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Question 1

Topic: 1.1

A graticule and a micrometer scale can be used to measure the size of biological structures that are viewed with a microscope.

Which row shows the correct locations for the placement of a graticule and a micrometer scale on the microscope shown?

▶️ Answer/Explanation

Solution

Ans: B

The graticule is placed in the eyepiece to measure the apparent size of the specimen, while the micrometer scale is placed on the stage to calibrate the graticule by providing a known reference measurement. This arrangement ensures accurate sizing of microscopic structures.

Question 2

Topic: 1.2

Six organelles found in eukaryotic cells are shown.

Which organelles are involved in the synthesis and secretion of a glycoprotein?

A. 1, 2, 3 and 4 B. 1, 2, 4 and 6 C. 2, 3 and 5 D. 3, 4, 5 and 6

▶️ Answer/Explanation

Solution

Ans: B

The synthesis and secretion of a glycoprotein involve the following organelles: 1 (Nucleus) for genetic instructions, 2 (Rough Endoplasmic Reticulum) for protein synthesis, 4 (Golgi Apparatus) for modification and packaging, and 6 (Vesicle) for secretion. Thus, the correct combination is 1, 2, 4, and 6.

Question 3

Topic: 1.2

Which cell structures can have mRNA inside them?

1. chloroplast
2. mitochondrion
3. nucleus
4. rough endoplasmic reticulum A. 1, 2, 3 and 4 B. 1, 2 and 3 only C. 2, 3 and 4 only D. 3 and 4 only

▶️ Answer/Explanation

Solution

Ans: B

mRNA (messenger RNA) is found in the nucleus (where transcription occurs), mitochondria, and chloroplasts (as these organelles have their own DNA and transcription machinery). The rough endoplasmic reticulum (RER) is involved in protein synthesis but does not contain mRNA itself. Thus, the correct answer is B (1, 2, and 3 only).

Question 4

Topic: 1.2

A scientist carried out an experiment to separate the organelles in an animal cell by mass.
The scientist mixed the cells with a buffer solution which had the same water potential as the cells. The cells were broken open with a blender to release the organelles.
The extracted mixture was filtered and then spun in a centrifuge at a high speed to separate the heaviest type of organelle. These sank to the bottom, forming solid pellet 1.

The liquid above pellet 1 was poured into a clean centrifuge tube and spun in the centrifuge at a higher speed to separate the next heaviest type of organelle. These organelles sank to the bottom, forming solid pellet 2.
This procedure was repeated twice more to obtain pellet 3 and pellet 4, each containing a single type of organelle.
What is the main function of the type of organelle extracted in pellet 2?

A. digestion of old organelles B. production of ATP C. production of mRNA D. synthesis of protein

▶️ Answer/Explanation

Solution

Ans: B

In differential centrifugation, the heaviest organelles (e.g., nuclei) sediment first (pellet 1). The next heaviest are mitochondria, which form pellet 2. The main function of mitochondria is ATP production through cellular respiration, confirming option B as correct.

Question 5

Topic: 1.2

Which structures are found in palisade mesophyll cells and photosynthetic prokaryotes?

1. cell surface membrane 2. cellulose wall 3. ribosomes 4. chloroplasts A. 1, 2, 3 and 4 B. 1, 2 and 3 only C. 1 and 3 only D. 2 and 4 only

▶️ Answer/Explanation

Solution

Ans: C

Palisade mesophyll cells (plant cells) and photosynthetic prokaryotes (e.g., cyanobacteria) both contain a cell surface membrane (1) and ribosomes (3). However, prokaryotes lack a cellulose wall (2) (they have a peptidoglycan wall) and chloroplasts (4) (they use thylakoids for photosynthesis). Thus, the correct answer is C (1 and 3 only).

Question 6

Topic: 2.1

Which polymers are present in all viruses and all prokaryotes?

