Topic 9.1 - 9.3 AHL Plant Science IB Test Bank Questions

  • View
    812

  • Download
    92

Embed Size (px)

DESCRIPTION

rncwxz,bjvkds;xa

Text of Topic 9.1 - 9.3 AHL Plant Science IB Test Bank Questions

2The answers/markschemes are below each question in italics. You can reorganize the document so that you do not see the answer/markschemes when you answer the questions. 1.Seed dispersal is important in the migration of plants from one area to another area. Plants have evolved many methods, both physical and biological, by which to disperse their seeds.50 maple seeds, which are wind dispersed, were dropped one at a time from two different heights, 0.54 m and 10.8 m respectively. The histograms below show the distribution of the distance the maple seeds travelled.[Source: student experiment, Guralnick] (a)For each height, identify the distance travelled by the greatest number of seeds.(i)Height = 0.54 m: ...............................................................................................(ii)Height = 10.8 m: ...............................................................................................(1) (b)State the effect of height on seed dispersal......................................................................................................................................(1) (c)Suggest two reasons for the effect of the drop height on the distance travelled by the seeds.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)The following graphs show the rate and timing of seed release from different species of grass in the same area during the summer.[Source: J L Harper, Population Biology of Plants, Academic Press (Harcourt Brace Jovanovich) 1997, page 57] (d)Identify the grass species which produces the most seeds in this area......................................................................................................................................(1) (e)Identify the grass species which produces the most seeds in June......................................................................................................................................(1)(f)Compare seed production for all species relative to the timing of their release...............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3) (g)Suggest two benefits for these plants in the timing of seed release.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)Biological seed dispersal is usually dependent on the nutritional content of the seed or fruit. The following table gives the nutritional content for fruits of different species in temperate and tropical climates.Percentage by DryWeightCommon Name (genus)ProteinLipidCarbohydrateDispersal AgentsTemperateCranberry (Vaccinium)3689BirdsHawthorn (Crataegus)2273BirdsPin cherry (Prunus)8384BirdsPokeberry (Phytolacca)14268BirdsStrawberry (Fragaria)6488BirdsTropicalBird palm (Chamaedorea)141655BirdsFig (Ficus)7479BatsMistletoe (Viscum)65338BirdsMonkey fruit (Tetragastris)1494MonkeysWild nutmeg (Virola)2639Birds[Source: H Howe and L Westley, Ecological Relationship of Plants and Animals, Oxford University Press 1988, page 121] (h)Compare tropical fruits to temperate fruits in relation to the mean values for lipid, carbohydrate and protein content.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)(i)Explain which fruit would have the highest energy content.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2) (j)Suggest one advantage and one disadvantage of dispersal of seeds by animals.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)(Total 17 marks)(a)height 0.54 m: 6079 cm / 0.600.79 m (from the plant)and height 10.8 m: 02.9 m(from the plant);1Units needed for both parts of the answer.(b)the greater the height from which the seed fell,the further it travelled from the parent plant1(c)at the greater height:seed can catch the wind to travel further / updrafts / more wind at greater height;farther to the ground and does not travel straight down / more time to be blown before hitting the ground;at lower height:seed can fall straight down;seed can hit downdraft and fall faster;2 maxAny point must explain the difference in distancetravelled from the two heights.(d)Agrostis stolonifera1(e)Poa trivialis1(f)Poa produces seed earliest in the summer / June;Holcus produces most seed in July;Agrostis and Festuca produce seed in (late July to) August;Holcus and Poa have a peak time of seed fall / short period of seed fall;Agrostis and Festuca may continue to increase inseed production to September;3 maxAccept any of these points made conversely as an alternative.