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Secondary 4 Pure Biology Practice Paper 2

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TuitionGoWhere Practice Paper - Pure Biology Secondary 4

TuitionGoWhere Practice Paper (AI) Version: 2 of 5

Subject: Pure Biology (6093) Level: Secondary 4 Paper: Practice Paper 2 Duration: 1 hour 45 minutes Total Marks: 80

Name: _________________________ Class: _________________________ Date: _________________________

Instructions to Candidates

This paper consists of two sections: Section A and Section B.
Answer all questions in Section A.
Answer any two questions in Section B.
Write your answers in the spaces provided.
The number of marks is given in brackets [ ] at the end of each question or part question.
You are advised to spend about 55 minutes on Section A and 50 minutes on Section B.

Section A: Structured Questions

Answer all questions in this section. [Total: 50 marks]

Question 1: Cell Structure and Function

Fig. 1.1 shows an electron micrograph of an animal cell.

(a) Identify the organelles labelled P, Q, and R. [3 marks]

P: ________________________________________________

Q: ________________________________________________

R: ________________________________________________

(b) Organelle P is described as the "powerhouse of the cell". Explain why this description is appropriate. [2 marks]

(c) A student observed that cells from the flight muscles of a migratory bird contain many more of organelle P than cells from the skin of the same bird. Suggest an explanation for this observation. [2 marks]

(d) Organelle Q is involved in the synthesis and transport of proteins. Describe the pathway of a protein molecule from its site of synthesis to its secretion from the cell. [3 marks]

[Total: 10 marks]

Question 2: Movement of Substances

A student investigated the effect of different concentrations of sucrose solution on the mass of potato strips. Five potato strips of equal dimensions were blotted dry, weighed, and placed in sucrose solutions of different concentrations for 30 minutes. The strips were then removed, blotted dry, and reweighed. The results are shown in Table 2.1.

Table 2.1

Sucrose concentration (mol/dm³)	Initial mass (g)	Final mass (g)	Change in mass (g)
0.0	2.50	2.75	+0.25
0.2	2.50	2.60	+0.10
0.4	2.50	2.50	0.00
0.6	2.50	2.35	–0.15
0.8	2.50	2.15	–0.35

(a) Calculate the percentage change in mass for each sucrose concentration. Write your answers in Table 2.1. [2 marks]

(b) Explain why the potato strip placed in 0.0 mol/dm³ sucrose solution increased in mass. [3 marks]

(c) Explain why the potato strip placed in 0.8 mol/dm³ sucrose solution decreased in mass. [2 marks]

(d) Using the results in Table 2.1, estimate the sucrose concentration that is isotonic to the potato cell sap. Explain your answer. [2 marks]

(e) State one variable, other than those already mentioned, that must be kept constant in this investigation to ensure valid results. [1 mark]

[Total: 10 marks]

Question 3: Biological Molecules and Enzymes

A student carried out food tests on four unknown solutions (W, X, Y, and Z). The results are shown in Table 3.1.

Table 3.1

Test	Solution W	Solution X	Solution Y	Solution Z
Benedict's test (with heating)	Blue solution remains	Brick-red precipitate	Blue solution remains	Brick-red precipitate
Iodine test	Brown solution remains	Brown solution remains	Blue-black colour	Brown solution remains
Biuret test	Purple colour	Blue solution remains	Blue solution remains	Purple colour
Ethanol emulsion test	Clear solution remains	Clear solution remains	White emulsion	White emulsion

(a) Identify the food substances present in solution Y. [1 mark]

(b) Identify the food substances present in solution Z. [2 marks]

(c) Explain why the Benedict's test must be heated. [2 marks]

(d) A student claims that solution W contains only water. Explain whether this conclusion is valid based on the results in Table 3.1. [2 marks]

(e) Fig. 3.1 shows the effect of pH on the activity of two enzymes, E1 and E2, found in different parts of the human digestive system.

[In the graph: Enzyme E1 shows maximum activity at pH 2. Enzyme E2 shows maximum activity at pH 8. Both enzymes show a bell-shaped curve.]

