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Question 1

Topic: 1.1

Fig. 1.1 is a transmission electron micrograph of a cell from the stem of sago pondweed, Stuckenia pectinata.

(a) (i) State the evidence from Fig. 1.1 that shows that the cell is from the stem of S. pectinata and not from the mesophyll of a leaf.
(ii) Complete each row in Table 1.1 to identify a cell structure shown in Fig. 1.1 that carries out the function listed.

(b) Plant vacuoles develop when vesicles fuse together. The vacuoles increase in size as more vesicles fuse.
Fig. 1.2 shows the movement of vesicles within a plant cell during the development of a vacuole.

(i) Name the process that is occurring at X.

(ii) Some of the vesicles formed by the Golgi body pass to the vacuole. These vesicles contain proteins that have been folded correctly and some that have not folded into their correct shapes. The proteins that have not folded correctly pass to the vacuole where they are broken down.

Explain how proteins that have not folded correctly are broken down in the vacuole.

(c) Small vacuoles in S. pectinata may have roles similar to lysosomes in animal cells.
Describe the role of lysosomes in animal cells in defence against pathogens.

▶️ Answer/Explanation

Solution

(a)(i)

Evidence: The cell lacks chloroplasts (which are present in mesophyll cells for photosynthesis) and contains many small vacuoles instead of a single large central vacuole. Additionally, the nucleus is centrally located, unlike in mesophyll cells where it is often pushed to the periphery.

(a)(ii)
Answer to Table 1.1

Explanation: The table is completed by identifying the correct cell structures: mitochondrion (ATP production), Golgi body (modification of proteins), rough endoplasmic reticulum (protein synthesis), and nucleus (DNA replication).

(b)(i) Endocytosis / Pinocytosis.

Explanation: The process at X is endocytosis, where the cell membrane engulfs extracellular material to form vesicles.

(b)(ii)

Explanation: Misfolded proteins are hydrolyzed in the vacuole by proteases, which break peptide bonds to form smaller peptides or amino acids. This process requires water and occurs in an acidic environment.

(c)

Explanation: Lysosomes defend against pathogens by fusing with phagocytic vesicles (forming phagolysosomes) and releasing hydrolytic enzymes (e.g., lysozyme, proteases) to digest the pathogen’s components (e.g., peptidoglycans, proteins). The breakdown products are harmless or reused by the cell.

Question 2

Topic: 2.1

Glycogen and cellulose are polymers.
Fig. 2.1 shows small, representative regions of a glycogen molecule and a cellulose molecule.

(a) Describe three ways, visible in Fig. 2.1, in which the molecule of glycogen differs from the molecule of cellulose.
(b) Glycogen is found in the form of granules in mammalian liver and muscle cells. Fig. 2.2 is a diagram of part of a molecule of glycogen isolated from a glycogen granule.

Explain how the structure of glycogen is related to its function in cells.
(c) Explain how the arrangement of cellulose molecules in plant cell walls is related to their function.

▶️ Answer/Explanation

Solution

(a)

1. Glycogen is branched, whereas cellulose is unbranched.

2. Glycogen consists of α-glucose monomers, whereas cellulose consists of β-glucose monomers.

3. Glycogen has 1,4- and 1,6-glycosidic bonds, whereas cellulose only has 1,4-glycosidic bonds.

Explanation: The differences are visible in Fig. 2.1. Glycogen’s branched structure allows for rapid glucose release, while cellulose’s linear structure provides rigidity. The type of glucose monomers and glycosidic bonds also differ, affecting their biological roles.

(b)

1. Glycogen serves as an energy store, releasing glucose when needed.

2. Its highly branched structure allows for quick glucose addition/removal.

3. It is compact and insoluble, preventing osmotic imbalance in cells.

Explanation: Glycogen’s structure is optimized for storage—branched for accessibility, compact for space efficiency, and insoluble to avoid disrupting cellular water potential.

(c)

1. Cellulose molecules are unbranched and form straight chains.

2. They align in parallel, forming hydrogen bonds for strength.

3. Microfibrils provide structural support to plant cell walls.

Explanation: Cellulose’s linear arrangement and hydrogen bonding create microfibrils, which resist turgor pressure and maintain cell shape, crucial for plant rigidity.

Question 3

Topic: 7.2

Fig. 3.1 shows some fruits (grapes) of the grapevine, Vitis vinifera.

Sucrose is transported in the phloem of the grapevine to the fruits. In the fruits, sucrose is hydrolysed by the enzyme sucrase, which is found in cell walls. The glucose and fructose produced by the hydrolysis of sucrose pass through membrane proteins, known as hexose transporters, into the cytoplasm of the fruit cells.

(a) State why membrane proteins are required for the movement of molecules, such as glucose, across cell surface membranes into cells.

Researchers investigated one type of hexose transporter, known as VvHT1, which is found in the fruit cells of V. vinifera. They used a mutant strain of yeast that has very few hexose transporters in its cell surface membranes to investigate the properties of VvHT1. The researchers inserted molecules of VvHT1 into the cell surface membranes of the mutant strain of yeast.