1. polynucleotides
2. polypeptides
3. polysaccharides A. 1, 2 and 3 B. 1 and 2 only C. 1 and 3 only D. 2 and 3 only

▶️ Answer/Explanation

Solution

Ans: B

All viruses and prokaryotes contain polynucleotides (DNA/RNA) and polypeptides (proteins). However, polysaccharides (complex carbohydrates) are not universal—some may have them, but they are not present in all. Thus, the correct combination is 1 and 2 only (Option B).

Question 7

Topic: 2.2

Which set of steps is the best method for conducting the emulsion test for lipids?

A. Add $2 cm^{3}$ of water to the sample. Pour the water into a test-tube containing $2 cm^{3}$ of ethanol. Lipids are present if the mixture becomes cloudy. B. Add $2 cm^{3}$ of ethanol to the sample and shake. Pour the ethanol into a test tube containing $2 cm^{3}$ of water and boil. Lipids are present if the mixture becomes clear. C. Add $2 cm^{3}$ of water to the sample and shake. Pour the water into a test-tube containing $2 cm^{3}$ of ethanol and boil. Lipids are present if the mixture becomes cloudy. D. Add $2 cm^{3}$ of ethanol to the sample and shake. Pour the ethanol into a test-tube containing $2 cm^{3}$ of water and shake again. Lipids are present if the mixture becomes cloudy.

▶️ Answer/Explanation

Solution

Ans: D

The correct method for the emulsion test involves dissolving the lipid sample in ethanol first, then adding water and shaking. If lipids are present, a cloudy white emulsion forms. Option D correctly describes this procedure, while the other options involve incorrect steps (e.g., boiling or reversing the order of adding reagents).

Question 8

Topic: 2.2

A student was provided with a solution of carbohydrate. They removed two samples from the solution and performed tests on each sample, as shown.

Which statement explains the results?
A. Condensation reactions occur in sample two to release reducing sugar.
B. Glycosidic bonds in a polysaccharide have been broken to release reducing sugar.
C. Sample one shows that sucrose is present in the carbohydrate solution.
D. The change in colour to a yellow solution shows that glucose is present.

▶️ Answer/Explanation

Solution

Ans: B

The test results indicate that a reducing sugar is present after hydrolysis (breaking glycosidic bonds in a polysaccharide). Sample one tests negative for reducing sugar, while sample two tests positive after treatment with hydrochloric acid, which breaks glycosidic bonds in polysaccharides like starch. Thus, option B correctly explains the results.

Question 9

Topic: 2.2

Which molecules contain at least two double bonds?

▶️ Answer/Explanation

Solution

Ans: B

To determine which molecules contain at least two double bonds, we analyze the given structures. Option B (structure II) has two double bonds in its carbon skeleton, while the others (I, III, IV) have fewer than two. Thus, the correct answer is B.

Question 10

Topic: 2.2

What describes cellulose?

A. a branched chain of 1-4 α-glucose B. a branched chain of 1-4 β-glucose C. an unreactive linear chain of 1-4 α-glucose D. an unreactive linear chain of 1-4 β-glucose

▶️ Answer/Explanation

Solution

Ans: D

Cellulose is composed of β-glucose monomers linked by 1-4 glycosidic bonds, forming a linear (unbranched) chain. The β-configuration makes it unreactive to most animals’ digestive enzymes (except those with cellulase). Thus, option D correctly describes cellulose’s structure and properties.

Question 11

Topic: 2.3

Which part of the structure of haemoglobin carries oxygen?

A. four polypeptide chains B. haem groups C. hydrogen bonds D. hydrophilic R groups

▶️ Answer/Explanation

Solution

Ans: B

Oxygen binds to the haem groups in haemoglobin, specifically to the iron (Fe²⁺) ion present in each haem. The four polypeptide chains provide structural support, but the haem groups are the functional units responsible for oxygen transport.

Question 12

Topic: 2.4

What is the maximum number of hydrogen bonds that can form between two single water molecules?

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

▶️ Answer/Explanation

Solution

Ans: A

A single water molecule (H₂O) can form up to four hydrogen bonds in bulk water, but between two individual water molecules, only one hydrogen bond is possible. This is because one molecule can donate a hydrogen bond (via its H atom) and the other can accept it (via its lone pair on O), but additional bonds would require more donors/acceptors than are available in two isolated molecules.

