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GCE AS MARKING SCHEME

SUMMER 2016 BIOLOGY - NEW AS UNIT 2 2400U20-1

© WJEC CBAC Ltd.

INTRODUCTION This marking scheme was used by WJEC for the Summer 2016 examination. It was finalised after detailed discussion at examiners' conferences by all the examiners involved in the assessment. The conference was held shortly after the paper was taken so that reference could be made to the full range of candidates' responses, with photocopied scripts forming the basis of discussion. The aim of the conference was to ensure that the marking scheme was interpreted and applied in the same way by all examiners. It is hoped that this information will be of assistance to centres but it is recognised at the same time that, without the benefit of participation in the examiners' conference, teachers may have different views on certain matters of detail or interpretation. WJEC regrets that it cannot enter into any discussion or correspondence about this marking scheme.

1

WJEC GCE AS BIOLOGY UNIT 2

MARK SCHEME

GENERAL INSTRUCTIONS

Recording of marks

Examiners must mark in red ink.

One tick must equate to one mark (apart from the questions where a level of response mark scheme is applied).

Question totals should be written in the box at the end of the question.

Question totals should be entered onto the grid on the front cover and these should be added to give the script total for each candidate.

Marking rules

All work should be seen to have been marked.

Marking schemes will indicate when explicit working is deemed to be a necessary part of a correct answer.

Crossed out responses not replaced should be marked.

Credit will be given for correct and relevant alternative responses which are not recorded in the mark scheme.

Extended response question

A level of response mark scheme is used. Before applying the mark scheme please read through the whole answer from start to finish. Firstly, decide which level descriptor matches best with the candidate’s response: remember that you should be considering the overall quality of the response. Then decide which mark to award within the level. Award the higher mark in the level if there is a good match with both the content statements and the communication statement. Award the middle mark in the level if most of the content statements are given and the communication statement is partially met. Award the lower mark if only the content statements are matched.

Marking abbreviations

The following may be used in marking schemes or in the marking of scripts to indicate reasons for the marks awarded.

cao = correct answer only ecf = error carried forward bod = benefit of doubt

© WJEC CBAC Ltd. 2

Question Marking Details Marks available

AO1 AO2 AO3 Total Maths Prac

1 (a) i 60 000 : 1

Accept correct ratio not to 1 e.g. 120 000:2

1 1 1

ii Advantage:

Any one from:

Reduces water loss/ allows them to live in arid conditions (1)

No {blood / circulatory system/ pigment} required (1)

Oxygen supplied directly to the cells/ muscles (1)

Tracheoles go directly into cells/ tissues (1)

Disadvantage:

Size/ shape limitation (1)

2 2

(b) Any 3 (x1) from:

Diffusion of gases related to Surface Area (1)

Oxygen use related to volume (1)

Amoeba has large Surface Area : volume ratio (1)

Therefore diffusion of gases sufficient (to supply demand) (1)

Short diffusion distance (1)

3 3

(c) Ventilation (movements/system)/ description of replacing oxygen in

alveoli/ ORA for carbon dioxide (1)

{Blood/ transport system} (takes oxygen away from respiratory surface)/

ORA for carbon dioxide/ or description of (1)

2 2

3

Question Marking Details Marks available


(d) Any two (x1) from:

Low body temperature (1)

Metabolic rate low / example of (metabolic) reactions not

required/ working at lower rate(1)

Less energy required for support / fish are buoyant (1)

ORA for dog

Reject dogs move more than fish/ dogs are more active than fish/ more

oxygen needed for respiration

2 2

(e) 1. Mammalian (red blood cells) evolved when O2 levels were lower/

ORA (1)

2. {Biconcave/shape} increases surface area (for increased O2

absorption) (1)

3. No nucleus + can carry more haemoglobin (so increased O2

transport) / ORA(1)

4. Thin centres/small/ biconcave so short diffusion distance (so

faster diffusion) (1)

5. Small(er) in size so have a higher sa : vol ratio (1)

6. Small(er), {so more of them/ total surface area larger} (1)

Accept ref. to smaller capillaries qualified/ biconcave shape

gives flexibility to fit through capillaries

4 4

Question 1 total 7 7 0 14 1 0

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Question Marking details Marks available


2 (a) i Hypothesis 2,

Because all three mutations only occur once. (In hypothesis 1 would

need to occur twice)/ mutations occur just before the branch of

mammals and dolphins/ OWTTE

1 1

ii Compare amino acid {sequence/ order} in a (specific) protein (1)

Greater similarity more closely related/ more recent common ancestor

(1)

OR

Antibody antigen precipitation test(1)

More precipitate the more related they are (1)

2 2 2

(b) i Any three (x1)from:

HIV 1 has evolved from chimp (SIV) (1)

HIV2 has evolved from monkey (SIV) (1)

HIV1 and HIV 2 have evolved several times/ 5 strains of HIV (1)

3 different origins for HIV1/ 2 different origins for HIV2 (1)

All forms of HIV from one common ancestor (1) must be correct

context

3 3

ii Predict when next mutation may occur 1 1

iii SIV must have been evolving for a longer period/ ORA 1 1


5



3 (a) i Dependent Variable = numbers of each species (1)

Independent Variable = (presence of) pollution (1)

2 2 2

ii Some easier to catch than others / misidentification/ recounting/ they

move around/ camouflaged

1 1 1

iii Calculation of N(N-1) =22350 (1)

Calculation ∑n(n-1) =6926 (1)

calculation diversity index = 0.69 (1)

3 3 3

iv biodiversity was low(er)/ the {number/types} of species were low(er). 1 1

v There had been a mathematical error (1)

highest possible diversity index is 1 (1)

2 2 2

(b) i Any two (x1) from:

The existence of a number of distinct (inherited) varieties (coexisting in

the same population in a single species)/ snails {are different colours/

have different bands}/ different morphology/ different phenotypes(1)

at frequencies too great to be explained by recurrent mutation (1)

multiple alleles for the same gene(1)

2 2

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3 b ii Different {colours / banding/ features} give (a selective)

advantage in different habitats/ different colours are

camouflaged in different habitats (1)

Habitats vary depending on times of the {year / seasons} and so

different colours will have an advantage (1)

{Main predator / thrush} will predate different forms of the snail

in {different areas / different seasons} (1)

3 3


7



4 (a) i Some water used by {photosynthesis/metabolic reactions}/

water produced in respiration/

measures the rate of absorption not the rate of transpiration

1 1

ii Any 3 (x1) from:

humidity/ or description (1)

wind/ air currents (1)

surface area of leaves (1)

age of leaves (1)

Accept air pressure

NOT same number/ mass of leaves/ length of stem/ plant

2 1 3 3

(b) Lower surface of oak leaf shaded/ or description of/ ORA (1)

so higher density of stomata to reduce water loss (1)

OR

neither surface of wheat shaded/ or description of (1)

equal distribution of stomata water loss equal both sides (1)

2 2

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(c) 1. (Potassium ions/ malate) reduce water potential in (guard) cell (1)

Accept osmotic pressure increases/ osmotic potential decreases/

solute potential decreases/ hypertonic to outside

2. water moves in by osmosis (down water potential gradient) (1)

3. (Turgor) pressure inside (guard) cell increases/ cells become turgid

(1) NOT cells expand

4. ends of guard cell have a thinner wall than centre/ ORA (1)

5. ends of guard cell expand and stomata opens (1)

3 2 5

(d) i 2 x π x 2 x 423 = 5312.9 (1)

= 5310 ( to 3 sig figs) (1)

Allow 5320 if they use value of π from calculator.

5310/ 5320 = 2 marks

5312.9/ 5313/ 5315.6/ 5316 = 1 mark

Evidence of 2 πr x 423 = 1 mark

2 2 2

ii (water molecules) escape more readily from) species B because it

has larger (total) circumference.

Ecf if calculation incorrect in (i)

1 1


9



5 (a) i % O2 saturation stays above 95% at altitudes/

O2 affinity stays high up to 1500m/

Enables humans to {live/ survive} at altitude

1 1

ii Increased red blood cell count/ more haemoglobin/ haemoglobin has an

increased affinity for oxygen

1 1

iii (gut lumen) highly anaerobic/ low concentration of oxygen (in gut) (1)

higher affinity (for O2 than humans)/ can absorb any available O2/ higher

saturation at lower partial pressures (1)

2

(b) i {Structures / molecules} with no close phylogenetic links / have evolved

from different origins/ different structures/ ref to analogous structures(1)

adapted to carry out {same/ similar} function (1)

2 2

ii As temp rises {higher ppO2 / higher concentration of oxygen} (needed

{to fully saturate/ reach saturation/ reach 96%}).

