Populations and sampling techniques — Paper 4 · A Level Marine Science 9693

1 The table shows the estimated populations of Atlantic cod and green sea urchins in an area of the Atlantic Ocean around New England from 1980 to 2010. estimated population / thousand year Atlantic cod green sea urchins 1980 157 0 1985 139 0 1990 142 12 1995 116 59 2000 117 64 2005 98 14 2010 54 8 (a) (i) Plot a graph to show the populations of Atlantic cod and green sea urchins from 1980 to 2010. (ii) Calculate the percentage decrease in population of Atlantic cod between 1980 and 2010. Show your working. … [2] (iii) The population estimates were based on the catches of fishermen. Suggest one reason why this estimate may be inaccurate. … … [1] (b) Fig. 1.1 shows part of a food web in the Atlantic Ocean near New England involving the Atlantic cod, crabs, green sea urchins and kelp. Atlantic cod kelp green sea urchin crab Fig. 1.1 • Atlantic cod was an important commercial catch throughout the twentieth century. • As the cod stocks decreased in the 1990s, fishermen began to catch green sea urchins. • Kelp now covers extensive areas of the sea bed. • Crab populations increased due to finding shelter amongst the kelp. (i) Use the information to explain the changes in populations of cod and green sea urchins shown in your graph on page 2. … … … … … … [3] (ii) Explain why conservationists have warned that it will be very difficult to return the balance of the species to pre-commercial fishing levels. … … … … [2] [Total: 13]

13 marks

Mark scheme: 1(a)(i) both axes (x and y) labelled ; suitable, linear scale ; plots correct ± 1 2 small square ; ; straight bars (that do not touch) / lines, identified by key ; 5 plots to cover at least half the grid A labels to lines 1(a)(ii) any 3 of: correct subtraction (157–54 OR 103) ; division by starting population, multiplied by 100 ((103÷157))*100) ; 2 A 66 (%) / 65.6 for two marks A negative percentage 1(a)(iii) any 3 of: idea of, fish populations from areas fished may not be representative / AW ; fishermen may lie / exaggerate catch / illegal catch / bias / different fishing methods may have been employed ; by-catch / discard not counted ; idea of, not random sample ; idea of, juveniles / small sizes not counted ; 1 Question Answer Marks Guidance 1(b)(i) any 3 of: cod population decreases because of (over)fishing / harvesting / AW ; so less predation of sea urchins, so sea urchin populations increased ; (over)fishing / harvesting of sea urchins decreases population ; fewer sea urchins / less food for cod, so cod population falls further ; high crab population reduces sea urchin population ; 3 1(b)(ii) any 3 of: loss of sea urchins increases kelp ; so crab population increases, due to shelter / kelp OR crabs can’t be removed ; idea of, more / high number of, crabs eating sea urchins, so sea urchin population can’t recover / idea of, positive feedback / AW ; idea of, cod has insufficient food / lack of prey for cod / lack of sea urchins for cod ; idea of, too few adult cod to breed, so population cannot recover ; 2 A population can’t recover as too few reach maturity A many years to replenish recruitment classes

This question in 9693/41 May/June 2017

1 An experiment was carried out to investigate the effect of temperature and mass of food on the growth of salmon. Young salmon were placed into tanks of seawater at six different temperatures between 0 °C and 25 °C. At each temperature, six different feeding regimes were used, ranging from no food given, to five feeds per day. The fish were given the same mass of food each time they were fed. The mean growth rate was determined by weighing the fish at the start and again after two weeks. This was expressed as mean percentage change in body mass compared to their initial mass. The results are shown in Fig. 1.1. 1.6 1.4 1.2 Y 1.0 0.8 0.6 X mean five feeds per daypercentage 0.4 change in four feeds per day body mass 0.2 three feeds per day two feeds per day 0.0 one feed per day no feeding –0.2 –0.4 –0.6 –0.8 0 5 10 15 20 25 temperature / °C Fig. 1.1 (a) (i) State the two independent variables in the investigation. … [1] (ii) Suggest two variables that should be controlled. … … [1] (b) The mean growth rate was expressed as the percentage change of initial body mass. (i) Calculate the difference between the mean percentage change in mass for the fish grown at 5 °C with one feed per day and two feeds per day. … [2] (ii) Suggest the factor that limits the growth of the fish with three feeds per day at each of points X and Y in Fig. 1.1. In each case, explain how you reached your conclusion. X … … … Y … … … [2] (c) Suggest explanations for the effects of temperature and mass of food on the growth rates of the salmon. … … … … … … … … … [4] (d) Salmon often inhabit waters with temperatures ranging between 5 °C and 7 °C. Current models of climate change predict a possible 2 °C to 3 °C rise in temperature of the waters where salmon are found. It is also predicted that the natural food of the salmon will be dramatically reduced. Use Fig. 1.1 to predict and explain the likely effects of a temperature rise of 2 °C to 3 °C on wild salmon populations. … … … … … … … … [3] [Total: 13]

13 marks

Mark scheme: 1(a)(i) feeding regime / number of feeds per day + temperature ; 1 A total mass food / amount of feeds per day 1(a)(ii) any two of: volume of water / water level / size of tank / salinity / type of food / oxygen concentrations / age of fish / pH ; 1 1(b)(i) 0.18 ;; 2 1(b)(ii) X: temperature + increasing temperature increases mass / increasing mass of food does not cause an increase ; Y: mass of food + feeding cause an increase / increasing temperature does not cause an increase ; 2 1(c) any four of: temperature increases respiration / metabolic rate ; respiration uses up glucose ; loss of body tissues / food reserves / AW ; ref. to enzymes ; more food is required for higher respiration rate / AW ; more food provides more energy for growth / AW ; no food causes a decrease in mass change as temperature increases ; another factor must limit rate at high temperature / digestibility as excess feeding decreases mass change ; credit manipulated numerical example ; 4 Question Answer Marks Guidance 1(d) any three of: populations will decrease ; (because) respiration / metabolic rates will increase ; (idea of) increased demand for food / AW ; insufficient food available (with temperature rise) / AW ; smaller fish breed less ; females produce less eggs / reduced fecundity ; reduced ability to migrate / reach spawning grounds ; 3 Question Answer Marks Guidance

This question in 9693/41 Oct/Nov 2017

2 Marine protection areas (MPAs) are regions of sea or ocean where fishing is prohibited or restricted in an effort to conserve species. The impact of MPAs on the lobster populations in the sea around Norway was investigated. In 2006, an MPA and a control area were established at site A. An MPA and a control area were also established at site B. The control areas were the same size as the MPAs. Commercial lobster fishing took place in the control areas. All fishing within the MPAs was restricted. Lobsters were sampled in all MPAs and control areas using identical traps placed at a depth of between 10 m and 30 m. Sampling of lobsters was carried out by inspecting 25 traps in each MPA and 25 traps in each control area every day for four separate days. This was repeated every year between 2006 and 2010. Sampling was conducted between 20 August and 10 September each year. (a) (i) Calculate the total number of samples taken each year. … [1] (ii) Suggest why the control areas were close to each of the MPAs. … … [1] (b) The mean number of lobsters caught per trap per day for both MPAs and control areas are shown in Fig. 2.1. The MPAs were established in 2006. 3.5 3.5 control area 3.0 3.0 MPA 2.5 2.5 indicates ± 1 mean standardnumber of 2.0 2.0 deviation lobsters per trap 1.5 1.5 per day 1.0 1.0 0.5 0.5 0.0 0.0 2006 2007 2008 2009 2010 2006 2007 2008 2009 2010 site A site B Fig. 2.1 The investigation led to the conclusion that MPAs assist in the recovery of the stocks of commercially fished marine species. Discuss the extent to which the results of this investigation support this conclusion. … … … … … … … … … … … … … … [5]

7 marks

Mark scheme: 2(a)(i) 400 ; 1 2(a)(ii) same temperature / food levels / substrate / water conditions ; 1 A any suitable biotic or abiotic factor Question Answer Marks Guidance 2(b) any five of: a always more lobsters in MPA than control areas ; b in both areas more lobsters are trapped after MPA is set up ; c in site B the increase in lobster trapped is higher than in site A / ORA ; d in site B lobsters caught in control areas have increased / ORA ; e (suggests) lobster move out from MPAs into control areas ; f correct reference to overlapping standard deviations ; g large sample size increases reliability ; h in 2007, in site A, the decrease (for MPA) is not significant / AW ; i (the data) is only for lobsters and other species may not do the same ; j the controls also increase in site B so another factor may be increasing lobsters trapped ; k credit correct manipulation of data ; 5

This question in 9693/41 Oct/Nov 2017

1 Surface longline fishing is a fishing method in which trawlers tow long fishing lines with baited hooks along the surface of the water. It is used to catch albacore tuna. The catch of albacore tuna from an area of the Indian Ocean around Mauritius was recorded each month, every year, from 2005 to 2011. This information was used to calculate the mean catch of albacore tuna per month. In order to determine fishing effort each month, the total number of hooks used to catch fish was recorded. This was used to calculate the mean number of hooks used for each month over the time period. The results are shown in Table 1.1. Table 1.1 standard mean deviation mean number catch per month albacore tuna of mean of hooks unit effort catch / kg albacore tuna used / kg hook–1 catch / kg Jan 6200 600 720 8.6 Feb 9000 2700 1100 8.2 Mar 5800 500 680 8.5 Apr 6200 600 700 8.9 May 5800 300 680 8.5 Jun 4000 300 550 Jul 3200 500 400 8.0 Aug 4000 800 500 8.0 Sep 5000 700 600 8.3 Oct 4500 600 550 8.2 Nov 9500 3000 1100 8.6 Dec 8500 1500 1000 8.5 (a) (i) Explain what is shown by the standard deviation of the mean albacore tuna catch. … … … … … [2] (ii) Plot a graph to show how the mean albacore tuna catch and mean number of hooks used changes. Label both y-axes fully and include the units. Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec month [5] (b) Catch per unit effort is considered to be a better comparative measure of the health of fish stocks than mean catch alone. mean albacore tuna catch catch per unit effort = mean number of hooks used Calculate the catch per unit effort for June. … kg hook–1 [1] (c) In order to determine whether the albacore tuna fishing was sustainable, the catch per unit effort for each month was calculated in 2015. Table 1.2 shows the catch per unit effort for each month during 2015. Table 1.2 catch per month unit effort / kg hook–1 Jan 9.2 Feb 9.4 Mar 9.1 Apr 8.9 May 8.6 Jun 9.2 Jul 8.9 Aug 8.2 Sep 8.3 Oct 9.1 Nov 8.6 Dec 8.9 Discuss whether the information in Tables 1.1 and 1.2 indicates that albacore tuna fishing is sustainable. … … … … … … … [3] [Total: 11]

11 marks

This question in 9693/40 Oct/Nov 2018

1 A marine reserve was set up in an area of the Mediterranean Sea to protect fish stocks. No fishing was allowed within the reserve. The mean daily fishing effort in areas surrounding the marine reserve was recorded and is shown in Fig. 1.1. Fishing effort was measured as the quantity of fishing gear used per square kilometre per day. fishing effort / quantity of fishing gear used per km2 per day marine reserve marine (no fishing) land reserve line 26–42 sampling 8km 18–25 14–17 10–13 7–9 <7 Fig. 1.1 (a) Suggest and explain one possible reason for the differences in fishing effort surrounding the marine reserve shown in Fig. 1.1. … … … … [2] (b) Scientists assessed the success of the marine reserve in conserving fish stocks. They measured the number of grouper eggs in different areas of the reserve, and also outside it. The areas sampled were along an 8 km line, which started in the centre of the reserve. The sampling line is shown on Fig. 1.1. The results are shown in Fig. 1.2. The scientists drew a line of best fit through the data points. inside marine outside marine reserve reserve 14 13 12 11 10 9 grouper egg 8 density / number of 7 eggs per 1000 m3 6 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 distance from centre of marine reserve / km Fig. 1.2 (i) Describe how grouper egg density changes with distance from the centre of the marine reserve. … … … … [2] (ii) Use the line of best fit on Fig. 1.2 to calculate the change in grouper egg density per kilometre between the centre of the reserve and a distance of 8 km from the centre. … eggs per 1000 m3 per km [3] (c) Discuss whether the information in Figs. 1.1 and 1.2 could be used to show that the marine reserve is beneficial to fish stocks in and around the reserve. … … … … … … … … … [4] [Total: 11]

11 marks

Mark scheme: 1(a) fish breed within the reserve / numbers increase in the reserve ; idea of: fish ‘spill over’ into areas surrounding reserve ; 2 1(b)(i) any 2 of: egg density decreases (with increasing distance) / negative correlation ; increase after 6 (km) / furthest from reserve ; idea of: wide variation (about the line of best fit) / weak correlation / (many) outliers ; 2 A ORA if explicitly stated A higher / goes at 7.6 (km) or 7.95/8 (km) A (many) anomalies A scattered / wide spread 1(b)(ii) (–)0.4375 to (–)0.4875 (or correct rounding of) ; ; ; 3 I minus 1(c) any 4 of: (supporting benefits) (a) egg density is higher within the reserve / more eggs in, reserve OR egg laying / breeding, is high in the reserve / high fecundity / high recruitment in reserve ; (b) fishing (effort), is high around the reserve showing fish, numbers are high / move out of reserve ; (against benefits) (c) there is only one set of data / no repeats / only one species ; (d) no mention of any controls / not compared to area without reserve ; (e) could be another (named), factor / variable, affecting (fish stocks / egg density / fishing patterns) ; (f) egg numbers, recover / increase, far away from the reserve (where fishing is less) ; 4 A egg density decreases with distance A spawning / breeding, is high in the, reserve / centre R idea of: having / needing to fish harder 1(c) (g) there is, over / excess, fishing near to the reserve ; (h) the correlation is weak with / many outliers ; idea of: fishing effort is not a reliable / accurate / precise, measure OR CPUE would be a better measure / catch would be a better measure ;

