Dame Alice Owen's School271.An investigation was carried out into the effect of light intensity on the uptake and release of carbon dioxide by two green plant species J and K.Single leaves of each species, still attached to plants, were sealed in glass vessels containing a known mass of carbon dioxide. The leaves in the vessels were exposed to light of known intensity for one hour. The change in mass of carbon dioxide in each vessel was determined and the change in mass of carbon dioxide per cm2 of leaf surface was then calculated. The experiment was repeated at a range of light intensities for both species.The results are shown in the graph below. Light intensity is expressed as a percentage of normal daylight.Adapted from Plant Physiology, Salisbury and Ross (1978)(a)(i)From the graph, determine the light intensity at which there is no net exchange of carbon dioxide by leaves of species J and K.J2.5 3.0%;K6%;(2 marks)(ii)Comment on the relationship between light intensity and the exchange of carbon dioxide in species J.CO2 decreases / OR uptake increases as light increases;gradient steepest up to 5%;little change after 15% / eq;(nett) CO2 produced / OR released up to 2.5/3% / below 3.0% / converse;correct ref to relative rate of photosynthesis and respiration;correct ref to compensation point;(3 marks)(b)(i)Give two differences between the curves for species J and K.Any 2 points allowed.gradient steeper / eq;levels off (at lower light intensity) / reaches a maximum;maximum rate of uptake is lower / eq / OR 33 units difference / higher uptake in K at high light intensities;change not directly proportional to light intensity;J has a higher uptake below 19.5%;J reaches its compensation point at lower light intensities than K;Allow converse if K stated.(2 marks)(ii)Suggest which species would be better adapted to a shaded habitat giving a reason to support your answer.J;compensation point / eq at lower light intensity / faster rate of photosynthesis at low light intensity;If K given, only allow if state that in the shade, light intensity is likely to be 20.5 55% of normal daylight; [So 2 or O!](2 marks)(c)(i)State one environmental factor, other than light intensity, which could also affect the uptake and release of carbon dioxide in the two plant species.temperature / light quality / wavelength / water availability / wind / air speed / oxygen;(1 marks)(ii)Explain why the factor you have chosen in (i) has an effect on carbon dioxide uptake and release.temp: affects (rate of) enzyme reactions; Calvin cycle / light independent reactions uses CO2 / OR respiration releases it / OR ref to photorespiration athigh temps.light quality/wavelength: pigments absorb specific wavelengths / eq;rate of photosynthesis changes so CO2 used - uptake changes;water: plant wilts / stomata close / converse;less photosynthesis so less CO2 used / CO2 cannot enter / converse;wind / air speed: increased water loss so stomata close;less photosynthesis so less CO2 used /CO2 cannot enter / converse;oxygen: reduces CO2 uptake / increases CO2 release;competitive inhibitor for RUBISCO / OR combines with RUBISCO / OR ref to photorespiration;(2 marks)[Total 12 marks]2.A belt transect was carried out from a grassy field through marsh and swamp into a pond. Along the transect the relative abundance of the plant species present was recorded.The diagram below shows the profile of the transect. The height of the bars indicates the relative abundance of species.(a)Describe how you would use a quadrat to determine the distribution and abundance of plants along this belt transect.size of quadrat stated as 0.5 / 1m square / point quadrat;placed along a (straight) line / the belt transect;at regular intervals / next to each other / flip it over;determine abundance by counting number of each species / % cover / ACFOR scale;(3 marks)(b)Name two abiotic factors, other than water, which could affect the distribution of the plant species along the transect shown in the diagram.soil type / eq; oxygen availability; minerals / ions / namedeg; pH; temperature; (2 marks)(c)(i)Compare the plant community in the grassy field with that in the marsh and suggest reasons for the differences.could have been planted / ref to agricultural use;maintained by mowing / eq; drainage;marsh plants compete better / more tolerant / OR adapted / prefer damp than those in field / OR converse for field; marsh plants not eaten by cattle / OR converse for field; marsh has different species from the field / OR greater diversity in the marsh;at least two specific examples of named species / OR ref to differences in abundance of named species;(3 marks)(ii)Comment on the changes in distribution of the plant species present along the transect