Abiotic Stress, Biomolecules, Secondary Metabolites, Plant Hormones, Senescence and Programmed Cell Death

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. An Arabidopsis mutant that is hypersensitive to NaCl accumulates more Na⁺ in its root cells than the wild type. The mutated gene encodes SOS1. SOS1 is best described as

   1. a tonoplast Na⁺/H⁺ antiporter that moves Na⁺ into the vacuole using the V-ATPase gradient
   2. a calcium-binding sensor that recruits a protein kinase to the plasma membrane after salt stress
   3. a plasma-membrane Na⁺/H⁺ antiporter that extrudes Na⁺ using the H⁺-ATPase gradient
   4. an HKT-type transporter that retrieves Na⁺ from the xylem sap before it reaches the leaves
   Show answer

   Answer: C — a plasma-membrane Na⁺/H⁺ antiporter that extrudes Na⁺ using the H⁺-ATPase gradient

   SOS1 is the plasma-membrane Na⁺/H⁺ antiporter of the salt-overly-sensitive pathway: the Ca²⁺ sensor SOS3 recruits the kinase SOS2, which activates SOS1 to pump Na⁺ out using the proton gradient. Loss of SOS1 traps Na⁺ in the cytosol. The tonoplast antiporter is NHX, the tempting look-alike; the Ca²⁺ sensor is SOS3; and xylem Na⁺ retrieval is done by HKT1.

2. Which of the following are typical responses of a crop plant to a slowly developing soil water deficit?

   1. Accumulation of compatible solutes such as proline, which lower Ψs without inhibiting cytosolic enzymes
   2. A rise in the root-to-shoot ratio, because root growth continues longer than shoot growth
   3. Wider stomatal opening to cool the leaves by transpiration as the soil dries
   4. A rise in leaf ABA, partly through increased expression of NCED
   Show answer

   Answer: A — Accumulation of compatible solutes such as proline, which lower Ψs without inhibiting cytosolic enzymes; B — A rise in the root-to-shoot ratio, because root growth continues longer than shoot growth; D — A rise in leaf ABA, partly through increased expression of NCED

   Under water deficit ABA rises, largely because NCED is induced; compatible solutes such as proline accumulate for osmotic adjustment; and shoot growth is inhibited before root growth, so the root:shoot ratio increases. Stomata close, not open, as the soil dries, because ABA drives closure to save water, even at the cost of a warmer leaf.

3. Two near-isogenic wheat lines differ in tolerance to acid soil. When roots are exposed to Al³⁺, only the tolerant line releases malate from its root apices, and only its root elongation is maintained. Which mechanism best explains the tolerance?

   1. Phytochelatins bind Al³⁺ through their thiol groups and sequester it in the vacuole
   2. Malate released through an Al³⁺-activated anion channel chelates Al³⁺ outside the root apex
   3. Malate is respired in the root apex to supply ATP for pumping Al³⁺ into the xylem
   4. Malate acidifies the rhizosphere further, converting Al³⁺ into a less toxic aluminate ion
   Show answer

   Answer: B — Malate released through an Al³⁺-activated anion channel chelates Al³⁺ outside the root apex

   Tolerant wheat opens the ALMT1 malate channel when Al³⁺ is present; the released malate forms a non-toxic complex with Al³⁺ in the rhizosphere, so less Al³⁺ reaches the elongation zone. Phytochelatins are thiol peptides that chelate soft metals such as Cd²⁺ and arsenic, not the hard cation Al³⁺. Respiring malate would not explain the dependence on its release, and acidification would increase Al³⁺ solubility; aluminate forms only at high pH.

4. Which step is the committed and usually rate-limiting step of ABA synthesis induced by water deficit?

   1. Oxidation of abscisic aldehyde to ABA by abscisic aldehyde oxidase in the cytosol
   2. Epoxidation of zeaxanthin to violaxanthin by zeaxanthin epoxidase in the plastid
   3. 8′-Hydroxylation of ABA to phaseic acid by a CYP707A cytochrome P450
   4. Cleavage of a 9-cis-epoxycarotenoid to xanthoxin by NCED in the plastid
   Show answer

   Answer: D — Cleavage of a 9-cis-epoxycarotenoid to xanthoxin by NCED in the plastid

   NCED (9-cis-epoxycarotenoid dioxygenase) cleaves the C40 precursor into C15 xanthoxin; it is the first step committed only to ABA and its expression rises sharply with dehydration. Zeaxanthin epoxidation also serves the xanthophyll cycle, so it is not committed. The aldehyde oxidase step is the last step, and CYP707A hydroxylation inactivates ABA.

