Secondary 3 Biology Practice Paper 4

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TuitionGoWhere Practice Paper - Biology Secondary 3

TuitionGoWhere Practice Paper (AI)
Subject: Biology
Level: Secondary 3 (G3/Express)
Paper: Practice Paper 4 (Cells & Biomolecules)
Duration: 1 hour 15 minutes
Total Marks: 50

Name: ________________________
Class: ________________________
Date: ________________________

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Instructions to Candidates

1. Write your name, class, and date in the spaces provided above.
2. Answer all questions in the spaces provided on the question paper.
3. The number of marks is given in brackets [ ] at the end of each question or part question.
4. The total number of marks for this paper is 50.
5. You are advised to spend approximately 1 minute per mark.
6. Diagrams are not drawn to scale unless stated otherwise.
7. For questions requiring calculations, show all working clearly.

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Section A: Multiple Choice Questions [10 marks]

Answer all questions. For each question, choose the correct answer and write the letter (A, B, C, or D) in the box provided.

Question 1 [1 mark]

An actively growing plant cell is supplied with radioactive amino acids. Which cell component would first show an increase in radioactivity?

A. Golgi body
B. Mitochondria
C. Rough endoplasmic reticulum
D. Nucleus

Answer: □

Question 2 [1 mark]

Which of the following correctly matches a biomolecule to its monomer and a function?

	Biomolecule	Monomer	Function
A	Starch	Glucose	Energy storage in animals
B	Glycogen	Glucose	Energy storage in plants
C	Cellulose	Glucose	Structural support in plant cell walls
D	Protein	Glycerol	Enzyme catalysis

Answer: □

Question 3 [1 mark]

The graph below shows the effect of temperature on the activity of an enzyme.

Image pending generation: graph for Q3.

Which statement best explains the shape of the curve between 40°C and 60°C?

A. The enzyme is denatured due to breakage of peptide bonds.
B. The enzyme is denatured due to breakage of hydrogen bonds and other weak interactions.
C. The substrate concentration becomes limiting.
D. The enzyme becomes saturated with substrate.

Answer: □

Question 4 [1 mark]

A student tests four unknown solutions for the presence of reducing sugars, starch, protein, and fat. The results are shown below.

Solution	Benedict's test	Iodine test	Biuret test	Ethanol emulsion test
W	Brick-red precipitate	Blue-black	Violet	Cloudy white emulsion
X	Blue (no change)	Blue-black	Blue (no change)	Clear
Y	Brick-red precipitate	Brown (no change)	Violet	Clear
Z	Blue (no change)	Brown (no change)	Blue (no change)	Cloudy white emulsion

Which solution contains only fat?

A. W
B. X
C. Y
D. Z

Answer: □

Question 5 [1 mark]

Which row correctly describes the structure and function of a triglyceride?

	Structure	Main Function
A	Three fatty acids + one glycerol; hydrophobic	Energy storage, insulation, buoyancy
B	Three fatty acids + one glycerol; hydrophilic	Cell membrane structure
C	Many glucose units; branched polymer	Energy storage in animals
D	Amino acids linked by peptide bonds	Enzyme catalysis

Answer: □

Question 6 [1 mark]

The diagram shows a section through a plant cell.

Image pending generation: diagram for Q6.

Which structure is the site of the light-dependent reactions of photosynthesis?

A. Cytoplasm
B. Mitochondria
C. Chloroplasts
D. Nucleus

Answer: □

Question 7 [1 mark]

A piece of potato tissue is placed in a concentrated sucrose solution. After 30 minutes, the cells are observed to be plasmolysed. Which statement correctly explains this observation?

A. Water moved out of the cell by osmosis, causing the cell membrane to pull away from the cell wall.
B. Sucrose moved into the cell by diffusion, causing the cell to swell.
C. Water moved into the cell by osmosis, causing the cell to burst.
D. The cell wall shrank due to loss of solutes.

Answer: □

Question 8 [1 mark]

Which of the following processes requires energy in the form of ATP?

A. Diffusion of oxygen into a cell
B. Osmosis of water across a partially permeable membrane
C. Active transport of nitrate ions into root hair cells
D. Facilitated diffusion of glucose through a carrier protein

Answer: □

Question 9 [1 mark]

The diagram shows the structure of a DNA nucleotide.

Image pending generation: diagram for Q9.

Which bond joins one nucleotide to the next in a DNA strand?

A. Glycosidic bond
B. Hydrogen bond
C. Phosphodiester bond
D. Peptide bond

Answer: □

Question 10 [1 mark]

A student carries out the Benedict's test on a solution of sucrose. The solution remains blue after heating. The student then hydrolyses the sucrose with dilute hydrochloric acid, neutralises the solution, and repeats the Benedict's test. A brick-red precipitate forms.

