301. A treatment triples the duration of \(G_1\) but does not alter the rate of cellular growth, chromosome number or the later durations of S, \(G_2\) and M phases. Immediately before the delayed cells enter S phase, they are expected to have:
ⓐ. greater cell mass, \(4C\) DNA and twice the chromosome number
ⓑ. greater cell mass, \(2C\) DNA and the original chromosome number
ⓒ. unchanged cell mass, \(2C\) DNA and half the chromosome number
ⓓ. greater cell mass, \(4C\) DNA and the original chromosome number
Correct Answer: greater cell mass, \(2C\) DNA and the original chromosome number
Explanation: \(G_1\) is the interval of growth and normal metabolism before nuclear DNA replication begins. Extending this phase while maintaining the same growth rate allows additional cellular mass to accumulate. Nuclear DNA remains at \(2C\), since the cells have not yet entered S phase. Chromosome number also remains unchanged. A longer \(G_1\) does not itself replicate DNA, create sister chromatids or alter ploidy. The prediction illustrates the difference between continuous or extended cellular growth and stage-restricted DNA synthesis. Only after the delayed cells enter S phase will DNA amount rise from \(2C\) to \(4C\), while chromosome number still remains constant until centromere division. The changed condition lengthens the pre-replication growth interval only. Since growth continues at the same rate, extra time in \(G_1\) increases mass, but DNA remains unreplicated at \(2C\) and chromosome number remains fixed by the unchanged centromere count. The later phase durations do not affect the immediate pre-S prediction.
302. Eight uninucleate cells each undergo one normal karyokinesis. Cytokinesis succeeds in five cells but fails completely in the remaining three. Immediately afterward, the preparation contains:
ⓐ. \(16\) nuclei in \(16\) cellular compartments
ⓑ. \(13\) nuclei in \(16\) cellular compartments
ⓒ. \(13\) nuclei in \(13\) cellular compartments
ⓓ. \(16\) nuclei in \(13\) cellular compartments
Correct Answer: \(16\) nuclei in \(13\) cellular compartments
Explanation: Every starting cell completes karyokinesis, so each of the \(8\) original nuclei produces two daughter nuclei:
\[
8\times2=16\ \text{nuclei}
\]
Cytoplasmic compartments must be counted separately. The \(5\) cells that complete cytokinesis each form two compartments:
\[
5\times2=10\ \text{compartments}
\]
The remaining \(3\) cells fail cytokinesis and therefore remain as \(3\) compartments, each containing two nuclei. The final compartment count is:
\[
10+3=13
\]
Failure of cytokinesis does not undo chromosome segregation or nuclear reconstitution, so those three cells still contribute six daughter nuclei. Karyokinesis determines nuclear-product number, whereas cytokinesis determines whether the products become physically separate cells. The outcome is therefore \(16\) nuclei distributed among \(13\) cytoplasmic compartments, including three binucleate compartments. A binucleate compartment is not equivalent to two cells: it contains two nuclear products within one continuous cytoplasm. Counting nuclei first and compartments second prevents cytokinesis failure from being mistaken for failure of karyokinesis.
303. A species has \(2n=10\). In Cell P, five X-shaped replicated chromosomes move towards each pole while sister centromeres remain intact. In each of two Cells Q, five V-shaped single-chromatid chromosomes move towards each pole with centromeres leading. Which identification is correct?
ⓐ. P is mitotic anaphase, while Q is anaphase I
ⓑ. P is metaphase I, while Q is mitotic metaphase
ⓒ. P is anaphase I, while Q is anaphase II
ⓓ. P is telophase I, while Q is prophase II
Correct Answer: P is anaphase I, while Q is anaphase II
Explanation: In Cell P, each moving unit remains X-shaped and replicated, showing that sister chromatids are still joined. Five such chromosomes move to each pole, representing separation of homologous chromosomes in anaphase I. In Cells Q, centromeres have divided and the moving structures are single-chromatid daughter chromosomes. Their centromere-leading V-shaped orientation is characteristic of anaphase-II movement. Five daughter chromosomes reach each pole in each haploid cell. The comparison distinguishes the two meiotic anaphases through chromosome structure rather than merely poleward direction. Anaphase I moves replicated homologues, whereas anaphase II resolves the sister chromatids retained after the first division. Metaphase II aligns individual replicated chromosomes in a haploid cell, unlike the bivalent alignment of metaphase I. Cell P began with five homologous pairs, so anaphase I sends one replicated chromosome from each pair to each pole. In each anaphase-II cell, the five centromeres split, temporarily producing ten daughter chromosomes in that cell, with five moving towards each pole.
