgi, trans-Golgi and the extracellular medium.
411. A treatment selectively prevents transport vesicles from budding from the ER. Secretory proteins are still synthesised, ATP remains available and Golgi cisternae retain their normal structure. Labelled secretory protein accumulates in the RER, while little label reaches the Golgi or extracellular medium. The strongest inference is:
ⓐ. Translation has stopped at RER-associated ribosomes
ⓑ. Golgi modification is occurring normally without incoming material
ⓒ. The plasma membrane has lost selective permeability
ⓓ. vesicular transport from ER to Golgi has been blocked
Correct Answer: vesicular transport from ER to Golgi has been blocked
Explanation: Continued synthesis and accumulation of labelled protein in the RER show that translation remains active. The decisive intervention is failure of vesicle budding from the ER. Without these carriers, secretory products cannot reach the Golgi cis face, so Golgi labelling and later extracellular appearance decline. Normal Golgi structure does not restore function when the required incoming transport step is absent. The experiment also provides no evidence for a primary plasma-membrane permeability defect. The treatment, accumulation site and downstream loss form a consistent pathway diagnosis: synthesis occurs, but export from the ER and delivery to the Golgi have failed. The preserved Golgi structure cannot compensate because the labelled protein never leaves the ER to reach the cis face. Label distribution thus locates the defect more precisely than the mere observation that secretion has declined.
412. Organelle P has a double membrane, an inner membrane folded into cristae and a matrix containing circular DNA and \(70S\) ribosomes. Organelle Q also has a double membrane and contains circular DNA and \(70S\) ribosomes, but its internal membranes form thylakoids and grana containing chlorophyll. P and Q are respectively:
ⓐ. Chloroplast and mitochondrion
ⓑ. Lysosome and chloroplast
ⓒ. Mitochondrion and chloroplast
ⓓ. Mitochondrion and Golgi apparatus
Correct Answer: Mitochondrion and chloroplast
Explanation: P is a mitochondrion. Its inner membrane forms cristae, and its matrix contains circular DNA and \(70S\) ribosomes. Q is a chloroplast, identified by thylakoids, grana and chlorophyll in addition to its circular DNA and \(70S\) ribosomes. Both organelles are double-membrane and semiautonomous, but their specialised internal membranes and principal functions differ. Cristae support mitochondrial aerobic energy production, while thylakoid membranes organise photosynthetic pigments and light-energy capture. A lysosome and Golgi apparatus lack the internal circular DNA and ribosome system described. The correct identification requires recognition of shared semiautonomous features followed by discrimination using organelle-specific architecture.
413. Match each feature in Column I with the organelle relation in Column II. A Column II entry is used once.
| Column I | Column II |
|---|
| P. Cristae | 1. Shared by mitochondria and chloroplasts |
| Q. Grana | 2. Mitochondrial inner-membrane folds |
| R. Circular DNA and \(70S\) ribosomes | 3. Chloroplast thylakoid stacks |
| S. Aerobic ATP production | 4. Principal mitochondrial function |
ⓐ. P-3, Q-2, R-4, S-1
ⓑ. P-2, Q-3, R-1, S-4
ⓒ. P-4, Q-1, R-2, S-3
ⓓ. P-1, Q-4, R-3, S-2
Correct Answer: P-2, Q-3, R-1, S-4
Explanation: Cristae are folds of the mitochondrial inner membrane, so P matches 2. Grana are stacks of chloroplast thylakoids, giving Q-3. Circular DNA and \(70S\) ribosomes occur in both mitochondria and chloroplasts and support their semiautonomous description, giving R-1. Aerobic ATP production is the principal mitochondrial function listed, giving S-4. The mapping separates shared genetic machinery from organelle-specific membrane architecture and activity. Similarity in DNA form and ribosome class does not make the two organelles functionally identical. Their internal membranes provide the major structural distinction between mitochondrial respiration and chloroplast photosynthesis. The four features identify different organelles by specialised structure. The correct mapping therefore combines two organelle-specific structures, one shared semiautonomous feature and one mitochondrial function.
414. In a green plant cell, a treatment destroys thylakoid membranes and chlorophyll but leaves mitochondrial cristae, matrix enzymes and oxygen supply initially normal. The most likely immediate pattern is:
ⓐ. Light capture declines, while mitochondrial ATP production may continue briefly
ⓑ. Mitochondrial aerobic ATP production stops first, while light capture remains normal
ⓒ. Both organelles lose their genetic material even though only thylakoids were damaged
ⓓ. Stromal enzymes fully replace chlorophyll-based light capture immediately
Correct Answer: Light capture declines, while mitochondrial ATP production may continue briefly
Explanation: Thylakoid membranes contain chlorophyll and organise the chloroplast's light-capturing system. Their destruction causes an immediate sharp decline in photosynthetic light-energy capture. Mitochondrial cristae, matrix enzymes and oxygen supply remain normal, allowing aerobic ATP production to continue initially. The treatment distinguishes the functions of two semiautonomous organelles rather than producing a general cellular collapse. Shared possession of circular DNA and \(70S\) ribosomes does not make chloroplasts and mitochondria respond identically to damage of a chloroplast-specific membrane. The immediate pattern is reduced light capture with preserved mitochondrial respiration under the supplied conditions. Chloroplast photochemical organisation and mitochondrial aerobic respiration occupy different membranes, allowing the stated selective effect. The selective chloroplast injury separates photochemical failure from initially preserved mitochondrial respiration. The word initially is important because prolonged loss of photosynthesis could later alter substrate supply and whole-cell metabolism.
