9700 Biology November 2025 Question Paper 24

1 There are similarities and differences between the structure of a typical plant cell and a typical animal cell.

(a) The nucleus of plant cells and animal cells contains chromosomes. In interphase of the cell cycle, individual chromosomes are present but cannot be seen. The chromosome material is known as chromatin.

(i) Changes occur in interphase, which result in a difference between the chromatin in the G1 phase compared with the chromatin in the G2 phase.

State and explain the difference in chromatin in the G1 phase compared with chromatin in the G2 phase [2]

(ii) Describe the features of a nucleus, other than containing chromatin.

You may use the space below the lines for a diagram [3] , ,

(b) Starch, cellulose and pectins are polysaccharides found in plant cells but not in animal cells.

Pectins are complex polysaccharides that are found in the cell wall.

(i) Describe the structural features of starch that are different from the structural features of cellulose [4]

(ii) Cell wall pectins can vary in different plant cell types and in different stages of cell development. Pectin molecules are released from cells as basic structures and then modified within the cell wall by adding side chains.

RG‑I is a pectin molecule with a variable structure. The basic structure is a repeated disaccharide made from two different monosaccharides. RG‑I has three different side chains that can be added in different positions.

Monoclonal antibodies (mAbs) are used to investigate the structure, location and role of cell wall pectins.

Suggest and explain why scientists need to use a number of different monoclonal antibodies when investigating a pectin such as RG‑I [3] [Total: 12] , ,

2 Cholera, malaria and tuberculosis (TB) are infectious diseases caused by unicellular organisms.

(a) For each of the diseases listed, state whether the disease is caused by a eukaryotic or prokaryotic organism.

cholera malaria TB [1]

(b) State the term given to an organism that causes diseases such as cholera, malaria and TB [1]

(c) Malaria is caused by species from the genus Plasmodium.

Ring cells are one stage in the complex life cycle of Plasmodium that are found within red blood cells. Fig. 2.1 is a scanning electron micrograph showing two ring cells, surrounded by the membrane of a red blood cell, which has just lysed (burst). X Y Fig. 2.1

(i) The actual diameter of the ring cell along the length X–Y is 2 µm.

Calculate the magnification of the image shown in Fig. 2.1.

Give your answer to 3 significant figures.

magnification = × [2] , ,

(ii) Describe how Plasmodium is transmitted from a person with malaria into the blood stream of an uninfected person [3]

(d) In a person with malaria, phagocytes destroy infected red blood cells. The phagocytes respond to the presence of particular molecules in the outer phospholipid bilayer of the cell surface membrane.

The cell surface membrane of a healthy, uninfected red blood cell shows an uneven distribution of types of phospholipid making up the bilayer (membrane asymmetry). For example, most of the phospholipid phosphatidylserine (PS) is located in the inner layer, facing the cytoplasm.

(i) After a red blood cell is infected, a much higher proportion of PS is found in the outer layer of the cell surface membrane, facing blood plasma.

Suggest how the presence of PS in the outer layer can cause a response in a phagocytic cell [2]

(ii) Suggest why some uninfected red blood cells in a person with malaria can also be destroyed by phagocytic cells [1] , ,

(e) Research has shown that Plasmodium uses cholesterol from the cell surface membrane of the red blood cell it has infected.

(i) Suggest why cholesterol is needed by Plasmodium [1]

(ii) The use of cholesterol by Plasmodium causes a decrease in the quantity of cholesterol in the cell surface membrane of the red blood cell.

Outline how a decrease in cholesterol could affect the cell surface membrane of the red blood cell [2]

(f) To achieve membrane asymmetry, which is an essential feature of a healthy cell, the red blood cell needs a supply of ATP and must maintain a very low concentration of calcium ions (Ca2+) within the cytoplasm.

Table 2.1 shows details of three membrane enzymes that are involved in the movement of phospholipids between the inner and outer layers.

Table 2.1

key

PS = phosphatidylserine

PE = phosphatidylethanolamine

PC = phosphatidylcholine enzyme enzyme action flippase hydrolyses ATP and moves PS and PE from the outer to the inner layer floppase hydrolyses ATP and moves PC from the inner to the outer layer scramblase after binding Ca2+, randomly moves PS, PE and PC between layers type of phospholipid , ,

A healthy red blood cell has most PS and PE located in the inner layer and most PC located in the outer layer.

(i) Phospholipids can occasionally move between layers without the action of an enzyme, so the continued activity of flippase and floppase is needed.

With reference to Table 2.1, explain why the action of the enzymes flippase and floppase involves the hydrolysis of ATP [1]

(ii) Blood plasma has a higher concentration of Ca2+ than the cytoplasm of red blood cells.

Suggest one way in which a red blood cell can have a very low concentration of Ca2+ when blood plasma has a higher concentration of Ca2+ [1]

(iii) The concentration of Ca2+ within the red blood cell increases when the cell is infected with the malarial parasite. This leads to the loss of membrane asymmetry.

With reference to Table 2.1, suggest and explain how membrane asymmetry is lost when the concentration of Ca2+ within the red blood cell increases [3] [Total: 18] , ,

3 Small interfering RNA (siRNA) is a short length of double‑stranded RNA (dsRNA), approximately 21 to 25 nucleotides long. siRNA helps to regulate protein synthesis in cells.

