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© WJEC CBAC Ltd.

GCE AS MARKING SCHEME

SUMMER 2016 BIOLOGY - NEW AS UNIT 1 2400U10-1

© 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 NEW AS BIOLOGY UNIT 1 – BASIC BIOCHEMISTRY AND CELL ORGANISATION

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) Telophase (1) Accept cytokinesis

Metaphase (1)

Any reference to meiosis – penalise 1 mark

2 2

(ii) Chromosomes

Accept chromatids

Reject chromatin/ spindle fibres

1 1

(iii)

DNA is found( in other parts of the cell)(1)

in mitochondria (1)

Reject chloroplast

2 2

(b) (4 days = 96 hours)

96/19 = 5 cycles (accept 5.05 cycles)

Evidence of 5 cycles e.g. 96/19 or 5 cycles/ 5.05 cycles/ 5.1

cycles = 1 mark

30 000 x 25 (accept 30 000 x 25.05/5.1)

= 960 000/ 996 000/ 1 028 902 = 2 marks

= 9.6 x 105 cm-3 (accept 9.96 x 105 cm-3/ 1.0 x 106 ) = 3 marks

3 3 3

Question 1 total 2 6 0 8 3 0

3

Question Marking details Marks Available


2 (a) genetic material in the epithelial cell is DNA (1)

Any 3 ( x1) from:

DNA/ epithelial cell RNA/ virus

double stranded/double helix single stranded (1)

deoxyribose ribose (1)

contains thymine contains uracil (1)

Larger/ longer molecule Smaller/ shorter molecule (1)

(Genetic material contained in

nucleus)

Genetic material not {contained in

nucleus/ membrane bound} (1)

Answers must be comparative (apart from fifth MP)

Reference to prokaryotic cells is neutral

4 4

(b) (i) {Endocytosis/phagocyctosis} Accept micropinocytosis 1 1

(ii) Any four (x1) from:

1. (Viral) RNA acts as a template (1) 2. (Free RNA nucleotides attach to their) complementary base

pairs (1) Reject DNA 3. Adenine with uracil and cytosine with guanine (1) Accept A:U, C:G 4. RNA polymerase catalyses the formation of the new RNA

molecule (1) 5. Forming the {sugar phosphate backbone/phosphodiester

bond} (1)

4 4

© WJEC CBAC Ltd. 4



(iii) Any five (x1) from:

1. Translation (1)

2. mRNA attaches to a ribosome (1)

3. tRNA brings a (specific) amino acid (1)

4. Complementary base pairing between mRNA codon and

tRNA anticodon/ codon anticodon interaction (or

description of) (1)

5. Ribosome has room for two tRNA molecules/Brings two

amino acids into close proximity/ or description of (1)

6. Condensation reaction occurs/peptide bond formed

(between amino acids)(1)

7. Ribosome moves to the next codon (or description of)/

reference to stop codon (1)

8. Post translation processing/ or description of e.g. folding

of polypeptide chain role of Golgi(1)

5 5

(iv) It does not possess {any ribosomes/tRNA / the organelles for

protein synthesis} (1) Reject: organelles unqualified

All other organelles = neutral

It does not possess mitochondria / cannot produce its own ATP

for the process (1)

2 2


5



3 (a) A: nucleus (1) Accept nucleolus

B: chloroplast (1) 1

1

2

(b) apparent length = 13mm / 13 x 1000 = 13000µm

magnification = 13000 / 32.3 µm or 13/0.0323 (1)

magnification = 402.48/ 402.5/ 403/ 402

(correct answer = 2 marks)

OR

apparent length = 13.5mm / 13.5 x 1000 = 13500µm

magnification = 13500 / 32.3 µm or 13.5/ 0.0323 (1)

magnification = 417.96/ 418 (correct answer = 2 marks)

Award 1 mark for evidence of figures showing image size/ actual

size

2 2 2

(c) 1. straight chains of β-glucose/ alternative molecules of β-glucose rotate through 180◦ / chains cross-linked / form microfibrils (1)

2. which provide {strength/rigidity/inelasticity} to the cell wall (1) support = neutral

3. When the {solute concentration/ solute potential} changes (water potential will change) causing water to {move into /out of} the cell (1)

4. cell wall prevents {osmotic lysis/ cell bursting}/ cell wall prevents cell shrinking (1) turgid/ plasmolysed = neutral Reject if direction of water movement is incorrect

1

1

1

1

4

© WJEC CBAC Ltd. 6



3 (d) Spirogyra – eukaryotic cells & Nostoc - prokaryotic cells (1)

Similarity: Both contain ribosomes / cell membranes / DNA/

genetic material (1)

Difference: any 2 from: (1 mark)

Spirogyra Nostoc

{Membrane bound organelles/

named organelle} present

{Membrane bound organelles/

named organelle} absent

DNA enclosed within nuclear

membrane

DNA free in cytoplasm

Linear DNA Loop of DNA Accept plasmid

Larger/80s ribosomes

Accept ribosomes are different

sizes

Smaller/70s ribosomes

Accept following

DNA associated with histones DNA not associated with

histones

Mesosome absent Mesosome present

1 1

1

3


7



4 (a) (i) Diagram shows the addition of water (1)

Glucose and fructose drawn with –OH groups on C1 and C2 in

down positions (1)

2

2

(ii) Glycosidic 1 1

(iii) They have the same {chemical formula/ molecular formula/

number of atoms of each element} but different {structural

formulae/ structures}/ both C6H12O6 but different {structural

formulae/ structures}

1 1

(iv) Add Biuret solution to a sample of both solutions (1)

Reject reference to heat

The sucrase will cause a colour change from blue to lilac (and

the sucrose will remain blue) (1)

OR

Heat both solutions with acid, then neutralise with alkali, then

(heat) with Benedicts (1)

The sucrose will cause a colour change from blue to brick red

(and the sucrase will remain blue)(1)

1

1

2 2

© WJEC CBAC Ltd. 8



(b) (i) Any two (x 1) from:

concentration of {sucrose / sucrase / DNS} (1)

temperature (1)

filter/ wavelength of light in the colorimeter (1)

method of mixing (1)

2 2 2

(ii) Linear axes with values at origin (1)

X-axis labelled pH and Y-axis labelled mean absorbance + au

(1)

Correct plots (1) tolerance ± ½ small square

Plots joined (1) no extrapolation

4 4 4 4

(iii) Hypothesis is incorrect (1)

The {fastest rate of reaction/ most product produced} is {pH 4/

acidic/ between pH3-5}/ the optimum pH is pH 4 (1)

Use of data e.g. mean light absorption was 0.87 au at pH 4 and

only 0.30 au at pH 7/ rate is approx three times faster at pH4

than at pH 7 (1)

3

3 3

9



4 (iv) 3.8 (µg cm-3) (1) Accept any figure between 3.7-3.8

3.8/2 = 1.9 (µg cm-3) (1)

OR

3.7 (µg cm-3) (1)

3.7/2 = 1.85(µg cm-3) (1)

OR

4.2 (µg cm-3) (1) Accept any figure between 4.2-4.3

4.2/2 = 2.1 (µg cm-3) (1)

OR

4.3 (µg cm-3) (1)

4.3/2 = 2.15 (µg cm-3) (1)

2 2 2 2

(c) There is {overlap / high variability} in the repeat readings at {pH

3/4/5/ low pH} (1)