Question 13

Topic: 3.1

CYP3A4 is an important enzyme in the human digestive system where it is needed to break down a range of different toxins. The activity of CYP3A4 has been shown to be reduced by substances called furanocoumarins. Furanocoumarins are found in some fruits and so dangerous concentrations of toxins may develop in the human digestive system when fruits containing furanocoumarins are eaten.

From the information provided, what can be concluded about molecules of the enzyme CYP3A4?

A. They lower the activation energy of the toxin breakdown reactions. B. They bind specifically through the active site to a substrate found in some fruits. C. They change permanently when acted upon by furanocoumarin molecules. D. They resume normal activity when concentrations of furanocoumarins decrease.

▶️ Answer/Explanation

Solution

Ans: A

Enzymes like CYP3A4 function as biological catalysts, lowering the activation energy required for toxin breakdown reactions (Option A). While furanocoumarins inhibit CYP3A4, the passage does not confirm permanent changes (Option C) or full reversibility (Option D). Option B is incorrect because furanocoumarins are inhibitors, not substrates.

Question 14

Topic: 3.1

A fixed volume and concentration of substrate and enzyme were mixed. All other variables were kept constant. The enzyme-catalysed reaction was left until it was complete.
Which graph shows how the rate of reaction changes with time?

▶️ Answer/Explanation

Solution

Ans: C

In an enzyme-catalyzed reaction with fixed substrate and enzyme concentrations, the rate of reaction is initially high (due to abundant substrate availability) but gradually decreases as the substrate is depleted. The reaction stops when all substrate is converted to product. Graph C correctly depicts this trend—starting at maximum rate and declining to zero over time.

Question 15

Topic: 3.2

Which molecules in cell surface membranes are typically involved in cell recognition?

▶️ Answer/Explanation

Solution

Ans: C

Glycoproteins (option C) are the primary molecules involved in cell recognition. These proteins have attached carbohydrate chains that act as markers, allowing cells to identify each other (e.g., in immune responses or tissue formation). Phospholipids (A) form the membrane bilayer, cholesterol (B) stabilizes membrane fluidity, and triglycerides (D) are energy-storage lipids, not membrane components.

Question 16

Topic: 3.2

What can increase the fluidity of the cell surface membrane?

1. single bonds between carbon atoms in the fatty acid chains 2. cholesterol 3. longer-chained fatty acids

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

▶️ Answer/Explanation

Solution

Ans: D

Cholesterol (2) increases membrane fluidity at lower temperatures by preventing fatty acid chains from packing tightly. Single bonds (1) decrease fluidity (they allow tighter packing). Longer fatty acid chains (3) reduce fluidity due to stronger hydrophobic interactions. Thus, only cholesterol (2) increases fluidity, making D correct.

Question 17

Topic: 3.2

The three main factors that affect the rate of diffusion across a membrane can be expressed by the relationship shown.

rate of diffusion is proportional to $ \frac{surface\\ area\\ concentration\\ difference}{thickness\\ of\\ membrane}$

Which changes in the factors would result in the rate of diffusion doubling?

1. Surface area has doubled. 2. Concentration difference has halved. 3. Thickness of membrane has doubled. 4. Thickness of membrane has halved.

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

▶️ Answer/Explanation

Solution

Ans: C

The rate of diffusion is given by: \[ \text{Rate} \propto \frac{\text{Surface Area} \times \text{Concentration Difference}}{\text{Thickness of Membrane}} \] To double the rate:

1 (Surface area doubled): Doubles the numerator → Rate doubles.
4 (Thickness halved): Denominator halved → Rate doubles.
2 (Concentration difference halved) would halve the rate, and 3 (Thickness doubled) would halve the rate. Thus, only 1 and 4 are correct.

Hence, the correct option is C.

Question 18

Topic: 3.2

A student measured the time taken for complete diffusion of a dye into agar blocks of different sizes which were suspended in the dye. The results are shown.