from the marsh through the into the pond.most / some plants occupy more than one area; yellow flag / jointed / flowering rush only in one area; need sweet grass occupies / eq 3 areas; swamp has 4 species that occur in the pond; marsh has widest variety / the most species / the greatest diversity; no species occupy the swamp only; species are adapted to different conditions along the transects / OR specific eg stated; there is a greater abundance of plants in the pond; due to fertiliser run-off / leaching from surrounding land;(3 marks)[Total 11 marks]3. No mark scheme4.An investigation was carried out into the effect of different concentrations of indoleacetic acid (IAA) on the growth of coleoptiles (shoots) from 5 day old barley seedlings.The tips of 70 coleoptiles were cut off, and the remainder of each coleoptile was trimmed to a length of 10 mm. Ten of these trimmed coleoptiles were placed in each of six solutions of IAA, and then incubated in the dark at 20 oC for 72 hours. A further ten trimmed coleoptiles were incubated for the same length of time in distilled water.The coleoptiles were then measured, and the mean increases in their lengths were plotted against the concentration of IAA used. The whole experiment was repeated at a temperature of 30 oC. The results are shown in the graph below.(a)Describe the effect of different concentrations of IAA on the growth of the coleoptiles.all concentrations of / IAA cause(s) an increase in length / eq;concentrations 0.0 to 0.01 / from 0 to 0.001 cause small increase in length;0.01.to 1.0 cause sleep increase in length;1.0to 100.0 cause lower increase in length/increase inhibited;greatest increase at 1.0 / 1.0 is the optimum;(3 marks)(b)Suggest an explanation for the differences between the results obtained at 20 oC and at30oC.IAA transported / eq. into coleoptiles more rapidly at 30oC / less rapidly at 20oc / increase in membrane permeability to IAA at 30oc;increased enzyme / metabolic / eq. activity at 30oc / decreased at 20oC;resulting in more / increased growth / elongation at 30oC / decreased at 20oC;(2 marks)(c)(i)Suggest why the tips were removed from the coleoptiles.ensures comparability of results / removes uncontrolled variable;since IAA (produced) in tips;different tips may produce different concentrations;(2 marks)(ii)Suggest why the investigation was carried out in the dark.light affects distribution of IAA / inactivates IAA / destroys IAA;(1 mark)(d)Explain one way in which auxins, such as IAA, may be used in horticulture and agriculture.spray on broad leaved weeds / use as selective weedkiller;stimulates growth of weeds to exhaustion / eq;dip cut stems / petiole into auxin / IAA powder.stimulates production / growth of (adventitious) roots;spray on to unpollinated / unfertilised flowers of fruit trees;produces seedless fruits;spray on to developing fruit / named e.g;to retard abscission / prevent fruit drop / eq;add to nutrient agar solution;to stimulate clove formation / callusgrowth / micropropagation;( = linked points)(2 marks)[Total 10 marks]5.(a)Increase in carbon dioxide increases the yield (not time / rate references ) ;(Because more) carbon dioxide available for photosynthesis / eq ;So more sugars are stored / fruit grows bigger ;Increase greatest between 0.03 and 0.06 / 0.09 ;Not much increase above 0.09 ;Suggests some other factor limiting ;max 4(b)Yield is always greater / greater at every concentration of carbon dioxide ;Follows similar pattern ;Credit manipulation of figures / must be comparative differences /eq ;max 2(c)Density at 28 000 causes overcrowding ;(Intraspecific) competition ;For light / mineral ions / water ;Plants could be smaller ;max 2(d)Speeds up the light-independent reactions /Calvin cycle of photosynthesis (not dark reaction) ;Which are enzyme controlled / rate of reaction istemperature dependent / reference to kinetic energy ;Sensible reference to fruit set / sucessful pollination / fertilisation at increased temperature ;2(e)(i)Uneconomical to heat the glasshouse for a verysmall increase in yield ;1(ii)Greatest increase in yield at all planting densities /at all temperatures / at 20C ;1[12]6.