Repeat experiment with a range of intermediate values between

pH 3-5/

Repeat data {to obtain a more reliable mean/ to improve

reliability} (1) Reject references to accuracy

2 2 2


© WJEC CBAC Ltd. 10



5 (a) 1. Oxygen crosses the membrane by (simple) diffusion (1)

NOT facilitated

2. oxygen passes through the phospholipid (bilayer) (1)

Membrane = neutral

3. Sodium ions pass through {channel/carrier} proteins (1)

intrinsic proteins = neutral

4. by {facilitated diffusion/active transport/ cotransport} (1)

Reject active transport in relation to channel proteins

If oxygen and sodium are not treated separately, allow one

mark max for correct linkage of diffusion with phospholipid

bilayer or {facilitated diffusion/ active transport/ cotransport}

with {channel/ carrier} proteins

4 4

(b) 1. The percentage ion composition of {fresh water/ habitat} is

lower than that in the carp’s blood plasma/ fresh water is

hypotonic to the blood/ ORA (1)

2. The water potential of fresh water is {higher/less negative}

than the water potential of the carp’s blood/ ORA (1)

3. The carp takes in water (through the gills) by osmosis (1)

4. (Producing large volumes of dilute urine) prevents {osmotic

lysis/ bursting} of cells (1)

5. Maintaining ion composition (1)

Accept reduces the loss of ions

1

1

1

1

4

11



5 (c) (i) 1. Percentage ion composition of sea water is greater than

that of the blood plasma/ sea water is hypertonic/ water

potential in the fish is higher than the sea water/ ORA (1)

2. ions will move into the blood plasma, {by (facilitated)

diffusion/ down a concentration gradient/ or description

of}(1)

3. therefore ions must be actively transported out (of the

blood plasma into the surrounding sea water) (1)

1

1

1

3

(c) (ii) Cell 1 is from the salmon in sea water (1)

Any one from:

as it is {highly folded to give a larger surface} /

contains {many/more} mitochondria to {provide ATP/ for

respiration/ for active transport } (1)

1

1

2




5 (d) 1. As the percentage ion composition of the blood

plasma is higher than the surrounding water (1)

2. Ions will move out of the {fish/gills/blood plasma} by

{(facilitated) diffusion/down a concentration gradient}

(1)

3. Acidification will {denature/ inactivate/ change the

shape of} the (carrier/ channel) proteins in the

specialised cells/ denature/ inactivate enzymes

(involved in ATP production) (1)

4. Therefore {no/fewer} ions are (actively) transported

into the {fish/gills/blood (plasma)}/ fewer ions can

diffuse out (1) (in context of damaged proteins)

1

1

1

1

4


13



6 Indicative content

Pyrophosphatase has tertiary structure

The active site has a specific shape

Pyrophosphate has a complementary shape

fits into the active site/ forms Enzyme Substrate Complex

lock and key theory/ induced fit

Phenylalanine must be a non-competitive inhibitor;

As it has a shape different to pyrophosphate;

Phenylalanine bonds to the allosteric site/site away from

the active site;

Causes a change in the shape of the active site;

Pyrophosphate’s active site is no longer

complementary/prevents pyrophosphate binding

fewer enzyme-substrate complexes are formed;

The maximum rate of reaction cannot be reached at any

concentration of pyrophosphate;

Phosphate must be a competitive inhibitor;

As it has a similar shape to {pyrophosphate/ substrate} /

phosphate has a complementary shape the active site;

Phosphate {binds to/ competes for} the active site;

Prevents pyrophosphate binding / fewer enzyme-

substrate complexes are formed;

The maximum rate of reaction can be reached at higher

concentrations of pyrophosphate;


7-9 marks Indicative content of this level is: Detailed explanation of lock and key/ induced fit theory. Detailed explanation of the action of inhibition of phenylalanine. Detailed explanation of the action of inhibition of phosphate. 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 Indicative content of this level: Any two from: Explanation of lock and key/ induced fit theory/ or description of. explanation of the action of inhibition of phenylalanine explanation of the action of inhibition of phosphate. 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.

15

1-3 marks Indicative content of this level is: Brief explanation of lock and key/ induced fit theory OR Brief explanation of the action of inhibition of phenylalanine OR Brief explanation of the action of inhibition of phosphate. 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 4 3 2 9


Unit 1: BASIC BIOCHEMISTRY AND CELL ORGANISATION

SUMMARY OF MARKS ALLOCATED TO ASSESSMENT OBJECTIVES

Question AO1 AO2 AO3 TOTAL MARK MATHS PRAC

1 2 6 0 8 3 0

2 10 6 0 16 0 0

3 3 6 2 11 2 0

4 5 10 4 19 6 15

5 4 5 8 17 0 0

6 4 3 2 9 0 0

Total 31 36 16 80 11 15

WJEC GCE AS Biology Unit 1 (New) MS Summer 2016

© WJEC CBAC Ltd.


SUMMER 2016 BIOLOGY COMPONENT 1 B400U10-1

© 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.

© WJEC CBAC Ltd. 1

EDUQAS AS BIOLOGY COMPONENT 1 – Basic Biochemistry and Cell Organisation

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.




© WJEC CBAC Ltd. 2



1 (a) (i) All 3 for 1 mark A Phosphate + B Pentose/ 5C sugar + C nitrogenous/organic base

1 1

(ii) DNA A 1 phosphate + B Deoxyribose (1)

RNA A 1 phosphate + B Ribose (1)

ATP A 3 phosphates + B Ribose(1)

3 3

(b) (i) Both for 1 mark Glucose + Galactose

1 1

(ii) Both decrease/or suitable figures (when cyanide is added) (1)

+ Any one from:

Cyanide stops active transport as {it stops ATP

production/ it requires energy}/

Other 2 sugars remain at the same level because {they

do not need {ATP/ energy} to be absorbed/ } (1)

2 2

(c) External concentration must be higher than internal

concentration (so that sugars enter the cells)/ diffusion is the

movement of molecules down a concentration gradient (1)

{All the sugars show some/ there is} uptake when cyanide is

present so this must be by diffusion/ diffusion is a passive

process and not affected by cyanide (1)

2 2


© WJEC CBAC Ltd. 3



2 (a) Protein + {Nucleic acid/DNA/RNA} (1) 1 1

(b) (Globular) protein (1) Accept Polypeptide

With a carbohydrate (chain)/polysaccharide/oligosaccharide

(attached to it) (1)

Reject Glycogen

2 2

(c) Binding to receptor/ glycocalyx/ carbohydrate (chains)/

glycoprotein/ protein (1)

Phagocytosis / endocytosis / (macro)pinocytosis (1)

Plasma/cell membrane {encloses/ engulfs} the virus (in a

vesicle) (1)

3 3


© WJEC CBAC Ltd. 4



3 (a) (i) That they had diabetes (1)

+ Any one from:

Because this is a reading of 0.75mg cm3 and it should be

0 for a non diabetic/

Because this is a reading of 0.75mg cm3 and it is above

0.18 mg/cm3 (1)

2 2

(ii) Blood is {red/coloured} so could not get a colorimeter reading/

light cannot pass through 1 1 1

(b) Any two (x1) from:

Only glucose will give a result/enzyme specific (to glucose) (1)

Detects very low concentrations (1)

Reading is more accurate /enzyme is more stable (1)