What does this show?

A. Sucrose is a reducing sugar.
B. Sucrose is a non-reducing sugar that can be hydrolysed to reducing sugars.
C. Hydrochloric acid is a reducing agent.
D. The Benedict's test is unreliable.

Answer: □

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Section B: Structured Questions [25 marks]

Answer all questions in the spaces provided.

Question 11 [4 marks]

The diagram shows a typical animal cell as seen under an electron microscope.

Image pending generation: diagram for Q11.

(a) Identify structures X and Y labelled on the diagram. [2]

X: _______________________________________________________________________

Y: _______________________________________________________________________

(b) State one function of structure X. [1]

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(c) Explain why structure Y is abundant in cells that secrete large amounts of protein. [1]

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Question 12 [5 marks]

A student investigated the effect of pH on the activity of the enzyme pepsin. Pepsin digests protein (albumen) into amino acids. The student set up five test tubes with the following contents:

Tube	Pepsin solution	Albumen suspension	Buffer solution (pH)	Water
1	2 cm³	2 cm³	2 cm³ (pH 1)	2 cm³
2	2 cm³	2 cm³	2 cm³ (pH 3)	2 cm³
3	2 cm³	2 cm³	2 cm³ (pH 5)	2 cm³
4	2 cm³	2 cm³	2 cm³ (pH 7)	2 cm³
5	2 cm³	2 cm³	2 cm³ (pH 9)	2 cm³

All tubes were incubated at 37°C for 20 minutes. The clarity of each tube was then measured using a colorimeter (absorbance units). Lower absorbance = more protein digested.

The results are shown below.

Image pending generation: graph for Q12.

(a) State the optimum pH for pepsin activity based on these results. [1]

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(b) Explain the results at pH 1 and pH 7 in terms of enzyme structure and function. [3]

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(c) Suggest one improvement to this investigation to increase the reliability of the results. [1]

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Question 13 [4 marks]

The diagram shows a phospholipid bilayer with embedded proteins.

Image pending generation: diagram for Q13.

(a) Name the part of the phospholipid molecule that is

<stage5_exam_md> (i) hydrophilic: _____________________________________________________________ [1]

(ii) hydrophobic: ____________________________________________________________ [1]

(b) State one function of cholesterol in the cell membrane. [1]

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(c) Explain how a channel protein differs from a carrier protein in its mechanism of transport. [1]

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Question 14 [6 marks]

A student investigated osmosis using potato cylinders. Five cylinders of equal length (50 mm) and mass were cut from the same potato. Each cylinder was placed in a different concentration of sucrose solution for 40 minutes. The cylinders were then removed, blotted dry, and their lengths and masses remeasured.

The results are shown below.

Sucrose concentration (mol dm⁻³)	Initial length (mm)	Final length (mm)	Change in length (mm)	Initial mass (g)	Final mass (g)	% Change in mass
0.00 (distilled water)	50	54	+4	2.50	2.85	+14.0
0.20	50	52	+2	2.50	2.65	+6.0
0.40	50	50	0	2.50	2.50	0.0
0.60	50	48	-2	2.50	2.35	-6.0
0.80	50	46	-4	2.50	2.20	-12.0

(a) Plot a graph of % change in mass against sucrose concentration on the grid below. [3]

Image pending generation: graph_paper for Q14.

(b) Use your graph to determine the sucrose concentration at which there is no net movement of water into or out of the potato cells. [1]

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(c) Explain why the potato cylinder in 0.00 mol dm⁻³ sucrose solution increased in mass. [2]

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Question 15 [6 marks]

The diagram shows the structure of a DNA molecule.

Image pending generation: diagram for Q15.

(a) Name the four nitrogenous bases found in DNA. [2]

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(b) State the base pairing rule in DNA. [1]

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(c) The two DNA strands are described as antiparallel. Explain what this means. [1]

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(d) During DNA replication, the enzyme DNA helicase breaks hydrogen bonds between base pairs. Explain why hydrogen bonds are broken rather than covalent bonds. [2]

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Section C: Free Response Questions [15 marks]

Answer one question only. Write your answer in the space provided.

