An original Thinka practice paper modelled on the structure and difficulty of the Jun 2023 (V2) Cambridge International A Level Biology (9700) paper. Not affiliated with or reproduced from Cambridge.
Paper 12
There are forty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C and D.
40 PastPaper.question · 40 PastPaper.marks
PastPaper.question 1 · multipleChoice
1 PastPaper.marks
A student calibrates an eyepiece graticule using a stage micrometer. The stage micrometer has divisions spaced at \(0.1\text{ mm}\). At a magnification of \(\times 100\), 20 divisions of the eyepiece graticule coincide exactly with 4 divisions of the stage micrometer. What is the value of 1 eyepiece graticule unit under this magnification?
A.\(2\,\mu\text{m}\)
B.\(20\,\mu\text{m}\)
C.\(50\,\mu\text{m}\)
D.\(200\,\mu\text{m}\)
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PastPaper.workedSolution
The stage micrometer divisions are spaced at \(0.1\text{ mm}\). Therefore, 4 divisions of the stage micrometer equal \(4 \times 0.1\text{ mm} = 0.4\text{ mm}\). Since \(1\text{ mm} = 1000\,\mu\text{m}\), \(0.4\text{ mm} = 400\,\mu\text{m}\). These \(400\,\mu\text{m}\) correspond exactly to 20 eyepiece graticule units. To find the value of 1 eyepiece graticule unit, divide \(400\,\mu\text{m}\) by 20, which equals \(20\,\mu\text{m}\).
PastPaper.markingScheme
Award 1 mark for the correct answer B. Deduce the total distance of the stage micrometer divisions in micrometers and divide by the number of eyepiece units. Reject other options due to incorrect unit conversions or division errors.
PastPaper.question 2 · multipleChoice
1 PastPaper.marks
A liquid sample extracted from a plant tissue is subjected to several chemical tests: 1. Adding Biuret reagent results in a blue solution. 2. Heating with Benedict's solution directly results in a blue solution. 3. Boiling with dilute hydrochloric acid, followed by neutralization with sodium hydrogencarbonate, and then heating with Benedict's solution results in a brick-red precipitate. 4. Adding iodine in potassium iodide solution results in a yellow-brown solution. Which biological molecules are present in this plant sample?
A.starch and reducing sugar
B.protein and non-reducing sugar
C.non-reducing sugar only
D.protein and reducing sugar
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PastPaper.workedSolution
The negative Biuret test (blue) indicates no protein is present. The negative direct Benedict's test (blue) indicates no reducing sugars are present. The positive test after acid hydrolysis (brick-red precipitate with Benedict's) indicates that a non-reducing sugar, such as sucrose, is present. The negative iodine test (yellow-brown) indicates no starch is present. Therefore, only non-reducing sugar is present.
PastPaper.markingScheme
Award 1 mark for the correct answer C. Correctly interpreting the negative results for protein, reducing sugar, and starch, and the positive result for non-reducing sugar.
PastPaper.question 3 · multipleChoice
1 PastPaper.marks
Which statement correctly describes the function of cholesterol in mammalian cell membranes?
A.It acts as a channel for the facilitated diffusion of small polar molecules.
B.It regulates membrane fluidity by preventing phospholipids from packing too closely at low temperatures.
C.It binds to external cell-signaling molecules to initiate a cytoplasmic response.
D.It forms covalent bonds with glycolipids to stabilize the lipid bilayer.
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PastPaper.workedSolution
Cholesterol regulates membrane fluidity. At low temperatures, it prevents the close packing of phospholipid hydrophobic tails, thereby preventing the membrane from freezing or becoming too rigid. At higher temperatures, it helps stabilize the membrane by reducing phospholipid movement.
PastPaper.markingScheme
Award 1 mark for the correct answer B. Cholesterol acts as a fluidity regulator. Options A, C, and D describe roles of transport proteins, glycoproteins/receptors, or are structurally incorrect.
