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Secondary 4 Pure Biology Practice Paper 3

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TuitionGoWhere Practice Paper - Pure Biology Secondary 4

TuitionGoWhere Practice Paper (AI)

Subject: Pure Biology (6093) Level: Secondary 4 Paper: Practice Paper 3 (Cells & Biomolecules) Duration: 1 hour 15 minutes Total Marks: 60

Name: _________________________ Class: _________________________ Date: _________________________

Instructions to Candidates

This paper consists of three sections: Section A, Section B, and Section C.
Answer all questions.
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 no more than 25 minutes on Section A, 25 minutes on Section B, and 25 minutes on Section C.

Section A: Multiple Choice and Short Answer [15 marks]

Answer all questions in this section.

Question 1 [1 mark]

Which organelle is the site of aerobic respiration in a cell?

A. Ribosome B. Nucleus C. Mitochondrion D. Chloroplast

Answer: ______

Question 2 [1 mark]

The diagram below shows an animal cell and a plant cell.

(Imagine two cells: one animal cell with irregular shape, no cell wall, small vacuoles; one plant cell with regular shape, cell wall, large central vacuole, chloroplasts.)

Which structure is present in the plant cell but absent in the animal cell?

A. Cell membrane B. Nucleus C. Cell wall D. Cytoplasm

Answer: ______

Question 3 [1 mark]

Which of the following correctly describes the function of the Golgi body?

A. Site of protein synthesis B. Modifies and packages proteins for secretion C. Produces ATP for cellular activities D. Contains digestive enzymes to break down worn-out organelles

Answer: ______

Question 4 [1 mark]

A student placed a drop of blood into a beaker of distilled water. After a few minutes, the red blood cells burst. Which process caused this to happen?

A. Diffusion of water out of the cells B. Active transport of water into the cells C. Osmosis of water into the cells D. Diffusion of solutes out of the cells

Answer: ______

Question 5 [1 mark]

Which of the following elements is found in proteins but not in carbohydrates?

A. Carbon B. Hydrogen C. Oxygen D. Nitrogen

Answer: ______

Question 6 [2 marks]

State two differences between the structure of a typical animal cell and a typical plant cell.

Question 7 [2 marks]

Define the term enzyme and state why enzymes are described as specific.

Question 8 [2 marks]

A student carried out a Benedict's test on an unknown solution and observed a brick-red precipitate after heating.

(a) What food substance is present in the solution? [1 mark]

(b) Explain why heating is necessary for the Benedict's test. [1 mark]

Question 9 [2 marks]

Explain why a plant wilts when it is not watered for several days.

Question 10 [2 marks]

State the chemical elements present in a fat molecule and give one function of fats in the human body.

Elements: __________________________________________________________

Function: _________________________________________________________

Section B: Structured Questions [25 marks]

Answer all questions in this section.

Question 11 [5 marks]

Fig. 11.1 shows the effect of temperature on the activity of enzyme X.

(Imagine a graph: x-axis "Temperature (°C)" from 0 to 70; y-axis "Rate of reaction / arbitrary units". The graph rises from 0 at 0°C to a peak of 10 units at 40°C, then drops sharply to 0 at 60°C.)

(a) Describe the trend shown in the graph between 0°C and 40°C. [1 mark]

(b) Explain why the rate of reaction increases between 0°C and 40°C. [2 marks]

Question 12 [5 marks]

A student investigated the effect of sucrose concentration on the mass of potato strips. Five potato strips of equal mass were placed in sucrose solutions of different concentrations for 30 minutes. The results are shown in Table 12.1.

Table 12.1

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	2.20	-0.30

(a) Explain why the potato strip gained mass in the 0.0 mol/dm³ sucrose solution. [2 marks]

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

Question 13 [5 marks]

Fig. 13.1 shows a diagram of part of a cell membrane with carrier proteins involved in active transport.

(Imagine a diagram showing a cell membrane with carrier protein molecules spanning the membrane. On one side, there are more solute molecules; on the other side, fewer. An arrow shows solute molecules moving from the lower concentration side to the higher concentration side through the carrier protein. ATP is shown being used.)

