Secondary 4 Pure Biology Preliminary Examination Paper 2

TuitionGoWhere Practice Paper - Pure Biology Secondary 4

PRELIMINARY EXAMINATION — VERSION 2 — ANSWER KEY

Section A: Multiple Choice Questions (10 marks)

Section B: Structured Questions (35 marks)

Question 11 [8 marks]

(a) Identify cell type P and cell type Q.

• Cell P: Plant cell [1]
• Cell Q: Animal cell [1]

Marking note: Accept "plant" and "animal" respectively. Do not accept vague answers like "cell with cell wall."

(b) Two structural differences between cell P and cell Q visible in Fig. 11.1.

1. Cell P has a cell cell wall; cell Q does not. [1]
2. Cell P has a large central vacuole; cell Q has small or no vacuoles. [1]

Alternative acceptable answer: Cell P has chloroplasts; cell Q does not.

Marking note: Award 1 mark per correct difference. The difference must be a visible structural feature, not a functional description.

(c) Organelle visible in electron micrograph of cell Q but NOT under light microscope. State its function.

• Organelle: Mitochondrion (accept also: Golgi body, endoplasmic reticulum, ribosome, lysosome) [1]
• Function: Site of aerobic respiration / produces ATP (energy) for the cell. [1]

Marking note: If mitochondrion is named, function must relate to respiration/energy production. If ribosome is named, function must relate to protein synthesis. Award 1 mark for correct organelle and 1 mark for correct corresponding function.

(d) Explain why the large central vacuole in cell P is important for maintaining the cell's shape.

The large central vacuole is filled with cell sap (a solution of sugars, salts, and other substances). [1] When the vacuole is full of water, it exerts turgor pressure against the cell wall, keeping the cell rigid/firm and maintaining its shape. [1]

Marking note: Must mention the vacuole contains fluid/cell sap AND that this creates internal pressure/turgor to maintain shape. Award 1 mark for each valid point.

Question 12 [9 marks]

(a) Independent and dependent variables.

• Independent variable: Temperature [1]
• Dependent variable: Time taken for starch to be completely broken down [1]

Marking note: Must be specific. Do not accept "time" alone — must specify "time for starch to be broken down."

(b) Describe the trend between 10 °C and 40 °C.

As the temperature increases from 10 °C to 40 °C, the time taken for starch to be broken down decreases. [1] This means the rate of enzyme activity increases with increasing temperature. [1]

Marking note: Award 1 mark for describing the decrease in time, and 1 mark for linking this to increased enzyme activity/rate. Accept "enzyme activity increases" or "reaction gets faster."

(c) Explain why the time increased at 50 °C and 60 °C.

At 50 °C and 60 °C, the temperature is too high (above the optimum). [1] The enzyme denatures — the active site changes shape so that the substrate (starch) can no longer fit into it. [1] This reduces enzyme activity, so starch takes longer to be broken down. [1]

Marking note: Award 1 mark for identifying that the temperature is above optimum/too high, 1 mark for stating the enzyme denatures, and 1 mark for explaining that the active site shape changes so substrate cannot bind.

(d) One way to improve reliability.

Repeat the experiment at each temperature and calculate the mean/average. [1]

Alternative acceptable answers: Use a water bath for more accurate temperature control; use a colorimeter for more objective measurement of starch remaining.

(e) Calculate the rate of enzyme activity at 40 °C.

Working: Rate = 1 / time = 1 / 3 [1]

Rate = 0.33 min⁻¹ [1]

Marking note: Award 1 mark for correct working (1/3) and 1 mark for correct answer (0.33 or 0.333 min⁻¹). Accept 0.3 if rounded to 1 decimal place.

Question 13 [7 marks]

(a) Identify the cell.

Neuron / Nerve cell [1]

Marking note: Accept "neurone." Do not accept "nerve" alone.

(b) Two adaptations of this cell.

1. Long axon — allows electrical impulses to be transmitted over long distances. [1]
2. Many dendrites / branching extensions — allow the neuron to receive impulses from many other neurons. [1]

Alternative acceptable answers: Myelin sheath for insulation; synaptic terminals for transmitting signals to the next cell; long length for long-distance transmission.

Marking note: Award 1 mark per correct adaptation. The adaptation must be structural.

(c) Explain how the structure relates to function in transmitting electrical impulses.

