401. A diploid angiosperm completes the following sequence successfully: meiosis in pollen and megaspore mother cells, gametophyte development, compatible pollination, pollen-tube growth, syngamy, triple fusion and normal seed and fruit maturation. Which integrated set of chromosome-set transitions is correct?
ⓐ. \(2n\) sporophyte → \(2n\) spores and gametophytes; syngamy → \(3n\) embryo; triple fusion → \(2n\) endosperm; seed coat and pericarp → maternal \(n\)
ⓑ. \(n\) sporophyte → \(2n\) spores and gametophytes; syngamy → \(n\) embryo; triple fusion → \(2n\) endosperm; seed coat and pericarp → maternal \(3n\)
ⓒ. \(2n\) sporophyte → \(n\) spores and gametophytes; syngamy → \(n\) embryo; triple fusion → \(2n\) endosperm; seed coat and pericarp → maternal \(3n\)
ⓓ. \(2n\) sporophyte → \(n\) spores and gametophytes; syngamy → \(2n\) embryo; triple fusion → \(3n\) endosperm; seed coat and pericarp → maternal \(2n\)
Correct Answer: \(2n\) sporophyte → \(n\) spores and gametophytes; syngamy → \(2n\) embryo; triple fusion → \(3n\) endosperm; seed coat and pericarp → maternal \(2n\)
Explanation: The angiosperm life cycle alternates a dominant diploid sporophyte with highly reduced haploid gametophytes. Meiosis in pollen and megaspore mother cells reduces \(2n\) to \(n\), and mitotic gametophyte development preserves that haploid condition. Syngamy combines two haploid gametes and restores \(2n\) in the zygote and embryo. Triple fusion combines three haploid nuclear contributions and normally establishes the \(3n\) primary endosperm nucleus and endosperm lineage. Meanwhile, seed coat and pericarp derive from maternal sporophytic integuments and ovary wall respectively, so they remain \(2n\). The integrated relation distinguishes reduction, restoration, triploid formation and persistence of maternal ploidy within one reproductive sequence. Chromosome-set arithmetic follows fusion history: two haploid sets form the zygote, while three haploid contributions establish the usual triploid endosperm nucleus.
402. A sample contains \(25\) viable pollen grains. At shedding, \(60\%\) are two-celled and the remainder are three-celled. Assuming the usual cell composition of each stage, what are the total number of cells present in the sample at shedding and the number of male gametes already formed at that moment?
ⓐ. \(60\) cells and \(20\) male gametes
ⓑ. \(50\) cells and \(20\) male gametes
ⓒ. \(60\) cells and \(50\) male gametes
ⓓ. \(75\) cells and \(20\) male gametes
Correct Answer: \(60\) cells and \(20\) male gametes
Explanation: Of the \(25\) pollen grains, \(60\%\) are two-celled:
\[
25\times0.60=15
\]
so \(10\) are three-celled. Each two-celled pollen grain contains one vegetative cell and one generative cell, contributing \(15\times2=30\) cells. Each three-celled pollen grain contains one vegetative cell and two male gametes, contributing \(10\times3=30\) cells. The sample contains \(60\) cells at shedding. Only the three-celled pollen grains already possess male gametes, and each contains two:
\[
10\times2=20
\]
The two-celled grains still require generative-cell mitosis before their two male gametes are produced. A second check separates total cell count from male-gamete count. The \(15\) two-celled grains contribute cells but no already formed male gametes, whereas the \(10\) three-celled grains contribute both. This is why the cell total is \(30+30=60\) while the male-gamete total is only \(10\times2=20\). The two totals measure different biological features of the mixed pollen sample. This boundary is essential when interpreting mixed shedding stages quantitatively.
403. A compatible pollen grain hydrates on a receptive stigma. Its exine remains intact except at a normal germ pore, and its intine is functional. Which structural event should occur during successful germination?
