Cambridge A Level Biology 9700 — 2009 May/June Paper 5 · Variant 1

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level This document consists of 7 printed pages and 1 blank page. SPA (GB/CG) V00656/4 © UCLES 2009 [Turn over 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 pen. You may use a soft 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. * 0 8 7 5 5 7 6 6 9 3 * BIOLOGY 9700/05 Paper 5 Planning, Analysis and Evaluation May/June 2009 1 hour 15 minutes Candidates answer on the Question Paper. For Examiner’s Use 1 2 3 Total

2 © UCLES 2009 For Examiner’s Use 9700/05/M/J/09 1 (a) Fig. 1.1 shows a simple apparatus that can be used to identify the tissue that conducts water up a stem. A cut shoot was placed in dye solution. After several hours the dye accumulated in the leaves. leafy shoot oil dye solution Fig. 1.1 (i) The stem was further investigated to locate the tissue that conducts water. Describe a method by which the tissue through which water is conducted in the stem can be identified. … … … …[2] (ii) Explain why a cut shoot is used rather than an intact plant with roots. … … … …[2]

3 © UCLES 2009 [Turn over For Examiner’s Use 9700/05/M/J/09 (b) (i) Describe how this apparatus could be set up and used to find the rate of movement of water up the stem. … … … … … … … … … … … … … … …[6] (ii) State how the rate of water movement is calculated. … …[1] (iii) Predict the effect on the rate of water movement of adding a respiratory inhibitor to the dye solution. … …[1] [Total: 12]

4 © UCLES 2009 For Examiner’s Use 9700/05/M/J/09 2 (a) A species of snail is found in a variety of habitats. The shells of these snails may have dark bands on the surface of the shell (banded) or may not have any bands on the shell (unbanded). These shells are shown in Fig. 2.1. banded snail unbanded snail Fig. 2.1 These snails are eaten by birds. The birds break the shell of the snail on a stone and eat the soft body. The shell remains are left by the stones. In an investigation all the snails were removed from two areas where trees were growing. One area had birds that feed on this snail, the other area did not. An equal number of banded and unbanded snails were released into both areas. • In the area with the predatory birds the types of broken snail shells found around the stones were counted over a two week period. • In the area without predatory birds the same number of living snails was collected and the number of banded and unbanded snails counted after two weeks. Table 2.1 shows the results of this investigation. Table 2.1 number unbanded snails banded snails broken snail shells found around stones in area with predatory birds 149 271 shells of living snails in area without predatory birds 206 214 (i) Identify the independent variable and the dependent variable in this investigation. independent … dependent …[2] (ii) Suggest an explanation for the results of this investigation. … … … … … …[3]

5 © UCLES 2009 [Turn over For Examiner’s Use 9700/05/M/J/09 (b) In another investigation in a different area, the population of this snail was investigated using a capture/recapture method. A sample of the snail population was collected and marked. These snails were then released and at a later time another sample was collected and the number of marked individuals counted. The population was estimated using the formula: number of animals marked in the first sample x total number of animals in the second sample number of marked animals in the second sample Table 2.2 shows the results of two different investigations of the same population. • In investigation 1, the second sample was collected 24 hours after release. • In investigation 2, the second sample was collected 72 hours after release. Table 2.2 number of snails investigation 1 investigation 2 total first sample all marked 255 200 total second sample 400 360 number of marked shells in second sample 150 30 estimate of total population 2 400 (i) Complete Table 2.2 by calculating the total population estimated from investigation 1. Write your answer in Table 2.2. [1] (ii) Suggest why the procedure used to estimate the population means that the estimate made from investigation 1 is more likely to be reliable than the estimate from investigation 2. … … … …[2] [Total: 8]

6 © UCLES 2009 For Examiner’s Use 9700/05/M/J/09 3 The yeast Rhodotorula glutinis produces an enzyme, α-arabinofuranosidase, that could be used in the production of compounds to enhance the flavour and smell of fruit juices. The effect of the initial pH of the culture medium on the growth rate of this yeast was tested. Three continuous culture systems were set up, each with a different initial pH. The cultures were sampled at hourly intervals for 20 hours at each pH. The mean growth rate was then calculated. The mean growth rates with their standard deviations are shown in Table 3.1. Table 3.1 pH mean growth rate /arbitrary units h–1 4.0 0.156 ± 0.001 5.2 0.197 ± 0.013 7.0 0.037 ± 0.011 (a) (i) Suggest why a continuous culture system was used to test the growth rate of this yeast. … …[1] (ii) Identify two variables that should be controlled during the fermentation. 1 … 2 …[2] (b) (i) Explain how the mean values in Table 3.1 were obtained. … …[1] (ii) State what the standard deviations tell you about these results. … … …[2] (iii) State whether or not the data in Table 3.1 indicate that the difference in growth rate of R. glutinis at pH 5.2 is significantly different from the growth rate at pH 4.0. Give a reason for your answer. … …[1]

7 © UCLES 2009 For Examiner’s Use 9700/05/M/J/09 (iv) A t-test was carried out on the results for pH 4.0 and pH 5.2 and gave the value, t = 2.4 State the number of degrees of freedom for this t-test. …[1] (c) Based on the findings of the t-test, a student concluded that pH 5.2 was optimum for the production of the enzyme α-arabinofuranosidase by R. glutinis. Suggest two reasons why this conclusion may not be valid. 1 … … 2 … …[2] [Total: 10]