(g)Award [1] each for any two of the following.to avoid predation / disperse at times when other species are dispersing their seeds;to avoid competition;late in the year to allow seeds to germinate over winter / better germination conditions;better dispersal conditions / more wind / animals for dispersal;photoperiod required day length for flowering;more energy stored at the end of the summer for seed production;more light / warmth / better conditions for seedling photosynthesis / growth;2 max(h)Award [1] each for any two of the following.tropical fruits have higher lipid content than temperate fruits;temperate fruits (80%) have greater carbohydratecontent than tropical fruits (55%);protein levels are similar in both groups of fruits / slightly higher in temperate fruitsthan tropical fruits;(must make it clear that the difference is slight)2 max(i)mistletoe;high proportion of lipid and carbohydrate(lipid has approximately twice the energy contentof protein and carbohydrate);2(j)Award [1] for advantage and [1] for disadvantage.animal dispersal advantage:travel further / digestion cracks seed coat for better germination / deposited in feces with organic matter / better in areas with little wind;animal dispersal disadvantage:predation / seeds eaten / deposited in poor environment / buried too deep / buried too shallow (if deposited with feces) / animal might become extinct / scarce;2 max[17]2.Describe the metabolic events of germination in a starchy seed.(Total 5 marks)absorption of water;(embryo) increases respiration;(embryo) secretes GA to (aleurone layer);(stimulates) production of amylase;digestion of starch to smaller sugars / maltose;mobilize to tissues / transport of foods / nutrients to embryo;3.Which would be an adaptation of xerophytes?A.Large air spacesB.Large numbers of stomataC.Hairs on the leavesD.Reduced rootsC(Total 1 mark)4.The leaves of plants are adapted to absorb light and use it in photosynthesis. Draw a labelled diagram to show the arrangement of tissues in a leaf.(Total 6 marks)Award [1] for each of the following structures, shown in the correct relative positionand labelled. Individual cells are not needed but do not penalize if they are shown.upper epidermis;palisade layer / mesophyll;spongy layer / mesophyll;lower epidermis;xylem (in a major or minor vein);phloem (in a major or minor vein);collenchyma (in the midrib);guard cells; (do not accept stoma / stomata only)[6]5.Explain how manipulation of day length is used in the production of flowers..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(Total 6 marks)some flowering plants are short-day plants;others are long-day plants;important variable is length of darkness / photoperiod;some plants grown in greenhouses with controlled light conditions;short-day plants kept in the dark during daylight hours; long-day plants artificially lit during the night;using an appropriate wavelength / far-red light / 730 nm;possible to expose only for brief periods to keep costs down but long enough to interrupt the dark period;involves interaction of phytochromes with metabolic reactions;controlled by the plants biological clock;[6]6.Glucosinolates are chemicals found in some vegetables, which are responsible for the taste of horseradish, wasabi and broccoli. There are two types of glucosinolate, aliphatic and indolyl. They have been found to have many positive health effects, including carcinogen detoxification and antioxidant properties. Different varieties of broccoli vary in their content of glucosinolates as shown in the graph below. Researchers have found that 61% of the variation in aliphatic glucosinolate concentration is due to genetic factors compared with 12% for indolyl glucosinolates.[Source: E H Jeffery et al., Nutrition Today, (2002), 37, page 208) (a)Using the graph, compare the amount of aliphatic glucosinolates among the different varieties of broccoli....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3)(b)Using the data, explain how outbreeding could be used to develop a new variety of broccoli with increased glucosinolate content....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3)(Total 6 marks) (a)in Brigadier and Packman, glucoraphanin higher in concentration than progoitrin;progoitrin approximately equal in Packman and Brigadier / greatest in Petro;Brigadier has highest total glucosinolate concentration;Brigadier has highest glucorphanin concentration;Packman has lowest total glucosinolate concentration;Packman has lowest glucorphanin concentration;3 max(b)(outbreeding is) reproduction involving fusion of gametes produced by genetically unrelated individuals;can increase aliphatic glucosinolate concentration because more variation is due to genetic factors;Brigadier and Petro have more aliphatic glucosinolates;cross Brigadier with Petro;because Brigadier has more glucoraphinin and Petro has more progoitrin;3 max[6]7.An experiment was carried out to investigate the effect of water stress on cucumber (Cucumis sativus) seedlings. Cotyledons were detached from four day old seedlings and treated with polyethylene glycol (PEG), a water absorbing compound. b-amylase activity was measured in cotyledons treated with PEG at concentrations of 0, 20, 30, 40 and 50%. This enzyme catalyses the conversion of starch into maltose. The mean results are shown in the graph.[Source: D Todak, et al., (2000), Journal of Experimental Botany, 51, pages 739745] (a)Identify the maximum activity of b-amylase in the 50% treatment......................................................................................................................................