PastPaper.question 4 · multipleChoice
1 PastPaper.marks
Which cellular structures or features are present in a typical plant cell during metaphase of mitosis? 1. Centrioles 2. Centromeres 3. Spindle microtubules 4. Nuclear envelope
A.1, 2 and 3
B.2 and 3 only
C.1 and 4 only
D.2, 3 and 4
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PastPaper.workedSolution
During metaphase, chromosomes align at the cell equator, held together by their centromeres (2), which are attached to spindle microtubules (3). Typical plant cells lack centrioles (1), and the nuclear envelope (4) breaks down completely during prophase and does not reform until telophase.
PastPaper.markingScheme
Award 1 mark for the correct answer B. Identify that plant cells do not contain centrioles during mitosis and the nuclear envelope is absent in metaphase.
PastPaper.question 5 · multipleChoice
1 PastPaper.marks
A sample of double-stranded DNA is analyzed and found to contain \(22\%\) adenine bases. What is the percentage of cytosine bases in this DNA sample?
A.\(22\%\)
B.\(28\%\)
C.\(44\%\)
D.\(56\%\)
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PastPaper.workedSolution
In double-stranded DNA, the percentage of adenine (A) equals the percentage of thymine (T). Thus, A = T = \(22\%\). Together, A + T = \(44\%\). The remaining bases must be cytosine (C) and guanine (G), which make up \(100\% - 44\% = 56\%\) of the DNA. Because G equals C, the percentage of cytosine is \(56\% / 2 = 28\%\).
PastPaper.markingScheme
Award 1 mark for the correct answer B. Use Chargaff's rules to calculate that C = (100 - 2A)/2.
PastPaper.question 6 · multipleChoice
1 PastPaper.marks
Which features are characteristic of mature companion cells in the phloem tissue of a flowering plant? 1. Presence of a nucleus 2. Presence of cellulose cell walls with plasmodesmata 3. Absence of mitochondria to maximize space for transport
A.1, 2 and 3
B.1 and 2 only
C.2 and 3 only
D.1 only
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PastPaper.workedSolution
Companion cells are metabolically active cells that retain their nucleus (1) and possess walls rich in plasmodesmata (2) to facilitate the movement of materials into sieve tube elements. They contain a high density of mitochondria to produce ATP for active loading of sucrose, so statement 3 is incorrect.
PastPaper.markingScheme
Award 1 mark for the correct answer B. Companion cells must have mitochondria for active transport; thus statement 3 is incorrect, leaving 1 and 2 as correct.
PastPaper.question 7 · multipleChoice
1 PastPaper.marks
Which row correctly identifies the presence of cartilage and smooth muscle in the walls of the trachea, bronchioles, and alveoli of a healthy human?
A.Trachea: cartilage and smooth muscle present; Bronchioles: cartilage absent, smooth muscle present; Alveoli: cartilage and smooth muscle absent
B.Trachea: cartilage present, smooth muscle absent; Bronchioles: cartilage and smooth muscle present; Alveoli: cartilage and smooth muscle absent
In the mammalian gas exchange system, the trachea contains both cartilage (to keep the airway open) and smooth muscle. Bronchioles lack cartilage but still contain smooth muscle to regulate airflow. Alveoli lack both cartilage and smooth muscle to ensure a very thin barrier for gas exchange.
PastPaper.markingScheme
Award 1 mark for the correct answer A. Students must correctly identify the anatomical distribution of tissues across the three regions.
PastPaper.question 8 · multipleChoice
1 PastPaper.marks
Which row correctly matches the infectious disease with its pathogen type and its primary method of transmission?
Cholera is caused by the bacterium Vibrio cholerae and is primarily transmitted via contaminated water or food (water-borne). Malaria is caused by a protoctist (not a virus), tuberculosis is caused by a bacterium (not a virus), and measles is caused by a virus (not a bacterium).
PastPaper.markingScheme
Award 1 mark for the correct answer A. Check the pathogen classification and transmission route for each disease.
PastPaper.question 9 · multipleChoice
1 PastPaper.marks
An eyepiece graticule is calibrated using a stage micrometer. At \(\times 100\) magnification (using a \(\times 10\) eyepiece lens and a \(\times 10\) objective lens), 40 eyepiece graticule units (epu) align exactly with 10 stage micrometer divisions. Each stage micrometer division is \(0.1\text{ mm}\) wide. If the objective lens is changed to \(\times 40\), what is the actual length represented by 1 eyepiece graticule unit?