ⓐ. Sporopollenin at the germ pore thickens and becomes the pollen tube wall
ⓑ. The entire exine dissolves before any tube can emerge
ⓒ. The generative cell forms a new germ pore by digesting the exine
ⓓ. Pollen tube emerges through the germ pore with inner-wall participation
Correct Answer: Pollen tube emerges through the germ pore with inner-wall participation
Explanation: The exine is the highly resistant outer pollen wall and contains sporopollenin, but germ pores are regions where this resistant material is absent or greatly reduced. Successful pollen germination does not require destruction of the entire exine. The pollen tube emerges through a germ pore, and the intine, the thinner inner wall composed mainly of cellulose and pectin, is continuous with the wall system involved in tube growth. The generative cell does not create the aperture; the germ pore is already a structural feature of the pollen wall. This arrangement solves two opposing requirements: the exine protects the male gametophyte, while the specialised aperture allows rapid reproductive outgrowth on a compatible stigma.
404. Anther components are assigned functions as shown below.
| Component | Assigned role |
|---|
| P. Epidermis | Outer protection |
| Q. Endothecium | Contribution to anther dehiscence |
| R. Tapetum | Nutrition of developing pollen |
| S. Sporogenous tissue | Source of cells that become pollen mother cells |
Which evaluation is correct?
ⓐ. All four assignments are correct
ⓑ. Only P and R are correctly assigned
ⓒ. Only Q and S are correctly assigned
ⓓ. R and S should be interchanged
Correct Answer: All four assignments are correct
Explanation: The anther contains both wall tissues and a central reproductive lineage, and their roles are distinct. The epidermis is the outer protective layer. The endothecium is part of the wall system associated with mechanical changes leading to dehiscence and pollen release. The tapetum is the innermost wall layer and supplies important nutritional support to developing microspores and pollen. Sporogenous tissue occupies the central region of the young microsporangium and gives rise to pollen mother cells that subsequently undergo meiosis. The assignments correctly distinguish structural protection, pollen release, nutritive support and production of meiotic precursor cells. Treating tapetum and sporogenous tissue as interchangeable would confuse support of the male lineage with the lineage itself.
405. Two parental lines flower in different seasons and cannot normally be crossed because the desired male parent finishes flowering months before the female parent becomes receptive. Viable pollen from the male line is cryopreserved successfully. Which breeding strategy is most appropriate?
ⓐ. Convert the stored pollen into seeds before the female parent flowers
ⓑ. Use only self pollen from the female parent because stored pollen cannot function after thawing
ⓒ. Thaw preserved pollen at stigma receptivity and use it for controlled pollination
ⓓ. Apply the preserved pollen to immature ovules before stigma formation
Correct Answer: Thaw preserved pollen at stigma receptivity and use it for controlled pollination
Explanation: Cryopreservation allows biologically functional pollen to be stored across a time interval in which the desired parents do not flower simultaneously. If viability is retained, the pollen can later be thawed and applied when the chosen female parent's stigma is receptive. This makes pollen banking useful not merely for storage but for overcoming temporal separation between parental flowering periods in controlled breeding. The pollen does not become a seed during storage and still requires normal pollen-pistil interaction and pollen-tube growth after application. Likewise, it should be applied to the receptive stigma rather than directly to immature ovules. The strategy links pollen viability, preservation and correct timing of controlled pollination. Bagging protects the receptive female flower from unwanted pollen, while deliberate application of the selected pollen establishes the intended male parent. Cryopreservation solves the timing mismatch only by preserving usable pollen until the female parent's stigma becomes receptive.
