Cambridge A Level Biology 9700 — 2006 May/June Paper 2 · Variant 1

This document consists of 12 printed pages and 4 blank pages. SP (CW/CGW) T08216/2 © UCLES 2006 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Subsidiary Level and Advanced Level BIOLOGY 9700/02 Paper 2 Structured Questions AS May/June 2006 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 in the spaces provided at the top of this page. Write in dark blue or black pen in the spaces provided on the Question Paper. You may use a soft pencil for any diagrams, graphs, or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all questions. The number of marks is given in brackets [ ] at the end of each question or part question. Centre Number Candidate Number Name For Examiner’s Use 1 2 3 4 5 TOTAL

2 9700/02/M/J/06 For Examiner’s Use © UCLES 2006 Answer all the questions. 1 Fig. 1.1 is a drawing made from an electron micrograph of a longitudinal section of a capillary in muscle tissue. A B C D E F × 8000 Fig. 1.1 (a) Complete the table below using the information in Fig. 1.1 to help you. cell A cell B cell C name of cell red blood cell function of cell ingest bacteria permit exchange of gases diameter / µm 20 7 [4] (b) Name the organelles D, E and F. D … E … F …[3]

3 9700/02/M/J/06 [Turn over For Examiner’s Use © UCLES 2006 (c) Explain how oxygen and glucose move from the blood inside the capillary to the tissue fluid in the muscle. oxygen … … … glucose … … …[3] (d) Describe how the structure of the wall of a vein differs from that of a capillary. … … … … … …[3] [Total: 13]

4 9700/02/M/J/06 For Examiner’s Use © UCLES 2006 2 Fig. 2.1 shows part of a summer squash, Cucurbita pepo. Fig. 2.2 is a high power drawing of an area of phloem from a transverse section of the stem of C. pepo. fruit Fig. 2.1 G H Fig. 2.2

5 9700/02/M/J/06 [Turn over For Examiner’s Use © UCLES 2006 (a) (i) Name G and H. G … H …[1] (ii) Describe three ways in which the structure of a xylem vessel differs from the structure of cell G. 1. … 2. … 3. …[3] (b) The liquid extracted from the phloem of C. pepo contains sucrose. Explain how sucrose is transported in the phloem along the stem from the leaf to the fruit. … … … … … … … …[4] (c) Most of the sucrose transported in the phloem enters the fruit. Suggest why summer squash fruits are not sweet. … …[1] [Total: 9]

6 9700/02/M/J/06 For Examiner’s Use © UCLES 2006 3 (a) Complete the table by indicating with a tick ( ) or a cross ( ) whether the statements apply to proteins, DNA, messenger RNA and cellulose. You should put a tick or a cross in each box of the table. statement protein DNA messenger RNA cellulose hydrogen bonds stabilise the molecule glucose is the subunit molecule subunits are joined by peptide bonds may be hydrolysed to amino acids contains uracil [5]

7 9700/02/M/J/06 [Turn over BLANK PAGE Question 3 continues on page 8

8 9700/02/M/J/06 For Examiner’s Use © UCLES 2006 During an immune response, B-lymphocytes become plasma cells and begin to make polypeptides that are assembled into antibodies. Fig. 3.1 is a diagram showing the formation of a polypeptide at a ribosome in a plasma cell. G A A G G G U U C A G tRNA ser gly lys ser val amino acid mRNA C C C U U U A G C movement of ribosome C J Fig. 3.1 (b) State the sequence of bases at J. …[1]

9 9700/02/M/J/06 [Turn over For Examiner’s Use © UCLES 2006 (c) Use the information in Fig. 3.1 to describe the role of transfer RNA molecules in translation. … … … … … … … … …[5] The bacterium that causes cholera, Vibrio cholerae, releases a toxin known as choleragen. During an immune response to cholera some B-lymphocytes produce antibodies that combine with choleragen so inactivating it. Antibodies that inactivate toxins are called antitoxins. (d) Explain how the structure of an antibody, such as the antitoxin for choleragen, makes it specific to one substance. … … … … … …[3] (e) Explain why cholera remains a significant infectious disease in some parts of the world. … … … … … …[3] [Total: 17]

10 9700/02/M/J/06 For Examiner’s Use © UCLES 2006 4 Fig. 4.1 is an electron micrograph of a chloroplast from a mesophyll cell in a leaf. Fig. 4.1 (a) Calculate the magnification of the electron micrograph in Fig. 4.1. Answer = …[1] 0.5 µm

