Questions on plant responses: tropisms, auxins and the classic experiments

Question 1

A student claims that the classic decapitation-and-auxin-paste experiment proves that auxin acting alone in the lateral bud is what keeps it dormant. Evaluate this claim, using the further evidence described in this lesson.

Accepted Answer

B1 in support: the paired result is genuine evidence for auxin's involvement, since removing the tip releases the buds and replacing the auxin with a paste on the cut stump keeps them dormant again, with a plain, auxin-free paste as the control that fails to do so B1 in support: because only the auxin supply differs between the treated and control stumps, the result does rule out the paste itself, or the act of cutting, as the cause B1 against: auxin does not accumulate inside the buds it inhibits, which would be expected if it acted there directly, so the evidence does not establish that auxin acts within the bud itself B1 against: cytokinin applied directly to a dormant lateral bud on an otherwise intact plant, with the apical bud left in place, causes that bud to grow out, showing a bud can be released from dormancy without any change in the auxin reaching it B1 judgement: the experiment shows convincingly that auxin from the apical bud contributes to keeping lateral buds dormant, but not that it is the sole cause or that it acts directly inside each bud; a balance of several signals, including cytokinins and strigolactones, is a more accurate account

Question 2

The tips of oat coleoptiles were cut off and replaced, with a thin layer separating each tip from its stump, and the coleoptiles were lit from one side. With a layer of gelatine, through which water and dissolved substances can diffuse, the coleoptiles bent towards the light. With a layer of mica, through which nothing can diffuse, they did not bend. In a further experiment, a thin plate of mica pushed part-way into one side of an intact coleoptile, just below the tip, stopped bending when it was on the shaded side but not when it was on the lit side. Explain what these results show about the signal that passes from the tip to the region that bends.

Accepted Answer

B1 the signal crosses a gap of non-living gelatine, so it does not need a continuous line of living cells or a mechanical link from the tip B1 it is stopped by mica, which nothing can diffuse through, so it is most likely a chemical that diffuses from the tip; wet gelatine also conducts ions, so these results alone do not rule out an electrical signal B1 blocking the shaded side stops bending while blocking the lit side does not, so the substance travels down the shaded side B1 the shaded side therefore elongates faster than the lit side, so the coleoptile bends towards the light; the experiments do not identify the substance

Question 3

Explain how auxin arriving at a cell in the growing region of a shoot causes that cell to become longer.

Accepted Answer

B1 auxin binds to receptors in the cell and stimulates the proton pumps in the cell surface membrane B1 the pumps actively transport hydrogen ions out into the cell wall, so the pH of the wall falls to about 5 B1 at that pH proteins called expansins become active and loosen the bonds between the cellulose microfibrils, so the wall becomes more plastic A1 the turgor pressure already present can then stretch the loosened wall, water enters by osmosis, and the cell lengthens along its axis

Question 4

A germinating seedling is laid on its side in complete darkness. Explain why its shoot grows upwards and its root grows downwards, even though auxin collects on the lower side of both.

Accepted Answer

B1 gravity is detected by statoliths, dense starch-filled organelles that sediment to the lower side of particular cells, such as the columella cells of the root cap B1 auxin is then redistributed to the lower side of both the shoot and the root B1 in the shoot that concentration stimulates elongation, so the lower cells elongate more than the upper ones and the shoot curves upwards, which is negative gravitropism A1 root cells are inhibited at concentrations that stimulate a shoot, so the lower cells of the root elongate less than the upper ones and the root curves downwards, which is positive gravitropism

Question 5

A gardener removes the growing tip from a young shrub and it becomes bushy within a fortnight. Explain this result, and explain what would happen if the cut end were smeared with a paste containing auxin.

Accepted Answer

B1 auxin made in the apical bud is transported down the stem and keeps the lateral buds below it dormant, which is apical dominance B1 removing the tip removes the source of that auxin, so the lateral buds are released and grow out into side shoots, and the shrub becomes bushy rather than tall B1 a paste containing auxin on the cut end replaces the supply the apical bud provided, so the lateral buds stay dormant and the shrub does not become bushy A1 the pair of treatments together is the evidence that auxin causes the inhibition; plain paste with no auxin in it is the control, and the buds grow out

Question 6

Tips were cut from oat coleoptiles and stood on small blocks of agar in the dark for some hours, then removed. Each block was placed on one side of the cut surface of a coleoptile whose own tip had been removed, and the coleoptiles were kept in the dark. Each stump curved away from the side carrying the block. A block that had never touched a tip caused no curvature, and blocks that had carried more tips caused greater curvature. Explain what these results show.

