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Q1.Algae are photosynthesising organisms. Some grow on rocky shores. Scientists investigated the abundance of different species of algae at two sites, A and B, on a rocky shore. Site A was on the upper shore and site B was on the lower shore. The diagram shows the location of sites A and B on the rocky shore.Table 1 shows some of the results the scientists obtained.???Table 1??Site AUpper shoreSite BLower shore?Species of algae with percentage cover more than 1%????Gigartina leptorhynchosGigartina canaliculataGelidium coulteriRhodoglossum affine???Gigartina spinosaRhodoglossum affineLaurencia pacificaGastroclonium coulteriCentroceros clavulatumGigartina canaliculataCorallina vancouveriensis(a)???? The scientists recorded data from 40 large rocks at each site.Describe one method that the scientists could have used to ensure that the large rocks were chosen without bias.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)(b)???? The scientists used percentage cover rather than frequency to record the abundance of algae presentSuggest why...........................................................................................................................................................................................................................................................(1)(c)???? Apart from availability of water, describe and explain how two abiotic factors may have caused differences in the species of algae growing at sites A and B.Factor 1 ..............................................................................................................Explanation ......................................................................................................................................................................................................................................Factor 2 ..............................................................................................................Explanation ......................................................................................................................................................................................................................................(2)(d)???? Use the information provided in Table 1 to explain why the diversity of consumers will be greater at site B.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)(e)???? The scientists also investigated the algae eaten by two consumers found on the rocky shore, the sea slug and the shore crab. The scientists carried out their investigation in a laboratory.???????? They put each consumer into a separate tank through which aerated seawater flowed slowly.???????? Each tank contained 5 grams of one species of alga.???????? After 50 hours, they measured the mass of the alga remaining in each tank.???????? They repeated this procedure several times using a different sea slug and a different shore crab each time.The scientists then calculated the mean mass of each species of alga eaten by the consumers. They used a statistical test to determine the P value.Table 2 shows some of the results they obtained.??Table 2?Species of algaMean mass eaten / gP value?Sea slugShore crab?Laurencia pacifica4.420.22<0.01?Egregia leavigata0.120.08>0.05?Microcystis pyrifera0.190.14>0.05?Cystoseira osmondacea0.170.04<0.05 (i)????? The consumers were starved for 5 days before the investigation.Explain why.............................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)(ii)?????The data in Table 2 for the mean mass of alga eaten were adjusted for loss of mass by the alga due to respiration.Suggest how the scientists were able to determine the loss of mass due to respiration of a sample of alga...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3)(iii)???? Suggest what conclusions the scientists could have made from this investigation when using the probability values in Table 2...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3)(Total 15 marks)Q2.Species richness and an index of diversity can be used to measure biodiversity within a community.(a)???? What is the difference between these two measures of biodiversity?................................................................................................................................................................................................................................................