Accept works at different temperatures

2 2

(c) (i) On carbon 4 glucose has {the H above the ring/ OH below} in

galactose it is reverse/ ORA 1 1

(ii) Lactose and water 1 1


© WJEC CBAC Ltd. 5



4 (a) (i) Substitution (1),

calculation (1),

1dp (1)

using 3.14

(2 x 3.14 x 16) + (2 x 3.14 x 4 x 45)

(100.48 + 1130.4)

=1230.9 (3 marks)

using π from calculator

(2π x16) + (2π 4x45)

(100.43 + 1130.97)

= 1231.5 (3 marks)

2662.7 (using 3.14 and diameter) (2 marks)

2789.7 (using and diameter) (2 marks)

3 3 3

(b) Correct axes with linear scale using at least 6 large

squares (1)

Fully labelled axes (sucrose concentration /M on x axis

and Percentage change in mass / % on y axis) (1)

Correct plots (1) ±½ small square

3 3 3

© WJEC CBAC Ltd. 6



(c) Figure from candidates graph where it crosses the x axis

converted to kPa (1) Must have unit

(normally a range between -540 and -680kPa)

{Where the line intercepts the X axis/ as the line passes

0} is where there is no change in mass. (1)

This is where the water potential of the external solution

= water potential of tissue/ It is the isotonic solution/

equal water potentials (in context) (1)

2 1 3 3

(d) In 0.0 M sucrose

{the external ψ is higher than ψcell/ OWTTE}, water

moves in by osmosis (1)

So gaining mass (1)

In 0.8M sucrose

{the external ψ is lower than ψcell/ OWTTE}, water

moves out by osmosis (1)

So decreasing the mass (1)

Osmosis only needs to be mentioned once

Reject gain/ loss of mass not linked to water movement

4 4 1

© WJEC CBAC Ltd. 7



4 (e) (i) Sweet potato has a {very high sugar concentration/ very low

water potential}/ OWTTE

1 1 1

(ii) Increase the sucrose concentration (1)

Accept use a wider range

(Until a point is reached) where water will flow out of the potato

(1)

2 2 2


© WJEC CBAC Ltd. 8



5 (a) Hydrolysis/ break down of lipids (1)

Causing production of fatty acids (1)

Fatty acids cause the milk to become more acidic/lower pH, (so

becomes colourless) (1)

1 2 3 3

(b) Result depends on pH change in test tube so cannot change pH

of experiment/ OWTTE

e.g. acid pHs start colourless therefore cannot see change/

pH buffer would prevent changes in colour

1 1 1

(c) Any 3 from:

1. Drug is a similar shape to the substrate/ complementary to

active site (1)

2. Drug {binds to/ blocks/ occupies} the active site of the

{enzyme/lipase}/ the {substrate/lipid} cannot enter active site/

forms enzyme inhibitor complex (1)

3. Fewer enzyme substrate complexes/less lipid molecules are

hydrolysed by the lipase/ fewer successful collisions (1)

3 3

(d) (i) Lipids are undigested/not broken down (and cannot be

absorbed, so pass out of body) (1)

Reduced {energy/calorie uptake} so body use{stored/ body} fat

(1)

2 2

(ii) Any 1 from:

{Continue to eat fatty food/ same diet as before} but now enzyme is not inhibited/

so lipids now {digested/broken down/hydrolysed} and absorbed

1 1

© WJEC CBAC Ltd. 9



5 (d) (iii) Any 3 from:

{More rapid / 4-5% / greater} weight loss {in first year/

initially/ at start} in drug treated/ORA (1)

No further weight loss after 1 year in drug treated / no

benefit after 4 years/ORA (1)

Long term treatment could affect health due to the effect

of diarrhoea/ deficiency diseases/ weight may be

regained/ psychological effect(1)

Additional cost of drug/ have to take vitamin tablets (1)

3 3





6 (a) {RNA polymerase/ helicase} {unwinding/ unzips} DNA (1)

complementary RNA nucleotides base pair (or indicated

with A-U, C-G) with DNA {nucleotides/ template} (1)

Correct role of RNA polymerase in joining nucleotides (1)

3 3

(b) (i) Introns do not code for the {amino acid (sequence) in / primary

structure of} the protein/ only exons code for the amino acid

(sequence) in the protein / OWTTE(1)

if introns are included a different protein (structure) would be

produced/ OWTTE (1)

2 2

(ii) Intron – no effect (as not translated) (1)

Exon- (may) change the {amino acid sequence/ primary

structure} or there would be a different amino acid in the

polypeptide chain (1)

2 2

(c) (i) Both lines

Met – phe – gln – trp – (stop) 1 1

(ii) codes for the same amino acid/ still codes for phe/ no change in

protein structure 1 1

(d) (i) 0.98:1 OR 1:1.02 + 2 decimal places (1) Accept 0.98

1.03:1 OR 1:0.97 + 2 decimal places (1) Accept 1.03 2 2 2




6 d (ii) Any 4 (x1) from:

Pauling

had 3 {helices/strands} {which is incorrect/should be 2}

(1)

Which does not allow 1:1 base pairing (1)

Has the phosphates on the inside instead of the outside /

correct in that sugar groups are not in the core/ has

bases on the periphery rather than the centre (1)

Franklin

Closer to modern knowledge than Pauling (1)

Correct in that DNA is a helical structure but did not

determine that it contained 2 chains (1) NOT alpha helix

Correct that phosphates are on the outside(1)

4 4





7 ESSAY

A is meiosis, B is mitosis; The significance of mitosis

daughter cells are genetically identical; growth; repair/healing following damage and disease; repeated cell renewal/continuous cell division – with e.g.

such as skin/gut lining/bone marrow and R & WBC production;

maintains chromosome number;

The significance of meiosis

produces non identical/genetically different daughter cells; Gamete production ; Raw material for evolution/survival of the fittest; Some comment on advantage of sexual reproduction in the

event of environmental change/disease; Some mention of sources of variation – crossing over in

prophase 1 and random assortment; Haploid cells produced so that at fertilisation the diploid

number is regained;

Tumour formation

Reference to a genetic change in B which allows mitosis to continue in an unrestricted way. Solid mass of cells which prevent normal cells from functioning. Reference to benign/malignant. Cell division stops after meiosis;

5

4




7-9 marks Indicative content of this level is…

Detailed coverage of significance of mitosis, meiosis and tumour formation.

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 Indicative content of this level is

Some coverage of significance of mitosis and meiosis. Reference to tumours may be missing/ incorrect

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.





Some coverage of significance of mitosis or meiosis. 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.




COMPONENT 1: BASIC BIOCHEMISTRY AND CELL ORGANISATION



1 4 5 0 9 0 0

2 6 0 0 6 0 0

3 3 2 2 7 0 1

4 4 8 4 16 6 7

5 1 9 3 13 0 4

6 4 5 6 15 2 0

7 5 4 0 9 0 0

TOTAL 27 33 15 75 8 12

B400U10-1 Eduqas AS Biology - Component 1 MS Summer 2016

© WJEC CBAC Ltd.


SUMMER 2017 AS (NEW) BIOLOGY - UNIT 1 2400U10-1

1 © WJEC CBAC Ltd.

Unit 1 – Basic Biochemistry and Cell Organisation

MARK SCHEME


Recording of marks





Marking rules









2 © WJEC CBAC Ltd.