1 2

iii As temperature rises more O2 released (to tissues). 1 1


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6 (a) i K+ (mainly) transported (upwards) in xylem (1)

K+ only moves {laterally / sideways} into phloem from xylem/

K+ does not move up or down in the phloem (1)

2 2

ii To show that the waxed paper stops the movement/ owtte (1)

Shows that separating the xylem and phloem (and then putting back

into contact) does not affect movement (1)

If when separated they return back together

Values would be the same at all points (1)

If stay separated

Expect values to be the same in xylem and phloem as in previous

experiment (1)

2 2 2

(b) i No / little transport in xylem (1)

transport in phloem in both directions (1)

2 2

ii Analyse samples from (above and) below the lower leaf (1)

If both positive then movement in both directions (1)

2 2

11



(c) Any 5 from:

1. In summer, leaves {photosynthesise / make sugars} {loaded/

moved into} {phloem / sieve tubes} (1)

2. Water potential falls water absorbed by osmosis (1)

Must be in phloem or sieve tubes

3. Reference to hydrostatic pressure (1)

4. Flow (in sieve tubes) from high to low pressure regions (1)

5. tubers (sucrose) converted into starch (1)

6. water potential rises + water lost (1)

7. (Sugars in tubers)allows (rapid) growth (in spring) / sugars used

to form cellulose for growth/ sugars for respiration (1)

3 2 5





7 Pepsin in stomach, hydrolyses peptide bonds breaking down polypeptides into shorter chains of amino acids.

The pancreas produces proteases such as trypsin which breaks down polypeptide chains into shorter chains.

Cells in small intestine secrete peptidases which complete the breakdown of polypeptides into amino acids.

Ref to exopeptidases and endopeptidases.

Amino acids absorbed into the blood from small intestine transported to the muscles.

Cows large numbers of bacteria are produced in the first three chambers of the ‘stomach’

make protein using urea

when the bacteria pass into the true stomach they are killed by the acid.

Proteins {in/ from} the bacteria are then digested and absorbed

Horses do not have saliva containing urea

This explains why horses need more protein in food than cows

In horses the bacteria are found in the caecum/ large intestine

Protein in these bacteria are lost in the faeces because no digestion or absorption takes place in the large intestine.

This explains why horse manure has a higher levels of organic nitrogen.

3

4

2

9

7-9 marks Detailed explanation of protein digestion Explanation of use of urea by bacteria in cow/ digestion of bacteria Explanation of increased protein in diet of horse/ nitrogen content in manure of horse The candidate constructs an articulate, integrated account, correctly linking relevant points, such as those in the indicative content, which shows sequential reasoning. The answer fully addresses the question with no irrelevant inclusions or significant omissions. The candidate uses scientific conventions and vocabulary appropriately and accurately.

13



4-6 marks Any two from: Explanation of protein digestion Brief explanation of use of urea by bacteria in cow/ digestion of bacteria Brief explanation of increased protein in diet/ nitrogen content in manure The candidate constructs an account correctly linking some relevant points, such as those in the indicative content, showing some reasoning. The answer addresses the question with some omissions. The candidate usually uses scientific conventions and vocabulary appropriately and accurately.

1-3 marks. Any one from: Brief explanation of protein digestion Brief explanation of ruminant digestion Brief explanation of increased protein in diet/ nitrogen content in manure The candidate makes some relevant points, such as those in the indicative content, showing limited reasoning. The answer addresses the question with significant omissions. The candidate has limited use of scientific conventions and vocabulary.

0 marks The candidate does not make any attempt or give a relevant answer worthy of credit.



SUMMARY OF MARKS ALLOCATED TO ASSESSMENT OBJECTIVES

Question AO1 AO2 AO3 TOTAL MARK MATHS PRAC

1 7 7 0 14 1 0

2 3 1 4 8 0 2

3 5 9 0 14 5 3

4 5 7 2 14 2 3

5 2 6 0 8 0 0

6 3 2 8 13 0 2

7 3 4 2 9 0 0

TOTAL 28 36 16 80 8 12

GCE AS Biology Unit 2 (New) MS Summer 2016

© WJEC CBAC Ltd.


SUMMER 2016 BIOLOGY - COMPONENT 2 B400U20-1

© WJEC CBAC Ltd.

INTRODUCTION This marking scheme was used by WJEC for the 2016 examination. It was finalised after detailed discussion at examiners' conferences by all the examiners involved in the assessment. The conference was held shortly after the paper was taken so that reference could be made to the full range of candidates' responses, with photocopied scripts forming the basis of discussion. The aim of the conference was to ensure that the marking scheme was interpreted and applied in the same way by all examiners. It is hoped that this information will be of assistance to centres but it is recognised at the same time that, without the benefit of participation in the examiners' conference, teachers may have different views on certain matters of detail or interpretation. WJEC regrets that it cannot enter into any discussion or correspondence about this marking scheme.

1 © WJEC CBAC Ltd.

EDUQAS AS BIOLOGY

COMPONENT 2 – Biodiversity and Physiology of Body Systems

MARK SCHEME


Recording of marks





Marking rules





Extended response question

A level of response mark scheme is used. Before applying the mark scheme please read through the whole answer from start to finish. Firstly,

decide which level descriptor matches best with the candidate’s response: remember that you should be considering the overall quality of the response. Then

decide which mark to award within the level. Award the higher mark in the level if there is a good match with both the

content statements and the communication statement. Award the middle mark in the level if most of the content statements are given and the communication

statement is partially met. Award the lower mark if only the content statements are matched.



cao = correct answer only

ecf = error carried forward

bod = benefit of doubt

2 © WJEC CBAC Ltd.



1 (a) (i) Any one from:

{(Bi)concave shape/ or description of}/

no nucleus /

thin /

flexible

contains haemoglobin (1)

1 1

(ii) (at sea level) more red blood cells mean more oxygen can be

delivered (to the muscles) (1) 1 1

(b) (i) Any three from:

CO2 enters rbc because {CO2 is lower inside than outside /

by diffusion} (1)

(Combines with water) to form carbonic acid (H2CO3) (1)

(due to) presence of carbonic anhydrase inside red blood

cells (1)

carbonic acid dissociates into hydrogen (ions) and

hydrogen carbonate (ions) (1)

3 3

(ii) The {H+/ hydrogen ions} from (the dissociation of carbonic

acid) combine with (oxy-)haemoglobin/ lowers the pH/

forms haemoglobinic acid(1)

It {displaces/ releases} the oxygen / lowers affinity of

haemoglobin (for oxygen) (1)

H+ displaces oxygen from haemoglobin = 2 marks

2 2


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2 (a) Polecat ( Mustela putorius) and bat (Pipistrellus pipistrellus)are

(more closely related) (1)

They are both carnivores/ ref to eating meat or prey/ in the

order carnivora (1)

Correct reference to dentition e.g. presence of {canines/ sharp

incisors/ carnassials/ pointed molars} (1)

Accept reference to dormouse having flat molars/ diastema/

absence of canines

1

1

1

3

(b) (i) noctule 1 1

(ii) Should use at least 4 pulses to gain two marks

Accept figures in range 40-45 = 2 marks

Award 1 mark if number of pulses/ time is seen but incorrect

answer (reject number of pulses higher than 10)

2 2 1 1

(c) 514 (1) Accept 512 if justified by start and stop codons

(Divide by 3) because 3 {bases/ nucleotides} code for one amino

acid/ it is a triplet code (1)

Reject 3 bases make up one amino acid

Ref to base pairs = neutral

2 2 1

(d) (i) Pipistrellus (pygmea) (1)

Plecotus (auritus) (1)

Muscardinus (avellanarius) (1)

3 3

(ii) Accept answers in the range 80-84 1 1

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2 (d) (iii)

X placed anywhere on line shown above

1 1


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3 (a)

(i) Phloem sieve tube correctly labelled (1)

(ii) Xylem vessel correctly labelled (1)

2

(b) because mass flow is from {sources to sinks/ one direction/

same direction} (1)

