Cambridge A Level Biology 9700 — 2011 Oct/Nov Paper 5 · Variant 3

This document consists of 8 printed pages and 4 blank pages. DC (CB (SE/SW)) 34655/4 © UCLES 2011 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level * 4 3 7 7 2 0 7 6 7 2 * BIOLOGY 9700/53 Paper 5 Planning, Analysis and Evaluation October/November 2011 1 hour 15 minutes Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST Write your Centre number, candidate number and name on all the work you hand in. Write in dark blue or black ink. You may use a pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. DO NOT WRITE IN ANY BARCODES. Answer all questions. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 2 Total

2 9700/53/O/N/11 © UCLES 2011 For Examiner’s Use 1 A student collected leaves from three varieties of the same species of a garden plant that has different coloured leaves. Variety A dark red leaves Variety B yellow leaves Variety C green and white striped leaves The student made a chloroplast extract from the leaves of each variety and measured the rate of photosynthesis for each extract in different wavelengths of light. The following procedure was used: Extracting the chloroplasts 1 place 10 g of leaves in a mortar with an ice cold phosphate buffer solution 2 place the mortar in an ice bath and crush the leaves using a pestle 3 filter the extract through a fine mesh into a small beaker as quickly as possible 4 completely cover the beaker with metal foil and place in an ice bath. Measuring the rate of photosynthesis 5 add 1 cm3 of methylene blue to each chloroplast extract 6 re-cover the extracts and keep in an ice bath 7 collect samples of each extract in capillary tubes and keep covered with metal foil 8 for each extract, remove the metal foil and measure the time taken for the blue colour to disappear in different wavelengths of light. Fig. 1.1 shows how the apparatus was used to measure the rate of photosynthesis for each extract. bright light source filter allowing light of known wavelength to pass through capillary tubes each containing 50 mm length of chloroplast extract metal foil covering capillary tubes before exposing to light 1 2 3 Fig. 1.1

3 9700/53/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use (a) (i) State two independent variables in this investigation. 1. … 2. … [2] (ii) State the dependent variable in this investigation. … … [1] (iii) Identify one variable that has been standardised in this investigation. … … [1] (b) State a reason for each of the following procedures: (i) using a phosphate buffer solution … … [1] (ii) keeping the extract ice cold … … [1] (iii) using a fine mesh to filter the extract. … … [1]

4 9700/53/O/N/11 © UCLES 2011 For Examiner’s Use Table 1.1 shows the student’s results. Table 1.1 time taken for blue colour to disappear / seconds (s) source of chloroplasts wavelength of light / nm dark red leaf yellow leaf green and white striped leaf 440 9 10 12 28 26 26 13 12 12 500 14 15 13 29 31 33 16 17 15 530 45 44 43 52 45 44 45 43 52 570 32 34 33 34 34 44 34 33 33 650 25 18 17 25 18 16 17 17 18 750 stayed blue after 100 seconds Table 1.2 shows the rates of photosynthesis calculated by the student. Table 1.2 rate of photosynthesis / s–1 source of chloroplasts wavelength of light / nm dark red leaf yellow leaf green and white striped leaf 440 0.097 0.038 0.081 500 0.071 0.032 0.063 530 0.023 0.023 570 0.030 0.029 0.039 650 0.057 0.049 0.058 750 0.000 0.000 0.000 (c) (i) On Table 1.1 indicate, by placing a circle around each value, two results that may be anomalous. … [2] (ii) Complete Table 1.2 by calculating the rate of photosynthesis for the green and white striped leaf at a wavelength of 530 nm. Show your working. [2]

5 9700/53/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use (d) The student analysed these results and proposed two hypotheses: • There is a pigment in red leaves which is not present in the other leaves. • Yellow leaves do not have a pigment that is found in the other leaves. To test these hypotheses the student used chromatography to separate the pigments in the leaves. (i) Outline a procedure the student could use to extract the leaf pigments and separate them by chromatography. … … … … … … … … … … … … … … … … … … … … [7]