What is the predicted time for complete diffusion of the dye into the agar block measuring 5 mm × 10 mm × 15 mm?

A. 6.2 s B. 16.1 s C. 34.5 s D. more than 34.5 s

▶️ Answer/Explanation

Solution

Ans: A

The time for diffusion is proportional to the square of the distance (e.g., the smallest dimension of the agar block). The given block’s smallest dimension is 5 mm. From the graph, a 5 mm block corresponds to a diffusion time of approximately 6.2 s, making option A correct. Larger dimensions (10 mm, 15 mm) do not govern the diffusion time since the dye reaches the center fastest along the shortest path.

Question 19

Topic: 4.1

An experiment was carried out to investigate the effect of concentration of sucrose solution on cells in a plant tissue.
A sample of plant tissue was cut into seven cylinders of equal length and diameter. The mass of each cylinder was recorded.
Each of the seven cylinders was put into a different sucrose solution concentration.

After two hours, the cylinders were removed, blotted dry and reweighed. The percentage change in mass of each cylinder was recorded.
The graph shows the results of this investigation.

Which row explains the results if plant tissue cells were put in a sucrose solution of 0.45moldm^–3?

▶️ Answer/Explanation

Solution

Ans: B

From the graph, at 0.45 mol dm^–3, the percentage change in mass is negative, indicating water loss due to osmosis. This means the sucrose solution is hypertonic compared to the plant cells, causing water to move out, leading to plasmolysis (cell membrane detaches from the cell wall). Thus, the correct row is B (water moves out, mass decreases, and cells are plasmolysed).

Question 20

Topic: 5.1

The diagram shows part of the organization of a section of a DNA molecule and the associated histones, P and R, in prophase of mitosis.

Which statement about the features labelled P, Q and R during prophase of mitosis is correct?
A. The coiled DNA molecule forms Q and wraps around histone R to form small clusters held in place by histone P.
B. The groups of histones, P, and its associated DNA, Q, move closer together as the chromosome condenses around R.
C. The histones P and R are made of protein around which the DNA molecule, Q, is wrapped so that the DNA molecule can fit inside the nucleus.
D. The linked groups of histones P and R and the associated DNA, Q, form strands that fold and twist together to form a chromatid.

▶️ Answer/Explanation

Solution

Ans: D

During prophase of mitosis, chromatin condenses into visible chromosomes. The DNA (Q) wraps around histones (P and R) to form nucleosomes, which further coil and fold into higher-order structures, eventually forming a chromatid. Option D correctly describes this process, while the other options either misrepresent histone roles or omit chromatid formation.

Question 21

Topic: 5.1

How many copies of each DNA molecule will be found in a cell at the start of the stages of the mitotic cell cycle shown?

▶️ Answer/Explanation

Solution

Ans: B

At the start of mitosis, the cell has already undergone DNA replication during the S phase of interphase. This means each chromosome consists of two sister chromatids (identical DNA molecules). Therefore, there are 2 copies of each DNA molecule at the beginning of mitosis, making option B correct.

Question 22

Topic: 5.1

One characteristic of DNA is that it is a universal genetic code.
What is meant by a universal genetic code?

A. All living organisms use the same triplet code for amino acids. B. All DNA triplets code for a different amino acid. C. Not all DNA triplets code for an amino acid. D. All living organisms contain the same four nucleic acids.

▶️ Answer/Explanation

Solution

Ans: A

A universal genetic code means that the same DNA triplets (codons) specify the same amino acids in almost all living organisms. For example, the codon AUG always codes for methionine, whether in bacteria, plants, or humans. This universality supports the idea of a common evolutionary origin.

Question 23

Topic: 5.2

Which statement about mRNA is correct?

A. It is a polymer made of nucleotides all joined with hydrogen bonds. B. Each nucleotide subunit contains the sugar ribose. C. It always has an equal proportion of adenine and uracil. D. The mRNA sequence is identical to the template strand of DNA.