(a)(Some) trees / specific examples [usually deciduous /broad leaved / ash / hazel /birch / willow] cut down to ground level / leave stool / leave stump ;Allowed to regrow / new shoots will regrow / appear ;2(b)(i)Look for comparative points:Same amount (of light) reaches both until April / until May / in March and April ;Coppiced reaches peak / increases in May while non-coppiced peaks in April / decreases in May ;Any good comparative use of figures ;Both fall after May ;Amount of light reaching coppiced floor remains higher thannon-coppiced from April / May to July ;max 3(ii)(Coppiced wood) more light for photosynthesis ;More light leads to more flowers / earlier flowering ; More successful pollination / seed production ;More time for ground flora to complete life cycle/eq ;Different light levels produce different plant communities /reference to shade tolerant plants ;[Allow converse points for non-coppiced.]max 2(c)Greater range of habitats / niches in coppiced ;Each area / age of coppice has different environment /seres reference / conditions ;Conditions always changing to suit a different range of species ;Greater diversity of plants / greater range of food sources /eq ;[Allow converse points for non-coppiced]max 2[9]7.(a){chemical energy / carbohydrates stored} / {energy fixed/ eq};in {producers / green plants/ by {photosynthesis / autotrophic nutrition} ;2(b)(i)180 145 = 35;(5 / 35) 100 or other intermediate stage ;= 14.3 (%) ;3(ii)1.{decompose / breakdown / rot /eq} {dead bodies / remains/eq} ;2.using external digestion ;3.to release {nutrients / nitrates / eq} / recycling of nutrients ;2(iii)1.temperature lower ;2.lower enzyme activity ;3.shorter growing season / less sunlight / less suitablewavelength of light ;4.less photosynthesis ;5.less water ;2(c)1.replanting after harvesting trees ;2.selective felling of timber trees, leaving rest of forest intact ;3.{pollarding/ coppicing} / harvesting/eq on rotation ;4.(coppicing) trees cut at ground level and allowed to regrow /(pollarding) cut leaving short trunk idea and regrow;5.for 425 years and then harvested ;6.plant fast growing species of trees ;3[12]8.(a).oxygen has low solubility in water;given off as gas / as bubbles // allow converse;easy to / can collect // easy to / can measure volume // allow converse;max 2(b)use hydrogen carbonate / bicarbonate / HCO3 /eq;several / 3 or more stated concentrations;measure length / volume of bubble;over a specified / stated time;repeat at each concentration /eq;maintain same temperature / any named factor /eq;equilibration/eq at each concentration;max 4(c)(i)rate (of photosynthesis / oxygen production)increases as current increases;greatest effect between no current and 2 cm s1 /suitable manipulation of figures;reaches a plateau / little change after 2 or 3 cm s1 /shallower gradient / suitable manipulation of figures;max 2(ii)carbon dioxide dissolved in water;plant will use up all local /eq supplies if no current;current / water movement will bring fresh supplies / more carbon dioxide /ref.. to equivalent to ventilation of gas exchange surface;maintaining concentration gradient of carbon dioxide /eq;plateau when rate of supply equals / is greater than rate of use/eq // another stated factor is limiting / CO2 is no longer limiting /eq;oxygen produced is removed /eq;max 3[11]9.(a)total/eq cover increases [ignore refs. to bare ground]1increase in named e.g.: Yorkshire fog / moss / clover / yarrow 1decrease in named e.g.: cock'sfoot / buttercup1one new species / yarrow appears / increase in species diversity1more main/eq species / more common species / more even cover1one instance of manipulation of figurese.g. moss increase by 41%, clover 15 /eq [not yarrow]1(max 3)(b)(i)reduction in competition from grasses / buttercup / taller plants /eq 1increase in / more light (intensity) / more photosynthesis1increase in availability of water / nutrients / minerals / ions/eq1growth unaffected by mower 1(max 2)[ignore refs. to seeds, flowers](ii)EITHER Yorkshire Fog1increased in cover by 33% / 34 / 3,400% / from 1 to 341OR Yarrow1Annual mowing is 9% higher than frequent /all of the percentages / 10% higher than not mown e.g. 25% is > 14% & 15%1(2)(c)(vegetative) growth by plants growing at edge /eq1spores / seeds carried by wind / blown in [not pollination]1spores / seeds carried / dropped by animals / faeces ref.1dormant seeds / seed bank / seeds left in soil [not bits of roots etc.]1(max 2)(d)lack of / competition for: light 1water1nutrients / minerals /eq1shoots eaten by rabbits / sheep / (grazing) animals 1tree species unsuited to soil / soil not suitable for tree growth / ref to lack of mycorrhizal symbionts/eq wrong pH / soil too shallow / rocky1(max 2)[ignore refs. to climate / altitude][11]10.A student noticed that the density of some plant species appeared to differ depending on how far the plants were from a main road.The mean density (plants per m2) of three plant species A, B and C was measured at different distances from the main road. The mean density of the same three plant species was also determined at the side of a narrower secondary road in the same locality.The results of the investigations are shown in the diagrams below.