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Question 1

Topic: 1.1

The electron micrograph shows onion root cells prepared using a freeze-fracture technique. The cells were quickly frozen and then physically broken apart. Freeze fracture breaks apart cells along weak areas, such as membranes and the surfaces of organelles.

Which statement best explains the appearance of the electron micrograph?

A. The cells were broken apart at the endoplasmic reticulum; structure X is a ribosome. B. The cells were broken apart at the nuclear envelope; structure X is a nuclear pore. C. The cells were broken apart at the nuclear envelope; structure X is a ribosome. D. The cells were broken apart at the tonoplast; structure X is a plasmodesma.

▶️ Answer/Explanation

Solution

Ans: B

The freeze-fracture technique typically splits membranes, such as the nuclear envelope, revealing embedded structures. Structure X appears as a pore-like feature, consistent with nuclear pores, which allow transport between the nucleus and cytoplasm. Ribosomes (A, C) are granular and smaller, while plasmodesmata (D) are plant cell junctions, not found in this context.

Question 2

Topic: 1.1

Which cell structures can form vesicles?

▶️ Answer/Explanation

Solution

Ans: A

Vesicles are small membrane-bound sacs that transport materials within the cell. The endoplasmic reticulum (ER) and Golgi apparatus are primary structures involved in vesicle formation. The ER synthesizes proteins and lipids, while the Golgi modifies and packages them into vesicles for transport. Thus, option A is correct.

Question 3

Topic: 1.1

Four students were asked to match the function with the appearance of some cell structures in an animal cell.
The functions were listed by number.

1. synthesis of polypeptides
2. synthesis of lipids
3. packaging of hydrolytic enzymes that will remain in the cell

The appearances were listed by letter.
V membranes which surround an enclosed inner cavity
W non-membrane-bound, spherical structures
X a double membrane interspersed with pores
Y non-membrane-bound, cylindrical structures
Z membrane-bound sacs, arranged as a flattened stack

Which student correctly matched the numbered function with the appearance of the cell structure?

▶️ Answer/Explanation

Solution

Ans: A

1. Synthesis of polypeptides is performed by ribosomes (non-membrane-bound, spherical structures – W).
2. Synthesis of lipids occurs in the smooth endoplasmic reticulum (membranes surrounding an enclosed inner cavity – V).
3. Packaging of hydrolytic enzymes is done by the Golgi apparatus (membrane-bound sacs arranged as a flattened stack – Z).
Student A correctly matched these functions with their corresponding structures.

Question 4

Topic: 1.1

Which cells contain a tonoplast?

▶️ Answer/Explanation

Solution

Ans: D

A tonoplast is the membrane surrounding the vacuole in plant cells. It regulates the movement of ions and other molecules in and out of the vacuole. Since the image (not explicitly labeled here) likely represents plant cells, the correct answer is D, indicating that tonoplasts are present in plant cells.

Question 5

Topic: 1.1

Which organelles found in animal or plant cells are surrounded by double membranes?

A. chloroplasts, mitochondria, vacuoles B. chloroplasts, mitochondria, nuclei C. chloroplasts, nuclei, vacuoles D. mitochondria, nuclei, vacuoles

▶️ Answer/Explanation

Solution

Ans: B

Chloroplasts, mitochondria, and nuclei are organelles surrounded by double membranes. Chloroplasts and mitochondria have double membranes due to their endosymbiotic origin, while the nucleus has a double-membraned nuclear envelope. Vacuoles, however, are typically single-membraned, eliminating options A, C, and D.

Question 6

Topic: 1.1

Some scientists think that mitochondria evolved from bacteria that entered the cytoplasm of a different cell and were able to survive there. Which structural features of mitochondria support this hypothesis?

▶️ Answer/Explanation

Solution

Ans: D

Mitochondria share several structural features with bacteria, supporting the endosymbiotic theory. These include:

Double membrane: Similar to bacterial cell walls.
Circular DNA: Found in mitochondria and bacteria but not in eukaryotic nuclei.
70S ribosomes: Present in mitochondria and bacteria, unlike eukaryotic 80S ribosomes.

