Jan 2008 W1,w2,u5,6

Mark Scheme January 2008

GCE

GCE Biology (8040/9040) GCE Biology (Human) (8042/9042) International Supplement

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January 2008 Publications Code UA 019777 All the material in this publication is copyright © Edexcel Ltd 2008

Contents

MARK SCHEMES page Unit 3

6103/02 W1

Written Alternative

1

Unit 5B

6105/01

Genetics, evolution and biodiversity

5

Unit 6

6106/02 W2

Written Alternative

14

6106/03

Synoptic paper

17

Please note that this document is an International Supplement to the GCE Biology / Biology (Human) Mark Schemes (8040/8042/9040/9042) and provides Mark Schemes for Units 5 and 6, which were only available to International centres.

GENERAL INTRODUCTION Mark schemes are prepared by the Principal Examiners and revised, together with the relevant questions, by a panel of senior examiners and subject teachers. The schemes are further amended at the Standardisation meetings attended by all examiners. The Standardisation meeting ensures as far as possible that the mark scheme covers the candidates' actual responses to questions and that every examiner understands and applies it in the same way. The schemes in this document are the final mark schemes used by the examiners in this examination and include the amendments made at the meeting. They do not include any details of the discussions that took place in the meeting, nor do they include all of the possible alternative answers or equivalent statements that were considered to be worthy of credit. It is emphasised that these mark schemes are working documents that apply to these papers in this examination. Every effort is made to ensure a consistent approach to marking from one examination to another but each marking point has to be judged in the context of the candidates' responses and in relation to the other questions in the paper. It should not be assumed that future mark schemes will adopt exactly the same marking points as this one. Edexcel cannot under any circumstances discuss or comment informally on the marking of individual scripts. Any enquiries about the marks awarded to individual candidates can be dealt with only through the official Enquiry about Results procedure.

Unit 3 (6103/02 W1) Question Number 1 (a)

Answer

Mark

Time of day 08.00 10.00 12.00 14.00 16.00 18.00

Cushion temp. / 0C 8,9,8 17,15,15 20,22,23 21,21,22 19,17,20 6,5,5

Mean cushion temp. / 0C 8.3 (8.33) 15.7 (15.67) 21.7(21.67) 21.3 (21.33) 18.7 (18.67) 5.3 (5.33)

Air temp. / 0C 4 7 9 11 9 3

max 4

suitable table with time, mean cushion temp and air temp ; correct rows and columns with headings and units ; means correct to consistent sig. figs. (all correct = 2 marks, one error = 1 mark);;

Question Number 1 (b)

Answer

Mark

F line graph format with both lines on same axes ; A axes correct orientation and labelled with suitable scale ; P points plotted correctly ; L straight line joins points or good curve through all points ;

Question Number 1 (c)

Answer

4

Mark

1. cushion temperature always above air temperature ; 2. as air temperature increases AND decreases so cushion temp increases AND decreases ; 3. some relevant use of figures e.g. maximum cushion temp. 10.3 0C higher than maximum air temp. ; 4. temp / max cushion temp. almost double max air temp. / eq ;

1

max 2

Question Number 1 (d)

Answer

Mark

1. no evidence of any significant anomalies / cushion temp. falls even though air temp. continues to rise ; 2. random cushion readings do not vary by more than 30C / cushion air temp. anomaly quantified ;

Question Number 1 (e)

Answer

2

Mark

1. reference to higher temperature promoting metabolism (e.g. enzyme activity, photosynthesis) ; 2. growing points / meristems can be protected from temperature extremes inside cushion ; 3. ref to reduction in surface area / eq ;

2

max 2

Question Number

Answer

Mark

2 (a)

Points 1 & 2: reference to initial seeds/seedlings at start of study Points 3 & 4: extraction of root nodules material (from established plants) Points 5 & 6: selection of seedlings for inoculation Points 7 to 13: design, growth parameters and measurement 1.

use of stated leguminous plant / member of Papilionaceae for study (e.g. clover) ;

2.