Question 16 [15 marks]

Either

Describe the structure and function of the following organelles in a plant cell:

• Nucleus
• Chloroplast
• Mitochondrion
• Rough endoplasmic reticulum
• Golgi body
• Cell wall

In your answer, explain how these organelles work together in the production and secretion of a protein (e.g., an enzyme) in a plant cell. [15]

OR

Question 17 [15 marks]

Or

(a) Describe the lock-and-key hypothesis and the induced-fit hypothesis of enzyme action. [6]

(b) Explain how temperature, pH, and substrate concentration affect the rate of enzyme-catalysed reactions. [9]

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End of Paper

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Rough Work Space

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TuitionGoWhere Practice Paper - Biology Secondary 3: Answer Key & Marking Scheme

Paper: Practice Paper 4 (Cells & Biomolecules)
Total Marks: 50

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Section A: Multiple Choice Questions [10 marks]

Question	Answer	Explanation
1	C	Rough endoplasmic reticulum (RER) has ribosomes attached where protein synthesis occurs. Radioactive amino acids are incorporated into polypeptides at ribosomes on the RER first.
2	C	Cellulose is a polymer of glucose that provides structural support in plant cell walls. Starch stores energy in plants (not animals), glycogen stores energy in animals (not plants), proteins are made of amino acids (not glycerol).
3	B	High temperatures cause denaturation by breaking hydrogen bonds and other weak interactions (hydrophobic interactions, ionic bonds) that maintain the enzyme's 3D shape. Peptide bonds (covalent) are not broken by heat alone.
4	D	Solution Z shows only a positive ethanol emulsion test (cloudy white emulsion = fat present). All other tests are negative: Benedict's (blue = no reducing sugar), Iodine (brown = no starch), Biuret (blue = no protein).
5	A	Triglycerides consist of three fatty acids + one glycerol, are hydrophobic, and function in energy storage, insulation, and buoyancy. B is wrong (triglycerides are hydrophobic, not hydrophilic; phospholipids form membranes). C describes glycogen. D describes proteins.
6	C	Light-dependent reactions of photosynthesis occur in the thylakoid membranes of chloroplasts.
7	A	Concentrated sucrose solution has lower water potential than cell sap. Water moves out by osmosis, causing the cell membrane to pull away from the rigid cell wall (plasmolysis).
8	C	Active transport requires ATP to move substances against their concentration gradient. Diffusion, osmosis, and facilitated diffusion are passive processes.
9	C	Phosphodiester bonds join the phosphate group of one nucleotide to the 3' carbon of the next nucleotide's deoxyribose, forming the sugar-phosphate backbone. Glycosidic bonds join base to sugar; hydrogen bonds join complementary bases; peptide bonds join amino acids.
10	B	Sucrose is a non-reducing disaccharide. Hydrolysis breaks it into glucose and fructose (both reducing sugars), which then give a positive Benedict's test.

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Section B: Structured Questions [25 marks]

Question 11 [4 marks]

(a) Identify structures X and Y labelled on the diagram. [2]

• X: Golgi body / Golgi apparatus (accept: Golgi complex)
• Y: Rough endoplasmic reticulum / RER (accept: rough ER with ribosomes)

(b) State one function of structure X. [1]

• Modifies, sorts, and packages proteins (and lipids) for secretion or delivery to other organelles.
• Accept: Adds carbohydrate groups to form glycoproteins; packages proteins into secretory vesicles.

(c) Explain why structure Y is abundant in cells that secrete large amounts of protein. [1]

• RER has ribosomes on its surface where protein synthesis occurs; newly synthesised proteins enter the RER lumen for folding and initial modification before transport to Golgi for secretion.

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Question 12 [5 marks]

(a) State the optimum pH for pepsin activity based on these results. [1]

• pH 1–2 (accept pH 1 or pH 2; lowest absorbance = most digestion)

(b) Explain the results at pH 1 and pH 7 in terms of enzyme structure and function. [3]

• pH 1 (optimum): Pepsin is a stomach enzyme; its tertiary structure is stable and active at low pH. The active site shape is complementary to the substrate (albumen), allowing maximum enzyme-substrate complex formation and rapid digestion → lowest absorbance.
• pH 7: The pH is far from optimum. High pH disrupts hydrogen bonds and ionic bonds maintaining pepsin's tertiary structure → active site shape changes → enzyme denatured → substrate cannot bind → little/no digestion → high absorbance (cloudy suspension remains).
• Key terms: tertiary structure, active site, denaturation, hydrogen/ionic bonds, enzyme-substrate complex.

(c) Suggest one improvement to this investigation to increase the reliability of the results. [1]

• Repeat the experiment at each pH (e.g., 3 replicates) and calculate mean absorbance.
• Accept: Use more pH values around the optimum (e.g., pH 1.5, 2.0, 2.5) to pinpoint optimum more precisely; use a buffer with smaller pH intervals; control temperature more precisely with a water bath.