8 9700/05/M/J/09 BLANK PAGE Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. University of Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2009 question paper for the guidance of teachers 9700 BIOLOGY 9700/05 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. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the May/June 2009 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 A/AS LEVEL – May/June 2009 9700 05 © UCLES 2009
; 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 Question Expected answer Extra guidance Mark AO 1 (a) (i) ref. to cutting sections of the stem; ref. to use of microscope and to find the location of the dye (in water conducting tissue/xylem); must be the idea of a section, not just a cut or looking at the cut end. allow abbreviations TS/LS allow idea of magnification allow idea of colour If any other marker used, then must have a method of locating the marker [2] M (ii) 2 of: ref. to root system not fitting into apparatus; ref. to idea that dye may not be able to pass into/across roots (to xylem)/the solution enters the xylem directly; ref. to partially permeable membrane/aw; ignore ref. to time/easier ignore ref. to other tissues in root do not allow ora in the context of time e.g. it is faster allow any ref. to a membrane transport system/property not available for dye – must be cell/root hair NOT root ignore drowned/killed roots [2] M

Page 3 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – May/June 2009 9700 05 © UCLES 2009 Question Expected answer Extra guidance Mark AO (b) (i) 6 of: Method of measuring independent variable 1. ref. to suitable method of measuring time and distance; e.g. cut sections/observing dye through stem, at known time interval and known distances / known distance and record time for dye to reach it. 2. ref. to accuracy measuring distance; e.g. using thread to measure stem and ruler in cm/mm, vernier callipers Procedure: 3. ref. to using several shoots/sequential measurements on the same shoot; 4. ref. to cutting under water/dye (to avoid air entering); Method for controlling external variables: max 2 examples 5. ref. to number/surface area of leaves; 6. temperature and suitable method e.g. temperature controlled room; 7. (light and) suitable method e.g. dark room with light of fixed illumination/ light at fixed distance; 8. (air flow and) suitable method e.g. fan set at constant speed; 9. ref to volume/concentration (dye) solution: Reliability: 10. ref. to making at least 3 measurements and taking a mean; Safety: 11. ref. to a low risk investigation; The question specifies this apparatus – if any other is used max 2. for external variables standardised/controlled plus 1 mark for correct ref. to reliability allow: measuring distance to leaf/cutting all stems to same length and timing for appearance of dye allow timing dye movement visually do not allow dye exuding from leaves do not allow dye concentration in leaf if any other marker used then must have a way of locating it Assume the name of the variable if the method of controlling a variable is correct ignore mass allow incubator do not allow water bath/air conditioning allow any standard method of standardising light allow keep out of drafts/close windows/ doors do not allow amount do not allow unqualified repeats allow ref. to a safety issue and prevention e.g. toxic dye + wear gloves [6] M (ii) (mean) distance moved by dye; (mean) time allow as a description allow ecf for respirometer/fall in dye methods [1] D (iii) the rate of water movement would be unchanged; do not allow if qualify unchanged [1] P [Total: 12] 10M 1D 1P

Page 4 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – May/June 2009 9700 05 © UCLES 2009 Question Expected answer Extra guidance Mark AO 2 (a) (i) Independent – predation (by birds); Dependant – (number of) snails of each type/banded and unbanded (eaten); allow number of broken shells each type [2] P (ii) 3 of: idea that birds predate by sight/find their prey by vision; banded snails more strongly predated/killed than non-banded; banded snails more obvious to predators/unbanded able to camouflage better; ref. to figures/control; must be a feature of the birds 271 banded and 149 unbanded/approx. double banded/no difference number of banded and unbanded in the unpredated area [3] C (b) (i) [1] D (ii) marking may make the molluscs more obvious to predators; 1 of: 72 hours means that more molluscs will be eaten (so the proportion is too low); immigration/emigration/migration may have occurred/72 hours is long enough for more movement of snail population; allow the marks may have faded ignore references to numbers marked in the two investigations allow reference to breeding [1] [1] E D [Total: 8] 2P 2D 3C 1E number of molluscs Investigation 1 Investigation 2 first sample 255 200 total second sample 400 360 marked second sample 150 30 total population 680 2 400

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – May/June 2009 9700 05 © UCLES 2009 Question Expected answer Extra guidance Mark AO 3 (a) (i) yeast maintained in exponential phase/most rapid growth; allow reference to primary metabolite [1] E (ii) 2 of: temperature; nutrient concentration; flow rate through fermenter; oxygen/air supply; allow rate of nutrient supply; do not allow amount/level of nutrient do not allow pH [2] P (b) (i) mean value calculated by adding all the values and dividing by the number in the sample; allow formula n x x Σ = allow if describe in terms of the question information, e.g. add up all the growth rates and divide by 20 [1] D (ii) ref. to spread of data around the mean; ref. to difference between the data and reliability, e.g. 5.2 is less reliable as the spread is greater; allow variation from the mean do not allow if refer to accuracy [2] D (iii) idea of: there is no overlap between the sets of data; (assume it’s yes if the answer is correct.) allow if given in terms of standard deviation/standard deviation used to show no overlap/confidence intervals/error bars do not allow percentage values [1] D (iv) 38; ignore any formulae [1] D (c) 2 of: only three pH values tested/only 2 pH values (used for T-test); no data between pH 4 and 5.2/5.2 and 7; only growth measured; yield of enzyme might be higher at different pH than optimum growth; allow the range of values is too small allow did not measure the enzyme do not allow errors in measurement due to small values/differences [2] E [Total: 10] 2P 5D 3E