(1) (b)Compare the b-amylase activity in the cotyledons treated with 20% PEG with those treated with 30% PEG.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2) (c)Deduce the relative free sugar content of the cotyledons treated with 20% PEG compared to those treated with 30% PEG...........................................................................................................................................................................................................................................................................(1) (d)Suggest reasons for the change in activity of b-amylase during water stress..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)(Total 6 marks)(a)0.29 ( 0.01) mmol maltose min1 (units needed)1(b)no substantial increase in activity in the 20% treatment over the five days of the experiment / 30% sample reaches maximum activity two days after start of treatment; higher level of activity at all times for 30% treatment compared with 20% treatment; activity approximately six times higher for 30% treatment than for 20% treatment on day two;any other appropriate numerical example;2 max(c)30% treatment will have greater amount of free sugars (maltose) than 20% treatment1(d)water stress will trigger synthesis of proteins / b-amylase;maltose and other free sugars attract water so will reduce the effect of water stress;maltose and other free sugars may prevent water loss (by osmosis);starch exerts less osmotic pressure / solute potential than free sugars;30% PEG creates an optimum environment for / b-amylase activity;2 max[6]8.Use and conservation preferences for savanna trees were investigated in a West African country. Residents from different villages evaluated the importance of 20 woody species for eight different uses: edible fruits, vegetable sauce, construction, firewood, medicine, commerce, field trees and conservation. The following data is based on answers from 200 residents.[Source: Economic Botany; Kristensen and Lykke, reprinted by permission from Economic Botany, vol. 57 (2), pages 203217, Kristensen and Lykke. Copyright 2003, The NewYork Botanical Garden, Bronx, New York] (a)Identify the most important tree species to the villagers......................................................................................................................................(1) (b)State the category of use for which villagers have most difficulty in finding useful species......................................................................................................................................(1)(c)Compare the usefulness of species in providing edible fruit with their usefulness for vegetable sauce..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2) (d)Determine the percentage of species that are valued entirely as very important in at least three categories...........................................................................................................................................................................................................................................................................(1) (e)Suggest a property of the wood from P. erinaceus that makes it one of the preferred species for use in building houses...........................................................................................................................................................................................................................................................................(1)(Total 6 marks)(a)Vitellaria paradoxa1(b)construction1(c)3 species / A. digitata, P. biglobosa, V. doniana"very important" in both categories;10 species ranked "very important" for ediblefruits and 7 species for vegetable sauce;overall the 20 tree species were slightly more valuablefor edible fruits than for vegetable sauce;Ximenia africana more useful as ediblefruit but not as vegetable sauce / vice versa for Bombax costatum;in both categories there were no "do not know" responses;2 max(d)Do not need to show working1(e)strength;resistance to insect attack eg termites;attractiveness;resistant to decay;flexible;not too heavy;easy to cut;tradition;1 max[6]9.In order to prevent transfer of pollen from an anther of one plant to the stigma of the same plant (self-pollination), the sunflower (Helianthus spp) anther sheds its pollen before the stigma is mature enough to receive it. Early in the morning the anther is exposed by elongation of the filaments. The anthers open at this time to release their pollen (anthesis). The stigma appears above the anthers by late afternoon, and by the following morning it is fully receptive.To see how the filament (F) and the style (S) are affected by light, their lengths were measured at time intervals starting 12 hours before anthesis (-12). Some plants were grown in continuous white light (L24) and some plants grown under cycles of 16 hours white light followed by 8 hours dark (L16/D8). The results are shown in the graph.