{germinate / grow} {seeds / seedlings} in two named standard conditions ;

3.

remove nodules (from established plant of same species) ;

4.

suitable method of extraction of constant concentration of inoculum (e.g. {grind / eq} stated mass of root nodules in stated volume of water) ;

5.

stated number (5 or more) of {seedlings / seeds} for each trial ;

6.

seedlings / eq / seeds {same size / same mass / measured parameter} at start ;

7.

preparation of concentrations described (e.g. named nitrogen source / nitrate / ammonium salt / fertiliser e.g. 1 g in 100 cm3) ;

8.

same volume of nitrogen source (in every case) ;

9.

5 or more concentrations of nitrogen source used ;

10.

suitable (control) with no nitrogen added (e.g. using distilled water) ;

11.

seedlings grown for stated period (minimum 1 week) ;

12.

{count number / obtain {wet / dry} mass of root nodules} ;

13.

minimum of 2 repeats (i.e. three trials in total) ;

3

max 9

Question Number 2 (b)

Answer

Mark

1. suitable table of raw data, with correctly labelled rows and columns, including units to match the method ; 2. line graph / bar chart ; 3. correctly orientated and labelled axes including units ; max 3

4. calculation of means ;

Question Number

Answer

2 (c)

Limitations:

Mark [Sub-maximum 4]

1. difficult to know if (original nodules / extract) contain living bacteria; 2. difficult to know if {seeds / growth medium} already contain Rhizobium ; 3. difficult to get {seeds / seedlings} {to take up bacteria to develop nodules} ; 4. environmental conditions inhibit {formation of nodules (if not controlled) / enzymes (of Rhizobium) ; 5. difficult to collect / count all root nodules ; Further work:

[Sub-maximum 4]

6. the effects of root nodules / leguminous plants on the growth of crop plants per se ; 7. test effect of different concentrations of inoculant ; 8. the effects of different species of legume and fertiliser on plant growth / crop production ; 9. the effect of root nodules and types of fertiliser on the growth of other legumes / eq ; 10. a named environmental condition on root nodule formation ; 11. undertaking field trials into the relationship between Rhizobium, fertilisers and crop production ;

4

max 6

Unit 5B (6105/01) Question Number

Answer

Mark

1 (a)

(biotype) B ;

1

Question Number

Answer

Mark

1 (b)

1. (populations of biotype A and B) are separated by {behavioural / reproductive / physiological} isolation ; 2. isolation described e.g. mate at different times ; 3. no gene flow between the populations / each population genetically isolated ; 4. natural selection described e.g. biotype B flies with the gene for resistance to insecticide survive ; 5. the two gene pools become different / eq ; 6. reference to sympatric speciation ;

Question Number 1 (c)

Answer

max 4

Mark

1. reference to resistance to insecticides ; 2. no contamination (by chemicals) ; 3. no need to reapply / it is longer lasting / reference to resurgence ; 4. correct reference to specificity of control ; 5. reference to organic status ;

5

max 3

Question Number 2 (a)

Answer

Mark

A = reverse transcriptase ; B = DNA polymerase ;

Question Number 2 (b)

2

Answer

Mark

1. plasmid {opened / cut / eq} using {endonuclease / eq} ; 2. reference to sticky ends ; 3. (ends of) DNA complementary to plasmid /eq ; 4. joins by hydrogen bonding ; 5. correct reference to (DNA) ligase / formation of phosphodiester bonds ;

Question Number 2 (c)

Answer

max 3

Mark

1. idea of marker gene ; 2. (marker gene) {fluorescence / can be for antibiotic resistance / eq} ;

6

2

Question Number 3 (a)(i)

Answer

Mark

1. idea of both alleles (in heterozygote) contributing (equally) to expression (in phenotype) ; 2. { I A / allele for A} is codominant with { I B / allele for B} / I A and I B are both dominant over I O ;

Question Number 3 (a)(ii)

