Secondary 4 Pure Biology Preliminary Examination Paper 5

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TuitionGoWhere Practice Paper – Pure Biology Secondary 4

PRELIMINARY EXAMINATION – Version 5

TuitionGoWhere Secondary School (AI)

Subject: Pure Biology (6093)
Level: Secondary 4
Paper: Preliminary Examination – Paper 2 (Structured and Free Response)
Duration: 1 hour 30 minutes
Total Marks: 80

Name: ________________________
Class: ________________________
Date: ________________________

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INSTRUCTIONS TO CANDIDATES

1. This paper consists of two sections: Section A and Section B.
2. Answer all questions in Section A. Write your answers in the spaces provided.
3. Section B consists of two questions. Answer any one question. Indicate your choice clearly.
4. The number of marks is given in brackets [ ] at the end of each question or part question.
5. You are advised to spend no more than 55 minutes on Section A and 35 minutes on Section B.
6. You may use a calculator.

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SECTION A (60 marks)

Answer all questions in this section. Write your answers in the spaces provided.

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Question 1: Cell Structure and Function (8 marks)

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

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

P: ________________________________________________
Q: ________________________________________________
R: ________________________________________________

(b) State the function of organelle Q. [1]

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(c) Explain why organelle P is abundant in muscle cells. [2]

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(d) A student observed that organelle R appears as small dots either free in the cytoplasm or attached to a membrane system. Name the membrane system to which R can be attached and state the role of this combined structure. [2]

Membrane system: ________________________________________________

Role: _______________________________________________________________________________

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Question 2: Movement of Substances (7 marks)

A student investigated the effect of different concentrations of sucrose solution on 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 2.1.

Table 2.1

Sucrose concentration (mol/dm³)	Initial mass (g)	Final mass (g)	Change in mass (g)	Percentage change in mass (%)
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) Complete Table 2.1 by calculating the percentage change in mass for each concentration. Show your working for the 0.0 mol/dm³ concentration. [3]

Working for 0.0 mol/dm³:

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Sucrose concentration (mol/dm³)	Percentage change in mass (%)
0.0
0.2
0.4
0.6
0.8

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

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(c) Using the data, estimate the water potential of the potato tissue. Explain your reasoning. [2]

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Question 3: Biological Molecules and Enzymes (9 marks)

Fig. 3.1 shows the effect of temperature on the rate of an enzyme-catalysed reaction.

Rate of
reaction
  ^
  |        /\
  |       /  \
  |      /    \
  |     /      \
  |    /        \
  |   /          \
  |  /            \
  | /              \
  |/                \
  +----------------------> Temperature (°C)
  0   10   20   30   40   50   60

(a) Describe the relationship between temperature and the rate of reaction between 10 °C and 40 °C. [2]

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(b) Explain the shape of the graph between 10 °C and 40 °C in terms of molecular movement and enzyme–substrate interactions. [3]

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(c) Explain why the rate of reaction decreases rapidly above 40 °C. [2]

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(d) A student suggested that the enzyme used in this investigation was extracted from a bacterium living in hot springs. State whether you agree with this suggestion and explain your answer. [2]

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Question 4: Cell Specialisation (6 marks)

Fig. 4.1 shows three specialised cells: a red blood cell, a root hair cell, and a muscle cell.

(a) State one structural adaptation of each cell and explain how the adaptation helps the cell perform its function. [6]

(i) Red blood cell:

Adaptation: _______________________________________________________________________________

Explanation: _______________________________________________________________________________

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(ii) Root hair cell:

Adaptation: _______________________________________________________________________________

Explanation: _______________________________________________________________________________

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(iii) Muscle cell:

Adaptation: _______________________________________________________________________________

Explanation: _______________________________________________________________________________

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Question 5: Diffusion and Active Transport (8 marks)

A student set up an experiment to investigate the uptake of mineral ions by root hair cells. Two sets of root tissue were placed in solutions containing potassium ions. Set A was kept in a warm, aerated solution. Set B was kept in a cold solution without aeration. The concentration of potassium ions in the root tissue was measured after 2 hours.

Table 5.1

Conditions	Potassium ion concentration in root tissue (arbitrary units)
Set A (warm, aerated)	85
Set B (cold, no aeration)	30
Initial concentration	25

(a) Calculate the increase in potassium ion concentration for Set A and Set B. [1]

Set A: _______________
Set B: _______________

(b) Explain why the potassium ion concentration increased in Set A. [2]

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(c) Explain why the increase in potassium ion concentration was lower in Set B. [3]

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(d) Suggest why aeration of the solution is necessary for active transport to occur. [2]

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Question 6: Osmosis in Plant and Animal Cells (7 marks)

(a) Define the term osmosis. [2]

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(b) A red blood cell is placed in a solution with a lower water potential than the cell contents. Describe and explain what happens to the cell. [3]

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(c) A plant cell is placed in a solution with a lower water potential than the cell contents. Describe and explain what happens to the cell. Explain why the outcome differs from that of the red blood cell in part (b). [2]

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Question 7: Food Tests (5 marks)

A student carried out food tests on an unknown solution. The results are shown in Table 7.1.