(i) State the optimum pH for enzyme E1. [1 mark]

(ii) Suggest where in the human digestive system enzyme E1 is likely to be found. Explain your answer. [2 marks]

(iii) Explain why enzyme E2 shows very low activity at pH 2. [3 marks]

[Total: 13 marks]

Question 4: Cell Specialisation and Organisation

Fig. 4.1 shows two specialised cells: a red blood cell and a root hair cell.

(a) State one structural feature of a red blood cell and explain how this feature adapts it for its function of transporting oxygen. [2 marks]

(b) State one structural feature of a root hair cell and explain how this feature adapts it for its function of absorbing water and mineral ions. [2 marks]

(c) Red blood cells lack a nucleus when mature. Explain the advantage and disadvantage of this feature. [2 marks]

(d) Fig. 4.2 shows a diagram of a villus from the small intestine.

(i) Name the structure labelled S. [1 mark]

(ii) Explain how the structure of a villus is adapted for the absorption of digested food substances. [3 marks]

[Total: 10 marks]

Question 5: Active Transport and Diffusion

A student set up an experiment using visking tubing to model absorption in the small intestine. The visking tubing contained a solution of starch and glucose. The tubing was placed in a beaker of distilled water at 37°C. After 30 minutes, samples of the water outside the tubing were tested with Benedict's solution and iodine solution.

(a) Predict the results of the Benedict's test and the iodine test on the water outside the visking tubing. Explain your predictions. [4 marks]

Benedict's test result: ________________________________________________

Explanation: ________________________________________________

Iodine test result: ________________________________________________

Explanation: ________________________________________________

(b) Explain why the water in the beaker was maintained at 37°C. [1 mark]

(c) The absorption of glucose by the villi in the small intestine involves active transport. Explain why active transport, rather than diffusion, is necessary for the complete absorption of glucose from the small intestine. [2 marks]

[Total: 7 marks]

Section B: Free-Response Questions

Answer any two questions in this section. Each question carries 15 marks. [Total: 30 marks]

Question 6

(a) Describe the structure of a typical animal cell as seen under a light microscope. Explain the functions of three of the structures you have described. [6 marks]

(b) Explain how the structure of a motor neurone is adapted to its function of transmitting nerve impulses rapidly over long distances. [4 marks]

(c) A student placed some red blood cells in a solution of 0.5% sodium chloride and observed them under a microscope. The cells appeared shrunken and had a crinkled appearance. Explain this observation. [5 marks]

[Total: 15 marks]

Question 7

(a) Describe the lock-and-key model of enzyme action. Use a labelled diagram to support your answer. [5 marks]

(b) An investigation was carried out to study the effect of temperature on the activity of catalase, an enzyme found in potato tissue. Potato discs were placed in hydrogen peroxide solution at different temperatures, and the volume of oxygen produced in one minute was measured.

(i) Predict the shape of the graph that would be obtained if the volume of oxygen produced was plotted against temperature (from 0°C to 60°C). Sketch the graph and explain its shape. [5 marks]

(ii) Explain why the enzyme activity decreases at temperatures above the optimum. [3 marks]

(c) State two differences between the effects of a competitive inhibitor and a non-competitive inhibitor on enzyme activity. [2 marks]

[Total: 15 marks]

Question 8

(a) Explain how the structure of the cell membrane is related to its function as a selectively permeable barrier. [4 marks]

(b) A plant was watered with a solution containing a high concentration of fertiliser salts. The next day, the plant appeared wilted. Explain why the plant wilted. [4 marks]

(c) Describe the process of protein synthesis in a cell, beginning with the genetic code in the nucleus and ending with the secretion of the protein from the cell. [7 marks]

[Total: 15 marks]

END OF PAPER

TuitionGoWhere Practice Paper (AI) – Version 2 of 5 – Pure Biology Secondary 4

Answers

TuitionGoWhere Practice Paper - Pure Biology Secondary 4

Answer Key and Marking Scheme

Version: 2 of 5 Subject: Pure Biology (6093) Level: Secondary 4 Total Marks: 80

Section A: Structured Questions [50 marks]

Question 1: Cell Structure and Function [10 marks]

(a) Identify organelles P, Q, and R. [3 marks]

Label	Organelle	Mark
P	Mitochondrion / Mitochondria	1
Q	Rough endoplasmic reticulum / Rough ER / Granular ER	1
R	Golgi body / Golgi apparatus	1

Award 1 mark for each correct identification. Accept singular or plural forms.