Reject down unqualified

this flow is {two way / bidirectional / up and down/ in both

directions } (in the sieve tubes/ in the phloem) (1)

named alternative hypothesis e.g. cytoplasmic streaming /

electro-osmosis / active phloem loading / diffusion/ active

transport/ protein filaments (1)

1

1 1 3

(c) (i) (The mean) time for radioactivity (which moves upwards) is less

(than the mean time for fluorescence which moves downwards)/ it

took longer to reach maximum value in fluorescence

Reject values alone without an attempt at comparison

Reference to standard deviation = neutral

1 1 1 1

(ii)

(i)

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(ii) 1. The results for fluorescence are more {consistent/ repeatable /

reliable}/ show less variation/ show less deviation from the

mean/ ORA (1)

2. The standard deviation for the radioactivity results is larger than

the standard deviation for the fluorescence results (even

though the mean is smaller) (1)

Accept ref to higher or longer

Use of data is neutral

Reject marking point 2 if candidate has stated results for

fluorescence are less consistent or radiation more consistent

2 2 2 2

(iii) Any 2 (x1) from:

(The method is not precise because) the smallest

difference which can be measured is 20 minutes (the time

intervals samples taken) (1)

Attaching a label to the sucrose might affect {the rate of

transport/ uptake by leaf} (1)

(reciprocal) experiment with radioisotope above and

fluorescent below was not carried out (1)

Reference to variability in aphids/ plants e.g. age/ size of

leaves/ species of plants/ size of stylet (1)

2 2

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3 (d) Volume exuded by 50 tubes in 6 hours = 0.015 mm3

Volume exuded by 1 tube in 1 hour = 0.015 (1) 6x50 = 5 x10-5 (1)

2 marks for correct answer in correct standard form

If incorrect

Award 1 mark for 0.00005 or Award 1 mark for sight of = 0.015 6x50

2 2 2


8 © WJEC CBAC Ltd.

Question Marking details Marks Available


4 (a) (i) 7 (1)

A smaller number would mean missing species (1)

A larger number will { not improve results/not provide more

meaningful results/ take longer} (1)

2 1 3 1 1

(ii) Any 2(x1) from:

Mark out a grid in the wood/use tape measures at right angles/ or

description of (1)

Use random number {generator/dice/ table} (1)

to pick coordinates/ or description (1)

2 2 2

(b) (i) Calculation of N(N-1)

=14X13=182 (1)

Calculation of ∑n(n-1)

= 28 (1)

Calculation of Diversity Index

= 0.85 to 2 places of decimal (1)

Correct answer = 3 marks

3 3 3 3

(ii) (Diversity index is greater in coppiced area)

The effect of coppicing is to {increase biodiversity/ increase

number of species/ increase in number of individuals/ increase

species {richness/ evenness}}.

Accept {improves/ higher/ beneficial} as alternatives for increase

Allow ECF from (i) if answer is less than 0.62

1 1

9 © WJEC CBAC Ltd.



(c) Coppicing has {eliminated/ killed off} Dog’s Mercury (1)

Reject reduced/ extinct/ not found

Coppicing has allowed more {light / water} through/ coppicing

results in less shade(1)

Accept no trees to climb/need trees to grow on

Reject to survive

In higher levels of light other plants {are better adapted/out-

compete Dog's Mercury}/ Dog's Mercury cannot survive in high

light intensity(1)

3 3


10 © WJEC CBAC Ltd.



5 (a) (i) absorbance/ light absorbed

Accept absorption

Reject 'absorption of red or infra- red light'

1 1

(ii) 0% = 0.26

100% = 0.02 1 1

(iii) Reference to level of oxygen in {blood/ body}

1 1

(b) In single systems there one circuit in double system there are two

(separate) circuits /blood passes through heart once compared

with twice (1)

It allows (oxygenated) blood to be delivered (to organs) under

(high) pressure/ there is no sustained drop in pressure (after

passing through lungs) (1)

2 2

(c) (i) Labels pointing to any two from:

hole between the right and left ventricles (1)

connection between the pulmonary artery and the aorta (1)

aortic valve should be three cusps (1)

2 2

x

x x




5 c (ii) Oximeter will detect less oxygenated /blood containing less

oxyhaemoglobin/give a lower reading (for % sat)/ ORA (1)

Any two from:

(The hole in the septum will allow) deoxygenated blood

(from right ventricle) to mix with oxygenated blood in (left

ventricle)/ aorta) (1) Must not be in incorrect context

The connection will allow deoxygenated blood into aorta/

allows deoxygenated and oxygenated blood to mix (1)

Faulty valve will allow backflow of blood into the left

ventricle (1)

1

1

1

3





6 (a) (i) Any three (x1) from:

1. Reference to active site of α amylase/ lock and key

hypothesis/ ref to enzyme substrate complexes (1)

2. complementary shape only to starch (not cellulose)(1)

3. Starch contains α glucose but cellulose contains β glucose/

reference to α bonds and β bonds (1)

4. Reference to {coiling in starch /straight chains in cellulose/

microfibrils in cellulose/ cross linking in cellulose / alternate

(glucose) molecules rotated by 1800 in cellulose/ molecules

not rotated in starch (1)

3

3

(ii) Reference to the {enzyme/ cellulase} being produced by {bacteria/

micro-organisms} 1 1

(b) (i) Food substance Reagent

Starch

Iodine (solution)/

(Potassium) iodide (1)

Reducing sugar

Benedict’s (reagent) (1)

2 2 2

(ii) Soft faeces contain more (reducing) sugar (than hard faeces)

Accept named reducing sugar

reject nutrients/ starch

1 1 1

(iii) (Soft faeces) would {taste/ smell} sweeter/ ORA 1 1




(c) 1. Small intestines come before the caecum in the digestive system

(1)

2. (When eating grass) cellulose digestion takes place after

absorption so sugar is {not absorbed/ passed out in soft faeces}

(1) Reject nutrients

3. When eating soft faeces the food is passing through the

alimentary canal a second time (1)

4. After eating the soft faeces the {sugar/ nutrients} can then be

absorbed (1)

4 4





7 I large surface area due to presence of gill filaments/ plates/ lamellae

permeable

rich blood supply to maintain concentration gradient between water and blood

reduced diffusion distance

ventilation mechanism

5 4 9

II separate buccal and opercular/gill cavities

operculum/bony plates can close gill cavity

lowering floor of buccal cavity increases volume and decreases pressure in buccal cavity

mouth opens and water enters

mouth closes floor of buccal cavity is raised

increased pressure forces water over the gills

water and blood flow in opposite direction – countercurrent flow

maintain a concentration gradient for O2 across whole of gill surface

O2 absorption is more efficient/ blood and water do not reach equilibrium/more O2 is absorbed

III Axolotls have external gills so no ventilation mechanism/rely on water currents/whole body movements to ventilate gills

Reference to parallel flow

Cannot absorb as much / less efficient at O2 absorption which is needed for aerobic respiration





7-9 marks Detailed explanation of properties of gas exchange surfaces. Detailed explanation of ventilation including reference to pressure/ volume and counter current mechanism in fish. Explanation of slower movement in axolotl.

The candidate constructs an articulate, integrated account, correctly linking relevant points, such as those in the indicative content, which shows sequential reasoning. The answer fully addresses the question with no irrelevant inclusions or significant omissions. The candidate uses scientific conventions and vocabulary appropriately and accurately.

4-6 marks Any two from: Explanation of properties of gas exchange surfaces Explanation of ventilation and counter current mechanism in fish Explanation of slower movement in axolotl

The candidate constructs an account correctly linking some relevant points, such as those in the indicative content, showing some reasoning. The answer addresses the question with some omissions. The candidate usually uses scientific conventions and vocabulary appropriately and accurately.




1-3 marks Any one from: Description of properties of gas exchange surfaces Description of ventilation or counter current mechanism in fish Attempt at explanation of slower movement in axolotl

The candidate makes some relevant points, such as those in the indicative content, showing limited reasoning. The answer addresses the question with significant omissions. The candidate has limited use of scientific conventions and vocabulary.

0 marks The candidate does not make any attempt or give a relevant answer worthy of credit



COMPONENT 2: BIODIVERSITY AND PHYSIOLOGY OF BODY SYSTEMS



1 7 0 0 7 0 0

2 2 9 2 13 2 1

3 4 3 5 12 5 3

4 2 5 5 12 4 6

5 2 5 3 10 0 0

6 5 7 0 12 0 3

7 5 4 0 9 0 0

TOTAL 27 33 15 75 11 13

B400U20-1 Eduqas AS Biology - Component 2 MS Summer 2016

© WJEC CBAC Ltd.