5. The dominant abi1-1 mutation alters a PP2C so that the ABA–PYR/PYL receptor complex can no longer bind it, although its phosphatase activity is intact. Plants carrying abi1-1 would be expected to

   1. close their stomata constitutively, because the PP2C now permanently activates SnRK2 kinases
   2. wilt readily and show reduced seed dormancy, despite normal or higher ABA content
   3. respond normally to ABA, because the wild-type PP2C copy is still bound and inhibited
   4. be fully rescued by spraying ABA, because the mutation lies in an ABA-synthesis enzyme
   Show answer

   Answer: B — wilt readily and show reduced seed dormancy, despite normal or higher ABA content

   PP2C keeps SnRK2 inactive; ABA works by letting the receptor inhibit PP2C. A PP2C the receptor cannot capture keeps dephosphorylating SnRK2 whatever the ABA level, so stomata fail to close and seeds lose dormancy: an ABA-insensitive phenotype with normal or raised ABA. It is dominant because the uninhibitable mutant enzyme suffices to keep SnRK2 off, so the wild-type copy cannot restore the response. PP2C inactivates SnRK2 rather than activating it, and the gene is not a biosynthetic enzyme.

6. An enzyme has Km = 10 µM for its substrate. A competitive inhibitor with Ki = 2 µM is present at 4 µM, and the substrate concentration is 45 µM. Using v = Vmax[S]/(αKm + [S]) with α = 1 + [I]/Ki, the rate as a percentage of Vmax is ____.

   Numerical answer — type the value.

   Show answer

   Answer: 60

   α = 1 + 4/2 = 3, so the apparent Km = 3 × 10 = 30 µM. v/Vmax = 45/(30 + 45) = 45/75 = 0.60, which is 60%. Without the inhibitor it would be 45/55 = 82%. Using α = [I]/Ki = 2 (forgetting the 1) gives 69%.

7. An enzyme obeying Michaelis–Menten kinetics has Km = 25 µM. The substrate concentration (in µM) at which v = 0.8 Vmax is ____.

   Numerical answer — type the value.

   Show answer

   Answer: 100

   0.8 = [S]/(Km + [S]) gives 0.8Km + 0.8[S] = [S], so 0.2[S] = 0.8Km and [S] = 4Km = 4 × 25 = 100 µM. Taking 0.8 × Km = 20 µM confuses the fraction of Vmax with a multiple of Km.

8. Which lipid class makes up most of the lipid of thylakoid membranes?

   1. Phosphatidylcholine and other phospholipids, as in the plasma membrane
   2. Monogalactosyl- and digalactosyldiacylglycerol, the galactolipids
   3. Triacylglycerols packed with oleosins, as in seed oil bodies
   4. Sterols and sphingolipids that form ordered membrane domains
   Show answer

   Answer: B — Monogalactosyl- and digalactosyldiacylglycerol, the galactolipids

   Thylakoids are dominated by the galactolipids MGDG and DGDG, rich in α-linolenic acid, with phospholipid (mainly phosphatidylglycerol) a minor part. Phospholipids dominate the plasma membrane, which makes that option tempting. Triacylglycerols are storage lipids, not membrane lipids, and sterols and sphingolipids are typical of the plasma membrane.

9. Phytoene, the first carotenoid of the pathway, is formed by joining two molecules of geranylgeranyl diphosphate (GGPP, C20). How many C5 isoprene units (IPP plus DMAPP) are incorporated into one molecule of phytoene?

   Numerical answer — type the value.

   Show answer

   Answer: 8

   Each GGPP is C20, built from one DMAPP and three IPP, which is four C5 units. Two GGPP give a C40 phytoene, so 2 × 4 = 8 C5 units. Counting a single GGPP gives 4, and dividing 40 by 10 (confusing C5 units with C10 monoterpene units) also gives 4.

10. Chalcone synthase condenses one p-coumaroyl-CoA, whose acyl part has 9 carbons (a C6–C3 unit), with several malonyl-CoA units. Each malonyl-CoA contributes two carbons to the product after losing one carbon as CO₂. The product, naringenin chalcone, has a C6–C3–C6 skeleton. How many molecules of malonyl-CoA are used per chalcone?

    Numerical answer — type the value.