(b) Explain why the mitochondrion is described as the "powerhouse of the cell". [2 marks]

Answer:

The mitochondrion is the site of aerobic respiration [1 mark]
During aerobic respiration, glucose is broken down to release energy in the form of ATP (adenosine triphosphate), which is used for cellular activities [1 mark]

Accept: "produces/releases energy for the cell" for 1 mark if linked to respiration.

(c) Explain why flight muscle cells contain more mitochondria than skin cells. [2 marks]

Answer:

Flight muscle cells require more energy than skin cells because they are involved in active, sustained contraction during flight [1 mark]
More mitochondria allow a higher rate of aerobic respiration to produce more ATP to meet the high energy demands of flight muscles [1 mark]

Accept any reasonable explanation linking energy demand to mitochondrial number.

(d) Describe the pathway of a protein molecule from synthesis to secretion. [3 marks]

Answer:

The protein is synthesised by ribosomes on the rough endoplasmic reticulum (RER) [1 mark]
The protein is transported through the RER and packaged into vesicles that bud off and travel to the Golgi body [1 mark]
At the Golgi body, the protein is modified, sorted, and packaged into secretory vesicles. These vesicles move to and fuse with the cell membrane, releasing the protein outside the cell by exocytosis [1 mark]

Award marks for correct sequence. Accept "Golgi apparatus" for Golgi body.

Question 2: Movement of Substances [10 marks]

(a) Calculate percentage change in mass. [2 marks]

Table 2.1 (completed):

Sucrose concentration (mol/dm³)	Percentage change in mass (%)
0.0	+10.0
0.2	+4.0
0.4	0.0
0.6	–6.0
0.8	–14.0

Formula: (Change in mass ÷ Initial mass) × 100% Award 1 mark for correct method/formula; 1 mark for all five values correct. Deduct 0.5 marks for each incorrect value if method is correct.

(b) Explain why the potato strip in 0.0 mol/dm³ sucrose increased in mass. [3 marks]

Answer:

The 0.0 mol/dm³ solution is distilled water, which has a higher water potential than the potato cell sap [1 mark]
Water molecules move from the solution into the potato cells by osmosis, down the water potential gradient [1 mark]
The cells gain water, become turgid, and the mass of the potato strip increases [1 mark]

Do not award marks for "water moves to a higher concentration". Must mention water potential or correct direction of osmosis.

(c) Explain why the potato strip in 0.8 mol/dm³ sucrose decreased in mass. [2 marks]

Answer:

The 0.8 mol/dm³ sucrose solution has a lower water potential than the potato cell sap [1 mark]
Water moves out of the potato cells by osmosis, down the water potential gradient, causing the cells to lose turgor and the mass to decrease [1 mark]

(d) Estimate the isotonic concentration and explain. [2 marks]

Answer:

The isotonic concentration is approximately 0.4 mol/dm³ [1 mark]
At this concentration, there is no net change in mass (0.00% change), indicating that the water potential of the sucrose solution is equal to the water potential of the potato cell sap. Therefore, there is no net movement of water by osmosis [1 mark]

Accept 0.4 mol/dm³ or any value between 0.35 and 0.45 with valid reasoning.

(e) State one variable that must be kept constant. [1 mark]

Answer (any one):

Temperature of the solutions
Volume of sucrose solution used
Time of immersion
Surface area of potato strips (ensure all strips are cut to the same dimensions)
Variety/type of potato used

Award 1 mark for any valid controlled variable.

Question 3: Biological Molecules and Enzymes [13 marks]

(a) Identify food substances in solution Y. [1 mark]

Answer:

Starch and fats (lipids) are present [1 mark]

Both must be stated for 1 mark. Accept "starch and oil" or "starch and lipid".