The neuron has a long axon that allows electrical impulses to travel over long distances within the body without losing signal strength. [1] The branching dendrites at one end allow the cell to receive signals from multiple other neurons, while the axon terminals at the other end allow it to pass the signal on to the next neuron or effector. [1]

Marking note: Award 1 mark for explaining the role of the long axon in long-distance transmission, and 1 mark for explaining how the branching structure (dendrites/terminals) enables communication with other cells.

(d) Name the sheath and explain its function.

• Name: Myelin sheath [1]
• Function: Insulates the axon and speeds up the transmission of electrical impulses. [1]

Marking note: Accept "speeds up nerve impulse transmission" or "prevents loss of signal." Award 1 mark for correct name and 1 mark for correct function.

Question 14 [7 marks]

(a) Identify biomolecules in each solution.

• Solution X: Starch [1]
• Solution Y: Lipid and protein [1]
• Solution Z: Reducing sugar [1]

Marking note: For Solution Y, both lipid AND protein must be stated for the mark. Solution X: iodine test is blue-black → starch. Solution Y: Biuret is violet → protein; ethanol emulsion is cloudy → lipid. Solution Z: Benedict's is brick-red → reducing sugar.

(b) Explain why Benedict's test requires heating.

Heating provides the activation energy needed for the reaction between Benedict's reagent and reducing sugars to occur. [1] The reaction only proceeds at an elevated temperature to produce the coloured precipitate.

Marking note: Accept "to speed up the reaction" or "the reaction requires heat to occur." The key idea is that heating is necessary for the chemical reaction to take place.

(c) One biological function of the biomolecule in solution Y.

Protein: growth and repair of cells / acts as enzymes / forms antibodies. [1]

OR

Lipid: long-term energy storage / insulation / forms part of cell membranes. [1]

Marking note: Award 1 mark for any valid biological function of either protein or lipid.

(d) Explain the observation after incubating solution X with amylase.

Amylase is an enzyme that breaks down starch into maltose (a reducing sugar). [1] After incubation, the starch has been hydrolysed/digested, so there is no starch left to react with iodine, giving a brown-yellow result. [1]

Marking note: Award 1 mark for stating that amylase breaks down starch, and 1 mark for explaining that starch is no longer present so the iodine test is negative.

Question 15 [6 marks]

(a) Name the condition.

Plasmolysis [1]

Marking note: Accept "plasmolysed" or "flaccid." Do not accept "crenation" (which refers to animal cells).

(b) Explain why the cell appears as shown.

The concentrated salt solution has a lower water potential (more negative / more concentrated) than the cell sap. [1] Water moves out of the cell by osmosis from a region of higher water potential (inside the cell) to a region of lower water potential (the salt solution). [1] The vacuole and cytoplasm shrink, pulling the cell membrane away from the cell wall. [1]

Marking note: Award 1 mark for identifying the water potential difference, 1 mark for stating water moves out by osmosis, and 1 mark for describing the result (cytoplasm/vacuole shrinks, membrane pulls away from wall).

(c) Describe and explain what happens when placed in distilled water.

The cell will become turgid / return to its original shape. [1] Distilled water has a higher water potential than the cell sap, so water moves into the cell by osmosis. [1] The vacuole and cytoplasm expand, pushing the cell membrane against the cell wall. [1]

Marking note: Award 1 mark for describing the outcome (cell becomes turgid/returns to normal), 1 mark for identifying the water potential gradient, and 1 mark for explaining osmosis into the cell.

(d) Explain why animal cells burst in distilled water but plant cells do not.

Animal cells do not have a cell wall, so when water enters by osmosis, the cell membrane stretches until it bursts (lysis). [1] Plant cells have a rigid cell wall that prevents the cell from bursting — it becomes turgid but the cell wall provides structural support. [1]

Marking note: Award 1 mark for explaining that animal cells lack a cell wall and burst, and 1 mark for explaining that the plant cell wall prevents bursting.

Section C: Free Response Question (15 marks)

Question 16 [13 marks]

(a) Plot a graph of mass of protein remaining (g) against pH. [3]

Marking scheme for graph:

• [1] Correctly labelled axes (x-axis: pH; y-axis: Mass of protein remaining (g)) with appropriate scales
• [1] All five points correctly plotted
• [1] Smooth curve or line drawn through the points

Expected shape: A curve that decreases from pH 1.0 to pH 2.0, then increases sharply from pH 2.0 to pH 7.0, with the lowest point at pH 2.0.