These features suggest mitochondria evolved from free-living bacteria that formed a symbiotic relationship with host cells.

Question 7

Topic: 2.1

What is present in all viruses, all prokaryotes and all eukaryotes?

A. ribose B. deoxyribose C. cytosine D. thymine

▶️ Answer/Explanation

Solution

Ans: C

Cytosine is a nitrogenous base found in all three groups: viruses (either DNA or RNA), prokaryotes (DNA), and eukaryotes (DNA). While ribose (A) and deoxyribose (B) are sugar components, they are not universal (e.g., RNA viruses lack deoxyribose). Thymine (D) is absent in RNA viruses, making cytosine the only common component.

Question 8

Topic: 2.2

The table shows some steps that can be made in carrying out the Benedict’s test.
Which combination of steps is required to carry out a semi-quantitative test on a reducing sugar solution?

▶️ Answer/Explanation

Solution

Ans: C

For a semi-quantitative Benedict’s test, we need to measure the initial volume of sugar solution (step 1), add excess Benedict’s solution (step 3), and compare the color against standards (step 5). Heating (step 4) is necessary for the reaction, while step 2 is irrelevant for semi-quantitative analysis. Thus, the correct combination is 1, 3, 4, 5.

Question 9

Topic: 2.1

The diagrams show three examples of different bonds.

Which bonds hold the secondary structure of proteins together?

A. 1, 2 and 3 B. 1 and 3 only C. 1 only D. 2 and 3 only

▶️ Answer/Explanation

Solution

Ans: C

The secondary structure of proteins (e.g., alpha-helices and beta-sheets) is stabilized primarily by hydrogen bonds (1). Ionic bonds (2) and disulfide bonds (3) contribute to tertiary or quaternary structures, not secondary. Thus, only 1 (hydrogen bonds) is correct.

Question 10

Topic: 2.1

Insulin is a globular protein involved in cell signalling. It is transported in the blood plasma from the cells that synthesise it to its target cells. A molecule of insulin contains six sulfur-containing amino acids and has two polypeptide chains.

Which statements about insulin are correct?

An insulin molecule has a quaternary structure.
Insulin polypeptides are held together by six disulfide bonds.
Amino acids with hydrophobic R groups would be found in the centre of an insulin molecule.

A. 1, 2 and 3 B. 1 and 2 only C. 1 and 3 only D. 2 and 3 only

▶️ Answer/Explanation

Solution

Ans: C

Statement 1 is correct because insulin consists of two polypeptide chains (A and B), forming a quaternary structure.

Statement 2 is incorrect. Insulin has three disulfide bonds (two interchain and one intrachain), not six.

Statement 3 is correct as hydrophobic R groups are typically buried in the core of globular proteins like insulin.

Thus, the correct statements are 1 and 3 only (Option C).

Question 11

Topic: 2.3

Which statement correctly explains why blood plasma can be maintained at a stable temperature?

A. It has a low specific heat capacity. B. It has a high specific heat capacity. C. It has a low latent heat of vaporisation. D. It has a high latent heat of vaporisation.

▶️ Answer/Explanation

Solution

Ans: B

Blood plasma maintains a stable temperature because it has a high specific heat capacity. This means it can absorb or release a large amount of heat with minimal temperature change, ensuring thermal stability in the body. Options A, C, and D are incorrect as they describe properties that do not contribute to temperature regulation.

Question 12

Topic: 3.1

Which region on the graph shows the activation energy of an enzyme-catalysed reaction?

▶️ Answer/Explanation

Solution

Ans: B

The activation energy is the energy required to initiate a reaction, represented by the peak of the curve before the reaction proceeds. In the graph, Region B corresponds to the energy barrier between reactants and products. Enzymes lower this barrier, but the activation energy is still the difference between the initial energy level and the transition state (highest point).

Question 13

Topic: 3.1

The graph shows the effect of an increasing substrate concentration on the rate of an enzyme-catalysed reaction.

Line P represents the result when the enzyme is used at its optimum pH and optimum temperature and without an inhibitor.
Line Q represents the result when the reaction conditions are changed.
Which descriptions of changes to the reaction conditions could result in line Q if all other conditions were kept the same?