(b) Identify food substances in solution Z. [2 marks]

Answer:

Reducing sugar and protein are present [1 mark]
Fats (lipids) are also present [1 mark]

Award 1 mark each for identifying reducing sugar/protein and fats. Accept "glucose" for reducing sugar.

(c) Explain why Benedict's test must be heated. [2 marks]

Answer:

Heating provides activation energy for the reaction between reducing sugars and copper(II) sulfate in Benedict's solution [1 mark]
Without heating, the reaction would be too slow for the colour change (brick-red precipitate) to be observed within a reasonable time [1 mark]

Accept any explanation linking heating to reaction rate or activation energy.

(d) Explain whether the conclusion that solution W contains only water is valid. [2 marks]

Answer:

The conclusion is not valid [1 mark]
The tests only show that reducing sugar, starch, protein, and fats are absent. Solution W could contain other substances not tested for, such as non-reducing sugars (e.g., sucrose), mineral ions, or vitamins [1 mark]

Award 1 mark for stating the conclusion is invalid; 1 mark for valid reasoning.

(e)(i) State the optimum pH for enzyme E1. [1 mark]

Answer:

pH 2 [1 mark]

(e)(ii) Suggest where enzyme E1 is found and explain. [2 marks]

Answer:

Enzyme E1 is likely to be found in the stomach [1 mark]
The stomach contains gastric juice with hydrochloric acid, creating a highly acidic environment (around pH 2). Enzyme E1 (likely pepsin, a protease) has an optimum pH of 2, which matches the stomach environment [1 mark]

Accept "stomach" with valid explanation linking pH 2 to gastric acid.

(e)(iii) Explain why enzyme E2 shows very low activity at pH 2. [3 marks]

Answer:

Enzyme E2 has an optimum pH of around 8, indicating it functions best in alkaline conditions [1 mark]
At pH 2, the highly acidic conditions disrupt the hydrogen bonds and ionic bonds that maintain the specific three-dimensional (tertiary) structure of the enzyme [1 mark]
The active site of enzyme E2 loses its specific shape and can no longer bind to its substrate. The enzyme is denatured, so enzyme-substrate complexes cannot form, resulting in very low activity [1 mark]

Award marks for: identifying pH 2 is far from optimum, explaining denaturation (bonds broken, shape change), and linking to active site function.

Question 4: Cell Specialisation and Organisation [10 marks]

(a) Structural feature of red blood cell and adaptation for oxygen transport. [2 marks]

Answer (any one feature with explanation):

Biconcave disc shape: Increases surface area to volume ratio for faster diffusion of oxygen into and out of the cell [1 mark for feature, 1 mark for explanation]
Contains haemoglobin: Haemoglobin binds reversibly with oxygen to form oxyhaemoglobin, enabling efficient oxygen transport [1 mark for feature, 1 mark for explanation]
Lacks a nucleus (when mature): Provides more space for haemoglobin, increasing oxygen-carrying capacity [1 mark for feature, 1 mark for explanation]

(b) Structural feature of root hair cell and adaptation for absorption. [2 marks]

Answer (any one feature with explanation):

Long, narrow protrusion (root hair): Greatly increases surface area to volume ratio for faster absorption of water and mineral ions by osmosis and active transport [1 mark for feature, 1 mark for explanation]
Thin cell wall: Reduces the distance for diffusion/osmosis, allowing faster absorption [1 mark for feature, 1 mark for explanation]
Contains many mitochondria: Provides ATP/energy for active transport of mineral ions against the concentration gradient [1 mark for feature, 1 mark for explanation]

(c) Advantage and disadvantage of red blood cells lacking a nucleus. [2 marks]

Answer:

Advantage: More space is available for haemoglobin, so the red blood cell can carry more oxygen [1 mark]
Disadvantage: The red blood cell cannot repair itself or divide/reproduce, so it has a limited lifespan (about 120 days) and cannot be replaced by cell division [1 mark]

Accept any valid advantage and disadvantage.

(d)(i) Name structure S. [1 mark]

Answer:

Lacteal [1 mark]

Accept "lymph vessel" or "lymphatic capillary".