406. Which lineage statement best describes formation of the common monosporic embryo sac?
ⓐ. Its eight nuclei are descendants of all four megaspores produced by one meiosis
ⓑ. Its nuclei arise by mitotic divisions of a single functional haploid megaspore
ⓒ. Its egg arises from one megaspore while its polar nuclei arise from the other three
ⓓ. Its antipodal nuclei are sporophytic while the remaining nuclei are gametophytic
Correct Answer: Its nuclei arise by mitotic divisions of a single functional haploid megaspore
Explanation: Meiosis of the diploid megaspore mother cell produces four haploid megaspores, but in the common monosporic pathway only one survives as the functional megaspore. The other three normally degenerate and do not contribute nuclei to the mature embryo sac. The surviving functional megaspore undergoes three successive mitotic nuclear divisions, producing the eight nuclei from which the mature \(7\)-celled female gametophyte is organised. The egg, synergids, antipodals and polar nuclei are all descendants of the same functional megaspore. Their different positions and functions arise through later organisation, not because different meiotic products separately contribute different parts of the embryo sac. Three free-nuclear mitoses of one functional megaspore produce eight haploid nuclei before the mature embryo sac is organised into seven cells.
407. During normal female-gametophyte development, nuclear number is plotted against the number of mitotic divisions completed by the functional megaspore. The observed sequence is \(1,\ 2,\ 4,\ 8\) nuclei after \(0,\ 1,\ 2,\ 3\) mitoses respectively. Which relationship best describes the graph?
ⓐ. Nuclear number doubles with each successive mitotic division
ⓑ. Nuclear number increases linearly by one after each mitosis
ⓒ. Nuclear number quadruples after every successive mitotic division
ⓓ. Nuclear number remains constant until cellularisation
Correct Answer: Nuclear number doubles with each successive mitotic division
Explanation: Each mitotic division produces two daughter nuclei from every nucleus that divides. Starting with one nucleus in the functional megaspore, the first mitosis gives \(2\), the second gives \(4\), and the third gives \(8\). The pattern is repeated doubling rather than addition of one nucleus per division. Symbolically, after \(k\) synchronous mitotic rounds the nuclear number follows \(2^k\), giving \(2^3=8\) after three divisions. These divisions are mitotic and preserve the haploid chromosome-set number established by meiosis. Cellular organisation occurs after the free-nuclear divisions and changes how those nuclei are partitioned among cells without creating the initial doubling sequence. Each completed mitosis doubles the nuclear count, so the plotted points follow \(1,2,4,8\) rather than a linear increase of one nucleus per division. The graph represents multiplicative nuclear proliferation during the free-nuclear phase and stops at the normal eight-nucleate stage before cellular organisation is considered.
408. A functional megaspore undergoes all three normal mitotic nuclear divisions, producing \(8\) haploid nuclei, but a mutation prevents subsequent cellular organisation. Which structure is expected immediately afterward?
ⓐ. A normal \(7\)-celled, \(8\)-nucleate embryo sac
ⓑ. Four separate embryo sacs, each containing two nuclei
ⓒ. An \(8\)-nucleate incompletely cellularised embryo sac
ⓓ. A diploid \(8\)-celled embryo sac
Correct Answer: An \(8\)-nucleate incompletely cellularised embryo sac
Explanation: The three mitotic divisions have occurred normally, so the female gametophyte has reached the expected \(8\)-nucleate condition. The defect arises afterward, during the organisational process that normally partitions those nuclei into the characteristic \(7\) cells. Blocking cellular organisation does not remove nuclei, alter their haploid ploidy or create four independent embryo sacs. Instead, the structure remains an \(8\)-nucleate incompletely cellularised gametophytic stage. This changed condition highlights the distinction between nuclear multiplication and cell formation. The mature \(7\)-celled, \(8\)-nucleate organisation cannot be inferred merely from the presence of eight nuclei; correct spatial partitioning of those nuclei is also required. Three mitoses have already established the normal eight-nuclear complement, so the defect occurs after nuclear multiplication rather than during it. Failure of cellular organisation prevents those nuclei from being partitioned into the usual seven-cell arrangement. The observation separates nuclear number from cell number and locates the mutation at a later organisational step.