Accepted Answer

B1 a substance passed out of the tips into the agar, since a block that had never touched a tip had no effect, so the agar itself is not the cause B1 the side beneath the block grew faster, so the substance stimulates elongation on the side it reaches, which makes the stump curve away from that side B1 the bending happened in darkness and with no tip present, so the substance alone is enough to cause curvature and light is not needed for it to act B1 greater curvature with more tips shows that the response depends on the amount of substance, so the angle of curvature can be used to measure how much substance a block contains

Question 7

Explain how abscisic acid causes a stoma to close when a plant is short of water.

Accepted Answer

B1 abscisic acid is released when a plant begins to lose more water than it is taking up B1 it causes solutes to leave the guard cells, so the water potential of the guard cells rises, becoming less negative B1 water then leaves the guard cells by osmosis down that water potential gradient, and the cells become flaccid A1 flaccid guard cells can no longer hold the stoma open, so it closes, which reduces transpiration but also reduces the rate of photosynthesis because less carbon dioxide can enter

Question 8

Compare the effects of gibberellin with the effects of abscisic acid on seed dormancy and on germination.

Accepted Answer

B1 gibberellin promotes germination, diffusing to the aleurone layer and causing transcription of the amylase gene, whereas abscisic acid maintains dormancy and prevents germination from starting B1 in a seed about to germinate, gibberellin overcomes the dormancy abscisic acid had been maintaining, so the relative concentrations of the two substances decide whether germination proceeds B1 gibberellin's effect here is on gene expression in the aleurone, producing an enzyme, whereas abscisic acid opposes this in the aleurone, suppressing amylase production A1 the two substances are broadly antagonistic here: one promotes growth and germination, and the other suppresses them, in the way that many pairs of plant growth substances balance each other

Question 9

Groups of a short-day plant were given the daily light and dark periods in the table, and whether each group flowered was recorded. The results are constructed to show the classic pattern. Explain the results of groups 2, 3 and 4, using what happens to phytochrome.

Accepted Answer

B1 in group 2 the night is longer than the critical night length, so Pfr reverts to Pr for long enough that little is left by dawn B1 Pfr inhibits flowering in a short-day plant, so with little Pfr the plant flowers B1 in group 3 the red light converts Pr to Pfr in the middle of the night, so neither part of the night is longer than the critical length and Pfr inhibits flowering B1 in group 4 the far-red converts the Pfr made by the red light straight back to Pr, so the night acts as one unbroken long night and the plant flowers

Question 10

Describe the appearance of a bean seedling that has grown in complete darkness, and describe how phytochrome brings about the changes when the seedling is moved into light.

Accepted Answer

B1 the stem is long and thin with long internodes, often with a hook at the top B1 the leaves are small, folded and yellow because no chlorophyll has been made; this is etiolation B1 in light, red light converts Pr into Pfr, the active form, which changes the transcription of genes B1 stem elongation slows, the hook straightens, the leaves expand and chlorophyll is made, which is photomorphogenesis

Question 11

Explain how IAA arriving at a cell in the growing region of a shoot changes the transcription of genes in that cell.

Accepted Answer

B1 without IAA, transcription factors called auxin response factors are bound to the auxin-responsive genes but held inactive by repressor proteins B1 IAA binds to a receptor protein in the nucleus, and the receptor with IAA bound causes the repressor proteins to be broken down B1 the freed auxin response factors activate transcription, so RNA polymerase transcribes the auxin-responsive genes B1 the mRNA is translated into proteins, such as expansins, that loosen the cell wall and sustain elongation

Question 12

Explain how a synthetic auxin sprayed over a field can kill broad-leaved weeds without seriously harming the cereal crop growing among them.

Accepted Answer

B1 broad-leaved weeds absorb and transport more of the synthetic auxin than the cereal, which intercepts less spray on its narrow upright leaves and breaks down more of what it absorbs; credit either the uptake or the breakdown point B1 the high concentration inside the weed causes rapid and uncontrolled elongation and growth A1 the weed cannot supply the water and nutrients that growth demands and it dies, while the concentration in the cereal's tissues stays too low to have that effect

Question 13

Maltsters soak barley grains in water and add gibberellin to the steeping tank. Suggest why this makes the starch in the grains turn into sugars sooner and more evenly across the batch.

Accepted Answer

B1 gibberellin diffuses to the aleurone layer surrounding the starchy endosperm and causes transcription of the gene for amylase B1 the amylase produced hydrolyses the stored starch to maltose, so sugars become available sooner than if the maltster waited for each embryo to secrete its own gibberellin A1 every grain receives gibberellin at the same time and in the same amount, so grains that would otherwise germinate at different rates produce amylase together and the batch is more uniform

Question 14

Suggest why a gardener treats the cut end of a stem cutting with a rooting powder containing auxin, and suggest what result they would expect if they used a paste with no auxin in it instead, as a comparison.