6 9700/53/O/N/11 © UCLES 2011 For Examiner’s Use Fig. 1.2 shows the chromatograms obtained by the student. origin red leaf yellow leaf green and white leaf origin origin pigment 1 pigment 2 pigment 3 pigment 4 pigment 5 pigment 6 pigment 7 pigment 8 Key: Fig. 1.2 (ii) Use Fig. 1.2 to describe the evidence that supports the student’s hypotheses. … … … … [2] Total: [20]

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8 9700/53/O/N/11 © UCLES 2011 For Examiner’s Use 2 A variety of domestic cat does not have a tail. This condition is controlled by a single gene with two alleles. These alleles are • ‘with tail’ • ‘without tail’ Table 2.1 shows the results of four crosses between cats with tails and cats without tails Each male was crossed with several females. Table 2.1 parental phenotype offspring phenotype male female cross male females with tail without tail with tail without tail 1 without tail without tail 21 32 19 40 2 with tail with tail 65 0 70 0 3 with tail without tail 40 25 25 36 4 without tail with tail 35 27 38 36 (a) (i) Explain how the results of crosses 1 and 2 show that the allele ‘without tail’ is dominant. … … … (ii) Explain how the results of crosses 3 and 4 show that the gene for this condition is not sex linked. … … … [3] (b) Based on the results of cross 1 in Table 2.1 a student decided that: • the cats without tails are always heterozygous • the ratio of their offspring ‘with tails’ and ‘without tails’ is not as expected for this type of cross. (i) The student carried out a chi- squared test (χ2 test) to determine if the offspring ratio in cross 1 differed significantly from that expected of a cross between two heterozygotes. State one reason why the student chose this statistical test. … … [1]

9 9700/53/O/N/11 © UCLES 2011 For Examiner’s Use (ii) Complete Table 2.2 to calculate the value of χ2 for the results of cross 1. The equation for the calculation of χ2 is: (O – E) 2 O = Observed result E E = Expected result χ² = Σ Table 2.2 offspring phenotype O E (O – E)2 E offspring with tail 40 offspring without tail 72 χ² = [3] Table 2.3 shows some critical values for χ² at different probability levels. Table 2.3 degrees of freedom probability (p) 0.10 0.05 0.01 0.001 1 2.71 3.84 6.64 10.83 2 4.61 5.99 9.21 13.82 3 6.25 7.82 11.34 16.27 (iii) State why the student should look for the critical value at 1 degree of freedom in this investigation. … … [1] (iv) State the conclusion from the χ2 value calculated in (ii). … … [1] (v) Cats without tails are always heterozygous for this gene. Homozygous dominant cats are never born. Suggest what could account for the effect of the dominant allele in the homozygous condition. … … [1] Total: [10]

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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2011 question paper for the guidance of teachers 9700 BIOLOGY 9700/53 Paper 5 (Planning, Analysis and Evaluation), maximum raw mark 30 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes must be read in conjunction with the question papers and the report on the examination. • Cambridge will not enter into discussions or correspondence in connection with these mark schemes. Cambridge is publishing the mark schemes for the October/November 2011 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Page 2 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – October/November 2011 9700 53 © University of Cambridge International Examinations 2011 Mark scheme abbreviations: ; separates marking points / alternatives answers for the same point R reject A accept (for answers correctly cued by the question, or guidance for examiners) AW alternative wording (where responses vary more than usual) underline actual word given must be used by candidate (grammatical variants excepted) max indicates the maximum number of marks that can be given ora or reverse argument mp marking point (with relevant number) ecf error carried forward I ignore AVP alternative valid point (examples given as guidance)

