Plant Development: Gametophytes, Fertilisation, Embryo and Meristems, Patterning, Flowering, Light and Hormones

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

1. A diploid angiosperm with Polygonum-type embryo sacs is to produce 80 seeds. Assume each seed develops from one ovule containing one megaspore mother cell, each fertilisation uses the sperm of one pollen grain, each microspore mother cell gives four pollen grains, and nothing is wasted. The minimum number of meiotic divisions required is ____.

   Numerical answer — type the value.

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   Answer: 100

   Female side: one meiosis per ovule, because one megaspore mother cell gives the single functional megaspore that forms the embryo sac: 80 meioses. Male side: each seed needs one pollen grain (its two sperm serve syngamy and triple fusion), and one meiosis gives four grains, so 80 ÷ 4 = 20 meioses. Total = 80 + 20 = 100. Counting 160 pollen grains because two sperm are used is the usual slip; both sperm come from the same grain.

2. In a diploid angiosperm with a Polygonum-type embryo sac, add the ploidy levels (as multiples of n) of the following: the zygote, the primary endosperm nucleus, one antipodal cell, one synergid and one cell of the seed coat. The sum is ____.

   Numerical answer — type the value.

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   Answer: 9

   Zygote 2n (egg n + sperm n); primary endosperm nucleus 3n (two polar nuclei + one sperm); antipodal n and synergid n (gametophytic cells); seed coat 2n, because it develops from the integuments of the maternal sporophyte. Sum = 2 + 3 + 1 + 1 + 2 = 9. Treating the seed coat as part of the new generation, or the antipodals as diploid, are the common slips.

3. Use the classic ABC model and an idealised Arabidopsis flower with 4 organs in whorl 1, 4 in whorl 2, 6 in whorl 3 and 2 in whorl 4, and assume each whorl keeps this number of organs in a mutant. In a plant that has completely lost A function (an apetala2 null allele, with C activity spreading into the outer whorls), how many stamens does one flower bear?

   Numerical answer — type the value.

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   Answer: 10

   Without A, C spreads into whorls 1 and 2 because A and C are antagonistic. Whorl 1: C alone = carpels (4). Whorl 2: B + C = stamens (4). Whorl 3: B + C = stamens (6). Whorl 4: C alone = carpels (2). Stamens = 4 + 6 = 10. Answering 6 (only the wild-type stamens) or reading whorl 2 as sepals are the usual slips.

4. A leaf peel shows 45 stomata and 255 other epidermal cells in the same field. Taking stomatal index = 100 × S/(S + E), where S is the number of stomata and E the number of other epidermal cells, the stomatal index is ____.

   Numerical answer — type the value.

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   Answer: 15

   S + E = 45 + 255 = 300, so SI = 100 × 45/300 = 15. Dividing by E alone (100 × 45/255 ≈ 17.6) is the usual slip: the denominator is all counted cells, stomata included.

5. In the classic tobacco pith callus experiments of Skoog and Miller, a medium with a high ratio of cytokinin to auxin caused the callus to

   1. form roots, while a high auxin-to-cytokinin ratio gave shoot buds
   2. stop dividing and senesce, because cytokinin alone blocks the cell cycle
   3. grow only as undifferentiated callus, whatever the absolute concentrations
   4. form shoot buds, while a high auxin-to-cytokinin ratio gave roots
   Show answer

   Answer: D — form shoot buds, while a high auxin-to-cytokinin ratio gave roots

   Skoog and Miller (1957) showed that the ratio of the two hormones decides organ formation: high cytokinin:auxin favours shoots, high auxin:cytokinin favours roots, and intermediate ratios keep the callus growing unorganised. The reversed pairing is the tempting error. Cytokinin promotes cell division when auxin is present; it does not block it.