A.\(1.56\ \mu\text{m}\)
B.\(6.25\ \mu\text{m}\)
C.\(25.0\ \mu\text{m}\)
D.\(100\ \mu\text{m}\)
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PastPaper.workedSolution
At \(\times 100\) magnification: 10 stage micrometer divisions = \(10 \times 0.1\text{ mm} = 1.0\text{ mm} = 1000\ \mu\text{m}\). Since 40 epu align with 10 divisions, \(40\text{ epu} = 1000\ \mu\text{m}\), so \(1\text{ epu} = \frac{1000}{40} = 25\ \mu\text{m}\). When the objective lens is changed from \(\times 10\) to \(\times 40\), the magnification increases by a factor of 4 (from \(\times 100\) to \(\times 400\)). Therefore, the actual distance represented by each eyepiece graticule unit decreases by a factor of 4: New \(1\text{ epu} = \frac{25\ \mu\text{m}}{4} = 6.25\ \mu\text{m}\).
PastPaper.markingScheme
Award 1 mark for the correct answer (B). - Reject A: calculated by dividing by 16 instead of 4. - Reject C: did not adjust for the change in magnification. - Reject D: multiplied by 4 instead of dividing by 4.
PastPaper.question 10 · multipleChoice
1 PastPaper.marks
A student carries out three biochemical tests on a sample of a liquid food mixture to identify its carbohydrate content.
- Test 1: Boil directly with Benedict's solution. The solution remains blue. - Test 2: Boil a fresh sample with dilute hydrochloric acid, neutralize with sodium hydrogencarbonate, then boil with Benedict's solution. A brick-red precipitate forms. - Test 3: Add iodine in potassium iodide solution. A blue-black color is observed.
Which carbohydrates are present in the liquid food mixture?
A.glucose and starch
B.sucrose and starch
C.sucrose and glycogen
D.maltose and starch
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PastPaper.workedSolution
Test 1 shows no direct reducing sugars are present (ruling out glucose or maltose). Test 2 shows that after acid hydrolysis, reducing sugars are produced, confirming the presence of a non-reducing sugar like sucrose. Test 3 shows that starch is present due to the blue-black color reaction with iodine.
PastPaper.markingScheme
Award 1 mark for the correct answer (B). - Reject A and D: Test 1 would be positive (brick-red) if glucose or maltose were present. - Reject C: Glycogen gives a red-brown color with iodine, not blue-black.
PastPaper.question 11 · multipleChoice
1 PastPaper.marks
Which change to the phospholipid bilayer of cell surface membranes helps single-celled organisms maintain membrane fluidity when environmental temperatures drop significantly?
A.an increase in the proportion of saturated fatty acid tails and an increase in cholesterol
B.an increase in the proportion of unsaturated fatty acid tails and an increase in cholesterol
C.a decrease in the proportion of unsaturated fatty acid tails and a decrease in cholesterol
D.an increase in the proportion of saturated fatty acid tails and a decrease in cholesterol
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PastPaper.workedSolution
At low temperatures, membranes can freeze or become too rigid. Organisms adapt by increasing the proportion of unsaturated fatty acids (which contain double bonds that create kinks, preventing the tight packing of hydrophobic tails) and increasing the proportion of cholesterol (which prevents the membrane from packing too tightly together, maintaining fluidity).
PastPaper.markingScheme
Award 1 mark for the correct answer (B). - Reject A, C, and D: Saturated fatty acid tails pack tightly together and would decrease fluidity, making the membrane more rigid at low temperatures.
PastPaper.question 12 · multipleChoice
1 PastPaper.marks
Which row correctly describes the behavior of chromosomes and spindle microtubules during anaphase of mitosis?
D.Centromeres do not split; sister chromatids move to opposite poles; spindle microtubules shorten.
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PastPaper.workedSolution
During anaphase of mitosis, centromeres split and sister chromatids (now referred to as chromosomes) are pulled apart to opposite poles of the cell. This movement is achieved as the spindle microtubules shorten.