Page 3 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – October/November 2011 9700 53 © University of Cambridge International Examinations 2011 Question Expected answer Extra guidance Mark 1 (a) (i) 1. the pigments (present in the leaves) / varieties of leaf / types of leaves / colours of leaves / source of chloroplasts / chloroplasts from different types of leaves / AW ; 2. the wavelength of light ; 1. A named pigments. R chlorophyll 2. A colours of light [2] (ii) time taken for the decolourising of methylene blue / time for loss of blue (colour) ; A idea of allows the methylene blue to work as a hydrogen acceptor A rate of photosynthesis [1] (iii) one of: mass of leaves ; volume of methylene blue ; start time of exposure to light ; intensity of light source / distance from lamp / wattage of bulb ; volume / length of extract ; (same) species of plant ; R temperature A pH R amount for methylene blue or leaves ignore length of capillary R mass of extract [max 1] (b) (i) idea of keeping the organelles intact / AW ; A explanations in terms of osmosis or water potential or pH or enzymes. ignore ref. to phosphate needed to make ATP [1] (ii) idea of inhibiting enzymes / slowing or stopping reactions ; R prevents denaturing A if answer in terms of slowing / stopping photosynthesis [1] (iii) idea of mesh traps cell debris but allows organelles through / AW ; ora that paper may not let chloroplasts through R impurities unqualified / chloroplast molecules / precipitate [1]

Page 4 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – October/November 2011 9700 53 © University of Cambridge International Examinations 2011 Question Expected answer Extra guidance Mark 2 of: time taken / seconds (s) source of chloroplasts wave length of light / nm dark red leaf yellow leaf green and white striped leaf 440 9 10 12 28 26 26 13 12 12 500 14 15 13 29 31 33 16 17 15 530 45 44 43 52 45 44 45 43 52 570 32 34 33 34 34 44 34 33 3 650 25 18 17 25 18 16 17 17 1 (c) (i) 750 Remained blue after 100 seconds If more than 2 identified : If all correct, allow both marks incorrect answers cancel correct e.g. 2 right and 1 wrong = 1 mark [max 2] (ii) 2 45 43 + = 44 ; 44 1 = 0.023 ; A alternative ways of setting out working as long as some correct working shown correct mean, no working = 1 mark A 0.021 if 52 is included in the calculation – max 1 A 0.021 in box if no calculation – max 1 A 0.022 for adding 1 divided by each value and then dividing by 3 – max 1 [2]

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – October/November 2011 9700 53 © University of Cambridge International Examinations 2011 Question Expected answer Extra guidance Mark (d) (i) 7 of: independent variable: 1. ref. to using (a sample from) all three leaves ; 2. ref to same quantity / amount of each (leaf type) ; dependent variable: 3. ref. to observing / measuring / marking / finding, position of the pigments / colours / spots (on the chromatogram) ; procedure: 4. ref. to a method of extracting pigments (from the leaves) ; 5. ref. to filtering / centrifuging to, remove debris / obtain pigments ; 6. ref. to method of concentrating extract ; 7. ref. to a method of applying sample ; 8 ref. to suitable placing in solvent ; 9. ref. to using the same solvent (if separate chromatograms) / spots at same level if all on one paper ; 10. ref. to running (chromatograms) to a set distance ; 11. ref. to covering container (prevents evaporation) ; Some points might be gained from a diagram e.g. mp 8. 2. A in terms of mass or number ecf if only two leaf types mentioned 3. A if refer to Rf values or measure distance to the pigments. A pattern for position A results for pigments, etc. 4. e.g. grind / crush / AW, leaves (separately or with solvent) / use a blender. A crushing directly onto paper A boil / heat in ethanol / alcohol / solvent 5. A ‘extract’ / supernatant, for pigments 6. e.g. by evaporating ,heating, partitioning with different solvents or (many spots) at the same point or pressing with a coin several times 7. e.g. capillary tube / fine or small dropper / small or fine paint brush / pin head A ref to keeping spot small / thin line on origin 8. e.g. solvent level below, sample / origin ignore names of solvents 9. ignore names of solvents including water, but must use water as a solvent for all chromatograms 10. e.g. before solvent front reaches the end / pre-marked line. A running for same times for 2 or more chromatograms but not if all on one, ignore any specific time 11. A airtight container. close with a stopper / cork

