SPEC/4/BIOLO/HP2/ENG/TZ0/XX –2– Section a Answer all questions. Write your answers in the boxes provided. 1. GPR120 is a trans-membrane protein that functions as a receptor for long-chain unsaturated fatty acids. When fatty acids outside the cell bind to the receptor, the receptor changes shape and as a result a signal is passed to the interior of the cell. This causes an increase in the intracellular calcium concentration (Ca2+ ), which has a wide range of effects on cell activity. A strain of mice was developed that did not produce GPR120 protein. Groups of these GPR120 deficient mice were fed from the age of 5 weeks until they were 16 weeks old either on a high fat diet containing 60 % fat or on a normal diet containing 13 % fat. Control groups of mice that did produce GPR120 protein were fed on the same diets. The bar chart shows the mean body mass of each group when the mice were 16 weeks old. 50 40 Mean body mass / gram 30 20 10 0 GPR deficient Control normal diet GPR deficient Control high fat diet [Source: adapted from A Ichimur, et al., (2012), Nature, 483, pages 350–354, Reprinted by permission from Macmillan Publishers Ltd] (This question continues on the following page) 20EP02 –3– SPEC/4/BIOLO/HP2/ENG/TZ0/XX (Question 1 continued) (a) Compare and contrast the body mass of GPR120 deficient mice and the control mice on a normal diet and a high fat diet. [3] ....................................................................... ....................................................................... ....................................................................... ....................................................................... ....................................................................... ....................................................................... (This question continues on the following page) Turn over 20EP03 SPEC/4/BIOLO/HP2/ENG/TZ0/XX –4– (Question 1 continued) The gene for GPR120 protein is located on Chromosome 10. The alleles for GPR120 protein from 312 extremely obese adults and children were base sequenced. Six alleles were discovered that differed by one base from the wild-type allele. They are likely to have been produced by a base substitution mutation. Two of the mutations cause a change in the amino acid sequence of the GPR120 protein (missense mutation) but the other four do not (synonymous mutation). The table gives details of the six mutant alleles. Mutant allele nucleotide change chromosome 10 position type of mutation R67C C→T 95 316 666 Missense R270H G →A 95 337 031 Missense V38V G →A 95 316 581 Synonymous S192S G →A 95 325 846 Synonymous V243V C→T 95 328 938 Synonymous S264S G →A 95 337 014 Synonymous [Source: adapted from A Ichimur, et al., (2012), Nature, 483, pages 350–354, Reprinted by permission from Macmillan Publishers Ltd] (b) Explain how only some of the base substitution mutations cause a change in the amino acid sequence of the gene for GPR120. [2] ....................................................................... ....................................................................... ....................................................................... ....................................................................... (c) Using the positions on chromosome 10, identify which two mutations were closest together in the gene for GPR120. 1. .................................................................. 2. .................................................................. [1] (This question continues on the following page) 20EP04 SPEC/4/BIOLO/HP2/ENG/TZ0/XX –5– (Question 1 continued) The frequency of the two alleles that caused a change in amino acid sequence was measured in 6942 unrelated obese humans and 7654 controls. The results are shown in the table. Allele Allele frequency / % obese humans controls R67C 5.5 4.3 R270H 2.4 1.3 [Source: adapted from A Ichimur, et al., (2012), Nature, 483, pages 350–354, Reprinted by permission from Macmillan Publishers Ltd] (d) Outline the reasons for using large numbers of obese humans and controls in this research. [2] ....................................................................... ....................................................................... ....................................................................... ....................................................................... (e) The association of one of the alleles with obesity was statistically significant. Deduce, with a reason, which allele this was. [1] ....................................................................... ....................................................................... (This question continues on the following page) Turn over 20EP05 –6– SPEC/4/BIOLO/HP2/ENG/TZ0/XX (Question 1 continued) A clone of human cells that contained the wild type allele for GPR120 was genetically modified by inserting either another wild-type allele (WT) or one of two mutant alleles (R67C and R270H). The intracellular concentration of calcium was measured in these cells at varying levels of linolenic acid. Linolenic acid is a long-chain unsaturated fatty acid. The graph shows the results. The scale on the x-axis is logarithmic. 3 Key: WT/WT WT/R67C WT/R270H 2.5 2 Intracellular calcium 1.5 concentration / arbitrary units 1 0.5 0 1 (f) 10 100 Linolenic acid concentration / mol dm–3 Suggest advantages of genetically modifying a clone of human cells for use in this experiment rather than using cells from obese people naturally containing the mutant alleles. [2] ....................................................................... ....................................................................... ....................................................................... ....................................................................... (This question continues on the following page) 20EP06 –7– SPEC/4/BIOLO/HP2/ENG/TZ0/XX (Question 1 continued) (g) Outline the effect of linolenic acid concentration on intracellular calcium concentration in the WT/WT cells. [2] ....................................................................... ....................................................................... ....................................................................... ....................................................................... (h) Compare and contrast the effect of the two mutant alleles on intracellular calcium concentration. [2] ....................................................................... ....................................................................... ....................................................................... ....................................................................... Biologists have often debated whether characteristics in humans are due to genes, the environment, or a combination of both. (i) Discuss the evidence provided by the data for the relative role of genes and diet in causing obesity. [3] ....................................................................... ....................................................................... ....................................................................... ....................................................................... ....................................................................... ....................................................................... Turn over 20EP07 –5– SPEC/4/BIOLO/HP2/ENG/TZ0/XX/M SECTION A Question 1. a Marking point a Answers Notes no difference on normal diet ‹between control and GPR120 deficient› b b c both higher on a high fat diet than a normal diet GPR120 deficient higher than control on a high fat diet a b c base substitution changes a codon amino acids are coded for by different codons several codons can code for the same amino acid 1. 95 337 031 AND 2. 95 337 014 c d a b c d e e Total 3 2 Both needed 1 increase reliability identify anomalous results some allele frequencies are very low because there is much genetic variation among obese people OR different causes of obesity to allow statistical testing of results 2 R27OH because of larger percentage difference between obese and control 1 –6– SPEC/4/BIOLO/HP2/ENG/TZ0/XX/M (Question 1 continued) Question f g h Marking point Answers a b c d control variables cells from obese people will have lots of differences only difference will be the genes that have been introduced repeatable experiment with the culture of the clone a intracellular calcium concentration increases as linolenic acid concentration increases b increases become smaller ‹given the logarithmic x-axis› a both mutant alleles reduce calcium concentration by the same amount at low linolenic acid concentrations still lower with high linolenic acid with R270H but as high as WT with R67C b Notes Total 2 max Do not accept answers stating that there is a slow initial increase. 2 2 –7– SPEC/4/BIOLO/HP2/ENG/TZ0/XX/M (Question 1 continued) Question Marking point Answers Notes Total Do not accept answers that are unrelated to the data eg: overeating, sedentary lifestyle. i Arguments for both factors having an effect: a b c d e f g h ‹all› mice on a high fat diet had higher body mass than on a normal diet mass of GPR deficient mice was higher than control mice on the high fat diet high fat diet will give high blood concentrations of linolenic acid responses in WT humans are mediated via an increase in intracellular Ca2+ less Ca2+ release with mutant alleles so less response Argument for diet having more effect: more of mass increase on the high fat diet was due to diet than to the genetic difference Argument for genes not being the only factor: there are differences in allele frequency between obese and non-obese groups but some non-obese people have the same alleles as obese people 3 max –2– M16/4/BIOLO/HP2/ENG/TZ0/XX Section A Answer all questions. Write your answers in the boxes provided. 1. Type I diabetes is a leading cause of death in advanced countries and is associated with various severe or fatal complications, including blindness, kidney failure, heart disease, stroke, neuropathy, and amputations. Embryonic stem cells are considered to be a powerful tool in the treatment of diabetes. In a study, embryonic stem cells were grown in culture and tested for insulin mRNA. A drug was injected into two groups of healthy mice in order to simulate type I diabetes 15 days prior to the transplant of embryonic stem cells. The mice in the transplant group received embryonic stem cells that produce insulin mRNA. The control group did not receive the transplant. The graph shows the blood glucose concentration in both groups. Transplant 600 500 Blood glucose / mg dL–1 400 300 200 100 0 –15 –7 Key: control group 0 7 Day 14 21 28 transplant group [Source: Reprinted from The American Journal of Pathology, Vol 106, no. 6, Takahisa Fujikawa et al., “Teratoma Formation Leads to Failure of Treatment for Type I Diabetes Using Embryonic Stem Cell-Derived Insulin-Producing Cells”, pp. 1781–1791, Copyright © 2005 American Society for Investigative Pathology. Published by Elsevier Inc. All rights reserved.] (a) State the highest mean concentration of blood glucose in the mice with transplants. . . . . . . . . . . . . . . . . . . . . . . . . . . mg dL–1 (This question continues on the following page) 20EP02 [1] –3– M16/4/BIOLO/HP2/ENG/TZ0/XX (Question 1 continued) (b) Outline the cause of type I diabetes in humans. [1] .......................................................................... .......................................................................... (c) Describe the reason for testing for insulin mRNA in the embryonic stem cell cultures. [1] .......................................................................... .......................................................................... (d) Compare and contrast the concentration of blood glucose resulting from the embryonic stem cell transplant with the control. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (e) Evaluate the effectiveness of the embryonic stem cell treatment in controlling blood glucose. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) Turn over 20EP03 –4– M16/4/BIOLO/HP2/ENG/TZ0/XX (Question 1 continued) In a second study, a group of patients recently diagnosed with type I diabetes received a transplant of stem cells. Based on their need for insulin after the transplant, participants were divided into two groups. Their C-peptide production levels were measured for 24 months as the levels indicate the degree of pancreatic beta-cell function. Group 1 did not require insulin and group 2 required insulin occasionally during the study. The graphs show the levels of C-peptides in each individual of both groups 1 and 2. Graphs removed for copyright reasons (f) State the highest rate of production of C-peptides after 24 months in group 2. . . . . . . . . . . . . . . . . . . . . . . . ng ml–1 2h–1 (This question continues on the following page) 20EP04 [1] –5– M16/4/BIOLO/HP2/ENG/TZ0/XX (Question 1 continued) (g) Insulin is produced by cutting C-peptide from the precursor molecule proinsulin. Suggest why group 1 has a greater level of C-peptide than group 2. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) Turn over 20EP05 –6– M16/4/BIOLO/HP2/ENG/TZ0/XX (Question 1 continued) A few years later, a third study used a treatment with umbilical cord stem cells on patients who had suffered from moderate or severe type I diabetes for an average of 8 years. They were divided into two groups: group 1 had moderate diabetes and group 2 had severe diabetes. The patients’ blood was circulated outside the body and exposed to umbilical cord stem cells before returning to the patients’ circulation. The control group had moderate diabetes and received the same treatment but without umbilical cord stem cells. 1.0 0.8 0.6 C-peptide / ng ml–1 0.4 0.2 0.0 Key: 0 Pre treatment 4 control group moderate, no exposure 12 Post treatment / weeks 24 group 1 moderate, exposure to stem cells lower limit for normal C-peptide group 2 severe, exposure to stem cells [Source: doi:10.1186/1741-7015-10-3 Zhao et al.: Reversal of type 1 diabetes via islet ß cell regeneration following immune modulation by cord blood-derived multipotent stem cells. BMC Medicine 2012 10:3.] (This question continues on the following page) 20EP06 –7– M16/4/BIOLO/HP2/ENG/TZ0/XX (Question 1 continued) (h) Compare and contrast the results of the treatment on group 1 with the results of the treatment on group 2. [3] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (i) Suggest an ethical advantage of using this type of therapy over embryonic stem cell therapy. [1] .......................................................................... .......................................................................... (j) Using the data from all three studies, evaluate the use of embryonic stem cells as a treatment for type I diabetes. [4] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... Turn over 20EP07 –3– M16/4/BIOLO/HP2/ENG/TZ0/XX/M Section A Question 1. Answers Notes a 470 Accept answers in the range of 460 to 480 «mg dL–1». b a. «autoimmune» destruction of beta/β cells Accept B cells instead of β cells. b. reduced/insufficient/no production of insulin c a. indicates «stem» cells can produce insulin OR is needed for insulin production OR shows insulin gene is working/being translated Total 1 1 max Answers must relate to insulin mRNA. 1 max b. insulin is needed to treat type I diabetes OR insulin is needed to bring blood glucose level down d a. decrease in transplant group «after treatment» in contrast to control group which does not decrease/decreases only very slightly «initially»/increases/is higher than treatment group b. glucose «remains» lower in transplant group «than control group» for 2 weeks/3 weeks/for a time c. «in the 4th week» transplant group rises back to level before transplant/to higher level than before transplant/to «near» level of control group The answer must include some indication of time or non-permanency. 2 max –4– Question e Answers M16/4/BIOLO/HP2/ENG/TZ0/XX/M Notes Total a. glucose level still higher than normal/higher than 100 «mg»/higher than it was before the drug injection b. effective/lowers blood glucose for 3 weeks/temporarily/for a short time OR glucose level rises back in 4th week/by day 28 OR rises back to level of control group OR rises again but not above control group This can either be positive (the treatment is effective for a while) or negative (it isn’t effective permanently). f 700 No other answer accepted. g a. more pancreatic beta-cell/β-cell function/more insulin «production»/less severe diabetes in group 1 Do not accept answers suggesting that only Group 1 produces insulin, There must be a correct indication of the timing of the effects. b. stem cell transplant is more successful/more effective/more stem cells survived in group 1 c. «group 1» produces more proinsulin 2 1 2 max Reject more C-peptide. –5– Question h Answers a. C-peptide increases after treatment in both groups OR treatment effective in both groups OR both groups rose higher than the control M16/4/BIOLO/HP2/ENG/TZ0/XX/M Notes Total There must be an explicit comparison. b. similar/same overall/total increase «in both groups» OR quoted figures to show this c. smaller percentage/% increase «pre to post treatment» in group 1 «than group 2» OR quoted figures to show this Reject answers relating to rates of increase. 3 max d. initial increase is greater in group 1 OR increases slowing/finished/rate of increase reduced by end of study/by week 24 in group 1 but continuing in group 2 e. group 1 rose above lower limit «by week 12» and group 2 remained below it «even at week 24» i umbilical cord «stem» cells are discarded/die if not harvested OR harvesting umbilical cord cells does not harm the baby OR taking «stem» cells from an embryo may harm/kill it Do not accept answers relating to consent. 1 max –6– Question j M16/4/BIOLO/HP2/ENG/TZ0/XX/M Answers Notes a. study 1/study with mice/embryonic stem cell study shows treatment can cause increased insulin production/ reduce blood glucose levels Strength b. «insulin production/reduction in blood glucose in study 1 was» only temporary/did not reduce glucose to normal levels Limitation c. study 2 shows increases in C-peptide/insulin OR some type I diabetes patients required no insulin after treatment Strength d. study 2 shows treatment effective for a long time/2 years Strength e. «stem cell treatment in study 2» was more successful in some patients than others OR more successful for moderate «than more severe» diabetes Limitation f. study 3 shows that stem cells can cause C peptide/insulin levels to double/rise significantly/rise above lower limit «for normal C-peptide»/rise and stay raised Strength g. «study 3» does not give evidence for embryonic stem cells OR used umbilical cord rather than embryonic stem cells Limitation Total 4 max –2– N16/4/BIOLO/HP2/ENG/TZ0/XX Section A Answer all questions. Write your answers in the boxes provided. 1. Coral reefs are among the most spectacular ecosystems on Earth. They support a rich diversity of life and provide economic benefits to the people who use them. In Papua New Guinea in the Pacific Ocean north of Australia the following data were collected. Coral cover is the percentage of the reef surface covered by live hard coral. 80 24.70 70 24.65 60 24.60 50 24.55 Coral cover 40 /% 30 24.50 24.45 20 24.40 10 24.35 0 1994 1996 1998 2000 2002 Mean annual ocean temperature / °C 24.30 2004 Year Key: percentage coral cover ocean temperature [Source: adapted from Jones et al. (2004), The Encyclopedia of Earth, Patterns of Coral Loss] (a) Calculate the difference in coral cover in 1996 and 2002. No working required. [1] ..................... % (b) Describe the evidence that the ocean temperature has an effect on coral cover. .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) 20EP02 [2] –3– N16/4/BIOLO/HP2/ENG/TZ0/XX (Question 1 continued) (c) Suggest causes for the changes in ocean temperature. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on page 5) Turn over 20EP03 N16/4/BIOLO/HP2/ENG/TZ0/XX –5– (Question 1 continued from page 3) In order to test the effect of temperature, live samples of a species of coral, Pocillopora damicornis, were placed in an experimental chamber at a constant pH, water depth and low light. All the coral samples were started at 26 °C and half of them were rapidly increased to 30 °C. Light source Coral sample pH meter Heat pad kept at 26 °C Containers holding the coral samples Heat pad rapidly increased from 26 °C to 30 °C The pie charts show the percentage of live and dead tissues at the end of the experiment. Key: live dead Constant temperature 26 °C Rapidly increasing temperature 26 °C to 30 °C [Source: Adapted from Mace G. Barron, Cheryl J. McGill, Lee A. Courtney, and Dragoslav T. Marcovich, “Experimental Bleaching of a Reef-Building Coral Using a Simplified Recirculating Laboratory Exposure System,” Journal of Marine Biology, vol. 2010, Article ID 415167, 8 pages, 2010. doi:10.1155/2010/415167] (d) Identify one advantage of conducting this experiment in the laboratory rather than in the ocean. [1] .......................................................................... (e) Comment on whether the experimental data supports the observed data from the ocean. [1] .......................................................................... .......................................................................... (This question continues on the following page) Turn over 20EP05 –6– N16/4/BIOLO/HP2/ENG/TZ0/XX (Question 1 continued) Acidification of the world’s oceans is an increasing threat to the health of oceanic life including coral reefs. Corals perform calcification to create their calcium carbonate exteriors. An experiment was conducted on Heron Island, Southern Great Barrier Reef, Australia. For the experiment the pH was altered by dissolving carbon dioxide in the water. Three different coral species were used, with each test group at two different temperature ranges and three different pH values. The white line in each photograph represents 5 cm. Porolithon onkodes Acropora intermedia Porites lobata 16 12 Calcification / percentage mass increase per month 8 4 0 –4 8.2 7.9 7.6 8.2 7.9 7.6 8.2 7.9 7.6 pH Key: 25 –26 °C 28 – 29 °C [Source: Adapted from K. R. N. Anthony, D. I. Kline, G. Diaz-Pulido, S. Dove, and O. Hoegh-Guldberg, “Ocean acidification causes bleaching and productivity loss in coral reef builders,” PNAS, vol. 105 no. 45, 17442–17446, Copyright 2008 National Academy of Sciences, U.S.A.] (f) (i) Describe the trend in calcification when the pH is decreased at 25 –26 °C. .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) 20EP06 [1] –7– N16/4/BIOLO/HP2/ENG/TZ0/XX (Question 1 continued) (ii) In environmental studies, a critical value is the level at which a population declines or shows signs of poor health. Suggest a critical pH for P. onkodes. [1] .......................................................................... .......................................................................... .......................................................................... (iii) Using all of the data, comment on the hypothesis that ocean acidification in warming seas will have the same effect on all species of coral. [1] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (g) Suggest another marine animal that has parts made of calcium carbonate and may therefore be damaged due to ocean acidification. [1] .......................................................................... (h) Outline causes of ocean acidification. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) Turn over 20EP07 –8– N16/4/BIOLO/HP2/ENG/TZ0/XX (Question 1 continued) (i) Discuss the need for international cooperation to solve the problems of declining coral populations. [3] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... 2. (a) The image is an electron micrograph of the lining of the small intestine. Goblet cell [Source: adapted from A. L. Mescher (2009), Junqueira’s Basic Histology: Text and Atlas, 12th Edition, © 2009 McGraw-Hill Education] (i) Label the microvilli using the letter M and a nucleus using the letter N. (This question continues on the following page) 20EP08 [1] – – N16/4/BIOLO/HP2/ENG/TZ0/XX/M Section A Question 1. Answers a 60 (%) b a. coral cover decreases as temperature rises (between 1996 and 1998/2000 and 2002) / negative correlation between temperature and coral cover / coral cover highest when temperature is lowest/vice versa Notes Total 1 Do not award this mark for “inversely proportional”, but the mark can still be awarded if other parts of the answer give one of the alternative parts of the mark point. 2 max b. coral cover remains constant when temperature drops (between 1998/1999 and 2000)/remains (nearly) constant when temperature stops rising (between 2002 and 2003) c. no proof of causation / only a correlation / other factors could be affecting the coral c a. increased carbon dioxide/methane in the atmosphere / carbon dioxide emissions from burning of fossil fuels / other specific source of a named greenhouse gas b. increased greenhouse effect / more heat/long wave radiation trapped in the atmosphere No marks for increased CO2 in the oceans, global warming or climate change. The idea of an increase must be included, not just greenhouse effect or heat trapping. 2 max c. heat transfer from atmosphere to ocean / ocean absorbs heat from atmosphere d control of variables/pH/light/temperature / no predators of coral e a. supports because there is more dead coral/less % cover at the higher temperature b. (experimental data) does not support (observed data) because experimental temperatures were (all) higher/rose much faster 1 The answer must make it clear whether or not the data provides support. 1 max – Question – Answers N16/4/BIOLO/HP2/ENG/TZ0/XX/M Notes f i less calcification in all three/each species (as pH decreased) f ii 7.6 / 7.7 / 7.8 Accept any pH that is 7.6 or higher, but lower than 7.9. f iii a. greater reduction in calcification as pH drops at the higher temperature in P. onkodes than on the other two species (so hypothesis not supported) This answer is based on the larger drop in calcification between 8.2 and 7.6 at both temperatures in onkodes than the other two species. This answer is based only on whether there are positive values for calcification or negative. b. net loss in calcification at lowest pH and highest temperature in P. onkodes whereas there is still calcification in the other two species (so hypothesis not supported) c. warming reduces calcification at all pH levels in A. intermedia but not in the other two species (so hypothesis not supported) d. combined effect of acidification and warming is a larger reduction in calcification in A. intermedia than in the other two species (so hypothesis not supported) e. more calcification as temperature rises at lower pH/pH 7.9 and 7.6 in P. lobata whereas there is less in the other two species (so hypothesis not supported) f. more calcification as pH drops from 8.2 to 7.9 at higher temperature in P. lobata whereas there is a drop/no rise in the other two species (so hypothesis not supported) Total 1 This answer is based on the drop in calcification at each pH when the temperature rises in intermedia, whereas in the other species there is a rise at one or more of the pHs. This answer is based on the larger overall drop in calcification between pH 8.2 at 25/26°C and 7.6 at 28/29°C. The answer must either state pHs 7.9 and 7.6 or specify lower pH or greater acidification. The answer must state the two pH values and state higher temperature or 28-29°C. 1 1 max – Question g h – Answers Mollusca/named marine mollusc with a shell/crustacean/named marine crustacean/Porifera/sponges/named calcareous marine sponge N16/4/BIOLO/HP2/ENG/TZ0/XX/M Notes Total Reject terrestrial examples. Reject sea shells, shellfish. Specific named examples must be verified if it is uncertain whether they have calcified parts. 1 a. carbon dioxide makes an acid/carbonic acid in water Do not award a mark for stating only that carbon dioxide causes ocean acidification. b. (carbon dioxide from) burning fossil fuels/forest fires Do not award marks for methane sources or sources of unspecified greenhouse gases or statements about increased carbon dioxide in the atmosphere. c. carbon dioxide forms solution with/dissolves into water/oceans/rain 2 max – Question i – Answers N16/4/BIOLO/HP2/ENG/TZ0/XX/M Notes Total a. international cooperation needed to reduce carbon dioxide emission/concentrations b. carbon dioxide produced anywhere increases the greenhouse effect/global warming/ocean acidification/health of coral everywhere c. ocean currents/tides/wind move carbon dioxide/acid/heat around the world / oceans of the world are interconnected/part of one overall system d. (some) coral reefs are in international waters (or words to that effect) / coral reefs cannot be protected by single national governments alone e. the more groups of people/nations/corporations that reduce their carbon emissions, the lower the impact on coral will become / not enough for one country/group/corporation to reduce carbon dioxide emissions f. sharing of technology/research/information/resources g. aid to poorer/developing countries (to help with coral conservation) h. reference to an economic/ecological benefit of conserving coral reefs 3 max –2– M17/4/BIOLO/HP2/ENG/TZ1/XX Section A Answer all questions. Write your answers in the boxes provided. 1. The Chinese soft-shelled turtle, Pelodiscus sinensis, lives in salt water marshes. The turtle can live under water and out of water. These turtles have fully developed lungs and kidneys, however, many microvilli have been discovered in the mouth of P. sinensis. A study was undertaken to test the hypothesis that oxygen uptake and urea excretion can simultaneously occur in the mouth. Initial experiments involved collecting nitrogen excretion data from P. sinensis. The turtle urinates both in water and out of water. When in water it allows waste products to be washed out of its mouth. When out of water it regularly dips its head into shallow water to wash its mouth. The table shows the mean rates of ammonia and urea excretion from the mouth and kidney over six days. Ammonia Urea Excretion of nitrogen by the mouth / µmol day-1 g-1 turtle Turtle submerged Turtle out of water in water 0.29 0.30 0.90 1.56 Excretion of nitrogen by the kidney / µmol day-1 g-1 turtle Turtle submerged Turtle out of water in water 0.63 0.54 0.07 0.73 [Source: Reproduced with permission, Y. Ip et al. (2012) The Journal of Experimental Biology, 215, pages 3723—3733. jeb.biologists.org. doi: 10.1242/jeb.068916] (a) Deduce whether the excretion of ammonia or urea changes more when a turtle emerges from water. .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) 24EP02 [2] –3– M17/4/BIOLO/HP2/ENG/TZ1/XX (Question 1 continued) (b) Compare and contrast the changes in urea excretion in the mouth with the changes in urea excretion in the kidney when a turtle emerges from the water. [3] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) Turn over 24EP03 –4– M17/4/BIOLO/HP2/ENG/TZ1/XX (Question 1 continued) Percentage saturation of water with dissolved oxygen / % It was noted that during long periods out of water, turtles rhythmically moved their mouths to take in water from a shallow source and then discharge it. Changes in the dissolved oxygen and the quantity of accumulated urea in the rinse water discharged by the turtles were monitored over time as shown in this graph. Key: Head out of water Head dipped into water Accumulated urea 4 µmol 100 Accumulated urea 20 µmol 95 90 Accumulated urea 38 µmol 85 80 75 70 210 230 250 270 290 310 330 350 370 390 Time / mins [Source: adapted with permission from Y. Ip et al. (2012) The Journal of Experimental Biology, 215, pages 3723–3733.] (c) (i) Describe the trends shown by the graph for dissolved oxygen in water discharged from the mouth. [1] .......................................................................... .......................................................................... (ii) Suggest reasons for these trends in dissolved oxygen. .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) 24EP04 [2] –5– M17/4/BIOLO/HP2/ENG/TZ1/XX (Question 1 continued) In order to test whether a urea transporter was present in the mouth tissues of the turtles, phloretin (a known inhibitor of membrane proteins that transport urea) was added to the water in which a further set of turtles submerged their heads. The results of that treatment are shown. 1.0 0.8 Mouth urea excretion rate / µmol urea-N day−1 g−1 turtle 0.6 0.4 0.2 0.0 Control Phloretin [Source: Reproduced with permission from Y. Ip et al. (2012) The Journal of Experimental Biology, 215, pages 3723–3733. jeb.biologists.org.] (d) Deduce with a reason whether a urea transporter is present in the mouth of P. sinensis. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on page 7) Turn over 24EP05 –7– M17/4/BIOLO/HP2/ENG/TZ1/XX (Question 1 continued from page 5) Further research was conducted to determine where mRNA expression of a urea transporter gene might be occurring in P. sinensis. Gel electrophoresis was used to analyse different tissue samples for mRNA activity. Roof of the mouth Tongue Esophagus Intestine Kidney Bladder [Source: Reproduced with permission from Y. Ip et al. (2012) The Journal of Experimental Biology, 215, pages 3723–3733. jeb.biologists.org.] (e) Outline the additional evidence provided by the gel electrophoresis results shown above. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) Turn over 24EP07 –8– M17/4/BIOLO/HP2/ENG/TZ1/XX (Question 1 continued) Expression of the urea transporter gene by cells in the turtle’s mouth was assessed by measuring mRNA activity. Turtles were kept out of water for 24 hours and then injected with either a salt solution that matched the salt concentration of the turtle, dissolved ammonia or urea, followed by another 24 hours out of water. 4.0 3.5 3.0 2.5 Relative amount of mRNA expression of urea transporter 2.0 1.5 1.0 0.5 0.0 Salt solution Ammonia Urea [Source: © International Baccalaureate Organization 2017] (f) (i) Identify which of these turtle groups represent the control, giving a reason for your answer. .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) 24EP08 [1] –9– M17/4/BIOLO/HP2/ENG/TZ1/XX (Question 1 continued) (ii) Suggest a reason for the greater expression of the gene for the urea transporter after an injection with dissolved ammonia than an injection of urea. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (g) The salt marshes where these turtles live periodically dry up to small pools. Discuss the problems that this will cause for nitrogen excretion in the turtles and how their behaviour might overcome the problems. [3] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... Turn over 24EP09 – – M17/4/BIOLO/HP2/ENG/TZ1/XX/M Section A Question Answers Mouth 1. a Notes Total Kidney In water Out In water Out Ammonia 0.29 0.30 0.63 0.54 Urea 0.90 1.56 0.07 0.73 a. urea b. for both mouth and kidney c. percentage change/change in μmol day−1 g−1 greater with urea/other acceptable numerical comparison 2 b a. both higher/increased on emergence from/with turtle out of water b. both increased by 0.66 «µmol−1 g−1 when turtle emerges from water» c. % increase is higher in kidney / kidney 940 % versus mouth 73/75 % / increase is higher proportionately higher in kidney / kidney x10 versus mouth nearly double/x1.73 d. urea excretion by mouth greater than kidney out of water «despite larger % increase in kidney excretion» 3 – Question c – Answers i decrease «when head is submerged» and increase when head is out of water ii a. oxygen absorbed from water/exchanged for urea when head dipped in water «so oxygen concentration decreases» b. lungs cannot be used with head in water / can «only» be used with head out of water M17/4/BIOLO/HP2/ENG/TZ1/XX/M Notes Total 1 2 max c. oxygen from water «in mouth» used in «aerobic cell» respiration d. oxygen from air dissolves in water when head out of water «so oxygen concentration increases» d a. urea transporter is present b. less urea «excreted»/ lower rate «of urea excretion» / excretion almost zero when phloretin/inhibitor was present 2 – Question 1. e – Answers M17/4/BIOLO/HP2/ENG/TZ1/XX/M Notes Total a. mRNA only in mouth and tongue/in mouth and tongue but not esophagus intestine kidney or bladder b. bands / lines indicate mRNA for/expression of urea transporter gene c. urea transporter gene expressed / urea transporters in mouth/tongue / not expressed/made in esophagus/intestine/kidneys/bladder d. mRNA/transcription/gene expression/urea transporters higher in tongue/more in tongue «than mouth» f i ii salt solution is control because it does not contain a nitrogenous/excretory waste product / it matches the salt concentration of the turtle / the turtle’s body already contains salt / because the turtle lives in salt water/salt marshes / because nothing has been altered 2 max 1 a. ammonia is «highly» toxic/harmful b. ammonia is more toxic than urea/converse c. ammonia converted to urea d. urea concentration raised «by injecting ammonia» e. difference between ammonia and urea «possibly» not «statistically» significant 2 max – Question g – Answers M17/4/BIOLO/HP2/ENG/TZ1/XX/M Notes Total Problems: a. urea becomes more concentrated «in small pools» / lower concentration gradient «between tongue/mouth and water» b. less water available for urine production/excretion by kidney OR less water in ponds for mouth rinsing/more competition for pools (to use for mouth rinsing) Behaviour to overcome problems: c. «still able to» dip mouth into/mouth rinse in water/pools d. «still able to» excrete urea «though the mouth» in the small pools e. more conversion of ammonia to urea/urea excretion rather than ammonia f. more urea transporters/expression of urea transporter gene g. urea excreted «in mouth/via microvilli» by active transport/using ATP h. excretion with little/no loss of water 3 max –2– M17/4/BIOLO/HP2/ENG/TZ2/XX Section A Answer all questions. Answers must be written within the answer boxes provided. 1. Auxin can be used to promote the development of roots from stem and leafy cuttings in some plants. In a study into the distribution of auxin in the development of these roots, scientists measured the amount of auxin in different leaves of a shoot tip of Petunia hybrida. The figure indicates the numbering of leaves on the shoot, from L1 as the youngest and smallest to L6 as the largest and oldest leaf. The developmental stage of L5 and L6 was very similar, so L5 was not analysed. The stem base is the lowest part of the cutting where roots may form. 1 cm L5 L3 L2 L1 L6 L4 Stem base [Source: A. Ahkami et al. (2013) Planta, 238, pages 499–517] The graph shows the auxin concentration in the different leaves. 80 70 60 Auxin concentration / pmol g–1 50 40 30 20 10 0 L1 L2 L3 Leaf L4 [Source: A. Ahkami et al. (2013) Planta, 238, pages 499–517] (This question continues on the following page) 20EP02 L6 –3– M17/4/BIOLO/HP2/ENG/TZ2/XX (Question 1 continued) (a) Calculate the difference in the concentration of auxin found in L1 and L6. [1] . . . . . . . . . . . . . . . . . pmol g–1 (b) Identify the relationship between the concentration of auxin and the age of the different leaves. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... N-1-naphthylphthalamic acid (NPA) is an inhibitor used to block auxin transport. NPA was sprayed onto the leaves of a set of cuttings for 14 days. Development of the roots in control (non-treated) and NPA-treated cuttings was measured 14 days after taking the cuttings. The table shows the influence of NPA on rooting. Control NPA-treated Mean number of roots per cutting Mean root length / cm Mean total root length per cutting / cm 53.2 1.4 47.7 8.0 0.6 1.0 [Source: adapted from A Ahkami, et al., (2013), Planta, 238, pages 499–517] (c) Analyse the effect of NPA on the formation of roots. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) Turn over 20EP03 –4– M17/4/BIOLO/HP2/ENG/TZ2/XX (Question 1 continued) The scientists also measured the changes in auxin concentration in L6 and the stem base during the early period of root formation. They recorded the concentration in the control and NPA-treated cuttings for 24 hours after taking the cuttings. 50 40 Auxin concentration in L6 / pmol g–1 30 20 10 0 0 2 4 6 12 24 Time after cutting / hours 400 Auxin concentration in stem base / pmol g–1 300 200 100 0 0 2 4 6 12 24 Time after cutting / hours Key: control NPA-treated [Source: adapted from A Ahkami, et al., (2013), Planta, 238, pages 499–517] (This question continues on the following page) 20EP04 –5– M17/4/BIOLO/HP2/ENG/TZ2/XX (Question 1 continued) (d) (i) Compare and contrast the changes in auxin concentration in the stem base over time for the control and NPA-treated cuttings. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (ii) Deduce the effect of NPA on auxin transport between L6 and the stem base. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (e) Based on all the data presented and your knowledge of auxin, discuss the pattern of auxin production and distribution in the leaves and the possible relationship to root formation in leafy cuttings of Petunia hybrida. [3] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) Turn over 20EP05 –6– M17/4/BIOLO/HP2/ENG/TZ2/XX (Question 1 continued) The scientists wanted to know whether the accumulation of auxin over time in the stem base of the controls affected expression of the GH3 gene, known to have a role in growth regulation in different plants. The technique that was used to quantify the level of transcription of the GH3 gene was Northern blotting. In this procedure the darkness and thickness of the band is an indicator of the level of transcription of a particular gene. The image shows the result of the Northern blot from 2 hours to 24 hours after cutting. 2 4 6 12 24 Time after cutting / hours [Source: adapted from A Ahkami, et al., (2013), Planta, 238, pages 499–517] (f) (i) State the name of the molecule which is produced by transcription. [1] .......................................................................... (ii) Compare the pattern of GH3 transcription with the pattern of auxin concentration in the stem base control cuttings. You may use the table provided to help you to record the patterns before you compare them. (Please note: a simple comparison in the table will not gain marks) 2 – 4 hours 4 – 6 hours 6 –12 hours 12–24 hours Auxin concentration GH3 bands .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) 20EP06 [2] –7– M17/4/BIOLO/HP2/ENG/TZ2/XX (Question 1 continued) (iii) The scientists concluded that auxin activates the transcription of the GH3 gene. Using the information on the auxin concentration in the stem base in the graph on page 4 and the Northern blot, evaluate whether this conclusion is supported. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... 2. (a) The sketch shows the relationship between the reaction rate and substrate concentration in the presence and the absence of a competitive inhibitor. Reaction rate No inhibitor With inhibitor Substrate concentration Explain the effect of the competitive inhibitor on the reaction rate. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) Turn over 20EP07 –3– M17/4/BIOLO/HP2/ENG/TZ2/XX/M Section A Question 1. Answers Notes a 45 «pmol g–1» Allow answers in the range of 44 «pmol g–1» to 46 «pmol g–1». b less auxin as the leaves become older/larger Vice versa Total 1 OR negative correlation from L1 to L4 2 L4 and L6 leaves have least auxin concentration OR L4 and L6/older leaves have about the same concentration of auxin/do not have significantly different concentrations c Accept other correct statements of overall changes in values. The word “mean” is not required. a. NPA decreased the «mean» number of roots per rooted cutting «by about 5» OWTTE b. NPA decreased the «mean» length per root «by more than half» c. NPA decreased the «mean» total root length per planted cutting «to about 2 % of control» OWTTE 2 max d. NPA inhibited the formation of roots OR decreased all three measures (continued…) – – M17/4/BIOLO/HP2/ENG/TZ2/XX/M (Question 1 continued) Question d i Answers Notes Total a. both decrease up to 6 hours/initially b. NPA-treated decrease more/at a faster rate than control «up to 6 hours» 2 max c. after 6 hours, control increases while NPA treated continues to fall ii a. NPA «appears to have» no effect on concentrations/transport of auxin in L6 as control and NPA-treated remain at same «low» level OWTTE A valid reason must be given for the mark. b. NPA «probably» inhibits the auxin efflux pumps/transport «in the leaves» as the levels drop in NPA-treated in stem base «but not in control» OWTTE A valid reason must be given for the mark. c. the transport of auxin to the stem base must occur from younger leaves 2 max OR L6 is not the source of auxin in the stem base d. NPA inhibits the auxin pumps/transport «in the leaves» as the levels drop in NPA-treated in stem base e a. L1 has the highest concentration of auxin so appears to be/is the main source/the producer of auxin b. as leaves age, they «appear to» decrease the production of auxin Vice versa c. the stem base is an auxin sink as seen by the accumulation in the control stem base «where roots form» OWTTE d. high concentration of auxin «in the stem base» promotes root formation Vice versa 3 max (continued…) – – M17/4/BIOLO/HP2/ENG/TZ2/XX/M (Question 1 continued) Question f Answers i mRNA/RNA ii a. at 2 and 24 hours, auxin levels are similar and at 2 and 24 hours GH3 levels are similar b. the pattern for the formation of auxin is similar to the pattern of transcription of the GH3 gene Notes Total 1 A comparison must be made to award marks. Do not award marks for simple completion of the table. 2 max OR both decrease and then increase c. «however» there is a lag between the peaks of the GH3 transcription and the peaks of auxin iii a. the data «partially» supports the conclusion OR the relationship is not clear b. the auxin concentration «seems to» rise before the transcription level increases OR To award mp b, awareness of the lag should be demonstrated there is a lag between auxin concentration changing and transcription level changing OR the auxin concentration falls before the transcription level falls c. more data is needed «before two hours/after 24 hours» OWTTE 2 max –2– N17/4/BIOLO/HP2/ENG/TZ0/XX Section A Answer all questions. Answers must be written within the answer boxes provided. 1. Hypoxia is a condition in which tissues of the body are deprived of an adequate oxygen supply. A study was carried out in rats to examine the effects of continuing hypoxia on the structure of the diaphragm, and to determine whether nitric oxide is implicated in adaptation of the diaphragm to hypoxia. The diaphragm helps to supply oxygen to tissues and organs in the body by ventilating the lungs. A group of 36 adult male rats were kept for 6 weeks in low oxygen while 36 adult male rats were kept in normal oxygen levels. Erythrocytes / % of total blood volume Body mass /g Mass of right ventricle muscle / mg Control 305.7 ! 7.4 39.3 ! 1.7 154.3 ! 7.4 Hypoxia *238.3 ! 5.0 *62.6 ! 1.9 *194.8 ! 8.9 Control 302.3 ! 5.0 39.6 ! 1.1 157.8 ! 3.4 Hypoxia *229.7 ! 4.6 *70.1 ! 1.0 *204.7 ! 11.2 45.0 ! 0.7 166.8 ! 3.6 1 week 2 weeks Control 3 weeks 325.0 ! 10.3 Hypoxia *255.0 ! 8.3 *71.3 ! 1.0 *238.7 ! 18.9 Control 369.8 ! 5.9 43.0 ! 2.6 164.7 ! 3.9 Hypoxia *277.5 ! 7.9 *75.1 ! 1.4 *251.3 ! 8.0 6 weeks Key: * indicates significant difference from corresponding control value (student’s t-test, p < 0.05) [Source: Reproduced with permission of the © ERS 2011. European Respiratory Journal June 2011, 37 (6) 1474–1481; DOI: 10.1183/09031936.00079810] (a) Outline the effect of hypoxia on body mass and erythrocyte percentage. .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) 16EP02 [1] –3– N17/4/BIOLO/HP2/ENG/TZ0/XX (Question 1 continued) (b) The graph shows the effect of hypoxia on the endurance of rats’ diaphragm muscle after 6 weeks. Endurance is the change in force measured as a percentage of the initial force. 100 80 Endurance / % 60 40 20 0 0 1 2 3 4 5 Testing time / mins Key: indicates significant difference from control (p < 0.0001) hypoxia control [Source: Reproduced with permission of the © ERS 2011. European Respiratory Journal June 2011, 37 (6) 1474–1481; DOI: 10.1183/09031936.00079810] Using the data in the graph, deduce whether hypoxia increases or decreases the endurance of the rats’ diaphragm muscle. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... (c) Using the data so far presented in this question, explain the effect of hypoxia on the body. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) Turn over 16EP03 –4– N17/4/BIOLO/HP2/ENG/TZ0/XX (Question 1 continued) (d) The sodium–potassium pump plays a role in muscle activity. Nitric oxide may have a role in the recovery of hypoxic muscles. The production of nitric oxide can be blocked with an inhibitor of the enzyme nitric oxide synthase. The graph shows the concentration of sodium–potassium pumps in the diaphragm of control and hypoxic rats without and with nitric oxide synthase inhibitor. 500 400 300 Concentration of sodium–potassium pumps / pmol g–1 200 100 0 Control Hypoxia Without nitric oxide synthase inhibitor Control Hypoxia With nitric oxide synthase inhibitor [Source: Reproduced with permission of the © ERS 2011. European Respiratory Journal June 2011, 37 (6) 1474–1481; DOI: 10.1183/09031936.00079810] (i) 1. Analyse the graph to obtain two conclusions about the concentration of sodium–potassium pumps. ..................................................................... ..................................................................... 2. ..................................................................... ..................................................................... (This question continues on the following page) 16EP04 [2] –5– N17/4/BIOLO/HP2/ENG/TZ0/XX (Question 1 continued) (ii) Muscle fibres are stimulated to contract by the binding of acetylcholine to receptors in their membranes and the subsequent depolarization. Suggest a reason for increasing the concentration of sodium–potassium pumps in the membranes of diaphragm muscle fibres. [1] .......................................................................... .......................................................................... .......................................................................... (e) Skeletal muscle contractions can take two different forms: if they are stimulated by a single action potential they take the form of a twitch and if they are stimulated by a series of action potentials the contraction is longer lasting (tetanic). The table shows the effects of hypoxia on the force of twitch and peak tetanic contraction in the diaphragm. Twitch contraction / N cm–2 Peak tetanic contraction / N cm–2 Control 4.0 ! 0.7 20.0 ! 2.3 Hypoxia 2.8 ! 0.4 14.2 ! 1.8 Diaphragm [Source: Reproduced with permission of the © ERS 2011. European Respiratory Journal June 2011, 37 (6) 1474–1481; DOI: 10.1183/09031936.00079810] (i) Outline the effect of hypoxia on the force of contraction of the diaphragm. [1] .......................................................................... .......................................................................... (ii) Hypoxia caused a 13 % increase in the surface area to volume ratio of the diaphragm. Suggest a reason for this change. [1] .......................................................................... .......................................................................... (This question continues on the following page) Turn over 16EP05 –6– N17/4/BIOLO/HP2/ENG/TZ0/XX (Question 1 continued) (f) Using all relevant data in the question, evaluate the effectiveness of the rats’ adaptation to hypoxia. [3] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (g) Discuss the advantages and disadvantages of using rats as models in this investigation. .......................................................................... .......................................................................... .......................................................................... .......................................................................... 16EP06 [2] –3– N17/4/BIOLO/HP2/ENG/TZ0/XX/M Section A Question Answers 1. a erythrocyte percentage increased AND body mass reduced/smaller increase in mass ✔ 1. b a. increases endurance «in relation to the control» ✔ Notes Total 1 b. higher force/endurance at every testing time/throughout OR smaller decreases in force «over time» ✔ 2 max c. the magnitude of the difference is similar throughout the five minutes experiment/testing ✔ d. differences are «statistically» significant ✔ 1. c e. endurance of control is «approximately» 35 % versus endurance of hypoxia «approximately» 55 % «after 5 minutes» ✔ Accept 5 % for both percentages a. diaphragm more endurance/stronger/generates more force for more ventilation/inspiration ✔ Reject “loss of body mass” b. right ventricle mass increases to pump more blood ✔ The physiological reason is required for each mark c. erythrocyte percentage increases to transport oxygen ✔ 2 max d. less growth/body mass which reduces oxygen demand ✔ 1. 1. d d i ii a. hypoxia increases the concentration of sodium–potassium pumps ✔ b. nitric oxide needed for/stimulates «production of» sodium-potassium pumps ✔ Award up to [1] for a conclusion on lines labelled 1 c. nitric oxide synthase inhibitor reduces the concentration of pumps OR concentration of pumps reduced by inhibiting nitric oxide production ✔ and up to [1] for a conclusion on the lines labelled 2 a. resting potential restored faster ✔ Accept shorter refractory period for mpa b. increases the «maximum» frequency/rate of contractions OR can contract again sooner ✔ Do not accept faster contraction/depolarization/ repolarization 2 max 1 max (continued…) –4– N17/4/BIOLO/HP2/ENG/TZ0/XX/M (Question 1 continued) Question 1. Answers e i reduces «force of» twitch AND peak tetanic contraction ✔ e ii a. decrease in volume/atrophy/loss of cells/less muscle fibres/less tissue in the diaphragm ✔ Notes Total 1 Do not accept reduction in area of diaphragm 1 max b. SA to volume ratio increased to make oxygen uptake into muscle/cells faster ✔ 1. f a. not effective because body mass lost ✔ b. effective because body mass still increases/rats still grow ✔ c. not effective because contractions/force exerted by diaphragm decreases d. effective because more sodium-potassium pumps so more/faster rate of diaphragm/muscle contractions ✔ e. effective because endurance of diaphragm increases ✔ For each marking point the candidate must make it clear whether they are arguing for adaptation being effective or not. This can be done by giving the physiological benefit of a change, for example greater mass of right ventricle so more blood pumped. 3 max f. effective because mass of right ventricle increases ✔ g. effective because erythrocyte percentage increases ✔ (continued…) –5– N17/4/BIOLO/HP2/ENG/TZ0/XX/M (Question 1 continued) Question 1. g Answers advantages: Notes Total Accept any one of the advantages a. small size OR easy to look after in research labs ✔ b. short lifespan OR study can extend over several generations ✔ c. can be killed «to get experimental results» if benefits of research justify it ✔ 2 max d. «mammalian» so similarities with humans ✔ e. fewer ethical objections than if humans are used/not ethical to subject humans to hypoxia/does not cause harm to humans ✔ disadvantages: f. ethical objections OR wrong to cause suffering to animals/rats ✔ g. rat physiology/anatomy not same as human ✔ Accept any one of the disadvantages –2– M18/4/BIOLO/H P2/ENG/TZ1/XX Section A Answer all questions. Answers must be written within the answer boxes provided. 1. White-nose syndrome (WNS) is a disease caused by the fungus Geomyces destructans. This kills bats from many species during their winter hibernation. Scientists used echolocation to record the number of bat flights over a station in 10 minute sample periods during the summers of 2007 to 2009. The graph shows the mean number of flights and number of recording samples for little brown bats (Myotis lucifugus), a species affected by the fungus, along with those of hoary bats (Lasiurus cinereus), which are not affected. Mean number of bats flying in 10 minute sample periods 40 Key: 35 Myotis lucifugus (affected) 30 Lasiurus cinereus (unaffected) 25 20 15 10 5 0 Year: Samples: 2007 79 2008 151 2009 144 [Source: Adapted from Dzal, Yvonne, et al., Going, going, gone: the impact of white-nose syndrome on the summer activity of the little brown bat (Myotis lucifugus), Biology Letters (2011), 7, p. 393. http://rsbl.royalsocietypublishing.org/content/7/3/392. Permission conveyed through Copyright Clearance Center, Inc.] (a) Calculate the total number of M. lucifugus flights that were recorded in the summer of 2007. [1] ................. (b) Suggest one limitation of this recording method in determining the accurate mean number of individual bats flying. .......................................................................... .......................................................................... (This question continues on the following page) 20EP02 [1] –3– M18/4/BIOLO/HP2/ENG/TZ1/XX (Question 1 continued) (c) (i) Calculate the percentage decline in the mean number of M. lucifugus flights for 2009 when compared to 2008. [1] ............... % (ii) Evaluate the conclusion that the decline in the population of M. lucifugus is due to infection by the fungus. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) Turn over 20EP03 –4– M18/4/BIOLO/HP2/ENG/TZ1/XX (Question 1 continued) Scientists have hypothesised that bats affected by WNS could be more likely than unaffected bats to emerge from hibernation during the winter. Bats occasionally undergo short interruptions in hibernation accompanied by an increase in body temperature. Scientists used sensors attached to M. lucifugus to monitor the temperature of a group infected with the fungus over the winter months. A second uninfected group was also monitored. The graphs show the results for two typical individuals. Month Dec 2010 Jan 2011 Feb 2011 Uninfected 30 Body temperature / °C 20 10 0 Infected 30 20 10 0 0 10 20 30 40 50 60 70 80 Days of study [Source: L Warnecke, et al. (2012), Inoculation of bats with European Geomyces destructans supports the novel pathogen hypothesis for the origin of white-nose syndrome, PNAS 109, p. 7000.] (d) Distinguish between the patterns of hibernation of the uninfected and infected bats. .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) 20EP04 [1] –5– M18/4/BIOLO/HP2/ENG/TZ1/XX (Question 1 continued) (e) Based on the data and your biological knowledge, suggest how the infection could lead to premature death in a bat. [3] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... The graph shows data from a small group of infected bats that died during hibernation. The average time interval between hibernation emergence periods and the date of death were recorded using temperature sensors for these bats. 14 12 Mean interval between hibernation emergence periods / days 10 8 6 4 2 0 28 Nov 18 Dec 7 Jan 27 Jan 16 Feb 8 Mar Date of death [Source: Reeder, D M, et al. (2012), Frequent Arousal from Hibernation Linked to Severity of Infection and Mortality in Bats with White-Nose Syndrome. PLoS ONE, 7(6), p7.] (f) Outline the relationship between date of death and the mean interval between hibernation emergence periods. [1] .......................................................................... .......................................................................... (This question continues on the following page) Turn over 20EP05 –6– M18/4/BIOLO/HP2/ENG/TZ1/XX (Question 1 continued) (g) Discuss whether the data in the graph show that there is a causal link between the date of death and the interval between hibernation emergence periods. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... (h) Suggest one reason, other than the interval between hibernation emergence periods, for some infected bats surviving longer than others. [1] .......................................................................... (i) Using all of the data, predict the effect of WNS on bat populations. .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... 20EP06 [3] – – M18/4/BIOLO/HP2/ENG/TZ1/XX/M Section A Question Answers 1. a (32 × 79 =) 2528 ✔ 1. b a. same bat may be recorded more than once ✔ b. some bats may not fly over [the recording station] OR only bats flying over the station are recorded ✔ c. two bats flying close/together might be recorded as one ✔ 1. c i Notes Total 1 1 max 82 / 82.1 / 82.14 (% decline) ✔ 1 (continued…) – – M18/4/BIOLO/HP2/ENG/TZ1/XX/M (Question 1 continued) Question 1. c ii Answers Conclusion supported (2008 to 2009) M. lucifugus declines more (than L. cinereus) OR (2007 to 2009) M. lucifugus declines whereas L. cinereus increases/fluctuates/did not decline OR more affected than unaffected bats in 2007 and 2008 but more unaffected in 2009 ✔ Conclusion not supported Notes Total Award one mark maximum for an argument supporting the conclusion and one mark maximum for an argument against the conclusion. Marks should only be awarded for statements that make an explicit or clearly implied comparison between the species. 