SUMMER 2017 AS (NEW) BIOLOGY - UNIT 2 2400U20-1

1 © WJEC CBAC Ltd.

UNIT 2 - Biodiversity and Physiology of Body Systems

MARK SCHEME


Recording of marks





Marking rules





Extended response question A level of response mark scheme is used. Before applying the mark scheme please read through the whole answer from start to finish. Firstly, decide which level descriptor matches best with the candidate’s response: remember that you should be considering the overall quality of the response. Then decide which mark to award within the level. Award the higher mark in the level if there is a good match with both the content statements and the communication statement. Award the middle mark in the level if most of the content statements are given and the communication statement is partially met. Award the lower mark if only the content statements are matched.




2 © WJEC CBAC Ltd.

Question Marking details

Marks Available


1 (a) (i) An organism that lives on (surface of) another organism / host(1) {Obtains nutrition from/ obtains food from/ feeds on} the host {at the expense of/ causes harm to} the host (1) ignore death

2 2

(ii) (Absence of long legs) so cannot leap / jump (like flea)/ORA (1) ignore fly

1 1

(iii) Domain – eukaryotic/ eukaryote/ description of eukaryotic characteristics (1) Kingdom – nervous {coordination/ system} / multicellular / heterotrophic / lacks cell wall / develop from a blastula (1)

2 2

(b) (i)

Pthirus pubis diverged from its common ancestor later than the common ancestor of gorillas and humans existed/ common ancestors were not alive at the same time/ pubic louse evolved after the common ancestor/ gorilla and human common ancestor died out before the pubic louse evolved(1) Reference to data (1)

2 2

(ii) Similarity in {nucleotide / base / codon/ base pairs } sequences Ignore DNA sequence Accept DNA hybridisation with explanation of similarity in bases needing a higher temperature to separate the strands

1 1

(iii) 5 1 1 1

(c) Able to interbreed to produce fertile offspring/ OWTTE (1) 1 1


3 © WJEC CBAC Ltd.


Marks Available


2 (a) Reduce {water/ heat} loss (1) NOT prevent

Filter air / Trap {dirt / dust / particles} (1)

Ignore bacteria/ viruses Reject reference to function of cilia

Short diffusion {pathway/distance} (1)

Reduce surface tension / prevent collapse (of alveoli and

tracheoles during expiration)/ prevent (alveoli/tracheoles)

sticking together (1)

1 3 4

(b) (i)

Allows trachea to collapse slightly when food passes down the oesophagus/ allows peristalsis/ increase in size of oesophagus caused by passage of food/ owtte

1 1

(ii) Xylem 1 1

(c) (i)

Any four (x1) from A. (expanding the rib cage / pulling on outer pleural membrane)

Lowers {pressure in pleural cavity / pleural pressure} (1) B. Inner pleural membrane pulls on the lungs (1) C. Which increases volume of {lungs / alveoli / thorax} (1) D. Which decreases pressure in {lungs / alveoli} (1) E. Below atmospheric pressure / so air moves in / credit correct

reference to negative pressure breathing (1) Answer must be in correct sequence (although some mark points may be omitted)

4 4

(ii) Any one from

{Greater / higher} pressure/ faster pressure changes {Greater / higher} volume/ faster changes in volume

More {rapid/ shorter} {inspiration/ expiration}/ more rapid breathing/ deeper breaths/ increased breathing rate

1 1


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3 (a) (i) Any four (x1) from A. Xylem transports {water/minerals} {from roots / up the plant/

to the leaves}/ xylem allows transpiration (1) NOT nutrients

B. Lack of water causes {leaf/ cells} to {lose turgor / become

flaccid}/ cells become plasmolysed (1)

C. Reduced surface area (1)

D. {No/reduced} photosynthesis/ less water for photosynthesis

(1)

E. Less {products / named products} of photosynthesis (1)

F. For respiration / growth (1)

2 2 4

(ii) Phloem transport is bidirectional / up and down plant (1) (so) Symptoms appear in both roots and leaves (in potato leaf roll) (1)

2 2

(b) (i) A – sieve tube (element) NOT sieve plate B – companion cell (1) 1 mark for both

1 1

(ii) A - lacks {nucleus / nucleolus } / {fewer/ no} mitochondria / ER / Golgi/ organelles/ less cytoplasm (1) ORA Must be clear which cell is referred to

1 1

(iii) Plasmodesmata (1) {Loading / unloading/ transport/ transfer} of sucrose / Transport of {ATP / proteins / enzymes / macromolecules} (1)

2 2


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4 (a) A – palisade mesophyll/ parenchyma (1) {Vertical/ elongated/ long/ columnar} cells/ high number of chloroplasts (1) B – spongy mesophyll (1) Air {spaces/ pockets}/ large surface area (1)

4 4

(b) (i) Any 2 (x1) from Humidity (1) Air currents/ wind speed/ air movement (1) Light intensity / wavelength of light(1)

2 2

2

(ii)

Any one from mm

3 cm

-2 min

-1

mm3 min

-1 cm

-2

mm3/cm

2/min

mm3/ min/ cm

2

1 1

1

1

(iii) {gap/ difference} between {range/ error} bars/ range bars do not overlap (1)

1 1 1 1

(iv)

Any suitable xeromorphic adaptation (1) correct explanation (1)

Adaptation Explanation

Fewer stomata Fewer spaces through which water vapour can be lost

Thick cuticle Waterproof/ reduce evaporation

Hairs/ sunken stomata/ stomata in pits/ rolled (curled) leaves

Trap {humid /moist} air/ trap water vapour}reduce {water potential/ diffusion gradient}

2 2

(v) Answer= ±2 % = 2 marks Accept ±4% = 2 marks (error at both ends of distance) If answer incorrect allow 1 / 50 x 100 = 1 mark 100/ 50= 1mark

2 2

2

2

(vi) Time over a longer distance / greater than 50 mm (1) 1 1 1


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5 (a) Bohr effect / shift (1) 1 1

(b) (i) 70 % (1) 10 % (1)

2 2 2

(ii) Reduced affinity of haemoglobin for oxygen/ (oxy)haemoglobin dissociates at a higher partial pressure (of oxygen) (1) More oxygen {released / unloaded/ dissociates}/ oxygen released (more) readily (1) Allow oxygen more readily dissociated from haemoglobin = 2 marks

2 2

(c) (i) CO2 removed faster/ more CO2 exhaled (1) Increases {diffusion / concentration} gradient (from blood into alveoli) (1)

2 2

(ii) increased affinity of haemoglobin for oxygen / decreased dissociation of oxyhaemoglobin (1) Less oxygen {released/ available} (for respiration) (1)

2 2

(d) (i) X = carbonic anhydrase Y = carbonic acid / H2CO3 Z = hydrogen ion / H

+ / proton

0/1 = 0 marks 2 correct = 1 mark, 3 correct = 2 marks

2 2

(ii) {In solution/dissolved} in plasma / as carbamino compounds / as carbaminohaemoglobin (1)

1 1

(iii) Any three from:

Chloride shift (1) must be in context of red blood cells

(chloride ions) {enter/diffuse} into red blood cells in exchange

for hydrogen carbonate ions (1)

One to one exchange / reference to figures (1)

To maintain (electrochemical) neutrality (in red blood cell)/ to

{equalise/ balance/ maintain} charge (1)

1 2 3


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AO1 AO2 AO3 Total Maths Prac 6 (a) 667 / 10 or 66.7 (1)

= 67 (1) allow ecf for 1 mark 2 marks for correct answer

2 2 2

2

(b) All {results / replicates / groups} show the same trend pattern}/ or suitable description (1)

1 1 1

(c) Any two (x2) from: (Variation in) temperature (1) Use a {thermostatically/ temperature} controlled water bath (1) Difficulty deciding the end-point / subjective decision/ owtte (1) Use a colorimeter (1) Difficulty in mixing the solutions (1) Agitate at regular intervals / or reasonable suggestion (1) Difficulty deciding when to start the stop clock (1) Start as soon as solutions are completely mixed / or reasonable suggestion (1)

4 4

4

(d) Endopeptidase – {hydrolyses / breaks} (peptide) bonds

within{polypeptide / protein/ amino acid chain}/ {hydrolyses /

breaks} non-terminal bonds (1)

Exopeptidase – {hydrolyses / breaks} (peptide) bonds at

ends of {polypeptide / protein/ amino acid chain}/ {hydrolyses

/ breaks} terminal bonds (1)

Enzyme mixture – endopeptidase creates more ends for

exopeptidase/ owtte (1)

3 3

(e) Pepsin {optimum/ works best} in {acid conditions/ at pH{1/2/3} (1) Would be {inactive / denatured}/ would not work at pH 8 (1)

1 1 2


8 © WJEC CBAC Ltd.