11 9700/02/M/J/06 [Turn over For Examiner’s Use © UCLES 2006 (b) State two features visible in Fig. 4.1 that identify the organelle shown as a chloroplast. 1. … 2. …[2] (c) Chloroplasts absorb phosphate ions from the surrounding cytoplasm. Suggest one way in which chloroplasts use phosphate ions. …[1] (d) Starch grains in plant cells contain both amylose and amylopectin. Explain how both of these substances are formed from glucose in plant cells. … … … … … … … …[4] (e) State three functions of the water stored in the vacuoles of plant cells. 1. … 2. … 3. …[3] [Total: 11]

12 9700/02/M/J/06 For Examiner’s Use © UCLES 2006 5 Some bacteria that are found in soils contain the enzyme urease. Urease catalyses the hydrolysis of urea to form ammonia and carbon dioxide: urea + water carbon dioxide + ammonia Some fertilisers added to soils to help crop growth contain urea. Although some crop plants can absorb ammonium ions, most obtain their source of nitrogen as nitrate ions. (a) Describe how urea from fertilisers becomes available to plants as nitrate ions. … … … … … …[3] The activity of urease can be measured by following the increase in pH as ammonia is produced in the reaction. A student was provided with urease extracted from bacteria and solutions of urea and two chemical inhibitors, thiourea and lead nitrate. The student prepared six reaction mixtures (1 to 6) as shown in Table 5.1 in order to investigate the effect of the two chemical inhibitors on the activity of urease. Table 5.1 reaction mixture urea water thiourea lead nitrate urease boiled urease 1 2 3 4 5 6 Key = present in reaction mixture = absent from reaction mixture The student recorded an increase in pH in reaction mixtures 1 and 2. The reaction was faster in 1 than in 2. The pH in the other reaction mixtures did not change.

13 9700/02/M/J/06 For Examiner’s Use © UCLES 2006 (b) The student made some conclusions about the results from the test-tubes. Match the statements to the reaction mixtures, 1 to 6. You may use the numbers once, more than once or not at all. (i) ‘No reaction took place because urease was denatured.’ [1] (ii) ‘There was no reaction because there was no substrate for urease.’ [1] (iii) ‘The reaction did not occur because there was an inhibitor present.’ [1] Thiourea has a molecular structure that is very similar to that of urea. The student designed an experiment to find out whether thiourea is a competitive inhibitor. The student set up several reaction mixtures like 1 using increasing concentrations of urea. The student determined the initial rate of the reaction for urease at each concentration of urea. The results are shown in Fig. 5.1. initial rate of reaction concentration of urea Fig. 5.1 The student then repeated the experiment using the same concentrations of urea. However, the student added the same volume and concentration of a thiourea solution to each test-tube in place of the water. (c) Sketch a curve on Fig. 5.1 to show the results that the student would expect if thiourea acts as a competitive inhibitor of urease. [2] (d) Explain why it is important to determine the initial rate of reaction when investigating the effect of a competitive inhibitor on an enzyme. … … … …[2] [Total: 10]

16 9700/02/M/J/06 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 University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level and GCE Advanced Subsidiary Level MARK SCHEME for the May/June 2006 question paper 9700 BIOLOGY 9700/02 Paper 2 Maximum raw mark 60 This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which Examiners were initially instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began. Any substantial changes to the mark scheme that arose from these discussions will be recorded in the published Report on the Examination. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes must be read in conjunction with the question papers and the Report on the Examination. The minimum marks in these components needed for various grades were previously published with these mark schemes, but are now instead included in the Report on the Examination for this session. • CIE will not enter into discussion or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the May/June 2006 question papers for most IGCSE and GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Page 1 Mark Scheme Syllabus Paper GCE A/AS Level – May/June 2006 9700 02 © University of Cambridge International Examinations 2006 1 (a) cell A cell B cell C name of cell phagocyte / neutrophil / AW; squamous epithelial (cell) / endothelial (cell); function of cell transports, oxygen / carbon dioxide; diameter / µm to be added [4] (b) D mitochondrion; E lysosome / (Golgi) vesicle; R vacuole F nucleus; [3] (c) oxygen diffuses, down concentration gradient / from high concentration to low concentration; through, phospholipid bilayer; R protein channels glucose (pressure) filtration / AW; e.g. ‘forced out by blood pressure’ through pores, in capillaries / between capillaries; facilitated diffusion; through channel proteins / idea; through cytoplasm; [max 3] (d) assume answer is about vein unless told otherwise thicker wall / more cells / more than one cell thick; A more, squamous epithelium / endothelium valve(s); three layers / described; to max 2 (smooth) muscle; collagen; elastic tissue / elastin; R references to size, width, size of lumen, amount of blood etc. [max 3] [Total: 13]