(a) Using information from Fig. 13.1, explain how active transport differs from diffusion. [2 marks]

(b) Explain the importance of active transport in the absorption of glucose by the villi in the small intestine. [2 marks]

Question 14 [5 marks]

A student carried out food tests on four unknown solutions, A, B, C, and D. The results are shown in Table 14.1.

Table 14.1

Test	Solution A	Solution B	Solution C	Solution D
Benedict's test	Blue	Brick-red	Blue	Blue
Iodine test	Brown	Brown	Blue-black	Brown
Biuret test	Purple	Blue	Blue	Blue
Ethanol emulsion test	Clear	Clear	Clear	Cloudy white

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

(b) Identify the food substances present in solution B. [1 mark]

(d) Identify the food substances present in solution D. [1 mark]

(e) A student claims that solution A contains both protein and starch. Explain whether this claim is supported by the results. [1 mark]

Question 15 [5 marks]

Fig. 15.1 shows an electron micrograph of an animal cell.

(Imagine an electron micrograph showing a cell with labelled structures: A – organelle with folded inner membrane (cristae); B – network of flattened membrane sacs with ribosomes attached; C – organelle consisting of stacked flattened sacs with vesicles budding off.)

(a) Identify structures A, B, and C. [3 marks]

A: _______________________________________________________________

B: _______________________________________________________________

C: _______________________________________________________________

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

(c) Explain why cells that secrete large amounts of enzymes, such as pancreatic cells, contain many copies of structure C. [1 mark]

Section C: Data-Based and Extended Response Questions [20 marks]

Answer all questions in this section.

Question 16 [5 marks]

A student investigated the effect of pH on the activity of enzyme Y. The enzyme was added to a substrate solution at different pH values, and the time taken for the substrate to be completely broken down was recorded. The results are shown in Table 16.1.

Table 16.1

pH	Time taken for complete breakdown (seconds)
2	120
4	60
6	20
7	15
8	18
10	55
12	130

(a) Calculate the rate of reaction at pH 7. Assume the initial substrate concentration is 1.0 arbitrary unit. Show your working and give the correct units. [2 marks]

(b) Describe and explain the effect of pH on the activity of enzyme Y. [3 marks]

Question 17 [5 marks]

Fig. 17.1 shows two red blood cells placed in solutions of different concentrations.

(Imagine a diagram: Cell P in a hypotonic solution – swollen and spherical, about to burst. Cell Q in a hypertonic solution – shrunken with a spiky, crenated appearance.)

(a) Explain what has happened to cell P. [2 marks]

(b) Explain what has happened to cell Q. [2 marks]

Question 18 [5 marks]

Describe the lock-and-key model of enzyme action. In your answer, explain how this model accounts for enzyme specificity and the effect of high temperature on enzyme activity. [5 marks]

Question 19 [5 marks]

A student set up an experiment using Visking tubing to model the absorption of digested food in the small intestine. The Visking tubing contained a solution of starch and amylase. 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) State the expected results of the Benedict's test and the iodine test on the water outside the tubing. Explain your answers. [3 marks]

(b) Explain why the Visking tubing is a good model for the small intestine in this experiment. State one limitation of this model. [2 marks]

Question 20 [5 marks]

Discuss the importance of diffusion, osmosis, and active transport in the survival of living organisms. In your answer, give one specific example for each process. [5 marks]

END OF PAPER

Answers

TuitionGoWhere Practice Paper - Pure Biology Secondary 4

Answer Key and Marking Scheme

Paper: Practice Paper 3 (Cells & Biomolecules) Total Marks: 60

Section A: Multiple Choice and Short Answer [15 marks]

Question 1 [1 mark] Answer: C. Mitochondrion Explanation: The mitochondrion is the site of aerobic respiration, where glucose is broken down to release ATP. Ribosomes are for protein synthesis, the nucleus contains genetic material, and chloroplasts are for photosynthesis (in plant cells only).