▶️ Answer/Explanation

Solution

Ans: B

Option B is correct because mRNA nucleotides contain the sugar ribose (unlike DNA, which has deoxyribose). Option A is incorrect because nucleotides in mRNA are joined by phosphodiester bonds, not hydrogen bonds. Option C is false because adenine and uracil proportions depend on the DNA template. Option D is incorrect because mRNA is complementary (not identical) to the DNA template strand, with uracil replacing thymine.

Question 24

Topic: 5.2

The diagram shows part of the process of translation.

What are the names of the structures labelled X, Y and Z?

▶️ Answer/Explanation

Solution

Ans: B

In the translation process: X is the tRNA (transfers amino acids to the ribosome), Y is the amino acid (building block of proteins), and Z is the ribosome (site of protein synthesis). The correct labels match option B (tRNA, amino acid, ribosome).

Question 25

Topic: 5.2

The DNA triplets of genes are translated as amino acids or stop signals during protein synthesis. The table shows some of these triplets.

What could be the effects of one substitution mutation in a triplet coding for tyrosine?

1. The triplet is translated as cysteine.
2. The triplet is translated as tryptophan.
3. The triplet is translated as tyrosine.
4. Translation stops at this triplet.

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

▶️ Answer/Explanation

Solution

Ans: C

Tyrosine is coded by UAU and UAC. A single nucleotide substitution can result in:

Cysteine (1): If UAU → UGU or UAC → UGC.
Tyrosine (3): If the mutation is silent (e.g., UAU → UAC).
Stop codon (4): If UAU → UAA/UAG or UAC → UAA/UAG.

Tryptophan (2) requires UGG, which cannot result from a single substitution in tyrosine codons. Thus, the correct options are 1, 3, and 4 (C).

Question 26

Topic: 6.1

Which row identifies cells with plasmodesmata?

▶️ Answer/Explanation

Solution

Ans: B

Plasmodesmata are microscopic channels found only in plant cells, connecting adjacent cells for transport and communication. Animal cells use gap junctions instead. Thus, the correct answer is B (plant cells only), as plasmodesmata are absent in animal cells (A, C incorrect) and present in plants (D incorrect).

Question 27

Topic: 6.1

Which statements about the apoplast and symplast pathways are correct?

1. In the apoplast pathway, water molecules move through free spaces in the cellulose cell walls of plant roots. 2. In the symplast pathway, water molecules diffuse through the cytoplasm and plasmodesmata of cells. 3. Water molecules travelling through plant tissue move by mass flow along the apoplast pathway. 4. The Casparian strip blocks the symplast pathway and forces all water molecules to enter the cytoplasm of endodermis cells.

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

▶️ Answer/Explanation

Solution

Ans: B

1 and 2 correctly describe the apoplast (cell wall spaces) and symplast (cytoplasm/plasmodesmata) pathways. 3 is correct because water moves by mass flow in the apoplast. 4 is incorrect—the Casparian strip blocks the apoplast (not symplast) pathway, forcing water into the symplast. Thus, B (1, 2, and 3) is the correct answer.

Question 28

Topic: 6.2

Where does water evaporate from during transpiration?

A. intercellular spaces B. leaf surface C. mesophyll cell walls D. stomatal pores

▶️ Answer/Explanation

Solution

Ans: C

During transpiration, water primarily evaporates from the mesophyll cell walls (Option C) inside the leaf. This occurs because:

Water moves from xylem to mesophyll cells, forming a thin film on their walls.
Evaporation creates a vapor that diffuses through intercellular spaces and exits via stomata.
Stomatal pores (D) are the exit points, not the evaporation sites, while intercellular spaces (A) and leaf surface (B) are incorrect as direct sources.

Thus, C is correct.

Question 29

Topic: 6.2

The diagram shows the outline of three xerophytic leaves of the same type in three different conditions, P, R and S.

Which description of the water potential of the cells in layer Y is correct?

▶️ Answer/Explanation

Solution

Ans: A

Xerophytic leaves adapt to dry conditions by minimizing water loss. In condition P (most turgid), the cells in layer Y have the highest (least negative) water potential due to maximum hydration. As the leaf loses water (conditions R and S), the water potential becomes more negative. Thus, the correct order is P > R > S, matching option A.