PastPaper.markingScheme
Award 1 mark for the correct answer (A). - Reject B: Homologous chromosomes separate during anaphase I of meiosis, not mitosis. - Reject C and D: Centromeres must split and spindle microtubules must shorten for chromatids to move to opposite poles.
PastPaper.question 13 · multipleChoice
1 PastPaper.marks
A double-stranded DNA molecule contains a total of 1200 nucleotides. If \(28\%\) of the bases in this DNA molecule are adenine (A), what is the total number of hydrogen bonds holding the two strands together?
A.1200
B.1464
C.1536
D.3000
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PastPaper.workedSolution
Since the DNA molecule is double-stranded: 1. Adenine (A) = \(28\%\), so Thymine (T) must also be \(28\%\). 2. Total A-T nucleotides = \(28\% + 28\% = 56\%\) of 1200 nucleotides = 672 nucleotides, which forms 336 A-T base pairs. 3. Each A-T base pair is held by 2 hydrogen bonds: \(336 \times 2 = 672\) hydrogen bonds. 4. The remaining nucleotides are Guanine (G) and Cytosine (C): \(100\% - 56\% = 44\%\) of 1200 nucleotides = 528 nucleotides, which forms 264 G-C base pairs. 5. Each G-C base pair is held by 3 hydrogen bonds: \(264 \times 3 = 792\) hydrogen bonds. 6. Total hydrogen bonds = \(672 + 792 = 1464\).
PastPaper.markingScheme
Award 1 mark for the correct answer (B). - Reject A: this is simply the total number of nucleotides. - Reject C: incorrect calculation of G-C and A-T pairs or bonds per pair. - Reject D: an overestimation of the total bond capacity.
PastPaper.question 14 · multipleChoice
1 PastPaper.marks
Which row correctly describes the structural features of a mature xylem vessel element and a mature phloem sieve tube element?
A.Xylem: lignified walls present, end walls absent, cytoplasm absent; Phloem: lignified walls absent, end walls present (sieve plates), cytoplasm present as a thin peripheral layer.
B.Xylem: lignified walls absent, end walls present, cytoplasm absent; Phloem: lignified walls present, end walls absent, cytoplasm present as a thin peripheral layer.
C.Xylem: lignified walls present, end walls absent, cytoplasm present; Phloem: lignified walls absent, end walls present (sieve plates), cytoplasm absent.
D.Xylem: lignified walls present, end walls present, cytoplasm absent; Phloem: lignified walls present, end walls present (sieve plates), cytoplasm present as a thin peripheral layer.
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PastPaper.workedSolution
A mature xylem vessel element is a dead cell; it has lignified walls to prevent collapse under tension, its end walls are completely lost to form a continuous tube, and it lacks cytoplasm. In contrast, a mature phloem sieve tube element is a living cell; it lacks lignified walls, has end walls modified into sieve plates, and retains a thin peripheral layer of cytoplasm (though it lacks a nucleus and many organelles).
PastPaper.markingScheme
Award 1 mark for the correct answer (A). - Reject B: Xylem elements have lignified walls, not phloem elements. - Reject C: Mature xylem vessel elements do not contain cytoplasm, and mature phloem elements do contain a thin layer of cytoplasm. - Reject D: Phloem walls are not lignified, and xylem vessel elements do not have end walls at maturity.
PastPaper.question 15 · multipleChoice
1 PastPaper.marks
Which row correctly matches the pathogen type, the method of transmission, and the main site of replication in the human host for the infectious disease cholera?
A.Pathogen: bacterium; Transmission: water-borne; Site of replication: epithelial cells of the small intestine.
B.Pathogen: bacterium; Transmission: vector-borne; Site of replication: red blood cells.
C.Pathogen: virus; Transmission: water-borne; Site of replication: epithelial cells of the large intestine.
D.Pathogen: protoctist; Transmission: water-borne; Site of replication: liver cells.
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PastPaper.workedSolution
Cholera is caused by the bacterium Vibrio cholerae. It is transmitted via water-borne or food-borne routes (fecal-oral pathway). Once inside the host, the bacteria multiply and replicate in the lumen and epithelial cells of the small intestine, secreting choleragen (cholera toxin).