6. Transgenic Arabidopsis plants express both APETALA3 and PISTILLATA from a constitutive promoter in all four floral whorls, while A and C functions remain normal. The expected flowers bear, from the outside inward,

   1. sepals, petals, stamens, carpels, because extra B protein is inactive outside its normal domain
   2. petals, petals, stamens, stamens, because B combines with A in whorl 1 and with C in whorl 4
   3. stamens, stamens, stamens, stamens, because B overrides the A and C functions in every whorl
   4. sepals, sepals, carpels, carpels, because surplus B protein competes away the A and C proteins
   Show answer

   Answer: B — petals, petals, stamens, stamens, because B combines with A in whorl 1 and with C in whorl 4

   Adding B everywhere turns whorl 1, which has A, into A + B, a petal, and whorl 4, which has C, into B + C, a stamen, giving petals, petals, stamens, stamens. B does not override A or C; organ identity follows from the combination present in each whorl. Sepals, sepals, carpels, carpels is the B-loss phenotype, the opposite manipulation.

7. Which of the following statements about pollen-tube growth and double fertilisation are correct?

   1. In chalazogamy, the pollen tube enters the ovule through the micropyle
   2. Synergid cells secrete LURE peptides that attract the pollen tube towards the ovule
   3. The primary endosperm nucleus is formed when a sperm fuses with the egg nucleus
   4. The pollen tube discharges its sperm into one of the synergids, which then degenerates
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   Answer: B — Synergid cells secrete LURE peptides that attract the pollen tube towards the ovule; D — The pollen tube discharges its sperm into one of the synergids, which then degenerates

   LUREs, cysteine-rich peptides secreted by the synergids, guide the tube at short range, and the tube bursts inside a receptive synergid, which degenerates as the two sperm are released. In chalazogamy (as in Casuarina) the tube enters through the chalaza; entry through the micropyle is porogamy. Sperm–egg fusion gives the zygote; the primary endosperm nucleus comes from a sperm fusing with the two polar nuclei.

8. A short-day plant is kept on an inductive cycle of 9 h light and 15 h darkness. In one set, a 5-min flash of red light is given in the middle of each dark period; in a second set, the red flash is followed at once by 5 min of far-red light. What is expected?

   1. Both sets flower, because the total duration of darkness is almost unchanged
   2. Neither set flowers, because any light given during the night resets the clock
   3. The first set flowers and the second does not, because far-red lengthens the effective photoperiod
   4. The first set does not flower and the second does, because far-red reverses the red night break
   Show answer

   Answer: D — The first set does not flower and the second does, because far-red reverses the red night break

   Short-day plants require an uninterrupted long night. A red night break converts Pr to Pfr and splits the night into two short ones, so flowering is prevented; far-red given immediately afterwards converts Pfr back to Pr, cancelling the break, so flowering proceeds. ‘Both flower’ is the tempting answer if one thinks only of total darkness, but the plant responds to the continuity of the dark period, not its total length.

9. In grafting and transport experiments on Arabidopsis and rice, the long-distance florigen signal that moves from induced leaves to the shoot apex is

   1. CONSTANS protein, exported from the leaf to activate floral genes at the apex
   2. FLOWERING LOCUS T protein, made in phloem companion cells and acting with FD at the apex
   3. FLOWERING LOCUS C protein, which travels in the phloem to repress floral repressors
   4. phytochrome B, which moves from the leaf blade to the apex once it has absorbed red light
   Show answer

   Answer: B — FLOWERING LOCUS T protein, made in phloem companion cells and acting with FD at the apex

   CO acts inside the leaf, in companion cells, where it switches on FT; the FT protein (Hd3a in rice) travels through the phloem to the apex and partners the bZIP factor FD to activate floral genes. CO is the tempting choice because it is the long-day sensor, but it is not the mobile signal. FLC is a repressor of FT, and phytochrome does not travel to the apex.

10. A winter-annual Arabidopsis plant given six weeks at 4 °C flowers soon after it is returned to warm long days, yet seedlings raised from its seeds again flower only after a cold treatment. The best explanation is that

    1. cold destroys FRIGIDA protein, which the plant cannot make again until the next generation
    2. cold mutates the FLC gene in the shoot apex, and the mutation is never passed on to the seeds formed later
    3. cold silences FLC by Polycomb-deposited H3K27me3, kept through mitosis but reset in the next generation
    4. cold induces FT protein that is stored in the leaves and is used up before the seeds are formed
    Show answer

    Answer: C — cold silences FLC by Polycomb-deposited H3K27me3, kept through mitosis but reset in the next generation

    Vernalisation silences FLC epigenetically: VIN3 and PRC2 (with VRN2) lay down H3K27 trimethylation, which is copied through cell divisions after warming, a cellular memory of winter, but erased when FLC is reactivated in the next generation. A mutation in the apex would be carried into the flowers and passed to the seeds, so it cannot explain the reset. FRIGIDA is not destroyed by cold, and FT is made in leaves in response to long days, not stored from the cold period.