(1)Scientists investigated the biodiversity of butterflies in a rainforest. Their investigation lasted several months.The scientists set one canopy trap and one understorey trap at five sites.?????????The canopy traps were set among the leaves of the trees 16–27 m above ground level.?????????The understorey traps were set under trees at 1.0–1.5 m above ground level.The scientists recorded the number of each species of butterfly caught in the traps. The table below summarises their results.??Species of butterflyMean number of butterfliesP value?In canopyIn understorey?Prepona laertes150< 0.001?Archaeopreponademophon1437< 0.001?Zaretis itys2511> 0.05?Memphis arachne8923< 0.001?Memphis offa213< 0.001?Memphis xenocles328< 0.001(b)???? The traps in the canopy were set at 16–27 m above ground level. Suggest why there was such great variation in the height of the traps.................................................................................................................................................................................................................................................(1)(c)???? By how many times is the species diversity in the canopy greater than in the understorey? Show your working.Use the following formula to calculate species diversity.d = where N is the total number of organisms of all species and n is the total number of organisms of each species.???Answer = ...................................(3)(d)???? The scientists carried out a statistical test to see if the difference in the distribution of each species between the canopy and understorey was due to chance. The P values obtained are shown in the table.Explain what the results of these statistical tests show.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(Extra space) ................................................................................................................................................................................................................................................................................................................................................(3)(Total 8 marks)Q3.Scientists investigated the effect of a mycorrhizal fungus on the growth of pea plants with a nitrate fertiliser or an ammonium fertiliser. The fertilisers were identical, except for nitrate or ammonium.The scientists took pea seeds and sterilised their surfaces. They planted the seeds in soil that had been heated to 85 °C for 2 days before use. The soil was sand that contained no mineral ions useful to the plants.(a)???? Explain why the scientists sterilised the surfaces of the seeds and grew them in soil that had been heated to 85 °C for 2 days.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)(b)???? Explain why it was important that the soil contained no mineral ions useful to the plants.................................................................................................................................................................................................................................................(1)The pea plants were divided into four groups, A, B, C and D.?????????Group A – heat-treated mycorrhizal fungus added, nitrate fertiliser?????????Group B – mycorrhizal fungus added, nitrate fertiliser?????????Group C – heat-treated mycorrhizal fungus added, ammonium fertiliser?????????Group D – mycorrhizal fungus added, ammonium fertiliserThe heat-treated fungus had been heated to 120 °C for 1 hour.(c)???? Explain how groups A and C act as controls.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)After 6 weeks, the scientists removed the plants from the soil and cut the roots from the shoots. They dried the plant material in an oven at 90 °C for 3 days. They then determined the mean dry masses of the roots and shoots of each group of pea plants.(d)???? Suggest what the scientists should have done during the drying process to be sure that all of the water had been removed from the plant samples.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)The scientists’ results are shown in the table below.??TreatmentMean dry mass / g per plant( standard deviation)?RootShoot?A – heat-treated fungus and nitrate fertiliser0.40(±0.05)1.01(±0.12)?B – fungus and nitrate fertiliser1.61(±0.28)9.81(±0.33)?C – heat-treated fungus and ammonium fertiliser0.34(±0.03)0.96(±0.26)?D – fungus and ammonium fertiliser0.96(±0.18)4.01(±0.47)(e)???? What conclusions can be drawn from the data in the table about the following?The effects of the fungus on growth of the pea plants.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................The effects of nitrate fertiliser and ammonium fertiliser on growth of the pea plants.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(4)The scientists determined the dry mass of the roots and shoots separately. The reason for this was they were interested in the ratio of shoot to root growth of pea plants. It is the shoot of the pea plant that is harvested for commercial purposes.(f)?????Explain why determination of dry mass was an appropriate method to use in this investigation.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)(g)?????Which treatment gave the best result in commercial terms? Justify your answer.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)(Total 15 marks)Q4.