Answer

Mark

1. idea of more than two alleles available at a locus ; 2. idea of three alleles in blood grouping / reference to IA, IB or Io being available ;

Question Number 3 (b)(i)

Question Number 3 (b)(ii)

2

Answer

max 1

Mark

C1 = I A I B C2 = I B I O C3 = I A I B ;

1

Answer

Mark

1. I O I O is identified as O group blood ; 2. {I A I O / Ao / AO} and {I B I O / Bo / BO} ; 3. gametes from each parent shown correctly ; 4. correct use of diagram or Punnett square to show possible combinations of offspring genotypes ; 5. (this gives) one in four chance / eq ;

7

max 4

Question Number 4 (a)(i)

Answer

Mark

2250 – 240 ; (2010 / 2250 x 100 =) 89.3 ;

Question Number 4 (a)(ii)

2

Answer

Mark

1. not all the primary consumer is eaten / some die and are not eaten / eq ; 2. some (of the eaten primary consumers) {undigested / egested / lost as faeces / eq} ; 3. losses from respiration / eq ; 4. loss from {excretion / urine / urea / eq} ;

Question Number 4(b)

Answer

max 2

Mark

1. blue and red light absorbed ; 2. (blue and red) absorbed by chlorophyll ; 3. blue light absorbed by carotene ; max 3

4. green light reflected / eq ;

Question Number 4 (c)

Answer

Mark

1. acid rain {damages / eq} cuticle of leaves / damage to guard cells ; 2. causes plants to transpire more / more stressed in drought / eq ; 3. causes {leaf drop / die back / crown loss / eq} ; 4. less photosynthesis / reduced surface area for absorbing light ; 5. damages root hairs ; 6. plants unable to absorb as much {water / nutrients / minerals / eq} ;

8

max 4

Question Number 5 (a)(i)

Question Number 5 (a)(ii)

Answer

Mark

A = (mono)nucleotide ;

1

Answer

Mark

1. phosphate 2. deoxyribose 3. {nitrogenous / organic} base / eq ;; [3 correct = 2 marks, 2 correct = 1 mark]

2

Question Number

Answer

Mark

5 (b)(i)

{inter / synthesis / S} (phase) ;

1

Question Number

Answer

Mark

5 (b)(ii)

1. reference to each strand as template (for synthesis of new strands) ; 2. idea that each {daughter / eq} molecule contains one of the {parental / eq} DNA strands ; 3. and one new strand ;

Question Number 5 (b)(iii)

Answer

max 2

Mark

1. enzyme ; 2. ref to {links nucleotides / formation of phosphodiester bonds / eq} ; 3. to form new strand / eq ; 4. use of {ATP / GTP / CTP / TTP} ;

9

max 2

Question Number

Answer

Mark

5 ( c)

(mitosis) 1. reference to any of the following events not occurring ; 2. allele sequence unchanged (on daughter chromosomes) ; (meiosis) 3. reference to {chromosome pairing / formation of bivalents} ; 4. during prophase 1 ; 5. reference to exchange of {alleles / genetic material} ; 6. idea of recombinant {chromatids / chromosomes} formed / new sequences of alleles ;

10

max 4

Question Number 6 (a)(i)

Answer

Mark

1. recognisable as synapse ; 2. two correct pairs of labels ;;

Question Number

Answer

6 (a)(ii)

(mitochondria) 1. release energy / produce ATP / through aerobic respiration / oxidative phosphorylation ;

3

Mark

2. energy used in active transport / synthesis of transmitter substance / movement of vesicles ; (synaptic vesicles) 3. contain {transmitter / named transmitter} ; 4. fuse with pre-synaptic membrane / releases transmitter (into synapse);

Question Number 6 (b)

Answer

max 3

Mark

1. into blood stream ; 2. through {skin / alveoli / nose / mouth / eq} ; 3. (carried) in the plasma ; 4. correct reference to diffusion of nicotine ; 5. (diffusion from blood) into tissue fluid / eq ; 6. nicotine mimics acetylcholine / eq ; 7. binds to receptors / eq ; 8. on post-synaptic membrane ; 9.