7

Indicative content In systemic circulation:

{High pressure/ pressure increase} (to 120 mmHg) in {aorta/ arteries} is due to ventricular systole/ contraction of left ventricle.

High pressure needed to pump blood {round body/ long distance}

Pressure decrease (to 80 mmHg) in {arteries / aorta} is due to {diastole/ ventricular relaxation}.

High pressure is maintained due to elastic recoil of artery walls / closure of {semilunar / aortic} valve.

Pressure decrease in {arterioles/ capillaries} is due to {frictional resistance / blood flowing through a greater (total) {cross sectional/ surface area}.

Pressure decrease in capillaries is also due to tissue fluid formation.

low pressure/ flow rate in capillaries necessary for diffusion

Blood flow in veins due to massaging effect of skeletal muscles.

In pulmonary circulation:

Pressure increase in pulmonary arteries due to {ventricular systole/ contraction of thinner walled right ventricle}.

Lower pressure (than systemic) due to {shorter distance /

(only) to lungs}.

Pressure decrease (to about 8 mmHg) in pulmonary arteries

is due to diastole / ventricle relaxation.

Pressure in capillaries is lower than systemic to prevent

tissue fluid formation in the alveoli / reduce damage to alveoli

/ allow time for gas exchange.

6

3

9

9 © WJEC CBAC Ltd.



Benefit over single circulation:

Correct description of single and double circulation

Pressure is lost when blood passes through (gill) capillaries.

so blood flow through the systemic circulation of a fish is slower.

Double circulation maintains high pressure (in systemic circulation)

to meet the high metabolic demands of a mammal.




7-9 marks Detailed explanation of systemic circulation And Detailed explanation of pulmonary circulation And Detailed explanation of benefit over single circulation The candidate constructs an articulate, integrated account, correctly linking relevant points, such as those in the indicative content, which shows sequential reasoning. The answer fully addresses the question with no irrelevant inclusions or significant omissions. The candidate uses scientific conventions and vocabulary appropriately and accurately.

4-6 marks Any two from Explanation of systemic circulation Explanation of pulmonary circulation Explanation of benefit over single circulation

The candidate constructs an account correctly linking some relevant points, such as those in the indicative content, showing some reasoning. The answer addresses the question with some omissions. The candidate usually uses scientific conventions and vocabulary appropriately and accurately.




1-3 marks Brief explanation of systemic circulation OR Brief explanation of pulmonary circulation OR Brief explanation of benefit over single circulation The candidate makes some relevant points, such as those in the indicative content, showing limited reasoning. The answer addresses the question with significant omissions. The candidate has limited use of scientific conventions and vocabulary. 0 marks The candidate does not make any attempt or give a relevant answer worthy of credit.



UNIT 2: BIODIVERSITY AND PHYSIOLOGY OF BODY SYSTEMS



1 6 2 2 10 1 0

2 1 9 1 11 0 0

3 1 7 2 10 0 0

4 6 5 2 13 4 7

5 7 4 4 15 2 0

6 1 6 5 12 2 7

7 6 3 0 9 0 0

TOTAL 28 36 16 80 9 14

WJEC GCE AS Biology Unit 2 (New) MS Summer 2017/GH


SUMMER 2017 AS (NEW) BIOLOGY - COMPONENT 2 B400U20-1

1 © WJEC CBAC Ltd.

COMPONENT 2 - Biodiversity and Physiology of Body Systems

MARK SCHEME


Recording of marks





Marking rules









2 © WJEC CBAC Ltd.



1 (a) (i) Electro cardiogram/ electrocardiograph (accept ecg) 1 1

(ii) 15 x 0.04 = 100 beats per minute / bpm = 2 marks

60

One beat takes 15 x 0.04 = 0.6 seconds = 1 mark

2 2

2

(b) (i) All correct for 2 mark

2 correct 1 mark

Cardiac cycle events Section of trace

Ventricular diastole T

Atrial systole P

Ventricular systole QRS

2 2

(ii) I SAN:

Initiates {the cardiac cycle / an impulse/ wave of excitation/

depolarisation/ electrical signal} /(acts as) pacemaker (1)

NOT message/ signal

causes atria to contract/ atrial systole (1)

2 2

II AVN:

passes {impulse/ electrical signal} down {Purkinje/ Purkyne

fibres/ Bundle of His} (1)

Results in delay (1)

2 2

III Bundle of His and Purkyne:

{Carries/conducts} {impulse/ electrical signal} (from AVN) down

to apex of heart (1)

causes ventricles to contract from {apex/ bottom} upwards (1)

2 2


3 © WJEC CBAC Ltd.



2 (a) (i) Potometer 1 1 1

(ii) {Air bubbles block/ water column interrupted in} xylem (vessels)/ OWTTE Ignore phloem

1 1

(iii) Any 2 (x1) from: Cut shoot under water (1) assemble {apparatus/ named apparatus} underwater (1) seal all {gaps / joints} with grease (1)

2 2 2

(b) (i) Temperature: Any one from:

increase in temperature increases kinetic energy/ or description of /

decrease in temperature decreases kinetic energy Light intensity: Any one from:

increase in intensity increases stomatal opening/ (increased light intensity) increased rate of photosynthesis

increases {use/ uptake} of water/ decrease in intensity decreases stomatal opening / (decreased light intensity) decreased rate of photosynthesis

decreases {use/ uptake} of water increase evaporation/ increases (rate of) diffusion of water vapour/ increased loss of water vapour/ increases transpiration/ORA (1) (credit in either temperature or light intensity) must be linked to correct explanation

3 3 2

(ii) Axes assigned with correct labels (time + total volume of water taken up by the leafy shoot) + units ({minutes/min} + cm3) (1)

Appropriate linear scales, including origin + use of ½ graph paper (1)

All plots correctly plotted (1) [tolerance ±½ small square] Plotted points joined with dot to dot with ruler or a curve +

key/ label (1) No extrapolation with dot to dot, allow 5 small squares extrapolation on a lobf. Reject sketchy/ thick lines/lobf, if line does not pass through centre of more than one plotted point

4 4 4

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2 (b) (iii) Any 5 (x1) from:

A. Polythene bag: Initial increase but (after 20 mins) {no

further increase / curve plateaus} and no bag: {(steady)

increase /constant rate of uptake} (1)

B. water vapour {trapped inside polythene bag / cannot

diffuse away from leaves} / increased humidity inside

bag (1)

C. {no diffusion gradient / no concentration gradient/

equilibrium reached / correct description of equilibrium} in

the bag(1)

D. water uptake stopped / no more water vapour diffuses out

of stomata (1) must be linked to the idea of no diffusion

gradient

E. with no bag, there is a {diffusion/ concentration/ water

potential} gradient (1)

F. because {diffusion shells {are removed/ do not build up}/

water vapour removed from (lower) surface of leaf/ ORA}

(1)

4

1

5


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Marks Available


3 (a) 1 epu= 200/90 = 2.2 µm = 2 marks (ignore additional decimal

places)

90 epu = 20 smd / 45 epu = 10 smd for 1 mark

Allow 22.2/ 22 for 1 mark (incorrect reading from scale)

2 2

2

2

(b) (i) Serosa (1)

Mucosa (1)

circular muscle (1)

longitudinal muscle

4 4

(ii) 20 x 2.2 = 44 µm (1)

Accept 440/444 (ECF) 1 1

1

(iii) not in proportion to original section (1) must be attempt at

numerically qualification

epu shows ratio of 4:5 or equivalent (1)

diagram shows ratio of 2:3 or equivalent (1)

3 3

3

(c) (i) Active form would {digest/ break down} cells/ autolysis/ ORA 1 1

(ii) Alkaline secretion/ bile (1)

from Brunner’s glands / pancreas/ liver cells/ gall bladder (1)

Neutralise acid from stomach / increase pH / act as buffer (1)