1. Add an inhibitor that attaches to a site other than the active site.
2. Add an inhibitor that has a similar shape to the substrate.
3. Add an inhibitor that blocks the active site of the enzyme.
4. Carry out the reaction at a higher temperature.

A. 1, 3 and 4B. 1 and 4 only C. 2, 3 and 4 D. 2 and 3 only

▶️ Answer/Explanation

Solution

Ans: D

Line Q shows a reduced maximum reaction rate (V_max) but the same substrate concentration at half V_max (K_m) as Line P. This indicates:

Non-competitive inhibition (Option 1): Changes K_m but not V_max → Incorrect.
Competitive inhibition (Options 2 & 3): Reduces V_max without altering K_m → Correct.
Higher temperature (Option 4): Denatures the enzyme, altering both V_max and K_m → Incorrect.

Thus, only 2 and 3 (competitive inhibitors) fit the graph, making D the correct answer.

Question 14

Topic: 4.1

Which statement about cell signalling is correct?

A. One type of receptor molecule will recognise all ligands in the body. B.. The binding of a ligand may cause a change to the shape of the receptor. C. The receptors for ligands are always found on the inside of cells. D. The same ligand is made by all of the cells in the body.

▶️ Answer/Explanation

Solution

Ans: B

(A) is incorrect because receptors are specific to particular ligands.
(B) is correct—ligand binding often induces a conformational change in the receptor, triggering downstream signalling.
(C) is false; receptors can be on the cell surface (e.g., G-protein-coupled receptors) or inside cells (e.g., steroid hormone receptors).
(D) is incorrect as different cells produce distinct ligands for specialised functions.

Question 15

Topic: 4.2

Four students, A, B, C and D, observed plant epidermal cells that had been placed in a concentrated sucrose solution for 30 minutes. They were asked to identify the partially permeable layer and to explain the appearance of the cells in terms of water potential and movement of water. Which student is correct?

▶️ Answer/Explanation

Solution

Ans: B

When plant cells are placed in a concentrated sucrose solution, water moves out of the cells due to osmosis, causing plasmolysis (shrinkage of the cytoplasm away from the cell wall). The partially permeable layer is the plasma membrane, not the cell wall (which is fully permeable). Student B correctly identifies the plasma membrane and explains the movement of water from a higher water potential (inside the cell) to a lower water potential (in the sucrose solution).

Question 16

Topic: 4.2

The table compares the surface area to volume ratios of five agar blocks that differ in dimensions but which all have the same volume. The agar blocks can be used to measure the efficiency of diffusion, where efficiency is measured as the time taken for a dye to reach all parts of the block.

Which prediction can be made about the way in which size and dimensions of these blocks affect the efficiency of diffusion?

A. The efficiency of diffusion will decrease as the width of a block increases. B. The efficiency of diffusion will increase as the height of a block increases. C. The efficiency of diffusion will increase as a block of fixed volume is flattened. D. The efficiency of diffusion will decrease as a block of fixed volume is elongated.

▶️ Answer/Explanation

Solution

Ans: C

Diffusion efficiency depends on the surface area to volume ratio (SA:V). A higher SA:V means faster diffusion. When a block of fixed volume is flattened, its surface area increases relative to its volume, improving diffusion efficiency. In contrast, increasing width (A) or height (B) reduces SA:V, slowing diffusion. Elongation (D) may not always increase SA:V significantly. Thus, option C is correct.

Question 17

Topic: 5.1

The cell cycle includes mitosis. What are features of this type of nuclear division?

Forms cells of equal size to the parent cell
Forms genetically identical nuclei
Semi-conservative replication of DNA

A. 1, 2 and 3 B. 1 and 2 only C. 1 and 3 only D. 2 only

▶️ Answer/Explanation

Solution

Ans: A

Mitosis is a type of nuclear division with the following key features:

Equal-sized daughter cells: Mitosis typically produces cells of similar size to the parent cell.
Genetically identical nuclei: Ensures genetic consistency between parent and daughter cells.
Semi-conservative DNA replication: Occurs during the S-phase of the cell cycle before mitosis begins.

Since all three statements (1, 2, and 3) are correct, the answer is A.