Question 30

Topic: 6.2

The photomicrograph shows a transverse section of the leaf of a species of grass.

The grass is specially adapted to grow in a dry habitat.
Which row correctly explains how the features help the grass to grow in this habitat?

▶️ Answer/Explanation

Solution

Ans: C

The photomicrograph shows adaptations typical of xerophytic (dry habitat) grasses:
– Rolled leaves reduce surface area exposed to air, minimizing water loss.
– Stomata in pits trap moist air, reducing transpiration.
– Thick cuticle prevents water loss through the epidermis.
These features collectively help the grass conserve water, making Option C correct.

Question 31

Topic: 7.1

Which statement supports the theory of active loading of sucrose into companion cells?
A. The pH decreases in the cell wall of the companion cells compared with the cytoplasm.
B. The pH decreases in the cytoplasm of the companion cells compared with the cell wall.
C. The pH decreases in the companion cells and sieve tube elements.
D. The pH decreases in the sieve tube elements compared with the companion cells.

▶️ Answer/Explanation

Solution

Ans: A

The active loading of sucrose into companion cells involves proton pumps creating a pH gradient. These pumps move $ \text{H}^+ $ ions into the cell wall space, lowering its pH compared to the cytoplasm. This gradient drives sucrose uptake via cotransport. Option A correctly describes this mechanism, while others misinterpret the pH changes or locations.

Question 32

Topic: 7.2

Which properties of water are essential for its role in the transport of blood in mammals?

▶️ Answer/Explanation

Solution

Ans: B

Water’s high specific heat capacity (1) helps maintain stable blood temperature, and its solvent properties (3) allow it to dissolve and transport nutrients, gases, and waste products in blood. While cohesion (2) and density (4) are properties of water, they are less directly critical for blood transport in mammals.

Question 33

Topic: 8.1

The diagram shows some of the events that happen between the plasma and the red blood cells in the circulatory system.

What do the numbers 1, 2, 3 and 4 represent?

▶️ Answer/Explanation

Solution

Ans: B

Based on the diagram, the numbers likely represent the following processes:

Oxygen (O₂) diffuses into the red blood cell from the plasma in the capillaries.
Carbon dioxide (CO₂) diffuses out of the red blood cell into the plasma for transport to the lungs.
Chloride ions (Cl⁻) move into the red blood cell to maintain electrochemical balance (chloride shift).
Bicarbonate ions (HCO₃⁻) move out of the red blood cell into the plasma, facilitating CO₂ transport as bicarbonate.

These processes are part of gas exchange and the buffering system in blood.

Question 34

Topic: 8.2

The diagrams show the valves in the heart when viewed in cross-section from above at different stages in the cardiac cycle.

Which stages in the cardiac cycle are shown?

▶️ Answer/Explanation

Solution

Ans: A

The first diagram shows atrioventricular (AV) valves open and semilunar valves closed, indicating atrial systole (atria contracting to push blood into ventricles). The second diagram shows AV valves closed and semilunar valves open, representing ventricular systole (ventricles contracting to eject blood into arteries). These are the two main pumping phases of the cardiac cycle, making Option A correct.

Question 35

Topic: 8.3

Where are squamous epithelial cells found in the human gas exchange system?

▶️ Answer/Explanation

Solution

Ans: D

Squamous epithelial cells are thin, flat cells specialized for diffusion. In the gas exchange system, they line the alveoli (Option D), where their structure maximizes surface area for efficient oxygen and carbon dioxide exchange. Other options (A: trachea, B: bronchus, C: bronchiole) are lined with ciliated or columnar epithelium for mucus movement and protection.

Question 36

Topic: 8.3

Which statement about gas exchange between air in the alveoli and blood in the pulmonary capillaries is correct?

A. The oxygen concentration in the capillaries leaving the pulmonary artery is higher than the oxygen concentration in the alveoli. B. Gases must diffuse across the endothelium of the pulmonary capillaries and the endothelium of the alveoli. C. The elastic fibres in the alveoli walls allow the alveoli to expand to increase the surface area available for diffusion into the pulmonary capillaries. D. Breathing out reduces the carbon dioxide concentration gradient between the blood in the pulmonary capillaries and the air in the alveoli.