(b) Optimum pH and explanation.

• Optimum pH: pH 2.0 [1]
• Explanation: At pH 2.0, the least mass of protein remains (0.1 g), meaning the most protein has been digested, so pepsin is most active at this pH. [1]

Marking note: Award 1 mark for correct pH (2.0) and 1 mark for explaining that the least protein remaining indicates the highest enzyme activity.

(c) Explain why mass of protein remaining is highest at pH 7.0.

At pH 7.0, the pH is far from the optimum for pepsin. [1] The enzyme is denatured / the active site changes shape, so it can no longer bind effectively to the protein substrate. [1] Very little protein is digested, so most of the protein remains. [1]

Marking note: Award 1 mark for stating pH 7.0 is far from optimum, 1 mark for stating the enzyme denatures/active site changes shape, and 1 mark for linking this to reduced digestion.

(d) Predict the mass of protein remaining at pH 1.5.

• Prediction: Between 0.1 g and 0.2 g (accept any value in this range, e.g., 0.15 g) [1]
• Reasoning: pH 1.5 is between pH 1.0 and pH 2.0, and the optimum is at pH 2.0, so the mass remaining should be slightly higher than at pH 2.0 but lower than at pH 1.0. [1]

Marking note: Award 1 mark for a prediction between 0.1 and 0.2 g, and 1 mark for explaining interpolation between the two known data points.

(e) Predict and explain the result with boiled pepsin at pH 2.0.

• Prediction: Mass of protein remaining will be high / close to the original mass (e.g., ~5.0 g) [1]
• Explanation: Boiling denatures the enzyme — the active site is permanently altered so the substrate can no longer bind. [1] The enzyme is no longer functional, so the protein will not be digested. [1]

Marking note: Award 1 mark for predicting high protein remaining (little or no digestion), 1 mark for stating boiling denatures the enzyme, and 1 mark for explaining the active site is permanently changed.

(f) Two variables to keep constant.

1. Volume/concentration of protein solution [1]
2. Volume/concentration of pepsin solution [1]

Alternative acceptable answers: Temperature; incubation time; total volume of mixture; source of pepsin.

Marking note: Award 1 mark per correct controlled variable. Do not accept "pH" (this is the independent variable) or "mass of protein remaining" (this is the dependent variable).

Question 17 [6 marks]

(a) Two functions of lysosomes.

1. Break down/digest worn-out organelles. [1]
2. Break down/digest engulfed bacteria / foreign particles. [1]

Alternative acceptable answers: Break down macromolecules; autophagy (digesting the cell's own components); digest materials taken in by endocytosis.

Marking note: Award 1 mark per correct function. Must relate to digestion/breakdown of materials.

(b) Explain why lysosomal enzymes function best at pH 5.0 but are less effective at pH 7.2.

Lysosomal enzymes have an optimum pH of 5.0 — at this pH, the active site has the correct shape to bind to the substrate. [1] At pH 7.2 (cytoplasm pH), the pH is too high / too alkaline for these enzymes, causing the active site to change shape so the enzyme-substrate complex cannot form effectively. [1]

Marking note: Award 1 mark for stating pH 5.0 is the optimum, and 1 mark for explaining that at pH 7.2 the active site shape changes, reducing enzyme activity.

(c) Explain why release of lysosomal enzymes into the cytoplasm would cause cell death.

If lysosomal enzymes are released into the cytoplasm, they would digest the cell's own organelles and structures. [1] This would break down essential components of the cell, leading to cell death.

Marking note: Award 1 mark for stating that the enzymes would digest the cell's own components, and 1 mark for linking this to cell death. Accept "autolysis" or "self-digestion."

(d) Suggest why plant cells do not require lysosomes for digestion.

Plant cells have a large central vacuole that contains hydrolytic enzymes which perform the same digestive function as lysosomes. [1] The vacuole can break down worn-out organelles and macromolecules, so lysosomes are not needed.

Marking note: Award 1 mark for stating that the central vacuole performs the digestive role in plant cells. Must mention the vacuole contains digestive/hydrolytic enzymes.

Mark Summary

Note: The paper totals 66 marks. In an actual exam, the total would be scaled or adjusted to 60 marks. For practice purposes, students should aim to score as high as possible out of 66.

© TuitionGoWhere Secondary School (AI) — Preliminary Examination Practice, Version 2 — Answer Key