2012 H1 JC2 Prelim - Core (Mark Scheme)

Civics Group

Index Number

Name (use BLOCK LETTERS)

ST. ANDREW’S JUNIOR COLLEGE2012 Preliminary Examination

H1 BIOLOGY 8875/2

Paper 2: Core (Mark Scheme)

Friday 14 September 2012 2 hours

Additional Materials: Answer PaperCover Sheet for Section B

READ THESE INSTRUCTIONS FIRST

Write your name, civics group and index number on all the work you hand in.Write in dark blue or black pen on both sides of the paper.You may use a soft pencil for any diagram, graph or rough working.Do not use staples, paper clips, highlighters, glue or correction fluid.

Section AAnswer all the questions.

Section BCompulsory question to be answered on writing paper provided.

At the end of the examination, fasten all your work securely together.The number of marks is given in brackets [ ] at the end of each question or part question.

For Examiner’s Use

Section A

1 /12½

2 /9

3 /10

4 /8½

5/6 /20

Total /60

This document consists of 11 printed pages.[Turn over

H1

2

Section A

Answer all questions.

QUESTION 1

Figure 1.1 is a diagram of an animal cell as seen using a transmission electron microscope.

Figure 1.1

(a) Name the structures of the cell labelled A, B, C and D

……………………………………………………………………………………....….……..[2]1 A – Smooth endoplasmic reticulum ;2 B – Nuclear, membrane / envelope ;3 C – Mitochondrion ;4 D – Nucleolus ;

(b) Structures C and E are examples of the same organelle. Suggest 2 reasons why E looks so different to C.

………………………………………………………………………..…………....…..……..[1]1 Mitochondria vary in shape ;2 Cut in different planes / angles ; NEED comparative statement

e.g. C has been cut in longitudinal plane, E has been cut in transverse section e.g. one cut horizontally, one cut vertically e.g. in different positions / one viewed from above the other from the side

SAJC / H1 Biology 8875/2 JC2 Preliminary Examinations 2012

3

3 Just divided / growing ;4 Artefact / deformed during preparation of section ;(c) Proteins are produced by the structure labeled F. Some of these proteins may be

extracellular proteins that are released from the cell.

Outline the sequence of events following the production of extracellular proteins that leads to their release from the cell.

……………………………………………………………………………………....….……..[4]1 Proteins undergo post-translational modifications (e.g. proteolysis,

biochemical modifications) in the rough ER / F ;2 Packaged into transport vesicles which bud off ER / F ;

3 Membrane of transport vesicle fuse with cis face of Golgi ;4 Proteins undergo further post-translational modifications

(e.g. biochemical modifications) ;5 Packaged into secretory vesicles; 6 Bud off trans face of Golgi ;

7 Secretory vesicles move towards cell / plasma membrane ;8 Vesicle membrane fuse with cell membrane ;9 Release of proteins by exocytosis ;10 Microtubules of cytoskeleton involved in transport of vesicles between

organelles ;

(d) The division of stem cells by mitosis produces genetically identical cells.

(i) Explain what is meant by the term pluripotency.……………………………………………..………………………...…………………….....[2]1 Reference to embryonic stem cells ;2 Ability of cells to differentiate into almost any cell type ;

/ to form any organ or type of cell ;3 Except the extra-embryonic tissues / placenta ;4 Not totipotent but are multipotent ;

(ii) Explain the normal functions of blood stem cells (found in the bone marrow) in a living organism.

……………………………………………..………………………...…………………......[1½]1 Differentiates to form platelets, red blood cells and white blood cells ;2 Replaces worn-out blood cells / daily turnover and for fighting infections ;3 Lost through normal wear and tear, disease, injury ;

(iii) Suggest how a blood stem cell could differentiate into its specialized cells.……………………………………………..………………………...…………………….....[2]1 Internal and external chemical signals ;2 Required to switch on and switch off certain genes ;3 Different proteins are produced ;4 To form cell-specific structures (for specialization) ;


4

[Total : 12½]


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QUESTION 2

(a) Explain the natural function of restriction endonucleases

…………………………………………………………………………………………..…….[1]1 Cuts foreign DNA that enters bacterial cell ;2 Protects bacteria from viral / bacteriophage infections ;

Figure 2.1 shows the structure of a functional insulin molecule. It is composed of two polypeptide chains: an A chain of 21 amino acids and a B chain of 30 amino acids. These two chains are held together by two disulfide bonds formed between two amino acid residues.