11. Arabidopsis embryos homozygous for loss-of-function monopteros (ARF5) alleles lack a root and hypocotyl. The best explanation is that

    1. auxin reaching the embryo base cannot switch on the response that makes the hypophysis the root founder
    2. the zygote divides symmetrically, so neither a basal cell nor a suspensor is ever formed
    3. PIN proteins cannot be trafficked to the plasma membrane, so the embryo becomes a ball without polarity
    4. the shoot stem cells are not maintained, so the apical half of the embryo is converted into root
    Show answer

    Answer: A — auxin reaching the embryo base cannot switch on the response that makes the hypophysis the root founder

    MP is an auxin-response factor. When auxin, delivered by PIN1 and PIN7, degrades its inhibitor BODENLOS, MP activates genes in the basal embryo cells that signal to the hypophysis and specify the root; without MP the root and hypocotyl are missing, while the shoot pole forms. Failure of PIN trafficking describes gnom, the tempting confusion, whose embryos lose apical–basal polarity altogether. The zygote division and the shoot stem cells are normal in mp.

12. Which of the following statements about plant photoreceptors are correct?

    1. Pr absorbs red light maximally and is converted to Pfr, the biologically active form of phytochrome
    2. Cryptochromes use phytochromobilin as their chromophore to sense far-red light
    3. Phototropins carry two LOV domains that bind FMN, linked to a serine/threonine kinase domain
    4. A low red:far-red ratio under a canopy raises the Pfr fraction and slows stem elongation
    Show answer

    Answer: A — Pr absorbs red light maximally and is converted to Pfr, the biologically active form of phytochrome; C — Phototropins carry two LOV domains that bind FMN, linked to a serine/threonine kinase domain

    Red light converts Pr to the active Pfr form, and phototropins are blue-light receptor kinases with two FMN-binding LOV domains (phototropism, chloroplast movement, stomatal opening). Cryptochromes are blue/UV-A receptors with FAD and a pterin; phytochromobilin belongs to phytochrome. Canopy shade lowers red:far-red, which lowers Pfr and speeds elongation (shade avoidance).

13. Dark-grown seedlings of the Arabidopsis mutant det2 have short hypocotyls and open cotyledons, and the defect is rescued by applying brassinolide. This shows that

    1. DET2 is a photoreceptor whose loss makes the seedling behave as though it were in light
    2. brassinosteroids inhibit hypocotyl growth, so the mutant must overproduce them in darkness
    3. DET2 is needed to make brassinosteroids, which sustain etiolated growth in darkness
    4. DET2 encodes the receptor kinase BRI1, which binds brassinolide at the plasma membrane
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    Answer: C — DET2 is needed to make brassinosteroids, which sustain etiolated growth in darkness

    Rescue by the hormone means the mutant lacks the hormone, so DET2 acts in brassinosteroid biosynthesis (it encodes a steroid 5α-reductase), and BRs are needed for the long hypocotyl and closed cotyledons of etiolated seedlings. A receptor mutant such as bri1 could not be rescued by brassinolide. DET2 is not a photoreceptor; the light-grown look resembles cop1, whose cause is a failure to degrade HY5.

14. Loss-of-function clavata3 mutants of Arabidopsis have an enlarged shoot apical meristem and flowers with extra organs, especially extra carpels. The best explanation is that

    1. the WUSCHEL domain expands because the CLV3 peptide no longer represses WUS through its receptor
    2. SHOOT MERISTEMLESS is lost, so the cells in the centre of the meristem divide much faster
    3. auxin maxima stop forming in the peripheral zone, so that no primordia ever leave the meristem at all
    4. AGAMOUS is overexpressed, so the floral meristem never stops producing new carpels
    Show answer

    Answer: A — the WUSCHEL domain expands because the CLV3 peptide no longer represses WUS through its receptor

    CLV3, secreted by the stem cells, signals through CLV1 to restrict WUS expression; without it WUS spreads, more cells are kept as stem cells, the meristem enlarges and floral meristems make extra organs. Loss of STM causes the opposite, a missing meristem; failure of auxin maxima gives pin-like stems without primordia; and AGAMOUS actually terminates the floral meristem by repressing WUS.