1 (a) (i) Cilia (1) Ignore ciliated cell

{Move/ sweep/ push/ clear} mucus (1) 2 2

(ii) An organ is an aggregation of several tissues but this only

shows one/

{It/ Image 1} only contains one type of {cell/ tissue}

Ignore similar cells

1 1

(b) (i) Self-replication/ Stage (of division) / stage of development / size/

not fully formed/ age 1 1

(ii) 25 x a measurement

1.45/ 1.475/ 1.48/ 1.5 um = 3 marks

If incorrect allow (250 x a measurement) for 2 marks

1450/1475/1480/1500 (nm)

If incorrect allow a measurement for 1 mark units must be

shown

58/59/60mm or 5.8/5.9/6.0 cm

3 3 3

(iii) Explanation must match adaptation

Increased/ larger surface area : volume (ratio)(1)

Increased gas exchange (1)

Cylindrical (1)

decreases diffusion distance / large surface area for gas

exchange(1)

Folding of inner membrane/ cristae (1)

{Increased/ large} SA for {enzymes/ membrane proteins /

(aerobic) respiration/ / ATP production} (1)

NOT energy production/ release

2 2


3 © WJEC CBAC Ltd.



2 (a) Any three (x1) from: Temperature Time Rate of spin Volume/ mass/ amount of DNA/ sample relevant condition of tube ie {density / concentration / volume / height of/ type of } medium Not solution unqualified 3 correct = 2 marks, 2/1 correct = 1 mark, 0 correct = 0 marks

2 2 2

(b) (i) A: 14N /light/14 B: 14N + 15N /light and heavy/14 and 15 C: 15N / heavy/ 15 All correct = 1 mark

1 1 1

(ii) A. {Generation 0 /{band/peak} at C} is all {heavy isotope/ N15}

(1)

B. Generation 1 {band/ peak} at B is {a mixture of heavy and

light isotope/ an intermediate}(1)

C. Because each molecule retains one strand of heavy isotope

and one strand of newly formed DNA of light isotope (1)

Accept hybrid DNA for intermediate DNA

2

1

3

(iii) A. (Only) {14N / light isotope} (nucleotides ) are available for

replication (1)

B. Generation 2 has equal {height scans/ peaks / density} (1)

C. because one strand of heavy and one strand of light DNA

have been used as templates for the formation of new DNA

molecules/ owtte (1)

D. Generation 3 has a higher peak of {14N / light isotope/ A}

because more light DNA (strands) are used as template/

owtte (1)

3

1

4

4 © WJEC CBAC Ltd.



(iv) Any two for 1 mark from: DNA polymerase / (DNA) helicase / (DNA) ligase (1) Any two (x1) from:

DNA polymerase – join free nucleotides (to the ends of the

new DNA strand)(1)

(DNA) helicase – {unzip/ unwind} {DNA/ helix} / breaks H-

bonds {between nucleotides/ within the double strand/

between the bases} / separate (DNA) strands (1)

(DNA) ligase – joins fragment of DNA to an existing

fragment(1)

3 3


5 © WJEC CBAC Ltd.



3 (a) (i) Pentose/ 5C (sugar) + phosphate (group) + nitrogenous /

organic base NOT named base 1 1

(ii) {All reactions/ source of energy} in all {cells/ organisms/ living

things} NOT energy produced 1 1

(iii) Exergonic (1)

{Energy/ 30.6kJ mol} {released/ given out} (1)

NOT produced

2 2

(b) 0.01 % = 2 marks

Allow correct conversion of units for one mark

5 / 1000 OR 50 x 1000 (1)

2 2 2

(c) A: exercise / contracting and B: relaxing/ at rest (1)

At rest: CP is created to be a store of phosphate (1)

During exercise ATP can be generated quickly (1)

(When) phosphate group from CP is given to ADP/ is used to

make ATP(1)

1

3 4


6 © WJEC CBAC Ltd.



4 (a) Intracellular 1 1

(b) (i) Independent: temperature (1)

Dependent: time (taken) to reach zero absorbance (1) 2 2 2

(ii) Acclimatisation/ owtte

reject stated temperature 1 1 1

(iii) A. {Colour change/ indicator} is dependent on a change in pH

(1)

B. (Hydrolysis produces) fatty acids {that lower the pH/ make it

acid} (1)

C. Buffer solution would {stabilise/ maintain/ control} pH (to

pH10/ alkaline conditions) / 'mop-up' H+ (1)

3 3 3

7 © WJEC CBAC Ltd.



(c) (i) Using 18 seconds

2.78 x 10-2 / 2.8 x 10-2 /2.63 x 10-2 / 2.6 x 10-2 = 3 marks

0.0277777777777778/ 0.028/ 0.02631579/ 0.03 = 2 marks (not

standard form)

2.7 x 10-2 = 2 marks (incorrectly rounded)

2.78 x 10-1 / 2.63 x 10-1 = 2 marks (incorrect conversion)

Using 19 seconds

2.63 x 10-2 / 2.6 x 10-2 = 3 marks

0.027 = 1 mark (not standard form and incorrect rounding)

NOT taking volume into account

Using 18 seconds

1.39 x 10-2/ 1.4 x 10-2 = 2 marks

1.38 x 10-2 = 1 mark (incorrectly rounded) 0.0139/0.014 = 1 mark ( not taking standard form into account )

1.3 x 10-2 = 1 mark (incorrectly rounded)

Using 19 seconds

1.32 x 10-2 /1.3 x 10-2 = 2marks 0.0132/0.013 = 1mark (not taking standard form into account) 1.31 x 10-2 = 1 mark (incorrect rounding)

OR

Allow 2 x 0.25 = 0.5 for 1 mark

3 3 3

(ii) Cooling in the colorimeter/

Accuracy of equipment used to measure volumes/

not all volumes transferred (1)

1 1 1

8 © WJEC CBAC Ltd.



(iii) A. Inappropriate (temperature) range (1)

B. {time taken/ graph} {is still decreasing/ has not levelled off}/

all enzymes still active (1)

Improvement :

C. Needs to test increase temperatures above 50°C, (to

determine how well the enzyme is able to work at higher

temperatures) (1) ignore any temperatures below 50°C

D. smaller increments (to determine the optimum accurately)

(1)

4 4 4


9 © WJEC CBAC Ltd.



5 (a) Any 4 (x1) from: Similarities: A. Have phospholipid bilayer/ hydrophobic tails & hydrophilic

heads (1) B. Correct location of heads and tails: (hydrophobic tails) inside

& (hydrophilic heads) outside (1) C. Have protein (1)

Differences: D. {Proteins not embedded in the lipid bilayer / proteins coat outer

surface / continuous layer of protein / no intrinsic proteins} in Davson Danielli model (1)

E. Ref to absence of {glycocalyx/ glycoproteins/ glycolipids/ cholesterol} in Davson Danielli model (1) Accept reverse for Fluid Mosaic

4 4

(b) (i) A. As the number of carbon atoms in the chain increases, the (transition) temperature increases (1) Accept increased chain length

B. Because the longer the chain length the greater the intermolecular forces (1) Accept increased number of C atoms

C. Therefore more energy needed to break the {bonds / overcome the intermolecular forces}(1)

Accept reverse argument

3 3

10 © WJEC CBAC Ltd.