409. A mature embryo sac retains two normal synergids with intact filiform apparatus, but the egg cell is selectively destroyed before a compatible pollen tube arrives. Which outcome is most directly expected?
ⓐ. Pollen-tube guidance and zygote formation both proceed normally
ⓑ. Triple fusion cannot occur because the polar nuclei lie inside the egg cell
ⓒ. Pollen-tube guidance may still occur, but normal syngamy cannot produce a zygote
ⓓ. One antipodal cell replaces the missing egg and undergoes normal syngamy
Correct Answer: Pollen-tube guidance may still occur, but normal syngamy cannot produce a zygote
Explanation: The two synergids are specialised cells of the egg apparatus, and their filiform apparatus helps guide the arriving pollen tube toward the embryo sac. Therefore, retaining functional synergids can preserve the guidance step even when the egg cell itself has been lost. Syngamy, however, specifically requires fusion of one male gamete with the egg cell to form the diploid zygote. Without an egg, that fusion cannot occur normally. The polar nuclei belong to the central cell rather than the egg, so their presence does not restore syngamy; an arriving second male gamete may still encounter the central-cell fusion partners for triple fusion. Antipodals do not normally replace a destroyed egg. The changed condition therefore separates pollen-tube guidance from the distinct gametic role of the egg and shows why successful guidance alone is insufficient for complete double fertilisation.
410. Pollen is transferred by an insect from one flower to a different flower on the same plant. Why is this event often described as functionally cross-pollination but genetically similar to autogamy?
ⓐ. It uses pollen from another species but does not require a pollinating agent
ⓑ. It occurs within one flower but requires two different stigmas
ⓒ. It always produces more genetic variation than xenogamy
ⓓ. It transfers pollen between flowers of the same individual genotype
Correct Answer: It transfers pollen between flowers of the same individual genotype
Explanation: The described event is geitonogamy. Functionally, it resembles cross-pollination: pollen must be physically transported from one flower to another and may require an external agent such as an insect. Genetically, however, both flowers occur on the same individual plant, so the pollen and ovule come from the same sporophytic genotype. In that sense, geitonogamy shares the genetic selfing character of autogamy more closely than the outcrossing character of xenogamy. Xenogamy transfers pollen between different individual plants of the same species and introduces a different parental genotype. The distinction illustrates why mode of pollen transport and genetic relationship between parents are separate dimensions of pollination biology. Changes in pollen transport can alter deposition without changing pollen ploidy, embryo-sac organisation or the later fusion products of double fertilisation.
411. A wind-pollinated flower still produces light, non-sticky pollen and retains a large feathery stigma, but a mutation keeps its anthers enclosed so that pollen is poorly exposed to moving air. The most likely consequence is:
ⓐ. pollen grains become heavier because the stigma remains feathery
ⓑ. fewer pollen grains enter air currents even though the pollen-capturing surface remains suitable
ⓒ. the stigma becomes physiologically unreceptive to any pollen that reaches it
ⓓ. pollination necessarily changes from wind pollination to water pollination
Correct Answer: fewer pollen grains enter air currents even though the pollen-capturing surface remains suitable
Explanation: Wind pollination depends on complementary adaptations for both release and capture. Light, non-sticky pollen can travel effectively once airborne, and a large feathery stigma provides a broad surface for intercepting wind-borne grains. In the changed flower, those two traits remain intact, but the anthers are no longer well exposed to air movement. That defect acts at the pollen-release stage: fewer grains are likely to enter the moving air in the first place. It does not make individual pollen grains heavier, nor does it directly alter stigma receptivity. Hydrophily would require pollen transport through water rather than merely poor exposure of anthers. The example therefore shows that efficient anemophily is not determined by one isolated pollen or stigma character; successful transfer requires coordinated presentation of pollen to the air as well as effective interception at the recipient surface.
412. A water lily grows rooted in an aquatic habitat, but its flowers project above the water surface and are visited by insects that transfer pollen between flowers. Which statement is most accurate?