This question in 9693/41 May/June 2019

1 Scientists genetically engineered zebrafish for rapid growth. They inserted an additional growth hormone gene into the DNA of zebrafish eggs. The growth hormone gene was injected into 366 zebrafish eggs. 273 of these eggs survived, but only 2 of these eventually developed into genetically engineered fish with the growth hormone gene switched on. (a) (i) Calculate the percentage of injected eggs that developed into genetically engineered zebrafish with the growth hormone gene switched on. … [1] (ii) Only 2 of the 273 zebrafish were successfully genetically engineered and showed rapid growth. Suggest why the success rate of genetic engineering is so low. … … [1] (b) The scientists carried out an experiment to investigate whether the genetically engineered zebrafish grew faster than non-genetically engineered zebrafish. They produced large numbers of genetically engineered zebrafish and calculated their mean mass every week from 4 weeks until 10 weeks. This experiment was repeated with non-genetically engineered zebrafish. The results of the experiment are shown in Table 1.1. Table 1.1 mean mass of zebrafish / g age of zebrafish / weeks genetically engineered non-genetically engineered 4 0.12 0.12 5 0.18 0.15 6 0.21 0.18 7 0.28 0.22 8 0.31 0.25 9 0.33 0.28 10 0.34 0.30 (i) Plot a graph to show the growth of both groups of zebrafish from 4 weeks to 10 weeks. Join your points with ruled, straight lines. [5] (ii) Compare the growth rates of the non-genetically engineered and genetically engineered zebrafish. … … … … [2] (c) Salmon produced by aquaculture are often fed with pellets made from wild fish such as anchovies and herring. Genetically engineered salmon have recently been produced that convert food into growth very efficiently. (i) 1000 kg of wild fish are required to make 250 kg of pellets. 5000 kg of wild fish are required to make the pellets to produce 1000 kg of non-genetically engineered salmon. 1125 kg of pellets are required to produce 1000 kg of genetically engineered salmon. Calculate the difference in the mass of wild fish used to produce 1000 kg of non-genetically engineered salmon compared to 1000 kg of genetically engineered salmon. … kg [1] (ii) Suggest and explain the environmental benefits of using the genetically engineered salmon. … … … … … … [3] [Total: 13]

13 marks

This question in 9693/40 May/June 2020

1 Gold mining causes the release of heavy metals, such as mercury, into estuaries. These heavy metals sink to the bed of the estuary. Some scientists have suggested that dredging of estuaries in areas where gold mining has occurred causes the release of mercury into the water. In January 2009, dredging was banned in an estuary in North America to reduce the release of mercury. A mining company objected to the ban on dredging and investigated the effect of the ban on the release of mercury into the water. They randomly sampled the concentrations of mercury in four species of organisms in the estuary, before and after dredging was stopped. The results are shown in Table 1.1. Table 1.1 species mean concentration of mercury in organism / parts per million 2007 2008 2012 A 42 30 61 B 60 41 50 C 65 50 81 D 190 101 120 (a) (i) State why the organisms were sampled randomly. … … [1] (ii) Plot a line graph to show the changes in concentration of mercury over time for each of the species in Table 1.1. Join your points with ruled, straight lines. Use a separate line for each species. [5] (iii) Species A consumes plants. Species D is a predator. Use this information to explain the differences between the concentrations of mercury in species A and species D in 2007. … … … … [2] (b) The mining company stated that dredging prevents mercury accumulating in the water and that sudden flooding of the estuary causes the release of mercury from the bed of the estuary. They provided data to show the concentration of mercury in the estuary in 2008 and 2009. The results are shown in Fig. 1.1. 300 250 flood flood 200 mercury concentration 150 in water / a.u. dredging 100 50 0 2008 2008 2008 2008 2008 2008 2009 2009 2009 2009 2009 2009 Jan Mar May Jul Sep Nov Jan Mar May Jul Sep Nov month Fig. 1.1 Use the information in Table 1.1 and Fig. 1.1 to discuss the claims of the mining company that banning dredging has led to increased release of mercury from the bed of the estuary. … … … … … … … … [4] [Total: 12]

12 marks

This question in 9693/40 Oct/Nov 2020

4 Fig. 4.1 shows part of the life cycle of the orca (killer whale). adult calf born juveniles live in groups with adult females Fig. 4.1 (a) Use Fig. 4.1 to explain why the life cycle of an orca is classed as a simple life cycle. … … … … [2] (b) Orcas belong to a group of organisms called cetaceans. In some species of cetaceans, females become infertile at a certain age and stop breeding. This is called a menopause. The time after the menopause until the animal dies is called the post-menopause period. Scientists compared the maximum life expectancy with the maximum length of the post-menopause period of a range of cetacean species. The results are shown in Fig. 4.2. Each plotted point indicates the mean value for one species. 35 30 25 20 maximum length of post-menopause period / years 15 10 5 0 40 50 60 70 80 90 100 maximum life expectancy / years Fig. 4.2 The scientists made a hypothesis that there is a positive correlation between the maximum life expectancy of cetacean species and the maximum length of the post-menopause period. Evaluate how strongly the graph shown in Fig. 4.2 supports the scientists’ hypothesis. … … … … … … [3] (c) The scientists carried out a Spearman’s rank correlation test to determine if there was a significant correlation between the maximum life expectancy and the maximum length of the post-menopause period. Their data and rankings are shown in Table 4.1. Table 4.1 cetacean maximum rank maximum rank D D 2 species life maximum length maximum expectancy life of post- length / years expectancy menopause of post- period menopause / years period fin whale 98 1 2 orca 95 2 30 1 1 1 long-finned 60 6.5 1 8.5 –2 4 pilot whale short-finned 64 4 20 2 2 4 pilot whale false killer 58 8 4 5 3 9 whale North Atlantic 49 9 2 6.5 2.5 6.25 right whale sperm whale 60 6.5 1 8.5 –2 4 beluga whale 62 5 12 4 1 1 narwhal 80 3 18 3 0 0 ∑D 2 = ∑ = sum of (total) D = difference in rank between each pair of measurements (i) Complete Table 4.1 by completing the values for: • fin whale • ∑D 2 [1] (ii) Give a null hypothesis for the statistical test. … … [1] (iii) Use the formula to calculate the Spearman’s rank correlation coefficient, rS, for the data in Table 4.1. 6 × ∑D 2 rS = 1 –  n3 – n  rS = Spearman’s rank correlation coefficient Σ = sum of (total) D = difference in rank between each pair of measurements n = number of pairs of items in the sample … [1] (iv) Table 4.2 is a critical values table for Spearman’s rank correlation coefficient. Table 4.2 number of pairs, n rS (p < 0.05) 5 1.000 6 0.886 7 0.786 8 0.738 9 0.700 10 0.648 11 0.618 Use your calculated value from 4(c)(iii), and Table 4.2, to assess whether there is a significant correlation between maximum life expectancy and maximum length of the post-menopause period. Justify your conclusion. … … … … … … [3] (d) Female orcas live in groups with other females they are related to. Use your knowledge of animal life cycles to suggest an explanation for the extended menopause of female orcas. … … … … … [2] [Total: 13]

13 marks

Mark scheme: 4(a) any 2 from: there is no, intermediate different form / larva / AW ; no metamorphosis ; no form that, occupies different niche / lives in different area / has another habitat / AW ; 4(b) any 3 from: (supported because) the line (of best fit) shows a positive correlation ; so as life expectancy increases, so does post-menopause length ; (not supported because) some / many / several of the points are far from the line of best fit / there are several anomalous values / outliers / points are very scattered ; no idea how large sample size was used ; 3 Question Answer Marks 4(c)(i) species maximum life expectancy rank maximum life expectancy maximum length of post menopause period rank maximum length of post menopause period D D2 fin whale 98 1 2 6.5 (–)5.5 30.25 ∑D2 59.5 ; 1 4(c)(ii) there is no, correlation / relationship / association, between the life expectancy and length of (post-) menopause period ; 1 4(c)(iii) 0.5(04…) ; 1 4(c)(iv) any 3 from: If the calculated number is smaller than the critical value… 1 there is no (significant) correlation / there is no (significant) association between life expectancy and (post) menopause length ; 2 the calculated value is not larger than the critical value ; 3 (identification) of 0.7(00) ; 4 so there is a probability of greater than, 0.05 / 5%, that the correlation / association is due to chance / AW ; 5 the null hypothesis is not rejected / null hypothesis is accepted ; If calculated number is larger, then accept: 1 there is a correlation ; 2 the calculated value is larger than the critical value ; 3 (identification) of 0.7(00) ; 4 so there is a probability of less than 0.05 / 5%, that the association is due to chance / AW ; 5 the null hypothesis is rejected ; 3 Question Answer Marks 4(d) any 2 from: 1 (helping to) feed / protect calves / nurture / protect the group / lots of parental care needed / investment needed / AW ; 2 increase survival of calves ; 3 orcas have few calves / are k-selectors / AW ; 4 (and need) calves stay with group for years / calves stay with mothers / group for a long time / AW ; 2

This question in 9693/41 May/June 2022

5 Lionfish are a species of carnivorous fish naturally found in the Pacific Ocean near Indonesia. Lionfish are now found in the western Atlantic Ocean and Caribbean Sea and are classed as an invasive species. (a) State what is meant by an invasive species, as defined by the IUCN. … … [1] (b) Fig. 5.1 shows the change in population of lionfish between 2004 and 2010 on an area of coral reef near the Bahamas in the western Atlantic Ocean. 50 45 40 35 mean 30 number of lionfish 25 counted per survey 20 15 10 5 0 2004 2005 2006 2007 2008 2009 2010 year Fig. 5.1 (i) Suggest explanations for the changes in population of lionfish on this coral reef between 2004 and 2010 shown in Fig. 5.1. … … … … … … [3] (ii) Fig. 5.1 has error bars that represent the standard deviation. Explain what the standard deviations in Fig. 5.1 show about the data. … … … … [2] (c) Fig. 5.2 shows the percentage change in abundance of other species of fish and algae growing on the same coral reef between 2004 and 2010. The species of fish were classed as: • small fish species that are prey of lionfish • small fish species that are not prey of lionfish • large fish species that are competitors of lionfish • large fish species that are not competitors of lionfish +100 +80 +60 +40 +20 percentage change 0 in abundance –20 –40 –60 –80 –100 small prey small non-prey large large algae species species competitor non-competitor species species Fig. 5.2 Discuss the change in abundance of the different species of fish and algae shown in Fig. 5.2. … … … … … … [3] (d) Scientists investigated the effect of removing lionfish on the population of damselfish on a reef. Divers physically removed lionfish from an area of reef each week for a period of six months. The population of damselfish on the reef was then recorded. The population of damselfish on an area of identical reef where lionfish were not removed was also recorded. Both reefs had the same initial populations of damselfish. The scientists carried out a chi-squared test to see if removing the lionfish caused a change in the population of damselfish. They made the following null hypothesis: Removing the lionfish did not affect the number of damselfish on the reef. The results are shown in Table 5.1. Table 5.1 reef area number of expected (O – E) (O – E)2 (O – E)2 damselfish number of E (O) damselfish (E) lionfish 420 390 30 900 2.308 removed no lionfish 360 390 removed (i) Complete Table 5.1. [1] (ii) Use the formula to calculate the chi-squared value for the results. (O – E)2 chi-squared = Σ E Σ = sum of (total) O = observed values E = expected values … [1]

11 marks

Mark scheme: 5(a)(i) a species that has been introduced outside its natural past or present distribution and has become problematic / AW ; 1 5(b)(i) any 3 from: between 2004 and 2006 / up to 2006, (slow population increase) as lionfish, adapt to area / have few to breed / take time to establish / AW ; from 2004, increase as lionfish have abundant food / few predators / high breeding rate / outcompete other species / AW ; plenty of niches available for lionfish ; from 2007 / 2008 population stabilises / levels off / falls, due to competition / as food limiting / predators arrive / control methods / harvesting / AW ; 3 Question Answer Marks 5(b)(ii) any 2 from: wide variation in number of lionfish sightings in years between 2007 and 2010 ; less / little variation in number of lionfish sightings in years 2004 / 2005 / 2006 ; no overlap between 2006 and 2007 shows (significant) difference / AW ; overlap between 2007 and 2010 shows no (significant) difference / AW ; 2 5(c) any 3 from: decrease in (small) prey species as they are eaten / predated (by lionfish) ; increase in (small) non-prey species due to less competition (from other fish) / more algae to eat ; large competitors decrease as lionfish consume food / take territory / AW ; large non-competitors show similar numbers so maintain niche / are unaffected / have enough food ; algae increases, as small herbivores / small prey species, are, consumed / are fewer ; 3 5(d)(i) Reef area Number of damselfish (O) Expected number of damselfish (E) (O–E) (O–E)2  2 ( ) O E E Lionfish removed 420 390 30 900 2.308 No lionfish removed 360 390 –30 900 2.308 ; 1 5(d)(ii) 4.616 ; 1 5(d)(iii) any 3 from: the null hypothesis is rejected ; there is a significant difference (in the number of damselfish) ; the calculated value is greater than the critical value ; correct reference to 3.841 / critical value for 0.05 and 1 degree of freedom ; there is a probability of less than, 0.05 / 5, that the difference is due to chance ; 3 Question Answer Marks 5(d)(iv) any 2 from: marine ecosystems are large ; species cross (national) national borders ; species are migratory / breed in different locations ; ocean currents move species / pollutants around the globe / AW ; trade of species occurs between countries / ref to trade restrictions ; some areas of ocean / High Seas have no national ownership 2