2 max other factors could be causing the difference between the species/the decrease in M. lucifugus OR there will be differences between the two bat species apart from WNS infection OR both species decreased from 2008 to 2009 ✔ 1. d a. more (frequent) interruptions/emergences from hibernation/shorter periods of hibernation/more spikes in temperature (indicating emergence) in infected bats✔ b. more fluctuation in (body) temperature (during hibernation) in infected (than uninfected bats) ✔ c. emergences/interruptions become more frequent during the hibernation period in infected bats versus (about) about the same frequency in uninfected ✔ 1 max (continued…) – – M18/4/BIOLO/HP2/ENG/TZ1/XX/M (Question 1 continued) Question 1. 1. 1. e f g Answers Notes a. energy needed to raise body temperature / lost during temperature spikes ✔ b. energy/heat released by/comes from (cell) respiration/metabolism ✔ c. food/fat (stores/reserves) used in (cell) respiration/in generating energy/heat/raising body temperature ✔ d. bats die/starve if fat/energy/food stores used up ✔ e. hibernation conserves food stores/reduces use of energy✔ f. no/little food available/food harder to find (in winter/during hibernation period) ✔ g. (more) energy/food used when flying/hunting ✔ h. (more) heat loss/hypothermia (in winter/cold weather) ✔ i. higher chance of being killed by predators when flying/emerged from hibernation ✔ later date of death with longer/bigger intervals (between hibernation emergence)/with less frequent interruptions (to hibernation) ✔ Total 3 max The correlation must be described. 1 Arguments for a causal link a. there is a trend/correlation/relationship (shown by the data in the graph) ✔ b. explanations of how more frequent emergence from hibernation could cause earlier death (are plausible)/example of an explanation ✔ Arguments against a causal link c. there is a correlation but this does not show a causal link / correlation does not prove causation d. more data/further research is needed to show the causes ✔ e. there is (much) variation/spread in the data ✔ f. other factors can affect the date of death ✔ 2 max (continued…) – – M18/4/BIOLO/HP2/ENG/TZ1/XX/M (Question 1 continued) Question 1. 1. h i Answers Notes Total a. differences in body mass OR differences in reserves/stores of food/energy/fat ✔ b. bats may be predated during a flight / chance events might affect the date of death ✔ c. more effective/stronger immune system/immunity (in some bats) ✔ d. more resistance to cold (in some bats) ✔ e. larger bats lose heat less rapidly ✔ f. infected at a different/later date ✔ 1 max higher mortality/more deaths ✔ shorter life expectancy/premature death/death before reproduction ✔ extinction/reduction in (size of) of bat populations ✔ L. cinereus/species of bats not affected by WNS may increase OR L. cinereus/species of bats not affected by WNS may experience less competition ✔ infection may affect birth rates/fertility ✔ bats will emerge more from hibernation/in winter ✔ bats will use up food/energy reserves faster in winter/faster due to (more) interruptions ✔ bat (populations) develop/evolve greater resistance to WNS ✔ 3 max a. b. c. d. e. f. g. h. –3– M18/4/BIOLO/HP2/ENG/TZ2/XX Section A Answer all questions. Answers must be written within the answer boxes provided. 1. Arabidopsis is a small flowering plant in the mustard family (Brassicaceae) that is widely used in basic research. It has a short life cycle, flowers quickly producing a large number of seeds and is easy to cultivate. It forms a circle of leaves known as a rosette that lies close to the soil. Flowers form at the end of short stems. Flower Leaf Rosette [Source: Adapted from a reproduction of a painting by the Swedish botanist C. A. M. Lindman (1856–1928), taken from his book(s) Bilder ur Nordens Flora (first edition published 1901–1905, supplemented edition 1917–1926), https://commons.wikimedia.org/wiki/File:Arabidopsis_thaliana_backtrav.jpg.] (This question continues on the following page) Turn over 20EP03 –4– M18/4/BIOLO/HP2/ENG/TZ2/XX (Question 1 continued) A study was carried out of differences in development between Arabidopsis plants grown in long days (16 hours light, 8 hours dark) or short days (8 hours light, 16 hours dark). The sixth leaf (L6) to emerge in the rosette of each plant was used in all investigations. New leaves are initiated by the meristem and go through four stages as they develop. • Stage 1 (S1) – rapid cell division • Stage 2 (S2) – cell division has ceased, cell expansion continues • Stage 3 (S3) – decreasing cell expansion rate • Stage 4 (S4) – leaf growth complete The start of each stage of leaf development for plants grown in long days and short days is shown above the first graph. 150 Long days S1 S2 S3 Short days S1 S2 S4 S3 S4 Grown in long days Mean leaf area of L6 / mm2 100 Grown in short days 50 0 0 5 10 15 20 25 30 Days after initiation of L6 35 Grown in short days 30 25 Mean number of leaves in rosette 20 15 Grown in long days 10 5 0 0 5 10 15 20 25 30 Days after initiation of L6 [Source: Adapted from K Baerenfaller, et al, (2015), “A long photoperiod relaxes energy management in Arabidopsis leaf six,” Current Plant Biology, 2, pp. 34–45. http://dx.doi.org/10.1016/j.cpb.2015.07.001. © 2015. Open access article under the CC BY license (http://creativecommons.org/licenses/by/4.0).] (This question continues on the following page) 20EP04 –5– M18/4/BIOLO/HP2/ENG/TZ2/XX (Question 1 continued) (a) (i) Calculate the difference in the mean leaf area of L6 at the start of stage 4 between the leaves of plants grown in long days and short days. [1] . . . . . . . . . . . . . . . . . . . . . . . . . . . . mm2 (ii) Distinguish between plants grown in long days and short days in the timing of the four stages of leaf development. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... (b) Distinguish between plants grown in long days and short days in the mean number of leaves per rosette during the experimental period. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) Turn over 20EP05 –6– M18/4/BIOLO/HP2/ENG/TZ2/XX (Question 1 continued) Leaves were removed from Arabidopsis plants that had been grown in long day and short day conditions and the concentration of starch within them was measured. This was done both at the end of the day (D) and at the end of the night (N) in each of the four stages of development (S1, S2, S3, S4). 120 100 Key: grown in short days grown in long days 80 Starch concentration / mol g–1 60 40 20 0 D N S1 D N S2 D N S3 D N S4 [Source: Adapted from K Baerenfaller, et al, (2015), “A long photoperiod relaxes energy management in Arabidopsis leaf six,” Current Plant Biology, 2, pp. 34–45. http://dx.doi.org/10.1016/j.cpb.2015.07.001. © 2015. Open access article under the CC BY license (http://creativecommons.org/licenses/by/4.0).] (c) Discuss the evidence provided in the bar chart for the hypothesis that plant leaves use up starch reserves for cell respiration during the night. .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) 20EP06 [2] –7– M18/4/BIOLO/HP2/ENG/TZ2/XX (Question 1 continued) (d) (i) For each of the stages, identify whether the starch concentration at the end of the day is higher in the leaves grown in long day or short day conditions. [1] .......................................................................... .......................................................................... .......................................................................... .......................................................................... (ii) Suggest reasons for the difference in end of day starch concentrations in stage 2 (S2) for the plants grown in long days and short days. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) Turn over 20EP07 –8– M18/4/BIOLO/HP2/ENG/TZ2/XX (Question 1 continued) To account for the observed phenotypic and metabolic differences, researchers analysed mRNA transcript data. They found certain transcripts over-represented in Arabidopsis plants grown in long days (dark grey) compared with the amount expected due to chance. Other types of transcripts were over-represented in Arabidopsis plants grown in short days (light grey). Biotic stress Calcium signalling Flowering hormone metabolism Nucleotide metabolism Secondary metabolism Oxidation/reduction reactions Phototrophic response Photosystem proteins Sugar transport RNA processing Amino acid transport Protein degradation Long days Short days 0 1 2 3 4 5 6 7 Probability that transcripts are over-represented / arbitrary units [Source: Adapted from K Baerenfaller, et al, (2015), “A long photoperiod relaxes energy management in Arabidopsis leaf six,” Current Plant Biology, 2, pp. 34–45. http://dx.doi.org/10.1016/j.cpb.2015.07.001. © 2015. Open access article under the CC BY license (http://creativecommons.org/licenses/by/4.0).] (e) Using the data in the bar chart, discuss the evidence for Arabidopsis plants adapting to different daylight regimes by changing the pattern of gene expression. .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) 20EP08 [3] –9– M18/4/BIOLO/HP2/ENG/TZ2/XX (Question 1 continued) (f) Using all relevant data in this question, deduce with reasons whether Arabidopsis is a long day plant or a short day plant in terms of flowering. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... Turn over 20EP09 –4– M18/4/BIOLO/HP2/ENG/TZ2/XX/M Section A Question Answers Notes Total 1 1. a i « 130 − 85 » = 45 « mm2 » ✔ Allow answers in the range of 40 to 50 «mm2» 1. a ii a. S1/S2 is longer in short day plants Accept vice versa. OR the stages in long day plants are more variable in length ✔ b. leaves of plants grown in long day reach S2 / S3 /S4 stages sooner 2 max OR S1/S2/S3 completed earlier in plants grown in long days ✔ c. leaves of plants grown in long day reach S1 later than plants grown in short days ✔ 1. 1. b c a. rosette of plant grown in long day has fewer leaves ✔ Accept vice versa. b. rosette leaf number of plant grown in long day plateaus/stays constant while the number continues to increase for plants grown in short days ✔ OWTTE. 2 max a. lower starch levels at end of night in all stages ✔ b. lower starch levels at end of night in both plants grown in short day and long days;; c. no evidence that starch is being used for respiration 2 max OR starch may have been exported/stored in other tissues/example tissue «rather than used in respiration» ✔ (continued…) – – M18/4/BIOLO/HP2/ENG/TZ2/XX/M (Question 1 continued) Question Answers Notes 1. d i higher in plants grown in short days in S1 and higher in plants grown in long days for all other stages/S2, S3 and S4 ✔ Candidates must mention all stages for the mark. 1. d ii a. leaves in plants grown in long day receive longer period of light / more leaf surface area so more photosynthesis occurs resulting in more starch ✔ Accept vice versa. b. plants in short days using starch to produce more leaves/for growth/S2 a period of rapid increase in number of leaves ✔ Accept vice versa. a. «mRNA» transcripts differ in plants grown in long days and short days ✔ Accept an example of such a transcript from the bar chart b. indicates different genes are being expressed ✔ Accept other valid reason. 1. e Total 1 2 c. plants adapt to different daylight regimes by altering gene expression ✔ 3 max d. short day length causes struggle to get enough light to photosynthesize and more «mRNA» transcripts related to photosynthesis OR plants produce large leaves rapidly when grown in long days which may result in more transcripts for biotic stress ✔ 1. f a. long day plant ✔ b. flowering hormone metabolism gene over represented in long day exposure c. fewer leaves produced «rapidly» by plant in long day as energy shifted to flower formation ✔ Accept other valid reasons from the data d. plants grown in short days produce more leaves over longer period before beginning to flower/need to use light more efficiently to photosynthesize ✔ Allow ECF if student indicates short day plant. 2 max –3– N18/4/BIOLO/HP2/ENG/TZ0/XX Section A Answer all questions. Answers must be written within the answer boxes provided. 1. Tobacco smoke contains a number of mutagens that are known to induce lung tumours in rodents, including NNK, a nitrosamine. The graph shows the relationship between NNK and lung tumour incidence in male rats. NNK was administered by subcutaneous injection for 20 weeks. Data points on the graph show percentage incidence of lung cancer in treatment groups of between 20 and 80 rats. Graph removed for copyright reasons (a) State the relationship between the dose of NNK and lung tumour incidence. [1] .......................................................................... .......................................................................... (b) Explain the effects of mutagens such as NNK. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) Turn over 20EP03 –4– N18/4/BIOLO/HP2/ENG/TZ0/XX (Question 1 continued) Mutagens can be removed from the body by converting them to readily excreted metabolites. NNK is converted to a metabolite, NNAL-Gluc, which can be used as a biomarker for the uptake of NNK. Cotinine, a metabolite of nicotine, is an indicator of tobacco smoke uptake. The graph shows the relationship between these two metabolites in the urine of 233 smokers. 45 40 35 30 Cotinine / nmol ml –1 urine 25 20 15 10 5 0 0 2 4 6 NNAL-Gluc / pmol ml 8 –1 10 12 urine [Source: Stephen S. Hecht; Tobacco Smoke Carcinogens and Lung Cancer, JNCI: Journal of the National Cancer Institute 1999; 91 (14): 1194–1210, doi:10.1093/jnci/91.14.1194. Reproduced by permission of Oxford University Press. OUP is not responsible or in any way liable for the accuracy of the translation. The International Baccalaureate Organization is solely responsible for the translation in this publication.] (c) State the highest concentration of cotinine in the urine, giving the units. .......................................................................... (This question continues on the following page) 20EP04 [1] –5– N18/4/BIOLO/HP2/ENG/TZ0/XX (Question 1 continued) (d) (i) Deduce, with a reason, whether the concentrations of cotinine and NNAL-Gluc would be higher in the urine or in the blood plasma of a smoker. [1] .......................................................................... .......................................................................... (ii) Suggest one advantage of using the urine concentration of cotinine rather than NNAL-Gluc to give a measure of the amount of tobacco smoke inhaled by a person. [1] .......................................................................... .......................................................................... (e) Rates of lung cancer in smokers are high. Discuss whether it can be concluded from the evidence in the two graphs that NNK causes lung cancer in smokers. [3] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) Turn over 20EP05 –6– N18/4/BIOLO/HP2/ENG/TZ0/XX (Question 1 continued) Nicotine addiction is the reason that people continue to smoke. Nicotine replacement therapy (NRT) is often used to help people quit smoking. Concerns about the safety of NRT led to a study where mice were given nicotine in drinking water and NNK was administered by subcutaneous injection. The table shows the effect of nicotine consumption on NNK-induced lung tumours in the mice. Table removed for copyright reasons (f) Describe the results when the mice were injected with NNK. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... (g) Evaluate the hypothesis that nicotine is not a mutagen. .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) 20EP06 [3] –7– N18/4/BIOLO/HP2/ENG/TZ0/XX (Question 1 continued) (h) Discuss the concerns about the safety of NRT. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... 2. (a) (i) H2N Label a peptide bond in the diagram of a polypeptide. H O C C H H O N C C H CH3 H O N C C H CH3 [1] H O N C C H CH2 H N C H H O C OH OH (ii) Outline the primary structure of proteins. [1] .......................................................................... .......................................................................... (b) Identify two hydrolysis reactions that occur in the small intestine. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... Turn over 20EP07 – – N18/4/BIOLO/HP2/ENG/TZ0/XX/M Section A Question 1. 1. Answers positive correlation OR lung tumour incidence increases as dose increases/OWTTE ✔ a Notes Total 1 a. NNK/mutagens alter «base» sequence of DNA/alter genes/create new alleles ✔ b b. increases rate/frequency/incidence of mutations ✔ c. in oncogenes/in genes that control cell division/mitosis ✔ 2 max d. tumors/cancers «develop/grow» if cell division/mitosis is uncontrolled ✔ 1. 42 nmol ml–1 ✔ b 1 1. d. d d i ii higher in urine due to concentration of waste products «during the process of urine production» OR higher in urine because water is reabsorbed «from glomerular filtrate/in the collecting duct»/ because «waste products» are not reabsorbed ✔ 1 higher «concentrations» so easier to measure/identify/find OR wider spread/greater range «of values/concentrations» OR units are larger/nmol rather than pmol ✔ 1 (continued...) – – N18/4/BIOLO/HP2/ENG/TZ0/XX/M (Question 1 continued) Question 1. e Answers Notes Total Answers supporting the conclusion a. first graph/data/research shows that NNK induces lung tumors/causes cancer ✔ b. second graph shows that smokers have absorbed NNK «from smoke»/shows that there is NNK in the blood of smokers ✔ Answers giving reservations about the conclusion 3 max c. results are for rats not humans ✔ d. injection of NNK and not inhalation of tobacco smoke ✔ e. dosage of NNK much larger than amounts likely in smokers ✔ f. other chemicals in smoke could cause lung cancer «in addition to those caused by NNK»/no proof that NNK is the only cause ✔ 1. f a. all/100% «of mice/in treatment group 3/in treatment groups 3 to 6» had tumours ✔ b. tumours per «significantly» increased «by NNK» ✔ 2 max c. no/little difference when nicotine was added «to mice with NNK» ✔ (continued...) – – N18/4/BIOLO/HP2/ENG/TZ0/XX/M (Question 1 continued) Question 1. g Answers Notes Total hypothesis supported (relevant treatment groups indicated with T numbers) a. «by itself» nicotine did not increase percentage «of mice» with tumours «T1 versus T2» OR percentage with tumours went down from 31 to 26 with nicotine «T1 versus T2» OR second treatment group does not have a higher percentage than first group ✔ b. «by itself» nicotine did not increase the number of tumours per mouse «T1 + T2» OR second treatment group does not have more tumours per mouse than first group ✔ c. in mice given NNK nicotine did not increase tumours «significantly» «T3 versus T4/5/6» OR 100% of rats already had tumours with NNK only «T3» OR no significant difference/more tumours per mouse in 5th treatment group than 4th/6th even though nicotine was given for a shorter time/for only 2 weeks «T5 versus T4/6» ✔ 3 max d. if nicotine was mutagenic there would have been more tumours ✔ . hypothesis not supported e. mice and humans may react to/metabolize nicotine differently/OWTTE ✔ f. nicotine from tobacco smoke may have different effects «from ingested nicotine» ✔ (continued...) – – N18/4/BIOLO/HP2/ENG/TZ0/XX/M (Question 1 continued) Question 1. Answers Notes Total a. effects of long term/longer than 46 weeks NRT not known ✔ h b. NRT/nicotine «in NRT» may have negative effects «other than cancer» on health/named other health effect/may affect unborn children ✔ c. tests needed on humans ✔ 2 max d. nicotine is addictive/causes dependency/NRT does not cure the addiction ✔ e. this research gives no grounds for concern ✔ f. data in second graph/previous study shows that nicotine is not a mutagen/does not cause cancer ✔ 2. a i Award [1] for any one of the four peptide bonds identified in this markscheme. 1 2. a ii number/sequence/order of amino acids «in a protein/polypeptide chain» ✔ 1 (continued...) –2– M19/4/BIOLO/HP2/ENG/TZ1/XX Section A Answer all questions. Answers must be written within the answer boxes provided. 1. Ebola virus disease (EVD) is the disease in humans and other primates that is caused by the Ebola virus. Fruit bats are the reservoir for the virus and are able to spread the disease without being affected. Humans can become infected by contact with fruit bats or with people infected by the virus, their body fluids or equipment used to treat them. The stacked bar graph shows the epidemiological data for the EVD cases in Conakry, the capital city of Guinea, surrounding suburbs and rural areas in Guinea from the beginning of January 2014 to the end of March 2015. 