Question 2 [1 mark] Answer: C. Cell wall Explanation: Plant cells have a cellulose cell wall that provides structural support. Animal cells lack a cell wall. Both cell types have a cell membrane, nucleus, and cytoplasm.

Question 3 [1 mark] Answer: B. Modifies and packages proteins for secretion Explanation: The Golgi body receives proteins from the rough endoplasmic reticulum, modifies them, and packages them into vesicles for secretion out of the cell. Protein synthesis occurs at ribosomes, ATP production occurs in mitochondria, and lysosomes contain digestive enzymes.

Question 4 [1 mark] Answer: C. Osmosis of water into the cells Explanation: Distilled water has a higher water potential than the cytoplasm of red blood cells. Water moves into the cells by osmosis down the water potential gradient, causing the cells to swell and eventually burst (haemolysis).

Question 5 [1 mark] Answer: D. Nitrogen Explanation: Proteins contain carbon, hydrogen, oxygen, and nitrogen (and sometimes sulfur). Carbohydrates contain only carbon, hydrogen, and oxygen.

Question 6 [2 marks] Answer: Award 1 mark for each correct difference, up to 2 marks:

Plant cells have a cellulose cell wall; animal cells do not.
Plant cells have a large central vacuole; animal cells have small, temporary vacuoles (if present).
Plant cells have chloroplasts; animal cells do not.
Plant cells have a regular/fixed shape; animal cells have an irregular shape.

Question 7 [2 marks] Answer:

An enzyme is a biological catalyst that speeds up the rate of a chemical reaction without being chemically changed at the end of the reaction. [1 mark]
Enzymes are specific because each enzyme has an active site with a specific shape that is complementary to only one type of substrate. Therefore, each enzyme can only catalyse one specific reaction. [1 mark]

Question 8 [2 marks] (a) Reducing sugar (or glucose) is present. [1 mark] (b) Heating provides activation energy for the reaction between the reducing sugar and copper(II) sulfate in Benedict's solution, allowing the reduction of Cu²⁺ ions to Cu⁺ ions, which form the brick-red precipitate of copper(I) oxide. [1 mark]

Question 9 [2 marks] Answer:

When a plant is not watered, the soil water potential decreases. [1 mark]
Water moves out of the root hair cells by osmosis, and the plant cells lose turgor pressure. Without turgor pressure pushing the cell contents against the cell wall, the cells become flaccid, and the plant wilts. [1 mark]

Question 10 [2 marks] Answer:

Elements: Carbon, hydrogen, and oxygen. [1 mark]
Function (any one): Long-term energy storage / insulation (reduces heat loss) / protection of vital organs (e.g., around kidneys) / component of cell membranes / solvent for fat-soluble vitamins. [1 mark]

Section B: Structured Questions [25 marks]

Question 11 [5 marks]

(a) [1 mark] The rate of reaction increases as temperature increases from 0°C to 40°C, reaching a maximum at 40°C.

(b) [2 marks]

As temperature increases, the kinetic energy of enzyme and substrate molecules increases. [1 mark]
This increases the frequency of collisions between enzyme and substrate molecules, leading to more enzyme-substrate complexes formed per unit time, so the rate of reaction increases. [1 mark]

(c) [2 marks]

At 60°C, the high temperature causes the enzyme to denature. [1 mark]
The high temperature breaks the hydrogen bonds (and other bonds) that maintain the enzyme's specific three-dimensional tertiary structure. The active site loses its specific shape and is no longer complementary to the substrate. Substrate molecules cannot bind to the active site, so no enzyme-substrate complexes can form, and the rate of reaction drops to zero. [1 mark]

Question 12 [5 marks]

(a) [2 marks]

The 0.0 mol/dm³ sucrose solution (distilled water) has a higher water potential than the potato cell sap. [1 mark]
Water moves into the potato cells by osmosis, down the water potential gradient. The cells gain water, become turgid, and increase in mass. [1 mark]

(b) [2 marks]