PastPaper.markingScheme
Award 1 mark for the correct answer (A). - Reject B: cholera is not vector-borne or replicating in red blood cells (that is malaria). - Reject C: Vibrio cholerae is a bacterium, not a virus, and replicates mainly in the small intestine, not the large intestine. - Reject D: the pathogen is a bacterium, not a protoctist (such as Plasmodium).
PastPaper.question 16 · multipleChoice
1 PastPaper.marks
Which process describes how antibodies can make pathogens more easily recognized and engulfed by phagocytes during an immune response?
A.Neutralisation, by blocking the binding sites on the surface of the pathogen.
B.Opsonisation, by binding to antigens on the pathogen surface and acting as marker proteins for phagocytes to bind to.
C.Lysis, by directly punching holes in the bacterial cell wall.
D.Agglutination, by binding to antigens and dissolving the bacterial capsule.
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PastPaper.workedSolution
Opsonisation is the process where antibodies bind to antigens on the pathogen surface. This coats the pathogen and provides a chemical marker ('flag') that is easily recognized by phagocytes, which have surface receptors that bind to the constant region (Fc portion) of the antibodies, stimulating phagocytosis.
PastPaper.markingScheme
Award 1 mark for the correct answer (B). - Reject A: Neutralisation blocks host cell entry or toxic actions, but does not describe the specific flagging process for phagocytosis. - Reject C: Lysis is the direct destruction of a cell wall/membrane, not a flagging process for engulfment. - Reject D: Agglutination clumps pathogens together to reduce their spread and aid mass engulfment, but it does not dissolve the bacterial capsule.
PastPaper.question 17 · multipleChoice
1 PastPaper.marks
A student calibrated an eyepiece graticule using a stage micrometer. The stage micrometer had divisions of 0.01 mm. Under high power, 20 divisions of the stage micrometer aligned with 80 units of the eyepiece graticule. An animal cell was measured under the same high-power lens and was found to be 6 eyepiece graticule units wide. What is the actual width of the animal cell?
A.1.5 \(\mu\text{m}\)
B.15 \(\mu\text{m}\)
C.150 \(\mu\text{m}\)
D.240 \(\mu\text{m}\)
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PastPaper.workedSolution
1. Calculate the distance represented by 20 divisions of the stage micrometer: 20 multiplied by 0.01 mm = 0.2 mm. 2. Convert this distance to micrometres: 0.2 mm = 200 micrometres. 3. Find the value of 1 eyepiece graticule unit: 200 micrometres divided by 80 units = 2.5 micrometres per unit. 4. Calculate the width of the cell: 6 units multiplied by 2.5 micrometres = 15 micrometres.
PastPaper.markingScheme
Award 1 mark for the correct calculation showing that the cell width is 15 micrometres (B). Reject alternative values derived from incorrect unit conversions.
PastPaper.question 18 · multipleChoice
1 PastPaper.marks
A mixture containing starch and sucrose is incubated with amylase at 37 °C for 30 minutes. Which observations would be made if tests are performed on the mixture after this incubation?
A.Iodine test: blue-black; Benedict's test: red; Biuret test: blue
D.Iodine test: blue-black; Benedict's test: blue; Biuret test: blue
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PastPaper.workedSolution
Amylase hydrolyses starch into reducing sugars (maltose), so the starch is fully broken down, resulting in a yellow-brown (negative) iodine test. The newly formed reducing sugars yield a red (positive) Benedict's test. Since amylase is a protein, the Biuret test detects its presence and turns purple (positive).
PastPaper.markingScheme
Award 1 mark for identifying that the iodine test is negative, Benedict's is positive, and Biuret is positive (B).
PastPaper.question 19 · multipleChoice
1 PastPaper.marks
A polypeptide contains only the amino acids alanine, glycine, leucine, and valine. Which bonds or interactions can form between the R-groups of these amino acids to stabilise the tertiary structure of the protein?