Figure 2.1

Currently, the production of recombinant insulin involves artificially synthesising the genes for the A chain and the B chain and then inserting each into 2 separate plasmids such that two recombinant plasmids are constructed (Figure 2.2), one carrying the gene for the A chain and one the gene for the B chain.

The A chains and B chains are synthesised separately in two different batches of E.coli. The two protein chains are then extracted and purified. Finally, the A and B protein chains are mixed under special conditions allowing disulphide bonds to form in vitro.

Figure 2.2

(b) State the function of the β-galactosidase gene in the plasmids.

…………………………………………………………………………………………..…….[1]1 Genetic marker ;2 Enables the selection of cells that have taken up recombinant plasmids ;


gene for chain A

gene for chain B

gene for β-galactosidase

gene for β-galactosidase

lac promoter lac promoter

6

(c) Explain the addition of the lac promoter in both vectors used for the insertion of genes for chain A and B.

…………………………………………………………………………………………….......[1] 1 lac promoter being prokaryotic in origin ; 2 Recognized by (prokaryotic) RNA polymerase ;

3 Transcription initiation can occur / ensures the expression / transcription of the gene of interest ;

(d) Explain how the cells must be first treated in order for transformation to occur.

…………………………………………………………………………………………..…….[2]1 Cells are made competent ;

(i.e. rendering them capable of taking up naked DNA from the environment / solution)

2 Addition of calcium ions (in the form of calcium chloride solution) ;

3 Cells are then subjected to brief heat shock / electroporation ;4 To create transient pores in the bacterial cell membranes ;5 For entry of recombinant plasmid into cells ;

(e) Suggest how the above introduction of recombinant DNA molecules into host cells is similar to mutations.

…………………………………………………………………………………………..…….[1]1 Both introduce new genetic variation into the cell;2 Only one gene or small segment of DNA representing a small fraction of the

entire genome is changed or added to the genome;3 Leads to changes in phenotype of the host cells;

(f) State two advantages of treating diabetics with human insulin produced by genetic engineering.

…………………………………………………………………………………………..…….[1]1 Identical to human insulin produced in human body and less likely to

trigger a possible immune / allergic response ;2 Eliminates risk of zoonosis (infectious disease transmitted from animals to

humans) ;3 Cheap / fast method of production of large quantities of the human protein ;4 Overcomes ethical objections / does not kill animals (vs. production from

live animals) ;5 Reduces the need to acquire proteins from other sources ;


7

(g) The recombinant chain A and B protein produced as described above is identical to naturally occuring human insulin chains A and B.

However, the synthetic gene inserted into the plasmid may possess a different nucleotide sequence to the actual human gene. Give reasons why this is so.

…………………………………………………………………………………………..…….[2]1 Some amino acids are encoded by more than one codon

/ ref degenerate genetic code ;2 cDNA used for cloning ;3 has no introns ;4 prokaryotes do not have spliceosomes ; REJECT splicing machinery

[Total : 9]


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QUESTION 3

Many varieties of rice, Oryza sativa, die quickly when totally submerged by flood water during the monsoon season.

Figure 3.1 Source : Wikipedia

In 2006, a gene, Sub1A, on rice chromosome 9 was found to be involved in tolerance of prolonged submergence.

An allele, Sub1A-1, was found only in submergence-tolerant rice. Varities that are submergence-intolerant either have a second allele, Sub1A-2, or lack the gene altogether.

(a) Sub1A-1 codes for a protein that controls the transcription of the gene for alcohol dehydrogenase. This gene is activated when the plant is in anaerobic conditions.

Explain the importance to submerged rice plants of producing alcohol dehydrogenase.