This question in 9693/42 May/June 2022

5 Lionfish are a species of carnivorous fish naturally found in the Pacific Ocean near Indonesia. Lionfish are now found in the western Atlantic Ocean and Caribbean Sea and are classed as an invasive species. (a) State what is meant by an invasive species, as defined by the IUCN. … … [1] (b) Fig. 5.1 shows the change in population of lionfish between 2004 and 2010 on an area of coral reef near the Bahamas in the western Atlantic Ocean. 50 45 40 35 mean 30 number of lionfish 25 counted per survey 20 15 10 5 0 2004 2005 2006 2007 2008 2009 2010 year Fig. 5.1 (i) Suggest explanations for the changes in population of lionfish on this coral reef between 2004 and 2010 shown in Fig. 5.1. … … … … … … [3] (ii) Fig. 5.1 has error bars that represent the standard deviation. Explain what the standard deviations in Fig. 5.1 show about the data. … … … … [2] (c) Fig. 5.2 shows the percentage change in abundance of other species of fish and algae growing on the same coral reef between 2004 and 2010. The species of fish were classed as: • small fish species that are prey of lionfish • small fish species that are not prey of lionfish • large fish species that are competitors of lionfish • large fish species that are not competitors of lionfish +100 +80 +60 +40 +20 percentage change 0 in abundance –20 –40 –60 –80 –100 small prey small non-prey large large algae species species competitor non-competitor species species Fig. 5.2 Discuss the change in abundance of the different species of fish and algae shown in Fig. 5.2. … … … … … … [3] (d) Scientists investigated the effect of removing lionfish on the population of damselfish on a reef. Divers physically removed lionfish from an area of reef each week for a period of six months. The population of damselfish on the reef was then recorded. The population of damselfish on an area of identical reef where lionfish were not removed was also recorded. Both reefs had the same initial populations of damselfish. The scientists carried out a chi-squared test to see if removing the lionfish caused a change in the population of damselfish. They made the following null hypothesis: Removing the lionfish did not affect the number of damselfish on the reef. The results are shown in Table 5.1. Table 5.1 reef area number of expected (O – E) (O – E)2 (O – E)2 damselfish number of E (O) damselfish (E) lionfish 420 390 30 900 2.308 removed no lionfish 360 390 removed (i) Complete Table 5.1. [1] (ii) Use the formula to calculate the chi-squared value for the results. (O – E)2 chi-squared = Σ E Σ = sum of (total) O = observed values E = expected values … [1]

11 marks

Mark scheme: 5(a)(i) a species that has been introduced outside its natural past or present distribution and has become problematic / AW ; 1 5(b)(i) any 3 from: between 2004 and 2006 / up to 2006, (slow population increase) as lionfish, adapt to area / have few to breed / take time to establish / AW ; from 2004, increase as lionfish have abundant food / few predators / high breeding rate / outcompete other species / AW ; plenty of niches available for lionfish ; from 2007 / 2008 population stabilises / levels off / falls, due to competition / as food limiting / predators arrive / control methods / harvesting / AW ; 3 Question Answer Marks 5(b)(ii) any 2 from: wide variation in number of lionfish sightings in years between 2007 and 2010 ; less / little variation in number of lionfish sightings in years 2004 / 2005 / 2006 ; no overlap between 2006 and 2007 shows (significant) difference / AW ; overlap between 2007 and 2010 shows no (significant) difference / AW ; 2 5(c) any 3 from: decrease in (small) prey species as they are eaten / predated (by lionfish) ; increase in (small) non-prey species due to less competition (from other fish) / more algae to eat ; large competitors decrease as lionfish consume food / take territory / AW ; large non-competitors show similar numbers so maintain niche / are unaffected / have enough food ; algae increases, as small herbivores / small prey species, are, consumed / are fewer ; 3 5(d)(i) Reef area Number of damselfish (O) Expected number of damselfish (E) (O–E) (O–E)2  2 ( ) O E E Lionfish removed 420 390 30 900 2.308 No lionfish removed 360 390 –30 900 2.308 ; 1 5(d)(ii) 4.616 ; 1 5(d)(iii) any 3 from: the null hypothesis is rejected ; there is a significant difference (in the number of damselfish) ; the calculated value is greater than the critical value ; correct reference to 3.841 / critical value for 0.05 and 1 degree of freedom ; there is a probability of less than, 0.05 / 5, that the difference is due to chance ; 3 Question Answer Marks 5(d)(iv) any 2 from: marine ecosystems are large ; species cross (national) national borders ; species are migratory / breed in different locations ; ocean currents move species / pollutants around the globe / AW ; trade of species occurs between countries / ref to trade restrictions ; some areas of ocean / High Seas have no national ownership ; 2

This question in 9693/43 May/June 2022

1 Fig. 1.1 shows a diagram of a cell from the leaf of a species of seagrass. B A C Fig. 1.1 (a) (i) Give the names of the structures labelled A and B. A … B … [2] (ii) The magnification of the diagram is × 5000. Calculate the maximum length of the structure labelled C. Show your working. State the unit. … [3] (b) The effect of salinity on the growth of seagrass leaves was investigated. Seagrass plants were placed in different salinities for two weeks. The increase in length of 9 leaves at each of the salinities was measured. The mean rate of leaf growth per day was then calculated. The results are shown in Fig. 1.2. The error bars represent ± 1 standard deviation. 0.40 0.35 0.30 0.25 mean rate of leaf 0.20 growth / cm day–1 0.15 0.10 0.05 0.00 0 5 10 15 20 25 30 35 40 45 50 salinity / ppt Fig. 1.2 (i) Describe the effect of increasing salinity on the mean rate of leaf growth per day. … … … … [2] (ii) A student concluded that the results show that the optimum salinity for growth of the seagrass is 25 ppt. Use the data in Fig. 1.2 to discuss whether their conclusion is correct. … … … … … … [3] (iii) Desalination plants are used to produce fresh water from sea water. Use the information in Fig. 1.2, and your own knowledge, to explain why outflow from a desalination plant might harm seagrass growth. … … … … … … … … [4] [Total: 14]

14 marks

Mark scheme: 1(a)(i) A: Golgi, (body / apparatus) ; B: rough endoplasmic reticulum / RER ; 2 1(a)(ii) 10 (mm) ; 10 / 5000 = 0.002 ; correct unit ; ONLY AWARD UNIT MARK IF ANSWER IS CORRECT (e.g. 2 m, 0.002 mm) 3 1(b)(i) increase and decrease; (increase up to) 25 (ppt) / from 30 (ppt); 2 1(b)(ii) any 3 of: 1 (correct because) the highest / fastest rate of growth is at 25 (ppt) / AW ; 2 carried out with ten / many leaves so reliable ; (incorrect because) 3 no significant difference between 20 / 30 (ppt) and 25 ) ppt) ; 4 (because) standard deviations overlap (between 20 / 30 (ppt) and 25 (ppt)) / AW ; 5 optimum may be between 20 (ppt) and 30 (ppt) ; 6 no measurements taken between 20 (ppt) and 25 (ppt) / 25 (ppt) and 30 (ppt) ; 3 1(b)(iii) any 4 of: 1 reduced photosynthesis ; 2 (because) desalination plants release high salinity water / brine / concentrated brine / increase salinity / return salt to sea / AW ; 3 growth is reduced / lower growth, at high salinities / concentrations / above 25 (ppt) / AW ; 4 osmotic effects harm the seagrass / osmosis will occur / AW ; 5 water leaves cells / plant / seagrass ; 6 cells / plant / seagrass, no longer supported / cells plasmolyse / cells are not turgid / AW ; 7 desalination plants stir up sediment / increase turbidity ; 8 AVP ; 4

This question in 9693/41 May/June 2023

5 (a) Describe how microplastics are formed. … … … … [2] (b) A scientist investigated the effect of human population density near coastal areas on the number of microplastic particles in the sand of 12 beaches. The scientist made the following null hypothesis: There is no correlation between human population density and the number of microplastic particles. They carried out a Spearman’s rank correlation test on their data. Table 5.1 shows some of their calculations. Table 5.1 human r1, number of r2, D D2 population rank of microplastic rank of (r1 – r2) density / people human particles number of per km2 population per 250 cm3 microplastic density sand particles 0 1 5 1 0 0 5 2 7 3 –1 1 10 3.5 19 6 –2.5 6.25 15 5 6 2 3 9 155 11 85 11 0 0 75 8.5 75 9.5 –1 1 65 7 25 7 0 0 75 … 65 8 … … 10 3.5 11 4.5 –1 1 20 6 11 4.5 1.5 2.25 120 10 75 9.5 0.5 0.25 175 12 115 12 0 0 ∑ D2 = … (i) Complete Table 5.1 by determining the missing values. Write your answers in Table 5.1. [2] (ii) Use the formula to calculate the Spearman’s rank correlation coefficient for the data in Table 5.1. 6 × ∑D 2 rS = 1 –  n3 – n  rS = Spearman’s rank correlation coefficient ∑ = sum of (total) D = difference in rank between each pair of measurements n = number of pairs of items in the sample rS = … [1] (iii) Table 5.2 is a critical values table for Spearman’s rank correlation coefficient. Table 5.2 number of pairs, n rS (p < 0.05) 5 1.000 6 0.886 7 0.786 8 0.738 9 0.700 10 0.648 11 0.618 12 0.587 13 0.560 14 0.538 15 0.521 Use your calculated value from (b)(ii), and Table 5.2, to assess whether the null hypothesis can be accepted or rejected. … … … … … … [3] (c) Use the information in this question to suggest why regularly eating mussels from shores near areas with high human population density may be harmful to humans. … … … … [2] [Total: 10]

10 marks

Mark scheme: 5(a) any 2 of: 1 plastic broken into pieces of less than 5 mm ; 2 due to action of UV (radiation) ; 3 due to wave (action) ; 4 due to high temperature ; 5 due to wind (action) ; 2 Question Answer Marks 5(b)(i) human population density / people per km2 r1, rank of human population density number of microplastic particles per 250 cm3 sand r2, rank of number of microplastic particles D (r1 – r2) D2 0 1 5 1 0 0 5 2 7 3 –1 1 10 3.5 19 6 –2.5 6.25 15 5 6 2 3 9 155 11 85 11 0 0 75 8.5 75 9.5 –1 1 65 7 25 7 0 0 75 8.5 65 8 0.5 0.25 ; 10 3.5 11 4.5 –1 1 20 6 11 4.5 1.5 2.25 120 10 75 9.5 0.5 0.25 175 12 115 12 0 0 2 D   21 ; 2 5(b)(ii) 0.93 ; 1 Question Answer Marks 5(b)(iii) any 3 of: the null hypothesis is rejected ; the calculated value is greater than the critical value ; critical value is 0.587 ; there is a significant positive correlation / there is a significant association between human population and microplastic density ; 3 5(c) any 2 of: (area with high human populations have) high densities of microplastic / AW ; microplastics absorb toxins / AW ; mussels, take in / eat, microplastics ; humans get toxins from eating mussels / humans get microplastics from eating mussels / AW ; 2