Question 18

Topic: 5.1

A student observed the cells in the growing region (meristem) of an onion root and obtained the data shown.

Which percentage of cells contains chromosomes that appear as two chromatids?

A. 7.3% B. 8.9% C. 95.9% D. 97.5%

▶️ Answer/Explanation

Solution

Ans: B (8.9%)

Chromosomes appear as two chromatids during the metaphase and anaphase stages of mitosis. From the given data, the percentage of cells in these phases is 8.9% (metaphase + anaphase). The other phases (prophase, telophase, interphase) do not show distinct chromatid pairs, making option B the correct choice.

Question 19

Topic: 6.1

Which statement about messenger RNA is correct?

A. In eukaryotic cells, mRNA is made by removing exons from the primary RNA transcript. B. mRNA is a single-stranded polynucleotide containing a different purine base than DNA. C. mRNA molecules contain ribose sugars joined to phosphate groups by phosphodiester bonds. D. The monomers of mRNA contain a phosphate group, deoxyribose sugar, and a nitrogenous base.

▶️ Answer/Explanation

Solution

Ans: C

Option C is correct because mRNA contains ribose sugars (not deoxyribose) linked by phosphodiester bonds. Option A is wrong (introns, not exons, are removed). Option B is incorrect (both DNA and mRNA use adenine and guanine as purines). Option D is false (mRNA has ribose, not deoxyribose).

Question 20

Topic: 6.1

Which structures are involved in transcription only?

▶️ Answer/Explanation

Solution

Ans: B

Transcription is the process of synthesizing RNA from a DNA template. The key structures involved only in transcription are RNA polymerase and promoter regions (marked as B in the diagram). Other structures (e.g., ribosomes, tRNA) participate in translation, not transcription.

Question 21

Topic: 6.1

One gene provides the code for the production of which type of molecule?

A. amino acid B. DNA C. nucleotide D. polypeptide

▶️ Answer/Explanation

Solution

Ans: D

A gene is a segment of DNA that contains the instructions for synthesizing a specific polypeptide (a chain of amino acids).

While genes are made of DNA and involve nucleotides, their functional product is a polypeptide (or protein).

Thus, the correct answer is D (polypeptide).

Question 22

Topic: 6.1

The table shows some mRNA codons that code for certain amino acids.

A DNA template strand has the base sequence shown.
ACAGTATTATTTGCAACG

What would the change in the amino acid be if the first base in the fifth DNA triplet was substituted for an A base?

A. alanine to cysteine B. alanine to threonine C. arginine to cysteine D. arginine to threonine

▶️ Answer/Explanation

Solution

Ans: C

1. Original DNA triplet (5th triplet): TTT (coding for AAA in mRNA → Lysine, but table shows TTT codes for Arginine in DNA template strand context).

2. After substitution (A replaces first T): ATT (coding for UAA in mRNA → Stop codon, but in DNA template, ATT codes for Cysteine).

3. Change: Original amino acid (Arginine) → New amino acid (Cysteine).

Thus, the correct change is Arginine to Cysteine (Option C).

Question 23

Topic: 7.1

Which structures contain cytoplasm with mitochondria and a nucleus?

▶️ Answer/Explanation

Solution

Ans: A

The structures labeled A (likely cells) contain cytoplasm, mitochondria, and a nucleus, which are key features of eukaryotic cells. Structures B and C appear to be non-cellular (e.g., extracellular fibers or acellular components) as they lack these organelles. Only A matches the given criteria.

Question 24

Topic: 2.3

What is the correct term to describe intermolecular hydrogen bonding between water molecules?

A. adhesion B. cohesion C. osmosis D. diffusion

▶️ Answer/Explanation

Solution

Ans: B

Cohesion refers to the attraction between like molecules, such as hydrogen bonding between water molecules. This property gives water its high surface tension and allows it to form droplets.

Adhesion (A) describes attraction between different molecules (e.g., water and glass), while osmosis (C) and diffusion (D) involve movement of molecules, not bonding. Thus, B is correct.

Question 25

Topic: 7.2

The diagram shows the outline of a xerophytic leaf that had been left for 45 minutes in different conditions, P and Q.