(a)Describe a procedure the student could have used to determine the mean density of the three plant species.stated area of sample / quadrat / reference to use of point quadrat;at measured distance away from road;several samples at each distance;count number of each of the plant species in each quadrat;calculate mean, total plants total area / eq;carry out investigation at same season / on same day / at same time;(4 marks)(b)(i)Comment on the relationships between plant density and the distance from the main and secondary road for species A and B.density of both / A / B decreases as more further from main road / orconverse;little change in density for A / B / both as distance increases fromsecondary road;mean plant density of A less than B / or converse on both main and secondary roads;lower (overall) density/eq of A / B / both on 2 road / converse;same density of A / B / both at 2.5m on main and 2 roads;allow credit for reference to figures;(4 marks)(ii)Comment on the ways in which the distribution of plant species C differs from that of plant species A.mean density of C increases further from the main road;more or less constant density along secondary road;greater density of C than A at side of secondary road / twice as much of C as of A;(2 marks)(c)In addition to determining the plant densities, the student measured the pH of soil samples taken at the same distances from each of the roads.The results are shown in the graph below.(i)Suggest an explanation for the differences between the pH of the soil at the side of the main road and the pH at the side of the secondary road.acidic gases / carbon dioxide / nitrogen oxides from vehicle exhausts / eq;get washed into soil by rain / ref to gases dissolving;reference to different soil types / road material / road treatment on main andsecondary;more traffic on main road than secondary road (so pH lower at road side)ORconverse;(2 marks)(ii)Using the data given for pH, suggest an explanation for the distribution of the three species A, B and C.A and B can tolerate / eq low pH;C not tolerant of eq low pH / allow converse;so A and B greater density alongside main road / allow converse;further away from main road pH tolerance not so important / competition from other species / C increases;little/less variation in pH along 2 road, C compares more successfully;(2 marks)(iii)Suggest one factor, other than pH, which could account for the differences in density distribution of the plant species at the side of the main road.other toxic chemicals / lead / dust / herbicide / salt / carbon dioxide / presence of rubbish / trampling / cutting management / herbivores / tree cover / shading / qualified light reference / eq;(1 mark)[Total 15 marks]11.An experiment was carried out to investigate the uptake of different mineral ions by barley plants. A large number of barley seedlings was grown in a nutrient solution containing a rangeof mineral ions including potassium (K+), calcium (Ca2+), magnesium (Mg2+) and nitrate(NO3).The experiment was set up as shown in the diagram below.The concentration of these ions in the solution were measured at the beginning and at the end of the experiment.The results are shown in the table below.(a)Calculate the percentage difference between the concentration of nitrate ions at the beginning and at the end of the experiment. Show your working.7.0 1.8 = 5.2;Answer 74.29 / 74.3 (%); (3 marks)(b)What do the results suggest about the mechanism of absorption of potassium ions?Explain your answer.involvement of active transport;(since) all (potassium) ions absorbed;absorption against concentration gradient has occurred;energy / ATP / respiration needed;(3 marks)(c)Suggest an explanation for the changes in concentrations of magnesium and calcium ions during the experiment.plant takes in water;ref water loss / transpiration (from barley plants);(2 marks)(d)State two precautions which should have been taken to ensure that results for all the barley seedlings were comparable.1seedlings all kept at same temperature;same light intensity / same amount of light;duration of experiments the same:same humidity;2ref to plants of same age / size / stage of development / same mass / same no. of leaves / same sized roots / same variety / eq;exclusion of light from roots / wrapping black paper around tube;(2 marks)(e)Describe the pathway taken by mineral ions as they pass from the nutrient solution to the xylem in the roots of the seedlings.root hair / root hair region / epidermal / piliferous layer cell;cortex;endodermis / pericycle;ref apoplast / symplast / vacuolar pathways;(3 marks)[Total 13 marks]12.An investigation was carried out into the effects of carbon dioxide concentration on yield. Tomato plants were cultivated in glasshouses, where it was possible to control the concentration of carbon dioxide in the atmosphere.The carbon dioxide concentrations ranged from 50 to 1200 parts per million (ppm) by volume. The yield of tomatoes was measured in kg per m. The temperature and light intensity conditions were constant for all concentrations of carbon dioxide.The results are shown in the graph below(a)(i)describe the effects of increasing the carbon dioxide concentration on the yield of tomatoes.rapid increase initially / at low levels of CO2 / up to 430 ppm;then levels off / eq.;not much increase in yield above 700/800 ppm.