(a)???? What is the name of a position of a gene on a chromosome?.............................................................................................................................(1)(b)???? What is meant by genetic diversity?..........................................................................................................................................................................................................................................................(1)A geneticist investigated genetic diversity in four different breeds of dog. She compared DNA base sequences of the same genes from a large number of dogs from each breed.The geneticist calculated the mean genetic diversity for each breed of dog. The value of this mean was between 0 and 1.???????? A mean value of 1 shows maximum genetic diversity.???????? A mean value of 0 shows no genetic diversity.Her results are shown in the table??Breed of dogMean genetic diversityStandard deviation?Airedale terrier0.51± 0.03?Bull terrier0.38± 0.02?Jack Russell terrier0.76± 0.01?Miniature terrier0.47± 0.02 (c)???? What do these data show about the differences in genetic diversity between these breeds of dog?.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3)(d)???? Miniature terriers were first bred from bull terriers in the 19th century.Suggest one explanation for the observed difference in genetic diversity between miniature terriers and bull terriers.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)(Total 7 marks)Q5.Impala and wildebeest are species of herbivore that live in large groups. They spend most of their time feeding with their heads near the ground.Scientists investigated the relationship between the number of predators in an area and the mean proportion of time these herbivores spent with their heads up, looking around rather than feeding. They obtained data from groups of impala and wildebeest in two areas. In one area there were few predators and in the other area there were many predators.The graph shows their results. The bars show standard deviations.?(a)???? The scientists observed both groups of animals for 75 hours.Use data from the graph to calculate the difference in the mean number of hours spent by each species looking around in the area where there were many predators.Show your working.Difference ........................... hours(2)(b)???? The scientists concluded that these herbivores spend more time looking for predators in areas where there are many predators.Do these data support this conclusion? Give reasons for your answer.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(4)(c)???? The behaviour of the herbivores in having their heads up has a benefit but it also has costs. The benefit is being able to see, and escape from, predators.Suggest and explain one cost to the herbivores of this behaviour.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)(Total 8 marks)?M1.(a)???? 1.?????(Use) coordinates / number the rocks/sites/squares;Ignore: references to grid, tape measures, metre rulers etc.2.??????Method of generating/finding random numbers e.g. calculator/computer/random number generator/random numbers table;Accept: numbers out of a hat / use of dice.2(b)???? Difficult/too many to count / individual organisms not identifiable / too small to identify/count / grows in clumps;Ignore: easier/quicker/representative/ more accurate, unless qualified.1(c)???? Any suitable factor with valid explanation = 1 mark1.??????Wave action - firmer grip on rock is necessary (at either site);2.??????Wind/air movement/less humid - more evaporation at site A / more (physical) damage;3.??????Light – (linked to) photosynthesis (at either site);4.??????Temperature – (linked to) photosynthesis/respiration/enzymes/ evaporation (at either site);5.??????pH – (linked to) enzymes/proteins;Note: other common factors include salt (salinity) linked to water potential / named nutrient e.g. nitrate linked to protein/DNA.Ignore: carbon dioxide/oxygen/pollution/rainfall/food/nutrients.Reject: biotic factors e.g. predation.2 max(d)???? 1.??????Greater variety of food / more food sources;Ignore: more food.2.??????More/variety of habitats/niches;Ignore: homes/shelters.Accept: different habitats.2(e)???? (i)????? 1.??????(So they were) hungry/not full;Accept: description of hunger e.g. appetite / ‘empty stomach’/’so they eat’.2.?????? (Allows) comparison;2(ii)????? 1.??????Alga without consumer/named consumer/animal;Accept: repeat experiment without consumer.Accept: in separate tank / in tank where not eaten.2.?????? (Find change in mass) in dark;3.?????? For 50 hours;Accept: ‘same time as in experiment’.Accept: For lower time period then scaled up to 50.3(iii)????? 1.?????For Laurencia pacifica and Cystoseira osmondacea (difference in results) significant /reject null hypothesis / not due to chance / less than 5%/0.05 probability due to chance;Accept: for Laurencia pacifica ‘less than 1%/0.01 probability’.2.??????For Egregia leavigata and Microcystis pyrifera no significant (difference in results)/accept null hypothesis / is due to chance/more than 5%/0.05 probability due to chance;Accept: ‘insignificant’ for ‘no significant difference’.3.??????