415. Assertion: Mitochondria and chloroplasts are completely autonomous cells living inside eukaryotic cells.
Reason: Both organelles contain circular DNA, RNA and \(70S\) ribosomes, yet they still depend substantially on the surrounding cell.
ⓐ. Both Assertion and Reason are true, and Reason correctly explains the Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain the Assertion
ⓒ. Assertion is true, but Reason is false; the Reason cannot explain the Assertion
ⓓ. Assertion is false, but Reason is true; the Reason cannot explain the Assertion
Correct Answer: Assertion is false, but Reason is true; the Reason cannot explain the Assertion
Explanation: The Assertion is false: mitochondria and chloroplasts are organelles, not independently living cells within the eukaryotic cytoplasm. They are described as semiautonomous rather than completely autonomous. The Reason is true and captures both sides of that qualification. Circular DNA, RNA and \(70S\) ribosomes permit limited internal genetic expression and protein synthesis, but many organellar components and regulatory requirements depend on the wider cell. Their ability to divide and synthesise some proteins establishes partial independence without eliminating cellular dependence. The Reason contradicts the claim of complete autonomy and supplies the evidence for the more accurate semiautonomous description. Internal genetic machinery supports semiautonomy, but dependence on the host cell makes the Assertion false.
416. Which classification is correct?
ⓐ. Endomembrane system: mitochondria and chloroplasts; non-membranous structures: lysosomes and vacuoles
ⓑ. Endomembrane: ER, Golgi, lysosomes and vacuoles; non-membranous: ribosomes, centrioles and nucleolus
ⓒ. Semiautonomous organelles: Golgi apparatus and lysosomes; microbodies: ribosomes and centrioles
ⓓ. Endomembrane system: cell wall and plasma membrane; non-membranous structures: mitochondria and chloroplasts
Correct Answer: Endomembrane: ER, Golgi, lysosomes and vacuoles; non-membranous: ribosomes, centrioles and nucleolus
Explanation: The endomembrane system is defined through functional coordination among ER, Golgi apparatus, lysosomes and vacuoles. Ribosomes, centrioles and the nucleolus lack surrounding membranes, although each has an organised structure and important cellular function. Mitochondria and chloroplasts are double-membrane semiautonomous organelles and are excluded from the endomembrane system despite their location within the cytoplasm. Microbodies are minute enzyme-containing membrane-bound vesicles, not ribosomes or centrioles. The cell wall is an extracellular rigid layer rather than a component of the endomembrane system. Correct classification requires attention to membrane status and functional coordination, not merely to whether a structure occurs inside a eukaryotic cell. A surrounding membrane alone is insufficient for endomembrane membership, while lack of a membrane is sufficient to exclude the named non-membranous structures.
417. Two enzyme-containing vesicles occur in the same cytoplasm. Vesicle P is Golgi-derived, contains acidic hydrolases and participates in intracellular digestion. Vesicle Q is a minute membrane-bound microbody containing various enzymes but is not part of the coordinated ER-Golgi-lysosome-vacuole pathway. Which conclusion is most accurate?
ⓐ. P is a lysosome and Q a microbody; only the Golgi-derived P is endomembrane-related
ⓑ. P and Q are both lysosomes because membrane enclosure and enzyme content define endomembrane membership
ⓒ. P is a lysosome and Q a microbody, so both are derived from the Golgi apparatus
ⓓ. P and Q are both microbodies because neither participates in controlled intracellular digestion
Correct Answer: P is a lysosome and Q a microbody; only the Golgi-derived P is endomembrane-related
Explanation: P is identified as a lysosome by its Golgi-related origin, acidic hydrolytic enzymes and digestive function. Q fits the biological description of a microbody: a minute membrane-bound enzyme-containing vesicle. Both occur within the cytoplasm and possess membranes, but endomembrane membership depends on functional coordination, not on location or membrane enclosure alone. Lysosomes participate in the ER-Golgi-associated system, whereas microbodies are classified separately. Enzyme content is also insufficient to make every vesicle a lysosome, since different compartments contain different enzyme sets and perform different functions. The case demonstrates why boundary type, origin and pathway relation must be considered together. Golgi origin and acidic digestion identify the lysosome, whereas enzyme content and separate classification identify the microbody. The decisive distinction is functional coordination. P participates in the Golgi-linked digestive pathway, whereas Q is classified separately despite sharing membrane enclosure and enzyme content. A shared cytoplasmic location cannot override those different origins and roles.