15. Arabidopsis roots homozygous for a null shortroot allele have

    1. an extra layer of ground tissue, because the cortex initials divide twice
    2. a single ground-tissue layer that lacks endodermal identity
    3. a normal endodermis but no cortex, because SHR acts only in the cortex
    4. a single ground-tissue layer with both cortical and endodermal features
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    Answer: B — a single ground-tissue layer that lacks endodermal identity

    SHR moves from the stele into the adjacent layer, where with SCR it drives the periclinal division that separates endodermis from cortex and specifies the endodermis. Without SHR there is one layer, and it lacks endodermal markers (it is cortex-like); the meristem also runs down over time. A single layer with mixed features is the scarecrow phenotype, the tempting confusion.

16. Which sequence correctly describes how an isolated Zinnia mesophyll cell, cultured with auxin and cytokinin, becomes a tracheary element?

    1. Loss of the nucleus and organelles, building callose-lined sieve plates, then support by a companion cell
    2. Programmed cell death first, then deposition of lignified walls around it by neighbouring parenchyma cells
    3. Repeated asymmetric divisions to form a meristemoid, then one symmetric division into two guard cells
    4. Dedifferentiation, patterned lignified secondary walls, then programmed death with vacuole rupture
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    Answer: D — Dedifferentiation, patterned lignified secondary walls, then programmed death with vacuole rupture

    In the Zinnia system the mesophyll cell dedifferentiates, builds a patterned secondary wall that is lignified, and then dies by programmed cell death, in which the vacuole ruptures and hydrolases clear the protoplast, leaving a hollow tube. The cell builds its own wall before dying, which rules out the death-first sequence. Enucleation with sieve plates describes a sieve element, and the meristemoid route describes stomatal development.

17. Which of the following statements about the spacing of trichomes and stomata in Arabidopsis are correct?

    1. FAMA is required for the asymmetric entry division that first creates a meristemoid
    2. speechless mutants make no stomata because the asymmetric entry division does not occur
    3. STOMAGEN secreted from the mesophyll lowers stomatal density
    4. triptychon mutants form clustered trichomes because a mobile inhibitor of trichome fate is missing
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    Answer: B — *speechless* mutants make no stomata because the asymmetric entry division does not occur; D — *triptychon* mutants form clustered trichomes because a mobile inhibitor of trichome fate is missing

    SPCH drives the entry division, so spch leaves have only pavement cells, and TRY is a small MYB protein that moves to neighbours and blocks the GL1–GL3–TTG1 complex, so try leaves have clustered trichomes. FAMA acts last, in the guard mother cell, to make guard cells, and STOMAGEN (EPFL9) raises stomatal density.

18. The Green Revolution semi-dwarf wheats carry the Rht-B1b or Rht-D1b alleles. Their short stature is explained by

    1. loss of a GA 20-oxidase, so that far too little active gibberellin is made in the elongating stem
    2. loss of the GID1 receptor, so gibberellin cannot bind at all in the elongating cells
    3. DELLA proteins that resist GA-triggered degradation, so growth stays repressed despite GA
    4. overproduction of brassinosteroids, which divert growth from the stem into the grain
    Show answer

    Answer: C — DELLA proteins that resist GA-triggered degradation, so growth stays repressed despite GA

    Rht-B1b and Rht-D1b encode N-terminally truncated DELLA proteins that cannot be targeted for degradation after GA binds GID1, so they keep repressing growth: the plants are GA-insensitive, with normal or raised GA levels. Loss of GA 20-oxidase is the tempting choice because it causes the rice Green Revolution semi-dwarf sd1, a GA-deficient mutant that applied GA would rescue, unlike Rht wheat. Brassinosteroid overproduction is not involved.