Page 2 Mark Scheme Syllabus Paper GCE A/AS Level – May/June 2006 9700 02 © University of Cambridge International Examinations 2006 2 (a) (i) G sieve tube (element), H companion cell; [1] (ii) vessels have thicker walls; thickening in walls (e.g. spiral, annular, reticulate); wider lumen; no cytoplasm; R dead (not structure) pits; no cross walls / no sieve plates / no sieve pores; lignin; [max 3] (b) (sucrose) loaded at, source / leaf; role of companion cells; further detail, e.g. H+ pumped out, sucrose moves in through co-transporter; absorption of water / water enters by osmosis; hydrostatic pressure builds up; mass flow; (sucrose) unloaded at, sink / fruit / root / AW; gives a difference in pressure (between source and sink); [max 4] (c) sucrose used in respiration; stored as starch; used to make, cellulose; A used to make cell walls stored as / converted to, organic acids (in vacuoles); converted into named other substances; e.g. lipid / protein / AW [max 1] [Total: 9]

Page 3 Mark Scheme Syllabus Paper GCE A/AS Level – May/June 2006 9700 02 © University of Cambridge International Examinations 2006 3 (a) one mark per row statement protein DNA messenger RNA cellulose hydrogen bonds stabilise the molecule

x
; glucose is the subunit molecule x x x
; subunits are joined by peptide bonds
x x x ; may be hydrolysed to amino acids
x x x ; contains uracil x x
x ; [5] (b) CAG; [1] (c) tRNA, combines with amino acid / carries amino acid to ribosome; idea of specificity; e.g. each type of tRNA is specific to an amino acid anticodon matches amino acid idea; example from Fig. 3.1; codon on messenger RNA pairs with anticodon on tRNA; example from Fig. 3.1; two sites on ribosome; further detail; e.g. P and A site (and E) leave ribosome after amino acid joins polypeptide; continually reused; [max 5] (d) variable region; binding region to antigen; shape is specific to, choleragen / antigen; complementary; ref to R groups on amino acids (in polypeptide / protein); different, sequences of amino acids / primary structures; ref to, folding of the molecule / secondary structure / tertiary structure; [max 3] (e) poor sanitation / no treatment of faecal waste; contamination of (drinking) water supply; poverty / poor living conditions / poor hygiene / poor (health) education; ref to natural disasters; e.g. assistance / aid / medical help / AW, cannot arrive in time no rehydration therapy available (at time when needed); no (effective) vaccine; further detail; (bacteria live in gut, where immune system is not effective) [max 3] [Total: 17]

Page 4 Mark Scheme Syllabus Paper GCE A/AS Level – May/June 2006 9700 02 © University of Cambridge International Examinations 2006 4 (a) (15,000 / 0.5) x 30,000; [1] (b) starch grain; grana / thylakoids / internal membranes; shape, qualified; ‘typical chloroplast shape’ is minimum acceptable length; A range of appropriate lengths, e.g. 5 to 10 µm [max 2] (c) make ATP; A combine with ADP phospholipids; DNA / RNA / nucleotides / named nucleotide; phosphorylated sugars / triose phosphate; [max 1] (d) condensation (reaction) / described as elimination of water; glycosidic, bond / link; 1:4 in, amylose / amylopectin / both; amylose, helix / unbranched; A curved chain R straight chain amylopectin, branched; 1:6 links (to give branches); [max 4] (e) (raw material) for photosynthesis; A for photolysis maintains turgidity / provides support; pushes chloroplasts to edge of cell; used in hydrolysis reactions; solvent for, ions / named ion / pigment / named pigment; [max 3] [Total: 11]

Page 5 Mark Scheme Syllabus Paper GCE A/AS Level – May/June 2006 9700 02 © University of Cambridge International Examinations 2006 5 (a) (bacterial urease converts) urea → ammonia; ammonia → nitrite; Nitrosomonas; nitrite → to nitrate; Nitrobacter; nitrification; oxidation / chemosynthesis; [max 3] (b) (i) 6 ; [1] (ii) 5 ; [1] (iii) 3 ; [1] (c) curve starting at 0; but lower; reaches same plateau but at higher concentration of urea; [2] (d) inhibition is reversible; enzyme is still active; inhibitor fits into active site temporarily; substrate is broken down (reaction does proceed); same end point; just takes longer / reaction is slower with inhibitor; [max 2] [Total: 10] [Total mark for paper: 60]