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Question 13 [4 marks]

(a) Name the part of the phospholipid molecule that is

• (i) hydrophilic: phosphate head / phosphate group (accept: glycerol-phosphate head)
• (ii) hydrophobic: fatty acid tails / hydrocarbon tails [2]

(b) State one function of cholesterol in the cell membrane. [1]

• Regulates membrane fluidity: prevents phospholipids packing too tightly at low temperatures (increases fluidity) and restricts excessive movement at high temperatures (decreases fluidity / adds stability).
• Accept: Provides mechanical stability; reduces permeability to small water-soluble molecules/ions.

(c) Explain how a channel protein differs from a carrier protein in its mechanism of transport. [1]

• Channel protein: Forms a hydrophilic pore/tunnel; allows specific ions/molecules to pass through by facilitated diffusion (down concentration gradient); does not change shape.
• Carrier protein: Binds specific molecule; undergoes conformational shape change to move molecule across membrane; can do facilitated diffusion or active transport.
• Key difference: channel = fixed pore; carrier = shape change / binding.

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Question 14 [6 marks]

(a) Plot a graph of % change in mass against sucrose concentration. [3]

• Axes: x-axis: Sucrose concentration (mol dm⁻³), 0.00 to 0.80; y-axis: % Change in mass, -15% to +15% (or suitable scale).
• Points plotted accurately: (0.00, +14.0), (0.20, +6.0), (0.40, 0.0), (0.60, -6.0), (0.80, -12.0).
• Line: Smooth curve or straight line of best fit through points.
• Marking: 1 mark for labelled axes with units, 1 mark for correct plotting of all 5 points, 1 mark for suitable line/curve.

(b) Use your graph to determine the sucrose concentration at which there is no net movement of water into or out of the potato cells. [1]

• 0.40 mol dm⁻³ (where % change in mass = 0%; isotonic point / water potential of potato cells)

(c) Explain why the potato cylinder in 0.00 mol dm⁻³ sucrose solution increased in mass. [2]

• Distilled water has a higher water potential (less negative) than the potato cell sap.
• Water moves into the cells by osmosis down the water potential gradient (from high to low water potential) across the partially permeable cell membrane.
• Cells become turgid; mass increases.
• Key terms: water potential, osmosis, partially permeable membrane, turgid.

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Question 15 [6 marks]

(a) Name the four nitrogenous bases found in DNA. [2]

• Adenine (A), Thymine (T), Guanine (G), Cytosine (C) (1 mark for purines A & G, 1 mark for pyrimidines T & C; all four required for 2 marks)

(b) State the base pairing rule in DNA. [1]

• Adenine pairs with Thymine (A-T) by two hydrogen bonds; Guanine pairs with Cytosine (G-C) by three hydrogen bonds. (Accept: complementary base pairing: A with T, G with C)

(c) The two DNA strands are described as antiparallel. Explain what this means. [1]

• The two strands run in opposite directions: one strand runs 5' → 3', the other runs 3' → 5'. The 5' end has a free phosphate group; the 3' end has a free hydroxyl (-OH) group on the deoxyribose.

(d) During DNA replication, the enzyme DNA helicase breaks hydrogen bonds between base pairs. Explain why hydrogen bonds are broken rather than covalent bonds. [2]

• Hydrogen bonds are weak compared to covalent bonds, so they can be broken easily and reversibly with less energy input, allowing the strands to separate for replication.
• Covalent bonds (phosphodiester bonds in the backbone) are strong; breaking them would destroy the nucleotide sequence / primary structure of DNA, causing permanent damage/mutations.
• Key: weak vs strong; reversible separation vs permanent damage; energy requirement.

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Section C: Free Response Questions [15 marks]

Question 16 [15 marks] – Plant Cell Organelles & Protein Secretion

Marking Guidance (Level of Response):

Level	Marks	Descriptor
3	11–15	Comprehensive, accurate descriptions of all 6 organelles. Clear explanation of coordinated protein production/secretion pathway (nucleus → RER → Golgi → secretion). Correct terminology, logical flow.
2	6–10	Good descriptions of most organelles. Protein secretion pathway described but may miss steps or have minor inaccuracies. Some correct terminology.
1	1–5	Basic descriptions, limited detail. Pathway incomplete or confused. Limited terminology.
0	0	No relevant content.