Table 7.1

Test	Observation
Benedict's test	Blue solution remained blue
Iodine test	Yellow-brown solution remained yellow-brown
Biuret test	Blue solution turned purple
Ethanol emulsion test	Clear solution remained clear

(a) Identify the food substance present in the unknown solution. Explain your answer. [2]

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(b) Describe how the Benedict's test is carried out. [2]

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(c) Explain why the ethanol emulsion test gives a cloudy white emulsion when fats are present. [1]

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Question 8: Enzyme Specificity (5 marks)

(a) Explain why enzymes are described as specific. [2]

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(b) Using the lock-and-key model, explain how an enzyme catalyses the breakdown of its substrate. [3]

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Question 9: Carbohydrates, Fats, and Proteins (5 marks)

(a) State the chemical elements present in carbohydrates. [1]

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(b) State the chemical elements present in proteins. [1]

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(c) Describe how large carbohydrate molecules such as starch are formed from smaller units. [1]

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(d) State two functions of fats in living organisms. [2]

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SECTION B (20 marks)

Answer any one question from this section. Write your answers on the lined pages provided. Indicate the question number clearly.

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Question 10: Enzymes – Extended Response (20 marks)

(a) Describe the structure of an enzyme and explain how its structure is related to its function. [6]

(b) A student investigated the effect of pH on the activity of amylase. Describe how the student could carry out this investigation, including how the rate of reaction could be measured. [8]

(c) Explain why biological washing powders containing enzymes are more effective at removing stains than non-biological washing powders. Discuss one limitation of using enzyme-based washing powders. [6]

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Question 11: Cells and Transport – Extended Response (20 marks)

(a) Compare the structure of a typical plant cell with that of a typical animal cell. Explain how the differences relate to the different functions of plant and animal cells. [8]

(b) Describe the processes of diffusion, osmosis, and active transport. For each process, provide one example of where it occurs in a living organism and explain its importance. [8]

(c) A patient with kidney failure undergoes dialysis. Explain how the principles of diffusion are applied in kidney dialysis to remove urea from the blood. [4]

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END OF PAPER

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TuitionGoWhere Practice Paper – Pure Biology Secondary 4

PRELIMINARY EXAMINATION – Version 5 – ANSWERS

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SECTION A (60 marks)

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Question 1: Cell Structure and Function

(a)
P: Mitochondrion / Mitochondria
Q: Rough endoplasmic reticulum / RER
R: Ribosome / Ribosomes

(b)
Synthesises / transports proteins;
(accept: folds/modifies proteins, transports proteins to Golgi body)

(c)
Muscle cells require a lot of energy / ATP for contraction;
Mitochondria are the site of aerobic respiration / release energy;

(d)
Membrane system: Rough endoplasmic reticulum;
Role: Synthesises / transports proteins;

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Question 2: Movement of Substances

(a)
Working for 0.0 mol/dm³:
Percentage change = (change in mass / initial mass) × 100
= (+0.25 / 2.50) × 100 = +10%

Sucrose concentration (mol/dm³)	Percentage change in mass (%)
0.0	+10
0.2	+4
0.4	0
0.6	–6
0.8	–12

(b)
Water potential of solution is higher than water potential of potato cells / potato cells have lower water potential;
Water enters cells by osmosis;
Cells become turgid / gain mass;

(c)
Water potential of potato tissue is equivalent to 0.4 mol/dm³ sucrose solution;
Because there is no net movement of water / no change in mass;
Indicating water potentials are equal;

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Question 3: Biological Molecules and Enzymes

(a)
As temperature increases from 10 °C to 40 °C, the rate of reaction increases;
(accept: rate increases proportionally / steadily)

(b)
As temperature increases, kinetic energy of molecules increases;
Molecules move faster, increasing frequency of collisions between enzyme and substrate;
More enzyme–substrate complexes formed per unit time;

(c)
Above 40 °C, the enzyme is denatured;
The active site loses its specific shape;
Substrate can no longer bind / enzyme–substrate complex cannot form;

(d)
Disagree;
Enzyme from hot spring bacteria would have a higher optimum temperature / would still be active above 40 °C;
The graph shows the enzyme denatures above 40 °C, indicating it is not from a hot spring bacterium;

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Question 4: Cell Specialisation

(a) (i) Red blood cell:
Adaptation: Contains haemoglobin;
Explanation: Binds to oxygen for transport around the body;
(accept: biconcave shape – increases surface area for oxygen diffusion; no nucleus – more space for haemoglobin)

(ii) Root hair cell:
Adaptation: Long / elongated projection / root hair;
Explanation: Increases surface area for absorption of water and mineral ions;

(iii) Muscle cell:
Adaptation: Contains many mitochondria;
Explanation: Provides energy / ATP for contraction;
(accept: contains protein fibres – can contract/shorten)

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Question 5: Diffusion and Active Transport