200 Key: rural areas suburbs Conakry 180 160 140 120 Number of cases 100 80 60 40 20 0 1 5 10 15 20 25 30 2014 35 40 45 50 1 5 10 2015 Time / weeks [Source: Adriana Rico, et al. “Epidemiology of Epidemic Ebola Virus Disease in Conakry and Surrounding Prefectures, Guinea, 2014–2015.” Emerging Infectious Diseases 22.2 (2016): 178–183. PMC. Web. 23 Mar. 2017. https://wwwnc.cdc.gov/eid/article/22/2/15-1304_article] (a) Identify the week and year in which the first cases were recorded in the suburbs. Week: ................................................................... Year: ................................................................... (This question continues on the following page) 20EP02 [1] –3– M19/4/BIOLO/HP2/ENG/TZ1/XX (Question 1 continued) (b) Based on the graph, compare and contrast the progress of the epidemic in the suburbs and rural areas. [3] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (c) Suggest two reasons for the overall decline in the epidemic after week 51. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) Turn over 20EP03 –4– M19/4/BIOLO/HP2/ENG/TZ1/XX (Question 1 continued) The table summarizes epidemiological data from Guinea during the Ebola outbreak in 2014. The data are based on figures supplied by Ebola treatment centres. The last column refers to people who died in places other than Ebola treatment centres. Number of cases Total Male Female Fatal cases at the Ebola treatment centres /% Conakry 553 307 246 40 18 Coyah 236 112 124 47 19 Forecariah 335 155 180 53 27 Kindia 108 45 63 60 16 Location Capital city Suburbs Fatal cases outside Ebola treatment centres /% [Source: Adriana Rico, et al. “Epidemiology of Epidemic Ebola Virus Disease in Conakry and Surrounding Prefectures, Guinea, 2014–2015.” Emerging Infectious Diseases 22.2 (2016): 178–183. PMC. Web. 23 Mar. 2017. https://wwwnc.cdc.gov/eid/article/22/2/15-1304_article] (d) Compare and contrast the data for Conakry with the data for the three suburbs. .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) 20EP04 [2] –5– M19/4/BIOLO/HP2/ENG/TZ1/XX (Question 1 continued) (e) Suggest reasons for the high percentage of fatal cases at Ebola treatment centres. [3] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... An antiviral drug, T-705, was tested in order to establish whether it has potential to treat EVD. The graph shows the data from an in vitro trial of T-705 on cells that had been infected with Ebola virus five days previously. Virus concentration and live cells are shown as a percentage of the control. 100 10 % Key: 1 virus concentration live cells 0.10 0.01 0.1 1.0 10 100 T-705 concentration 1000 M [Source: Oestereich, Lisa & Rieger, Toni & Neumann, Melanie & Bernreuther, Christian & Lehmann, Maria & Krasemann, Susanne & Wurr, Stephanie & Emmerich, Petra & de Lamballerie, Xavier & Ölschläger, Stephan & Günther, Stephan. (2014). Evaluation of Antiviral Efficacy of Ribavirin, Arbidol, and T-705 (Favipiravir) in a Mouse Model for Crimean-Congo Hemorrhagic Fever. PLoS neglected tropical diseases. 8. e2804. 10.1371/journal.pntd.0002804.] (f) Based on these data, outline the evidence that T-705 has potential to be used as a treatment for EVD. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) Turn over 20EP05 –6– M19/4/BIOLO/HP2/ENG/TZ1/XX (Question 1 continued) In 2015, an experimental vaccine was trialled in Guinea in an area where new Ebola cases continue to develop. Among the nearly 6000 people who accepted the vaccine, no cases were recorded after vaccination. In comparison, there were 23 cases among those who did not accept the vaccine. (g) Explain how vaccination can lead to the production of B cells specific to the Ebola virus. [3] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (h) Suggest possible reasons for the difficulty of preventing or controlling a viral epidemic such as the 2014 EVD epidemic in a remote rural region. .......................................................................... .......................................................................... .......................................................................... .......................................................................... 20EP06 [2] – – M19/4/BIOLO/HP2/ENG/TZ1/XX/M Section A Question Answers 1. a week 34 AND 2014 ✔ 1. b a. start of epidemic/first cases in rural areas OR epidemic spread to suburbs later ✔ Notes both needed Total 1 b. higher maximum number of cases/greater increase in rural areas OR converse for suburbs ✔ c. increase came earlier in rural areas «than suburbs» OR number of cases peaked earlier in rural areas OR more cases in rural areas «than suburbs» in 2014 ✔ 3 max d. decrease came earlier in rural areas «than suburbs» OR decreasing in rural areas but not in suburbs in 2015/by end of study period OR more cases in suburbs than rural areas in 2015 ✔ e. «large» fluctuations in both ✔ (continued…) – – M19/4/BIOLO/HP2/ENG/TZ1/XX/M (Question 1 continued) Question 1. c Answers Notes Total a. «overall decline due to» fewer cases in rural areas ✔ Answers relating to people who died from the disease or develop immunity to it: b. fewer cases due to deaths of people who had the disease/people recovering OR more people vaccinated/became immune/made antibodies/were not vulnerable to infection ✔ Answers relating to health care workers or availability of resources: c. more doctors/nurses/medical equipment/treatment centers/hospitals/spending/aid/NGOs ✔ Answers relating to medical techniques used to tackle the epidemic: d. better treatments/infection control/hygiene/quarantine/new vaccine/new antiviral drugs ✔ 2 max Answers relating to the public and patients: e. education/better awareness/avoidance of infection/taking precautions/vaccination accepted ✔ Answers relating to reservoirs of infection: f. fewer infected people «who could spread infection»/fewer bats/less contact with bats ✔ (continued…) – – M19/4/BIOLO/HP2/ENG/TZ1/XX/M (Question 1 continued) Question 1. d Answers Notes Total differences: a. Conakry has more cases than any of the suburbs OR more cases in total in the suburbs than in Conakry ✔ b. more male cases in Conakry whereas more female cases in suburbs ✔ c. higher «% of» fatal cases at Ebola treatment centers in suburbs than in Conakry ✔ 2 max similarity: d. in both Conakry and suburbs «% of» fatal cases in treatment centers is higher than outside ✔ (continued…) – – M19/4/BIOLO/HP2/ENG/TZ1/XX/M (Question 1 continued) Question 1. e Answers Notes Total a. most serious cases are in/are taken to treatment centers OR treatment centers are set up where there are most cases/most serious cases ✔ b. long time/distance to travel between contracting disease and arrival at treatment center OR travel to treatment center weakens/upsets/harms the patient ✔ c. Ebola is a virulent disease/Ebola virus mutated «to become virulent» OR little known about Ebola/new disease so treatments not yet developed ✔ d. no/not enough vaccine/antiviral drug available «in 2014/15» OR antibiotics do not work against viral diseases ✔ 3 max e. secondary infections/Ebola patients infected with other diseases/other Ebola strains OR ineffective hygiene/cleaning/sterilization/use of contaminated equipment/disposal of corpses ✔ f. small number of staff relative to patients/treatment centers overcrowded/swamped with patients OR insufficient equipment/supplies for large number of patients/with the rapid rise in patients ✔ g. better reporting at Ebola centers/deaths due to Ebola not reported in rural areas ✔ (continued…) – – M19/4/BIOLO/HP2/ENG/TZ1/XX/M (Question 1 continued) Question 1. f Answers Notes Total a. cells not killed/few cells killed «even at high concentrations» ✔ b. «T-705» effective/viruses reduced/viruses killed at 100 μM OR «T-705» very effective/viruses much reduced/nearly all viruses killed at 1000 μM ✔ 2 max c. virus concentration decreases as T-705 concentration increases ✔ d. drug has «high» potential for treatment «at high enough concentration» ✔ 1. g a. vaccine contains Ebola antigens ✔ b. vaccine «could» contain weakened/attenuated/dead/killed form of «Ebola» virus/virus genetically modified to express an Ebola/viral protein ✔ c. phagocyte/macrophage engulfs the antigen/presents the antigen to T cell ✔ d. antigen recognized by «specific» T cells/binds to T cells ✔ 3 max e. «activated» T cells activate «specific) B cells ✔ f. «activated» B cells make the antibodies «against Ebola» ✔ g. B cells divide forming «clone of» plasma cells/producing more B cells specific to Ebola ✔ (continued…) – – M19/4/BIOLO/HP2/ENG/TZ1/XX/M (Question 1 continued) Question 1. h Answers Notes Total a. poor transport infrastructure/poor communication/bad roads/difficult access/no maps/support slow arriving/scattered population ✔ b. poor education/understanding of disease amongst health workers/local population OR continued contact with infected people / other example of unsafe actions ✔ c. more sources of infection such as bats/difficult to find sources of infection ✔ d. lack of/limited access to medical care/doctors/health care workers ✔ e. lack of/no access to/unaffordability of treatment centers/medical supplies/equipment/antivirals/drugs/vaccine/treatments ✔ f. refusal/reluctance in local population to be vaccinated OR difficult to find/reach everyone to vaccinate them/repeat the vaccination ✔ g. migration of people spreads the infection ✔ h. poor sanitation/lack of clean water ✔ 2 max –3– M19/4/BIOLO/HP2/ENG/TZ2/XX Section A Answer all questions. Answers must be written within the answer boxes provided. 1. Organisms often release chemicals when attacked as part of their defence system. Scientists studied lima bean plants (Phaseolus lunatus) infested with either an armyworm, Spodoptera exigua, or a herbivorous mite, Tetranychus urticae. Both organisms feed on lima bean leaves, causing the leaves to release chemicals. Herbivorous mite Armyworm Lima bean [Source: https://commons.wikimedia.org/wiki/File:Phaseolus_lunatus_Blanco2.369.png] The study was conducted to see which defence chemicals were produced by lima bean leaves when infested by armyworms or herbivorous mites. The scientists identified a mixture of compounds (C) released by the plant when attacked. Two of the chemicals in this mixture were identified (C1 and C2). The scientists hypothesized that the defence chemicals in C act as signals to produce other chemicals (X, Y and Z) that are also involved in the defence of the plant. (This question continues on the following page) Turn over 20EP03 –4– M19/4/BIOLO/HP2/ENG/TZ2/XX (Question 1 continued) 10 5 0 15 Amount produced / arbitrary units Amount produced / arbitrary units 15 Leaves infested with armyworm Leaves treated with C1 only 5 5 X Y Z Chemicals produced by lima bean leaves Leaves treated with C2 only 10 5 0 X Y Z Chemicals produced by lima bean leaves Leaves infested with herbivorous mite 10 15 10 0 15 0 X Y Z Chemicals produced by lima bean leaves Amount produced / arbitrary units Amount produced / arbitrary units The graphs show the amounts of chemicals X, Y and Z produced when the plants were infested by either one of the two herbivores or treated with the different chemicals C1 or C2. X Y Z Chemicals produced by lima bean leaves [Source: R Ozawa and G Arimura, Involvement of Jasmonate- and Salicylate-Related Signaling Pathways for the Production of Specific Herbivore-Induced Volatiles in Plants, Plant and Cell Physiology, 2000, 41, 4, 391–398, by permission of Oxford University Press] (a) Distinguish between the data for the leaves infested with the armyworm and the leaves infested with the herbivorous mite. .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) 20EP04 [3] –5– M19/4/BIOLO/HP2/ENG/TZ2/XX (Question 1 continued) (b) Compare and contrast the effects of treatment of the leaves using C1 and C2 with the effects of infestation. [3] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... RNA was collected from leaves of the plants after each treatment (armyworm, herbivorous mite and the chemicals C1 and C2). DNA copies of the extracted RNA were made by a process called reverse transcription. Targeted genes in the DNA were then amplified. (c) Identify the process that was used to amplify the targeted genes. [1] .......................................................................... (This question continues on the following page) Turn over 20EP05 –6– M19/4/BIOLO/HP2/ENG/TZ2/XX (Question 1 continued) The scientists then used the transcribed DNA of each treatment to study the activation of three genes of the plants’ defence system. The DNA was separated by gel electrophoresis. The activation was tested one hour after treatment and again after 24 hours. C1 1 Herbivorous mite infestation C2 24 1 24 1 24 Armyworm infestation 1 24 (hours) Gene 1 642 bp Gene 2 412 bp Gene 3 302 bp [Source: R Ozawa and G Arimura, Involvement of Jasmonate- and Salicylate-Related Signaling Pathways for the Production of Specific Herbivore-Induced Volatiles in Plants, Plant and Cell Physiology, 2000, 41, 4, 391–398, by permission of Oxford University Press] (d) Deduce, with a reason, which gene is first transcribed when exposed to C2. .......................................................................... .......................................................................... (This question continues on the following page) 20EP06 [1] –7– M19/4/BIOLO/HP2/ENG/TZ2/XX (Question 1 continued) (e) Each gene is activated by one or more of the treatments. From the gel electrophoresis data, discuss the impact of the herbivorous mite infestation on gene activation compared to treatment with C1 and C2. [3] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (f) Using the gene activation data, deduce, giving two reasons, whether the armyworm or the herbivorous mite has infested lima bean plants over a longer period of time. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... Turn over 20EP07 – – M19/4/BIOLO/HP2/ENG/TZ2/XX/M Section A Question 1. Answers a Notes Total Clear distinction required not simple lists of values. a. armyworm «infestation» produced more X than Y than Z/decreasing amounts AND herbivorous mite showed the opposite pattern/more Z than Y than X ✔ 3 max b. armyworm «infestation» produced more X than herbivorous mite ✔ c. armyworm «infestation» produced more Y than herbivorous mite / Y is the middle value for both ✔ Accept OWTTE. d. armyworm «infestation» produced less Z than the herbivorous mite ✔ e. other valid distinction ✔ For mp b-d accept vice versa. (continued…) – – M19/4/BIOLO/HP2/ENG/TZ2/XX/M (Question 1 continued) 1. b Clear comparison required between herbivore infestation and chemical treatment not simple lists of values. a. C1 caused the leaf to produce two of the same chemicals/Y and Z as the attack of herbivorous mites in a similar pattern «but in lower quantities» ✔ OWTTE 3 max b. C1 produces the least «total» amount of chemicals of all the treatments ✔ c. C2 has very similar pattern to those caused by the armyworms «but in lower quantities» ✔ OWTTE d. both herbivores caused a greater production of chemicals/all three chemicals compared to either C1 or C2 ✔ e. armyworms cause the greatest total amount of chemical production of any of the other treatments ✔ f. other valid comparison of chemical effect versus herbivore effect ✔ 1. c PCR 1. d gene 1 is first transcribed «after C2 treatment» as it shows activation after one hour ✔ Accept RT-PCR. 1 1 (continued…) – – M19/4/BIOLO/HP2/ENG/TZ2/XX/M (Question 1 continued) Question 1. e Answers Notes Total a. herbivorous mites induce activation of gene 2 first «at 1 hour» and then also gene 1 and gene 3 «at 24 hours» OR herbivorous mite «infestation» is the only treatment to affect all three genes/leads to greater gene expression overall ✔ b. gene 2 activation similar for mite and C1 «at both 1 and 24 hours» ✔ c. gene 3 3 max activation similar for mite and C2 «both at 24 hours» ✔ d. gene 1 activation slower for mite compared to C2 but more intense (than C2 at Both parts OWTTE required for mpd. 24 hours) ✔ e. gene 1 and gene 3 expressed in higher amounts «after 24 hours» in mite infestation compared to C2 ✔ 1. f a. the greater «gene expression» response of the lima bean plant to the mite infestation indicates a longer evolutionary relationship ✔ b. herbivorous mites cause more genes to be expressed/higher intensity of gene activation ✔ c. herbivorous mites cause a more immediate/earlier response in gene activation ✔ OWTTE. 2 max –2– N19/4/BIOLO/HP2/ENG/TZ0/XX Section A Answer all questions� Answers must be written within the answer boxes provided� 1. The black-legged tick (Ixodes scapularis) is an arthropod which sucks blood from humans and other mammals� It is encountered mainly in wooded and semi-wooded areas� Some ticks can be infected by the bacterium Borrelia burgdorferi� When a tick bites a human, the bacterium is often introduced, causing Lyme disease� Lyme disease is a public health problem in North America and, if left untreated, can cause important neurological impairment� The diagram represents the two-year life cycle of a tick� Spring Eggs Second year Eggs ear st y r i F Larva feeds on small mammals and birds Adult feeds on medium to large animals, including humans Autumn [Source: Cary Institute of Ecosystem Studies / Leslie Tumblety] (This question continues on the following page) 20EP02 Summer Winter tion fec f in ko is tr fo s an um rh Nymph feeds on mammals, including humans Hi gh es –3– N19/4/BIOLO/HP2/ENG/TZ0/XX (Question 1 continued) (a) State the domain into which ticks are classified. [1] �������������������������������������������������������������������������� (b) Using information from the text, identify one possible simple treatment for Lyme disease� [1] �������������������������������������������������������������������������� �������������������������������������������������������������������������� (This question continues on the following page) Turn over 20EP03 –4– N19/4/BIOLO/HP2/ENG/TZ0/XX (Question 1 continued) Scientists fear that global warming will change the distribution range of ticks� The graphs show the developmental stages of ticks throughout seasons in a densely human-populated area of south-eastern Canada, surrounded by woods (circled on the map)� Values are already established for 2000 and are predicted for 2080� 60 Proportion of the annual total number of ticks (of each stage) present each month / % 50 2000 – statistics 40 30 20 10 0 J F M A winter M J J spring A S O summer N D autumn Month 60 Larva Nymph Adult 50 40 Key for development stages: 2080 – prediction 30 20 10 0 J F M winter A M J J spring A S summer O N D autumn Month [Source: reprinted from International Journal for Parasitology, 36(1), N�H� Ogden, A� Maarouf, I�K� Barker, M� Bigras-Poulin, L�R� Lindsay, M�G� Morshed, C�J� O’Callaghan, F� Ramay, D� Waltner-Toews, D�F� Charron, Climate change and the potential for range expansion of the Lyme disease vector Ixodes scapularis in Canada, 63–70, Copyright (2006), with permission from Elsevier] (c) Identify the month when small birds had the greatest chance of being infected by B. burgdorferi bacteria in the year 2000 and the month when they would be most likely to become infected according to the 2080 predictions� 2000: � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � 2080: � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � (This question continues on the following page) 20EP04 [1] –5– N19/4/BIOLO/HP2/ENG/TZ0/XX (Question 1 continued) (d) Using the life cycle diagram and the graph for the year 2000, analyse the distribution of adult ticks throughout the different seasons. [2] �������������������������������������������������������������������������� �������������������������������������������������������������������������� �������������������������������������������������������������������������� �������������������������������������������������������������������������� (e) Evaluate the effect of the change in distribution of the different life stages of ticks on the spread of Lyme disease in south-eastern Canada� [3] �������������������������������������������������������������������������� �������������������������������������������������������������������������� �������������������������������������������������������������������������� �������������������������������������������������������������������������� �������������������������������������������������������������������������� �������������������������������������������������������������������������� (This question continues on the following page) Turn over 20EP05 –6– N19/4/BIOLO/HP2/ENG/TZ0/XX (Question 1 continued) Antibodies to B. burgdorferi present in mice / arbitrary units White-footed mice (Peromyscus leucopus) in eastern North America’s wooded areas often host B. burgdorferi bacteria� To determine whether bacterial transmission from mice to tick nymphs could be prevented, mice were vaccinated with antigens from Lyme disease-causing B. burgdorferi. Scientists captured wild mice at two different sites in the woods once a month, over 4 