……………………………………………..………………………...…………………….....[3]1 Lack of oxygen, plants undergo anaerobic respiration ;

2 Ethanal produced from oxidative decarboxylation of pyruvate ;3 Alcohol dehydrogenase catalyse conversion of ethanal to ethanol ;4 NADH oxidized to NAD+ / regeneration of NAD+

/ hydrogen atoms from NADH donated to ethanol to form NAD+ ;

5 NAD+ function as electron / hydrogen acceptors in glycolysis ;6 For continued production of ATP (under anaerobic conditions) ;


9

(b) Alleles Sub1A-1 and Sub1A-2 differ by one nucleotide. The amino acid, serine, in the protein produced by submergence-tolerant plants is replaced by proline in submergence-intolerant plants.

Suggest why the protein produced by submergence-intolerant plants in inactive.

……………………………………………..………………………...…………………….....[2]1 Different R groups ;2 Different bonds formed (between amino acids in the proteins) ;3 Tertiary / 3D shape of protein changed ;4 Conformation of DNA-binding site changed ; REJECT active site references

A variety of submergence-intolerant rice, which lacked the gene Sub1A, was genetically engineered to express Sub1A-1.

Typically, submergence-intolerant plants respond to submerged conditions by elongating rapidly.

The mean heights of three types of rice plants were compared before and after 10 days of submergence in water. The types of rice plants were:

submergence-intolerant plants, I transgenic submergence-intolerant plants expressing Sub1A-1, I+

submergence-tolerant plants, T.

The results are shown in Figure 3.2

Figure 3.2


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(c) With reference to Figure 3.2,

(i) Describe the effect of submergence on the three types of rice plants.……………………………………………..………………………...…………………….....[3]1 All three plants elongate during submergence ;2 Expression of Sub1-A1 in I+

and T resulted in reduced elongation ;

3 I+ much smaller before submergence ;4 Elongation by 5cm in I+ ; 5 Insignificant / 2cm, elongation in T ;6 Elongation by 28cm / x 2.4 ;

(ii) Suggest a reason why the response shown by the submergence-intolerant plants, I, may lead to their death.

……………………………………………..………………………...…………………….....[1]1 Rapid growth will require increased respiration ;2 Plants deleted of ATP ;

3 Oxygen deficit ;4 Built-up of toxic ethanol ;

5 Rapid elongation of cells ;6 Cell walls too thin / too little cellulose / poor support / collapse ;

(d) The allele Sub1A-1 could be introduced into high-yielding variety of submergence-intolerant rice either by genetic engineering or by selective breeding.

State two advantages of using genetic engineering, rather than selective breeding, to introduce the allele.

……………………………………………..………………………...…………………….....[1]1 Desired traits conferred by the allele are transferred ;2 Does not dilute desirable traits of high yield variety ;3 Does not dilute alleles of background genes ;4 Can use genes from different species can be combined (to produce a

transgenic organism) ;5 Shorter time required to see results compared to traditional selective

breeding which may take several generations ;

[Total : 10]


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QUESTION 4

Scientists have been studying finches on an island in the Galapagos for over thirty years.

Before 1982, the only type of ground finch living on the island was the medium ground finch, Geospiza fortis.

In 1982, the large ground finch, Geospiza magnirostris settled on the island. The large ground finch is almost twice the size of the medium ground finch.

Figure 4.1

The key food for both species is the large seed from the plant Tribulus cistoides. Both species use their beaks to crack open the seeds.

During the period of the study, there have been two major droughts in 1977 and 2003. Some of the observations made by the scientists are listed below :

The drought in 1977 wiped out mainly plant species producing small seeds, leaving Tribulus cistoides plants that produced large seeds with hard coats.

Following the 1977 drought, the population of the medium ground finch was greatly reduced and the average beak size had increased in both length and depth by approximately 4%.

The 2003 drought greatly reduced seed production by all plants including Tribulus cistoides.

Measurements were made of the beak mean lengths and mean beak depths of the medium ground finch population in both 2002 and 2005. These are shown in Table 4.1

2002 2005 % decreaseMean beak length /mm 11.2 10.6 5.4%Mean beak depth /mm 9.4 8.6 8.5%

Table 4.1

(a) Calculate the percentage decreases in the mean length and mean depth of the beaks of the medium ground finch population between 2002 and 2005. Write your answers in Table 4.1.

……………………………………………..………………………...…………………….....[1]


1 – Geospiza magnirostris2 – Geospize fortis

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(b) State the two main selection pressures acting on the medium ground finch population following the 2003 drought.