This question in 9693/41 May/June 2023

4 Mangrove forests are areas of environmental importance. Replanting mangroves has been suggested as a method to increase the biodiversity of the areas they are grown in. (a) Fig. 4.1 shows a section through a leaf from a holly mangrove bush. Fig. 4.1 Make a large drawing of the part of the mangrove leaf in the circle in Fig. 4.1. Do not label your drawing. [4] (b) A student investigated if the size of mangrove forests affects biodiversity. The student collected data from different-sized areas of mangrove forest. The student then carried out a Spearman’s rank correlation coefficient test to test the null hypothesis. The null hypothesis was that there was no correlation between the area of mangrove forest and Simpson’s index of diversity of organisms. Table 4.1 shows the results. Table 4.1 area of rank of area Simpson’s rank of D (r1–r2) D2 mangrove of forest (r1) index of Simpson’s forest / km2 diversity index of diversity (r2) 0.02 8 0.44 7 1 1 0.04 5.5 0.56 5 0.5 0.25 0.02 0.34 10 0.11 1 0.82 1 0 0 0.02 0.38 8 0.05 4 0.61 3 1 1 0.06 3 0.58 4 –1 1 0.09 2 0.75 2 0 0 0.01 10 0.35 9 1 1 0.04 5.5 0.54 6 –0.5 0.25 ΣD2 = … (i) Complete Table 4.1 to calculate the value of ΣD2. [2] (ii) Use your answer to (b)(i) and the formula to calculate the Spearman’s rank correlation coefficient, rs. Give your answer to three decimal places. 6 × ΣD2 rs = 1 –  n3 – n  Σ = sum of (total) n = number of pairs of items in the sample D = difference in rank between each pair of measurements … [2] (iii) Table 4.2 is a critical values table for Spearman’s rank correlation coefficient. Table 4.2 number of P<0.05 paired items, n 5 1.000 6 0.886 7 0.786 8 0.738 9 0.700 10 0.648 11 0.618 12 0.587 Use your answer to (b)(ii) and Table 4.2 to evaluate the null hypothesis. … … … … … … [3] (iv) Suggest why the size of the area of mangrove forest affects the Simpson’s index of diversity. … … … … [2] (c) In a further study in a coastal area of Indonesia, local people were asked to complete a questionnaire about how willing they were to take part in replanting mangroves. The results of the questionnaire are shown in Table 4.3. Table 4.3 question percentage of people with response / % strongly disagree unsure agree strongly disagree agree Are you willing to participate 12 12 0 40 36 in mangrove replanting in the area? Do you currently volunteer to 38 42 8 7 5 help mangrove replanting? Should you have appropriate 9 18 3 44 26 pay if you participate in the mangrove replanting? Do you have skills in mangrove 35 28 5 18 14 replanting? Use the results of the questionnaire in Table 4.3 to discuss how a government could successfully engage local communities in replanting mangrove forests. … … … … … … [3] [Total: 16]

16 marks

Mark scheme: 4(a) 1 thin, unbroken clear lines in pencil ; 4 2 correct proportions and takes up at least two-thirds of space ; 3 six cells drawn, cell walls for all cells, nucleus in middle cell, double line for outer layer ; 4 no shading and only draw what is inside circle ; 4(b)(i) 2 area of rank of area Simpson’s rank of D (r1–r2) D2 mangrove of forest (r1) index of Simpson’s forest / km2 diversity index of diversity (r2) 0.02 8 0.44 7 1 1 0.04 5.5 0.56 5 0.5 0.25 0.02 8 0.34 10 –2 4 0.11 1 0.82 1 0 0 0.02 8 0.38 8 0 0 ; 0.05 4 0.61 3 1 1 0.06 3 0.58 4 -1 1 0.09 2 0.75 2 0 0 0.01 10 0.35 9 1 1 0.04 5.5 0.54 6 -0.5 0.25  D2 = 8.5 ; 4(b)(ii) 0.948 ;; 2 4(b)(iii) any 3 of: 3 null hypothesis is rejected ; calculated value is greater than the critical value ; critical value identified as 0.648 ; there is a (significant) positive correlation between size of mangrove forest area and species diversity / there is probability of less than 0.05 that the association is due to chance ; 4(b)(iv) any 2 of: 2 more producers / more productivity / more photosynthesis ; more energy enters ecosystem ; more (trophic) levels possible in food chains ; more niches ; more habitats / more (variety of) food sources / more nesting sites / more shelter from predators / AW ; 4(c) any 3 of: 3 1 few people are currently involved / AW ; 2 but local people, want / are willing to get involved / AW ; 3 need to pay people / compensate for time spent working / AW ; 4 education / training needed (as people do not already have skills) ; 5 train local people to teach other locals (peer education) / encourage social activities around planting / AW ; 6 local community involvement could be more successful than ‘top down’ government work / make local people stakeholders / AW ;

This question in 9693/41 Oct/Nov 2023

4 Mangrove forests are areas of environmental importance. Replanting mangroves has been suggested as a method to increase the biodiversity of the areas they are grown in. (a) Fig. 4.1 shows a section through a leaf from a holly mangrove bush. Fig. 4.1 Make a large drawing of the part of the mangrove leaf in the circle in Fig. 4.1. Do not label your drawing. [4] (b) A student investigated if the size of mangrove forests affects biodiversity. The student collected data from different-sized areas of mangrove forest. The student then carried out a Spearman’s rank correlation coefficient test to test the null hypothesis. The null hypothesis was that there was no correlation between the area of mangrove forest and Simpson’s index of diversity of organisms. Table 4.1 shows the results. Table 4.1 area of rank of area Simpson’s rank of D (r1–r2) D2 mangrove of forest (r1) index of Simpson’s forest / km2 diversity index of diversity (r2) 0.02 8 0.44 7 1 1 0.04 5.5 0.56 5 0.5 0.25 0.02 0.34 10 0.11 1 0.82 1 0 0 0.02 0.38 8 0.05 4 0.61 3 1 1 0.06 3 0.58 4 –1 1 0.09 2 0.75 2 0 0 0.01 10 0.35 9 1 1 0.04 5.5 0.54 6 –0.5 0.25 ΣD2 = … (i) Complete Table 4.1 to calculate the value of ΣD2. [2] (ii) Use your answer to (b)(i) and the formula to calculate the Spearman’s rank correlation coefficient, rs. Give your answer to three decimal places. 6 × ΣD2 rs = 1 –  n3 – n  Σ = sum of (total) n = number of pairs of items in the sample D = difference in rank between each pair of measurements … [2] (iii) Table 4.2 is a critical values table for Spearman’s rank correlation coefficient. Table 4.2 number of P<0.05 paired items, n 5 1.000 6 0.886 7 0.786 8 0.738 9 0.700 10 0.648 11 0.618 12 0.587 Use your answer to (b)(ii) and Table 4.2 to evaluate the null hypothesis. … … … … … … [3] (iv) Suggest why the size of the area of mangrove forest affects the Simpson’s index of diversity. … … … … [2] (c) In a further study in a coastal area of Indonesia, local people were asked to complete a questionnaire about how willing they were to take part in replanting mangroves. The results of the questionnaire are shown in Table 4.3. Table 4.3 question percentage of people with response / % strongly disagree unsure agree strongly disagree agree Are you willing to participate 12 12 0 40 36 in mangrove replanting in the area? Do you currently volunteer to 38 42 8 7 5 help mangrove replanting? Should you have appropriate 9 18 3 44 26 pay if you participate in the mangrove replanting? Do you have skills in mangrove 35 28 5 18 14 replanting? Use the results of the questionnaire in Table 4.3 to discuss how a government could successfully engage local communities in replanting mangrove forests. … … … … … … [3] [Total: 16]

16 marks

Mark scheme: 4(a) 1 thin, unbroken clear lines in pencil ; 4 2 correct proportions and takes up at least two-thirds of space ; 3 six cells drawn, cell walls for all cells, nucleus in middle cell, double line for outer layer ; 4 no shading and only draw what is inside circle ; 4(b)(i) 2 area of rank of area Simpson’s rank of D (r1–r2) D2 mangrove of forest (r1) index of Simpson’s forest / km2 diversity index of diversity (r2) 0.02 8 0.44 7 1 1 0.04 5.5 0.56 5 0.5 0.25 0.02 8 0.34 10 –2 4 0.11 1 0.82 1 0 0 0.02 8 0.38 8 0 0 ; 0.05 4 0.61 3 1 1 0.06 3 0.58 4 -1 1 0.09 2 0.75 2 0 0 0.01 10 0.35 9 1 1 0.04 5.5 0.54 6 -0.5 0.25  D2 = 8.5 ; 4(b)(ii) 0.948 ;; 2 4(b)(iii) any 3 of: 3 null hypothesis is rejected ; calculated value is greater than the critical value ; critical value identified as 0.648 ; there is a (significant) positive correlation between size of mangrove forest area and species diversity / there is probability of less than 0.05 that the association is due to chance ; 4(b)(iv) any 2 of: 2 more producers / more productivity / more photosynthesis ; more energy enters ecosystem ; more (trophic) levels possible in food chains ; more niches ; more habitats / more (variety of) food sources / more nesting sites / more shelter from predators / AW ; 4(c) any 3 of: 3 1 few people are currently involved / AW ; 2 but local people, want / are willing to get involved / AW ; 3 need to pay people / compensate for time spent working / AW ; 4 education / training needed (as people do not already have skills) ; 5 train local people to teach other locals (peer education) / encourage social activities around planting / AW ; 6 local community involvement could be more successful than ‘top down’ government work / make local people stakeholders / AW ;

This question in 9693/42 Oct/Nov 2023

4 Mangrove forests are areas of environmental importance. Replanting mangroves has been suggested as a method to increase the biodiversity of the areas they are grown in. (a) Fig. 4.1 shows a section through a leaf from a holly mangrove bush. Fig. 4.1 Make a large drawing of the part of the mangrove leaf in the circle in Fig. 4.1. Do not label your drawing. [4] (b) A student investigated if the size of mangrove forests affects biodiversity. The student collected data from different-sized areas of mangrove forest. The student then carried out a Spearman’s rank correlation coefficient test to test the null hypothesis. The null hypothesis was that there was no correlation between the area of mangrove forest and Simpson’s index of diversity of organisms. Table 4.1 shows the results. Table 4.1 area of rank of area Simpson’s rank of D (r1–r2) D2 mangrove of forest (r1) index of Simpson’s forest / km2 diversity index of diversity (r2) 0.02 8 0.44 7 1 1 0.04 5.5 0.56 5 0.5 0.25 0.02 0.34 10 0.11 1 0.82 1 0 0 0.02 0.38 8 0.05 4 0.61 3 1 1 0.06 3 0.58 4 –1 1 0.09 2 0.75 2 0 0 0.01 10 0.35 9 1 1 0.04 5.5 0.54 6 –0.5 0.25 ΣD2 = … (i) Complete Table 4.1 to calculate the value of ΣD2. [2] (ii) Use your answer to (b)(i) and the formula to calculate the Spearman’s rank correlation coefficient, rs. Give your answer to three decimal places. 6 × ΣD2 rs = 1 –  n3 – n  Σ = sum of (total) n = number of pairs of items in the sample D = difference in rank between each pair of measurements … [2] (iii) Table 4.2 is a critical values table for Spearman’s rank correlation coefficient. Table 4.2 number of P<0.05 paired items, n 5 1.000 6 0.886 7 0.786 8 0.738 9 0.700 10 0.648 11 0.618 12 0.587 Use your answer to (b)(ii) and Table 4.2 to evaluate the null hypothesis. … … … … … … [3] (iv) Suggest why the size of the area of mangrove forest affects the Simpson’s index of diversity. … … … … [2] (c) In a further study in a coastal area of Indonesia, local people were asked to complete a questionnaire about how willing they were to take part in replanting mangroves. The results of the questionnaire are shown in Table 4.3. Table 4.3 question percentage of people with response / % strongly disagree unsure agree strongly disagree agree Are you willing to participate 12 12 0 40 36 in mangrove replanting in the area? Do you currently volunteer to 38 42 8 7 5 help mangrove replanting? Should you have appropriate 9 18 3 44 26 pay if you participate in the mangrove replanting? Do you have skills in mangrove 35 28 5 18 14 replanting? Use the results of the questionnaire in Table 4.3 to discuss how a government could successfully engage local communities in replanting mangrove forests. … … … … … … [3] [Total: 16]

16 marks

Mark scheme: 4(a) 1 thin, unbroken clear lines in pencil ; 4 2 correct proportions and takes up at least two-thirds of space ; 3 six cells drawn, cell walls for all cells, nucleus in middle cell, double line for outer layer ; 4 no shading and only draw what is inside circle ; 4(b)(i) 2 area of rank of area Simpson’s rank of D (r1–r2) D2 mangrove of forest (r1) index of Simpson’s forest / km2 diversity index of diversity (r2) 0.02 8 0.44 7 1 1 0.04 5.5 0.56 5 0.5 0.25 0.02 8 0.34 10 –2 4 0.11 1 0.82 1 0 0 0.02 8 0.38 8 0 0 ; 0.05 4 0.61 3 1 1 0.06 3 0.58 4 -1 1 0.09 2 0.75 2 0 0 0.01 10 0.35 9 1 1 0.04 5.5 0.54 6 -0.5 0.25  D2 = 8.5 ; 4(b)(ii) 0.948 ;; 2 4(b)(iii) any 3 of: 3 null hypothesis is rejected ; calculated value is greater than the critical value ; critical value identified as 0.648 ; there is a (significant) positive correlation between size of mangrove forest area and species diversity / there is probability of less than 0.05 that the association is due to chance ; 4(b)(iv) any 2 of: 2 more producers / more productivity / more photosynthesis ; more energy enters ecosystem ; more (trophic) levels possible in food chains ; more niches ; more habitats / more (variety of) food sources / more nesting sites / more shelter from predators / AW ; 4(c) any 3 of: 3 1 few people are currently involved / AW ; 2 but local people, want / are willing to get involved / AW ; 3 need to pay people / compensate for time spent working / AW ; 4 education / training needed (as people do not already have skills) ; 5 train local people to teach other locals (peer education) / encourage social activities around planting / AW ; 6 local community involvement could be more successful than ‘top down’ government work / make local people stakeholders / AW ;