304. During plant-cell cytokinesis, the distance from the centre of the cell to the advancing edge of a partition is plotted against time. The distance rises steadily until the partition meets the lateral walls. Which conclusion best integrates the graph with the later fate of the partition?
ⓐ. A cleavage furrow grows from the surface and later forms the nuclear envelope
ⓑ. A spindle fibre grows outward and later becomes the daughter-cell wall
ⓒ. A cell plate grows outward and contributes to the daughter-separating wall
ⓓ. A metaphase plate expands outward and becomes the middle lamella
Correct Answer: A cell plate grows outward and contributes to the daughter-separating wall
Explanation: The partition begins centrally and its edge moves progressively towards the lateral walls, which is the spatial pattern of plant-cell cytokinesis. The structure is the cell plate, not the metaphase plate. When it reaches the existing walls, it completes cytoplasmic separation and contributes to formation of the new partition between the daughter cells, including the future middle-lamella region. An animal cleavage furrow shows the opposite direction, beginning at the periphery and progressing inward. Spindle fibres organise chromosome movement and do not become wall material. The graph tests both direction and developmental fate: central initiation identifies the mechanism, while contact with the lateral walls marks completion of the separating partition. A plateau indicates no change in the plotted quantity during that interval, while a rise marks the phase in which the quantity is increasing.
305. Two metabolically active adult-cell populations, P and Q, contain \(2C\) DNA and are not dividing. After a replacement signal, P incorporates a labelled DNA precursor and its cell number later doubles. Q remains unlabelled and its cell number stays constant. The strongest inference is:
ⓐ. P re-entered from \(G_0\), whereas Q remained non-proliferative after stimulation
ⓑ. both populations were cycling through S phase before the replacement signal
ⓒ. Q completed S phase but failed to progress through mitosis after stimulation
ⓓ. P entered \(G_2\), whereas Q remained arrested in mitotic metaphase
Correct Answer: P re-entered from \(G_0\), whereas Q remained non-proliferative after stimulation
Explanation: Before stimulation, both populations were metabolically active but non-dividing, a condition compatible with \(G_0\). Population P incorporated the labelled DNA precursor after the replacement signal, showing entry into S phase. Its later increase in cell number confirms completion of proliferation and demonstrates reversible quiescence. Population Q neither incorporated label nor increased in number, so it did not re-enter the cycle under the tested condition. The experiment does not prove that Q can never divide under any circumstance, but it does distinguish its response from that of P. DNA content, labelling and population change together separate active quiescence from cycling interphase and reveal variation in proliferative capacity among adult cell types. A cell returning from \(G_0\) resumes the ordered cycle through \(G_1\); it does not bypass genome duplication and move directly into M phase.
306. During S phase in animal cells, Treatment P permits centriole duplication but prevents nuclear DNA from rising above \(2C\). Treatment Q permits nuclear DNA to reach \(4C\) but prevents centriole duplication. Which conclusion is most strongly supported?