months� Each time, they measured the levels of antibodies to B. burgdorferi present in the captured and re-captured mice, inoculated all of them, and released them into the woods� The control group was not vaccinated with B. burgdorferi antigen� 12 Key: 11 Vaccinated 10 Control 9 8 7 6 5 4 3 2 1 Month: M J J A M J J A Site 1 M J J A M J J A Site 2 [Source: Copyright (2004) National Academy of Sciences, U�S�A� An ecological approach to preventing human infection: Vaccinating wild mouse reservoirs intervenes in the Lyme disease cycle, Jean I� Tsao, J� Timothy Wootton, Jonas Bunikis, Maria Gabriela Luna, Durland Fish, Alan G� Barbour, Proceedings of the National Academy of Sciences Dec 2004, 101 (52) 18159–18164; DOI: 10�1073/pnas�0405763102] (f) (i) State the reason for performing the experiment in the months of May to August� �������������������������������������������������������������������������� �������������������������������������������������������������������������� (This question continues on the following page) 20EP06 [1] –7– N19/4/BIOLO/HP2/ENG/TZ0/XX (Question 1 continued) (ii) Suggest possible reasons for the observed pattern of presence of antibodies in vaccinated mice� [3] �������������������������������������������������������������������������� �������������������������������������������������������������������������� �������������������������������������������������������������������������� �������������������������������������������������������������������������� �������������������������������������������������������������������������� �������������������������������������������������������������������������� The summer after vaccination, the prevalence of B. burgdorferi infection in tick nymphs collected on mice from the two sites was measured� Host mice Control mice Vaccinated mice State of infection of tick nymphs with B. burgdorferi Site 1 Site 2 Infected Not infected Infected Not infected 90 87 315 288 57 49 89 121 [Source: Copyright (2004) National Academy of Sciences, U�S�A� An ecological approach to preventing human infection: Vaccinating wild mouse reservoirs intervenes in the Lyme disease cycle, Jean I� Tsao, J� Timothy Wootton, Jonas Bunikis, Maria Gabriela Luna, Durland Fish, Alan G� Barbour, Proceedings of the National Academy of Sciences Dec 2004, 101 (52) 18159–18164; DOI: 10�1073/pnas�0405763102] (g) Analyse the data on the state of infection of tick nymphs with B. burgdorferi in control and vaccinated mice� [2] �������������������������������������������������������������������������� �������������������������������������������������������������������������� �������������������������������������������������������������������������� �������������������������������������������������������������������������� �������������������������������������������������������������������������� �������������������������������������������������������������������������� (This question continues on the following page) Turn over 20EP07 –8– N19/4/BIOLO/HP2/ENG/TZ0/XX (Question 1 continued) (h) Using all the data, discuss whether inoculating mice with the antigen to B. burgdorferi could be an effective method of controlling the spread of Lyme disease. [3] �������������������������������������������������������������������������� �������������������������������������������������������������������������� �������������������������������������������������������������������������� �������������������������������������������������������������������������� �������������������������������������������������������������������������� �������������������������������������������������������������������������� 2. The diagram shows one of Thomas Hunt Morgan’s crosses of Drosophila in the early 20th century� Parents: Offspring: Grey body – normal wings heterozygous for both genes Grey body – normal wings n = 965 Black body – vestigial wings n = 944 × Black body – vestigial wings homozygous for both genes Grey body – vestigial wings n = 206 Black body – normal wings n = 185 Total offspring = 2300 (a) State the type of inheritance shown� [1] �������������������������������������������������������������������������� (b) Identify the recombinants� [1] �������������������������������������������������������������������������� �������������������������������������������������������������������������� (This question continues on the following page) 20EP08 – – N19/4/BIOLO/HP2/ENG/TZ0/XX/M Section A Question 1. Marking point Answers Notes a eukaryote ✔ b antibiotics / named antibiotics ✔ c «2000» August AND «2080» July ✔ Both required. a adults present through autumn and winter «according to the life cycle diagram» OR some adults «must» survive winter «despite graph suggesting zero» ✔ b adults peak in October «& November»/in autumn/between September and December ✔ Each mark point, requires month or season. Jan - Mar = winter Apr - Jun = spring Jul - Sep = summer Oct - Dec = autumn = fall d c e Accept eukaryotes. Total 1 1 1 2 max adults die after laying eggs in winter/beginning of spring ✔ d smaller peak/10% versus 55% peak/smaller numbers of adults in April/spring ✔ e adults absent from June to September/summer ✔ a nymphs present through most of year/longer period/from March to November/through spring and summer «so more risk of infection» ✔ b more adults in winter/in January/February so more risk of infection then ✔ c infection will be possible through more/most months of/throughout the year ✔ d Lyme disease likely to/will increase ✔ Do not accept that there are the lowest number or no adults in winter. 3 max (continued...) – – N19/4/BIOLO/HP2/ENG/TZ0/XX/M (Question 1 continued) f i ii g because nymphs are present/numbers of nymphs rise «in these months» OR build up immunity/antibodies in mice before nymphs «peak» ✔ Ignore references to larvae. a low antibody level initially as mice not previously exposed to antigen/bacteria ✔ b vaccination causes antibody production/development of immunity ✔ c increased proportion of mice have been vaccinated in each successive month ✔ d second vaccination/booster shot increases antibody level/speeds up antibody production ✔ e memory cells produced so greater/faster antibody production ✔ Ignore any references to non-vaccinated/control mice – this means that no marks are awarded for them because the question is about vaccinated mice, but there is no penalty for including this information in an answer. f many/rising numbers of nymphs which may spread the bacteria/antigens to mice ✔ a at Site 1 there is little/no significant difference in the proportion of infected nymphs/numbers of infected and uninfected nymphs collected from both control and vaccinated mice ✔ 1 3 max 2 max (continued...) – – N19/4/BIOLO/HP2/ENG/TZ0/XX/M (Question 1 continued) h b at Site 2 the proportion of infected nymphs is lower in those collected from vaccinated than control mice OR at Site 2 «significantly» more nymphs are not infected from vaccinated than control mice ✔ c at both sites there are fewer infected than uninfected nymphs in those collected from both vaccinated and control mice ✔ d proportion of infected nymphs is lower at Site 1 than Site 2 in nymphs collected from both control and vaccinated mice OR 22% of control mice and 23% of vaccinated mice with infected nymphs at Site 1 AND 39% of control mice and 29% of vaccinated mice with infected nymphs at Site 2 ✔ a Site 2 suggests that vaccination could reduce «nymph» infection rate «so method might be effective» ✔ b Site 1 suggests that vaccination does not reduce «nymph» infection rate «so method probably not effective» ✔ c effective «to some extent» as vaccination increases antibodies/immunity in mice ✔ Accept “ticks” instead of “tick nymphs” or “nymphs” Do not accept quoting of untransformed numerical data. Percentages are required for the second alternative of mpd. For mpb and mpd, accept converse answers that give the proportions/percentages of uninfected nymphs rather than infected. 3 max – – d high antibody levels needed/ many mice need to be vaccinated «for the method to be effective» ✔ e some nymphs are still infected / «absolute» numbers «rather than proportions» of infected nymphs are similar in those collected from control and vaccinated mice ✔ f there are other hosts/mammals/birds ✔ g difficult/expensive «to vaccinate many small mammals/mice» OR cheaper to use protective clothing/tick repellant/avoid wooded areas/other method ✔ N19/4/BIOLO/HP2/ENG/TZ0/XX/M –2– 8820 – 6002 Section A Answer all questions. Answers must be written within the answer boxes provided. 1. White clover (Trifolium repens) is native to Eurasia but is now a common plant found worldwide in lawns, next to roads, in pastures and similar habitats. Some T. repens plants are able to produce the toxin hydrogen cyanide (HCN) by cyanogenesis. A study at 128 sites (2509 plants) in Toronto (Canada) looked at the proportion of T. repens plants producing HCN. The sites were at regular intervals from the city centre towards rural areas. Toronto 1.0 0.8 Proportion of plants with HCN 0.6 0.4 0.2 0.0 0 10 20 30 40 50 Distance from the city centre / km (a) (i) State the distance from the city centre at which the highest proportion of plants sampled contained HCN. .......................................................................... (This question continues on the following page) 20EP02 [1] –3– 8820 – 6002 (Question 1 continued) (ii) Outline the relationship shown in the graph. [1] .......................................................................... .......................................................................... To determine whether similar patterns in cyanogenesis were seen in other cities, studies were also carried out around New York City and Boston (USA) and Montreal (Canada). Proportion of plants with HCN New York City Boston Montreal 1.0 1.0 1.0 0.8 0.8 0.8 0.6 0.6 0.6 0.4 0.4 0.4 0.2 0.2 0.2 0.0 (b) 0 10 20 30 40 Distance from the city centre / km 0.0 0 10 20 30 40 Distance from the city centre / km 0.0 0 10 20 30 40 Distance from the city centre / km Deduce whether the pattern of cyanogenesis was the same in all of the areas around all four cities. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) Turn over 20EP03 –4– 8820 – 6002 (Question 1 continued) The researchers considered two possible ecological causes for the observed gradients in cyanogenesis. First, the researchers assessed the extent of herbivory of T. repens plants in Toronto by measuring the percentage of leaf area that was eaten. The graph shows the results for both cyanogenic and non-cyanogenic plants. Mean herbivory of T. repens / % 60 40 20 Key: Non-cyanogenic Cyanogenic 10 (c) 20 30 40 50 Distance from the city centre / km Discuss whether the data supports the hypothesis that the gradient in cyanogenesis is due to its benefits against herbivory in rural areas. .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) 20EP04 [2] –5– 8820 – 6002 (Question 1 continued) The researchers then investigated a possible correlation between cyanogenesis and exposure to freezing conditions. It had been proposed that when a cyanogenic plant freezes, its cells burst, releasing HCN which is toxic to the plant. Snow can insulate the ground and plants from freezing temperatures. However, snow is more likely to melt in cities, which then exposes plants to freezing temperatures. All four of the cities studied receive below freezing temperatures and winter snowfall. Researchers looked at the number of days below freezing (0 °C) that did not have snow cover in these cities. 16 12 Mean number of days below 0 °C without snow cover 8 4 0 Toronto (d) (i) New York City Boston Montreal Identify with a reason the city where the plants were more insulated from freezing temperatures. [1] .......................................................................... .......................................................................... (This question continues on the following page) Turn over 20EP05 –6– 8820 – 6002 (Question 1 continued) (ii) Using all of the data so far, suggest whether exposure to freezing temperatures in the four cities is supported as a reason for the differences in HCN production in T. repens. .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) 20EP06 [2] –7– 8820 – 6002 (Question 1 continued) Removed for copyright reasons (e) (i) (This question continues on the following page) Turn over 20EP07 –8– 8820 – 6002 (Question 1 continued) (i) (ii) Removed for copyright reasons 2. (a) (i) Outline how the amphipathic properties of phospholipids play a role in membrane structure. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... (ii) State the role of cholesterol in animal cell membranes. .......................................................................... (This question continues on the following page) 20EP08 [1] – – N20/4/BIOLO/HP2/ENG/TZ0/XX/M Section A Question Answer Notes Total 1 a i 32 (km); a ii positive correlation / (proportion with) HCN increases as distance increases; b a Toronto NYC and Boston show same pattern/all show positive correlation/relationship/WTTE; b Montreal shows negative correlation/negative relationship/WTTE so is different; c a (hypothesis) not supported; b large overlap/little difference between cyanogenic and non-cyanogenic (in herbivory); c smaller difference between cyanogenic and non-cyanogenic in rural areas; d both show negative correlation between herbivory and distance from city center/same trend; e some support/hypothesis partly supported by lower herbivory in cyanogenic (at all distances); Accept answers in the range of 31 to 33 (km) 1 1 2 max Do not accept ‘No’ unqualified as an answer. Do not award mpe if the answer states that the hypothesis is supported without doubt/fully 2 max (continued...) – – N20/4/BIOLO/HP2/ENG/TZ0/XX/M (Question 1 continued) d i Montreal because it has the lowest number of days below 0°C without snow cover; d ii a cities with more days without snow cover have positive correlation between distance from city center and HCN / vice versa for Montreal; b fewer plants with HCN within cities that have more days without snow cover/have more exposure to freezing temperatures / converse for Montreal which has fewer days without snow cover; c HCN is 0.2 (or less) HCN in cities that have more days without snow cover proportion whereas city with fewer days/Montreal it is 0.5/more than 0.4; d in Toronto cyanogenic and non-cyanogenic plants show little difference in herbivory; e support for hypothesis/exposure to freezing temperatures as reason; Do not award the mark for Montreal if the reason is not given 1 2 max Do not accept ‘Yes’ unqualified as an answer, but accept it if supported by reasoning e i hours of daylight/light intensity/soil pH/soil nutrients/mineral nutrients/watering regime/humidity/ /population plants came from/location plants were adapted to/duration of cold period/ /size of plant pot/size of plant/wind speed/carbon dioxide concentration; Mark the first answer given only. Reject nutrition and sunlight unqualified. Reject number of plants. 1 e ii Lowest: non-cyanogenic and produces A; Highest: non-cyanogenic and produces neither (chemical); Do not accept genotypes 2 (continued...) – – N20/4/BIOLO/HP2/ENG/TZ0/XX/M (Question 1 continued) e iii Comparing the first and second bar with the third and fourth: a higher survival with aa than Aa / allele/gene A lowers survival; b higher survival if substrate/chemical A not produced / substrate/chemical A lowers survival; Comparing the first bar with the second: c no significant difference between A_L_ and A_ll / if A is present L/l doesn’t affect survival; d conversion of substrate/chemical A to cyanide does not affect survival; Comparing the fourth bar with the third: e higher survival with aall than aaL_ / if A is not present L reduces survival; f lowers survival if chemical L/enzyme produced even if substrate/chemical A not produced; Comparing the fourth bar with the other three: g highest survival with aall/homozygous recessive h chemicals A and L/substrate and enzyme both lower survival; Do not accept L decreases survival unqualified as it does not for plants with allele A. Do not accept lower survival in A_ll than A_Ll as the difference is unlikely to be significant. Do not accept ranking the four genotypes in order of survival probability without more analysis. Accept any of these points if given as the converse, for example for mph ‘highest survival without either chemical. Accept ‘withstand freezing’ in place of ‘survival’ in any mark point. 3 max –2– 2221 – 6008 Section A Answer all questions. Answers must be written within the answer boxes provided. 1. Mitosis in cancerous tissues is uncontrolled and the number of cells undergoing mitosis increases exponentially. Rapidly dividing cells in root tips can be used as a model for studying the effects of anticancer drugs. Aqueous extracts of the fruit of avocado (Persea americana) and the leaves of crepe jasmine (Tabernaemontana divaricata) have been shown to be toxic to some human cancer cell lines. Root tips of the broad bean (Vicia faba) were exposed to these aqueous plant extracts over a range of concentrations. The mitotic index is the percentage of cells examined undergoing mitosis. The table shows the numbers of broad bean root cells examined and the mitotic indices over a range of plant extract concentrations. Avocado (P. americana) Crepe jasmine (T. divaricata) Extract concentration / ppm Number of cells examined Mitotic index /% Number of cells examined Mitotic index /% Control 3017 8.30 3017 8.30 100 3516 8.08 3000 8.00 1250 3450 7.68 3076 6.83 2500 4322 6.94 3089 6.31 5000 4200 4.29 3014 4.90 10 000 4023 3.11 3020 4.00 20 000 3697 2.11 3009 3.72 [Source: Republished with permission of SPRINGER-VERLAG DORDRECHT, from Do cancer cells in human and meristematic cells in plant exhibit similar responses toward plant extracts with cytotoxic activities?, Cytotechnology, Noha S. Khalifa, Hoda S. Barakat, Salwa Elhallouty, Dina Salem, Volume 67, 2015; permission conveyed through Copyright Clearance Center, Inc.] (a) (i) Calculate the number of cells undergoing mitosis using 100 ppm of crepe jasmine extract. [1] .......................................................................... (ii) Describe the effect of avocado extract on mitotic index. .......................................................................... .......................................................................... (This question continues on the following page) 20EP02 [1] –3– 2221 – 6008 (Question 1 continued) (b) Compare and contrast the effect of increasing extract concentration on the mitotic indices for the two plant extracts. [3] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) Turn over 20EP03 –4– 2221 – 6008 (Question 1 continued) The percentage of cells in different stages of mitosis in the root tips was also recorded. 60 50 40 Stages of mitosis 30 /% 20 10 0 Control 100 1250 2500 5000 10 000 20 000 Concentrations of avocado extract / ppm 60 50 40 Stages of mitosis 30 /% 20 10 0 Key: 1250 2500 5000 10 000 20 000 Control 100 Concentrations of crepe jasmine extract / ppm prophase metaphase anaphase and telophase [Source: Republished with permission of SPRINGER-VERLAG DORDRECHT, from Do cancer cells in human and meristematic cells in plant exhibit similar responses toward plant extracts with cytotoxic activities?, Cytotechnology, Noha S. Khalifa, Hoda S. Barakat, Salwa Elhallouty, Dina Salem, Volume 67, 2015; permission conveyed through Copyright Clearance Center, Inc.] (c) Outline the effect of avocado extract concentration on the percentage of cells in the different stages of mitosis. .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) 20EP04 [2] –5– 2221 – 6008 (Question 1 continued) (d) Using both the table of mitotic indices and the bar charts, deduce whether these extracts contain chemicals that block mitosis in broad bean root tips. [3] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) Turn over 20EP05 –6– 2221 – 6008 (Question 1 continued) The mechanism of action of vinblastine, an anticancer drug, was investigated over a range of concentrations. Vinblastine is an alkaloid isolated from the periwinkle plant (Catharansus roseus). The percentage of cells in mitosis and ratio of anaphase to metaphase in cells exposed to this drug in vitro for a fixed time were recorded. The data are displayed in two graphs. 