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Question 1

Topic: 4.1 (Fluid mosaic membrane)

(a) Fig. 1.1 is a diagram representing part of the phospholipid bilayer of a cell surface membrane.

Phospholipid Bilayer Diagram

(i) Identify the part of a phospholipid molecule, labelled A in Fig. 1.1, that forms bonds with the phosphate heads and with the fatty acid tails.

(ii) Cholesterol is an important lipid component of many cell surface membranes. Fig. 1.2 shows the structure of a cholesterol molecule.

Cholesterol Molecule Structure

Using the information in Fig. 1.2, explain the orientation (positioning) of cholesterol molecules in the phospholipid bilayer, as shown in Fig. 1.1.

(iii) State one role of cholesterol in phospholipid bilayers.

(b) (i) Explain why sodium ions cannot cross phospholipid bilayers by simple diffusion.
(ii) Ions and some molecules move across cell surface membranes by facilitated diffusion and active transport.

Compare facilitated diffusion and active transport by stating one way in which they are similar and two ways in which facilitated diffusion is different from active transport.

(c) Prostaglandins are small lipids produced in many tissues of the body. One role of prostaglandins is to cause inflammation at the site of an injury or infection. Inflammation is the normal first response of the immune system to injury or infection.
Cyclooxygenase (COX) is an enzyme that catalyses one of the steps in the reaction pathway for the formation of prostaglandins from phospholipids. The reaction pathway occurs in the smooth endoplasmic reticulum (SER) of cells. Part of the reaction pathway is shown in Fig. 1.3.

Prostaglandin Reaction Pathway

(i) Suggest an advantage for this reaction pathway occurring in the smooth endoplasmic reticulum of a cell rather than in the cytoplasm.
(ii) Sometimes inflammation can have side-effects, such as pain. Aspirin is a drug that can be used to reduce these side-effects.

Aspirin reduces the catalytic activity of the COX enzyme by modifying the R-group of one of the amino acids.

Suggest how modifying the R-group of an amino acid in the COX enzyme can reduce the catalytic activity of the enzyme.

(iii) Prostaglandins are examples of cell-signalling molecules.
Outline the process of cell signalling that leads to a response by the cells involved in inflammation.

▶️ Answer/Explanation
Solution

(a)(i) Glycerol

Explanation: The part labelled A is the glycerol backbone, which forms ester bonds with both the phosphate heads (hydrophilic) and fatty acid tails (hydrophobic).

(a)(ii) The hydroxyl (polar) group of cholesterol interacts with the phosphate heads (hydrophilic region), while its non-polar hydrocarbon ring aligns with the fatty acid tails (hydrophobic region).

Explanation: Cholesterol’s amphipathic nature allows it to embed in the membrane, with its polar group facing outward and non-polar region inward.

(a)(iii) Regulates membrane fluidity.

Explanation: Cholesterol stabilizes the membrane by reducing fluidity at high temperatures and preventing solidification at low temperatures.

(b)(i) Sodium ions are charged and cannot pass through the hydrophobic core of the bilayer.

Explanation: The non-polar fatty acid tails repel charged particles like Na+, requiring transport proteins for movement.

(b)(ii)

Comparison Table

Explanation: Both processes use membrane proteins, but facilitated diffusion is passive (no energy) and follows the concentration gradient, while active transport requires ATP and moves against the gradient.

(c)(i) Compartmentalization in the SER provides a lipid-rich environment for prostaglandin synthesis.

Explanation: The SER’s membrane-bound structure ensures proximity to phospholipid substrates and isolates the pathway from cytoplasmic reactions.

(c)(ii) Modifying the R-group alters the enzyme’s active site shape, reducing substrate binding efficiency.

Explanation: Aspirin’s modification disrupts ionic/hydrogen bonds, making the active site less complementary to arachidonic acid.

(c)(iii) Prostaglandins bind to target cell receptors, triggering signal transduction (e.g., secondary messengers) that induces inflammation.

Explanation: This paracrine signaling involves ligand-receptor binding and intracellular cascades to elicit a physiological response.

Question 2

(a) Topic: 8.1 (The circulatory system)

(a) Table 2.1 shows descriptions of three types of white blood cell.
Complete Table 2.1 by stating the names of these three types of white blood cell.

Table 2.1

(b) Dromedary camels are classified in the family Camelidae and live in desert habitats of North Africa and Asia. In these hot, dry environments, dromedary camels can lose up to 30% of their body mass from dehydration, causing their blood to become more viscous (thicker). Fig. 2.1 shows a drawing of red blood cells of a dromedary camel. Fig. 2.2 is a drawing of human red blood cells.

Fig. 2.1 and Fig. 2.2

Fig. 2.1 and Fig. 2.2 show differences between the red blood cells of dromedary camels and the red blood cells of humans. Suggest how these differences adapt dromedary camels for living in hot, dry environments.

(c) The llama is also classified in the family Camelidae. Llamas live in mountainous areas of South America, often at altitudes of 3500m or higher. As the altitude above sea level increases, the air pressure decreases. The partial pressure of oxygen in the lungs of mammals at 3500m is 6.4kPa. Fig. 2.3 shows the oxygen dissociation curve of adult human haemoglobin and adult llama haemoglobin.