ⓐ. Every aquatic flowering plant must use water as its pollinating agent
ⓑ. Aquatic habitat does not imply hydrophily when insects carry the pollen
ⓒ. The plant shows submerged hydrophily because its roots are underwater
ⓓ. The pollination is anemophilous because flowers emerge above water
Correct Answer: Aquatic habitat does not imply hydrophily when insects carry the pollen
Explanation: Pollination is classified by the agent that transports pollen, not simply by the habitat in which the plant grows. In the described water lily, the flowers rise above the water and insects perform the transfer between anthers and stigmas. The pollination mechanism is animal-mediated rather than hydrophilous. The functional contrast is important since true water pollination is relatively uncommon even among aquatic angiosperms. Species such as Vallisneria and Zostera provide specialised examples of water-mediated transfer, whereas many other aquatic plants expose their flowers to air and use insects or wind. Rooting in water does not determine the transport mechanism of pollen. Water pollination must be demonstrated by water-mediated pollen transport; merely living in an aquatic habitat does not establish hydrophily.
413. In an animal-pollinated species, normal flowers provide nectar as well as visual and scent signals. A mutant flower retains normal pollen, stigma, colour and scent but produces no nectar. Pollinator visits and pollen deposition both fall sharply in the mutant. The observations most strongly support the conclusion that:
ⓐ. nectar directly changes pollen from haploid to diploid
ⓑ. nectar replaces the stigma as the site of pollen reception
ⓒ. the absence of nectar converts the flower into an obligately wind-pollinated flower
ⓓ. nectar acts as a floral reward that promotes visits and thereby supports pollen transfer
Correct Answer: nectar acts as a floral reward that promotes visits and thereby supports pollen transfer
Explanation: Animal-pollinated flowers commonly use attraction signals together with rewards such as nectar or pollen. Here, the mutant retains the structures and signals needed to advertise the flower, and its pollen and stigma remain normal, yet removal of nectar is followed by fewer visits and reduced pollen deposition. The linked change supports a behavioural role for nectar as a reward that encourages pollinators to visit and contact the reproductive organs. Nectar does not alter the ploidy of pollen and it cannot substitute for a stigma, which is the receptive surface of the pistil. Reduced visitation also does not by itself establish a switch to wind pollination; the transport agent would have to change accordingly. The evidence therefore connects floral reward with pollinator behaviour and then with the probability of pollen transfer, rather than treating nectar as a reproductive cell or pollen-receiving structure.
414. A monoecious plant bears separate male and female flowers on the same individual. Pollen can readily move between these flowers, but the plant also possesses a recognition system that rejects pollen carrying the same incompatibility identity as the female tissue. What outcome is most likely for same-plant pollen transfer?
ⓐ. It becomes autogamy because both flowers belong to one plant
ⓑ. It automatically becomes xenogamy after stigma recognition
ⓒ. Geitonogamy can occur, but incompatibility may still prevent fertilisation
ⓓ. Monoecy always ensures seed formation even when pollen-pistil incompatibility is present
Correct Answer: Geitonogamy can occur, but incompatibility may still prevent fertilisation
Explanation: Monoecy determines floral arrangement: male and female flowers occur separately on the same individual. Pollen transfer between them is geitonogamy and can occur physically even though autogamy within one unisexual flower is impossible. The additional incompatibility system operates at a later recognition stage. If same-plant pollen carries an identity rejected by the female tissues, pollen deposition can occur without successful pollen-tube progression and fertilisation. The two mechanisms act at different levels. Monoecy governs which flowers can exchange pollen, whereas incompatibility governs whether deposited pollen is accepted. Physical transfer and successful fertilisation should not be treated as the same reproductive event. Geitonogamy is genetically similar to self-pollination but functionally requires a pollinating agent because pollen must move between two separate flowers.