This question in 9693/43 Oct/Nov 2023

2 The giant tubeworm, Riftia, lives near hydrothermal vents. Endoriftia bacteria live inside Riftia. (a) (i) Outline how Endoriftia produces glucose by chemosynthesis. … … … … [2] (ii) Explain the relationship between Riftia and Endoriftia. … … … … [2] (b) Some species of chemosynthetic bacteria live in the water and substrate close to hydrothermal vents. Scientists investigated the relationship between the population density of free-living chemosynthetic bacteria and the presence of hydrogen gas (H2) and methane gas (CH4). They took samples at three different hydrothermal vents and also at an area of the deep sea bed where there were no hydrothermal vents. The results are shown in Table 2.1. Table 2.1 factor location vent 1 vent 2 vent 3 deep sea bed concentration of H2 21.9 127.0 9.4 0.0 / μmol dm–3 concentration of CH4 23.7 10.1 38.4 0.0 / μmol dm–3 population density of 5.51 3.14 5.62 2.91 chemosynthetic bacteria / cells per cm3 × 104 Suggest an explanation for the different densities of chemosynthetic bacteria at these four locations. Use the information in Table 2.1 to support your answer. … … … … … … [3] (c) Chemosynthetic bacteria around hydrothermal vents are important parts of the ecosystem. Protoctists are microscopic single-celled organisms. Species of protoctist consume the bacteria. The protoctists are then consumed by other organisms in the ecosystem. In a further investigation, the scientists brought samples of water containing the bacteria and protoctists up to the surface. They measured the mean rates at which the protoctists consumed the bacteria in the samples of water from each vent and the deep sea bed. The results are shown in Fig. 2.1. 2500 2000 mean rate of 1500 consumption of bacteria by protoctists / cells cm–3 hr–1 1000 500 0 vent 1 vent 2 vent 3 deep sea bed location Fig. 2.1 (i) The error bars in Fig. 2.1 represent standard deviations. Explain what the error bars in Fig. 2.1 demonstrate about the reliability of the data. … … … … [2] (ii) The scientists concluded that chemosynthetic bacteria are a more important part of food chains on the sea bed around hydrothermal vents than in areas of the deep sea bed away from the vents. Discuss the scientists’ conclusion. Use the data in Fig. 2.1 to support your answer. … … … … … … [3] [Total: 12]

12 marks

Mark scheme: 2(a)(i) any 2 of: 1 (uses) hydrogen sulfide ; 2 (and) carbon dioxide ; 3 (hydrogen sulfide is a) source of energy ; 2 2(a)(ii) mutualistic relationship / mutualism ; Riftia obtains sugars / glucose from Endoriftia and Endoriftia gains minerals / carbon dioxide / shelter / protection / habitat from Riftia ; 2 2(b) any 3 of: 1 more bacteria around vents / fewer bacteria around seabed / AW ; 2 methane / hydrogen provides energy ; 3 methane has a more / bigger, effect (than hydrogen) / ORA / AW ; 4 data manipulation to support answer ; 3 2(c)(i) any 2 of: 1 there is a lot of variation (for the vents) because error bars are large (so less reliable) / AW ; 2 the error bars for the vents overlap so all vents have similar rates / AW ; 3 the error bar for deep sea bed does not overlap so is different (to vents) / AW ; 2 2(c)(ii) any 3 of: 1 feeding is high(er) around vents (compared with deep seabed) / ORA / AW ; 2 protoctists obtain energy from the (chemosynthetic) bacteria / AW ; 3 (chemosynthetic) bacteria are producers / AW ; 4 but there may be other organisms that consume the bacteria / AW ; 5 conditions around the vents / seabed / AW, may differ from the laboratory ; 6 only one experiment / need more repeats / AW ; 3

This question in 9693/41 May/June 2024

4 Foraminifera are microscopic, single-celled marine organisms. Some species of foraminifera produce shells made from calcium carbonate. Fig. 4.1 shows the shell of a species of foraminifera. Fig. 4.1 Scientists investigated the effect of carbon dioxide concentration in the water on the length of the shells and on population growth of foraminifera. Tanks of sea water were set up with different concentrations of carbon dioxide. Foraminifera were added to each of the tanks. Oxygen was bubbled into each tank and food was added daily. The mean shell lengths of the foraminifera and the population density were determined after eight weeks. The results are shown in Table 4.1. Table 4.1 carbon dioxide mean shell length / mm mean population density concentration / ppm / foraminifera per cm3 10 0.85 1800 25 0.82 1750 50 0.84 1500 110 0.52 950 125 0.25 250 (b) The foraminifera have shells made from calcium carbonate. Explain why increasing the carbon dioxide concentration affects the growth of the foraminifera. … … … … … … [3] (c) There are many different species of foraminifera in areas of coral reef. The species diversity of foraminifera is thought to be affected by factors such as temperature, carbon dioxide concentration, and calcium ion concentration. Some scientists think that global warming will affect the species diversity of foraminifera. Plan a laboratory-based investigation that you could do to investigate the effect of temperature on the species diversity of a sample of foraminifera that grows around coral. You are provided with a mixed starter culture of 20 different species of foraminifera, sea water, microscopes, coral around which the foraminifera live, and other standard laboratory equipment. Your plan should: • include a clear statement of the hypothesis • identify the independent, dependent and standardised variables • include full details of the method so that another person can follow it • describe how you would analyse your results • be safe and ethical. … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … [11] [Total: 19]

19 marks

Mark scheme: 4(a) 1 both y axes labelled with units ; 2 two linear y axes with suitable scales so that lines cover at least half grid ; 3 points correctly plotted ; 4 neat lines joining points with no extrapolation ; 5 key for both lines ; 5 Question Answer Marks 4(a) Question Answer Marks 4(b) any 3 of: 1 carbon dioxide dissolves in water ; 2 forms, carbonic acid / H2CO3 / acidification occurs / pH drops / AW ; 3 (carbonic dissociates) into, bicarbonate (ions) / HCO3- (and H+ ions) ; 4 (H+) reacts with, CO32- / carbonate ions ; 5 loss of calcium carbonate from shells / shells dissolve / weaker shells produced / poor shell growth / AW ; 3 4(c) Hypothesis (h): increase in temperature will reduce diversity / AW ; any 10 of: Independent variable (i) 1 independent variable identified as temperature / AW ; 2 at least five stated temperatures ; Dependent variable (d) 3 dependent variable identified as species diversity / AW ; 4 (count) number of different species / AW ; 5 (count) number of (individuals of) each species of foraminifera / AW ; Standardised variables (s) max 3 6 same stated time (more than one week) ; 7 same pH ; 8 same salinity / same calcium / mineral ions / AW ; 9 same number of foraminifera at start / same volume of starter culture / initial diversity / AW ; 10 same quantity of food / nutrients / AW ; 11 same volume of water / volume of tanks / AW ; 12 same carbon dioxide / oxygen ; 13 same size / mass / amount of coral / AW ; 11 Question Answer Marks 4(c) Method marks (m) max 2 14 method of changing temperature (e.g. water-bath) / AW ; 15 use of pipette / syringe / AW ; 16 count / identify foraminifera using, microscope / haemocytometer / AW ; 17 use of key to identify foraminifera / AW ; Analysis (a) max 2 18 use Simpson’s index of diversity / AW ; 19 plot graph of diversity index against temperature ; 20 calculation of correlation coefficient / Spearman’s rank ; 21 example of correct results table ; 22 repeats and calculate mean / median / standard deviation / standard error ; Ethical / Health and safety (e) 23 identification of any risk with method to minimise / statement that experiment is low risk / AW ; 24 treat foraminifera ethically by using acceptable range of temperatures / return foraminifera to sea / replace coral / AW ;

This question in 9693/41 May/June 2024

1 Fig. 1.1 shows some barnacles and a dogwhelk on a rock. Fig. 1.1 Adult barnacles are sessile and are often found attached to rocks on rocky shores. Barnacles have a complex life cycle with planktonic larvae. (a) Explain why a complex life cycle is an advantage for sessile organisms. … … … … [2] (b) The dogwhelk feeds on barnacles on rocky shores. Scientists investigated whether the settlement of barnacles on a rocky shore was affected by: • the presence of dogwhelks • the quantity of phytoplankton in the sea water next to the shore. The scientists placed 0.25 m2 plastic tiles onto four rocky shores: • one shore with a high quantity of phytoplankton in the sea and with dogwhelks present • one shore with a high quantity of phytoplankton in the sea and with no dogwhelks present • one shore with a low quantity of phytoplankton in the sea and with dogwhelks present • one shore with a low quantity of phytoplankton in the sea and with no dogwhelks present. All the tiles had a central area which dogwhelks could not access. Fig. 1.2 shows one of the tiles used. central area covered with mesh to stop dogwhelks getting in plastic tile Fig. 1.2 The plastic tiles were examined after two months. The population densities of the barnacles growing in the central area which dogwhelks could not access were calculated. (i) The central area of one plastic tile had 12 barnacles settled in an area of 0.0225 m2. Calculate the population density of barnacles for this plastic tile as the number of barnacles per m2. Give your answer to three significant figures. … per m2 [2] Fig. 1.3 shows the population density of barnacles that had settled on the plastic tiles on all four rocky shores. Key dogwhelks present dogwhelks not present 800 700 600 500 number of barnacles 400 per square metre 300 200 100 0 low quantities of high quantities of phytoplankton phytoplankton rocky shore Fig. 1.3 (ii) Summarise the results of the investigation shown in Fig. 1.3. … … … … [2] (iii) The scientists concluded that there was a strong probability that the presence of dogwhelks affects the settlement of barnacles onto rocks. Discuss the extent to which the error bars for standard deviation shown in Fig. 1.3 support the scientists’ conclusion. … … … … [2] (iv) Suggest explanations for the effect of dogwhelks and quantities of phytoplankton on the settling of the barnacles. … … … … … … [3] (c) The acorn barnacle has spread all around the world on ships and is now considered to be an invasive species. (i) Explain why invasive species are a risk to ecosystems. … … … … … … [3] (ii) It has been suggested that introducing dogwhelks into new areas colonised by acorn barnacles could be a method of control. Suggest a possible negative consequence of introducing dogwhelks into areas with acorn barnacles. … … [1] [Total: 15]

15 marks

Mark scheme: Question Answer Marks 1(a) any 2 of: 2 allows dispersal to wider area / other areas / moves to different locations / AW ; to reduce competition / larvae have different niches from adults / AW ; allows (larvae to) feed on plankton / AW ; 1(b)(i) 533 (to 3 sig figs) ;; 2 1(b)(ii) more barnacles are found on the shores where there are more phytoplankton / ORA ; 2 more barnacles where there are no dogwhelks / ORA ; 1(b)(iii) any 2 of: 2 supported because no overlap (with and without dogwhelks) on shore with low quantities of phytoplankton / AW ; less support as there is an overlap (with and without dogwhelks) with larger quantities of phytoplankton / AW ; other factors may be affecting result / correlation not causal ; 1(b)(iv) any 3 of: 3 barnacles consume phytoplankton / more food for barnacles (if more phytoplankton) / ORA / AW ; so more survival of larvae ; dogwhelks reduces, attachment / settling, of barnacle larvae ; due to presence of, chemicals / scent, from dogwhelks / AW ; less reproduction of barnacles (to reduce population) (when dogwhelks present / less phytoplankton) / AW ; 1(c)(i) any 3 of: 3 lack predators so increase in population / AW ; overconsume prey / AW ; out compete native species / AW ; AVP ; 1(c)(ii) may become an invasive species itself / may have no predator / consume other native species / AW ; 1

This question in 9693/41 Oct/Nov 2024

4 Fig. 4.1 shows a female green sea turtle returning to the sea after laying eggs in a nest. Fig. 4.1 Conservationists are concerned that global warming could lead to a reduction in populations of green sea turtles. (a) Explain how an enhanced greenhouse effect leads to global warming. … … … … … … [3] (b) The sex of turtles is controlled by the temperature at which the eggs develop within the nest. Fig. 4.2 shows the effect of temperature on the percentage of hatched turtles that are female. 120 100 80 percentage of hatched 60 turtles that are female 40 20 0 27 28 29 30 31 32 temperature / °C Fig. 4.2 (i) Use Fig. 4.2 to predict the percentage of turtles that would be female when eggs develop at a temperature of 29.5 °C. … % [1] (ii) Green sea turtles nest on beaches in Suriname between June and November. During the breeding period in 2001 the temperature of the sand ranged between 29 °C and 31 °C. Some scientists are predicting that the mean global temperature will rise between 2 °C and 4 °C by the year 2100. Use Fig. 4.2 and your own knowledge to suggest and explain the impact of a global temperature rise on green sea turtle populations. … … … … … … [3] (c) Conservationists have been running captive breeding and release programmes to help conserve green sea turtles since 1990. Eggs are taken from nests and incubated in protected areas. After hatching, the young are released into the sea from the beach that the eggs were taken from. Female turtles return to the beach that they were released from and lay their eggs. The number of eggs on a beach was estimated every month over a period of five months in 1990 and 2017. The results are shown in Table 4.1. Table 4.1 estimated number of eggs month 1990 2017 June 530 790 July 650 890 August 750 990 September 690 980 October 610 800 These data were used to calculate the mean estimated numbers of eggs per month and standard deviations. The results are shown in Table 4.2. Table 4.2 year mean estimated number standard deviation of eggs per month 1990 646 83 2017 890 95 (i) Calculate the standard error for the mean estimated number of eggs per month on the beach in 2017. Use the formula: s standard error, S = M n s = standard deviation n = sample size (number of months that samples were collected) … [1] (ii) Use your answer to (c)(i) to calculate the 95% confidence interval for the mean estimated number of eggs per month on the beach in 2017. Use the formula: 95% confidence interval (95% CI) = x ± (2 × SM) x = mean SM = standard error 95% confidence interval: … to … [1] (iii) The 95% confidence interval for the mean estimated number of eggs per month on the beach in 1990 is 646 ± 74. Use your answer to (c)(ii) to assess whether the captive breeding and release of the turtles has successfully led to an increase in the mean estimated number of turtle eggs per month in 2017 compared with 1990. … … … … … … [3] [Total: 12]