ⓐ. Nuclear DNA replication normally begins only after prophase starts
ⓑ. Centriole duplication is the direct trigger for nuclear DNA replication
ⓒ. The two S-phase events are coordinated but can be disrupted independently
ⓓ. Completion of either event alone guarantees a normal mitotic division
Correct Answer: The two S-phase events are coordinated but can be disrupted independently
Explanation: Both events normally occur during S phase, but the selective treatments show that completion of one does not automatically guarantee completion of the other. In P, centrioles duplicate despite failure of nuclear DNA replication. In Q, nuclear DNA doubles despite failure of centriole duplication. The observations support coordination in timing while also showing experimentally separable outcomes. Neither event alone is sufficient for a normal later division. The genome must be duplicated for equal genetic distribution, while duplicated centriole-associated structures contribute to organisation of the animal-cell mitotic apparatus. The data do not establish that one event directly causes the other, nor do they relocate DNA synthesis to prophase. The experimental design separates two coordinated events by allowing one to proceed while selectively disrupting the other. The conclusion remains limited to the supplied treatment and observations; broader causal claims would require additional evidence.
307. A graph follows one cellular compartment through four successive rounds. Nuclear number rises \(1\rightarrow2\rightarrow4\rightarrow8\rightarrow16\), while the number of cytoplasmic compartments remains \(1\) throughout. The graph most directly represents:
ⓐ. repeated cytokinesis without segregation, producing empty compartments
ⓑ. karyokinesis without cytokinesis, producing a multinucleate syncytium
ⓒ. ordinary mitosis with cytokinesis, producing \(16\) separate cells
ⓓ. meiotic divisions with fertilisation, producing one compartment
Correct Answer: karyokinesis without cytokinesis, producing a multinucleate syncytium
Explanation: The doubling nuclear series shows that each nucleus repeatedly completes nuclear division. The unchanged compartment curve shows that none of those rounds is followed by cytoplasmic separation. Multiple nuclei accumulate in one continuous cytoplasmic mass, producing a syncytial organisation. Four complete mitotic cycles with cytokinesis would increase cell compartments alongside nuclear number, yielding \(16\) separate cells. Cytokinesis without karyokinesis would create cytoplasmic partitions without the observed nuclear multiplication. The graph demonstrates why nuclear number and cell number cannot be treated as equivalent variables. Their curves remain coupled only when karyokinesis is followed by successful cytokinesis. Axes, interval labels and curve shape must be interpreted together before assigning a stage. A change in slope marks a change in rate or process, whereas a horizontal segment represents persistence of the current state.
308. A single uninucleate cell undergoes five successive rounds of karyokinesis without any cytokinesis. Assuming every nucleus divides once in each round, the final condition is:
ⓐ. \(10\) nuclei distributed among \(5\) cells
ⓑ. \(32\) nuclei within one cytoplasmic compartment
ⓒ. \(16\) nuclei within two cytoplasmic compartments
ⓓ. \(32\) nuclei distributed among \(32\) cells
Correct Answer: \(32\) nuclei within one cytoplasmic compartment
Explanation: Every round doubles the number of nuclei. Beginning with one nucleus, five rounds produce:
\[
1\times2^5=32\ \text{nuclei}
\]
No cytokinesis occurs at any stage, so the original cytoplasmic compartment is never divided. All \(32\) nuclei remain within one continuous cellular mass. The calculation models formation of a multinucleate syncytial condition such as that observed during free-nuclear development in liquid coconut endosperm. The number of nuclei reflects repeated karyokinesis, while the number of cellular compartments reflects cytokinesis. Exponential nuclear increase does not imply an equal increase in cell number when cytoplasmic division is absent. The exponent records five nuclear doublings, whereas the compartment count remains fixed since no membrane or cell-wall partition forms. This distinction is essential: karyokinesis changes nuclear number, but cytokinesis is the event that would convert the shared cytoplasm into separate daughter cells. The stated assumptions also require every existing nucleus to divide during each round.