Which statements about the cells in layer Y of the leaf in each of the conditions P and Q after 45 minutes are correct?

1. There is a less negative water potential in P than in Q. 2. The cells may be turgid in P and plasmolysed in Q. 3. The cells are less turgid in P than in Q. 4. There is no net diffusion of water into Y in either P or Q.

A. 1, 2 and 4 B. 1, 3 and 4 C. 1 and 3 only D. 2 and 4 only

▶️ Answer/Explanation

Solution

Ans: A

Statement 1 is correct because the leaf in condition P appears less shriveled, indicating higher water potential (less negative) than in Q.
Statement 2 is correct—cells in P likely remain turgid (maintaining structure), while those in Q may plasmolyze due to water loss.
Statement 3 is incorrect; P shows better turgidity than Q (opposite of the claim).
Statement 4 is correct—after 45 minutes, equilibrium is reached, halting net water movement.
Thus, 1, 2, and 4 are correct, matching option A.

Question 26

Topic: 7.2

How does sucrose move from chloroplasts to the phloem?

1. diffusion
2. apoplast pathway
3. symplast pathway

A. 1, 2 and 3
B. 1 and 2 only
C. 1 and 3 only
D. 2 and 3 only

▶️ Answer/Explanation

Solution

Ans: A

Sucrose moves from chloroplasts to the phloem through multiple mechanisms: 1. Diffusion (passive movement along a concentration gradient), 2. Apoplast pathway (movement through cell walls and intercellular spaces), and 3. Symplast pathway (movement via plasmodesmata through the cytoplasm of adjacent cells). All three methods contribute to sucrose transport, making option A (1, 2, and 3) correct.

Question 27

Topic: 7.2

How are companion cells involved in loading sucrose into phloem sieve tube elements?

A. actively through cotransporter proteins and passively through plasmodesmata B. actively through cotransporter proteins and plasmodesmata C. passively through cotransporter proteins only D. actively through plasmodesmata only

▶️ Answer/Explanation

Solution

Ans: A

Companion cells load sucrose into phloem sieve tube elements in two ways:
1. Actively via cotransporter proteins (e.g., H⁺-sucrose symporters), which use ATP to pump protons and facilitate sucrose uptake.
2. Passively via plasmodesmata, which allow sucrose to diffuse between cells without energy expenditure.
Option B incorrectly suggests plasmodesmata are active, while C and D exclude one of the mechanisms. Thus, A is correct.

Question 28

Topic: 8.1

The photomicrograph shows blood cells as seen using a high-power light microscope.

Which row correctly identifies the different types of white blood cell?

▶️ Answer/Explanation

Solution

Ans: C

Key identifying features of the white blood cells in the photomicrograph:

Lymphocyte (X): Small, round cell with a large nucleus occupying most of the cytoplasm.
Phagocyte (Y): Larger cell with a lobed nucleus (e.g., neutrophil) and granular cytoplasm.
Monocyte (Z): Largest white blood cell with a kidney-shaped nucleus.

These morphological characteristics match the labels in Option C.

Question 29

Topic: 8.1

Plan diagrams of two blood vessels are shown.

Which labels are correct?

▶️ Answer/Explanation

Solution

Ans: B (1 and 3)

The plan diagrams depict a vein (thin muscle layer, large lumen, presence of valves) and an artery (thick muscle layer, small lumen). Labels 1 (vein) and 3 (artery) correctly identify these vessels based on their structural features. Labels 2 and 4 are incorrect as they misrepresent the vessel types or their characteristics.

Question 30

Topic: 8.1

What are found in blood and tissue fluid?

1. carbon dioxide
2. fatty acids
3. white blood cells
4. proteins

A. 1, 2, 3 and 4 B. 1, 2 and 3 only C. 1, 2 and 4 only D. 3 and 4 only

▶️ Answer/Explanation

Solution

Ans: A

All four components are found in blood and tissue fluid: 1. CO₂ (waste product transported in blood), 2. Fatty acids (transported via plasma), 3. White blood cells (present in blood and migrate to tissues), and 4. Proteins (e.g., albumin in blood, smaller amounts in tissue fluid). Thus, the correct combination is 1, 2, 3, and 4 (Option A).

