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Question 1

Topic: 1.2 (Cell as the basic units of living organisms)

Fig. 1.1 is a diagram showing part of a cell surface membrane of an animal cell.

Cell Surface Membrane Diagram

(a) (i) State the approximate thickness of the membrane as shown by the line G–H.

(ii) Complete Table 1.1 to show:

• the names and functions of the components of the cell surface membrane
• the letters of the labels in Fig. 1.1 that identify each component.

Table 1.1

(b) Fig. 1.2 is a drawing of a transmission electron micrograph (TEM) of a cell from the palisade mesophyll of a leaf.
The drawing does not show all of the organelles visible in a transmission electron micrograph.

TEM of Palisade Mesophyll Cell

(i) Complete Fig. 1.2 by drawing and labelling:

• a mitochondrion
• rough endoplasmic reticulum
• smooth endoplasmic reticulum.

Your drawings should show the detail that can be seen in a transmission electron micrograph.

(ii) Identify the organelle labelled X and state one function of this organelle.

▶️ Answer/Explanation
Solution

(a)(i) 7 nm (within the range 5 to 10 nm).

Explanation: The thickness of the cell membrane (phospholipid bilayer) is approximately 7 nm, as measured between points G and H in the diagram.

(a)(ii)

Answer to Table 1.1

Explanation: The table identifies key membrane components: phospholipid bilayer (A, structural barrier), protein (B, transport/signaling), glycoprotein (C, cell recognition), and cholesterol (D, membrane stability).

(b)(i)

Explanation: The drawing should include: (1) a mitochondrion (double membrane with cristae), (2) rough ER (flattened sacs with ribosomes), and (3) smooth ER (tubular, no ribosomes). Labels must match the structures.

(b)(ii) Golgi body; modifies/packages proteins or lipids.

Explanation: Organelle X is the Golgi apparatus, which processes proteins (e.g., glycosylation) and packages them into vesicles for transport.

Question 2

Topic: 8.1 (The circulatory system)

(a) Water is the main component of blood. Explain how the properties of water make it suitable as the main component of blood.

(b) Fig. 2.1 is a diagram of the circulation in a mammal.

(i) Complete Fig. 2.1 by naming P and Q. Write your answers on Fig. 2.1.

(ii) Describe the functions of P and Q.

(iii) Explain why the mammalian circulation is described as a closed, double circulation.

(c) Fig. 2.2 is a transmission electron micrograph of a cross‐section of an arteriole. Blood flows from muscular arteries through arterioles into capillary networks. The lining of the arteriole is folded because the arteriole has constricted. This constriction causes the blood pressure to decrease from 12.7kPa in the muscular artery to 2.7kPa at the end of the arteriole.

(i) Explain why it is important that the pressure of blood decreases as it passes through arterioles.

(ii) Compare the structure of a muscular artery with the structure of the arteriole shown in Fig. 2.2.

▶️ Answer/Explanation
Solution

(a)

Explanation: Water is an excellent solvent, dissolving essential substances like glucose and ions for transport. Its high specific heat capacity helps maintain stable blood temperature, and its polarity aids in dissolving polar molecules, ensuring efficient nutrient and waste transport.

(b)(i) P: Aorta, Q: Vena cava

(ii) Explanation: The aorta (P) carries oxygenated blood at high pressure to the body, while the vena cava (Q) returns deoxygenated blood to the heart. This ensures efficient oxygen delivery and waste removal.

(iii) Explanation: Mammalian circulation is closed (blood flows in vessels) and double (blood passes through the heart twice per circuit—once for pulmonary and once for systemic circulation), enhancing oxygenation efficiency.

(c)(i) Explanation: Blood pressure decreases in arterioles to prevent capillary damage due to their thin walls and to allow slow blood flow for efficient exchange of substances like oxygen and carbon dioxide.

(ii) Explanation: Muscular arteries have a thicker tunica media (more smooth muscle) and a wider lumen compared to arterioles. Arterioles have a higher wall-to-lumen ratio and less elastic tissue, aiding in pressure regulation.

Question 3

Topic: 10.1 (Infectious diseases)

(a) A class of students was studying the features of some human pathogens. One of the students constructed a flow chart to identify four different human pathogens. The student used information about the structure and mode of transmission of each of these pathogens. Fig. 3.1 shows the partially completed flow chart.

Flow Chart of Pathogens

Complete the flow chart in Fig. 3.1 by identifying:
• the modes of transmission
• the scientific names of the pathogens
• the name of one of the diseases.

(b) HIV has a nucleic acid core of RNA. The virus also contains the enzyme reverse transcriptase. After HIV enters T lymphocytes, reverse transcriptase catalyses the formation of DNA using activated DNA nucleotides with the viral RNA as a template. Some drugs, such as tenofovir, have been developed to inhibit the action of reverse transcriptase. The structure of tenofovir is similar to the structure of deoxyribose adenosine monophosphate, as shown in Fig. 3.2.