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – October/November 2011 9700 53 © University of Cambridge International Examinations 2011 Question Expected answer Extra guidance Mark reliability: 12. ref. to repeating to, compare / find anomalies ; safety (max 1): 13. ref. to solvents / leaves + suitable precaution ; 14. ref. to safe disposal of solvent ; 12. ignore ref. to mean unqualified A means of Rf values / AW 13. e.g. flammable – no naked flames / AW toxic – in fume cupboard / ventilated space / covered containers / gloves / goggles corrosive or allergy to leaves or solvent – gloves / goggles ignore low risk / radiation [max 7] (ii) red leaf has pigment 2 (not present the other two leaves) ; yellow leaf does not have pigment 5 (found in the other two leaves) ; ora pigment 5 only in red and green and white leaf ignore ref. to pigment 7 [2] Total: [20]

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – October/November 2011 9700 53 © University of Cambridge International Examinations 2011 Question Expected answer Extra guidance Mark 2 (a) (i) 3 of: cross 1. offspring of cats ‘without tails’ (dominant) crossed together have offspring with tails’ (recessive) ; cross 2. offspring of cats ‘with tails’ (recessive) crossed together are always ‘with tails’ (recessive) ; A cross 1 – always more without tails / AW A ora ‘none without tails’ (ii) idea of ‘with tails’ (recessive) and ‘without tail’ (dominant) phenotypes occur in (approximately) equal numbers in each sex / 1:1 ratio in each sex / 1:1:1:1 / AW ; cross 4: male ‘without tail’ (dominant) and female ‘with tail’ (recessive). If it were sex linked all males would be tailed. (This is not so), so not sex linked / AW ; must have idea that ratio is between male and female not just that there are the same number (approximately) of ‘tail’ to ‘without tail unqualified as this seems to imply regardless of gender. 1:1:1:1 implies gender so does not need qualifying. Could apply to either cross 3 or 4 a male ‘without tail’ cannot pass this allele to the male offspring if it is sex linked [max 3] (b) (i) 1 of: the data is categoric / discrete ; looking for a ‘goodness of fit’ / idea of whether expected and observed results match or not / whether there is a significant difference between the expected and observed results ; A discontinuous data / discontinuous distribution / not continuous R discontinuous variation R stating there are O and E values, must have idea of matching ratios (of offspring) Ignore ref. to null hypothesis [max 1] offspring phenotype O E ( ) E E O 2 − offspring with tail 40 28 5.14 offspring without tail 72 84 ; 1.71 ; (ii) χ2 = 6.85 (/6) ; 1 mark E column 1 mark ( ) E E O 2 − column. ecf from E A as fractions ignore decimal places 1 mark correct addition to χ2 to 2 decimal places ecf from column ( ) E E O 2 − [3]

Page 8 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – October/November 2011 9700 53 © University of Cambridge International Examinations 2011 Question Expected answer Extra guidance Mark (iii) idea of: one less degree of freedom than the number of categories / AW ; A there are two, types of data / types of offspring / phenotypes / rows / (sets of) observations / categories / (sets of) results / samples ignore any formula unqualified e.g. 2 – 1 [1] (iv) 1 of: significant ; a factor other than chance is causing the deviation from the expected ratio ; ignore references to probability A reverse argument A ecf on candidates calculated chi squared value R answers which: quantify significance. e.g. more / less significant qualify significance. e.g. 'there is a significant difference between the means' 'it is significant which improves reliability / accuracy / AW’ [max 1] (v) Idea that (the homozygous genotype) stops development / lethal gene / AW ; A any ref. to die during development / will not develop / ref. to specific possible defects e.g. abnormal spine, provided clear they kill in utero / gamete incompatibility R (gene) mutation R infertile [1] Total: [10]