it causes the release of adrenalin in some synapses ;

11

max 6

Question Number

Answer

Mark

7 (a)

{α / alpha} ;

1

Question Number

Answer

Mark

7 (b)

1. idea of making molecule {more reactive / able to react more easily} ; 2. by lowering activation energy ; 3. maintains concentration gradient / prevents loss of glucose from cell ;

Question Number 7 (c)

Answer

max 2

Mark

{pyruvate / pyruvic acid} ATP {NADH / reduced NAD / eq} ;; [3 correct = 2 marks, 2 correct = 1 mark]

Question Number 7 (d)(i)

Answer

2

Mark

1. binds to {hexokinase / enzyme} {not at active site / at allosteric site} ; 2. changes shape of active site ; 3. glucose no longer fits / eq ; 4. idea of more molecules of glucose-6-phosphate (as concentration increases) ; 5. causes more molecules of enzyme to be inhibited ;

12

max 3

Question Number 7 (d)(ii)

Answer

Mark

1. {reaction / phosphorylation of} glucose {slows down / stopped} ; 2. glucose remains in {cytoplasm / cell} / glucose not removed from {cytoplasm / cell} ; 3. as (more) glucose diffuses into {cytoplasm / cell}, concentration increases ; 4. diffusion {stops / slows down} because {equilibrium has been reached / no concentration gradient / eq} ;

13

max 3

Unit 6 (6106/02 W2) Question Number 1 (a)

Question Number 1 (b)

Answer

Mark Number of barnacles / m-2

14

Mean no. of barnacles / 25 cm-2 38.4

2

6

14.4

5760

3

9

23.4

9360

4

1

14.4

5760

5

12

37.6

15040

6

4

14.0

5600

7

10

32.6

13040

8

9

21.6

8640

Quadrat No.

No. of dogwhelks / m-2

1

15360

neat table correctly formatted (shaded column optional) ; correctly labelled rows and columns ; barnacle numbers correct (all correct = 2 marks, one error = 1 mark) ;;

4

Answer

Mark

A – graph correctly orientated with labelled axes ; F – scatter graph format with or without sensible line of best fit ; P – all points plotted correctly ;

Question Number 1 (c)

Answer

3

Mark

correct substitution of data ; rs = 1- 6 x 4 512 – 8

2

correct rs value = 0.9683 ;

Question Number 1(d)

Answer

Mark

there is a significant correlation between numbers of dogwhelks and numbers of barnacles ; at the 5% confidence level (not simply p=0.05) ;

14

2

Question Number 2 (a)

Answer

Mark

1. collect roots of Hieracium and wash carefully ; 2. standard mass of root placed in standard volume of water ; 3. grind in pestle, and mortar filter extract ; 4. Festuca / grass seeds of same age and single species ; 5. stated number of Festuca seeds (min 10) used in each test ; 6. stated growth medium for germination test e.g. filter paper ; 7. seeds placed fixed distance apart ; 8. stated volume of extract added to each ; 9. use of same volume distilled / deionised water as control ; 10. placed in controlled light and temperature conditions ; 11. germinated seeds counted at least every 12 hours / total germinated counted after stated time ; 12. germination defined e.g. emergence of radicle ; 13. at least 5 tests for control and extract ;

max 10

SPG Account is concise and well-organised, there is good use of technical vocabulary and almost no spelling errors = 2 marks ;; There is some lack of organisation, limited vocabulary and a number of spelling errors = 1 mark ;

Question Number 2 (b)

The account lacks organisation, there is little or no technical vocabulary and many spelling errors = 0 marks

2

Answer

Mark

1. table of raw data with suitable accurately labelled rows and columns with units ; 2. calculation of mean % germination ; 3. suitable graphical format which matches hypothesis and method with correctly labelled axes ; 4. reference to suitable type of statistical test (for significant difference) ;

15

4

Question Number

Answer

2 (c)

Limitations

Mark [sub maximum = 4]