309. A meiocyte is observed at three successive times.
| Time | Observation |
|---|
| P | Homologues begin pairing and a synaptonemal complex forms |
| Q | Four chromatids are distinct and recombination nodules are present |
| R | The pairing complex has dissolved and homologues remain connected at X-shaped sites |
The stage sequence is:
ⓐ. Zygotene → pachytene → diplotene
ⓑ. Leptotene → zygotene → pachytene
ⓒ. Pachytene → diplotene → diakinesis
ⓓ. Zygotene → diplotene → metaphase I
Correct Answer: Zygotene → pachytene → diplotene
Explanation: Time P shows the initiation of synapsis and formation of the synaptonemal complex, identifying zygotene. At Q, synapsis is complete, all four chromatids of each tetrad are distinct and recombination nodules mark crossing-over sites, identifying pachytene. At R, the synaptonemal complex has dissolved and homologues have moved apart except at visible chiasmata, identifying diplotene. The observations form a mechanistic progression rather than a purely memorised stage list. Pairing must precede recombination, and dissolution of the pairing complex must precede the characteristic partial separation of homologues at chiasmata. The first biologically meaningful difference among the rows locates the process or stage represented by the data. Row interpretation requires the same counting convention and observation unit to be applied consistently across the table. The recorded features are sufficient for the stage inference, while any molecular explanation beyond them would require additional evidence.
310. Control and treated meiocytes complete synapsis normally. In treated cells, recombinase is inactive. No segment exchange is detected during pachytene, and homologues later separate without the usual crossover-associated connections. The treatment most directly explains:
ⓐ. failure of chromosome condensation during leptotene
ⓑ. premature centromere division during pachytene
ⓒ. repeated DNA replication during diplotene
ⓓ. reduced crossing over and fewer later chiasmata
Correct Answer: reduced crossing over and fewer later chiasmata
Explanation: Synapsis occurs normally, so homologous recognition and formation of the paired chromosome structure are intact. Recombinase activity is specifically required for exchange of genetic material between non-sister chromatids during pachytene. Its loss accounts for the absence of detected segment exchange. Chiasmata become evident later when homologues begin separating at sites associated with earlier crossing over. If exchange is reduced, the usual crossover-associated connections should also be reduced. The treatment does not directly affect leptotene condensation, centromere integrity or DNA replication. The experiment links an enzymatic event in pachytene with a later structural observation in diplotene, while preserving the distinction between synapsis and recombination. Preserved events act as internal controls, showing which earlier parts of the sequence remain functional. The strongest conclusion distinguishes the direct observation from the later consequence inferred from it.
311. In a metaphase-I bivalent, both homologous chromosomes become attached to spindle fibres leading towards the same pole, while sister centromeres remain intact. If this arrangement persists, the most direct risk is:
ⓐ. normal homologue segregation to opposite meiosis-I poles
ⓑ. premature sister-chromatid separation within metaphase I
ⓒ. failure of equal homologue segregation during meiosis I
ⓓ. conversion of the bivalent into a mitotic chromosome
Correct Answer: failure of equal homologue segregation during meiosis I
Explanation: Normal metaphase I requires the two homologues of a bivalent to be oriented towards opposite spindle poles. This geometry prepares one homologue to enter each daughter chromosome set while sister chromatids remain joined. Attachment of both homologues towards the same pole removes the normal bipolar relation. If the abnormal orientation persists, both homologues may move into the same daughter region, leaving the opposite region without that homologous representative. Sister chromatids are not expected to separate, since centromeres remain intact throughout meiosis I. The immediate problem concerns distribution of homologues and the reductional outcome, not conversion to a mitotic arrangement or completion of an equational division. Each pole receives one member of every homologous pair, so ploidy falls to \(n\) even though each chromosome still contains two sister chromatids.
312. A diploid meiocyte of a species with \(2n=8\) has completed premeiotic S phase. During pachytene, one crossover occurs between non-sister chromatids of a homologous chromosome pair. Immediately after the exchange is completed, what are the chromosome count, chromatid count and DNA content of the whole cell?