Indicative Content:

Organelle Structure & Function:

• Nucleus: Double membrane (nuclear envelope) with pores; contains chromatin (DNA + proteins); nucleolus makes ribosomes. Function: Stores genetic info; controls cell activities; transcription of mRNA for protein synthesis.
• Chloroplast: Double membrane; thylakoids (grana) with chlorophyll; stroma. Function: Photosynthesis (light-dependent in thylakoids, Calvin cycle in stroma); produces glucose.
• Mitochondrion: Double membrane; inner membrane folded (cristae); matrix. Function: Aerobic respiration; produces ATP for cellular processes.
• Rough ER: Flattened sacs (cisternae) with ribosomes on cytoplasmic surface. Function: Protein synthesis (on ribosomes); folding/modification in lumen; transport to Golgi.
• Golgi Body: Stack of flattened membrane-bound sacs (cisternae); vesicles bud off. Function: Receives proteins from RER; further modification (e.g., glycosylation); sorting/packaging into secretory vesicles.
• Cell Wall: Rigid layer outside cell membrane; cellulose microfibrils in matrix. Function: Structural support; maintains shape; prevents bursting (turgor); allows plasmodesmata for transport.

Coordinated Protein Production & Secretion (e.g., enzyme):

1. Nucleus: Gene transcribed → mRNA exits via nuclear pores.
2. Ribosomes on RER: mRNA translated → polypeptide chain enters RER lumen.
3. RER: Folding, initial modification (e.g., signal peptide cleavage, glycosylation start).
4. Transport vesicles: Bud off RER → fuse with cis face of Golgi.
5. Golgi: Further modification (carbohydrate chains trimmed/added); sorting.
6. Secretory vesicles: Bud off trans face of Golgi → move along cytoskeleton.
7. Cell membrane: Vesicles fuse (exocytosis) → enzyme secreted outside cell / into cell wall space.
8. Mitochondria: Provide ATP for translation, vesicle transport, exocytosis.
9. Chloroplasts: Provide carbon skeletons/energy (in photosynthetic cells) for amino acid synthesis.
10. Cell wall: Final destination for some secreted proteins (e.g., cell wall enzymes); plasmodesmata allow intercellular movement.

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Question 17 [15 marks] – Enzyme Hypotheses & Factors

Marking Guidance (Level of Response):

Level	Marks	Descriptor
3	11–15	Both hypotheses clearly described with diagrams/analogy. All three factors explained with correct mechanisms (kinetic theory, denaturation, active site saturation). Graph shapes described.
2	6–10	Hypotheses described but may lack distinction. Factors explained with some gaps (e.g., missing denaturation detail, or saturation not linked to active sites).
1	1–5	Basic statements only. Confusion between hypotheses. Factors listed without mechanism.
0	0	No relevant content.

Indicative Content:

(a) Lock-and-Key vs Induced-Fit Hypotheses [6 marks]

• Lock-and-Key (Fischer):

  • Enzyme active site has a fixed, rigid shape complementary to substrate.
  • Substrate fits exactly like a key in a lock.
  • Explains specificity (only one substrate fits).
  • Does not explain how enzyme stabilises transition state or why some inhibitors bind without reaction.

• Induced-Fit (Koshland):

  • Active site is flexible; substrate binding induces a conformational change in enzyme.
  • Enzyme 'moulds' around substrate → active site becomes fully complementary.
  • Strains substrate bonds (lowers activation energy); explains catalysis and allosteric regulation.
  • More widely accepted; explains why some molecules bind but don't react (wrong shape change).

(b) Factors Affecting Enzyme Rate [9 marks]

• Temperature:

  • Low temp: Low kinetic energy → few collisions → low rate.
  • Increasing temp: More kinetic energy → more frequent & energetic collisions → more ES complexes → rate increases (approx. doubles per 10°C, Q₁₀).
  • Optimum temp: Maximum rate (human enzymes ~37–40°C).
  • Above optimum: Excessive vibration breaks hydrogen bonds/ionic bonds → denaturation → active site shape lost → rate drops sharply.
  • Graph: Bell-shaped curve.

• pH:

  • Each enzyme has optimum pH (pepsin pH 2, trypsin pH 8, most ~pH 7).
  • Deviation from optimum: H⁺/OH⁻ ions disrupt hydrogen bonds and ionic bonds in tertiary structure → active site shape altered → denaturation.
  • Extreme pH → irreversible denaturation.
  • Graph: Bell-shaped curve (peaks at optimum).

• Substrate Concentration (at fixed enzyme concentration):

  • Low [S]: Many free active sites → rate ∝ [S] (first-order kinetics).
  • Increasing [S]: More ES complexes → rate increases but curve flattens.
  • High [S]: All active sites occupied → saturation → Vmax reached (zero-order kinetics).
  • Rate limited by enzyme concentration / turnover number.
  • Graph: Rectangular hyperbola (Michaelis-Menten).

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End of Marking Scheme