(ii) Unsaturated contain {double bonds between carbons/ C=C}(1) Any 3 (x1) from:

Produces a kink / bend, in the side chain (1) Less packing is possible/ increased distance between the

chains (1) Lowers intermolecular force /(phase) transition temperature (1) less energy needed {to break the bonds/ overcome the

intermolecular forces} (1)

1

3 4

(iii) Between the {fatty acids / hydrophobic tails} Accept labelled diagram

1 1

(iv) Raises level of LDLs/ causes LDLs to be made (1) Increase incidence of atheroma in arteries/ detailed description of atheroma (1)

2 2





6 (a) Magnesium - chlorophyll (1) NOT chloroplast Calcium – (structure of) cell walls in plants (1) Phosphate - nucleic acids/ nucleotides/ phospholipids/ ATP/ NADP/ NAD/ FAD(1)

3 3

(b) (i) Active transport (1) Against concentration gradient (1) Correct use of data (1)

1 2 3

(ii) A. anaerobic conditions / lack of oxygen/ oxygen is required(1)

B. {Less/ no} ATP produced (1)

C. active transport {cannot occur/ reduced}, fewer ions can be

transported (against concentration gradient) (1)

D. Stunted growth because of a lack of named ion {e.g.

phosphate, required to make DNA and proteins essential for

growth} (1)

3

1

4





7 Indicative content: A: Cell division

Mitosis

DNA mass doubles, then halves/ returns to same content

one cell cycle is within 24 hours

B: DNA mass

DNA replication occurs during interphase

Causing DNA to double

After interphase DNA mass remains constant

After interphase DNA present as chromosomes/ 2

chromatids

Mass DNA halves at telophase/ cytokinesis

when nuclear membrane reforms

C: Cell mass

Interphase: mass of cell increases

As does cell size

Mass of cell continues to increase

due to replication of DNA / organelles

and possibly due to formation of spindle fibres

cell undergoes cytokinesis and mass of cell halves

4 5 0 9

0 0


7-9 marks Indicative content of this level Correct identification of mitosis with explanation Detailed explanation of changes in DNA mass Detailed explanation of changes in cell mass

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 Indicative content of this level Correct identification of mitosis possibly with explanation Either explanation of changes in DNA mass Or explanation of changes in cell mass


1-3 marks Indicative content of this level Either Correct identification of mitosis possibly with explanation or brief explanation of changes in DNA mass Or Brief explanation of changes in cell mass


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.




Unit 1: BASIC BIOCHEMISTRY AND CELL ORGANISATION



1 5 4 0 9 3 0

2 5 7 1 13 0 3

3 4 3 3 10 2 0

4 3 7 5 15 3 11

5 3 5 6 14 0 0

6 4 5 1 10 0 0

7 4 5 0 9 0 0

TOTAL 28 36 16 80 8 12

WJEC GCE AS Biology Unit 1 (New) MS Summer 2017/GH

© WJEC CBAC Ltd.


SUMMER 2017 AS (NEW) BIOLOGY - COMPONENT 1 B400U10-1

1 © WJEC CBAC Ltd.

COMPONENT 1 – Basic Biochemistry and Cell Organisation

MARK SCHEME


Recording of marks





Marking rules









2 © WJEC CBAC Ltd.



1 (a) (i) triglyceride (1)

Any 2 from:

protection of vital organs (1)

{thermal/ electrical} insulation (1) NOT prevent heat loss

energy {storage / source}(1) NOT energy release

metabolic water (1)

buoyancy (1)

waterproofing (1)

3 3

(ii) mix sample thoroughly with ethanol (and water) (1)

emulsion (test) / goes {cloudy/ milky}(1)

Accept details of alternative tests, e.g. Sudan III - goes red,

brown paper test - translucent

2 2 2

(iii) X is saturated but {Y and Z are unsaturated /

Y monounsaturated + Z polyunsaturated} (1)

unsaturated fats decrease level of {LDL / low density

lipoprotein} / cholesterol/ ORA/ unsaturated fats increase

level of HDL (1)

reduces risk of{ heart disease / atherosclerosis/ atheroma

formation/ description of atheroma formation}/ ORA (1)

1

1

1

3

(b) (i) {phosphate/ head} is {hydrophilic/ polar} so attracted to

water/ owtte (1)

{fatty acids/ tails} are {hydrophic/ non-polar} so repelled from

water/ owtte (1)

Accept heads are hydrophilic, tails are hydrophobic (with no

explanation)

2 2

(ii) 140 x 2 = 280µm2 1 1 1

(iii) phospholipids arranged in bilayer in cell membrane (but in a

single layer on water)

1 1

3 © WJEC CBAC Ltd.



1 (c) ref to fluid mosaic model (1)

proteins can {move / diffuse} within membrane / proteins are

arranged randomly(1)

{Fluorescence restored in the area exposed to the laser after

5 minutes / {Other/ non affected} proteins can move into the

area exposed to the laser after 5 minutes (1) (must in context

of protein movement)

1

1

1

3


4 © WJEC CBAC Ltd.



2 (a) (i) domain = eukaryote/ eukarya (1)

kingdoms: human = Animal(ia) and yeast = fungi (1)

human has no cell wall and yeast has chitin cell wall (1)

1

1

1 3

(ii) Correct answer= 12000/ 12005/ 12009= 2 marks

Accept answers which would round to the above

If incorrect, accept either of following for 1 mark

Use of scale bar [(12 x 1000)/1]

width of image/ length of scale bar [110/12 = 9.16]

height of image/ [length of scale bar [91/12=7.58]

Accept measurements of either cell divided by 12/1.2 (must

be matching units)

2 2 2

(iii) I digestion will be internal / intracellular (1) 1 1

II Golgi body {processes/ produces/modifies} enzymes/

{packages into/ produces} lysosomes (1)

lysosomes fuse with phagocytic vesicle and release enzymes (1)

1

1

2

(b) (i) molecule of {nucleic acid / DNA / RNA} surrounded by a {protein

coat / capsid} (1)

(acellular as) {does not have membranes/ no organelles/ cell

membranes/ cytoplasm} (1)

Accept labelled diagram

1

1

2

(ii) = 0.13µm / 1.3 x 102 nm/ 130 nm = 3 marks

= 1.3 x 10-4mm/ 0.00013mm/ 1.3 x 10-7m/ 0.00000013m = 2

marks (inappropriate units for virus)

0.133333...µm/ 1.333333 x 102 nm = 2 marks (not 2 sig fig)

18 x 1000 / 135000 = 1 mark for calculation

Deduct one mark for missing units/ wrong units

3 3 3


5 0

5 © WJEC CBAC Ltd.



3 (a) A = chromatid (1) ignore sister / daughter

reject chromosome

B = centromere (1)

2 2

(b) (i) crossing over / synapsis / chiasmata formation (1)

reject chiasmata are exchanged/ crossed over

in Prophase I (of meiosis) (1)

2

2

(ii) Any two (x1) from:

{two (cell) divisions/ two named phases I and II} (producing

four cells) (1)

crossing over/ independent assortment/ or description of(1)

Only one chromatid from each pair of chromosomes in each

daughter cell

2 2

(iii) {more difficult for/ less likely that/ more rare that/ lower

probability that} crossing over to take place (1)

because Y chromosome shorter than X chromosome/OWTTE

(1)

1 1 2


6 © WJEC CBAC Ltd.