ⓐ. \(16\) chromosomes, \(16\) chromatids and \(4C\) DNA
ⓑ. \(8\) chromosomes, \(8\) chromatids and \(2C\) DNA
ⓒ. \(4\) chromosomes, \(8\) chromatids and \(2C\) DNA
ⓓ. \(8\) chromosomes, \(16\) chromatids and \(4C\) DNA
Correct Answer: \(8\) chromosomes, \(16\) chromatids and \(4C\) DNA
Explanation: Before premeiotic S phase, the diploid cell contains \(8\) chromosomes and \(2C\) DNA. Replication produces two sister chromatids for each chromosome, so the cell enters prophase I with \(8\) chromosomes, \(16\) chromatids and \(4C\) DNA. Chromosome number remains \(8\) because it is counted by centromeres, and replication does not create additional centromeres. During pachytene, crossing over exchanges corresponding DNA segments between non-sister chromatids of homologous chromosomes. This reciprocal exchange changes the combinations of genetic material carried by the participating chromatids, but it does not add or remove chromosomes, chromatids or DNA from the cell. Homologues remain paired as bivalents, and centromeres have not divided. Consequently, the whole-cell counts immediately after crossing over remain \(8\) chromosomes, \(16\) chromatids and \(4C\) DNA, although some chromatids now contain recombined segment arrangements.
313. Cell P is a haploid plant cell in mitotic metaphase. Cell Q is a metaphase-II cell produced by meiosis I. Both have the condition \(n,2C\) and align individual replicated chromosomes at the equator. Which evidence specifically identifies Q as meiotic?
ⓐ. Opposite-pole orientation of sister kinetochores at metaphase
ⓑ. Origin from a prior division that separated homologous chromosomes
ⓒ. Presence of two sister chromatids in each aligned chromosome
ⓓ. Persistence of undivided centromeres during metaphase alignment
Correct Answer: Origin from a prior division that separated homologous chromosomes
Explanation: The current chromosome state and spindle geometry of the two cells can be very similar. Both are haploid, contain replicated chromosomes and prepare to separate sister chromatids through opposite-pole kinetochore attachment. Those observations alone do not distinguish haploid mitosis from metaphase II. The decisive evidence is division history. Cell Q arose through meiosis I, in which homologous chromosomes were separated without sister-centromere division. Cell P reached the same \(n,2C\) state through S phase in a haploid mitotic cycle. This comparison shows that an \(n\)-\(C\) state describes chromosome-set number and DNA amount but does not uniquely specify the biological division programme. Whole-cell chromosome counts after centromere division can transiently double without restoring diploidy.
314. During mitotic anaphase, Curve P records chromosome number in the complete undivided cell, while Curve Q records chromosome number at one pole. P rises from \(2n\) to \(4n\), but Q remains \(2n\). The divergence occurs because:
ⓐ. each pole receives \(4n\) because both daughter copies move together
ⓑ. DNA replication raises the whole-cell count to \(4n\) during anaphase
ⓒ. centromere division gives \(4n\) per cell but \(2n\) at each pole
ⓓ. cytokinesis reduces the whole-cell count to \(n\) before chromosome movement
Correct Answer: centromere division gives \(4n\) per cell but \(2n\) at each pole
Explanation: Before anaphase, every replicated chromosome has one centromere and is counted once. When centromeres divide, the two sister chromatids become independent daughter chromosomes. The complete undivided cell contains twice as many counted chromosomes, producing the rise from \(2n\) to \(4n\). Equal poleward movement distributes half of those chromosomes to each side, so each pole receives \(2n\), the original parental chromosome count. No new DNA replication occurs during anaphase. The graph highlights the importance of observation level: whole-cell counts after centromere division differ from counts assigned to one future daughter nucleus. A change in slope marks a change in rate or process, whereas a horizontal segment represents persistence of the current state. The graph tests sequence as well as quantity, since the order of curve changes constrains the underlying cellular events. A plateau indicates no change in the plotted quantity during that interval, while a rise marks the phase in which the quantity is increasing.
315. A species has \(2n=12\). Compare two cells:
| Cell | Stage | Chromosomes in complete cell | Chromatids | DNA |
|---|
| P | Metaphase I | \(12\) | \(24\) | \(4C\) |
| Q | Anaphase II after all centromeres split | \(12\) | \(12\) | \(2C\) |
Why does the equal chromosome count not imply equal ploidy?