12 marks

Mark scheme: 4(a) any 3 of: 3 (increased) carbon dioxide / methane in atmosphere ; short wavelength, light / radiation, passes through atmosphere (to ground) / AW ; longer wavelength light reflects back into atmosphere ; heat / radiation / (IR) light, trapped in atmosphere / AW ; 4(b)(i) 54–56 % ; 1 4(b)(ii) any 3 of: 3 1 increased proportion of females / more female turtles / fewer males ; 2 less breeding / less reproduction / AW ; 3 decreased populations ; 4 flooding of beaches / sea level rise, reducing nesting sites ; 5 food chains affected / less food / food matures at different times / AW ; 4(c)(i) 42.5 ; 1 4(c)(ii) 805 to 975 ; 1 4(c)(iii) any 3 of: 3 there is no overlap between the confidence intervals ; there is a significant difference between means ; so there is an increase in population ; probability of <0.05 that the difference is due to chance / AW ;

This question in 9693/41 Oct/Nov 2024

1 Fig. 1.1 shows some barnacles and a dogwhelk on a rock. Fig. 1.1 Adult barnacles are sessile and are often found attached to rocks on rocky shores. Barnacles have a complex life cycle with planktonic larvae. (a) Explain why a complex life cycle is an advantage for sessile organisms. … … … … [2] (b) The dogwhelk feeds on barnacles on rocky shores. Scientists investigated whether the settlement of barnacles on a rocky shore was affected by: • the presence of dogwhelks • the quantity of phytoplankton in the sea water next to the shore. The scientists placed 0.25 m2 plastic tiles onto four rocky shores: • one shore with a high quantity of phytoplankton in the sea and with dogwhelks present • one shore with a high quantity of phytoplankton in the sea and with no dogwhelks present • one shore with a low quantity of phytoplankton in the sea and with dogwhelks present • one shore with a low quantity of phytoplankton in the sea and with no dogwhelks present. All the tiles had a central area which dogwhelks could not access. Fig. 1.2 shows one of the tiles used. central area covered with mesh to stop dogwhelks getting in plastic tile Fig. 1.2 The plastic tiles were examined after two months. The population densities of the barnacles growing in the central area which dogwhelks could not access were calculated. (i) The central area of one plastic tile had 12 barnacles settled in an area of 0.0225 m2. Calculate the population density of barnacles for this plastic tile as the number of barnacles per m2. Give your answer to three significant figures. … per m2 [2] Fig. 1.3 shows the population density of barnacles that had settled on the plastic tiles on all four rocky shores. Key dogwhelks present dogwhelks not present 800 700 600 500 number of barnacles 400 per square metre 300 200 100 0 low quantities of high quantities of phytoplankton phytoplankton rocky shore Fig. 1.3 (ii) Summarise the results of the investigation shown in Fig. 1.3. … … … … [2] (iii) The scientists concluded that there was a strong probability that the presence of dogwhelks affects the settlement of barnacles onto rocks. Discuss the extent to which the error bars for standard deviation shown in Fig. 1.3 support the scientists’ conclusion. … … … … [2] (iv) Suggest explanations for the effect of dogwhelks and quantities of phytoplankton on the settling of the barnacles. … … … … … … [3] (c) The acorn barnacle has spread all around the world on ships and is now considered to be an invasive species. (i) Explain why invasive species are a risk to ecosystems. … … … … … … [3] (ii) It has been suggested that introducing dogwhelks into new areas colonised by acorn barnacles could be a method of control. Suggest a possible negative consequence of introducing dogwhelks into areas with acorn barnacles. … … [1] [Total: 15]

15 marks

Mark scheme: Question Answer Marks 1(a) any 2 of: 2 allows dispersal to wider area / other areas / moves to different locations / AW ; to reduce competition / larvae have different niches from adults / AW ; allows (larvae to) feed on plankton / AW ; 1(b)(i) 533 (to 3 sig figs) ;; 2 1(b)(ii) more barnacles are found on the shores where there are more phytoplankton / ORA ; 2 more barnacles where there are no dogwhelks / ORA ; 1(b)(iii) any 2 of: 2 supported because no overlap (with and without dogwhelks) on shore with low quantities of phytoplankton / AW ; less support as there is an overlap (with and without dogwhelks) with larger quantities of phytoplankton / AW ; other factors may be affecting result / correlation not causal ; 1(b)(iv) any 3 of: 3 barnacles consume phytoplankton / more food for barnacles (if more phytoplankton) / ORA / AW ; so more survival of larvae ; dogwhelks reduces, attachment / settling, of barnacle larvae ; due to presence of, chemicals / scent, from dogwhelks / AW ; less reproduction of barnacles (to reduce population) (when dogwhelks present / less phytoplankton) / AW ; 1(c)(i) any 3 of: 3 lack predators so increase in population / AW ; overconsume prey / AW ; out compete native species / AW ; AVP ; 1(c)(ii) may become an invasive species itself / may have no predator / consume other native species / AW ; 1

This question in 9693/42 Oct/Nov 2024

4 Fig. 4.1 shows a female green sea turtle returning to the sea after laying eggs in a nest. Fig. 4.1 Conservationists are concerned that global warming could lead to a reduction in populations of green sea turtles. (a) Explain how an enhanced greenhouse effect leads to global warming. … … … … … … [3] (b) The sex of turtles is controlled by the temperature at which the eggs develop within the nest. Fig. 4.2 shows the effect of temperature on the percentage of hatched turtles that are female. 120 100 80 percentage of hatched 60 turtles that are female 40 20 0 27 28 29 30 31 32 temperature / °C Fig. 4.2 (i) Use Fig. 4.2 to predict the percentage of turtles that would be female when eggs develop at a temperature of 29.5 °C. … % [1] (ii) Green sea turtles nest on beaches in Suriname between June and November. During the breeding period in 2001 the temperature of the sand ranged between 29 °C and 31 °C. Some scientists are predicting that the mean global temperature will rise between 2 °C and 4 °C by the year 2100. Use Fig. 4.2 and your own knowledge to suggest and explain the impact of a global temperature rise on green sea turtle populations. … … … … … … [3] (c) Conservationists have been running captive breeding and release programmes to help conserve green sea turtles since 1990. Eggs are taken from nests and incubated in protected areas. After hatching, the young are released into the sea from the beach that the eggs were taken from. Female turtles return to the beach that they were released from and lay their eggs. The number of eggs on a beach was estimated every month over a period of five months in 1990 and 2017. The results are shown in Table 4.1. Table 4.1 estimated number of eggs month 1990 2017 June 530 790 July 650 890 August 750 990 September 690 980 October 610 800 These data were used to calculate the mean estimated numbers of eggs per month and standard deviations. The results are shown in Table 4.2. Table 4.2 year mean estimated number standard deviation of eggs per month 1990 646 83 2017 890 95 (i) Calculate the standard error for the mean estimated number of eggs per month on the beach in 2017. Use the formula: s standard error, S = M n s = standard deviation n = sample size (number of months that samples were collected) … [1] (ii) Use your answer to (c)(i) to calculate the 95% confidence interval for the mean estimated number of eggs per month on the beach in 2017. Use the formula: 95% confidence interval (95% CI) = x ± (2 × SM) x = mean SM = standard error 95% confidence interval: … to … [1] (iii) The 95% confidence interval for the mean estimated number of eggs per month on the beach in 1990 is 646 ± 74. Use your answer to (c)(ii) to assess whether the captive breeding and release of the turtles has successfully led to an increase in the mean estimated number of turtle eggs per month in 2017 compared with 1990. … … … … … … [3] [Total: 12]

12 marks

Mark scheme: 4(a) any 3 of: 3 (increased) carbon dioxide / methane in atmosphere ; short wavelength, light / radiation, passes through atmosphere (to ground) / AW ; longer wavelength light reflects back into atmosphere ; heat / radiation / (IR) light, trapped in atmosphere / AW ; 4(b)(i) 54–56 % ; 1 4(b)(ii) any 3 of: 3 1 increased proportion of females / more female turtles / fewer males ; 2 less breeding / less reproduction / AW ; 3 decreased populations ; 4 flooding of beaches / sea level rise, reducing nesting sites ; 5 food chains affected / less food / food matures at different times / AW ; 4(c)(i) 42.5 ; 1 4(c)(ii) 805 to 975 ; 1 4(c)(iii) any 3 of: 3 there is no overlap between the confidence intervals ; there is a significant difference between means ; so there is an increase in population ; probability of <0.05 that the difference is due to chance / AW ;

This question in 9693/42 Oct/Nov 2024

1 Fig. 1.1 shows some barnacles and a dogwhelk on a rock. Fig. 1.1 Adult barnacles are sessile and are often found attached to rocks on rocky shores. Barnacles have a complex life cycle with planktonic larvae. (a) Explain why a complex life cycle is an advantage for sessile organisms. … … … … [2] (b) The dogwhelk feeds on barnacles on rocky shores. Scientists investigated whether the settlement of barnacles on a rocky shore was affected by: • the presence of dogwhelks • the quantity of phytoplankton in the sea water next to the shore. The scientists placed 0.25 m2 plastic tiles onto four rocky shores: • one shore with a high quantity of phytoplankton in the sea and with dogwhelks present • one shore with a high quantity of phytoplankton in the sea and with no dogwhelks present • one shore with a low quantity of phytoplankton in the sea and with dogwhelks present • one shore with a low quantity of phytoplankton in the sea and with no dogwhelks present. All the tiles had a central area which dogwhelks could not access. Fig. 1.2 shows one of the tiles used. central area covered with mesh to stop dogwhelks getting in plastic tile Fig. 1.2 The plastic tiles were examined after two months. The population densities of the barnacles growing in the central area which dogwhelks could not access were calculated. (i) The central area of one plastic tile had 12 barnacles settled in an area of 0.0225 m2. Calculate the population density of barnacles for this plastic tile as the number of barnacles per m2. Give your answer to three significant figures. … per m2 [2] Fig. 1.3 shows the population density of barnacles that had settled on the plastic tiles on all four rocky shores. Key dogwhelks present dogwhelks not present 800 700 600 500 number of barnacles 400 per square metre 300 200 100 0 low quantities of high quantities of phytoplankton phytoplankton rocky shore Fig. 1.3 (ii) Summarise the results of the investigation shown in Fig. 1.3. … … … … [2] (iii) The scientists concluded that there was a strong probability that the presence of dogwhelks affects the settlement of barnacles onto rocks. Discuss the extent to which the error bars for standard deviation shown in Fig. 1.3 support the scientists’ conclusion. … … … … [2] (iv) Suggest explanations for the effect of dogwhelks and quantities of phytoplankton on the settling of the barnacles. … … … … … … [3] (c) The acorn barnacle has spread all around the world on ships and is now considered to be an invasive species. (i) Explain why invasive species are a risk to ecosystems. … … … … … … [3] (ii) It has been suggested that introducing dogwhelks into new areas colonised by acorn barnacles could be a method of control. Suggest a possible negative consequence of introducing dogwhelks into areas with acorn barnacles. … … [1] [Total: 15]

15 marks

Mark scheme: Question Answer Marks 1(a) any 2 of: 2 allows dispersal to wider area / other areas / moves to different locations / AW ; to reduce competition / larvae have different niches from adults / AW ; allows (larvae to) feed on plankton / AW ; 1(b)(i) 533 (to 3 sig figs) ;; 2 1(b)(ii) more barnacles are found on the shores where there are more phytoplankton / ORA ; 2 more barnacles where there are no dogwhelks / ORA ; 1(b)(iii) any 2 of: 2 supported because no overlap (with and without dogwhelks) on shore with low quantities of phytoplankton / AW ; less support as there is an overlap (with and without dogwhelks) with larger quantities of phytoplankton / AW ; other factors may be affecting result / correlation not causal ; 1(b)(iv) any 3 of: 3 barnacles consume phytoplankton / more food for barnacles (if more phytoplankton) / ORA / AW ; so more survival of larvae ; dogwhelks reduces, attachment / settling, of barnacle larvae ; due to presence of, chemicals / scent, from dogwhelks / AW ; less reproduction of barnacles (to reduce population) (when dogwhelks present / less phytoplankton) / AW ; 1(c)(i) any 3 of: 3 lack predators so increase in population / AW ; overconsume prey / AW ; out compete native species / AW ; AVP ; 1(c)(ii) may become an invasive species itself / may have no predator / consume other native species / AW ; 1