1. difficult to judge time / actual germination ; 2. reference to differences between laboratory and field conditions ; 3. inhibitor may be unstable when extracted / eq ; 4. extract not necessarily the same concentration as in soil ; Further Work 1. test extract on growth of mature plants ; 2. test extract on different (grassland) plant species ; 3. reference to further analysis of extract to identify inhibitor ; 4. test effect of different concentrations of extract ;

16

max 5

Unit 6 (6106/03) Question Number 1 (a)

Question Number 1 (b)(i)

Question Number 1 (b)(ii)

Answer

Mark

A

14 ;

B

21 ;

2

Answer

Mark

no (viable) gametes produced / meiosis cannot occur / no (pairing of) homologous chromosomes / hybrid B has an odd number of chromosomes ;

Answer

1

Mark

1. chromosomes replicate ; 2. spindle fibres do not form / eq ; 3. chromosomes do not separate / all go to one pole / ref to non-disjunction ; 4. during mitosis / anaphase ;

Question Number 1(b)(iii)

Answer

max 2

Mark

1. (doubling of chromosomes) results in an even number of chromosomes ; 2. (doubling of chromosomes) results in formation of homologous chromosomes / ref. to pairing of homologous chromosomes ; 3. meiosis can occur / reduction division can occur ; 4. ref. to formation of gametes ;

17

max 2

Question Number 1(c)

Answer

Mark

1. reference to larger gene pool / 2. (more) genetic variation / 3. one species only reduces genetic variation / 4. may have beneficial genes / 5. may increase chances of long-term survival / eq / 6. may be able to grow in mineral deficient soils / eq ;

18

max 1

Question Number 2 (a)

Answer

Mark

1. carbon dioxide diffuses into red cells ; 2. {combines / eq} with haemoglobin ; 3. forms {carbaminohaemoglobin / carbamino compounds} ; 4. correct ref. to presence of carbonic anhydrase ; 5. carbon dioxide {combines / reacts} with water and forms carbonic acid / correct equation ; 6. carbonic acid dissociates / eq into H+ and hydrogencarbonate / correct equation ; 7. ref. to hydrogencarbonate diffusing out of red cell (into plasma) ;

Question Number 2 (b)(i) Question Number 2 (b)(ii)

Answer

max 4

Mark

medulla (oblongata) ;

1

Answer

Mark

1. volume of air breathed increases as carbon dioxide concentration increases / eq ; 2. ref. to small increase from 0.03 to 2.00% / greater change from 2 to 8 / greatest change from 6 to 8 % ; 3. credit a manipulated quantitative comment ;

Question Number 2 (b)(iii)

Answer

max 2

Mark

1. they both increase as CO2 increases / eq ; 2. volume of air breathed by seal is higher up to 6.00 % / converse ; 3. at 8.00 % carbon dioxide, volume of air breathed by seal is less than human / converse ; 4. credit a manipulated quantitative comparison ;

19

max 3

Question Number 2 (c)(i)

Answer

Mark

1. ref. to more red blood cells ; 2. higher haemoglobin concentration ; 3. haemoglobin has a higher affinity for oxygen ; 4. therefore takes up more oxygen / more highly saturated (at a given partial pressure) ;

Question Number 2 (c)(ii)

Answer

Mark

1. seal muscle has more myoglobin (than human muscle) ; 2. myoglobin acts as an oxygen store ;

Question Number 2(d)

max 2

Answer

2

Mark

(lactic acid produced by) anaerobic respiration ;

20

1

Outline scheme for marking essay questions 3, 4B and 5H

Mark

11 available for Scientific content (S) 2 available for Balance (B) 2 available for Coherence (C) Scientific content (S) 11 (good)

9 (above average)

7 (average)

5 (below average)

Description The essay demonstrates a sound understanding of the topic and contains a significant amount of material from most areas of the mark scheme, including A2 content. Suitable examples are included and the candidate has clearly and coherently linked together information from different parts of the specification. An above average essay, with accurate content. The essay includes a good balance of material from several areas of the mark scheme, including A2 content, and examples where appropriate. There may be some minor factual errors. The essay includes relevant information from some areas of the mark scheme, including A2 content. The candidate links together some facts and principles. Some examples are included. There may be some minor factual errors. The essay includes some generally factually accurate and relevant material, and there is some attempt to link material from more than one area of the mark scheme. The A2 content, in particular, lacks depth and accurate details.