……………………………………………..………………………...…………………….....[1]1 Food / seed, shortage / (intraspecific) competition for food ;2 Competition from large ground finch / interspecfic competition ;

(c) State how selection pressure on the medium ground finch population was different following the 1977 drought.

……………………………………………..………………………...…………………….....[1]1 No competition from large ground finch / interspecific competition ;2 Seed size larger ;

(d) Explain the mechanism that has produced this change in mean beak size in the medium ground finch population between 2002 and 2005.

……………………………………………..………………………...…………………….....[4]1 Mechanism : natural selection ;

2 Heritable genetic variations in beak size ;3 (Selection pressure) Competition for food ;

4 Medium ground finch with larger beaks in competition with large ground finch ;

5 Medium ground finches with smaller beaks for opening small seeds at selective advantage ;

6 Less competition for small seeds ;

7 Medium ground finch with smaller beaks more likely to survive to maturity ;8 Reproduce ;9 Pass on alleles to offspring ; REJECT beneficial trait / genes10 Increase in frequency of alleles encoding for small beak, over a number of

generations ; CREDIT only if some mention of time was given

(e) It was suggested from the data in Table 4.1 that stabilizing selection was taking place in the medium ground finch population.

State whether you agree with this statement and explain your answer.….………………………………………..………………………...…………………….....[1½]1 Disagree ;

2 Directional selection is taking place ;3 Change in phenotypic feature of entire population is taking place;

/ More individuals with smaller beak size in the population ;

[Total : 8½]


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Section B

Answer the question.

Write your answers on the separate answer paper provided.Your answers should be illustrated by large, clearly labelled diagrams, where

appropriate.Your answers must be in continuous prose, where appropriate.

Your answers must be set out in sections (a), (b) etc., as indicated in the question.

5 (a) Distinguish between chromosomal and gene mutation. [5]

(b) With reference to specific examples, explain the significance of genetic engineering in improving the quality and yield of crop plants and animals in solving the demand for food in the world.

[8]

(c) Describe how the structure of the membrane affects the movement of substances into and out of a cell.

[7]

6 (a) Compare the structure of collagen and cellulose. [5]

(b) Describe how information on mature mRNA is used to synthesise polypeptides in eukaryotes.

[8]

(c) Describe the source of genetic variation in a population of sexually reproducing organisms.

[7]


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QUESTION 5

(a)Distinguish between gene and chromosomal mutation.

……………………………………………...……………………..…………….…..………[5]Features Gene Mutation Chromosome Mutation/

Chromosome AberrationExample Sickle cell anaemia due to substitution

of one nucleotide ;Down’s syndrome due to extra chromosome 21 ;

Definition Change in the base sequence/ nucleotide sequence of the DNA ;

Occurs at a single locus on a chromosome ;

Chromosomal aberration or mutation is defined as a change in the structure of a chromosome or the number of chromosomes ;

Several gene loci are involved ;

Mechanism Brought about by deletion, insertion, substitution or inversion of bases;

(Reject: Answers presented in a table (or in prose) that does not explain that deletion, insertion etc that are stated refers to changes in the bases.)

Brought about by:1. (Structure) Deletion, inversion,

translocation or duplication of the chromosome;

2. (Deletion) Non-disjunction (of homologous chromosomes or sister chromatids during mitosis or meiosis);

Results Only 1 gene is affected (and hence 1 polypeptide product is affected);

Effect is usually more profound as several genes are affected;

Base substitution may lead to the formation of new alleles;

Base sequence of individual gene usually not affected;

Affects the:1. Positions of several genes on a

chromosome;2. The number of genes on a

chromosome;3. Number of chromosomes in a cell;

(max of 1 mark for this point)a) Loss or gain of 1 chromosome/

aneuploidy;b) Loss or gain of sets of

chromosomes/ polyploidy;

Occurrence

Frequency

Occur anytime (could be due to mutagens or reading error in DNA replication);

(Note: For gene mutation as well as chromosomal aberration in chromosome structure, these are random events and therefore can occur anytime.)