Fig. 2.3

(i) With reference to Fig. 2.3, explain how the differences between the oxygen dissociation curves for humans and llamas show that llamas are better adapted for living at high altitudes than humans.

(ii) Sketch a curve on Fig. 2.3 to show the effect of an increased carbon dioxide concentration on the percentage saturation of adult human haemoglobin with oxygen.

(iii) Explain the importance of the Bohr shift in metabolically active organs, such as the liver.

▶️ Answer/Explanation
Solution

(a)

Answer to Table 2.1

Explanation: The three types of white blood cells are lymphocytes (produce antibodies), phagocytes (engulf pathogens), and platelets (involved in blood clotting). These are identified based on their functions described in the table.

(b) The elliptical shape and smaller size of camel red blood cells allow easier flow of thicker blood during dehydration. This adaptation ensures efficient circulation despite high viscosity.

(c)(i) Llama haemoglobin has a higher oxygen affinity (86% saturation at 6.4 kPa) compared to humans (79%), enabling better oxygen uptake in low-pressure high-altitude environments.

(c)(ii) A right-shifted curve (Bohr effect) would be drawn, showing reduced oxygen affinity at higher CO2 levels.

(c)(iii) The Bohr shift releases more oxygen in active tissues (e.g., liver) where CO2 levels are high, supporting increased aerobic respiration and ATP production.

Question 3

(a) Topic: 7.1 (Structure of transport tissues)

(a) Fig. 3.1 is a photomicrograph showing part of a transverse section through the root of an iris, Iris germanica. Irises are herbaceous monocotyledons. These plants have the same transport tissues as herbaceous dicotyledons, but the transport tissues are distributed differently. In monocotyledons, the central tissue in the root is parenchyma (packing tissue).

Transverse section of iris root

(i) Cells R, S and T in Fig. 3.1 are found in different tissues. Name the tissues in which the cells labelled R, S and T are found.
(ii) Outline the role of the tissue in which cell R is found in Fig. 3.1.
(iii) State an example of an organic compound that is translocated in the root of an iris.

(b) The electron micrograph in Fig. 3.2 shows a section through some root cells in an onion, Allium cepa.

Electron micrograph of onion root cells

On Fig. 3.2, draw a label line and label it with the letter P to identify one plasmodesma.

(c) Table 3.1 contains information about four polysaccharides found in animals or plants. Complete Table 3.1 by filling in the missing information.

Table 3.1
▶️ Answer/Explanation
Solution

(a)(i)

R: endodermis (endodermal tissue)

S: xylem

T: phloem

Explanation: Cell R is part of the endodermis, which regulates water movement. S is in the xylem (transports water and minerals), and T is in the phloem (transports organic nutrients).

(a)(ii)

The endodermis (where R is found) controls water entry into the vascular tissue via the Casparian strip, ensuring selective absorption.

Explanation: The endodermis forces water to pass through cell membranes (symplastic pathway), allowing regulation of mineral uptake.

(a)(iii)

Sucrose (or amino acids/proteins).

Explanation: Organic compounds like sucrose are translocated in the phloem for energy distribution in plants.

(b)

A correct label “P” should mark a plasmodesma (microscopic channel between plant cells).

Explanation: Plasmodesmata facilitate intercellular communication and transport.

(c)

Completed Table 3.1

Explanation: The table is completed with: Cellulose (role: structural support in cell walls), Glycogen (storage in liver/muscles), and Starch (energy storage in plants).

Question 4

(a) Topic: 6.2 (Protein synthesis)

(a) Table 4.1 shows a sequence of 12 nucleotides in the template strand of a short length of a DNA molecule, the corresponding primary transcript and the four amino acids coded for by the sequence. The table is incomplete.

(i) Complete Table 4.1 to show the sequence of nucleotides in the primary transcript that would result from transcription of this short length of DNA.

Table 4.1

(ii) Table 4.2 shows all the possible template strand DNA triplets that code for the amino acids labelled aa1, aa2, aa3 and aa4 in Table 4.1.

Table 4.2

Complete Table 4.3 to identify the four amino acids labelled AA1, AA2, AA3, and AA4 in Table 4.1.

Table 4.3

(iii) One type of gene mutation is caused by the substitution of a DNA nucleotide. Using the information in Table 4.2, state and explain the effect on the final protein structure of a substitution of the nucleotide at position 3 in Table 4.1.

(iv) A second type of gene mutation is caused by the deletion of a DNA nucleotide. Using the information in Table 4.2, state and explain the effect on the final protein structure of a deletion of the nucleotide at position 3 in Table 4.1.

(b) Replication of nuclear DNA occurs just once in every mitotic cell cycle. Six named events associated with the mitotic cell cycle are listed. The events are not listed in any particular order. Draw a circle around each event where replication of nuclear DNA occurs.

Mitotic Cell Cycle Events

(c) Outline how DNA is replicated inside the nucleus.

(d) Fig. 4.1 shows the structure of an ATP molecule.