415. Pollen from several unrelated donors is viable and germinates normally on isolated nutrient medium. When placed on the stigma of one mutant plant, however, both self and normally compatible cross pollen fail to sustain pollen-tube growth, whereas the same pollen performs normally on wild-type pistils. What is the strongest inference?
ⓐ. Every donor pollen grain carries the same self-incompatibility genotype
ⓑ. The mutant pistil likely has a pollen-recognition or support defect
ⓒ. All donor pollen has lost viability after reaching the stigma
ⓓ. The mutant plant must be dioecious
Correct Answer: The mutant pistil likely has a pollen-recognition or support defect
Explanation: The pollen performs normally in two independent contexts: it germinates on nutrient medium and can progress on wild-type pistils. These observations argue against a general pollen defect. The common failure occurs specifically when several otherwise functional pollen types encounter the mutant pistil. The most economical interpretation is a female-side defect in recognition, acceptance or physiological support of pollen-tube development. The result is broader than ordinary self-incompatibility since even normally compatible cross pollen is affected. Dioecy concerns distribution of male and female flowers among individuals and cannot explain selective failure of pollen tubes within an already functioning pistil. Experimental comparison localises the defect to the recipient reproductive tissue. A compatible grain can germinate and extend a pollen tube through stigma and style toward the ovary, whereas incompatible pollen may be rejected before completing that route.
416. A pollen tube enters one synergid and releases two male gametes into its cytoplasm. Which subsequent description correctly preserves the normal pathway?
ⓐ. Both male gametes remain permanently inside the synergid and form two embryos there
ⓑ. The two male gametes are available for separate fusion with the egg and central-cell nuclei
ⓒ. One male gamete fuses with a synergid nucleus and the other with an antipodal nucleus
ⓓ. The synergid itself becomes the primary endosperm cell after receiving the pollen tube
Correct Answer: The two male gametes are available for separate fusion with the egg and central-cell nuclei
Explanation: The synergid acts as the receiving site for the incoming pollen tube, but it is not the final fusion partner for either male gamete in normal double fertilisation. After tube discharge, one male gamete reaches the egg and participates in syngamy, producing the diploid zygote. The other reaches the central-cell nuclear system and participates in triple fusion, producing the primary endosperm nucleus. The two gametes diverge functionally after being delivered through the same synergid. This conceptual separation prevents a common spatial misconception: entry of the pollen tube into a synergid does not mean fertilisation takes place with the synergid nucleus. The synergid facilitates delivery; the egg and polar nuclear system are the actual fusion partners.
417. In a hypothetical mutant embryo sac, three separate polar nuclei are present in the central cell instead of the normal two. Each is haploid, and one normal haploid male gamete fuses simultaneously with all three. What ploidy would the resulting central-cell fusion product have?
ⓐ. \(4n\)
ⓑ. \(2n\)
ⓒ. \(3n\)
ⓓ. \(5n\)
Correct Answer: \(4n\)
Explanation: The mutant central cell contributes three separate haploid polar nuclei, and the male gamete contributes one additional haploid chromosome set. The total nuclear contribution is:
\[
n+n+n+n=4n
\]
The resulting fusion product would be tetraploid under the hypothetical condition. This differs from normal angiosperm triple fusion, in which one male gamete combines with two polar nuclei and produces the typical \(3n\) primary endosperm nucleus. The altered example tests whether ploidy is derived from the actual number of chromosome sets contributed rather than from memorising the word “endosperm” as automatically triploid. It also shows that normal triploidy depends on the usual two-polar-nucleus organisation of the central cell. Triple fusion joins the second male gamete with the two polar nuclear contributions in the central cell, establishing the usually triploid primary endosperm nucleus.
418. A normally apomictic line produces a diploid egg. In one experimental flower, that \(2n\) egg unexpectedly undergoes syngamy with a normal haploid male gamete instead of developing directly. If the fusion succeeds, what is the immediate ploidy of the resulting zygote?