(a)
Set A: 60 arbitrary units
Set B: 5 arbitrary units

(b)
Potassium ions are absorbed by active transport;
Active transport requires energy from respiration;
Warm, aerated conditions allow respiration to occur;

(c)
In Set B, low temperature reduces rate of respiration / enzyme activity;
Lack of aeration means less oxygen available for respiration;
Less energy / ATP produced, so less active transport;

(d)
Aeration provides oxygen;
Oxygen is needed for aerobic respiration;
Respiration releases energy / ATP required for active transport;

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Question 6: Osmosis in Plant and Animal Cells

(a)
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;

(b)
Water moves out of the cell by osmosis;
Cell shrinks / shrivels / becomes crenated;
Because water moves from higher water potential (inside cell) to lower water potential (outside solution);

(c)
Water moves out of the cell by osmosis;
Cell becomes flaccid / plasmolysed (cell membrane pulls away from cell wall);
Plant cell has a cell wall which prevents bursting / maintains shape;
Red blood cell has no cell wall, so it shrinks/crenates;

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Question 7: Food Tests

(a)
Protein is present;
Biuret test turned purple, which is a positive result for protein;
Other tests were negative;

(b)
Add Benedict's solution to the sample;
Heat in a water bath / boil;
Observe colour change (blue to green/yellow/orange/brick-red if reducing sugar present);

(c)
Ethanol dissolves fats;
When water is added, the fat comes out of solution / forms a fine suspension;
This suspension scatters light, appearing cloudy white;

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Question 8: Enzyme Specificity

(a)
Enzymes have an active site with a specific shape;
Only a substrate with a complementary shape can bind / fit into the active site;

(b)
Substrate molecule fits into the active site of the enzyme (like a key into a lock);
Forms an enzyme–substrate complex;
Reaction occurs / bonds in substrate are broken;
Products are released;
Enzyme remains unchanged and can be reused;

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Question 9: Carbohydrates, Fats, and Proteins

(a)
Carbon, hydrogen, oxygen;

(b)
Carbon, hydrogen, oxygen, nitrogen;
(accept: sometimes sulfur/phosphorus)

(c)
Many glucose / simple sugar units join together;
By condensation reactions / forming glycosidic bonds;

(d)

1. Energy storage / source of energy;
2. Thermal insulation;
3. Protection of organs;
4. Component of cell membranes;
   (any two)

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SECTION B (20 marks)

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Question 10: Enzymes – Extended Response

(a)
Enzymes are proteins;
They have a specific three-dimensional shape;
Contain an active site;
Active site has a shape complementary to the substrate;
This allows the enzyme to bind specifically to its substrate;
Enzyme–substrate complex forms, lowering activation energy;

(b)
Use starch solution and amylase;
Set up a series of buffer solutions with different pH values (e.g., pH 4, 5, 6, 7, 8, 9);
Add equal volumes of amylase to each;
Add equal volumes of starch solution;
Start timing;
At regular intervals, remove samples and test with iodine solution;
Record time taken for iodine to remain yellow-brown (starch no longer present);
Rate of reaction = 1/time;
Repeat and calculate mean;
Control variables: temperature, enzyme concentration, substrate concentration;

(c)
Biological washing powders contain enzymes (e.g., proteases, lipases);
Enzymes break down stains (e.g., proteins, fats) into smaller, soluble molecules;
These molecules can be washed away more easily;
More effective at lower temperatures (save energy);
Limitation: Enzymes can be denatured at high temperatures;
(accept: may cause allergic reactions; may damage certain fabrics like wool/silk)

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Question 11: Cells and Transport – Extended Response

(a)
Similarities: Both have cell membrane, cytoplasm, nucleus, mitochondria, ribosomes;
Differences: Plant cell has cell wall (animal cell does not);
Plant cell has chloroplasts (animal cell does not);
Plant cell has large central vacuole (animal cell has small/temporary vacuoles);
Plant cells often have a fixed/regular shape due to cell wall;
Cell wall provides structural support (important for plants which do not have skeletons);
Chloroplasts allow photosynthesis to produce food;
Large vacuole stores water/solutes, maintains turgor pressure;

(b)
Diffusion: Net movement of particles from a region of higher concentration to a region of lower concentration;
Example: Oxygen diffusing from alveoli into blood capillaries;
Importance: Allows gas exchange for respiration;

Osmosis: Net movement of water molecules from a region of higher water potential to a region of lower water potential through a partially permeable membrane;
Example: Water absorption by root hair cells from soil;
Importance: Allows plants to obtain water for photosynthesis/support;

Active transport: Movement of molecules/ions from a region of lower concentration to a region of higher concentration, against the concentration gradient, using energy;
Example: Absorption of mineral ions by root hair cells;
Importance: Allows plants to obtain minerals needed for growth even when soil concentration is low;

(c)
Patient's red blood cells have a lower water potential than the distilled water;
Water enters the cells by osmosis;
Red blood cells have no cell wall;
Cells swell and burst / haemolyse;
This would reduce the number of red blood cells / cause harm;
Therefore, distilled water should not be used;
A saline solution with the same water potential as blood plasma should be used (isotonic);