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Secondary 4 Pure Biology Preliminary Examination Paper 2

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TuitionGoWhere Exam Practice (AI) - Prelim Paper 2 (Version 2)

TuitionGoWhere Secondary School (AI)
Subject: Pure Biology
Level: Secondary 4
Paper: Preliminary Examination - Paper 2 (Section A Focus)
Duration: 1 hour 15 minutes
Total Marks: 40
Name: ________________________
Class: ________________________
Date: ________________________

Instructions to Candidates:

Write your name, class, and date in the spaces provided.
Answer all questions in the spaces provided.
The number of marks is given in brackets [ ] at the end of each question or part question.
You should use an HB pencil for all diagrams and graphs.

Section A: Cells and Biomolecules

Answer all questions in this section.

1. Fig. 1.1 shows an electron micrograph of a specialised animal cell involved in the secretion of enzymes.

(Note: In a real exam, Fig 1.1 would show a cell with abundant Rough Endoplasmic Reticulum, Golgi apparatus, and mitochondria.)

(a) Identify the structures labelled A and B in Fig. 1.1. [2] A: _______________________________________________________ B: _______________________________________________________

(b) Explain how the structure of organelle A is adapted for its function in protein synthesis. [2]

(d) State why this cell contains a high number of mitochondria. [1]

[Total: 7]

2. A student investigated the effect of temperature on the activity of the enzyme amylase. The time taken for starch to be completely broken down was recorded at different temperatures. The results are shown in Table 2.1.

Table 2.1

Temperature (°C)	Time taken for starch breakdown (seconds)	Rate of reaction (1/time)
10	120	0.008
20	60	0.017
30	30	0.033
40	15	0.067
50	45	0.022
60	>300	0.003

(a) Calculate the rate of reaction at 50°C. Show your working. [1] Rate = ____________________

(b) Describe the trend in the rate of reaction as the temperature increases from 10°C to 40°C. [2]

(d) Suggest why the student kept the pH constant throughout the investigation. [1]

[Total: 7]

3. Fig. 3.1 shows three plant cells (P, Q, and R) placed in solutions of different water potentials. The arrows indicate the net movement of water by osmosis.

(Note: Fig 3.1 would show: Cell P in hypertonic solution (water out), Cell Q in isotonic solution (no net movement), Cell R in hypotonic solution (water in).)

(a) Define osmosis. [2]

(b) Identify which cell (P, Q, or R) is in a solution with the highest water potential relative to the cell sap. [1] Cell: _______

(c) Explain what would happen to Cell P if it were left in the solution for a prolonged period. Use the terms plasmolysis and cell wall in your answer. [3]

(d) Root hair cells absorb water from the soil. Explain how the structure of a root hair cell increases the efficiency of water absorption. [2]

[Total: 8]

4. Biological molecules are essential for life. Table 4.1 lists four biological molecules and their constituent elements.

Table 4.1

Molecule	Constituent Elements
Starch	Carbon, Hydrogen, Oxygen
Protein	Carbon, Hydrogen, Oxygen, Nitrogen
Fat	Carbon, Hydrogen, Oxygen
DNA	Carbon, Hydrogen, Oxygen, Nitrogen, Phosphorus

(a) Name the test reagent used to identify the presence of reducing sugars and state the colour change observed if the test is positive. [2] Reagent: _________________________ Colour change: _________________________

(b) A food sample was tested with Biuret solution and remained blue. It was then tested with Iodine solution and turned blue-black. (i) State which biological molecule is present in the food sample. [1]

(ii) State which biological molecule is absent from the food sample. [1]

(ii) Describe how these units are joined together to form a protein. [1]

[Total: 6]

5. Active transport is a mechanism for moving substances across cell membranes.

(a) Define active transport. [2]

(b) Compare active transport with diffusion by stating two differences. [2]

[Total: 6]

6. Fig. 6.1 shows a diagram of a red blood cell and a white blood cell.

(a) State one structural difference between a red blood cell and a white blood cell. [1]

(b) Red blood cells are specialised for oxygen transport. Explain how the biconcave shape of the red blood cell aids this function. [2]

(c) Haemoglobin is a protein found in red blood cells. (i) State the element found in haemoglobin that binds to oxygen. [1]