3 3


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Marks Available


4 (a) (i) Total muscle mass = 70 x 60/100 = 42(kg) (1) O2 capacity of muscle = 42 x 69.8 = 2931.6/ 2932/ 2930 cm3 = 2 marks

2 2 2

(ii) Much higher myoglobin content in seal muscle than in human (1) Any one from

More O2 available to be {released/ dissociated} {when the partial pressure of oxygen is low/ when seal holds breath to dive/ when underwater} (1)

Therefore can respire aerobically for longer (1)

2 2

(b) A. More red blood cells in general circulation (1) B. more haemoglobin available (1) C. More oxygen supplied to {muscle / brain / tissue/ cells} (1) D. Aerobic respiration can continue for longer / seal does not

need to come to surface {to breathe / for oxygen/ for air} as often / anaerobic respiration is not required(1)

4 4

(c) (i)

A. Myoglobin curve well to left because myoglobin has a higher affinity for oxygen than haemoglobin(1)

B. Myoglobin retains high levels of O2 / is highly saturated at low ppO2 / reaches saturation at lower ppO2(1)

C. Only dissociates oxygen at low ppO2/ dissociates more readily at low partial pressures (1)

D. At low pp a small change in pp results in large change in saturation (1)

4 4

(ii) Llama haemoglobin has a higher affinity for O2 than adult human haemoglobin (1) Reference to llama habitat /higher altitude/ environment with has a Low pp O2 (1)

2 2


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5 (a) (i) 10 µm or less (1) Mesh size needs to be less than {widest part of the smallest organism/ Pseudo-nitzchia} (1) Second marking point can only be awarded if first is awarded

2 2

2

(b) (i) A. Site 1 is more biodiverse (1) B. (Species richness is same for) both sites as they have

the same {number/amount} of species (1) C. site 2 is dominated by one species / site 1 has more

evenness in the size of each population/ OWTTE (1) Accept correct calculations for 3 marks Site 1 = 0.74 Site 2 = 0.07

3 3

(ii) Simpson’s (index) 1 1

(iii) Sample in more locations along coastline (1)

Sample more than once in a day / more frequently (1)

Use a {larger diameter/ bigger} net / increase volume of water sampled / increase sample size (1)

3 3

3

(iv) {(water) temperatures / light intensities} higher (in summer) (1) Higher {growth/ reproduction} rate of phytoplankton and ref to relevant risk to humans/ ORA (1)

2 2


8 © WJEC CBAC Ltd.



6 Indicative content Gas exchange adaptations

{Skin/ body surface} is gas exchange surface

Outer body surfaces folded – further increases surface area for diffusion

Diameter of approximately 3 mm, length up to 45mm - narrow diameter gives short diffusion pathway

Cylindrical shape - high surface area to volume ratio Behaviour adaptations

Head and front part of body buried in mud – when highly active, less of the body is buried – more surface in contact with gas exchange medium

Part of the body exposed to the water moves vigorously -maintains steep concentration gradient

Haemoglobin

Red in colour due to presence of haemoglobin – presence of respiratory pigment increases ability to absorb oxygen

Presence of blood vessels to transport oxygen/ maintain diffusion gradient

Haemoglobin may have higher affinity for oxygen

6 3

9 © WJEC CBAC Ltd.



7-9 marks Indicative content of this level is… Detailed explanation of gas exchange adaptations Detailed explanation of behavioural adaptations Detailed explanation of role of haemoglobin To award 9 marks a comment linking adaptations to survival in a low oxygen environment is required The candidates constructs an articulate, integrated account. which shows sequential reasoning. The answer fully addresses the question with no irrelevant inclusions or significant omissions. The candidate uses scientific conventions and vocabulary appropriately and accurately. 4-6 marks Indicative content of this level is… Any two from: Explanation of gas exchange adaptations Explanation of behavioural adaptations Explanation of role of haemoglobin The candidate constructs an account correctly linking some relevant points, such as those in the indicative content, showing some reasoning. The answer addresses the question with some omissions. The candidate usually uses scientific conventions and vocabulary appropriately and accurately.




1-3 marks Indicative content of this level is… Brief explanation of gas exchange adaptations OR Brief explanation of behavioural adaptations OR Brief explanation of role of haemoglobin The candidate makes some relevant points, such as those in the indicative content, showing limited reasoning. The answer addresses the question with significant omissions. The candidate has limited use of scientific conventions and vocabulary. 0 marks The candidate does not make any attempt or give a relevant answer worthy of credit.






1 9 2 0 11 2 0

2 3 12 1 16 4 5

3 8 3 3 14 2 6

4 0 10 4 14 2 0

5 1 3 7 11 0 5

6 6 3 0 9 0 0

TOTAL 27 33 15 75 10 16

TARGET 27 33 15 75 8 12

B400U20-1 Eduqas AS Biology - Component 2 MS Summer 2017/GH

2 © WJEC CBAC Ltd.



1 (a) (i) Any two × (1) from:

Flow rate/speed/current (1)

temperature (1)

(named) {ion/mineral} concentration (1) NOT content

conductivity (1)

oxygen concentration (1)

light intensity/exposure (1)

vegetation {on banks/in water}(1) Accept algae in water

time of year/season/time of day (1)

Reject pH/width/depth/stream bed/any ref to method/technique

2 2

2

(ii) Any two × (1) from:

A. (same) sampling area (1) NOT distance

B. {time spent/force/power/speed/amount} of kicking/number of

kicks (1)

C. net mesh size (1) NOT area of net/same net

D. Distance from bank/distance of kicking from the net (1)

Reject any reference to stream conditions/time net is in water

2 2

2

(iii) would kill them/animals die(1)

preserve diversity/effect on food chains/endangered species(1)

Accept reverse answers for riverside identification

2

2 2

(b) (i)

Remove {bias/experimenter choice} (1)

Reject fair test/accuracy/reliability/validity 1 1 1

3 © WJEC CBAC Ltd.



(ii) Any two × (1) from: A. Animals not caught by net/ Missed net/passed through

net/net removed too soon (1) B. Animals not dislodged from stream bed/not kicked hard

enough to dislodge them/organisms moved away from sample area (1)

C. Not kicked entire area/uneven kicking (1) D. Unable to identify animals/animals too small to identify (1)

2

2

2

(iii) D = 0.7/0.699 = 3 marks

If incorrect rounding e.g. 0.69 award 2 marks

If incorrect answer award

D = 0.3/0.301 = 2 marks

OR

N(N-1) = 6972 (1)

n(n-1) = 2098 (1)

3 3

3

(iv) Forest stream:

Lower species richness/fewer species(1) Accept correct use of

numbers

Lower species evenness/dominated by {single (named)

species/stonefly nymph} (1)

ORA moorland stream

2 2

(c) (i)

More (than five) sample sites/{more/larger} area sampled(1)

More (than two) streams (1) Not forest unqualified 2 2 2

(ii) Create buffer zone between forest and stream/cut down trees

next to stream/{increase pH of/neutralise} stream (eg

liming)/OWTTE (1)

NOT cut down trees

Introduction of more species = neutral

1 1

1


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2 (a) (i) Holozoic (1) NOT heterotrophic unqualified 1 1

(ii) Hydra

Receives oxygen/glucose/products of

photosynthesis/sugars/amino acids(1)

Chlorella

Receives {carbon dioxide/glucose/sugars/amino acids/products

of digestion/minerals/protection/place to live/habitat} (1)

2 2

(b) Similarity: {Both/owtte} are autotrophic (1) Difference: For Nostoc Photosynthetic/photoautotrophic/phototrophic/ ref to use of light energy (1) For Nitrosomonas Chemosynthetic/chemoautotrophic/chemotrophic/ref to use of energy from chemical reactions (1) Correct use of photoautotroph and chemoautotroph = 3 marks

1 2 3

(c) Similarities:

{Both/owtte} are heterotrophic (1)

Difference:

For R. stolonifer

Saprotrophic/saprobiontic/saprophytic (1)

For R. oryzae

Parasitic (1)

Extracellular digestion/suitable description (1) Accept reference

to {either/both} species carrying out extracellular digestion

2 2 4

Question 2 total 1 5 4 10

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3 (a) Phylogenetic tree/Evolutionary tree/phylogenic (1)

Accept Cladogram

NOT Family tree/pedigree diagram

1 1

(b) (i) Diagram B (1)