(a) Fig. 3.1 is an outline summary of one way in which a primary transcript can be processed to produce a molecule of siRNA. The transcript does not code for a sequence of amino acids. primary transcript section with paired nucleotides loop with unpaired nucleotides 5' 3' Drosha 5' 3' Dicer 5' 3' 5' 3' 3' 5' 21 nucleotides siRNA step 1: a primary transcript is synthesised from one strand of DNA step 2: the primary transcript loops and a section of the RNA becomes double stranded step 3: the enzyme Drosha attaches and cleaves (cuts) the RNA to produce a shorter length step 4: in the cytoplasm, the enzyme Dicer attaches and cleaves the RNA step 5: siRNA, which is double stranded and has 2 unpaired nucleotides at each 3' end, is produced

Fig. 3.1 , ,

(i) In step 2 in Fig. 3.1, the single‑stranded primary transcript loops and forms a section where dsRNA is present.

Explain how it is possible for the double‑stranded section of RNA to be held together in step 2 and maintained this way in steps 3, 4 and 5 [2]

(ii) After step 3, the shorter dsRNA produced by the action of Drosha is transported to the cytoplasm, where it is cleaved further by Dicer to produce ds siRNA.

Suggest why two different enzymes, Drosha and Dicer, are needed to cut dsRNA into shorter lengths [2]

(b) From 2018, siRNA has been used as a therapeutic drug to treat a number of diseases.

The presence of molecules of siRNA in the cytoplasm can result in the cleavage of messenger RNA (mRNA) molecules coding for a protein involved in the disease. This prevents the synthesis of the protein.

Describe the differences between a molecule of mRNA and a molecule of siRNA, such as the siRNA shown in step 5 in Fig. 3.1 [2] , ,

(c) Fig. 3.2 summarises how mRNA coding for a protein involved in disease can be targeted and cleaved by siRNA. passenger strand (not required) siRNA guide strand (required) guide strand AGO2 proteins presence of siRNA activates a protein, AGO2, which binds siRNA and other proteins to form a complex known as RISC RISC uses guide strand to target and bind mRNA for cleavage passenger strand is separated from RISC RISC passenger strand released into cytoplasm site of cleavage mRNA

Fig. 3.2 , ,

(i) With reference to Fig. 3.2, state why the passenger strand needs to be separated and released from the RISC [1]

(ii) The aim of siRNA therapy is to prevent or decrease the synthesis of a protein involved in the disease being treated.

A target mRNA molecule can be cleaved in a different location by a RISC with a different siRNA.

Suggest how cleaving mRNA in different locations will have different effects on protein synthesis and explain how these different effects can result in a lack of functioning protein [3] [Total: 10] , ,

4 In mammals, the gas exchange system includes a set of branching airways that carry air to and from the gas exchange surface.

Air from the external atmosphere passes through different types of airway to reach the gas exchange surface.

(a) Name the type of airway of the gas exchange system that branches into airways known as bronchioles [1]

(b) Fig. 4.1 is a photomicrograph of a section through a bronchiole. T

Fig. 4.1 , ,

(i) The bronchiole shown in Fig. 4.1 is not part of the gas exchange surface.

Name one structure in the gas exchange system, visible in Fig. 4.1, where gas exchange is carried out [1]

(ii) Name the type of cell found in the tissue labelled T in Fig. 4.1 [1]

(iii) State the features that help to identify the type of airway shown in Fig. 4.1 as a bronchiole and not the other types of airway present in the gas exchange system [3]

(c) Blood is pumped to the lungs in the pulmonary circulation.

The lungs also receive a supply of blood from the systemic circulation.

Explain why the pulmonary circulation and the systemic circulation need to supply blood to the lungs [2] [Total: 8] , ,

5 Veins transport blood towards the heart. The structure of a vein is adapted to its function.

(a) The inner layer of a vein is the tunica intima, composed of a single layer of endothelial cells that form a protective barrier.

When the tunica intima of a blood vessel is damaged, endothelial cells can carry out mitosis to allow tissue repair to occur.

(i) The spindle that is formed during mitosis is composed of spindle fibres.

Name the cell structures that are organised to form spindle fibres during mitosis [1]

(ii) Describe the telophase stage of mitosis [2]

(b) Explain how the structure of a vein is related to its function.

You do not need to include details of the structure of the tunica intima and its function [3] [Total: 6] , ,

6 (a) State and explain why the same leaf of a plant can be described as a source or as a sink, depending on the stage of maturity (age) of the leaf [2]

(b) Fig. 6.1 lists seven types of plant cell found in leaves. 1 epidermal cell 2 guard cell 3 palisade mesophyll cell 4 phloem sieve tube element 5 sclerenchyma cell 6 spongy mesophyll cell 7 xylem vessel element Fig. 6.1

Match the correct type of cell from the list in Fig. 6.1 with each statement, A to E.

Each cell type can be used once, more than once, or not at all.

The first match has been done for you.

A This is a thick‑walled cell that provides support B This cell is one of a pair of cells that form a stoma C This cell receives water to build up hydrostatic pressure for mass flow D This cell needs water for photosynthesis and is columnar‑shaped E This cell secretes a waxy substance to help prevent water loss [4] [Total: 6] 5 , ,