(Difference in results) for Laurencia pacifica is the most significant;Note: reference to probabilities on their own is not sufficient.1, 2 and 3. Accept: abbreviations for all species.3[15]M2.(a) ????Species richness measures only number of (different) species / does not measure number of individuals.1(b) ????Trees vary in height.1(c) ????1.??????Index for canopy is 3.73;2.??????Index for understorey is 3.30;3.??????Index in canopy is 1.13 times bigger;If either or both indices incorrect, allow correct calculation from student’s values.3(d) ????1.??????For Zaretis itys, difference in distribution is probably due to chance / probability of being due to chance is more than 5%;2.??????For all species other than Zaretis itys, difference in distribution is (highly) unlikely to be due to chance;3.??????Because P < 0.001 which is highly significant / is much lower than 5%.3[8]M3.(a) ???? 1.??????To kill any fungus / bacteria on surface of seeds or in soil;2.??????So only the added fungus has any effect.2(b) ????So that only nitrate or ammonia / type of fertiliser affects growth.1(c) ????1.??????So that effects of nitrate or ammonium alone could be seen;2.??????So that effects of fungus can be seen.2(d) ????1.??????Weigh samples at intervals during drying;2.??????To see if weighings became constant (by 3 days).2(e) ????With live fungus – showing effects of the fungus:1.??????Fungus increases growth of roots and shoots in both;2.??????Produces greater growth with nitrate.With heat-treated fungus – showing effects of fertiliser:3.??????Similar dry masses for roots and shoots;4.??????(Probably) no significant difference because SDs overlap.4(f) ????1.??????Dry mass measures / determines increase in biological / organic material;2.??????Water content varies.2(g) ????1.??????Fungus with nitrate-containing fertiliser gave largest shoot: root ratio;2.??????And largest dry mass of shoot;3.??????6.09:1 compared with ammonium-containing fertiliser 4.18:12 max[15]M4.(a)???? Locus;Accept: loci1(b)???? Differences in DNA / differences in base sequence of DNA;Accept: number of different alleles / size/variation in gene poolReject: genes1(c)???? 1.??????Jack Russell (genetic) diversity is (significantly) greatest;2.??????Bull terrier (genetic) diversity is (significantly) smallest / is most inbred;3.??????Miniature terrier and Airedale terriers are similar;1-3: do not credit just a list of values4.?????? Standard deviations do not overlap / do overlap with correct ref to significance;Reference to significance must be relevant to examples givenMax 3(d)???? 1.??????(Bull terrier) breeding has included a genetic bottleneck/ small population/more inbreeding/ greater selection (pressure);Accept: founder effect2.??????Reduced number of different alleles/size of gene pool;Reject: decrease in number of genesIgnore ref to mutationsOR3.??????Miniature (terrier) breeding has included more outbreeding/less selection (pressure);4.??????Increased number of different alleles/larger gene pool/more variety of alleles;Reject if genes used instead of allelesReject: lower frequency of allelesIgnore ref to mutations2[7]M5.(a)???? 9 (hours);;If multiply 75 by 0.11 and 0.23 but wrong answer, then 1 markAccept for one mark if multiply 75 by two wrong proportions near to 0.11 ± 0.01 and 0.23 ± 0.01 or multiply by the difference between the two (wrong) proportions2(b)???? (Yes because)1.????? Both/Each species (mean) time spent looking around greater where many predators;2.????? Differences (appear to be) significant because SDs do not overlap;(No because)3.????? Wildebeest spend same (mean) time looking around where many predators as impalas where few predators;4.????? Don’t know what they are looking for (when heads up); 5.????? Habitats might be different in different areas (which could affect the behaviour);Accept ‘mean proportion’ means ‘time’1. Require idea of both, not just quoting numbers2. This point must be in the context of point 12. Do not accept results significant2. Accept ‘because bars do not overlap’2. Do not accept SE for SD3. Accept overlap in SD as equivalent to same time5. Ignore ‘other factors’ unqualified and discussions of experimental variables4 max(c)???? 1.????? Less time spent feedingORMore energy lifting head/looking round;2.????? (So) less food/biomass for respirationORless energy for growth/reproduction/care of young;OR3.????? Raising head makes them more visible to predators;4.????? So more likely to be attacked/eaten/killed;2. Accept any appropriate suggestion of less energy for something to do with life of the herbivore2. Allow less food/biomass for growth/reproduction2. Ignore references to energy for respiration2[8] ................
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