Hippo closest relative/shares most recent common ancestor with

the hippo (1)

Fewest differences/most similarities/or use of data (1)

1

2 3

(ii) Classification {may change/not fixed/not permanent} (1)

As more {information/knowledge/techniques} becomes

available/OWTTE (1)

2

2

(c) Family (1) 1 1

(d) (i) Increase from Antarctic circle to the tropics (1)

Accept as latitude decreases numbers increase

Reject references to latitude increasing

Reject reference to Arctic

levels off/slight decrease in the tropics (1)

2 2

(ii) Temperature/food availability (1) 1 1

(iii) Two peaks/OWTTE (1)

Accept two modes 1 1 1

Question 3 total 4 5 2 11 1

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4 (a) (i) Both needed for one mark

X = (gill) filaments

Y = (gill) arch/bar

1 1 1

(ii) {Filter/sieve/remove/catch} {food/particles/solids NOT

molecules}

/protection from debris/preventing damage to the gill filaments

(1)

1 1

(iii) A. normally water separates {filaments/lamellae/plates/gills}/Gills

dry out (when out of water) (1)

B. {Filaments/lamellae/plates/gills} {stick together/collapse} (1)

C. Reduced Surface Area for {diffusion/gas exchange} (1)

D. Water needed for diffusion (of gases)/{oxygen/gases} need to

be in solution to diffuse (1)

4

4

(b) Accept reverse argument for any point

A. Countercurrent more efficient (1)

B. Maintains {diffusion/concentration} gradient/or description

of/equilibrium not met (1)

C. Across {entire/whole} {gill/filament/lamellae/plate} (1)

D. {Higher/more} {saturation/oxygen absorbed} (1)

Accept correct references to higher percentages in counter

current/use of values e.g. 80% compared to 50%

4 4

(c) (i) C = parallel flow and A = countercurrent flow (1) 1 1

(ii) Correct insertion of arrows on A only (1)

NOT ECF 1 1

7 © WJEC CBAC Ltd.



(d) (i) Positive (1)

1 1 1

(ii) Range of m = 0.6 – 0.65 = 2 marks

Accept correct fraction on answer line

number of dp = neutral

If incorrect award one mark for sight of dy/dx (dy is smaller than

dx) (1)

2 2 2

(iii) Accept range 96-104 a.u. (1)

(value from (ii) x 160) + 0

ECF

1

1 1

(iv) Any two (x1) from

A. wider the { lumen/blood vessel} faster {velocity/flow

rate/speed} (1)

Reject faster velocity the wider the lumen

B. so blood can be returned to the heart as quickly as possible

(1)

C. without {losing velocity/slowing down} due to decreased

blood pressure/because no contractions to increase

pressure (1)

2 2


8 © WJEC CBAC Ltd.



5 (a) (i) electrocardiogram (1) reject electrocardiograph/echocardiogram 1 1

(ii) 75 (1) 1 1 1

(iii) I SAN generates {electrical impulse/wave of excitation/electrical

signals} (1)

(causes) depolarisation of atria (1)

(causes) contraction of atria/atrial systole (1)

3 3

II Any three (x1) from:

A. AVN {transmits/relays/passes on} {electrical impulse/electrical signals/wave of excitation}(1)

B. Passes through {Bundle of His/Purkinje tissue/septum}(1) C. Depolarisation of ventricles(1) D. Causes contraction of ventricles/ventricular systole(1)

3 3

III Repolarisation of ventricles (1)

Causes ventricular diastole/relaxation of ventricles (1) 2 2

(b) Shorter/closer together (1) NOT quicker/faster

(due to) shorter isoelectric line/flat part/PR segment/P-QRS

interval/T-P interval/less time between {atrial systole and ventricular

systole/ventricular systole and ventricular diastole/ventricular

systole and atrial systole} (1)

1 1 2

(c) (i) Circle at any point around the PR segment/flat section of the graph

between the endpoint of the P wave and the onset of the QRS

complex (1)

Circle may extend slightly into QRS complex (NOT further than

letter Q) but may extend to the start of the P wave ( NOT further

than start)

1

1

(ii) AVN (1) Accept bundle of His/Purkyne tissue 1 1

(iii) Slow heart rate/bradycardia/longer for a heart beat/longer between

atrial systole and ventricular systole (1) 1 1

Question 5 total 9 4 2 15 1

9 © WJEC CBAC Ltd.



6 Indicative content

Definitions:

mesophyte – adapted to conditions of

{adequate/moderate/sufficient/owtte} water supply

xerophyte – adapted to conditions where water is scarce/groundwater

is frozen/named example/dry conditions.

hydrophyte – adapted to aquatic conditions/live in water/float on

surface of water

Pinus/xerophyte

Adaptations to reduce transpiration/water (vapour) loss Comments must relate to image Reject adaptations not visible Linked points:

Sunken stomata/stomata in pits + trap humid air/to reduce {diffusion/concentration/water potential} gradient

Small leaves/needle shaped/compact + reduce surface area/SA:vol

{fewer/smaller} stomata + {fewer/smaller} gaps for water loss

Thick cuticle/epidermis + reduce evaporation/water loss

Fibres/sclerenchyma/lignified tissue + provides support/prevents wilting

Potamogeton/hydrophyte

Linked points:

Cuticle thin/absent + no need to reduce water loss/evaporation

Stomata on {upper/adaxial} surface/no stomata on {lower/abaxial} surface + for gas exchange with air

Air spaces/Aerenchyma/lacunae + provide { buoyancy/flotation} for {light/photosynthesis}/{act as gas reservoir/for gas exchange}

{Little/no} {lignified tissue/xylem} + water provides support

Poorly developed/little xylem + water provided by surroundings

3 6 9




7-9 marks Correct Definitions of xerophytes + hydrophytes + mesophytes. + Full description of Pinus xerophytic adaptations + Full description of Potamageton hydrophytic adaptations The candidate constructs an articulate, integrated account, correctly linking relevant points, such as those in the indicative content, which shows sequential reasoning. The answer fully addresses the question with few irrelevant inclusions or significant omissions. The candidate uses scientific conventions and vocabulary appropriately and accurately. 4-6 marks Any two from: definitions of mesophytes/xerophytes/hydrophytes description of adaptation of the xerophyte description of adaptation of the hydrophyte The candidate constructs an account correctly linking some relevant points, such as those in the indicative content, showing some reasoning. The answer addresses the question with some omissions. The candidate usually uses scientific conventions and vocabulary appropriately and accurately. 1-3 marks A definition of xerophyte/hydrophyte/mesophyte or an adaptation of xerophytes

or an adaptation of hydrophytes

The candidate makes some relevant points, such as those in the indicative content, showing limited reasoning. The answer addresses the question with significant omissions. The candidate has limited use of scientific conventions and vocabulary.

0 marks

The candidate does not make any attempt or give a relevant answer worthy of credit.

Question 6 total 3 6 9





1 1 11 5 17 3 12

2 1 5 4 10

3 4 5 2 11 1

4 10 5 3 18 4 1

5 9 4 2 15 1

6 3 6 9

TOTAL 28 36 16 80 9 13

2400U20-1 WJEC GCE AS Biology Unit 2 NEW MS Summer 2018

1 © WJEC CBAC Ltd.

EDUQAS AS COMPONENT 2 BIODIVERSITY AND PHYSIOLOGY OF BODY SYSTEMS

MARK SCHEME SUMMER 2018


Recording of marks Examiners must mark in red ink. One tick must equate to one mark (apart from the questions where a level of response mark scheme is applied). Question totals should be written in the box at the end of the question. Question totals should be entered onto the grid on the front cover and these should be added to give the script total for each candidate. Marking rules All work should be seen to have been marked. Marking schemes will indicate when explicit working is deemed to be a necessary part of a correct answer. Crossed out responses not replaced should be marked. Credit will be given for correct and relevant alternative responses which are not recorded in the mark scheme. Extended response question A level of response mark scheme is used. Before applying the mark scheme please read through the whole answer from start to finish. Firstly, decide which level descriptor matches best with the candidate’s response: remember that you should be considering the overall quality of the response. Then decide which mark to award within the level. Award the higher mark in the level if there is a good match with both the content statements and the communication statement. Award the middle mark in the level if most of the content statements are given and the communication statement is partially met. Award the lower mark if only the content statements are matched.

2 © WJEC CBAC Ltd.