[Source: Lobello et al, Journal of Experimental Botany, (2000), 51, pages 1403-1412] (a)Filaments of the plants grown in continuous white light increased in length by 0.25 mm in the 28 hours after anthesis. Calculate how much the filaments of the plants grown in alternating white light and dark increased during the same period.........................................................................................................................................................................................................................................................................(1)(b)Compare the increase in the length of the style in the plants grown in continuous white light with those grown in alternating white light and dark.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2) The table compares the percentage of ovules that have been fertilized and developed into seeds in sunflower plants grown under continuous white light with those grown under alternating light and dark. The numbers represent the mean one standard deviation.Light treatmentsPercentage of fertilized ovulesContinuous white light (L24)11.40 7.76Alternating light and dark (L16/D8)58.26 4.06 (c)Explain the differences in the percentages of ovules fertilized using the data in the graph about the growth of filaments and styles.............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3)(d)Explain how standard deviation (SD) shown in this table can be used to help in comparing the effect of light treatments on the fertilization of ovules.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3) To analyse the effect of growth regulators on filament elongation, further experiments were performed in the dark, white light and red light. The flowers were treated with auxin or with gibberellic acid and compared to a control with no growth regulator. The results are shown in the bar chart below.(e)Identify, with reasons, which factors promote and which factors inhibit the elongation of filaments.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3) (f)Explain the disadvantages to a plant of self-pollination.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)(Total 14 marks)(a)2.9 ( 0.2) mm (units required)1(b)cyclic light makes style grow almost immediately while with continuouslight it takes longer to start to grow / L16 / D8 starts growing in first hourwhile L24 style starts growing after 6 hours / growth is more gradual in L24;with continuous light the style grows less / continuous (L24) grows to 9.8 mmwhile cyclic (L16 / D8) grows to 10.2 mm / little difference after 28 hours;in both cases growth only starts with anthesis;2 max(c)47% / more fertilized ovules in cyclic light;filament grows more in cyclic light than continuous;pollen closer to stigma so pollination more probable;in continuous light anthers do not become exposed;3 maxAccept converse wording.(d)standard deviation is a measure of variability / indicates the spread ofvalues around the mean;continuous light data is more variable (because it has a higher standarddeviation);helps to decide whether the difference between two means is significant;68% of values are 1 SD from mean;difference between means is approximately 47 / appears to be significantlydifferent / light treatment makes a significant difference;3 max(e)darkness promotes / white light inhibits because filaments shorter than indarkness;red light inhibits because filaments shorter than in darkness;auxins promote because filaments are longer than in control / in whiteand red light;gibberellic acid inhibits because filaments are shorter in continuous whitelight / darkness;3 maxReason must be present to receive the mark. Accept if converse wording.(f)self-pollination reduces / does not promote variation / no new combinationof alleles;no variation for natural selection;more susceptible to infectious diseases;more prone to genetic disease / (inbreeding) more likely to be homozygousfor disease;2 max[14]10.Sweet pepper (Capsicum annuum) is an important widespread agricultural crop. Scientists studied the transport and distribution of sodium in sweet pepper by growing plants in sodium chloride solutions.The graph below shows the sodium ion concentration in plant parts of sweet pepper grown in 15 mM sodium chloride for three weeks.[Source: M Blom-Zandstra et. al., Sodium fluxes in sweet pepper exposed to varying sodium concentrations,Journal of Experimental Botany (1 November 1998), vol. 49, issue 328, pp. 18631868,by permission of Oxford University Press] (a)(i)State the concentration of sodium ions in fruits............................................................................................................................(1) (ii)Calculate the percentage increase in sodium ion concentration between root and stem............................................................................................................................