(d)(ii) Explain how villus structure is adapted for absorption. [3 marks]

Answer (any three adaptations with explanations):

The villus has a large surface area due to its finger-like projections, increasing the rate of absorption of digested food substances [1 mark]
The epithelium is one cell thick, providing a short diffusion distance for absorbed substances to enter the blood capillaries or lacteal [1 mark]
The villus contains a dense network of blood capillaries to rapidly transport absorbed glucose and amino acids away, maintaining a steep concentration gradient for continued absorption [1 mark]
The villus contains a lacteal for the absorption of fatty acids and glycerol (fats) into the lymphatic system [1 mark]
The epithelial cells have microvilli on their surface, further increasing the surface area for absorption [1 mark]

Award 1 mark for each valid adaptation with correct explanation, up to 3 marks.

Question 5: Active Transport and Diffusion [7 marks]

(a) Predict Benedict's and iodine test results and explain. [4 marks]

Answer:

Benedict's test result: Brick-red precipitate (positive result) [1 mark]

Explanation: Glucose molecules are small enough to diffuse through the pores of the visking tubing (which acts as a selectively permeable membrane) from inside the tubing into the distilled water in the beaker. The presence of glucose (a reducing sugar) in the external water gives a positive Benedict's test [1 mark]

Iodine test result: Brown solution remains (negative result) [1 mark]

Explanation: Starch molecules are too large to pass through the pores of the visking tubing. Therefore, starch remains inside the tubing and is not present in the external water, giving a negative iodine test [1 mark]

Award marks for correct prediction and correct explanation for each test.

(b) Explain why the water was maintained at 37°C. [1 mark]

Answer:

37°C is body temperature (or the optimum temperature for the processes being modelled). Maintaining this temperature simulates the conditions in the human body and ensures that the rate of diffusion is consistent with what occurs in the small intestine [1 mark]

Accept any answer linking 37°C to body temperature or providing a constant temperature for valid comparison.

(c) Explain why active transport is necessary for complete glucose absorption. [2 marks]

Answer:

After a meal, glucose is absorbed from the small intestine into the blood by diffusion until the concentrations are equal [1 mark]
To absorb the remaining glucose from the small intestine (when the concentration in the intestine is lower than in the blood), active transport is required to move glucose against the concentration gradient, using energy from ATP [1 mark]

Award 1 mark for identifying that diffusion alone cannot absorb all glucose; 1 mark for explaining active transport moves glucose against the concentration gradient using energy.

Section B: Free-Response Questions [30 marks]

Answer any two questions. Each question carries 15 marks.

Question 6 [15 marks]

(a) Describe the structure of a typical animal cell and explain functions of three structures. [6 marks]

Answer:

Structure description [3 marks]:

The cell is bounded by a cell membrane, a thin, partially permeable layer composed of phospholipids and proteins
The cytoplasm is a jelly-like substance that fills the cell, containing organelles and where most cellular activities occur
The nucleus is a large, spherical structure surrounded by a nuclear membrane; it contains chromatin (DNA) and a nucleolus
Mitochondria are rod-shaped organelles with folded inner membranes (cristae)
Ribosomes are small, spherical structures, either free in the cytoplasm or attached to the endoplasmic reticulum
Vacuoles are small, temporary, fluid-filled spaces in the cytoplasm

Functions of three structures [3 marks – 1 mark for each valid function]:

Nucleus: Contains genetic material (DNA) that controls all cellular activities, including cell division and protein synthesis
Cell membrane: Acts as a selectively permeable barrier, controlling the movement of substances into and out of the cell
Mitochondria: Site of aerobic respiration, where glucose is broken down to release energy in the form of ATP for cellular activities
Ribosomes: Site of protein synthesis, where amino acids are assembled into polypeptide chains
Cytoplasm: Site of many metabolic reactions; contains organelles and dissolved substances

Award up to 3 marks for describing at least three structures with some detail. Award up to 3 marks for explaining the functions of three different structures. Accept other valid structures (e.g., rough ER, Golgi body) if described correctly.