Question 31

Topic: 8.2

Which reactions would be slowed down by an inhibitor of carbonic anhydrase?

1 CO₂ + haemoglobin → carbaminohaemoglobin
2 CO₂ + H₂O → H₂CO₃
3 H₂CO₃ → H⁺ + HCO₃^–

A. 1, 2 and 3 B. 1 and 2 only C. 1 and 3 only D. 2 and 3 only

▶️ Answer/Explanation

Solution

Ans: D

Carbonic anhydrase catalyzes two key reactions:

CO₂ + H₂O ⇌ H₂CO₃ (reaction 2)
H₂CO₃ ⇌ H⁺ + HCO₃^– (reaction 3)

Inhibiting carbonic anhydrase would slow down both these reversible reactions. Reaction 1 (CO₂ binding to hemoglobin) is independent of this enzyme.

Question 32

Topic: 8.2

Which sequence of letters correctly identifies the order of events during the cardiac cycle?

T atrial walls contract
U impulse is delayed by a fraction of a second
V wave of excitation enters the atrioventricular node
W wave of excitation passes down the Purkyne tissue
X wave of excitation spreads from the sinoatrial node
Y ventricles contract

▶️ Answer/Explanation

Solution

Ans: C

The correct sequence of events during the cardiac cycle is:

X: Wave of excitation spreads from the sinoatrial node (SA node).
T: Atrial walls contract (atrial systole).
V: Wave of excitation enters the atrioventricular node (AV node).
U: Impulse is delayed briefly at the AV node to allow complete atrial emptying.
W: Wave of excitation passes down the Purkyne tissue (bundle of His and Purkinje fibers).
Y: Ventricles contract (ventricular systole).

Thus, the correct order is X → T → V → U → W → Y (Option C).

Question 33

Topic: 8.2

The diagram gives information about blood pressure in the left side of the heart during the cardiac cycle.

Valves open and close at the points numbered.

Which row identifies the valves opening or closing at the points numbered?

▶️ Answer/Explanation

Solution

Ans: D

1. Point 1: Atrial pressure exceeds ventricular pressure → Atrioventricular (AV) valves open (blood flows into ventricles).

2. Point 2: Ventricular pressure rises above atrial pressure → AV valves close (preventing backflow).

3. Point 3: Ventricular pressure exceeds aortic pressure → Semilunar valves open (blood ejected into aorta).

4. Point 4: Ventricular pressure drops below aortic pressure → Semilunar valves close (preventing backflow).

Thus, the correct sequence is D (AV open → AV close → Semilunar open → Semilunar close).

Question 34

Topic: 9.1

A person with no breathing conditions rests for an hour. Their breathing in this time is shallow and slow, so little air from outside the body reaches the alveoli. The person’s heart rate remains constant.

Which statement is correct?

A. The carbon dioxide concentration in the blood in the pulmonary vein will be higher than in the pulmonary artery. B. Carbon dioxide molecules in the air of the alveoli move out of the blood by active transport. C. The air in the alveoli has a lower concentration of oxygen than the blood in the pulmonary vein. D. Oxygen molecules diffuse from the air in the alveoli into the blood at a slower rate than when the person is active.

▶️ Answer/Explanation

Solution

Ans: D

During shallow breathing, less air reaches the alveoli, reducing the oxygen concentration gradient between alveolar air and blood. Since diffusion rate depends on the gradient, oxygen moves into the blood more slowly than during activity (where deeper breathing maintains a steeper gradient). Options A–C are incorrect because: (A) Pulmonary vein blood has less CO₂ (after gas exchange); (B) CO₂ moves by diffusion, not active transport; (C) Alveolar air has higher O₂ than pulmonary vein blood.

Question 35

Topic: 9.1

Which statements about all bronchioles are correct?

1. They have epithelium.
2. They have goblet cells.
3. They have muscle tissue.

A. 1, 2 and 3 B. 1 and 3 only C. 2 only D. 3 only

▶️ Answer/Explanation

Solution

Ans: B

Analysis of each statement:

Epithelium: All bronchioles are lined with epithelial cells (simple columnar to cuboidal epithelium). → Correct.
Goblet cells: Present only in larger bronchioles; absent in terminal/respiratory bronchioles. → Incorrect.
Muscle tissue: Smooth muscle is present in all bronchioles to regulate airflow. → Correct.