(ii) Explain why red blood cells lack a nucleus. [2]

[Total: 6]

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Answers

TuitionGoWhere Exam Practice (AI) - Prelim Paper 2 (Version 2) - Answer Key

Subject: Pure Biology
Level: Secondary 4
Paper: Preliminary Examination - Paper 2 (Section A Focus)

Section A: Cells and Biomolecules

1. (a) A: Rough Endoplasmic Reticulum (RER) [1] B: Golgi Apparatus / Golgi Body [1]

(b)

The RER has a large surface area due to its folded membrane structure. [1]
It is studded with ribosomes, which are the sites of protein synthesis. [1]

(c)

The Golgi apparatus receives proteins from the RER. [1]
It modifies, packages, and sorts these proteins into vesicles for secretion. [1]

(d)

Mitochondria provide the energy (ATP) required for protein synthesis, modification, and transport/secretion processes. [1]

[Total: 7]

2. (a) Rate = 1 / 45 [1] = 0.022 (accept 0.02 or 0.0222) [1] (Note: The table already provides 0.022, but showing working 1/45 is required for full credit if the value wasn't pre-filled. If pre-filled, award mark for correct calculation logic.)

(b)

As temperature increases from 10°C to 40°C, the rate of reaction increases. [1]
The rate reaches a maximum at 40°C (optimum temperature). [1]

(c)

At 60°C, the high temperature causes the enzyme (amylase) to denature. [1]
The shape of the active site changes/is destroyed. [1]
The substrate (starch) can no longer fit into the active site, so no enzyme-substrate complexes are formed. [1]

(d)

pH is a variable that affects enzyme activity; keeping it constant ensures that only temperature affects the rate of reaction (fair test). [1]

[Total: 7]

3. (a)

Osmosis is the net movement [1]
of water molecules from a region of higher water potential to a region of lower water potential [1]
through a partially permeable membrane. [1] (Award 2 marks for a concise correct definition including "net movement", "water potential gradient", and "partially permeable membrane".)

(b) Cell: R [1] (Water moves into R, meaning the external solution has higher water potential than the cell sap.)

(c)

Water leaves Cell P by osmosis because the external solution has a lower water potential. [1]
The vacuole shrinks and the cytoplasm pulls away from the cell wall. [1]
This state is called plasmolysis. The cell wall remains rigid but the cell becomes flaccid. [1]

(d)

Root hair cells have a long, narrow projection (the hair) which greatly increases the surface area to volume ratio. [1]
This allows for a faster rate of water absorption by osmosis. [1]

[Total: 8]

4. (a) Reagent: Benedict’s solution [1] Colour change: Blue to Brick-red / Orange / Yellow / Green (depending on concentration) [1] (Must mention heating for full context, but question asks for reagent and colour. Accept "Blue to Brick-red" as standard positive result.)

(b) (i) Starch [1] (ii) Protein [1]

[Total: 6]

5. (a)

Active transport is the movement of molecules/ions [1]
against a concentration gradient (from low to high concentration) [1]
using energy (ATP). [1] (Award 2 marks for "movement against concentration gradient using energy".)

(b) Any two of the following:

Active transport requires energy (ATP); diffusion does not. [1]
Active transport moves substances against the concentration gradient; diffusion moves down the concentration gradient. [1]
Active transport requires carrier proteins; simple diffusion does not (or occurs through the phospholipid bilayer). [1]

(c)

The concentration of mineral ions in the soil is often lower than in the root hair cell sap. [1]
Therefore, ions must be moved against the concentration gradient, which requires active transport. [1]

[Total: 6]

6. (a)

Red blood cells have no nucleus; white blood cells have a nucleus. [1]
OR: Red blood cells are biconcave; white blood cells are irregular/spherical. [1]
OR: White blood cells can change shape (amoeboid); red blood cells cannot. [1]

(b)

The biconcave shape increases the surface area to volume ratio. [1]
This allows for faster diffusion of oxygen into and out of the cell. [1]
It also allows the cell to be flexible to squeeze through capillaries. [1] (Award 2 marks for surface area/diffusion explanation.)

Lack of nucleus provides more space for haemoglobin. [1]
This increases the oxygen-carrying capacity of the cell. [1]

[Total: 6]