Accepted Answer

B1 auxin stimulates cell elongation and division in the tissue at the cut end, which is needed for new roots to form there, so a treated cutting roots sooner and more reliably than an untreated one B1 an auxin-free paste applied to a comparison group of cuttings acts as a control, showing what proportion of cuttings would root without any auxin being supplied at all A1 if far fewer of the untreated cuttings produced roots, or produced them much later, that difference would be the evidence that the auxin itself, rather than something else about the paste, caused the improvement

Question 15

Fruit is often picked while unripe, transported long distances, and then exposed to ethene gas just before it reaches the shop. Suggest why picking the fruit unripe is useful, and suggest why the ethene is applied only near the end of the journey rather than at the start.

Accepted Answer

B1 unripe fruit is firmer and less easily damaged than ripe fruit, so it survives the mechanical stress of transport far better B1 ethene promotes ripening, so exposing the fruit to it converts the firm, unripe fruit into ripe fruit ready for sale A1 applying the ethene only near the end of the journey means the fruit stays firm and undamaged for as much of the transport as possible, ripening just before it is needed

Question 16

Chrysanthemums are short-day plants that naturally flower in autumn. Suggest how a grower could produce flowering chrysanthemums for sale in June, and how they could stop plants flowering in autumn until they are wanted.

Accepted Answer

B1 in June, cover the plants with blackout sheeting for part of each day so that the night is longer than the critical night length B1 during the long night Pfr reverts to Pr, so little of the Pfr that inhibits flowering is left and the plants flower B1 in autumn, light the plants briefly in the middle of each night, which converts Pr to Pfr and breaks the long night, so they do not flower

Question 17

Seedlings growing beneath the leaves of taller plants have longer stems than seedlings of the same species growing in the open. Explain this difference using phytochrome.

Accepted Answer

B1 chlorophyll in the leaves above absorbs most of the red light but little of the far-red, so the light reaching the shaded seedlings is rich in far-red B1 far-red light converts Pfr into Pr, so a smaller proportion of the shaded seedlings' phytochrome is in the active Pfr form B1 less Pfr means less inhibition of stem elongation, so the shaded seedlings elongate faster, which tends to lift their leaves out of the shade

Question 18

In a starch agar assay, half barley grains without embryos were soaked in two concentrations of gibberellic acid. The mean diameter of the clear zones was 9.0 mm at the lower concentration and 15.0 mm at the higher; the results are invented for this question. Calculate the percentage increase in the area of the clear zone between the two concentrations.

Accepted Answer

M1 radii 4.5 mm and 7.5 mm, so areas π × 4.5² = 63.6 mm² and π × 7.5² = 176.7 mm² M1 (176.7 − 63.6) ÷ 63.6 × 100 A1 178%; accept 177% to 178%

Question 19

In an investigation of the effect of gibberellin on amylase production, explain why the embryo is cut off each barley grain, why the half grains are surface-sterilised, and why one group of half grains is soaked in distilled water only.

Accepted Answer

B1 the embryo is the grain's source of gibberellin, so removing it means any amylase made depends on the gibberellin supplied B1 bacteria and fungi on the grain surface make amylase, which would hydrolyse starch and give clear zones not caused by the aleurone layer B1 the water-only group is a control showing how much amylase is made with no gibberellin, so any increase can be put down to the gibberellin

Question 20

Give the name of the tropism shown by a shoot growing towards light from one side, and give the name of the plant growth substance responsible for the bending.

Accepted Answer

B1 it is positive phototropism, because growth is towards the stimulus B1 the substance is auxin, in practice indole-3-acetic acid

Question 21

Charles and Francis Darwin lit grass seedlings from one side. Seedlings with an opaque cap over the tip of the coleoptile did not bend. Seedlings with the tip uncovered but the region below it shielded from light bent towards the light, and the bend formed in the shielded region. Explain what these results show about where light is detected and where the response happens.

Accepted Answer

B1 covering only the tip stops bending, while shading the region below it does not, so light is detected at the tip B1 the bend forms below the tip, in a region that was itself in shade, so a signal must pass from the tip to the region that bends; these results do not show what that signal is

Question 22

Name the tropism shown by a root growing towards water, and name the tropism shown by a pollen tube growing down a style towards chemicals released by the ovule.

Accepted Answer

B1 it is positive hydrotropism B1 it is positive chemotropism

Question 23

State the colour of light that converts Pr into Pfr, and state what happens to Pfr when a plant is kept in darkness.

Accepted Answer

B1 red light, with a peak at about 660 nm, converts Pr into Pfr B1 in darkness Pfr slowly reverts to Pr, and some of it is broken down

Group	Light / h	Dark / h	Treatment in the middle of the dark period	Flowered
1	16	8	none	no
2	8	16	none	yes
3	8	16	5 minutes of red light	no
4	8	16	5 minutes of red, then 5 minutes of far-red	yes