Structure of Tenofovir

After tenofovir is absorbed into cells it is phosphorylated twice and can be used by reverse transcriptase in the synthesis of DNA. When a tenofovir molecule is added to the DNA strand being synthesised, the process stops. Suggest the mechanism of action of tenofovir to prevent the synthesis of DNA by reverse transcriptase. Use the information in Fig. 3.2 in your answer.

Pre‐exposure prophylaxis (PrEP) is the use of therapeutic drugs to prevent the replication of HIV in the body following infection. The drugs are taken by people who are at risk of becoming infected. Tenofovir is one of these therapeutic drugs. In 2016, the United Nations (UN) set a global target of 3 million PrEP users by 2020. Table 3.1 shows the number of people across the world who received a therapeutic drug for PrEP in each of the years between 2012 and 2019.

Table 3.1

(i) Calculate the percentage of people who received PrEP in 2019 as a percentage of the target set by the UN in 2016. Give your answer to the nearest whole number.

(ii) PrEP does not prevent transmission of HIV. State and explain how health authorities can reduce the transmission of HIV.

▶️ Answer/Explanation
Solution

(a)

Explanation: The flow chart is completed as follows:

  • Mode of transmission: Insect vector (Anopheles mosquito) for Plasmodium (malaria), contaminated water for Vibrio cholerae (cholera), airborne droplets for Mycobacterium tuberculosis (tuberculosis).
  • Scientific names: Plasmodium falciparum, Vibrio cholerae, Mycobacterium tuberculosis.
  • Disease name: Malaria, cholera, tuberculosis.

(b)

Explanation: Tenofovir acts as a competitive inhibitor of reverse transcriptase. Its structure mimics deoxyribose adenosine monophosphate, allowing it to bind to the enzyme’s active site. Once incorporated into the DNA strand, it lacks a 3′-OH group, preventing further elongation and stopping DNA synthesis.

(c)(i)

Calculation: The UN target was 3 million PrEP users by 2020. In 2019, 600,000 people received PrEP. The percentage is calculated as \(\frac{600,000}{3,000,000} \times 100 = 20\%\).

(c)(ii)

Explanation: Health authorities can reduce HIV transmission by:

  • Providing condoms and clean needles to prevent contact with contaminated bodily fluids.
  • Implementing education programs to raise awareness about HIV transmission and prevention.
  • Testing high-risk groups for early diagnosis and treatment to reduce viral load.
  • Screening donated blood to prevent transmission through transfusions.
Question 4

Topic: 9.1 (The gas exchange system)

Fig. 4.1 is a scanning electron micrograph showing the tissue that lines the bronchi in the gas

exchange system.
Fig. 4.2 is a transmission electron micrograph of a horizontal section made at the position indicated
by the two arrows in Fig. 4.1.

Scanning Electron Micrograph of Bronchial Lining
Transmission Electron Micrograph of Cilia

(a) (i) Name the cells labelled A and B in Fig. 4.1.

(ii) Describe how the tissue shown in Fig. 4.1 is adapted to its function in the gas exchange system.

(b) (i) The structures labelled X in Fig. 4.1 have a characteristic internal appearance, as seen in Fig. 4.2. Describe the internal appearance of the structures labelled X.

(ii) Explain how Fig. 4.2 shows that each of the structures labelled X are intracellular.

(c) Stem cells are found in the lining of the bronchi. Describe the function of centrioles and explain how they are involved in the cell cycle of a stem cell.

▶️ Answer/Explanation
Solution

(a)(i) A: Ciliated epithelial cell; B: Goblet cell.

Explanation: In Fig. 4.1, A is identified as a ciliated epithelial cell due to the presence of hair-like cilia, while B is a goblet cell, which secretes mucus to trap particles.

(a)(ii) The tissue is adapted for gas exchange by: (1) Goblet cells secreting mucus to trap pathogens, (2) Cilia moving mucus upward to clear airways, and (3) Maintaining a clean surface for efficient gas exchange.

Explanation: The mucus traps dust and microbes, while cilia sweep the mucus away from the lungs, ensuring unobstructed airflow.

(b)(i) The structures X (cilia) show a 9+2 microtubule arrangement—nine outer doublets and two central single microtubules.

Explanation: This arrangement is characteristic of cilia and flagella, providing structural support and enabling movement.

(b)(ii) Fig. 4.2 confirms X as intracellular because each cilium is enclosed by the cell membrane, indicating it is part of the cytoplasm.

Explanation: The membrane continuity in the micrograph shows the cilia are extensions of the cell, not external structures.