Equal volumes of mutant yeast cells with VvHT1 were kept in eight different concentrations of glucose solution.
The rate of uptake of glucose by the yeast cells in each solution was determined.
All the solutions were kept at the same temperature and pH.

The results are shown in Fig. 3.2.

(b) (i) The researchers concluded that VvHT1 is responsible for the facilitated diffusion of glucose into the cells. Explain how the results in Fig. 3.2 provide evidence to support this conclusion.

(b) (ii) The researchers thought that grapevines could be modified to have more hexose transporters to increase the size and quality of grapes. Explain why increasing the number of hexose transporters could be commercially important to growers of grapevines.

(c) Fig. 3.3 is a diagram of a protein in the cell surface membrane of a macrophage from a mouse. Macrophages use these proteins in antigen presentation. Non-self antigens bind to the proteins and are involved in the activation of specific T-lymphocytes during the immune response.

(c) (i) State what is meant by a non-self antigen.

(c) (ii) Some pathogens enter human cells. Macrophages partially digest these pathogens and present antigens to T-lymphocytes during immune responses. With reference to Fig. 3.3, explain how T-lymphocytes respond to infection by a specific type of pathogen.

▶️ Answer/Explanation

Solution

(a) Membrane proteins are required because glucose is polar/water-soluble and cannot pass through the hydrophobic phospholipid bilayer of the cell membrane. These proteins facilitate diffusion or active transport.

Explanation: The hydrophobic core of the membrane repels polar molecules like glucose, so transporter proteins are necessary for their movement.

(b)(i) The graph shows the rate of glucose uptake plateaus at higher concentrations (>150 μmol dm^–3), indicating that VvHT1 transporters are saturated and working at maximum capacity. This is characteristic of facilitated diffusion.

Explanation: The plateau suggests that the number of transporters (VvHT1) is the limiting factor, consistent with protein-mediated transport.

(b)(ii) Increasing hexose transporters could make grapes sweeter (more sugar uptake), improve water absorption via osmosis, enhance energy availability for growth, and increase yield/profit for growers.

Explanation: More transporters allow faster glucose accumulation, directly improving fruit quality and commercial value.

(c)(i) A non-self antigen is a foreign molecule (e.g., protein/polysaccharide) that triggers an immune response by binding to antibodies or activating lymphocytes.

Explanation: These antigens are recognized as “not belonging” to the host organism.

(c)(ii) T-lymphocytes with complementary receptors bind to the presented antigen, undergo clonal expansion, and differentiate into helper T-cells (secreting cytokines) or killer T-cells (destroying infected cells). Memory T-cells are also produced for future immunity.

Explanation: The response is highly specific to the pathogen’s antigens, ensuring targeted immunity.

Question 4

Topic: 6.1

The gene for the enzyme catalase is on chromosome 11 in humans.
(a) Explain the meaning of the term gene.
(b) Two enzymes, DNA polymerase and DNA ligase, are involved in the replication of DNA.
Fig. 4.1 shows the replication of part of human chromosome 11 by DNA polymerase. The arrows show the direction of synthesis of the new polynucleotides by DNA polymerase.

(i) Describe the roles of DNA polymerase and DNA ligase in the replication of DNA.
(ii) State the name of the stage of interphase in the cell cycle when DNA replication occurs.

(i) State the names and functions of structures A and B.
(ii) Complete Fig. 4.3 to show what happens to chromosome 11 in anaphase, so that the daughter nuclei are genetically identical.

▶️ Answer/Explanation

Solution

(a) A gene is a sequence of nucleotides (bases) that forms part of DNA and codes for a polypeptide (protein/enzyme).

Explanation: A gene is a functional unit of DNA that carries genetic information. It consists of a specific sequence of nucleotides that determines the sequence of amino acids in a protein.

(b)(i) DNA polymerase adds complementary nucleotides to the growing DNA strand, forming phosphodiester bonds and proofreading for errors. DNA ligase joins Okazaki fragments on the lagging strand by forming phosphodiester bonds.

Explanation: DNA polymerase synthesizes new DNA strands in the 5’→3′ direction, ensuring accuracy. DNA ligase seals gaps between Okazaki fragments, completing the lagging strand.

(b)(ii) S phase (synthesis phase).

Explanation: DNA replication occurs during the S phase of interphase, where the cell’s genome is duplicated before mitosis.

(c)(i) A – Centromere: Attaches chromatids to spindle fibres and holds sister chromatids together.
B – Spindle fibres: Align chromosomes at the equator and separate chromatids during anaphase.

Explanation: The centromere is crucial for chromosome movement, while spindle fibres ensure proper segregation of chromatids to daughter cells.

(c)(ii) In anaphase, sister chromatids separate and are pulled toward opposite poles by spindle fibres, forming V-shaped chromosomes.

Explanation: The centromere splits, allowing single chromatids to move to opposite poles, ensuring each daughter nucleus receives identical genetic material.

Question 5

Topic: 8.1

Fig. 5.1 is a longitudinal section of a capillary in muscle tissue as viewed with a transmission electron microscope.