More frequent as there are many more genes than chromosomes;(Note: it is necessary to qualify why the frequency of gene mutations appear to be higher)

For chromosome aberration in structure:Occur anytime (could be due to mutagens);

For chromosome aberration in number:Occurs when homologous chromosomes or sister chromatids fail to separate;During mitosis and/or meiosis;

Less frequent;


15

Evolutionary importance

Involves introduction of new alleles into the gene pool(Note: point awarded for new alleles once only)

Involves reshuffling of existing alleles in the gene pool

Consequence (Can have different consequences ranging from no effect, mutated protein produced to possible beneficial mutations.)

May have no effect eg. degenerate code i.e. substitution leads to same amino acid being coded for)/ in non-coding regions of the DNA;

May be drastic as in frame-shift mutations where a mutated polypeptide can be produced/ truncated polypeptide;

May have beneficial mutations (for natural selection to work on) opposed to no beneficial chromosomal mutations;

(Note: Not all gene mutations are less severe than chromosomal mutations. 1 gene can code for a protein that is required in a pathway that is crucial is providing material for several other pathways and hence the consequence can be fatal.)

Generally will cause the expression of a different phenotype eg. mental retardation as in Down’s syndrome


16

(b)With reference to specific examples, explain the significance of genetic engineering in improving the quality and yield of crop plants and animals in solving the demand for food in the world.

……………………………………………...……………………..…………….…..….……[8]Animals with faster growth rate1 Atlantic salmon / GMO salmon / AquAdvantage Salmon with faster growth rate ;2 (Modification) Active growth hormone regulating gene from the Pacific Chinook

salmon and regulatory sequences of the ocean pout ;3 (Result of modification) Produces higher levels of fish growth hormones ;4 Accelerated growth of fish / reaches its full length in a shorter period of time ;5 (Significance) More fish can be harvested in a year / allows salmon to grow all year

around (instead of only during Spring and Sunmmer)

Plants with improved nutritional qualities 6 Golden rice enriched with beta-carotene ;7 (Modification) Produced by transplanting genes from daffodils and bacteria ;8 (Result of modification) Helps prevent Vitamin A deficiency ;9 Which leads to blindness and susceptibility to disease in undeveloped countries ;10 (Significance) Improved nutritional qualities in staple crop plants ;

Pest-resistant plants11 Bt corn which produces the Bt toxin ;12 (Modification) Expression of Bt toxin gene from the bacteria Bacillus thuringiensis ;13 (Result of modification) Kills specifically insect pests ;14 (Significance) Reduce crop losses to increase yields ;

Herbicide resistant plants15 Glyphosate resistant soybean ;16 (Modification) Engineered with enzyme EPSP synthetase not inhibited by

glyphosate ;17 EPSP synthetase required to make essential aromatic amino acids ;18 (Result of modification) Can spray herbicides to kill non-resistant weeds without

affecting crops ;19 (Significance) Reduce crop losses to increase yields ;

Plants with delayed ripening20 Flavr-Savr tomato ;21 (Modification) Inserted antisense gene of enzyme polygalacturonase binds to the

mRNA, preventing it from being translated by the ribosomes ;22 Polygalacturonase normally responsible for the ripening process ;23 (Result of modification) Delay in fruit ripening / spoiling during transport / improved

shelf life ;24 (Significance) Larger fruit and has greater flavor ;OR25 (Modification) blocking the biosynthetic pathway for ethane ;26 ethene is hormone responsible for fruit ripening ;27 (Result of modification) fruits can be allowed to grow on the vine for a longer period

of time ;28 delay in fruit ripening / spoiling during transport / improved shelf life ;29 (Significance) larger and has greater flavor ;


17

(c)Describe how the structure of the membrane affects the movement of substances into and out of a cell.

……………………………………………...……………………..…………….…..….……[7]1 Membrane is selectively / partially permeable ;2 Due to phospholipid bilayer ;3 Hydrophilic phosphate heads face the aqueous environment on both sides ;4 Hydrophobic fatty acid tails face inwards, unable to interact with aqueous

environment ;

Non-polar substances5 Hydrophobic / non-polar substances diffuse across phospholipid bilayer directly ;6 Able to interact with the hydrophobic core / tails of fatty acid chains ;

Polar substances7 Polar molecules and ions are unable to pass through phospholipid bilayer

directly ;8 Require the presence of integral proteins on membrane ;

9 Channel proteins ;10 Allows selective ions and polar molecules to pass through water-filled pore ;11 Ref facilitated diffusion ;

12 Carrier proteins ;13 Selective binding and transport of polar molecules and ions across cell membrane

;14 Ref facilitated diffusion / active transport ;

15 Larger molecules may require bulk transport via endocytosis / exocytosis ;16 Involves formation of vesicle which encloses molecule to be transported ;


18

QUESTION 7

(a)Compare the structure of collagen and cellulose.