0.30 Anaphase : metaphase ratio Cells in mitosis / % 100 80 60 40 20 0 0.1 0.25 0.20 0.15 0.10 0.05 0 0.1 1 10 Vinblastine concentration / nM 1 10 Vinblastine concentration / nM [Source: Republished with permission of American Society for Pharmacology and Experimental Therapeutics, from Mechanism of Mitotic Block and Inhibition of Cell Proliferation by the Semisynthetic Vinca Alkaloids Vinorelbine and Its Newer Derivative Vinflunine, Molecular Pharmacology, Vivian K. Ngan, Krista Bellman, Bridget T. Hill, Leslie Wilson and Mary Ann Jordan, Volume 60 , Issue 1, 2001; permission conveyed through Copyright Clearance Center, Inc.] (e) By referring to both graphs, evaluate the hypothesis that vinblastine targets cells in mitosis and prevents them from completing the process. .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) 20EP06 [3] –7– 2221 – 6008 (Question 1 continued) (f) Some anticancer drugs inhibit mitosis by blocking the formation of the spindle. Suggest one other way in which vinblastine could block mitosis. [1] .......................................................................... .......................................................................... (g) Discuss one advantage and one disadvantage of using plant tissue to investigate drugs intended to treat cancer in humans. Advantage: [2] .............................................................. Disadvantage: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Turn over 20EP07 – – M21/4/BIOLO/HP2/ENG/TZ1/XX/M Section A Question Answers 1. a i 240 ✔ 1. a ii reduces it ✔ 1. b similarities a. both extracts reduce the mitotic index/percentage of cells undergoing mitosis OR Reject ‘about 240’ and any answer other than 240 1 1 Accept MI for mitotic index and accept crepe or jasmine for T. divaricata. For similarities in mpa the answer must refer to the different extracts together, not in separate parts of the answer. differences b. avocado (extract) more effective/reduces MI more (at higher concentrations) For mpb or mpc the differences can be expressed in the reverse, for example jasmine less effective at higher concentrations for mpb. lower MI with avocado (than jasmine) above 2500/at 5000 to 20000/at 20000 ✔ Do not award marks for quoting figures without a statement of the similarity or difference. c. at lower concentrations jasmine (extract) more effective/reduces MI more For the first alternatives of mpb and mpc, do not accept ‘greater rate’ instead of ‘more’. OR lower MI with jasmine (than avocado) below 5000/at 100 to 2500 ✔ c Total mitotic index decreases as concentration of both extracts increases / negative correlations ✔ OR 1. Notes a. % in prophase increased (at higher extract concentrations) ✔ b. % in metaphase decreased (slightly) ✔ c. % in anaphase with telophase decreased ✔ Do not award mpa for ‘prophase percentage is highest’ The initial rise of metaphase % is unlikely to be significant so this and other fluctuations are not included in the mark scheme. 3 max 2 max (continued…) – – M21/4/BIOLO/HP2/ENG/TZ1/XX/M (Question 1 continued) Question 1. d Answers a. yes / extracts (do contain chemicals that) block mitosis (in broad bean root tips) ✔ evidence from table of mitotic indices Notes Total Do not award mpa if no attempt at deduction is presented in the answer. This is a deduce question so mpb to b. lower MI shows (both) extracts prevent cells from undergoing/entering/starting mpd not awarded for only describing mitosis ✔ of the data. evidence from bar charts c. avocado increases prophase % indicating progression to metaphase/through mitosis slowed/blocked 3 max OR avocado decreases metaphase and anaphase-telophase % indicating progression from prophase/through mitosis slowed/blocked ✔ d. jasmine increases metaphase % so progression to anaphase/through mitosis slowed/blocked OR jasmine reduces prophase % so (entry to/start of) mitosis slowed/blocked ✔ 1. e evaluation of evidence in graph on left a. increase in (percentage of) cells in mitosis (as vinblastine concentration rises) ✔ b. supports hypothesis that cells get stuck in/cannot complete mitosis ✔ evaluation of evidence in graph on right Do not allow mpa if the candidate is arguing that the hypothesis is not supported. 3 max c. drop in anaphase-metaphase ratio due to fewer cells in anaphase/more cells in metaphase ✔ d. cells not progressing from metaphase to anaphase/get stuck in metaphase ✔ (continued…) – – M21/4/BIOLO/HP2/ENG/TZ1/XX/M (Question 1 continued) Question 1. f Answers a. causes microtubules/spindle fibres to break up / tubulin molecules to depolymerize ✔ b. prevents contraction of spindle microtubules/fibres ✔ c. disrupts/damages kinetochores/centromeres/microtubule motors/centrioles/centrosomes ✔ d. prevents separation/pulling apart of (sister) chromatids/chromosomes/centromeres ✔ Notes Total Mark the first suggestion only in the answer. Do not allow answers about DNA replication or other processes that precede mitosis. 1 max e. prevents microtubules/spindle binding to chromatids/chromosomes/centromeres/DNA ✔ 1. g advantage: avoids risks for humans/harm to humans / more ethical (than with human patients/volunteers) ✔ disadvantage: differences between plant and human cells so humans may not respond in same way OR plants have cell wall/no centrioles/other relevant difference between plant and human cells ✔ Not enough for mpa to say ‘not using humans’ For mpb there must be either a statement that differences between cells may cause a different response, or a specific example of a cell difference. 2 –2– 8821 – 6002 Section A Answer all questions. Answers must be written within the answer boxes provided. 1. A community living in the water of an estuary was used to investigate how climate change may affect ecological systems. The food web in this community included phytoplankton (producers), zooplankton (consumers) and saprotrophic bacteria. Small plastic mesocosms were set up with water from the estuary containing only these three groups of organisms. The mesocosms were subjected to four different temperatures and two nutrient levels (control and nutrients added) to replicate local variations of the conditions in the estuary during springtime warming. The graph shows the biomass of the community for each of the eight mesocosms at the end of the experimental period. Biomass was measured in terms of the amount of carbon present. The horizontal line indicates the initial biomass. Key: Nutrients added Control 1.5 1.0 Biomass / mg C dm–3 0.5 Initial biomass 0.0 20 21 22 23 24 25 26 27 28 Temperature / °C [Source: adapted from O’Connor, M.I., Piehler, M.F., Leech, D.M., Anton, A. and Bruno, J.F., 2009. PLOS Biology, [e-journal] 7(9). https://doi.org/10.1371/journal.pbio.1000178.] (This question continues on the following page) 20EP02 –3– 8821 – 6002 (Question 1 continued) (a) Describe the effect of temperature on the total biomass. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) Turn over 20EP03 –4– 8821 – 6002 (Question 1 continued) The graph shows the mean biomass of autotrophs and heterotrophs in the eight mesocosms. The horizontal lines indicate the initial biomasses. Key: Autotrophs with added nutrients Autotrophs with no added nutrients (control) 1.5 Heterotrophs with added nutrients Heterotrophs with no added nutrients (control) 1.0 Biomass / mg C dm–3 0.5 Autotrophs initial biomass Heterotrophs initial biomass 0.0 20 21 22 23 24 25 26 27 28 Temperature / °C [Source: adapted from O’Connor, M.I., Piehler, M.F., Leech, D.M., Anton, A. and Bruno, J.F., 2009. PLOS Biology, [e-journal] 7(9). https://doi.org/10.1371/journal.pbio.1000178.] (b) Compare and contrast the effects of temperature on the biomass of autotrophs and heterotrophs with added nutrients. .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) 20EP04 [2] –5– 8821 – 6002 (Question 1 continued) Chlorophyll concentration was used as an estimate of the photosynthetic capacity of the community. The rate of photosynthesis and mass of chlorophyll per unit volume were measured in a mesocosm at three different temperatures. 3.0 2.5 Rate of photosynthesis per mass of chlorophyll / mg C mg Chl–1 h–1 2.0 1.5 1.0 20 21 22 23 24 25 26 27 28 Temperature / °C [Source: adapted from O’Connor, M.I., Piehler, M.F., Leech, D.M., Anton, A. and Bruno, J.F., 2009. PLOS Biology, [e-journal] 7(9). https://doi.org/10.1371/journal.pbio.1000178.] (c) Explain the effect of temperature on the rate of photosynthesis in this mesocosm. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) Turn over 20EP05 –6– 8821 – 6002 (Question 1 continued) (d) Suggest reasons for the decreases in biomass of autotrophs as temperature rises, despite the increases in photosynthesis. .......................................................................... .......................................................................... .......................................................................... .......................................................................... In a larger study, California grassland was exposed to elevated temperature and nitrate concentration for five years. The graph shows the total biomass production in individual and in combined treatments. Error bars denote one standard error. 1200 Key: Initial conditions (C) 1000 Biomass / g m–2 Increased temperature (T) 800 Increased nitrate (N) 600 Increased temperature and increased nitrate (TN) 400 200 0 C T N TN Conditions [Source: adapted from Dukes, J.S., Chiariello, N.R., Cleland, E.E., Moore, L.A., Shaw, M.R., Thayer,S., Tobeck, T., Mooney, H.A. and Field, C.B., 2005. PLOS Biology, 3(10), e319. https://doi.org/10.1371/journal.pbio.0030319.] (This question continues on the following page) 20EP06 [2] –7– 8821 – 6002 (Question 1 continued) (e) Describe the effects of temperature and nitrate concentration on biomass. [3] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (f) Suggest two abiotic factors, other than temperature and nutrient supply, that may affect the production of biomass of the grasslands. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... (g) The first study used mesocosms and the second study was carried out in natural grassland. Discuss the use of mesocosms as opposed to a study in a natural environment. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... Turn over 20EP07 –4– N21/4/BIOLO/HP2/ENG/TZ0/XX/M Section A Question 1. a b Answers a negative correlation/decrease (in biomass) as temperature rises in added-nutrients (mesocosms); b little/no (significant) change in biomass as temperature increases in control (mesocosms); OR d Total 2 max a autotroph biomass decreases and heterotroph biomass increases with higher temperatures; b decrease in autotrophs is greater/larger/more than increase in heterotrophs little difference in biomass (between auto and heterotrophs) at highest temperature/27°C; c autotrophs show smaller and smaller gains in biomass from initial as temperature rises/WTTE; d heterotrophs no gain in biomass at 21°C then larger and larger gains as temperature rises; c Notes 2 max rate of photosynthesis increases as temperature rises because: a temperature is the limiting factor for photosynthesis; b higher temperatures increase enzyme activity; c faster molecular motion/more molecular kinetic energy/more frequent enzyme-substrate collisions; d Calvin cycle/light independent reactions (of photosynthesis) speed up; 2 max biomass of autotrophs decreases as temperature rises because of: a more herbivory/grazing/feeding by (zooplankton/heterotrophs); b higher populations/numbers/biomass of zooplankton/heterotrophs; c more mortality/more decomposition/decay of autotrophs/phytoplankton; d respiration (rate higher than photosynthesis rate in autotrophs/phytoplankton); 2 max (continued...) – – N21/4/BIOLO/HP2/ENG/TZ0/XX/M (Question 1 continued) Question 1. e f g Answers Notes a increased temperature raises biomass; b increased nitrate raises biomass more than increased temperature ; c increased nitrate and temperature raises biomass by same amount as nitrate alone; a b c d water availability/rainfall/humidity; light/sunlight (intensity) / daylength; salinity of soil / high/low soil pH; chemical pollution/herbicides/allelopathy/parasitic weeds; advantages of mesocosms/converse problems with studies in natural environments a easier to manipulate/control variables/conditions / less susceptible to outside influences OR easier to replicate OR take up less space; Total 3 max Mark the first two answers only. Do not accept carbon dioxide or weather conditions. 2 max Allow only one mark for an advantage and one mark for a disadvantage as this is a discuss question. 2 max disadvantages of mesocosms/converse opportunities with studies in natural environments b some trophic levels missing/incomplete food chains in mesocosms OR large animals cannot be included / ethical concerns about enclosing animals in mesocosms OR some variables lacking in mesocosms / doesn’t show what happens in natural ecosystems; –2– 2222 – 6008 Section A Answer all questions. Answers must be written within the answer boxes provided. 1. In winter when temperatures drop, brown bears (Ursus arctos) enter a cave and hibernate. The graph shows the mean values for the body temperature, heart rate and activity of 14 brown bears throughout the year. The grey shaded areas indicate the periods when the bears are in transition between hibernation and normal activity. 20 10 Ambient temperature / °C 0 –10 –20 38 Body temperature / °C 36 34 32 100 75 Heart rate / beats min–1 50 25 0 100 75 Activity / arbitrary units 50 25 0 1 July 2010 1 July 2011 1 January 2011 1 January 2012 Date 1 July 2012 1 January 2013 [Source: Adapted from Evans, A.L., Singh, N.J., Friebe, A., Arnemo, J.M., Laske, T.G., Fröbert, O., Swenson, J.E. and Blanc, S., 2016. Drivers of hibernation in the brown bear. Frontiers in Zoology, 13(7). This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/).] (This question continues on the following page) 20EP02 –3– 2222 – 6008 (Question 1 continued) (a) Estimate the difference between the highest and lowest mean body temperatures. [1] .......................................................................... (b) Compare and contrast the changes in mean ambient and body temperatures during 2012. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... (c) Explain the change in heart rate during the period of hibernation. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) Turn over 20EP03 –4– 2222 – 6008 (Question 1 continued) It was observed that during hibernation bears are not susceptible to osteoporosis, a condition characterized by a decrease in the density of bone, resulting in porous and fragile bones. This condition may develop in humans during long periods of inactivity and with increasing age. The graph shows the porosity of the tibia bones of black bears (Ursus americanus) and humans at different ages. Age is expressed as the proportion of normal life span. 10 8 Porosity /% Key: human black bear 6 4 2 0 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Age / proportion of normal life span [Source: Adapted from Journal of Biomechanics, 39(8), Donahue, S.W., McGee, M.E., Harvey, K.B., Vaughan, M.R. and Robbins, T., Hibernating bears as a model for preventing disuse osteoporosis, pp.1480–1488. Copyright (2006), with permission from Elsevier.] (d) Distinguish between the changes in porosity of the bones in humans and bears as age increases. [1] .......................................................................... .......................................................................... (e) The life expectancy of a human at the time of the study was 80 years. Estimate the porosity of the bones of the individual who was approximately 32 years old. [1] .......................................................................... (f) The researchers assessed age as a proportion of normal life span, rather than in years. Suggest one reason for this. .......................................................................... .......................................................................... (This question continues on the following page) 20EP04 [1] –5– 2222 – 6008 (Question 1 continued) Healthy bone is constantly being broken down (bone resorption) and at the same time being rebuilt (bone formation). Bone mass should therefore not normally change, nor bone diseases occur. To test whether these processes occurred during hibernation in black bears, blood serum was tested for the markers ICTP (indicating bone resorption) and PICP (indicating bone formation). 1.0 0.8 Concentration of markers 0.6 in blood serum / arbitrary units 0.4 0.2 0.0 Key: Hibernation period ICTP bone resorption PICP bone formation [Source: Republished with permission of Company of Biologists Ltd, from Parathyroid hormone may maintain bone formation in hibernating black bears (Ursus americanus) to prevent disuse osteoporosis. Donahue, Seth W; Galley, Sarah A; Vaughan, Michael R; Patterson-Buckendahl, Patricia; Demers, Laurence M; Vance, Josef L;McGee, Meghan E, Journal of experimental biology, 01 May 2006, Vol. 209, Issue Pt 9, pages 1630–1638; permission conveyed through Copyright Clearance Center, Inc.] (g) Describe what is happening to the bone during hibernation. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... (h) Suggest how the graph would differ for a human during a long period of inactivity. [1] .......................................................................... .......................................................................... (This question continues on the following page) Turn over 20EP05 –6– 2222 – 6008 (Question 1 continued) Osteocalcin is a peptide hormone that causes calcium to bind in the bones, so is involved in bone formation and regeneration. Research shows that changes in the mean concentration of osteocalcin in blood serum occur before and after hibernation in bears. In this research, concentration of both osteocalcin and parathyroid hormone were measured in the blood serum of bears. Results are shown in the bar chart and graph. 100 Osteocalcin / ng ml –1 Osteocalcin / ng ml –1 160 80 60 40 20 140 120 100 80 60 40 20 0 0 Prehibernation Hibernation 0 Posthibernation 20 40 60 80 100 120 Parathyroid hormone / pg ml –1 [Source: Republished with permission of Company of Biologists Ltd, from Parathyroid hormone may maintain bone formation in hibernating black bears (Ursus americanus) to prevent disuse osteoporosis. Donahue, Seth W; Galley, Sarah A; Vaughan, Michael R; Patterson-Buckendahl, Patricia; Demers, Laurence M; Vance, Josef L;McGee, Meghan E, Journal of experimental biology, 01 May 2006, Vol. 209, Issue Pt 9, pages 1630–1638; permission conveyed through Copyright Clearance Center, Inc.] (i) Calculate the percentage increase in the mean concentration of osteocalcin from pre-hibernation to hibernation. [1] .......................................................................... (j) A hypothesis has been proposed that an increase in parathyroid hormone concentration causes an increase in osteocalcin in bears. Evaluate the evidence for this hypothesis provided by the data. .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) 20EP06 [2] –7– 2222 – 6008 (Question 1 continued) (k) Discuss how helpful these studies of bears can be in developing an understanding of osteoporosis in humans. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... Turn over 20EP07 –5– M22/4/BIOLO/HP2/ENG/TZ1/XX/M Section A Question 1. 1. a b Answers 5 °C; Notes Total Units required. Accept answers in the range 4.5 to 5.5 °C Accept one similarity: a. both rise and then fall / both fall with hibernation and rise with activity b. both reach minimum during hibernation and maximum during activity; c. both lowest in January/February / both rise from January/February; 2 max Accept one difference: d. one peak of ambient temp but body temp has two peaks / OWTTE; e. body temp remains maximal for longer/plateaus whereas ambient peaks; f. body temperature is always higher than ambient temperature; g. ambient range is greater than body temperature range / OWTTE; 1. c 1 a. decreased/slower heart rate because bears less active/use less energy; b. less (cell) respiration / lower (rate of) metabolism; c. less oxygen/glucose required / less CO2 produced/needing to be removed; d. less muscle contraction/muscles require less blood; a. e. conserves energy; 2 max 1. d porosity increased in humans and decreased in bears; Both needed 1 1. e 6 %; Accept answers in the range 6.0 to 6.5 %. Percentage sign required 1 (continued…) –6– M22/4/BIOLO/HP2/ENG/TZ1/XX/M (Question 1 continued) Question 1. 1. 