(a) State the evidence, visible in Fig. 5.1, that identifies the cells inside the capillary as red blood cells.
(b) Explain how the structure of the capillary wall is related to its functions.

(c) Fig. 5.2 is a diagram showing some of the events that occur as blood flows through a capillary in a respiring tissue.

(i) An increase in respiration results in an increase in the carbon dioxide concentration in the blood and the release of more oxygen from red blood cells to tissues.

Explain how an increase in carbon dioxide in the blood leads to the release of more oxygen from red blood cells.

(ii) Chloride ions are a constituent of blood plasma. The concentration of chloride ions in the plasma of deoxygenated blood is between 2–4mmoldm–3 lower than in the plasma of oxygenated blood.

Explain why the concentration of chloride ions in the blood plasma of deoxygenated blood is lower than in the plasma of oxygenated blood.

▶️ Answer/Explanation

Solution

(a) any two from:
no nucleus ;
no organelles / uniform appearance / homogenous cytoplasm ;
(some are) biconcave shape / described ; A cells have different shapes
same, width / size, as the lumen of the capillary ;
A same width as capillary / diameter of cells is 6 – 7 μm

Explanation: Red blood cells (RBCs) are identified by their lack of a nucleus and organelles, uniform appearance, and biconcave shape. Their size matches the capillary lumen, confirming their identity.

(b) any three from:
1 (wall is) thin / one cell thick / 1–2 μm in thickness ;
2 ref. to, endothelial cells / endothelium ; I squamous / epithelium
3 short distance for diffusion ;
4 endothelial pores / fenestrations
5 for passage of, (named) small molecules / AW, from / to, plasma / blood / tissue fluid ;
6 AVP ; ref. to pinocytosis across endothelial cells

Explanation: The capillary wall is thin (one cell thick) to allow efficient diffusion. Endothelial cells with pores enable small molecule exchange, while the short diffusion distance ensures rapid nutrient and gas exchange.

(c)(i) any four from:
1 more carbon dioxide diffuses into red blood cells ;
2 more carbonic acid is formed by carbonic anhydrase ;
3 formation of more hydrogen ions ; A H+
4 haemoglobin has a high affinity for hydrogen ions ; A H+
5 haemoglobin binds more hydrogen ions to form, haemoglobinic acid (HHb) ; AH+
6 (formation of HHb) decreases affinity of haemoglobin for oxygen ;
7 haemoglobin releases more oxygen ;
8 carbon dioxide binds to -NH2 / N terminal, of, globin / polypeptides / α chains and β chains ;
9 forms carbaminohaemoglobin ; R carboxyhaemoglobin
10 allosteric effect / change in tertiary structure / AW, in (oxy)haemoglobin (causes, release / AW, more oxygen) ;

Explanation: Increased CO₂ lowers blood pH, forming more H⁺ ions. Haemoglobin binds H⁺, reducing its oxygen affinity (Bohr effect), releasing more oxygen to tissues.

(c)(ii) any two from:
1 hydrogencarbonate ions / HCO_{3 –} , pass out of red blood cells (into the plasma to increase concentration in deoxygenated blood) ;
2 chloride ions pass into red blood cells ;
3 to replace the, negatively-charged ions / anions / HCO_3-;
4 chloride shift ;
5 AVP ; chloride ions pass through, channel proteins / anion exchangers / transport proteins / by facilitated diffusion

Explanation: Chloride ions (Cl^–) move into RBCs to balance the loss of HCO₃^– (chloride shift). This maintains electroneutrality, reducing Cl^– concentration in deoxygenated blood plasma.

Question 6

Topic: 7.2

Fig. 6.1 is a diagram showing the passage of water through the tissues of a flowering plant from the soil to the atmosphere. The arrows show the direction of water movement.

(a) The structure labelled X is part of the symplast pathway. State the name of structure X.
(b) The structure labelled Y in the cell wall is a barrier to the apoplast pathway. State the name of structure Y.
(c) With reference to Fig. 6.1, complete the statements about the movement of water in the flowering plant.
Water moves from the soil solution to the cytoplasm of root hair cells by …………………….
Water moves from the xylem in the root to the leaf by …………………………………
Water moves from mesophyll cell walls to intercellular air spaces by …………………………..
Water vapour moves from intercellular air spaces to the atmosphere outside the leaf by ………………………….

▶️ Answer/Explanation

Solution

(a) plasmodesma

Explanation: Structure X is a plasmodesma, which connects plant cells and allows cytoplasmic (symplast) movement of water and nutrients.

(b) Casparian strip

Explanation: Structure Y is the Casparian strip, a waterproof layer made of suberin in the endodermis that blocks the apoplast pathway, forcing water into the symplast.

(c)

Water moves from the soil solution to the cytoplasm of root hair cells by osmosis.
Water moves from the xylem in the root to the leaf by transpiration pull (or cohesion-tension).
Water moves from mesophyll cell walls to intercellular air spaces by evaporation.
Water vapour moves from intercellular air spaces to the atmosphere outside the leaf by diffusion.

Explanation: Osmosis drives water uptake in roots, transpiration pull moves water upward, evaporation occurs in leaves, and diffusion releases water vapour into the air.

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