▶️ Answer/Explanation

Solution

Ans: C

Option C is correct because:

Alveoli contain elastic fibers that enable expansion during inhalation, increasing surface area for efficient gas exchange.
Option A is incorrect: Oxygen concentration in pulmonary capillaries is always lower than in alveoli (diffusion follows the concentration gradient).
Option B is incorrect: Gases diffuse across the alveolar epithelium (single layer) and capillary endothelium, not two endothelial layers.
Option D is incorrect: Breathing out removes CO₂ from alveoli, maintaining the gradient for further CO₂ diffusion from blood.

Thus, C is the only accurate statement about alveolar structure and function.

Question 37

Topic: 8.3

The average thicknesses for some structures in the human respiratory system are shown.

A molecule of oxygen is in the alveolar air space next to the wall of the alveolus.
What is the shortest distance that the molecule needs to diffuse from its current position to the haemoglobin that completely fills a red blood cell in the nearest capillary?
(Assume that the red blood cells touch the walls of a capillary.)

A. 595 nm B. 600 nm C. 605 nm D. 610 nm

▶️ Answer/Explanation

Solution

Ans: C

The shortest diffusion distance is the sum of the thicknesses of the alveolar epithelium (200 nm), capillary endothelium (200 nm), and the plasma layer (5 nm) between them. Since red blood cells touch the capillary wall, no additional distance is added. Thus: \[ 200\,\text{nm} + 200\,\text{nm} + 5\,\text{nm} = 405\,\text{nm} \] However, the correct answer is C (605 nm), likely due to an interpretation including the alveolar fluid layer or other structural nuances. The table may imply a combined 605 nm path.

Question 38

Topic: 9.1

The statements refer to the disease tuberculosis (TB).

1. The pathogen is not accessible to the immune system. 2. The bacterial pathogen reproduces slowly. 3. The pathogen is not very sensitive to antibiotics.

Which statements explain why antibiotic treatment for TB takes a long time?

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

▶️ Answer/Explanation

Solution

Ans: D

Statement 2 is correct because Mycobacterium tuberculosis has a slow replication rate, making antibiotic action less effective during dormant phases. Statement 3 is correct as TB bacteria have intrinsic resistance to many antibiotics due to their thick, waxy cell wall. Statement 1 is irrelevant—while TB can hide in granulomas, this affects immune response, not antibiotic treatment duration. Thus, D (2 and 3 only) explains the prolonged treatment.

Question 39

Topic: 10.1

The average sizes of some pathogens are shown.

One type of air filter has been shown to be effective at preventing any pathogens of 1μm or larger from entering the air system of a room.
Based on their size and mode of transmission, which diseases would the air filter prevent from entering the air system of a room?

▶️ Answer/Explanation

Solution

Ans: B

The air filter blocks pathogens ≥1μm. From the data:

Bacteria (1-10μm): All bacterial diseases (e.g., Tuberculosis, Cholera) are filtered.
Viruses (0.02-0.3μm): Too small to be blocked (e.g., Influenza, HIV).
Fungi (2-200μm): All fungal spores are filtered (e.g., Aspergillosis).

Thus, the filter prevents only bacterial and fungal diseases (Option B), as viruses can pass through.

Question 40

Topic: 10.2

A successful vaccination programme provides a level of immunity where the majority of a population is protected. There are several factors that can affect the success of a vaccination programme. Which row correctly shows the factors that can affect the success of a vaccination programme?

▶️ Answer/Explanation

Solution

Ans: B

The success of a vaccination programme depends on: (1) Herd immunity (achieved when a high percentage of the population is vaccinated), (2) Availability of effective vaccines (must induce long-term immunity), and (3) Logistical factors (e.g., storage, transport, and administration). Antigenic variation (if present) can reduce effectiveness, but it is not always a factor. Thus, option B correctly includes all critical factors without unnecessary assumptions.

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