    Show answer

    Answer: 3

    A C6–C3–C6 skeleton has 15 carbons. The p-coumaroyl unit supplies 9, so malonyl units supply 15 − 9 = 6 carbons; at two carbons each this is 6/2 = 3 malonyl-CoA, which release 3 CO₂ and build the A ring. Dividing 6 by 3 (counting all three malonyl carbons) gives 2, a slip.

11. Which of the following statements about the biosynthetic origin of plant metabolites are correct?

    1. Gibberellins and carotenoids are derived from IPP made by the plastid MEP pathway
    2. Sterols and brassinosteroids are derived from squalene made from cytosolic MVA-pathway IPP
    3. Nicotine is a terpenoid indole alkaloid made from tryptophan in tobacco leaves
    4. Caffeine is a benzylisoquinoline alkaloid derived from tyrosine
    Show answer

    Answer: A — Gibberellins and carotenoids are derived from IPP made by the plastid MEP pathway; B — Sterols and brassinosteroids are derived from squalene made from cytosolic MVA-pathway IPP

    Plastid MEP-pathway IPP feeds diterpenes such as gibberellins and tetraterpenes such as carotenoids, while cytosolic MVA-pathway IPP feeds squalene and hence sterols and brassinosteroids. Nicotine is a pyridine–pyrrolidine alkaloid from nicotinic acid and ornithine, made in roots and carried to leaves. Caffeine is a purine alkaloid from xanthosine; benzylisoquinolines such as morphine come from tyrosine.

12. A dominant dwarf mutant of Arabidopsis contains high levels of bioactive GA, and applying GA does not restore its height. The mutant protein is a DELLA protein lacking the DELLA motif. The best explanation is that

    1. the altered DELLA is not bound by GA–GID1, so it is not degraded and keeps repressing growth
    2. the plant cannot make bioactive GA, because the altered DELLA protein inhibits GA 3-oxidase
    3. the altered DELLA has lost its repressor activity, so growth genes are permanently switched off
    4. GID1 is missing, because the DELLA motif normally serves as the GA-binding site of the receptor
    Show answer

    Answer: A — the altered DELLA is not bound by GA–GID1, so it is not degraded and keeps repressing growth

    GA bound to GID1 binds the DELLA motif region, and the complex recruits SCF(SLY1) for DELLA ubiquitination and proteasomal degradation. A DELLA lacking that motif (as in gai) cannot be recognised, so it stays and represses growth whatever the GA level: a dominant, GA-insensitive dwarf with raised GA because feedback is lost. A GA-deficient dwarf would respond to GA; losing repressor activity would give a tall, slender plant; and GA binds GID1, not the DELLA motif.

13. Which of the following statements about hormone perception in plants are correct?

    1. Auxin acts as molecular glue that stabilises the binding of Aux/IAA repressors to the F-box protein TIR1
    2. JA-Ile promotes binding of JAZ proteins to COI1, leading to JAZ degradation and release of MYC2
    3. Ethylene binding activates the CTR1 kinase, which then switches on EIN2 and the ethylene response
    4. ABA binding to PYR/PYL receptors targets SnRK2 kinases for proteasomal degradation
    Show answer

    Answer: A — Auxin acts as molecular glue that stabilises the binding of Aux/IAA repressors to the F-box protein TIR1; B — JA-Ile promotes binding of JAZ proteins to COI1, leading to JAZ degradation and release of MYC2

    Auxin sits in the TIR1 pocket and fills the gap to Aux/IAA, and JA-Ile does the same between COI1 and JAZ; in both cases the repressor is ubiquitinated and destroyed, releasing ARF or MYC2. Ethylene does the reverse of activation: without ethylene the receptors keep CTR1 active and CTR1 keeps EIN2 off; ethylene switches the receptors and CTR1 off. ABA receptors inhibit PP2C, which frees and activates SnRK2; SnRK2 is not destroyed.

14. Seedlings of a loss-of-function ctr1 mutant are grown in darkness in air with no added ethylene and with an inhibitor of ethylene synthesis. They would show

    1. long, thin hypocotyls without an exaggerated hook, like ethylene-insensitive etr1-1 seedlings
    2. a normal etiolated phenotype that turns into the triple response only when ethylene is supplied
    3. the triple response, because CTR1 is a negative regulator that acts downstream of the receptors
    4. the triple response only if the receptors are also mutated, because CTR1 acts upstream of them
    Show answer

    Answer: C — the triple response, because CTR1 is a negative regulator that acts downstream of the receptors

    CTR1 is the kinase that, activated by ethylene-free receptors, keeps EIN2 off. Without CTR1, EIN2 signals constitutively, so dark-grown ctr1 seedlings show the triple response (short, thick hypocotyl, exaggerated apical hook, short root) even when ethylene synthesis is blocked. Long hypocotyls without a hook describe insensitive mutants such as etr1-1; the ethylene-requiring phenotype is the wild type; and CTR1 acts downstream, not upstream, of the receptors.