This question in 9693/43 Oct/Nov 2024

4 Fig. 4.1 shows a female green sea turtle returning to the sea after laying eggs in a nest. Fig. 4.1 Conservationists are concerned that global warming could lead to a reduction in populations of green sea turtles. (a) Explain how an enhanced greenhouse effect leads to global warming. … … … … … … [3] (b) The sex of turtles is controlled by the temperature at which the eggs develop within the nest. Fig. 4.2 shows the effect of temperature on the percentage of hatched turtles that are female. 120 100 80 percentage of hatched 60 turtles that are female 40 20 0 27 28 29 30 31 32 temperature / °C Fig. 4.2 (i) Use Fig. 4.2 to predict the percentage of turtles that would be female when eggs develop at a temperature of 29.5 °C. … % [1] (ii) Green sea turtles nest on beaches in Suriname between June and November. During the breeding period in 2001 the temperature of the sand ranged between 29 °C and 31 °C. Some scientists are predicting that the mean global temperature will rise between 2 °C and 4 °C by the year 2100. Use Fig. 4.2 and your own knowledge to suggest and explain the impact of a global temperature rise on green sea turtle populations. … … … … … … [3] (c) Conservationists have been running captive breeding and release programmes to help conserve green sea turtles since 1990. Eggs are taken from nests and incubated in protected areas. After hatching, the young are released into the sea from the beach that the eggs were taken from. Female turtles return to the beach that they were released from and lay their eggs. The number of eggs on a beach was estimated every month over a period of five months in 1990 and 2017. The results are shown in Table 4.1. Table 4.1 estimated number of eggs month 1990 2017 June 530 790 July 650 890 August 750 990 September 690 980 October 610 800 These data were used to calculate the mean estimated numbers of eggs per month and standard deviations. The results are shown in Table 4.2. Table 4.2 year mean estimated number standard deviation of eggs per month 1990 646 83 2017 890 95 (i) Calculate the standard error for the mean estimated number of eggs per month on the beach in 2017. Use the formula: s standard error, S = M n s = standard deviation n = sample size (number of months that samples were collected) … [1] (ii) Use your answer to (c)(i) to calculate the 95% confidence interval for the mean estimated number of eggs per month on the beach in 2017. Use the formula: 95% confidence interval (95% CI) = x ± (2 × SM) x = mean SM = standard error 95% confidence interval: … to … [1] (iii) The 95% confidence interval for the mean estimated number of eggs per month on the beach in 1990 is 646 ± 74. Use your answer to (c)(ii) to assess whether the captive breeding and release of the turtles has successfully led to an increase in the mean estimated number of turtle eggs per month in 2017 compared with 1990. … … … … … … [3] [Total: 12]

12 marks

Mark scheme: 4(a) any 3 of: 3 (increased) carbon dioxide / methane in atmosphere ; short wavelength, light / radiation, passes through atmosphere (to ground) / AW ; longer wavelength light reflects back into atmosphere ; heat / radiation / (IR) light, trapped in atmosphere / AW ; 4(b)(i) 54–56 % ; 1 4(b)(ii) any 3 of: 3 1 increased proportion of females / more female turtles / fewer males ; 2 less breeding / less reproduction / AW ; 3 decreased populations ; 4 flooding of beaches / sea level rise, reducing nesting sites ; 5 food chains affected / less food / food matures at different times / AW ; 4(c)(i) 42.5 ; 1 4(c)(ii) 805 to 975 ; 1 4(c)(iii) any 3 of: 3 there is no overlap between the confidence intervals ; there is a significant difference between means ; so there is an increase in population ; probability of <0.05 that the difference is due to chance / AW ;

This question in 9693/43 Oct/Nov 2024

3 Oysters are shelled molluscs that have a complex life cycle. The adults are sessile organisms that anchor to a substrate. Fig. 3.1 shows some adult oysters. Fig. 3.1 (a) Outline the importance of having a complex life cycle for organisms such as oysters. … … … … … … [3] (b) Scientists investigated the effects of temperature and pH on the survival of oyster larvae. Oyster larvae were placed into tanks of water at different temperatures and pHs. The percentages of larvae surviving were calculated after two days and then again after 15 days. Fig. 3.2 and Fig. 3.3 show the results. 100 Key 90 day 2 day 15 80 70 60 percentage of 50 larvae surviving 40 30 20 10 0 20 25 27 30 35 temperature / °C Fig. 3.2 100 Key 90 day 2 day 15 80 70 60 percentage of 50 larvae surviving 40 30 20 10 0 6.5 7.0 7.5 8.0 8.2 8.5 pH Fig. 3.3 (i) The scientists placed 500 larvae in each condition. Calculate the number of larvae that did not survive from day 2 to day 15 when placed at a temperature of 30 °C. … [2] (ii) Use Fig. 3.2 and Fig. 3.3 to state the optimum temperature and optimum pH for the survival of oyster larvae. optimum temperature … °C optimum pH … [1] (c) Fishers in India reported declining harvests of adult oysters from an area of coastal water after 2011. (i) Table 3.1 shows the temperature and pH of the water during 2009 in this area. Table 3.1 month temperature / °C pH February 25 8.0 April 27 7.9 June 34 8.2 August 30 7.0 October 32 6.5 December 28 7.0 Use Table 3.1 to plot a line graph to show the temperature and the pH from February to December. [5] (ii) Around the coasts of India, oysters spawn throughout the year but have two peak periods of breeding, in April and August. Discuss the reasons for the reduction in the number of oysters that the fishers harvested after 2011. Use information in Table 3.1, Fig. 3.2 and Fig. 3.3 to support your answer. … … … … … … … … [4] (d) Explain why the use of fossil fuels places future oyster populations at risk. … … … … … … [3] [Total: 18]

18 marks

Mark scheme: 3(a) any 3 from: 3 1 larvae can move to other areas / larvae allow distribution to other areas / AW ; 2 reduced competition / AW ; 3 (reduced competition for) food / nutrients / AW ; 4 idea that larvae and adults occupy different niches / AW ; 5 increased genetic diversity (if oysters spread to other areas) / AW ; 6 reduces spread of disease (as population density is lower) ; 3(b)(i) 180 = 2 marks 2 36 (%) = 1 mark OR 320 survived = 1 mark 3(b)(ii) (optimum temperature) = 27 (°C) 1 optimum pH = 8(.0) ; 3(c)(i) 1. linear y axes for both temperature and pH, labelled with units, and horizontal axis as month ; 5 2. all three scales enable plots to cover at least half grid ; 3. plots correct +/- ½ square for temperature ; 4. plots correct +/- ½ square for pH ; 5. points joined with straight lines and key ; 3(c)(ii) any 4 from: 4 1 in April, conditions enable survival / there are optimal conditions / oysters can breed / conditions are ideal for breeding / AW ; 2 August has a temperature of 30 oC and pH of 7(.0) / April has a temperature of 27 oC and pH of 7.9 / AW ; 3 in August, conditions reduce larvae settling / kill larvae / few larvae survive / will not settle in August / too acidic for survival / AW ; 4 there is only one successful breeding season (per year) / AW ; 5 idea that few oyster larvae become adults / it takes time to produce adults (so effects are only seen in 2011) / AW ; 6 few other months have ideal conditions for larvae / AW ; 7 overfishing / pollution / AW, may be causing the fall ; 3(d) any 3 from: 3 1 release of carbon dioxide / AW ; 2 causes (enhanced) greenhouse effect / increased temperature / global warming / AW ; 3 acidification of water / AW ; 4 (acid) reduces oyster shell formation / erodes shells / dissolves shells / AW ; 5 larvae do not survive / larvae cannot settle / fewer adult oysters to breed (in future) / AW ;

This question in 9693/41 May/June 2025

4 A student investigated if the surface area of fish gills has a correlation with the activity levels of the fish. (a) The student used a Spearman’s rank correlation to test if there was a significant correlation. Table 4.1 shows the activity level, gill surface area and the ranking for different species of fish. Table 4.1 activity rank gill surface rank gill level species activity area surface D D 2 (10 = highly level / cm2 g–1 area active) butterfish 5 7 461 6 1 1 fluke 2 11 247 11 0 0 mackerel 9 2 1040 2 0 0 menhaden 10 1 1241 1 0 0 mullet 8 3 1010 3 0 0 puffer 4 423 9 scup 6 5 498 4 1 1 sea robin 5 7 432 8 –1 1 sea trout 5 7 275 10 –3 9 sheepshead 7 4 467 5 –1 1 tautog 4 450 7 toadfish 1 12 151 12 0 0 ΣD 2 = Σ = sum of (total) D = difference in rank between each pair of measurements (i) Complete Table 4.1 for the puffer and tautog and calculate the value of ΣD 2. Write your answers in Table 4.1. [2] (ii) Calculate the Spearman’s rank correlation coefficient using the formula: 6 # R D 2 r = 1 - c m s n 3 - n where, rS = Spearman’s rank correlation coefficient Σ = sum of (total) D = difference in rank between each pair of measurements n = number of pairs of items in the sample. Show your working. rS = … [2] (iii) Table 4.2 shows the critical values for the Spearman’s rank correlation coefficient. Table 4.2 rS number of pairs, n P = 0.05 5 1.000 6 0.886 7 0.786 8 0.738 9 0.700 10 0.648 11 0.618 12 0.587 13 0.560 The student made the following null hypothesis for the data in Table 4.1. ‘There is no correlation between the activity level of the fish and the gill surface area.’ Use Table 4.2 and your answer to 4(a)(ii) to determine whether the student’s null hypothesis can be accepted or rejected. … … … … … … [3] (b) Explain why fish with different activity levels require different gill surface areas. … … … … … … [3] [Total: 10]

10 marks

Mark scheme: 4(a)(i) 2 species activity level rank activity gill surface rank gill D D2 (10 = highly level area / cm2 g–1 surface area active) butterfish 5 7 461 6 1 1 fluke 2 11 247 11 0 0 mackerel 9 2 1040 2 0 0 menhaden 10 1 1241 1 0 0 mullet 8 3 1010 3 0 0 puffer 4 9.5 423 9 0.5 0.25 scup 6 5 498 4 1 1 sea Robin 5 7 432 8 –1 1 sea Trout 5 7 275 10 –3 9 sheepshead 7 4 467 5 –1 1 tautog 4 9.5 450 7 2.5 6.25 ; toadfish 1 12 151 12 0 0 19.5 ; 4(a)(ii) 0.93(1818181818181818….) (2 marks) ;; 2 one mark for 1716 OR (123 – 12) OR 12(122 – 1) OR 12(144 – 1) OR 12(143) OR (1728 – 12) OR 0.068(….) in working 4(a)(iii) any 3 from: 3 1 the calculated value is greater than the critical value ; 2 of 0.587 ; 3 so the null hypothesis is rejected ; 4 there is a significant positive correlation ; 5 probability of less than 0.05, that the correlation is due to chance ; NOTE: If no calculated value, then only mp2 awarded for recognition of 0.587 4(b) any 3 from: 3 1 more active fish require a larger (gill) surface area / ORA / AW ; 2 for fast diffusion of oxygen (into blood) / AW ; 3 for fast diffusion of carbon dioxide (out) / AW ; 4 (more) (aerobic) respiration ; 5 produce ATP / release energy, for muscle contraction ;

This question in 9693/41 May/June 2025

2 Lionfish are an invasive species around coral reefs in many parts of the West Atlantic Ocean and Caribbean Sea. Fig. 2.1 shows a photograph of a lionfish. Fig. 2.1 In an area of Mexico, the government developed a conservation project to reduce the impact of lionfish and provide a sustainable lionfish industry for local people. As part of this conservation project, a fishery was set up to encourage local people to catch lionfish. Lionfish were sold at market and became a major source of income for local people. (a) The lionfish fishery was set up in 2011. Net fishing for the lionfish was banned. Spearfishing was the only permitted method. Table 2.1 shows the catches of lionfish from 2011 to 2017. Table 2.1 year number of lionfish caught 2011 1017 2012 2515 2013 17 250 2014 23 121 2015 12 022 2016 6032 2017 1035 (i) Suggest two reasons why spearfishing was the only method allowed to catch the lionfish. 1 … … 2 … … [2] (ii) Calculate the percentage increase of catch from 2011 to 2014. Give your answer to two significant figures. Space for working. … % [3] (b) In 2013, 2014 and 2015 the population densities of lionfish in three reef areas of the fishery were estimated. (i) Outline one method that could be used to estimate the population density of lionfish on a reef. … … … … [2] (ii) Fig. 2.2 shows the mean population densities of lionfish in the three reef areas during 2013, 2014, and 2015. 350 Key reef area 1 300 reef area 2 250 reef area 3 mean population 200 density of lionfish / number m–2 150 100 50 0 2013 2014 2015 year Fig. 2.2 Explain the effects of the conservation project on the changes in population of lionfish in the reef areas. … … … … … … [3] (c) Table 2.2 shows the mean price of lionfish at a market in Mexico from 2011 to 2017. Table 2.2 year mean price per kilogram / USD ($) 2011 3.8 2012 3.8 2013 2.5 2014 2.4 2015 2.7 2016 4.2 2017 6.8 Evaluate the success of the lionfish fishery as a method of controlling the lionfish population and providing an industry for local people. Use Table 2.1, Fig. 2.2 and Table 2.2 to support your answer. … … … … … … … … [4] [Total: 14]