(b) Explain how a motor neurone is adapted for transmitting impulses rapidly over long distances. [4 marks]

Answer:

The motor neurone has a long axon (nerve fibre), which allows nerve impulses to be transmitted over long distances from the central nervous system to effector organs (e.g., muscles) [1 mark]
The axon is covered by a myelin sheath (formed by Schwann cells), which acts as an electrical insulator [1 mark]
The myelin sheath increases the speed of nerve impulse transmission by allowing saltatory conduction, where the impulse jumps from one node of Ranvier to the next [1 mark]
The cell body has many dendrites that branch extensively, providing a large surface area to receive nerve impulses from other neurones and connect with many other neurones [1 mark]

Award 1 mark for each valid adaptation with correct explanation, up to 4 marks.

(c) Explain why red blood cells appeared shrunken and crinkled in 0.5% sodium chloride solution. [5 marks]

Answer:

The 0.5% sodium chloride solution has a lower water potential than the cytoplasm of the red blood cells [1 mark]
Water molecules move out of the red blood cells by osmosis, down the water potential gradient, from a region of higher water potential (inside the cells) to a region of lower water potential (the external solution) [2 marks]
As water leaves the cells, the red blood cells lose water, shrink, and become crinkled (crenated) [1 mark]
Unlike plant cells, animal cells do not have a cell wall, so they cannot maintain their shape when water is lost [1 mark]

Award marks for: identifying water potential difference, explaining direction of osmosis, describing the outcome (crenation), and noting the absence of a cell wall.

Question 7 [15 marks]

(a) Describe the lock-and-key model of enzyme action with a labelled diagram. [5 marks]

Answer:

Description [3 marks]:

An enzyme has a specific three-dimensional shape with a depression called the active site [1 mark]
The active site has a specific shape that is complementary to the shape of its specific substrate, like a lock and key [1 mark]
The substrate molecule binds to the active site, forming an enzyme-substrate complex. The reaction occurs, and the product(s) are formed and released. The enzyme remains unchanged and can be reused [1 mark]

Diagram [2 marks]:

Diagram should show: enzyme with active site, substrate with complementary shape, enzyme-substrate complex, and products released with enzyme unchanged [1 mark for correct shapes and labels]
Labels: Enzyme, Active site, Substrate, Enzyme-substrate complex, Products [1 mark for correct labelling]

Award up to 2 marks for a clear, labelled diagram showing the sequence of events.

(b)(i) Predict and sketch the graph of oxygen produced vs. temperature, and explain its shape. [5 marks]

Answer:

Graph sketch [2 marks]:

The graph should show a bell-shaped curve [1 mark]
X-axis: Temperature (0°C to 60°C); Y-axis: Volume of oxygen produced (cm³)
The curve rises from near zero at 0°C, reaches a peak at the optimum temperature (around 30–40°C for catalase), then falls sharply to near zero by 60°C [1 mark for correct shape and optimum temperature range]

Explanation [3 marks]:

From 0°C to optimum: As temperature increases, the kinetic energy of enzyme and substrate molecules increases. They move faster and collide more frequently, increasing the rate of formation of enzyme-substrate complexes. Therefore, the rate of reaction (volume of oxygen produced) increases [1 mark]
At the optimum temperature: The rate of reaction is at its maximum because the frequency of effective collisions is highest [1 mark]
Above the optimum temperature: The high temperature disrupts the hydrogen bonds and other bonds that maintain the specific three-dimensional shape of the enzyme. The active site loses its shape, and the enzyme is denatured. Substrate molecules can no longer bind, so the rate of reaction decreases sharply [1 mark]

Award marks for correct graph shape and clear explanation of each phase.

(b)(ii) Explain why enzyme activity decreases above the optimum temperature. [3 marks]

Answer:

Above the optimum temperature, the high kinetic energy causes the atoms within the enzyme molecule to vibrate violently [1 mark]
This disrupts the hydrogen bonds and ionic bonds that maintain the specific three-dimensional (tertiary) structure of the enzyme protein [1 mark]
The active site loses its specific shape and is no longer complementary to the substrate. The enzyme is denatured, and enzyme-substrate complexes can no longer form, so the rate of reaction decreases [1 mark]

Award marks for: identifying bond disruption, linking to shape change of the active site, and stating denaturation.