Thus, only 1 and 3 are universally true, making B the correct answer.

Question 36

Topic: 10.1

The diagram shows some of the pathogens that cause disease in humans and some of the ways they are transmitted.

What is the correct pathogen and method of transmission for malaria?

A. 1 and X B. 2 and W C. 2 and X D. 3 and Y

▶️ Answer/Explanation

Solution

Ans: C

Malaria is caused by Plasmodium (a protozoan, Pathogen 2) and transmitted via mosquito vectors (Method X).
Option A is incorrect because Pathogen 1 (bacteria) does not cause malaria.
Option B is wrong—Method W (direct contact) is not how malaria spreads.
Option D is invalid as Pathogen 3 (virus) and Method Y (airborne) do not apply to malaria.
Thus, the correct combination is 2 and X (Option C).

Question 37

Topic: 10.2

Some of the processes which result in the formation of a population of bacteria that are resistant to a new antibiotic are listed.

1. change in reproductive success of bacteria
2. increase in frequency of the resistance allele in the population
3. increase in genetic variation within the population
4. random mutation occurs in bacterial DNA

What is the correct order of these processes?

A. 1 → 3 → 2 → 4
B. 2 → 1 → 3 → 4
C. 3 → 4 → 1 → 2
D. 4 → 3 → 1 → 2

▶️ Answer/Explanation

Solution

Ans: D

The correct sequence of events in the development of antibiotic resistance is: 4. Random mutation (creating genetic variation) → 3. Increased genetic variation (including resistant alleles) → 1. Changed reproductive success (antibiotic kills non-resistant bacteria, favoring resistant ones) → 2. Increased allele frequency (resistant bacteria proliferate). This matches option D (4 → 3 → 1 → 2), illustrating natural selection in action.

Question 38

Topic: 10.2

When bacteria are grown in a Petri dish containing discs with antibiotics, there will be zones of inhibition of bacterial growth.

The chart shows the size of the zones of inhibition when a species of bacteria was incubated on five different plates of agar, each containing a disc with a different antibiotic.

Which conclusions can be made about the most and least effective antibiotics on this species of bacteria?

▶️ Answer/Explanation

Solution

Ans: A

The zone of inhibition indicates antibiotic effectiveness—larger zones mean greater effectiveness. From the chart:

Most effective: Antibiotic with the largest zone (e.g., 25 mm).
Least effective: Antibiotic with the smallest zone (e.g., 5 mm).

Option A correctly identifies these extremes based on the data. Other options either mislabel effectiveness or ignore the chart’s clear size hierarchy.

Question 39

Topic: 11.1

Which blood cell type does not recognise, engulf and digest non-self particles?

A. Macrophages B. Neutrophils C. Phagocytes D. T-killer cells

▶️ Answer/Explanation

Solution

Ans: D

Analysis of the options:

Macrophages (A), neutrophils (B), and phagocytes (C) are all specialized in phagocytosis (engulfing and digesting pathogens).
T-killer cells (D) are cytotoxic lymphocytes that destroy infected or cancerous cells by inducing apoptosis without phagocytosis.

Thus, T-killer cells (Option D) are the correct answer as they use non-phagocytic mechanisms.

Question 40

Topic: 11.2

Repeated infections with malaria result in more effective immunity to malaria.

Which type of immunity is responsible for the more effective immunity?

A. artificial active B. artificial passive C. natural active D. natural passive

▶️ Answer/Explanation

Solution

Ans: C (natural active)

Repeated malaria infections lead to natural active immunity because the body’s immune system produces its own antibodies and memory cells in response to the actual pathogen. This differs from:

Artificial immunity (A/B): Involves vaccines (active) or antibody injections (passive).
Natural passive immunity (D): Temporary protection from maternal antibodies (e.g., through placenta/breast milk).

Since the immunity arises naturally from exposure to malaria, option C is correct.

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