(2 marks)(ii)The normal concentration of carbon dioxide in the atmosphere is approximately 300 ppm. From the graph determine the yield of tomatoes when the concentration in the glasshouse was 300 ppm5.5/5.6 Kg m-2 .(1 mark)(iii)Calculate the percentage change in yield that would be expected if the tomatoes were grown in an atmosphere where the carbon dioxide concentration was increased to 800 ppm compared with the yield at 300 ppm. Show your working. 100 / eq.;Answer 47.3% / eq.(2 marks)(b)Explain why carbon dioxide concentration affects the yield of tomatoes.CO2 needed / raw material for photosynthesis / eq.,increase in CO2 increases rate of photosynthesis / eq.(2 marks)(c)A further experiment was carried out to investigate the effect of temperature on yield. In this experiment the carbon dioxide concentration was 300 ppm and the light intensity remained constant.The results are shown in the graph below.Compare the yield of tomatoes at 15C with that at 25C and suggest an explanation for the difference in yieldlarge / eq. increase in yield;yield 11 greater / increases from 0.5 to 5.5 Kg / eq.;increase in temperature increases rate of enzyme reactions;light independent stages of photosynthesis enzyme controlled / eq.(3 marks)(d)Name one factor other than light, temperature and carbon dioxide concentration, which could affect the yield of tomatoes grown in a glasshouse.water / mineral ions / named ion / presence of pollinating insects /disease / variety.(1 mark)[Total 11 marks]13.An experiment was carried out to investigate the effect of light intensity on the rate of photosynthesis of an aquatic plant, using the apparatus shown in the diagram below. (a)State two environmental conditions, other than light intensity, which would need to be controlled. For each condition, describe how control could be achieved. Condition 1 & 2 How controlled carbon dioxide 2 pairs as, condition; how controlled;temperature; use of a water bathcarbon dioxide; use of (hydrogen) carbonate;wavelength; use of same lamp/filter;pH; use of a buffer;(4 marks)(b)The plant was allowed to carry out photosynthesis for 10 minutes. Describe how you would use the apparatus to determine the volume of oxygen produced by the plant during this 10 minute period.switch off lamp/eq;use syringe to move oxygen/gas bubble to the scale;measure length of bubble on the scale;multiply by cross sectional area of tube/ ref to calibration of the tube as a volume;(4 marks)(c)Using this apparatus, the volume of oxygen produced after 10 minute periods of photosynthesis was determined at different light intensities. The results of this investigation are shown in the graph below.Comment on the results of this investigation.rate of photosynthesis/eq increases with light intensity;lens change between 2000 lux and 3000 lux/converse;(calculation from data) eg changes o.8mm3 and 0.3mm3 respectively/ref to gradients;ref to other factor/named fact or limits at high light intensity.(3 marks)[Total 11 marks]14.Avocets and gulls are both bird species that breed in coastal ecosystems. Gulls can feed on the eggs and young of the avocets.A study was made of the relationship between the numbers of these two species in a particular coastal area. The numbers of birds of each species were recorded at two-yearly intervals over a period of thirty years between 1951 to 1981.In 1965 some nests containing gulls eggs were removed from the area as a management technique attempting to maintain both species within this coastal area. The changes in the numbers of each species are shown in the graphs below.