(c) State two differences between competitive and non-competitive inhibitors. [2 marks]

Answer (any two differences):

Competitive Inhibitor	Non-Competitive Inhibitor
Has a shape similar to the substrate and competes for the active site	Has a different shape and binds to an allosteric site (not the active site)
Effect can be overcome by increasing substrate concentration	Effect cannot be overcome by increasing substrate concentration
Binds to the active site	Binds to a site other than the active site, changing the shape of the active site
Does not change the shape of the active site	Changes the shape of the active site

Award 1 mark for each valid difference, up to 2 marks. Accept any two clear, contrasting statements.

Question 8 [15 marks]

(a) Explain how the cell membrane structure is related to its function as a selectively permeable barrier. [4 marks]

Answer:

The cell membrane is composed of a phospholipid bilayer [1 mark]
The hydrophobic (fatty acid) tails face inwards, creating a barrier that prevents water-soluble substances (e.g., ions, glucose) from passing through freely, while allowing small, non-polar molecules (e.g., oxygen, carbon dioxide) to diffuse through [1 mark]
Protein molecules are embedded in the phospholipid bilayer. Some act as channel proteins or carrier proteins that allow specific ions and larger molecules to pass through the membrane by facilitated diffusion or active transport [1 mark]
The membrane is fluid (fluid mosaic model), allowing proteins to move and function, and enabling the membrane to change shape for processes like endocytosis and exocytosis [1 mark]

Award 1 mark for each valid structural feature linked to its function, up to 4 marks.

(b) Explain why a plant wilted after being watered with high-concentration fertiliser solution. [4 marks]

Answer:

The fertiliser solution has a high concentration of salts, giving it a lower water potential than the cell sap of the root hair cells [1 mark]
Water moves out of the root hair cells by osmosis, down the water potential gradient, instead of being absorbed from the soil [1 mark]
The root cells lose turgor, and water is not taken up by the plant. As water continues to be lost by transpiration from the leaves, the cells in the leaves and stem also lose water [1 mark]
The plant cells become flaccid (plasmolysed), and the plant loses its turgidity, causing it to wilt [1 mark]

Award marks for: identifying water potential difference, direction of osmosis, loss of turgor, and linking to wilting.

(c) Describe the process of protein synthesis from the genetic code to secretion. [7 marks]

Answer:

Transcription (in the nucleus) [3 marks]:

The section of DNA containing the gene for the specific protein unwinds and unzips, exposing the nucleotide bases on one strand (the template strand) [1 mark]
Free RNA nucleotides pair with complementary bases on the template strand (A with U, T with A, C with G, G with C). The enzyme RNA polymerase joins the RNA nucleotides together to form a molecule of messenger RNA (mRNA) [1 mark]
The mRNA molecule is a complementary copy of the gene. It detaches from the DNA and moves out of the nucleus through a nuclear pore into the cytoplasm [1 mark]

Translation (in the cytoplasm) [2 marks]:

The mRNA attaches to a ribosome. The ribosome reads the sequence of codons (triplets of bases) on the mRNA [1 mark]
Transfer RNA (tRNA) molecules, each carrying a specific amino acid, have anticodons that are complementary to the mRNA codons. The tRNA molecules bind to the mRNA codons in sequence, bringing their amino acids. The ribosome joins the amino acids together by peptide bonds to form a polypeptide chain [1 mark]

Secretion [2 marks]:

The polypeptide chain enters the rough endoplasmic reticulum (RER), where it folds into its specific three-dimensional shape to become a functional protein [1 mark]
The protein is transported through the RER and packaged into vesicles. The vesicles travel to the Golgi body, where the protein is modified and packaged into secretory vesicles. These vesicles move to and fuse with the cell membrane, releasing the protein outside the cell by exocytosis [1 mark]

Award marks for clear description of each stage: transcription (3 marks), translation (2 marks), and secretion pathway (2 marks). Accept alternative valid descriptions.

END OF ANSWER KEY

TuitionGoWhere Practice Paper (AI) – Version 2 of 5 – Pure Biology Secondary 4