4 (a) Staining/ add a dye/ add correct named dye

1 1 1

(b) Prophase B

Metaphase C and E

Anaphase F

Telophase A and D

6 correct answers = 3 marks

4/5 correct answers = 2 marks

2/3 correct answers = 1 mark

0/1 correct answers = 0 marks

3 3 3

(c) (i) 12.8 = 2 marks

If incorrect 1 mark only for any of:

6/47 x 100

12.76

rounded to 12.7

13.0 NOT 13

2 2 2

(ii) Answer must be comparative

higher mitotic index closer to the tip/ORA (1)

{higher rate of growth/higher rate of cell division/ shorter cell

cycle/ mitosis takes place more rapidly/ more cells undergoing

mitosis} / ORA (1)

2 2

(iii) {Repeat/ increase sample} and calculate a mean (1)

Accept average

1 1 1

7 © WJEC CBAC Ltd.



(d) Any four (x1) from:

A. mitotic index only gives information about cell division

OR

mitotic index does not take into account cell {length/ size} (1)

B. cells increase in size during growth/

OR

(cell) growth can also be measured by cell {size/ length}

OR

{cells get longer/ bigger} when they grow (1)

C. {cells are {longer/ bigger} {further from root tip/ at 1.8mm}

OR

cells are {shorter/ smaller} {closer to the root tip/ at 0.2mm}

OR

more {mitosis/ cell division} {at/ closer to} root tip

OR

less {mitosis/ cell division} further from root tip} (1)

D. Growth is a combination of cell division and increase in

size (1)

E. Use of data (1)

4

4

(e) Any three (x1) from:

cells may be damaged/ broken(1) NOT squashed

not stained enough (to see chromosomes)/

chromosomes not visible/ blurred image/ poor

resolution (1)

layers of cells overlap (1)

{non-random selection of /lack of consistency in how part

cells are included in} fields of view (1)

3 3 3


8 © WJEC CBAC Ltd.



5 (a) (i) {insertion/ addition} of water to break a bond NOT hydrogen 1 1

(ii) breaks (bonds) in the {middle/ inside} of {a molecule / DNA} 1 1

(iii) AGA¦TA¦TGGA¦CG¦TAA (1)

bonds broken on 5’ side of each nucleotide with a pyrimidine (1)

C and T are pyrimidines (1) Accept on diagram

1

1

1

3

(b) (i) Accept range 1.32 – 1.38 accept rounding to 1d.p. 1 1 1

(ii) higher initial rate as {substrate/ DNA} is at {maximum/ higher}

concentration (1)

(therefore) higher rate of collisions/ more ES complexes

formed/ owtte(1)

Accept reverse argument / explanation for 0.5 minutes

2 2

(iii) temperature and pH (1)

Any 2 from:

change in temperature changes kinetic energy and therefore changes {rate of reaction/ number of successful collisions} (1)

changes to pH {changes 3D structure of active site/ results in less enzyme substrate complexes/ successful collisions/ enzyme substrate complexes} (1)

extremes of temperature/ pH cause denaturation (1)

1

1

1

3

3

(c) Any three (x1) from:

DNA used in experiments was synthetic + DNase may not be effective against human DNA (1)

(DNA used in experiments) was single stranded + DNA from dead cells double stranded (1)

not tested on live {human cells / target organ}/ valid comment on experimental vs live conditions (1)

could damage (healthy) cells/ could cause side effects (1)

3

3


9 © WJEC CBAC Ltd.



6 Indicative content

Primary, secondary, tertiary, quaternary structure

primary – sequence of amino acids held by peptide bonds

secondary – coiling/ folding of polypeptide chain held by

hydrogen bonds

tertiary – further folding of secondary structures due to R

group interactions/ ionic/ covalent/ S-S/ hydrogen bonds/

hydrophobic interactions

quaternary – more than one polypeptide chain held together

Effect of changes on levels of protein structure

primary protein structure changed due to amino acid

sequences of both A and B chains are different

secondary structure could change how the α helices in A

and B chains would form

tertiary structure would change as different amino acids

would change R/ variable groups bonding between different

parts of polypeptide chains

e,g, Changing aa 1 and 19 of chain A could result in

additional / different disulphide bridges forming

Quaternary structure Changing Cys to Gly (aa B chain 19)

and Cys to Tyr (aa A chain 7) mean that disulphide bridges

may not form between the A and B chains.

Effect on Functionality

Insulin must be able to bind to receptors so must have a

specific shape

Any change to 3D shape of the insulin molecule could

affect how it binds to receptor molecules

{prevent/ reduce abiity of} cells to absorb glucose from the

plasma.

5

4

9




7-9 marks

Indicative content of this level is: Detailed explanation of levels of protein structure and Detailed explanation of effects of change on protein structure and Detailed explanation of the effect of functionality The candidate 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 levels of protein structure Explanation of effects of change on protein structure Explanation of the effect of functionality


1-3 marks

Indicative content of this level is: Brief explanation of levels of protein structure OR Brief explanation of effects of change on protein structure OR

Brief explanation of the effect of functionality


Question 6 total 5 4 9


COMPONENT 1 – AS BIOLOGY EDUQAS 2017


Q AO1 AO2 AO3 TOTAL MARK MATHS PRAC

1 9 6 0 15 1 2

2 4 9 0 13 5 0

3 7 0 1 8 0 0

4 4 5 7 16 2 7

5 3 8 3 14 1 3

6 0 5 4 9 0 0

TOTAL 27 33 15 75 9 12

B400U10-1 Eduqas AS Biology - Component 1 MS Summer 2017/GH

© WJEC CBAC Ltd.


SUMMER 2018 AS (NEW) BIOLOGY - UNIT 1 2400U10-1

© WJEC CBAC Ltd. 1

WJEC GCE AS BIOLOGY

UNIT 1 – Basic Biochemistry and Cell Organisation

SUMMER 2018 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.




© WJEC CBAC Ltd. 2



1 (a) A: Mitochondrion/mitochondria B: Golgi {body/complex/apparatus} C: Ribosome(s) D: Nuclear pore 2 or 3 correct for 1 mark All correct for 2 marks

2 2

(b) (i) Nucleus/ E: (contains DNA which) codes for the production of {proteins/polypeptides/sequence of amino acids in a polypeptide}/ transcription/ {pre-processing/ production/ synthesis} of mRNA (1)

Nuclear pores/ D:

allow {mRNA/rRNA} to leave the nucleus (1)

Ribosome(s)/C: carry out translation/ protein synthesis/or description of (1)

3 3

© WJEC CBAC Ltd. 3



1 (b) (ii) Any 3 × (1) from: Rough Endoplasmic Reticulum/ RER/ F:

transports proteins {through the cell/ through the cytoplasm}/ transports proteins to golgi body/ package proteins into vesicles (1)

Golgi body/ B: {Packaging/Modification} of protein/description of/ activation of enzyme (1)

(transport) vesicle/ G: transports {proteins/enzymes} to the {cell membrane/ plasma membrane} (1)

Exocytosis (of enzymes from the cell) (1)

3 3

(iii) Provide ATP : {for transcription / translation / protein synthesis} (1) exocytosis (1)

2 2


© WJEC CBAC Ltd. 4



2 (a) (i) X: phosphate

Y: ribose

Z: adenine NOT adenosine

(2 correct = 1 mark, 3 correct = 2 marks)

2 2

(ii)

Any two for one mark

Active transport / DNA replication / protein synthesis/ cell division (1)

Accept any correct function for a plant cell

1 1

(b) 40.5 = 3 marks

If incorrect sig fig

1162.8/2870 × 100 =40.5156794 = 2 marks Accept any correct

rounding

If answer incorrect

38 x 30.6 = 1162.8 = 1 mark

3 3 3

Question 2 total 3 3 0 6 3

© WJEC CBAC Ltd. 5



3 (a) J, K & M = 2 marks 2 correct = 1 marks 0/1 correct = 0 marks If use more than three letters then deduct one mark for each additional letter