ⓐ. P is haploid, whereas Q is diploid after centromere division
ⓑ. Both are haploid, but P underwent an extra S phase before metaphase I
ⓒ. Both are diploid, but Q has lost half of its DNA during anaphase II
ⓓ. P is diploid; Q is haploid with a transient doubled anaphase-II count
Correct Answer: P is diploid; Q is haploid with a transient doubled anaphase-II count
Explanation: Cell P contains six homologous pairs and is diploid. Its \(12\) chromosomes remain replicated and together contain \(24\) chromatids and \(4C\) DNA. Cell Q entered meiosis II with six replicated chromosomes in a haploid cell. Centromere division converted its \(12\) chromatids into \(12\) independently counted daughter chromosomes within the complete undivided cell. Despite the numerical count of \(12\), the chromosomes represent two separating copies of one haploid set, not six homologous pairs. Each pole will receive \(6\) chromosomes. Ploidy records chromosome sets and homologue representation, so it cannot be determined from a transient whole-cell chromosome count without considering stage and centromere state. The data support only the stated cellular inference; they do not identify an unstated molecular cause. Each column represents a separate condition, and the answer must satisfy them simultaneously. The table converts several observations into one stage decision by requiring their biological compatibility.
316. Arrange the following nuclear-envelope changes in their normal meiotic order when telophase-I reconstitution occurs.
P. The envelope disappears during late diakinesis.
Q. It may re-form around chromosome groups in telophase I.
R. It disappears again during prophase II.
S. It forms around the four chromosome groups in telophase II.
ⓐ. P → Q → R → S
ⓑ. P → R → Q → S
ⓒ. R → P → S → Q
ⓓ. Q → P → R → S
Correct Answer: P → Q → R → S
Explanation: Late diakinesis prepares bivalents for metaphase I, and the nuclear envelope loses its visible integrity. After homologues separate, telophase I may include re-formation of nuclear envelopes around the two chromosome groups. Interkinesis follows without DNA replication. During prophase II, those envelopes disappear again when present, permitting chromosomes to interact with the second meiotic spindle. Telophase II restores nuclear envelopes around the four final chromosome groups. The sequence demonstrates that envelope disappearance and reappearance can occur twice during meiosis. These structural cycles are coordinated with chromosome accessibility and nuclear reconstruction rather than with additional rounds of DNA synthesis. Failure of cytokinesis II can leave four haploid nuclei within two cytoplasmic compartments, separating nuclear outcome from cell number. At the end of meiosis II, each chromosome consists of one chromatid and each product contains one haploid set.
317. Twenty reversibly quiescent cells receive a replacement signal. They require \(6\,\text{hours}\) to re-enter the active cycle and then complete one division every \(18\,\text{hours}\). If all cells divide synchronously and no cell is lost, how many cells are present \(42\,\text{hours}\) after stimulation?
ⓐ. \(320\)
ⓑ. \(40\)
ⓒ. \(160\)
ⓓ. \(80\)
Correct Answer: \(80\)
Explanation: The first \(6\,\text{hours}\) are used for re-entry from quiescence and do not complete a division. The remaining time is:
\[
42-6=36\,\text{hours}
\]
With an \(18\,\text{hour}\) cycle, each cell completes:
\[
\frac{36}{18}=2\ \text{divisions}
\]
Each complete division doubles cell number, so the final population is:
\[
20\times2^2=80
\]
The calculation distinguishes re-entry delay from active cycle duration. Counting the full \(42\,\text{hours}\) as uninterrupted cycling would overestimate the number of completed divisions. The result assumes synchronous progression, successful cytokinesis and absence of cell loss. The first completed division occurs after the \(6\,\text{hour}\) re-entry delay plus one \(18\,\text{hour}\) cycle, at \(24\,\text{hours}\). The second completes at \(42\,\text{hours}\), exactly at the observation time, so two full doublings and no third doubling are counted. Only completed cycles increase the synchronous population, and the result depends on successful division of every cell. A third division would require another complete \(18\,\text{hour}\) interval and would finish only at \(60\,\text{hours}\), beyond the stated endpoint. The endpoint is included in the count.
318. Mitosis is selectively blocked in shoot apical meristems, while lateral cambium remains mitotically active. Mature cells continue normal metabolism. Which long-term growth pattern is most likely?