ATP Molecule Structure

State the name of the part of the ATP molecule labelled A in Fig. 4.1.

▶️ Answer/Explanation
Solution

(a)(i)

Answer to Table 4.1

Explanation: The primary transcript is formed by transcribing the DNA template strand, replacing thymine (T) with uracil (U). The sequence is complementary to the template strand.

(a)(ii)

Answer to Table 4.3

Explanation: The amino acids are identified using Table 4.2. AA1 is valine (CAA), AA2 is leucine (GAA), AA3 is aspartate (CTA), and AA4 is glutamate (CTT).

(a)(iii) No effect on the protein structure because all four triplets beginning with CA code for valine (CAA, CAG, CAT, CAC). This demonstrates the redundancy of the genetic code.

(a)(iv) The deletion causes a frameshift mutation, altering all subsequent codons. The first amino acid (valine) remains unchanged, but the rest of the sequence is disrupted, potentially leading to a non-functional protein.

(b) Replication occurs during interphase and specifically the S phase of the cell cycle.

(c) DNA replication involves unwinding the double helix (via helicase), using both strands as templates, adding complementary nucleotides (via DNA polymerase), and joining fragments (via ligase). The process is semi-conservative.

(d) The part labelled A is ribose, a pentose sugar in ATP.

Question 5

(a) Topic: 10.1 (Infectious diseases)

The pathogen that causes cholera is a prokaryote.
(a) Fig. 5.1 shows an electron micrograph of the pathogen that causes cholera.

(i) Name the type of electron microscope used to produce the image shown in Fig. 5.1.

(ii) Name the species of prokaryote that causes cholera.

(b) The passage contains a description of the main features of prokaryotic cells. There is one
factual error in the passage.

Prokaryotic cells are unicellular and generally between 1μm and 5μm in diameter.
Prokaryotes do not have organelles surrounded by double membranes. They do
have cell surface membranes, 70S ribosomes and a cellulose cell wall. The DNA of a
prokaryotic cell is circular and is found free in the cytoplasm rather than enclosed in a
nuclear envelope.

Identify and correct the factual error in the passage.

▶️ Answer/Explanation
Solution

(a)(i) Scanning electron microscope.

Explanation: The image shows a 3D surface view of the pathogen, which is characteristic of a scanning electron microscope (SEM). SEMs are used for detailed surface imaging.

(a)(ii) Vibrio cholerae.

Explanation: The causative agent of cholera is the bacterium Vibrio cholerae, a well-known prokaryotic pathogen.

(b) The error is “a cellulose cell wall.” The correction is “a peptidoglycan/murein cell wall.”

Explanation: Prokaryotic cell walls are composed of peptidoglycan (murein), not cellulose. Cellulose is found in plant cell walls, not bacteria.

Question 6

(a) Topic: 10.1 (Infectious diseases)

Fig. 6.1 is a simplified diagram representing a section through the human immunodeficiency virus (HIV) particle that causes HIV/AIDS. The diagram shows the virus particle about to attach to the cell surface membrane of a T-helper cell at a receptor protein called CD4. A second protein (coreceptor) called CCR5 is also necessary for the virus particle to enter and then infect the T-helper cell.

(a) Identify structure X in Fig. 6.1.
(b) Explain how the ability of the immune system to resist the damaging effects of a pathogen is affected by destruction of T-helper cells.
(c) Studies have shown that some individuals did not become infected with HIV even though they were repeatedly exposed to the virus. Later discoveries indicated that these individuals had a mutation in the gene for the CCR5 coreceptor protein. Suggest how mutation of the gene for the CCR5 coreceptor protein provided protection against HIV infection.
(d) The use of monoclonal antibodies against the CCR5 coreceptor protein (anti-CCR5) has been shown to be effective in the treatment of HIV infection. Outline how anti-CCR5 monoclonal antibodies can be synthesised in the laboratory using the hybridoma method.

▶️ Answer/Explanation
Solution

(a) capsid

Explanation: Structure X is the capsid, which is the protein coat surrounding the viral genetic material. It protects the viral RNA and plays a key role in the infection process.

(b) The destruction of T-helper cells weakens the immune response because:

  • Fewer cytokines are released, reducing immune cell activation.
  • Fewer plasma cells are produced, leading to reduced antibody production.
  • Fewer T-killer cells are stimulated, decreasing the ability to kill infected cells.

Explanation: T-helper cells coordinate the immune response, so their destruction impairs both humoral and cell-mediated immunity.

(c) The mutation in the CCR5 gene prevents HIV from entering T-helper cells.

Explanation: The mutated CCR5 protein does not function as a coreceptor, blocking HIV’s ability to trigger endocytosis and infect the cell.

(d) Steps to synthesize anti-CCR5 monoclonal antibodies:

  1. Inject a mammal (e.g., mouse) with CCR5 to trigger an immune response.
  2. Extract plasma cells from the spleen.
  3. Fuse plasma cells with myeloma cells to form hybridomas.
  4. Screen hybridomas for those producing the desired antibody.

Explanation: The hybridoma method ensures mass production of identical antibodies targeting CCR5, which can block HIV entry.

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