2 2

(b) (i)

4 correct for 2 marks 2/3 correct for 1 mark 0/1 correct = 0

2 2

(ii) Eggs/ female gametes/ ova are produced by meiosis (1) Sperm/male gametes are produced by mitosis (1)

2 2

(c) (i) 1. 3.4 units DNA {before replication / in early interphase}/

{quantity of DNA halves/ returns to original value} {following

cytokinesis / (at the end of )telophase} (1)

2. {6.8 units of DNA/ DNA doubles} due to DNA replication (1)

3. (6.8 units of DNA will be present during) (Late) interphase/

prophase/ metaphase/ anaphase(1)

3 3

© WJEC CBAC Ltd. 6



(ii) {mitosis is faster/ more mitosis} in young spider mites than older spider mites/ ORA (1) Young spider mite - mitosis required for growth (and repair of muscle tissue) / Older spider mite – mitosis required for repair (of muscle tissue only) (1)

2 2


© WJEC CBAC Ltd. 7



4 (a) (i) Molecules drawn with a peptide bond correct (1) Molecule of water/H2O also produced (1)

2 2

(ii) peptide bond 1 1

(b) (i) Quaternary 1 1

(ii) α helix (1) By hydrogen bonding (1) Ignore reference to peptide bonds

2 2

(iii) Two - one for each polypeptide chain (1) 1 1

(c) (i) 360 / 363 / 366 (1) 1 1 1

(ii) Any five (×1) from: 1. DNA/ (pre)mRNA contains both introns and exons (1)

2. The entire gene is transcribed (1)

3. Introns are non-coding/ only exons are coding (1)

4. The (pre)mRNA is then edited to remove the introns / introns

are removed from (pre)mRNA/ only leave exons(1)

5. 'pre-edited' (pre) mRNA molecules and 'post-edited' mRNA

molecules have different masses/ OWTTE (1)

6. only the 'edited' mRNA is used {in translation / to produce

the protein} (1)

2 3 5


© WJEC CBAC Ltd. 8



5 (a) (calculating the percentage change) allows the results to be comparable (1) Initial masses of worm would differ (1)

2 2 2

(b) (i) Linear scale (1) including figure at origin and no break line Correct plots (1) ± ½ small square Line drawn (with no extrapolation) (1) extrapolation is acceptable with a curve of best fit

3 3 3 3

(ii) Any four (x1) from: 1. Increase to 60 and then levels off(1)

2. (the body fluids of the marine) worms {are hypertonic to

/have a lower water potential than} {the dilute seawater/ the

solution}/ ORA(1)

3. The worms gain water by osmosis (resulting in an increase

in mass)(1)

4. (during the first 15 mins) the increase in mass is greater due

to the steepness of the water potential gradient(1)

5. (after 60 mins there is no change in mass as) equilibrium is

reached / isotonic/ no net movement of water/ OWTTE (1)

4 4 3

(c) As a control (1) to show that any change in mass was due to the worms being placed in diluted seawater (and no other variable).(1)

2 2 2

(d) (i) The percentage change in mass of Golfingia {returns to zero/ decreases} (after 45 mins) whereas the mass of Nereis {continued to increase/ plateaus} (1) The maximum percentage increase in mass in Golfingia {is much lower / is reached sooner} than that of Nereis/ equilibrium is reached quicker in Golfingia than in Nereis(1)

2 2 2

© WJEC CBAC Ltd. 9



5 (d) (ii) increase water potential of the cells of Golfingia (1) therefore water moves out by osmosis (1)

2 2 2





6 (a) (i) Activation energy 1 1

(ii) Curve drawn with a lower activation energy under existing curve –

energy state at beginning and end must be the same. 1 1

(b) {Alcohol dehydrogenase/ enzyme} has a specific {shaped active

site/ tertiary structure/ OWTTE} (1)

{Ethanol/ substrate} has a complementary (shape) (1)

(The two fit together) to form an enzyme-substrate complex (1)

3 3

(c) (i) Any answer between 0.57 - 0.63 = 2 marks

If incorrect award 1 mark for sight of :

attempted calculation of gradient

2 2 2

(ii) 1. P: the rate of reaction is higher as the concentration of

{ethanol/ substrate} is high (1)

2. The concentration of {alcohol dehydrogenase/ enzyme} is

limiting the rate of reaction; (1)

3. Q: the rate of reaction is lower the concentration of {ethanol/

substrate} decreases(1)

4. The concentration of {ethanol/ substrate} becomes the

limiting factor. (1)

2 2 4




6 (d) 1. Ethanol and ethylene glycol must have a similar structure /

Ethanol and ethylene glycol must both be complementary to

the shape of the active site of {alcohol dehydrogenase/ the

enzyme} (1)

2. ethanol acts as a competitive inhibitor (1)

3. When ethanol binds to the active site it prevents ethylene

glycol from attaching (1)

4. Fewer enzyme-substrate complexes form (1)

5. Which reduces the rate of production of {glycoaldehyde/

product} (1)

3

2

5





7 Starch

Polymer of α-glucose

Composed of amylose and amylopectin

Amylose contains only 1,4 glycosidic bonds

Forms a helical structure

Amylopectin contains 1,4 and 1,6 glycosidic bonds

Forms a branched structure

Triglycerides/lipids

Composed of glycerol and three fatty acids

Joined together by ester bonds

Saturated fatty acids contain only C-C single bonds

Unsaturated fatty acids contain at least one C=C double bond

Properties of triglyceride dependent upon the fatty acids they contain

Functions in the seed

Starch and triglycerides are insoluble so are osmotically inert

Starch’s helical/branched structure makes the molecule compact

Hydrolysis provides glucose readily

Required for respiration/produce ATP

Triglycerides also have a compact structure

Triglycerides have many high energy bonds/ provide

approximately twice the quantity of energy than starch

6 3





Detailed description of the structure of starch

Detailed description of the structure of lipids/triglycerides

Detailed explanation of how these structures and properties relate to their function in the seed

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 Indicative content of this level is… Any two from:

Description of the structure of starch

Description of the structure of lipids/triglycerides

Explanation of how these structures and properties relate to their function in the seed






Brief description of the structure of starch or

Brief description of the structure of lipids/tryglycerides or

Reference to how these structures and properties relate to their function in the seed

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 1: BASIC BIOCHEMISTRY AND CELL ORGANISATION



1 8 2 0 10 0 0

2 3 3 0 6 3 0

3 0 7 4 11 0 0

4 6 3 4 13 1 0

5 0 11 4 15 3 15

6 5 7 4 16 2 0

7 6 3 0 9 0 0

Total 28 36 16 80 9 15

2400U10-1 GCE AS level Biology Unit 1 MS Summer 2018/LG

1 © WJEC CBAC Ltd.

EDUQAS AS COMPONENT 1 BASIC BIOCHEMISTRY AND CELL ORGANISATION

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.