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. The cells lose water, become flaccid/plasmolysed, and decrease in mass. [1 mark]

(c) [1 mark]

The isotonic concentration is 0.4 mol/dm³. At this concentration, there is no net movement of water into or out of the cells, so the mass remains unchanged (0.00 g change). [1 mark]

Question 13 [5 marks]

(a) [2 marks]

Active transport moves substances from a region of lower concentration to a region of higher concentration (against the concentration gradient), whereas diffusion moves substances from a region of higher concentration to a region of lower concentration (down the concentration gradient). [1 mark]
Active transport requires energy in the form of ATP and involves carrier proteins, whereas diffusion is a passive process that does not require energy. [1 mark]

(b) [2 marks]

After a meal, the concentration of glucose in the lumen of the small intestine may be lower than inside the epithelial cells of the villi. [1 mark]
Active transport allows glucose to be absorbed into the blood even against the concentration gradient, ensuring that all glucose is absorbed and none is lost in the faeces. [1 mark]

(c) [1 mark] Any one example:

Absorption of mineral ions (e.g., nitrates) by root hair cells from the soil.
Reabsorption of glucose in the proximal convoluted tubule of the kidney nephron.

Question 14 [5 marks]

(a) [1 mark] Solution A contains protein.

(b) [1 mark] Solution B contains reducing sugar.

(c) [1 mark] Solution C contains starch.

(d) [1 mark] Solution D contains fats (lipids).

(e) [1 mark] The claim is not supported. The Biuret test was positive (purple), indicating protein is present. However, the iodine test was negative (brown), indicating starch is absent. Therefore, solution A contains protein but not starch.

Question 15 [5 marks]

(a) [3 marks] A: Mitochondrion [1 mark] B: Rough endoplasmic reticulum (RER) [1 mark] C: Golgi body (Golgi apparatus) [1 mark]

(b) [1 mark] Any one function:

The ribosomes on the RER synthesise proteins.
The RER transports proteins within the cell (to the Golgi body).

(c) [1 mark] The Golgi body modifies and packages proteins (enzymes) into secretory vesicles for secretion out of the cell. Pancreatic cells secrete large amounts of digestive enzymes, so they require many Golgi bodies to process and package these enzymes.

Section C: Data-Based and Extended Response Questions [20 marks]

Question 16 [5 marks]

(a) [2 marks]

Rate = Amount of substrate broken down / Time taken
Rate = 1.0 arbitrary unit / 15 seconds
Rate = 0.067 arbitrary units per second (or 0.067 s⁻¹) Award 1 mark for correct calculation and 1 mark for correct units.

(b) [3 marks]

The rate of reaction is low at very low pH (pH 2) and very high pH (pH 12). The rate increases as pH increases from 2 to 7, reaching a maximum at pH 7 (optimum pH), then decreases as pH increases from 7 to 12. [1 mark]
Enzyme Y has an optimum pH of 7. At this pH, the active site has the specific shape that is complementary to the substrate, allowing maximum enzyme-substrate complex formation. [1 mark]
At pH values far from the optimum (pH 2 and pH 12), the high concentration of H⁺ or OH⁻ ions disrupts the ionic and hydrogen bonds that maintain the enzyme's tertiary structure. The active site loses its specific shape (the enzyme denatures), so the substrate can no longer bind, and the rate of reaction decreases. [1 mark]

Question 17 [5 marks]

(a) [2 marks]

Cell P has been placed in a hypotonic solution (a solution with a higher water potential than the cell cytoplasm). [1 mark]
Water enters the cell by osmosis, down the water potential gradient. The cell swells and becomes spherical. Because animal cells lack a cell wall, the cell membrane cannot withstand the pressure, and the cell bursts (haemolysis). [1 mark]

(b) [2 marks]

Cell Q has been placed in a hypertonic solution (a solution with a lower water potential than the cell cytoplasm). [1 mark]
Water leaves the cell by osmosis, down the water potential gradient. The cell loses water, shrinks, and the cell membrane develops a spiky appearance. This process is called crenation. [1 mark]

(c) [1 mark] Plant cells have a strong cellulose cell wall that prevents the cell from bursting. When water enters by osmosis, the cell becomes turgid, and the cell wall exerts an equal and opposite pressure (turgor pressure) that prevents further water entry.