14 marks

Mark scheme: 2(a)(i) any 2 of: 2 1 prevents bycatch / AW ; 2 no loss of tackle on reefs / less damaging to reefs / habitats / seabed / AW ; 3 prevents over-fishing / AW ; 4 less chance of catching juveniles / immature fish / AW ; 2(a)(ii) 2200 (%) ;;; 3 Two marks for: 2173.45…. One mark for:  1017 OR 23 121 – 1017 OR 22014 2(b)(i) 1 practical method of counting lionfish ; 2 2 divide number by area ; 2(b)(ii) any 3 of: 3 1 populations fall in all reefs / AW ; 2 reef area 1 and 2 increase after 2014 / reef area 3 decreases less steeply after 2014 / AW ; 3 less breeding / reproduction / recruitment / loss of breeding stock / AW ; 4 financial reward encourages, high rate of capture / AW ; 5 fishing intensity may reduce after 2014 / AW ; 2(c) any 4 of: 4 1 lionfish have been successfully controlled / lionfish population has lowered / AW ; 2 total catch increases until 2014 then decreases / highest catch in 2014 / AW ; 3 (fewer lionfish means) reduced damage to native species / less damage to habitats / AW ; 4 (however) there is a slight rise in population in 2014 (in two reef areas) (suggesting they are recovering) / AW ; 5 price of lionfish decreases until 2014 / price of lionfish rises after 2014 / lowest prices between 2013 and 2014 / AW ; 6 low / falling, price if many fish caught / AW ; 7 fishing industry is not long-term / AW ; 8 (loss of fishery) could result in loss of income / employment / sociological issues / poverty / AW ; 9 AVP ;

This question in 9693/42 May/June 2025

2 Lionfish are an invasive species around coral reefs in many parts of the West Atlantic Ocean and Caribbean Sea. Fig. 2.1 shows a photograph of a lionfish. Fig. 2.1 In an area of Mexico, the government developed a conservation project to reduce the impact of lionfish and provide a sustainable lionfish industry for local people. As part of this conservation project, a fishery was set up to encourage local people to catch lionfish. Lionfish were sold at market and became a major source of income for local people. (a) The lionfish fishery was set up in 2011. Net fishing for the lionfish was banned. Spearfishing was the only permitted method. Table 2.1 shows the catches of lionfish from 2011 to 2017. Table 2.1 year number of lionfish caught 2011 1017 2012 2515 2013 17 250 2014 23 121 2015 12 022 2016 6032 2017 1035 (i) Suggest two reasons why spearfishing was the only method allowed to catch the lionfish. 1 … … 2 … … [2] (ii) Calculate the percentage increase of catch from 2011 to 2014. Give your answer to two significant figures. Space for working. … % [3] (b) In 2013, 2014 and 2015 the population densities of lionfish in three reef areas of the fishery were estimated. (i) Outline one method that could be used to estimate the population density of lionfish on a reef. … … … … [2] (ii) Fig. 2.2 shows the mean population densities of lionfish in the three reef areas during 2013, 2014, and 2015. 350 Key reef area 1 300 reef area 2 250 reef area 3 mean population 200 density of lionfish / number m–2 150 100 50 0 2013 2014 2015 year Fig. 2.2 Explain the effects of the conservation project on the changes in population of lionfish in the reef areas. … … … … … … [3] (c) Table 2.2 shows the mean price of lionfish at a market in Mexico from 2011 to 2017. Table 2.2 year mean price per kilogram / USD ($) 2011 3.8 2012 3.8 2013 2.5 2014 2.4 2015 2.7 2016 4.2 2017 6.8 Evaluate the success of the lionfish fishery as a method of controlling the lionfish population and providing an industry for local people. Use Table 2.1, Fig. 2.2 and Table 2.2 to support your answer. … … … … … … … … [4] [Total: 14]

14 marks

Mark scheme: 2(a)(i) any 2 of: 2 1 prevents bycatch / AW ; 2 no loss of tackle on reefs / less damaging to reefs / habitats / seabed / AW ; 3 prevents over-fishing / AW ; 4 less chance of catching juveniles / immature fish / AW ; 2(a)(ii) 2200 (%) ;;; 3 Two marks for: 2173.45…. One mark for:  1017 OR 23 121 – 1017 OR 22014 2(b)(i) 1 practical method of counting lionfish ; 2 2 divide number by area ; 2(b)(ii) any 3 of: 3 1 populations fall in all reefs / AW ; 2 reef area 1 and 2 increase after 2014 / reef area 3 decreases less steeply after 2014 / AW ; 3 less breeding / reproduction / recruitment / loss of breeding stock / AW ; 4 financial reward encourages, high rate of capture / AW ; 5 fishing intensity may reduce after 2014 / AW ; 2(c) any 4 of: 4 1 lionfish have been successfully controlled / lionfish population has lowered / AW ; 2 total catch increases until 2014 then decreases / highest catch in 2014 / AW ; 3 (fewer lionfish means) reduced damage to native species / less damage to habitats / AW ; 4 (however) there is a slight rise in population in 2014 (in two reef areas) (suggesting they are recovering) / AW ; 5 price of lionfish decreases until 2014 / price of lionfish rises after 2014 / lowest prices between 2013 and 2014 / AW ; 6 low / falling, price if many fish caught / AW ; 7 fishing industry is not long-term / AW ; 8 (loss of fishery) could result in loss of income / employment / sociological issues / poverty / AW ; 9 AVP ;

This question in 9693/43 May/June 2025

5 Pacific salmon are euryhaline fish that spend part of their life cycle in the sea and part of their life cycle in rivers. (a) State what is meant by the term euryhaline. … … [1] (b) Scientists investigated the effect of placing salmon into water of different salinities. Salmon were first kept in water with a salinity of 35 ppt for two weeks. Five of these salmon were selected and each one was placed into a separate tank of water with a salinity of 15 ppt. The decrease in oxygen concentration of the water in each of the tanks was determined over a period of one hour. This was repeated twice. In the first repeat, five different salmon were selected and each one placed into separate tanks with a salinity of 35 ppt. In the second repeat, five different salmon were selected and each one placed into separate tanks with a salinity of 47 ppt. Table 5.1 shows the results. Table 5.1 salinity number of mean decrease standard standard 2 × / ppt salmon (n) in oxygen deviation error standard concentration / mg dm–3 kg–1 error / mg dm–3 kg–1 15 5 2.3 0.50 0.22 0.44 35 5 2.1 0.20 0.09 0.18 47 5 3.8 0.40 (i) Calculate the standard error for the mean decrease in oxygen concentration of the water with a salinity of 47 ppt. Use the equation: s standard error, SE = n s = standard deviation n = sample size (number of observations) … [1] (ii) Use your answer to (b)(i) to calculate the 95% confidence interval for the mean decrease in oxygen concentration of the water with a salinity of 47 ppt. Use the equation: 95% confidence interval (95% CI) = x ± (2 × SE) x = mean SE = standard error … to … [1] (iii) Assess whether placing the salmon into water with salinities of 15 ppt and 47 ppt resulted in significant differences in the mean decrease in oxygen concentration of the water compared with the results at 35 ppt. Use the information in Table 5.1 and your answer to (b)(ii) to support your answer. … … … … … … [3] (iv) Explain why placing the salmon into the water with a salinity of 47 ppt caused a decrease in the oxygen concentration. … … … … … … [3] [Total: 9]

9 marks

Mark scheme: 5(a) can live in a range of salinities / can tolerate a wide range of salinity / AW ; 1 5(b)(i) 0.18 ; 1 5(b)(ii) 3.44 to 4.16 ; 1 5(b)(iii) there is a significant, decrease / difference, in oxygen when in 47 ppt ; 3 there is not a significant, decrease / difference, in oxygen when in 15 ppt ; because the ranges for 47 ppt do not overlap with 35 ppt / because the ranges when in 15 ppt the ranges do overlap with 35 ppt ; 5(b)(iv) any 3 of: 3 1 water potential of, body fluids / AW, is higher than solution ; 2 so water loss occurs / AW ; 3 osmoregulation occurs ; 4 active transport of salt / ions / AW ; 5 (more) respiration (using oxygen) ; 6 correct reference to ATP use ;

This question in 9693/41 Oct/Nov 2025

5 Pacific salmon are euryhaline fish that spend part of their life cycle in the sea and part of their life cycle in rivers. (a) State what is meant by the term euryhaline. … … [1] (b) Scientists investigated the effect of placing salmon into water of different salinities. Salmon were first kept in water with a salinity of 35 ppt for two weeks. Five of these salmon were selected and each one was placed into a separate tank of water with a salinity of 15 ppt. The decrease in oxygen concentration of the water in each of the tanks was determined over a period of one hour. This was repeated twice. In the first repeat, five different salmon were selected and each one placed into separate tanks with a salinity of 35 ppt. In the second repeat, five different salmon were selected and each one placed into separate tanks with a salinity of 47 ppt. Table 5.1 shows the results. Table 5.1 salinity number of mean decrease standard standard 2 × / ppt salmon (n) in oxygen deviation error standard concentration / mg dm–3 kg–1 error / mg dm–3 kg–1 15 5 2.3 0.50 0.22 0.44 35 5 2.1 0.20 0.09 0.18 47 5 3.8 0.40 (i) Calculate the standard error for the mean decrease in oxygen concentration of the water with a salinity of 47 ppt. Use the equation: s standard error, SE = n s = standard deviation n = sample size (number of observations) … [1] (ii) Use your answer to (b)(i) to calculate the 95% confidence interval for the mean decrease in oxygen concentration of the water with a salinity of 47 ppt. Use the equation: 95% confidence interval (95% CI) = x ± (2 × SE) x = mean SE = standard error … to … [1] (iii) Assess whether placing the salmon into water with salinities of 15 ppt and 47 ppt resulted in significant differences in the mean decrease in oxygen concentration of the water compared with the results at 35 ppt. Use the information in Table 5.1 and your answer to (b)(ii) to support your answer. … … … … … … [3] (iv) Explain why placing the salmon into the water with a salinity of 47 ppt caused a decrease in the oxygen concentration. … … … … … … [3] [Total: 9]

9 marks

Mark scheme: 5(a) can live in a range of salinities / can tolerate a wide range of salinity / AW ; 1 5(b)(i) 0.18 ; 1 5(b)(ii) 3.44 to 4.16 ; 1 5(b)(iii) there is a significant, decrease / difference, in oxygen when in 47 ppt ; 3 there is not a significant, decrease / difference, in oxygen when in 15 ppt ; because the ranges for 47 ppt do not overlap with 35 ppt / because the ranges when in 15 ppt the ranges do overlap with 35 ppt ; 5(b)(iv) any 3 of: 3 1 water potential of, body fluids / AW, is higher than solution ; 2 so water loss occurs / AW ; 3 osmoregulation occurs ; 4 active transport of salt / ions / AW ; 5 (more) respiration (using oxygen) ; 6 correct reference to ATP use ;

This question in 9693/42 Oct/Nov 2025

5 Pacific salmon are euryhaline fish that spend part of their life cycle in the sea and part of their life cycle in rivers. (a) State what is meant by the term euryhaline. … … [1] (b) Scientists investigated the effect of placing salmon into water of different salinities. Salmon were first kept in water with a salinity of 35 ppt for two weeks. Five of these salmon were selected and each one was placed into a separate tank of water with a salinity of 15 ppt. The decrease in oxygen concentration of the water in each of the tanks was determined over a period of one hour. This was repeated twice. In the first repeat, five different salmon were selected and each one placed into separate tanks with a salinity of 35 ppt. In the second repeat, five different salmon were selected and each one placed into separate tanks with a salinity of 47 ppt. Table 5.1 shows the results. Table 5.1 salinity number of mean decrease standard standard 2 × / ppt salmon (n) in oxygen deviation error standard concentration / mg dm–3 kg–1 error / mg dm–3 kg–1 15 5 2.3 0.50 0.22 0.44 35 5 2.1 0.20 0.09 0.18 47 5 3.8 0.40 (i) Calculate the standard error for the mean decrease in oxygen concentration of the water with a salinity of 47 ppt. Use the equation: s standard error, SE = n s = standard deviation n = sample size (number of observations) … [1] (ii) Use your answer to (b)(i) to calculate the 95% confidence interval for the mean decrease in oxygen concentration of the water with a salinity of 47 ppt. Use the equation: 95% confidence interval (95% CI) = x ± (2 × SE) x = mean SE = standard error … to … [1] (iii) Assess whether placing the salmon into water with salinities of 15 ppt and 47 ppt resulted in significant differences in the mean decrease in oxygen concentration of the water compared with the results at 35 ppt. Use the information in Table 5.1 and your answer to (b)(ii) to support your answer. … … … … … … [3] (iv) Explain why placing the salmon into the water with a salinity of 47 ppt caused a decrease in the oxygen concentration. … … … … … … [3] [Total: 9]

9 marks

Mark scheme: 5(a) can live in a range of salinities / can tolerate a wide range of salinity / AW ; 1 5(b)(i) 0.18 ; 1 5(b)(ii) 3.44 to 4.16 ; 1 5(b)(iii) there is a significant, decrease / difference, in oxygen when in 47 ppt ; 3 there is not a significant, decrease / difference, in oxygen when in 15 ppt ; because the ranges for 47 ppt do not overlap with 35 ppt / because the ranges when in 15 ppt the ranges do overlap with 35 ppt ; 5(b)(iv) any 3 of: 3 1 water potential of, body fluids / AW, is higher than solution ; 2 so water loss occurs / AW ; 3 osmoregulation occurs ; 4 active transport of salt / ions / AW ; 5 (more) respiration (using oxygen) ; 6 correct reference to ATP use ;

This question in 9693/43 Oct/Nov 2025