ⓐ. \(n\)
ⓑ. \(2n\)
ⓒ. \(3n\)
ⓓ. \(4n\)
Correct Answer: \(3n\)
Explanation: The egg in the stated line is already diploid since it was produced without the usual reduction to the haploid condition. If such a \(2n\) egg now fuses with a normal \(n\) male gamete, the chromosome-set relation is:
\[
2n+n=3n
\]
The immediate zygotic product is triploid. This hypothetical changed condition contrasts with two normal pathways. In ordinary sexual reproduction, an \(n\) egg fuses with an \(n\) male gamete to form a \(2n\) zygote. In the specified apomictic route, the \(2n\) egg would normally develop without syngamy and remain \(2n\). The experiment changes only the presence of fertilisation, revealing its direct ploidy consequence. The unreduced egg begins at \(2n\), while the normal male gamete contributes \(n\). Under the unusual syngamy described, chromosome sets add as \(2n+n=3n\). This differs from the usual apomictic route, in which the \(2n\) egg develops directly without any paternal contribution and remains diploid.
419. A primary endosperm nucleus undergoes many rounds of nuclear division, but a mutation blocks cytokinesis throughout the observed period. Which tissue condition should predominate?
ⓐ. Free-nuclear endosperm with many nuclei in common cytoplasm
ⓑ. A multicellular embryo with no endosperm nuclei
ⓒ. A mature seed coat divided into triploid cells
ⓓ. Perisperm produced directly from the primary endosperm nucleus
Correct Answer: Free-nuclear endosperm with many nuclei in common cytoplasm
Explanation: Free-nuclear endosperm develops when the primary endosperm nucleus divides repeatedly without corresponding cell-wall formation or cytokinesis. Blocking cytokinesis while allowing nuclear division prolongs exactly this condition: nuclear number increases, but the tissue is not partitioned into separate cells. The resulting structure contains many endosperm nuclei within a common cytoplasmic region. Cellular endosperm would require later wall formation. The embryo belongs to the zygotic lineage, seed coat to the integuments and perisperm to persistent nucellus, so none of those structures can arise simply from blocking cytokinesis in the endosperm lineage. The experiment distinguishes nuclear proliferation from cellularisation as separate developmental processes. Endosperm may later be retained or consumed, but its mature persistence is a separate issue from whether the primary endosperm lineage formed normally after triple fusion. Chromosome-set arithmetic follows fusion history: two haploid sets form the zygote, while three haploid contributions establish the usual triploid endosperm nucleus. The persistent multinucleate condition specifically reflects continued nuclear division without cellularisation, while the lineage remains endosperm.
420. In a dicot embryo, a mutation specifically damages the embryonal axis below the level of cotyledon attachment but leaves both cotyledons and the upper axis normal. Which structures are most directly affected?
ⓐ. Epicotyl and plumule
ⓑ. Cotyledons and seed coat
ⓒ. Scutellum and coleoptile
ⓓ. Hypocotyl and radicle
Correct Answer: Hypocotyl and radicle
Explanation: The cotyledon attachment level divides the typical dicot embryonal axis into upper and lower regions. The epicotyl lies above the cotyledons and terminates in the plumule, whereas the hypocotyl lies below the cotyledons and continues toward the radicle. A mutation restricted to the lower axis most directly affects the hypocotyl-radicle region. The cotyledons are separately stated to be normal, and the seed coat is maternal tissue derived from integuments rather than part of the embryo. Scutellum and coleoptile are characteristic terms associated with monocot embryo organisation. The supplied positional clue is sufficient to identify the affected dicot structures without requiring an external diagram. A typical dicot embryo carries two cotyledons on an embryonal axis whose epicotyl ends in the plumule and whose hypocotyl ends in the radicle. Plumule and radicle are terminal embryonic structures, whereas epicotyl and hypocotyl are broader axis regions defined relative to cotyledon attachment.