ⓐ. Continued increase in length, with reduced increase in thickness
ⓑ. Reduced increase in length, with continued increase in thickness
ⓒ. Normal growth in both dimensions through cell enlargement alone
ⓓ. Formation of haploid tissues in place of new meristematic cells
Correct Answer: Reduced increase in length, with continued increase in thickness
Explanation: Shoot apical meristems supply new cells for continued extension of shoots, so blocking their mitosis reduces growth in length. Lateral cambium remains active under the stated condition and can continue producing cells that contribute to increase in thickness. Metabolic activity of mature cells does not replace the sustained cell production normally supplied by the blocked apical meristem. Cell enlargement may contribute temporarily, but it cannot fully maintain organised long-term extension without new daughter cells. The treatment does not initiate meiosis or convert vegetative tissues into haploid products. The selective block separates the longitudinal role of apical meristems from the radial growth supported by lateral cambium, giving reduced increase in length with continued increase in thickness. Preserved events act as internal controls, showing which earlier parts of the sequence remain functional.
319. Assertion: Assuming no DNA loss, the four final products of one meiosis together contain the same total DNA as the original meiocyte immediately after premeiotic S phase.
Reason: The decreases from \(4C\) to \(2C\) and then \(1C\) are per-cell changes caused by partitioning DNA among two and then four products.
ⓐ. Both Assertion and Reason are true, and Reason correctly explains the Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not correctly explain the Assertion
ⓒ. Assertion is true and Reason is false, so Reason cannot correctly explain the Assertion
ⓓ. Assertion is false and Reason is true, so Reason cannot correctly explain the Assertion
Correct Answer: Both Assertion and Reason are true, and Reason correctly explains the Assertion
Explanation: Immediately after premeiotic S phase, the single meiocyte contains \(4C\) DNA. Meiosis I partitions this amount between two cells:
\[
2\times2C=4C
\]
Meiosis II partitions the same replicated material among four products:
\[
4\times1C=4C
\]
The assertion is true under the stated no-loss condition. The reason is also true and directly explains the assertion by distinguishing DNA content per cell from total DNA across all products. Meiotic divisions redistribute DNA while changing cell number, ploidy and chromatid state; they do not normally destroy half of the total DNA at each division. A fall from \(4C\) to \(2C\) and then \(1C\) describes each successive product, whereas the summed DNA remains \(4C\). The \(C\) value describes DNA amount per stated cell, nucleus or product, so the observation unit must remain explicit throughout the calculation. Replication doubles DNA and chromatid number without adding homologous sets, while centromere division changes chromosome counting without increasing DNA. Counts across all products can differ from counts within one cell as division redistributes material among an increasing number of cellular units.
320. Evaluate the following statements.
I. A \(2n,4C\) cell may be in \(G_2\), mitotic metaphase or metaphase I.
II. An \(n,2C\) cell may be in interkinesis, metaphase II or \(G_2\) of a haploid mitotic cycle.
III. A whole anaphase-II cell counted as \(2n\) must have restored diploidy.
IV. The same \(n\)-\(C\) state can occur in different biological histories.
ⓐ. I, II and IV only
ⓑ. II and III only
ⓒ. I, II and III only
ⓓ. I and III only
Correct Answer: I, II and IV only
Explanation: Statement I is valid since each listed stage can contain a replicated diploid chromosome complement. Statement II is also valid. After meiosis I, a cell is \(n,2C\), and the same symbolic condition occurs in metaphase II or after S phase in a haploid mitotic cell. Statement III is false. Centromere division during anaphase II transiently doubles the chromosome count within the complete undivided cell, but each pole still receives one haploid set. Statement IV expresses the central interpretive lesson: \(n\)-\(C\) notation specifies chromosome-set number and DNA amount, yet stage identity may also require information about chromosome pairing, centromere state, spindle arrangement and prior division history. Chromosome number is fixed by centromere count, whereas chromatid number records replicated DNA units; these values coincide only when chromosomes are unreplicated.