(c) Centrioles organize spindle fibers during cell division. In stem cells, they replicate in the S/G2 phase, move to poles in prophase, and regulate spindle formation for mitosis.

Explanation: Centrioles ensure accurate chromosome segregation by forming the mitotic spindle, critical for stem cell proliferation and tissue repair.

Question 5

Topic: 7.1 (Structure of transport tissues)

The pressure of water vapour inside and outside leaves can be measured. The difference between these pressures is known as the leaf vapour pressure deficit (LVPD). LVPD is one of the factors that influences the rate of transpiration. Scientists measured the effect of changing the LVPD on the rate of transpiration in several species of flowering plant that live in a variety of different habitats. Two of these species were:

  • Nerium oleander, a species that is adapted to grow in hot, dry conditions
  • Helianthus annuus, a species that is not adapted for survival in hot, dry conditions.

Fig. 5.1 shows the effect of increasing the LVPD on the transpiration rates of the two species. All other factors were kept constant. Key:

———— Helianthus annuus

_______ Nerium oleander

Effect of LVPD on Transpiration Rates

(a) Compare the results of the two species shown in Fig. 5.1.

(b) Fig. 5.2 shows part of a plant of N. oleander.

Nerium oleander plant part

Fig. 5.3 shows a cross-section of part of an oleander leaf.

Cross-section of Nerium oleander leaf

Fig. 5.4 is a drawing of a high-power view of region N on Fig. 5.3.

High-power view of region N

State and explain two adaptations shown by the leaves of N. oleander that are visible in Fig. 5.3 and Fig. 5.4.

one adaptation visible in Fig. 5.3 explanation

one adaptation visible in Fig. 5.4 explanation

▶️ Answer/Explanation
Solution

(a)

Comparison of transpiration rates:

  1. Both species show no transpiration when LVPD is 0 kPa.
  2. Transpiration rates increase for both until LVPD reaches 2.5 kPa.
  3. At higher LVPD (2.5–3.0 kPa), H. annuus maintains a constant rate, while N. oleander decreases slightly.
  4. H. annuus consistently has a higher transpiration rate than N. oleander at all LVPD values.

Explanation: N. oleander, being adapted to dry conditions, has mechanisms to reduce water loss, leading to a lower transpiration rate compared to H. annuus.

(b)

Adaptations of N. oleander leaves:

From Fig. 5.3: Thick cuticle and multiple epidermal layers reduce water loss by limiting cuticular transpiration.

Explanation: The thick waxy cuticle acts as a waterproof barrier, while multiple cell layers increase the diffusion distance for water vapour.

From Fig. 5.4: Sunken stomata in pits create a humid microclimate, reducing the water potential gradient.

Explanation: The pits trap moist air, slowing evaporation and minimising water loss under high LVPD conditions.

Question 6

Topic: 11.2 (Antibodies and vaccination)

Antibodies are produced by plasma cells.
Fig. 6.1 shows antigens bound to antigen‐binding sites of an antibody molecule.

(a) (i) Explain how the structure of an antigen‐binding site makes it specific to a particular antigen, as shown in Fig. 6.1.

(ii) State the function of the hinge region of the antibody shown in Fig. 6.1.

(iii) Antibodies can bind to membrane receptors on cells of the immune system, such as macrophages.

Suggest an advantage of antibodies binding to receptors on macrophages.
(b) It is estimated that the immune system of each person can make enough antibodies to bind to over 1012 different antigens.
When plasma cells make antibody molecules they combine the polypeptides produced by the expression of genes for heavy chains and the genes for light chains.
Research has shown that producing this very large number of antibodies is only possible by modifying the primary transcripts of the genes that code for heavy chains and the genes that code for light chains.
Suggest how this modification of the primary transcripts occurs in plasma cells.

▶️ Answer/Explanation
Solution

(a) (i) The antigen-binding site has a complementary shape to the antigen due to its unique amino acid sequence and tertiary structure. This specificity arises from the variable region’s 3D conformation, ensuring only matching antigens bind.

Explanation: The variable region’s unique folding and side chains (R-groups) create a precise fit for the antigen, like a lock-and-key mechanism.

(ii) The hinge region provides flexibility, allowing the antibody to bind antigens at varying angles.

Explanation: This flexibility enhances the antibody’s ability to attach to multiple antigens simultaneously, improving immune efficiency.

(iii) Antibody binding to macrophage receptors facilitates phagocytosis of pathogen-antibody complexes.

Explanation: This process, called opsonization, marks pathogens for destruction, speeding up immune response.

(b) Primary transcripts are modified via alternative splicing, where exons are rearranged or skipped, and introns are removed.

Explanation: This generates diverse antibody variants from a limited set of genes, enabling the immune system to target countless antigens.

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