……………………………………………...……………………..…………….…..………[5]

Similarities

1 Both are fibrous in nature ;2 Both are long-chained molecules with cross-linkages between molecules ;

Differences [4max]

Collagen Cellulose1 Made up of amino acids ; Made up of (β-)glucose;

2 Monomers held together by peptide bonds ;

Monomers held together by (β1,4-) glycosidic bonds;

3 Basic structural unit is tropocollagen; Basic structural unit is polymer/chain of (β-)glucose;

4 Involves formation of triple helix (ref. to tropocollagen);

No formation of helix / all chains exist in linear and non-helical form;

5 Tropocollagen molecules staggered and held together by covalent cross-linkages;

Chains of cellulose held together by hydrogen bonds;

6 Collagen fibres arranged in bundles; Macrofibrils laid down in layers at right angles to each other;


19

(b)Describe how information on mature mRNA is used to synthesise polypeptides in eukaryotes.

……………………………………………...……………………..…………….…..….……[8]

Amino acid activation1 attachment of an amino acid to its specific tRNA (amino acid activation)2 catalysed by aminoacyl tRNA synthetases

Initiation (max 2.5)3 initiator tRNA with anticodon UAC carrying methionine4 associates with the small ribosomal subunit

5 (initiator tRNA-small ribosomal subunit) binds to the 5’ cap of the mRNA6 (small ribosomal subunit) then moves downstream along the mRNA until it

reaches the start codon/AUG, 7 (initiator tRNA) binds with the start codon/AUG (through complementary base-

pairing)8 at P site of large ribosomal subunit

9 formation of translation initiation complex / assembly of ribosome + mRNA + translation initiation factors (TIFs)

Elongation (max 2.5)10 ref. complementary base pairing between anti-codon of incoming aminoacyl-

tRNA binds to codon on mRNA11 at A site of large ribosomal subunit12 peptide bonds formed between adjacent amino acids catalysed by peptidyl

transferase;

13 ribosomes translocates one codon / 3 nucleotides along the mRNA in the 5’ to 3’ direction

14 movement expose codon at A site of ribosome to receive another aminoacyl-tRNA

15 initial tRNA is relocated to E site and ejected from ribosome16 tRNA carrying the growing polypeptide chain is repositioned to P site

Termination17 chain termination occurs when stop codon (UAG, UGA, UAA) in A site;18 a protein called release factor binds directly to the stop codon in the ‘A’ site 19 addition of water instead of amino acid to release polypeptide chain;


20

(c)Describe the source of genetic variation in a population of sexually reproducing organisms.

……………………………………………...……………………..…………….…..….……[7]

1 Spontaneous mutation in germ-line cells ;2 Gene mutations and chromosomal mutations ;3 (Gene mutations) Gives rise to new alleles ;4 (Chromosomal mutations) Reshuffling of alleles on a chromosome ;

5 Crossing over during Prophase I ;6 Between non-sister chromatids of homologous chromosomes ;7 Exchange of sections of chromatids ;8 Genetic recombination occurs ;9 Creates new combinations of allelles on the chromosomes of the gametes;

10 Independent assortment during Metaphase I ;11 Random arrangement of homologous chromosomes at metaphase plate ; 12 Each daughter cell has 50% chance of receiving either of the homologous

chromosomes ;

13 Independent assortment of chromatids during Metaphase II ; REJECT sister chromatids

14 Forms gametes with unique combinations of paternal and maternal chromosomes ;

15 Random fusion of gametes ;16 Any male gamete can fuse with any female gametes ;

17 Fusion of genetically variable gametes contribute to genetically variable individuals ;


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2012 H1 JC2 Prelim - Core (Mark Scheme)