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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

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1 (a) (i) {A has/meiosis I results in} two {cells/nuclei} and {B has/ meiosis II results in} 4 / A results from one division and B results from two divisions (1)

1 1 1

(ii) plane/angle of section of through cell may not include a nucleus (1) (where nucleus visible) may have been cut at different {levels/planes} (1)

1 1

2 2

(b) anaphase II meiosis (1) Any two (x1) from: Cell is haploid as only 4 chromosomes / resulting cells will {only have one copy of each chromosome/be haploid} (1) if mitosis two copies of each chromosome / lack of homologous pairs (1) if anaphase I each chromosome would have 2 chromatids / (anaphase II) involves the separation of (sister) chromatids (1)

2

1 3


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S AO1 AO2 AO3 Total Maths Prac

2 (a) (i) F B D All 3 =2, 2 = 1;

2 2

(b) (i) ATP can be regenerated quickly/More ATP for muscle contraction (1) Not: more energy produced

1 1

(ii) lock and key active site shape already ‘fixed’/ 'perfectly complementary' (1) (3D) active site changes shape when substrate binds/OWTTE (1)

2 2

(c) (i) intra act within a cell + extra outside a cell (1) 1 1

(ii) contains {N/amine group} polysaccharides only CHO (1) 1 1

(iii) same {molecular/chemical} formula but different {structural formulae / structure / arrangement of atoms} (1)

1 1

(iv) beta because {–OH/-O-/glycosidic bond) is above C1/H is below C1 (1)

1 1

(v) alternate monomers rotated 180o / forms (long) straight chains (1) enables H-bonding between {adjacent/parallel} chains (1) H bonds weaker/ {peptide/covalent} bonds much stronger (1)

2

1 3


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3 (a) (i) ribosome + rRNA is ribosomal RNA/constituent of ribosomes (1) OR Nucleolus + synthesis of RNA

1 1

(ii) 15.5/100 x 50 000 000 000 = 1 mark = 7 750 000 000 / 7.75 x 109 = 2 marks = 7.8 x 109 = 3 marks

3 3 3

(iii) different mRNA for every {protein/polypeptide}/many different genes/one gene one polypeptide/OWTTE (1) ref to base triplet code hypothesis (in mRNA/DNA) giving {64/43} triplets (1) tRNA molecules have anticodons which have three bases (which are complementary to codons)/tRNA are specific to one amino acid(1)

3 3

(b) peptide (1) bond could form due to loss of OH from methionine or from glycine/amino end and carboxyl end / depends upon the {DNA/mRNA} base sequence (1)

1 1 2


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4 (a) (i) longitudinal (1) cells have been cut {vertically/down}/(in LS) cells appear rectangular/in TS cells would appear more ‘round’/shape of thickening / several rings of lignin can be seen (1)

2 2 2

(ii) correct label of a xylem vessel (1) only one type of cell/similar cells aggregated to perform the same function/OWTTE (1)

2 2 1

(b) (i) H δ+ / slight positive charge (1) O δ- / slight negative charge (1) Accept shown on diagram uneven distribution of charge/electrons / dipole / hydrogen has positive charge and oxygen has a negative charge = 1 mark

2

2

(ii) Any four (x1)from: cohesion between between H2O molecules (1) Hydrogen bonding (between Hδ+ and Oδ-) in context (1) {evaporation/loss} of water from leaves {places tension on water column in xylem / reduces pressure at the top of the xylem} (1) pulls water up the xylem/transpiration stream (1) adhesion of water molecules to xylem (1)

4 4

(c) (i) Mg2+ chlorophyll NO3

- (N source) for protein/nucleic acid/ATP/amino acids/

2 2

(ii) Hypothesis 1 higher [NO3

-] in xylem than phloem/moving down concentration gradient (1) Accept use of figures plasmodesmata provide (cytoplasmic) channels for movement 1) Hypothesis 2 higher [PO4

3-] in phloem than xylem (1) Accept use of figures movement is against concentration gradient (1) through carrier proteins embedded in cell membranes (1)

1

1

1

2

2

3

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(d) (i) 5s = distance2 x 1/5 x 104 distance2 = 250000 / distance2 = 5 x 5 x 104 = 1 mark (rearrangement of equation) distance = √ 250000 or √25 x 104 = 2 marks

= 500µm = 3 marks

3 3 3

(ii) increase in KE increases speed of movement of molecules (1) 1 1


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5 (a) chloroplast/cellulose cell wall/starch grains = plant (1) flagellum/ small vacuoles/eyespot = animal (1) Not mitochondria (golgi body neutral)

2 2

(b) (i) 4/3 x 3.142 x 13 = 4.189/4.19/4.2 (µm3)1 mark for volume volume of water released at -0.24MPa = 15.5µm3 1 mark for graph reading 15.5/4.189 = 3.7 = 3 marks ecf if incorrect volume used

3 3 3

(ii) Any four (x1)from: 1. water moves into {cell/contractile vacuoles}(1) 2. by osmosis (1) 3. down a water potential gradient / from a hyopotonic

solution to a hypertonic soltion / water potential inside (cell/contractile vacuoles) lower than water potential outside (1)

4. (as the water potential of the external solution increase) the water potential gradient becomes steeper (1)

5. Rate of osmosis increases / water moves into the cell faster (1)

6. contractile vacuoles empty to get rid of excess water from the cell (1)

1

3 4

(c) (i) chromosome broken down {into fragments/pieces of different sizes} (1) travel different distances from top of centrifuge tube (1)

2 2 2

(ii) No/less digestion of the {DNA/chromosomes} (1) fewer fragments/less variation in distances travelled (1) different peaks seen for each chromosome (1)

3 3 2

(iii) cold temperature – to reduce {activity/KE} of enzymes (1) buffer – to maintain pH (1)

2 2 3

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(iv) Any two (x1)from: single peak/spike for DNA from both organelles (1) Therefore there is only one molecule of DNA present (1) could be arranged as plasmids/circular DNA as found in bacteria (1)

2 2 1

Question 5 total 2 1 8 9 18 3 8




6. Indicative content

• description of experiment including details of use of heavy and light isotopes of N and use of ultracentrifugation

• explanation of how bands seen in centrifuge tube are formed giving details of heavy and light strands for each generation in terms of the heavy and light isotopes of N

• the relative amounts of DNA in each band are explained

conservative replication:

• always some heavy DNA present • increasing mass of light DNA • no intermediate DNA • because original heavy DNA is not split

Dispersive replication:

• apart from G0, all DNA would be intermediate in molecular mass

• getting lighter between Go and G3 • because original heavy DNA split between all new

molecules

2 2 5 9




7-9 marks

detailed description of experimental procedure and explanation of results

correct detailed explanation to reject conservative theory

correct detailed explanation to reject dispersive theory

The candidate 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

description of experimental procedure and explanation of results

correct explanation to reject conservative theory

correct explanation to reject dispersive theory OR

detailed description of experimental procedure and explanation of results

correct detailed explanation to reject either conservative theory or dispersive theory





1-3 marks

some description of experimental procedure and/or explanation of results

OR

correct explanation to reject conservative theory OR

correct explanation to reject dispersive theory




COMPONENT 1 – AS BIOLOGY EDUQAS 2017


Q AO1 AO2 AO3 TOTAL MARK MATHS PRAC

1 1 4 1 6 0 3

2 9 3 0 12 0 0

3 1 8 0 9 3 0

4 13 8 0 21 3 3

5 1 8 9 18 3 8

6 2 2 5 9 0 2

TOTAL 27 33 15 75 9 16

B400U10-1 EDUQAS AS BIOLOGY - COMPONENT 1 SUMMER 2018 MS

gce as marking scheme - thiacin.co.uk · 1 wjec new as biology unit 1 – basic biochemistry and...

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