3 (poor)

There are some correct facts, but the essay lacks depth and accuracy. The essay contains little or no relevant information from the A2 content.

1 (poor)

There are very few correct facts. The essay is generally superficial and inaccurate.

0 (poor)

No correct or relevant material is included.

Notes If a scientific content mark of 0, 1, or 3 is awarded, it is very unlikely that a balance mark of more than 1 is appropriate. An essay containing AS content only can be awarded a max of 3 for scientific content. An essay containing A2 content only can be awarded a max of 7 for scientific content.

21

S = 11

Balance (B) 2

Most of the main topic areas outlined are covered Some discussion of each of the areas chosen, illustrated with suitable examples where appropriate Material included is all relevant to the topic and the candidate has linked information from more than one area of the specification. Few, if any, errors

1

Some of the main topic areas outlined are covered. Some discussion of each of the areas chosen. Some irrelevant material included. There are some examples which link together different areas of the specification. Some errors

0

Very limited account, possibly only one aspect chosen Material mostly irrelevant No examples of the candidate linking information from different topics Large number of errors

B=2

Coherence (C) 2

Material logically presented, with little or no repetition Essay has coherence, ideas are developed well; continuous prose used throughout Essay has an introduction and a conclusion, summing up the main points Technical terms have been used correctly Spelling, punctuation and grammar are sound

1

Material is presented in an orderly way and some ideas developed Continuous prose used throughout The introduction and conclusion may be present, but brief Technical terms are used and generally in the correct context Spelling, punctuation and grammar are generally sound

0

Essay style not used Material in note form or numbered points Very poor standard of spelling, punctuation and grammar

22

C=2

Question Number

Answer

Mark

3 (a)

The structure and roles of carbohydrates in living organisms Introduction could include reference to the general nature of carbohydrates, monosaccharides, disaccharides, polysaccharides – structure of monosaccharides – formation of glycosidic bond – specific references to: pentoses – glucose, fructose, galactose – maltose, sucrose, lactose (disaccharides) – starch, cellulose, glycogen (polysaccharides) pentoses in nucleic acids – glucose as an energy source – roles of disaccharides – glycogen as an energy store – cellulose in plant cell walls – starch as an energy store in plants glucose as a substrate for respiration – glucose – glycogen metabolism – synthesis of carbohydrate from GP in photosynthesis – Could also expect references to glycoproteins in cell membranes – Notes Scientific content 11 marks Balance 2 marks Coherence 2 marks

15

23

Question Number

Answer

Mark

4B

The structure and roles of chloroplasts Introduction could include chloroplasts as organelles, as energy transducers, location in plants – Structure of chloroplasts: double membrane – stroma – DNA thylakoids – grana – location of pigments – starch grains – Functions: light-dependent reactions – cyclic and non-cyclic photophosphorylation – production of ATP, reduced NADP and production of oxygen – light-independent reactions – fixation of carbon dioxide onto ribulose bisphosphate formation of GP – reduction of GP to form triose phosphate – regeneration of ribulose bisphosphate Notes Scientific content 11 marks Balance 2 marks Coherence 2 marks

15

24

Question Number 5H

Answer

Mark

The regulation of the internal environment in humans Introduction could include explanation of homeostasis and negative feedback, factors that are kept constant – Specific references to: regulation of body temperature – structure and roles of the skin – roles of thermoreceptors – hypothalamus –

reproductive hormones – regulation of body water – ADH – regulation of blood glucose – insulin, glucagon and adrenaline Notes Scientific content 11 marks Balance 2 marks Coherence 2 marks

15

25

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