418. Which classification correctly separates endomembrane components, semiautonomous organelles, non-membranous structures and microbodies?
ⓐ. Endomembrane: mitochondria, chloroplasts, ribosomes; semiautonomous: ER, Golgi, vacuoles; non-membranous: lysosomes, wall; microbodies: centrioles
ⓑ. Endomembrane: wall, plasma membrane, ribosomes; semiautonomous: lysosomes, vacuoles, Golgi; non-membranous: chloroplasts, mitochondria; microbodies: nucleolus
ⓒ. Endomembrane: ribosomes, nucleolus, centrioles; semiautonomous: Golgi, ER, lysosomes; non-membranous: microbodies, vacuoles; microbodies: mitochondria
ⓓ. Endomembrane: ER, Golgi, lysosomes, vacuoles; semiautonomous: mitochondria, chloroplasts; non-membranous: ribosomes, centrioles, nucleolus; microbodies: enzyme-containing vesicles
Correct Answer: Endomembrane: ER, Golgi, lysosomes, vacuoles; semiautonomous: mitochondria, chloroplasts; non-membranous: ribosomes, centrioles, nucleolus; microbodies: enzyme-containing vesicles
Explanation: Endoplasmic reticulum, Golgi apparatus, lysosomes and vacuoles form the coordinated endomembrane system. Mitochondria and chloroplasts are semiautonomous through possession of circular DNA and \(70S\) ribosomes, although they still depend on the cell. Ribosomes, centrioles and the nucleolus lack their own surrounding membranes. Microbodies form a separate category of minute membrane-bound vesicles containing enzymes. These criteria are not interchangeable: endomembrane membership concerns coordinated compartmental function, semiautonomy concerns limited internal genetic and protein-synthetic machinery, and non-membranous classification concerns absence of an enclosing membrane. A microbody is membrane-bound but is not thereby placed in the endomembrane system. Applying each defining criterion consistently keeps the four categories distinct and avoids grouping organelles merely from sharing a cytoplasmic location.
419. A treatment blocks nuclear pores without rupturing the nuclear envelope. RNA export from the nucleus and import of selected cytoplasmic proteins both decline, while mitochondrial ATP production and existing cytoplasmic ribosomes remain initially functional. The most direct diagnosis is:
ⓐ. Selective destruction of mitochondrial cristae
ⓑ. Failure of Golgi trans-face packaging
ⓒ. Impaired nuclear-cytoplasmic exchange
ⓓ. Rupture of the tonoplast
Correct Answer: Impaired nuclear-cytoplasmic exchange
Explanation: Nuclear pores permit regulated movement in both directions across the nuclear envelope. RNA can move from the nucleus to the cytoplasm, while selected proteins move from the cytoplasm into the nucleus. Blocking the pores while preserving the envelope explains the simultaneous decline in both transport directions. Initial preservation of mitochondrial ATP production and existing cytoplasmic ribosomes rules out a primary crista or ribosome defect. Golgi packaging and tonoplast integrity involve different compartments and would not directly produce the paired nuclear transport pattern. The treatment identifies a specific boundary failure: the nucleus remains enclosed, but controlled molecular exchange through its pore system is impaired. Existing cytoplasmic ribosomes may continue translation briefly, but newly exported RNA and imported nuclear proteins are restricted by the blocked pores. The paired import-export defect is diagnostic because both movements use the same pore complexes in opposite directions.
420. A mutant cell shows a sharp decline in aerobic ATP production. Its mitochondrial inner membrane has very few cristae, but chlorophyll-containing thylakoids, Golgi secretion and cytoplasmic translation remain normal. The most strongly supported defect is:
ⓐ. Reduced inner-membrane area caused by sparse mitochondrial cristae
ⓑ. Loss of chloroplast stroma enzymes required for carbon fixation
ⓒ. Failure of protein synthesis on rough-ER-associated ribosomes
ⓓ. Rupture of lysosomal membranes with release of hydrolytic enzymes
Correct Answer: Reduced inner-membrane area caused by sparse mitochondrial cristae
Explanation: The major abnormality lies in the mitochondrial inner membrane, which has lost most of its cristae. Cristae expand the membrane area available for processes associated with aerobic ATP production. Their reduction provides a direct structural explanation for the measured energetic decline. Normal thylakoids and chlorophyll argue against a primary chloroplast defect, while preserved secretion and translation exclude major RER, Golgi or cytoplasmic-ribosome failure. Lysosomal rupture would principally disturb intracellular digestion rather than selectively reduce aerobic ATP output. The evidence links one altered organellar structure with its corresponding function: diminished crista surface limits mitochondrial energy-producing capacity even though other cellular systems remain active. The unchanged systems act as controls within the case: they show that the defect is not a general loss of organelles or protein synthesis but a structure-specific mitochondrial limitation.