A.hydrophobic interactions only
B.hydrogen bonds and hydrophobic interactions only
C.hydrogen bonds, disulfide bonds and ionic bonds only
D.hydrogen bonds, hydrophobic interactions and ionic bonds only
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PastPaper.workedSolution
Alanine, glycine, leucine, and valine all possess non-polar R-groups (or in the case of glycine, just a hydrogen atom). They do not contain polar, acidic, basic, or sulfur-containing side chains. Therefore, they cannot form hydrogen bonds, ionic bonds, or disulfide bonds. Only hydrophobic interactions can form between these non-polar R-groups.
PastPaper.markingScheme
Award 1 mark for selecting the option indicating that only hydrophobic interactions can form (A).
PastPaper.question 20 · multipleChoice
1 PastPaper.marks
An enzyme-catalysed reaction is carried out in the presence of a competitive inhibitor. How do the maximum velocity of the reaction (\(V_{\max}\)) and the Michaelis-Menten constant (\(K_m\)) change compared to the reaction without the inhibitor?
A competitive inhibitor binds to the active site of the enzyme, decreasing its affinity for the substrate. This increases the Michaelis-Menten constant (\(K_m\)) because a higher concentration of substrate is needed to reach half the maximum velocity. However, the maximum velocity (\(V_{\max}\)) remains unchanged because high substrate concentrations can completely overcome the competitive inhibition.
PastPaper.markingScheme
Award 1 mark for identifying that \(V_{\max}\) remains unchanged and \(K_m\) increases (B).
PastPaper.question 21 · multipleChoice
1 PastPaper.marks
Plant cells with a water potential of \(-600\text{ kPa}\) are placed in a sucrose solution with a solute potential of \(-900\text{ kPa}\) in an open beaker. Which statement correctly describes the movement of water and the state of the cells at equilibrium?
A.Water moves into the cells, the cells become turgid, and the water potential of the cells increases to \(0\text{ kPa}\).
B.Water moves out of the cells, the cells become plasmolysed, and the water potential of the cells decreases to \(-900\text{ kPa}\).
C.Water moves out of the cells, the cells become plasmolysed, and the water potential of the cells increases to \(-300\text{ kPa}\).
D.There is no net movement of water, the cells remain unchanged, and their water potential remains at \(-600\text{ kPa}\).
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PastPaper.workedSolution
Since the beaker is open, the pressure potential of the external solution is 0 kPa, meaning its water potential is \(-900\text{ kPa}\). This is lower (more negative) than the water potential of the plant cells (\(-600\text{ kPa}\)). Water moves out of the cells by osmosis down a water potential gradient, causing plasmolysis. At equilibrium, the water potential of the cells will equal that of the solution (\(-900\text{ kPa}\)).
PastPaper.markingScheme
Award 1 mark for the statement describing water leaving the cells, causing plasmolysis, and the cell water potential dropping to \(-900\text{ kPa}\) (B).
PastPaper.question 22 · multipleChoice
1 PastPaper.marks
A diploid eukaryotic cell contains 16 chromosomes (\(2n = 16\)). How many chromosomes, sister chromatids, and telomeres are present in this cell during metaphase of mitosis?
At metaphase, chromosomes are fully replicated but have not yet separated. Thus, there are still 16 chromosomes, each consisting of 2 sister chromatids (total of 32 chromatids). Since each chromatid is a single linear DNA molecule, it contains 2 telomeres (one at each end). Therefore, the total number of telomeres is 32 chromatids multiplied by 2 telomeres per chromatid = 64 telomeres.
PastPaper.markingScheme
Award 1 mark for 16 chromosomes, 32 sister chromatids, and 64 telomeres (C).
PastPaper.question 23 · multipleChoice
1 PastPaper.marks
A double-stranded DNA molecule is analysed and found to contain 34% cytosine. What is the percentage of adenine and the total percentage of purines in this DNA molecule?
A.16% adenine and 32% purines
B.16% adenine and 50% purines
C.34% adenine and 50% purines
D.34% adenine and 68% purines
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PastPaper.workedSolution
If cytosine (C) = 34%, then guanine (G) = 34%. Together, C + G = 68%, leaving 32% for adenine (A) + thymine (T). Since A = T, adenine = 16%. Purines are adenine (A) and guanine (G). In any double-stranded DNA molecule, purines make up exactly 50% of the total nitrogenous bases (A + G = 16% + 34% = 50%).