(a)Describe the changes in the numbers of gulls and of avocets between 1951 and 1961.No. of avocets rises / eq. or figures quoted to 1957; then avocet numbers fall and then rise to 1961;gull numbers fluctuate / eq. until 1955;then gull numbers rise and fall until 1961;(3 marks)(b)(i)Gulls nests and eggs were removed in 1965. Calculate the percentage decrease in the number of gulls over the period 1965 to 1967. Show your working.3600 700 (from graph) / 3.6 103 0.7 103; 100;Answer 80.6% / 80.56% / 80.5%;(3 marks)(ii)Describe and explain how the decrease in the number of gulls affected the avocet population between 1965 and 1971.population increases (rapidly) / almost doubles / figs quoted;due to reduced / less predation / eq. by gulls;(2 marks)(iii)Suggest two reasons for the changes in numbers of the avocets after 1971.1changes in food availability;changes in named climatic factorhigh / low rainfall levels;2(still some) predation / eq. by gullsintraspecific competition / OR competition for named factor eg nest sites / land / space;effects of other predators;qualified ref to pollution;(2 marks)(c)Name one other management technique and describe its effect on the ecosystem in which it is used.mowing; periodic cutting of grass; decrease in coarse grass / stops succession / eq;flooding / drainage / eq;controls water levels; maintains species / eq;coppicing; cut trees when young / eq;increase ground flora / eq;other suitable examples: eg grazing / use of pesticide / culling;(3 marks)[Total 13 marks]15.An investigation was carried out into the effect of cytokinin on cell division in cultures of soya bean tissue. Uniform pieces of soya bean tissue were added to artificial media either with or without cytokinin, and the number of cells in each culture was estimated each day for six days. The results are shown in the graph below. Each figure is the mean of eight measurements.Adapted from Fosket and Short, Physiol. Plant. 28: 14-23, 1973(a)(i)From the graph determine the difference in numbers of cells in the cultures with and without cytokinin on day 5. Show your working.3.2 105 with cytokinin and 1.37 105 without cytokinin; (allow 1.36 - 1.38)3.2 1.37 = 1.83 105(2 marks)(ii)Comment on the results as shown in the graph, including any conclusions that may be drawn concerning the effect of cytokinin on cell division.not much difference in first three days / converse;no increase / division stops after 3 days without cytokinin / converse; cytokinin stimulated cell division; reference to very high numbers of cells with cytokinin compared with no cytokinin / valid use of figures;(3 marks)(b)It was suggested that cytokinin might exert its effect on cell division by stimulating the replication of DNA.To investigate this suggestion, soya bean tissue cultures were grown on media containing different concentrations of cytokinin. The DNA content of each culture was measured after six days. The results are shown in the table below.(i)Do the data in the table support the suggestion that cytokinin exerts its effect on cell division by stimulating DNA replication? Explain your answer.No; higher concentration gives lower DNA content / converse / cytokinin has a negative effect on DNA content;(2 marks)(ii)Suggest an explanation for the effect of cytokinin concentration on the DNA content of the dividing cells in these tissue cultures.in absence of cytokinin DNA replication continues / but cytokinesis / cell division does not / cytokinin needed for cell division;so where no cytokinin doubling of DNA content of cells occurs;(2 marks)(c)Plant tissue culture media contain auxins in addition to cytokinins. Explain why the combination of the two plant growth substances is used.presence of auxins needed for cytokinin to work;reference to synergistic effect; high cytokinin; auxin ratio promotes shoot growth; low cytokinin auxin; ratio promotes root growth;(2 marks)[Total 11 marks]16.An investigation was carried out into the absorption of mineral ions by beech tree seedlings. The absorption of phosphate ions by beech roots was measured in moist air, and in an atmosphere of moist nitrogen.The results are shown in the graph below.(a)Calculate the rate of absorption of phosphate ions by beech roots in air between 10 hours and 20 hours. Show your working.8.3 or 8.2 5.5 or 5.62.6 to 2.8/10;Answer 0.26 to 0.28;arbitrary units hr1(3 marks)(b)(i)Compare the rates of absorption of phosphate ions by roots in air and roots in nitrogen.higher / eq. in air / lower in nitrogen;correct ref to figures / highest level of absorption in nitrogen level than lowest level in air;both rates approximately constant / linear / rise steadily;(2 marks)(ii)Suggest an explanation of the difference in rates of absorption.oxygen (from air) available for respiration;provides energy;for active uptake (of phosphate ions);ref. nitrogen preventing / inhibiting absorption;(2 marks)(c)(i)Suggest two reasons why the atmosphere in which the roots are kept has to be moist.uptake of oxygen / gas exchange / diffusion easier of moist;water needed for absorption of ions / ions are in solution / soluble;keeps cells turgid / prevents drying out / desiccation / wilting;(2 marks)(ii)Suggest two factors that should be kept constant in this experiment.temperature;concentration of phosphate ions / eq; pH;mass / size / surface area / volume of roots;(2 marks)[Total 11 marks]