(1)(b)Suggest why a high sodium ion concentration in the cells of the stem is important in providing support to this type of plant.............................................................................................................................................................................................................................................................................................................................................................................................................(1) (c)State one possible use of sodium in plants.....................................................................................................................................(1) (d)Scientists also found that the concentrations of sodium ion in cells of the stem and in xylem sap were the same. Explain why this led the scientists to believe there was no active transport between xylem and stem.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2) (e)Suggest one possible method of transport of sodium ions between xylem and stem.....................................................................................................................................(1)The graph below shows the sodium ion concentration of the xylem sap in relation to the distance from the base. Two experiments were performed.Experiment 1:plants grown in 15 mM sodium chloride solution onlyExperiment 2:plants grown in 15 mM sodium chloride solution and then transferred to a sodium-free solution for an extra week[Source: M Blom-Zandstra et. al., Sodium fluxes in sweet pepper exposed to varying sodium concentrations,Journal of Experimental Botany (1 November 1998), vol. 49, issue 328, pp. 18631868,by permission of Oxford University Press] (f)State the relationship between sodium ion concentration and distance from the base of the stem in experiment 1.....................................................................................................................................(1) (g)(i)State the sodium concentration when the stem is 50 arbitrary units from the base in plants for each of the following.Experiment 1:.................................................................................................Experiment 2:.................................................................................................(1)(ii)Explain the difference in the sodium ion concentration in this part of the stem in plants of experiment 1 and experiment 2.............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2) (h)Compare the distribution of sodium ions in the stems of plants in both experiments.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3) Water transported in xylem is transpired through stomata in leaves. The electron micrograph shows one stoma of the lower epidermis of the leaf of a Simon bamboo (Arundinaria simonii). Its magnification is 3000.[Source: Courtesy of Professor Zulma E Rgolo, Instituto Darwinion, Buenos Aires, Argentina] (i)(i)Draw a line showing the maximum length of the stomatal pore.(1)(ii)Calculate the real size of the stomatal pore. Show your working............................................................................................................................(1) (j)Explain two abiotic factors that affect transpiration rate in this leaf.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)(Total 18 marks)(a)(i)2 mM kg1(units required)1(ii)180 (%) or1(b)cells in stem absorb water (by osmosis) providing turgidity / turgor pressure1(c)maintain osmotic balance;help to maintain turgidity / assist active transport;1 max(d)active transport means movement against a concentration gradient;there is no concentration gradient / concentration in xylem should be lowerthan stem (but it is not);2(e)diffusion / facilitated diffusion1(f)sodium ion concentration decreases as you get further away from thebase / vice versa1(g)(i)experiment 1: 17mMexperiment 2: 6mM1Both answers must be correct to receive [1].(ii)in experiment 1 (concentration of sodium ions is high because) sodium is continually taken up;it is lower in experiment 2 because sodium is lost by diffusion intomedium (when no sodium in medium);in experiment 2 there is more water uptake (by osmosis);2 max(h)sodium ion concentration in experiment 1 is higher close to base;(moving away from base) sodium ion concentration decreases in experiment1 but remains constant / decreases slightly in experiment 2;sodium ion concentration rises in experiment 2 after 100 arbitrary units butcontinues to decline in experiment 1;same sodium ion concentration in both experiments at 105 arbitrary units;lower sodium ion concentration in experiment 1 compared with experiment 2when close to the top / far from base;3 max(i)(i)1Award mark if line drawn off the image but it must be 21(1) mm.(ii)real size = length of line drawn 3000 = 0.007(0.0005)mm1Units are required, allow for ECF. Also accept answer in cm.(j)humidity decreases transpiration rate because atmosphere is saturated;temperature increases transpiration rate because there is more kinetic energy;light increases photosynthesis which opens stomata increasing the rate oftranspiration;wind / air movement lowers concentration outside leaf;2 max[18]