1. f g h Answers Notes Total a. to allow comparison of bears and humans; b. bears have a different life span to humans / bears do not live to 80 years; c. because they age at different rates; 1 max a. resorption/breaking down occurs and formation/rebuilding occurs; b. at similar rates / more resorption at most times; c. no/little (overall) change (in bone mass); d. lag between bone resorption rising and formation rising / OWTTE; e. bone resorption rising towards end as formation dropping / OWTTE; 2 max a. more resorption than formation; b. PICP/bone formation (always) lower (than in bears); c. ICTP/bone resorption (always) higher (than in bears); d. ICTP above PICP by a greater amount in humans (than in bears); 1. i 250 % (Allow 240 to 260%) 1. j a. (hypothesis supported by) positive/direct correlation/direct relationship (between parathyroid hormone and osteocalcin) OR osteocalcin rises as parathyroid hormone rises/vice versa; b. no evidence for causal link / causal link cannot be assumed / correlation does not prove causation; OR no evidence that parathyroid hormone causes change in osteocalcin; OR other factors may cause change in osteocalcin; 1 max 1 2 (continued…) –7– M22/4/BIOLO/HP2/ENG/TZ1/XX/M (Question 1 continued) Question 1. k Answers Notes Accept one reason for the studies being helpful: a. (helps us understand how)) bears avoid osteoporosis; b. bone structure of bears and humans is similar / both are mammals; c. suggests that hormones/osteocalcin/parathyroid hormone might be a (preventative) treatment; Total 2 max Accept one reason for the studies not being helpful: d. humans do not hibernate / are not inactive for long periods; e. humans live for much longer; 2. a 0.87 2. b a. short-term reading could show global temperatures falling while the trend is rising; b. fluctuations from year to year may not show long-term trend; 2. c a. short wave radiation from sun passes through atmosphere/is not absorbed by CO2; b. infrared/long wave (radiation)/heat emitted from/released from (surface of) Earth; c. CO2 in the atmosphere absorbs infrared/long wave (radiation) / heat cannot pass through the greenhouse gases; d. this results in warm/increased temperatures on Earth/global warming; Accept values from 0.8 to 0.9 1 1 max 3 max –2– 2222 – 6014 Section A Answer all questions. Answers must be written within the answer boxes provided. 1. Increases in the frequency and severity of drought are part of climate change in many areas of the world. Drought represents one of the major threats to food security as it can drastically decrease crop yield. Water stress occurs when the demand for water exceeds its availability. A water stress index of 0.0 indicates non-water-stressed plants with normal transpiration and 1.0 is maximum water stress with much less transpiration. (a) Define transpiration. [1] .......................................................................... .......................................................................... A study was carried out on sorghum (Sorghum bicolor), an important cereal crop. The sorghum plants were grown for 15 weeks after the date of planting. Flowering occurred in week 9. There were 3 treatment groups in the study: y Control: plants were watered throughout the study y Pre-flowering drought: no water until week 9, followed by normal watering y Post-flowering drought: normal amounts of water until week 9, but none after. Removed for copyright reasons (This question continues on the following page) 20EP02 –3– 2222 – 6014 (Question 1 continued) (b) (i) Compare the changes in water stress of the pre- and post-flowering drought plants over the period shown on the graph. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (ii) Using the data, evaluate the hypothesis that sorghum plants are more vulnerable to drought after flowering. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) Turn over 20EP03 –4– 2222 – 6014 (Question 1 continued) It was known that plant growth under certain drought conditions is intimately linked to microbial communities in the root and in the soil around the root. The scientists took samples from both the root and soil, identified the bacterial phyla present and classified them into two groups: Gram-positive and Gram-negative bacteria. The graph shows the abundance in the root of the three most common Gram-positive phyla, a, b and c, and the three most common Gram-negative phyla, d, e and f, found at week 8 (before flowering), under control conditions and pre-flowering drought conditions. Removed for copyright reasons (c) Distinguish between pre-flowering drought plants and control plants in terms of the effect of water availability on the relative abundance of Gram-positive and Gram-negative bacteria in the root. .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) 20EP04 [1] –5– 2222 – 6014 (Question 1 continued) The relative abundance of Gram-positive and Gram-negative bacteria in pre-flowering drought conditions was compared over time inside the root and in the soil around the root. Removed for copyright reasons [Source: adapted from Xu, L., et al., 2018. PNAS, 115(18), Supporting information appendix. reference redacted.] (d) (i) Compare and contrast the relative abundance of Gram-negative bacteria in the soil and the roots of pre-flowering drought plants. [3] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (ii) Suggest a reason for the changes in relative abundance of bacteria in the soil around the root between week 8 and week 9. [1] .......................................................................... .......................................................................... (This question continues on the following page) Turn over 20EP05 –6– 2222 – 6014 (Question 1 continued) Scientists inoculated the roots of the sorghum plants with one of two different species of Gram-positive bacteria. One set of plants was grown under drought conditions and the control with normal water. They compared the fresh mass of the roots of these two groups of plants. Removed for copyright reasons [Source: adapted from Xu, L., et al., 2018. PNAS, 115(18), Supporting information appendix. reference redacted.] (e) (i) Deduce the effect of drought on the fresh mass of the roots that have not been inoculated (X). .......................................................................... .......................................................................... (This question continues on the following page) 20EP06 [1] –7– 2222 – 6014 (Question 1 continued) (ii) Compare and contrast the effect of the inoculations with Gram-positive I (Y) and Gram-positive II (Z) on the fresh mass of control and drought roots. [3] .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... .......................................................................... (iii) Suggest a reason for the observed effects of the inoculations in sorghum plants under drought conditions. [1] .......................................................................... .......................................................................... (iv) Suggest an advantage of using bacterial inoculation, as shown in this study, over traditional selective breeding to obtain crops that are more resistant to drought. [1] .......................................................................... .......................................................................... Turn over 20EP07 –5– M22/4/BIOLO/HP2/ENG/TZ2/XX/M Section A Question 1. a 1. b 1. b Answers Notes loss of water/evaporation through the stomata/leaves; i ii a. in both groups drought/lack of water causes (significant) increase in water stress index; b. in both groups, with water, water stress index close to values for the control/not significantly different; c. both groups have no changes between weeks 14 and 15 / values remain constant in weeks 14 and 15; a. (the hypothesis is supported as) more immediate response to drought in post-flowering plants than pre-flowering; b. at week 5 of pre-flowering drought the stress index has only reached 0.15 whereas after two weeks of post-flowering drought it is 0.5 and/or after five weeks it is nearly 0.8 Total 1 Only accept similarities; 2 OWTTE Accept OWTTE for valid contrasts OR larger/higher/greater level response to drought in post-flowering plants than pre-flowering; c. stress index reaches a maximum of 0.56 pre-flowering but 0.78 post-flowering / much higher at week 15/end of study; d. stress index remains high for post flowering; 1. c Gram-positive more common in pre-flowering drought while Gram-negative more common in control; Gram-negative e and f never present in (pre-flowering) drought; 2 max OWTTE 1 (continued…) –6– M22/4/BIOLO/HP2/ENG/TZ2/XX/M (Question 1 continued) 1. d i Similarities: [2 max] a. both groups have an increase from week 1 to week 2; b. there is an (overall) decrease (in the relative abundance of Gram-negative) in both after week 2 / lower abundance from weeks 3 to 8; c. both increase (greatly) after week 8/starting from week 9/flowering/end of drought period/with water onwards; For [3 max] answer must include a difference (mp g) d. both have similar abundance after week 8/from week 9/end of drought period onwards; 3 max e. both plateau in the last weeks; f. other correct similarity e.g., the overall pattern is similar for both roots and soil over the entire study / overall similar trend; Difference: g. the soil has more Gram-negative than the root in the drought period/up to week 8/until flowering / OWTTE; 1. d ii a. Gram-positive are more resistant/better adapted to drought conditions OR Gram-negative are more resistant/better adapted to conditions with water; b. Gram-positive outcompete the Gram-negative in drought conditions OR Gram-negative outcompete Gram-positive in conditions with water; Accept vice versa 1 max c. water allows for greater metabolism/reproduction of Gram-negative; 1. e i drought causes a (very large) drop in the fresh mass (compared to controls) / decrease in range of fresh mass; 1 (continued…) –7– M22/4/BIOLO/HP2/ENG/TZ2/XX/M (Question 1 continued) 1. e ii a. in drought condition root, both inoculations have a higher mean of root fresh mass compared to no inoculation; b. in control conditions root, a (slightly) lower mean (of the root mass) in both inoculations compared to no inoculation; Award [2 max] if only similarities or only differences c. II/Z has a (slightly) higher mean than I/Y in the drought root; d. Inoculation increases the range of fresh mass values (in both cases) / more in control conditions; 1. e iii 3 a. Gram-positive bacteria may have a symbiotic/positive relationship with the sorghum; b. Gram-positive bacteria may provide plants with oxygen/more nutrients/change pH; c. Gram-positive bacteria may help to retain/absorb more water; 1. e iv faster/cheaper to develop to resistance to drought / known way to increase yield during drought; 1 max OWTTE 1 –2– 8822 – 6002 Section A Answer all questions. Answers must be written within the answer boxes provided. 1. Common shrews (Sorex araneus) are small mammals found in Northern Europe. Their diet includes insects, slugs, spiders, worms and amphibians. They do not hibernate in winter because their bodies are too small to store sufficient fat reserves. [Source: [Shrew], n.d. [image online] Available at: https://www.pxfuel.com/en/free-photo-jslkw [Accessed 29 October 2021].] To study brain size in shrews, researchers anesthetize them, X-ray their skulls and measure the height of the braincase (BCH) where the brain is located. The graph shows the relationship between BCH and the brain mass of individual adult shrews. Brain mass / arbitrary units 0.039 0.036 0.033 P 0.030 0.027 0.75 0.80 0.85 0.90 BCH / arbitrary units [Source: adapted from Lázaro, J., Hertel, M., LaPoint, S., Wikelski, M., Stiehler, M. and Dechmann, D.K.N., 2018. Journal of Experimental Biology 221. http://doi.org/10.1242/jeb.166595.] (This question continues on the following page) 24EP02 –3– 8822 – 6002 (Question 1 continued) (a) State the relationship between BCH and brain mass of shrews. [1] .......................................................................... .......................................................................... (b) Outline how the shrew labelled P differs from the normal relationship between BCH and brain mass. [1] .......................................................................... .......................................................................... (c) Suggest a reason that researchers use BCH rather than brain mass to indicate brain size. [1] .......................................................................... .......................................................................... (This question continues on the following page) Turn over 24EP03 –4– 8822 – 6002 (Question 1 continued) The researchers found that the BCH of any individual adult shrew could vary seasonally. They collected shrews at different times of the year. The BCH of each shrew was compared with its body mass. The results are displayed in the chart. Key: 1.0 Summer Winter Spring 0.9 BCH / arbitrary units 0.8 0.7 5.0 7.5 10.0 12.5 15.0 17.5 Body mass / g [Source: adapted from Schaeffer, P.J., O’Mara, M.T., Breiholz, J., Keicher, L., Lázaro, J., Muturi, M., Dechmann, D.K.N., 2020. R. Soc. Open Sci. 7. http://dx.doi.org/10.1098/rsos.191989.] (This question continues on the following page) 24EP04 –5– 8822 – 6002 (Question 1 continued) (d) State the season when shrew brain mass is greatest. [1] .......................................................................... .......................................................................... (e) Compare and contrast the results for winter and spring. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... (f) Suggest a reason for the difference in BCH in summer and winter. [1] .......................................................................... .......................................................................... (This question continues on the following page) Turn over 24EP05 –6– 8822 – 6002 (Question 1 continued) Shrews were observed in different seasons and the time they spent on a particular activity was recorded and expressed as a percentage of the total observation time. The circles in the kite shapes represent the mean value of time for each activity. Walk 40 Time in activity / % Time in activity / % Run 30 20 10 0 40 30 20 10 0 Summer Winter Spring Summer Spring Eat and drink 100 Time in activity / % Time in activity / % Rest Winter 80 60 40 0 4 3 2 1 0 Summer Winter Spring Summer Winter Spring [Source: adapted from Schaeffer, P.J., O’Mara, M.T., Breiholz, J., Keicher, L., Lázaro, J., Muturi, M., Dechmann, D.K.N., 2020. R. Soc. Open Sci. 7. http://dx.doi.org/10.1098/rsos.191989.] (This question continues on the following page) 24EP06 –7– 8822 – 6002 (Question 1 continued) (g) State the activity and season that occupied the greatest mean percentage of observation time. [1] .......................................................................... .......................................................................... (h) Suggest a reason for the difference in the time observed eating and drinking. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... (This question continues on the following page) Turn over 24EP07 –8– 8822 – 6002 (Question 1 continued) The researchers were interested in the seasonal differences in searching for food. They set up a square arena with sides of 110 cm and four entrances (A, B, C and D). Containers were placed in the arena, some with food and others with no food. The diagram shows a top-down view of the arena. Key: A Container with no food Container with food D B C [Source: adapted from Lázaro, J., Hertel, M., LaPoint, S., Wikelski, M., Stiehler, M. and Dechmann, D.K.N., 2018. Journal of Experimental Biology 221. http://doi.org/10.1242/jeb.166595.] Each shrew was starved of food for two hours before its cage was opened at one of the entrances to the arena. The length of the path taken by the shrew to obtain food was measured. This was standardized by dividing the path length by the straight-line distance from the entrance to the containers with food. Each shrew was used for 10 trials. (This question continues on the following page) 24EP08 –9– 8822 – 6002 (Question 1 continued) The graph shows the standardized mean path length taken by all the shrews at different seasons of the year. The letters show where the cages were placed for each trial. 150 Standardized mean path length Key: Summer (n = 8) Winter (n = 8) 100 Spring (n = 7) 50 <0 1 D 2 B 3 A 4 B 5 D 6 C 7 B 8 A 9 C 10 A Trial and entrance [Source: adapted from Lázaro, J., Hertel, M., LaPoint, S., Wikelski, M., Stiehler, M. and Dechmann, D.K.N., 2018. Journal of Experimental Biology 221. http://doi.org/10.1242/jeb.166595.] (i) Calculate the percentage of containers that contained food. [1] .......................................................................... .......................................................................... (j) Outline a reason that the path length was standardized. [1] .......................................................................... .......................................................................... (This question continues on the following page) Turn over 24EP09 – 10 – 8822 – 6002 (Question 1 continued) (k) Compare and contrast the results for trials 2 and 9. [2] .......................................................................... .......................................................................... .......................................................................... .......................................................................... (l) With reference to all the data, suggest a reason for the difference in standardized mean path length for summer and winter. .......................................................................... .......................................................................... .......................................................................... .......................................................................... 24EP10 [2] –5– N22/4/BIOLO/HP2/ENG/TZ0/XX/M Section A Question 1. Answers a positive correlation/the greater the BCH the greater the brain mass; b a. high BCH but brain mass is low/lower than expected/lower than others with similar BCH; b. (fairly) low brain mass but BCH is high/higher than expected/higher than others with similar brain mass; c d easier to measure/doesn’t require dissection/non-invasive / shrew not harmed/killed/more ethical; Summer; Notes No mark for ‘positive relationship’ or for directly proportional Total 1 1 max The mark can awarded for one of these reasons even if it not the first reason given in the answer. 1 1 –6– N22/4/BIOLO/HP2/ENG/TZ0/XX/M Question Answers Notes 1. Compare part of answer = similarity: a. both have low BCH (compared with summer); For the second alternative in mpb, the answer must not state simply that BCH is higher in spring as there is much overlap. Do not accept quoted figures without the similarity or difference being stated. e Contrast part of answer: b. greater body mass in spring than winter; OR overall/mean/average BCH higher in spring than in winter; OR more variation in body mass in spring than winter; f a. b. c. d. e. large brain (indicated by large BCH) requires/uses much energy; shrews need/use much energy in winter (other than for the brain); much energy used in winter for keeping warm/searching for food; food/energy more abundant in summer/less abundant in winter; growth between winter and summer (so BCH larger in summer); g resting in spring; h a. b. c. d. e. more food/energy eaten/required in winter/cold; food needed to maintain temperature/stay warm/generate heat; more loss of body heat in cold conditions; more energy used hunting for food; food less available in winter/harder to find enough food; Total 2 Do not accept hibernation/lower metabolic rate/lower body temperature/lower activity levels/less food eaten in winter. Do not accept answers about body mass. 1 max 1 2 max –7– Question 1. Answers i 4(%); j compensates for the different distances between entrances and food/OWTTE; OR to enable (fair/valid) comparison/OWTTE; k N22/4/BIOLO/HP2/ENG/TZ0/XX/M Notes Total 1 Do not accept unspecific answers such as ‘to be consistent’ 1 Similarity between 2 and 9: a. winter path length longer (than spring and summer) in both (trials 2 and 9/from entrances B and C); Contrast between 2 and 9: b. path length longer in trial 2 than 9/from entrance B than entrance C (in all seasons); OR error/bar/standard deviation/variation in data greater in trial 2 than 9/from entrance B than entrance C (in all seasons); 2 –8– 1. l a. in winter shrews have smaller brains/smaller BCH / converse for summer; b. lower/poorer memory/thinking/cognitive skills/learning/intelligence/senses/sense of smell/ability to find food in winter/ converse for summer; N22/4/BIOLO/HP2/ENG/TZ0/XX/M 2