15. A dwarf, high-tillering rice mutant has its tillering restored to wild-type levels by treatment with the synthetic strigolactone GR24. Which gene is most likely mutated?

    1. D14, which encodes the α/β-hydrolase receptor that binds and hydrolyses strigolactone
    2. D3, which encodes the F-box protein that targets the repressor in strigolactone signalling
    3. D53, whose protein product is the repressor degraded in response to strigolactone
    4. D10, which encodes the carotenoid cleavage dioxygenase CCD8 of strigolactone synthesis
    Show answer

    Answer: D — D10, which encodes the carotenoid cleavage dioxygenase CCD8 of strigolactone synthesis

    Rescue by applied hormone means the plant can still perceive strigolactone but cannot make it, so the lesion is in biosynthesis: D10 encodes CCD8. Mutants in the receptor D14 or the F-box D3 are high-tillering but insensitive to GR24, because perception is broken. The known d53 mutant is a dominant, stabilised repressor, also high-tillering and GR24-insensitive, and loss of D53 would reduce tillering instead.

16. Which pairing of a plant peptide hormone with its role is correct?

    1. Systemin, a sulfated pentapeptide that promotes cell proliferation in culture
    2. EPF2, a peptide from the mesophyll that increases the number of stomata
    3. CLV3, a CLE peptide that limits WUSCHEL expression in the shoot meristem
    4. RALF, a peptide that acidifies the apoplast and so speeds up cell expansion
    Show answer

    Answer: C — CLV3, a CLE peptide that limits WUSCHEL expression in the shoot meristem

    CLV3 from the stem cells is perceived by the CLV1 receptor kinase and represses WUSCHEL, the feedback loop that holds the shoot meristem at a constant size. The sulfated pentapeptide is phytosulfokine; systemin is an 18-amino-acid wound signal. EPF2 restricts stomatal number, while stomagen (EPFL9) from the mesophyll increases it. RALF alkalinises the apoplast and slows expansion.

17. Tobacco plants carrying the cytokinin-synthesis gene IPT under the control of the senescence-activated SAG12 promoter keep their lower leaves green much longer than controls, yet otherwise grow normally. Why is the rest of the phenotype normal?

    1. Cytokinin is made only as a leaf begins to senesce, and the cytokinin then switches the promoter off
    2. IPT degrades ethylene in old leaves, so senescence is blocked without any change in hormone levels
    3. The SAG12 promoter is active only in roots, so only small amounts of cytokinin reach the leaves
    4. Cytokinin made in every leaf from germination onwards is destroyed by cytokinin oxidase in young tissue
    Show answer

    Answer: A — Cytokinin is made only as a leaf begins to senesce, and the cytokinin then switches the promoter off

    The SAG12 promoter switches on only at the onset of senescence, so IPT makes cytokinin just where and when a leaf starts to yellow; the cytokinin delays senescence, which in turn turns SAG12 off. This autoregulatory loop avoids the stunting and loss of root growth caused by constitutive cytokinin overproduction. IPT is an isopentenyltransferase, not an ethylene-degrading enzyme, and SAG12 is a leaf senescence promoter, not a root promoter.

18. Which of the following are examples of programmed cell death in plants?

    1. Death of the protoplast during differentiation of xylem vessel elements
    2. Death of the aleurone cells of cereal grains after germination, hastened by gibberellin
    3. Death of leaf cells crushed mechanically during a hailstorm
    4. Rapid death of cells at an infection site during the hypersensitive response
    Show answer

    Answer: A — Death of the protoplast during differentiation of xylem vessel elements; B — Death of the aleurone cells of cereal grains after germination, hastened by gibberellin; D — Rapid death of cells at an infection site during the hypersensitive response

    Tracheary-element differentiation, post-germination aleurone death and the hypersensitive response are all genetically controlled, executed by the cell's own machinery (vacuolar rupture, hydrolases, metacaspases, vacuolar processing enzyme). Cells crushed by hail die by uncontrolled injury, which is necrosis, not programmed death.