Marking abbreviations The following may be used in marking schemes or in the marking of scripts to indicate reasons for the marks awarded. cao = correct answer only ecf = error carried forward bod = benefit of doubt

3 © WJEC CBAC Ltd.



1 (a) Enzymes secreted outside of the body/extracellular digestion and the products of digestion are absorbed (1)

1 1

(b) (i) Scolex/hooks and suckers + attach to gut wall 1 1

(ii) Nutrients absorbed through its body surface (so it doesn't need a mouth)(1) Food is pre-digested (so does not need an alimentary canal) (1)

2

2

(iii) Oxygen levels are too low for aerobic respiration in the intestine/ It has a low metabolic rate and {does not require aerobic respiration/does not move much/owtte (1)

1

1

(c) (i) Any four (x1) from (Thick) cuticle protects the worm from the effects of {acids/ enzymes} (1) Lime cell secretion neutralises acid (1) Microtriches increase surface area for absorption of digested food (1) Glands secrete mucus to protect the worm from digestive enzymes (1) Muscles allow tapeworm to increase contact with digested food (1)

4 4

(ii) Enable the worm to absorb ions/amino acids/glucose/correct named product of digestion(1) Against concentration gradient (1)

2 2


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2 (a) raccoon (1) There is only 1 base difference between red panda and racoon/more bases in common between red panda and racoon/ there are more differences between red panda and giant panda and black bear(1)

2 2

(b) 17 differences/3.95 x 10 -7

= 43037974.68 1 mark = 43 000 000 2 marks = 4.3 x 107 years 3 marks

3

3

3

(c) Analogous structures (evolve separately) are different structures to perform a similar function/or example (1) Homologous structures (evolve from a common ancestor) similar structure performing different functions/or example (1) Analogous structures arise through convergent evolution/OWTTE (1)

3 3

(d) Eukaryota (1) Contains membrane bound organelles/DNA bound in nuclear envelope/80S ribosomes/multicellular (1)

1

1 2


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3 (a) Sunken stomata + Water vapour builds up in pits (1) Reduces diffusion gradient (1) Rate of transpiration reduced Ligustrum has no pits/stomata nearer surface + Moving air can easily disrupt diffusion shells/blow away water vapour. (1)

3 3

(b) (i) Obtain impression using nail varnish or PVA/owtte (1) Prepare a slide using impression and observe under microscope (1) Count number of stomata in known area (1) Repeat to obtain mean (1)

4 4

4

(ii) 13/14 stomata in field of view Area shown = 3.142 x 0.52 = 0.7855 mm2 If 13 Density = 16.5499682 mm-2 = 16.55 stomata mm-2 (2 d.p.) If 14 Density = 17.8230426 mm-2 = 17.82 stomata mm-2 (2 d.p.) Correct answer to 2 dp = 3 marks Correct answer not to 2 dp = 2 marks If use diameter instead of radius = 2 marks Sight of 3.142 x 0.52 = 1 mark

3 3

3

3

(iii) Stomata are sunken/in pits so will not show up on impression/AVP (1)

1 1 1

(c) Species A and has lowest stomatal density/least stomata/higher percentage of stomata on the lower surface (1) So lower rate of water loss/transpiration (1)

2 2


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4 (a) (i) = 27/5 (1) = x 5.4 (2)

2 2 2 2

(ii)

Ring of cartilage (1) Prevents the trachea collapsing (due to air pressure decrease during inspiration) (1)

2 2

(b) Any 3 (x1) from: External intercostal muscles contract + ribs move up and out + diaphragm contracts and flattens (1) Correct reference to role of pleural membranes (1) Volume of thorax increases therefore pressure decreases (1) air pressure in alveolus falls to {-1 a.u./below atmospheric pressure} and air rushes in (1)

3

3

(c) (i) I PP O2 in capillary blood rises from 5.5 kPa to 14 kPa then remains constant (1) because of initial large concentration gradient so oxygen diffuses into blood (1) plateau because of equilibrium at 14 kPa/OWTTE (1)

2

2

II 14 kPa because equilibrium has been reached between ppO2 in the alveoli and the capillary

1 1

(ii) Thin walls in capillary/walls one cell thick (1) Dense network of capillaries/Large total surface area (1) Low velocity of blood/low cross sectional area/small diameter (1)

2 2

(d) hydrostatic pressure is higher than osmotic pressure at the arterial end (1) Forces {water and small soluble molecules/fluid} out of capillary (1) Blood cells and larger protein molecules retained in capillary (1)

3 3


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5 (a) Keep all other conditions as experimental plants (1) But do not remove vascular tissue (1)

2 2 2

(b) (i) Removal of leaves above cut reduces photosynthesis (1) Reducing sugar production above the experimental region (1)

2 2 2

(ii) I The cylinder prevents the region from drying out (1) 1 1 1

II Replenishing the water stops the build-up of {chemicals/bacteria}/replenish oxygen (1)

1 1 1

(c) (i) Phloem is involved in the translocation of sugar (1) Removal of phloem in A almost completely stopped the upward transport of sugars past the cut/ORA. (1) Growth is reduced in A because of reduced supply of sugars/ORA (1)

1

2 3

(ii) Removal of xylem prevents the transport of water up the plant which will reduce growth of B (1) Less {growth/respiration/energy needed} therefore more sugar remaining in B(1)

2 2

(d) (i) Repeat the experiment and calculate the mean increase in stem length for each group of plants (1)

1 1 2

(ii) Any two for 1 mark from Light intensity/wavelength/Species/Temperature/Surface area of leaves/Age of plant/Size of cut/humidity/air movement

1 1

(e) Arrow pointing at the endodermis (1) 1 1

(f) Any 3 (x1) from: Stops apoplast pathway/forces ions into symplast pathway (1) Movement of Ions into xylem requires active transport (1) Cyanide is a respiratory inhibitor/prevents cells respiring/Stops ATP synthesis (1) Lower water potential gradient reduces root pressure/ORA (1)

3 3


8 © WJEC CBAC Ltd.



6 Indicative content

Wolves and dogs have carnivore dentition

Sharp incisors and long pointed canines Carnassial teeth for shearing through bone, ligaments and tendons

(Relatively) short gut adapted for protein digestion

Salivary amylase hydrolyses starch to maltose

Optimum pH maintained by mineral salts/buffers in saliva

Starch digestion resumes in duodenum with pancreatic amylase

Maltose digested to glucose by maltase in small intestine

Domestic dogs fed on human food waste containing large amounts of starch

Some dogs better at digesting starch than others

Dogs that could digest starch well more likely to survive and breed/can live on human food waste

Wolves/wild dogs cannot digest starch as there is no selective advantage/wolves cannot produce amylase

3 4 2 9

7-9 marks Detailed explanation of carnivore dentition Detailed account of starch digestion Logical explanation of advantage of starch digestion in domesticated dogs The candidate constructs an articulate, integrated account, correctly linking relevant points, such as those in the indicative content, which shows sequential reasoning. The answer fully addresses the question with no irrelevant inclusions or significant omissions. The candidate uses scientific conventions and vocabulary appropriately and accurately.

9 © WJEC CBAC Ltd.



4-6 marks Any two from: Explanation of carnivore dentition An account of starch digestion Attempt to account for advantage of starch digestion in early dogs The candidate constructs an account correctly linking some relevant points, such as those in the indicative content, showing some reasoning. The answer addresses the question with some omissions. The candidate usually uses scientific conventions and vocabulary appropriately and accurately. 1-3 marks Any one from: Brief explanation of carnivore dentition Brief account of starch digestion Attempt to give an advantage for starch digestion in early dogs The candidate makes some relevant points, such as those in the indicative content, showing limited reasoning. The answer addresses the question with significant omissions. The candidate has limited use of scientific conventions and vocabulary. 0 marks The candidate does not make any attempt or give a relevant answer worthy of credit.




SUMMARY OF MARKS ALLOCATED TO ASSESSMENT OBJECTIVE


1 4 5 2 11 0 0

2 4 6 0 10 3 0

3 7 4 2 13 3 8

4 8 6 1 15 2 2

5 1 8 8 17 0 8

6 3 4 2 9 0 0

TOTAL 27 33 15 75 8 18

B400U20-1 EDUQAS AS BIOLOGY - COMPONENT 2 SUMMER 2018 MS/ED

gce as marking scheme - home - thiacin€¦ · marking schemes will indicate when explicit working...

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