Question 18 [5 marks]

Answer: Award marks for the following points:

The lock-and-key model states that an enzyme has a specific three-dimensional shape with an active site that is complementary in shape to a specific substrate molecule (like a lock and key). [1 mark]
The substrate molecule fits into the active site to form an enzyme-substrate complex. [1 mark]
The reaction takes place at the active site, and products are formed. The products leave the active site, and the enzyme remains unchanged and can be reused. [1 mark]
Enzyme specificity: Because the active site has a specific shape, only a substrate with a complementary shape can bind. This explains why each enzyme catalyses only one specific reaction. [1 mark]
Effect of high temperature: High temperatures (above the optimum) break the hydrogen bonds and other bonds that maintain the enzyme's specific tertiary structure. The active site loses its specific shape (the enzyme denatures). The substrate can no longer fit into the active site, so enzyme-substrate complexes cannot form, and the enzyme loses its catalytic activity. The change is permanent and irreversible. [1 mark]

Question 19 [5 marks]

(a) [3 marks]

Benedict's test: A brick-red (or green/yellow/orange) precipitate will form. [1 mark]
Explanation: Amylase breaks down starch into reducing sugars (maltose/glucose). These reducing sugar molecules are small enough to pass through the pores of the Visking tubing into the surrounding water. Benedict's solution reacts with reducing sugars upon heating to form a coloured precipitate. [1 mark]
Iodine test: The iodine solution will remain brown (no blue-black colour). [0.5 marks]
Explanation: Starch molecules are too large to pass through the pores of the Visking tubing. Therefore, no starch is present in the water outside the tubing. [0.5 marks]

(b) [2 marks]

Visking tubing is a good model because it is a partially permeable membrane, similar to the membrane of the epithelial cells lining the small intestine. It allows small molecules (like reducing sugars) to pass through but prevents large molecules (like starch) from passing through. [1 mark]
Limitation (any one): Visking tubing does not have carrier proteins for active transport, unlike the living cell membranes of the small intestine / Visking tubing is non-living and does not have cellular processes / The small intestine has villi and microvilli to increase surface area, which Visking tubing does not model. [1 mark]

Question 20 [5 marks]

Answer: Award marks for the following points. Award up to 2 marks for each process (one for explanation, one for example), up to a maximum of 5 marks.

Diffusion:

Diffusion is the net movement of particles from a region of higher concentration to a region of lower concentration, down a concentration gradient. It is a passive process that does not require energy. [1 mark]
Example (any one): Oxygen diffuses from the alveoli (high oxygen concentration) into the blood capillaries (low oxygen concentration) in the lungs for transport to body cells / Carbon dioxide diffuses from respiring cells into the blood and then into the alveoli to be exhaled. [1 mark]

Osmosis:

Osmosis is the net movement of water molecules from a region of higher water potential to a region of lower water potential, through a partially permeable membrane. It is a passive process. [1 mark]
Example (any one): Water is absorbed by root hair cells from the soil by osmosis / Water is reabsorbed in the kidney nephrons (collecting ducts) by osmosis under the influence of ADH. [1 mark]

Active Transport:

Active transport is the movement of molecules or ions from a region of lower concentration to a region of higher concentration, against the concentration gradient, using energy from ATP and carrier proteins. [1 mark]
Example (any one): Absorption of glucose by the villi in the small intestine when glucose concentration in the lumen is low / Absorption of mineral ions (e.g., nitrates) by root hair cells from the soil / Reabsorption of glucose in the proximal convoluted tubule of the kidney nephron. [1 mark]

Conclusion (optional, for synthesis): All three processes are essential for the survival of living organisms as they enable the uptake of nutrients, removal of waste products, and maintenance of water balance within cells and organisms.

END OF ANSWER KEY