PastPaper.markingScheme
Award 1 mark for identifying 16% adenine and 50% purines (B).
PastPaper.question 24 · multipleChoice
1 PastPaper.marks
Which row correctly describes the presence (\(\checkmark\)) or absence (\(\times\)) of cilia and cartilage in the trachea, bronchioles, and alveoli of a human?
The trachea contains both cilia (on the ciliated epithelium) and cartilage (to keep the airway open). Bronchioles lack cartilage but still retain ciliated cells to move mucus. Alveoli lack both cilia and cartilage to maintain a very thin, flexible barrier for rapid gas exchange.
PastPaper.markingScheme
Award 1 mark for the row showing that trachea has cilia and cartilage, bronchioles have cilia but no cartilage, and alveoli have neither (A).
PastPaper.question 25 · multipleChoice
1 PastPaper.marks
A student calibrated an eyepiece graticule using a stage micrometer. The stage micrometer has divisions spaced \(0.01\text{ mm}\) apart.
At \(\times 10\) magnification, \(20\) divisions of the eyepiece graticule align with \(8\) divisions of the stage micrometer.
The student then observed a slide of plant epidermis at \(\times 40\) magnification using the same microscope and eyepiece graticule. An epidermal cell was measured to be \(45\) eyepiece graticule divisions in length.
What is the actual length of the epidermal cell?
A.\(11.25\ \mu\text{m}\)
B.\(45.0\ \mu\text{m}\)
C.\(180\ \mu\text{m}\)
D.\(450\ \mu\text{m}\)
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PastPaper.workedSolution
At \(\times 10\) magnification: \(8\) divisions of the stage micrometer \(= 8 \times 0.01\text{ mm} = 0.08\text{ mm} = 80\ \mu\text{m}\). Since \(20\) eyepiece graticule divisions (e.g.d.) align with \(8\) stage micrometer divisions: \(1\text{ e.g.d.} = 80\ \mu\text{m} / 20 = 4\ \mu\text{m}\).
At \(\times 40\) magnification: The magnification has increased by a factor of \(4\) (\(\times 40\) compared to \(\times 10\)). Therefore, the actual distance represented by each eyepiece graticule division decreases by a factor of \(4\): \(1\text{ e.g.d.} = 4\ \mu\text{m} / 4 = 1\ \mu\text{m}\).
The cell measures \(45\text{ e.g.d.}\) in length at \(\times 40\) magnification: \text{Actual length} \(= 45 \times 1\ \mu\text{m} = 45\ \mu\text{m}\).
PastPaper.markingScheme
Award 1 mark for the correct answer (B). - Reject incorrect calculation of calibration at \(\times 10\) (e.g., \(4\ \mu\text{m}\) per division). - Reject failure to adjust for the change in magnification from \(\times 10\) to \(\times 40\) (which leads to C, \(180\ \mu\text{m}\)). - Reject multiplying by the magnification factor instead of dividing (which leads to \(720\ \mu\text{m}\)).
PastPaper.question 26 · multipleChoice
1 PastPaper.marks
Four liquid food samples, W, X, Y and Z, were tested for the presence of biological molecules. The results of the tests are shown in the list below:
* **Sample W**: Benedict's test on original solution = blue; Benedict's test after acid hydrolysis and neutralisation = brick-red; Biuret test = blue; Emulsion test = cloudy emulsion * **Sample X**: Benedict's test on original solution = blue; Benedict's test after acid hydrolysis and neutralisation = blue; Biuret test = purple; Emulsion test = cloudy emulsion * **Sample Y**: Benedict's test on original solution = brick-red; Benedict's test after acid hydrolysis and neutralisation = brick-red; Biuret test = blue; Emulsion test = clear * **Sample Z**: Benedict's test on original solution = blue; Benedict's test after acid hydrolysis and neutralisation = brick-red; Biuret test = purple; Emulsion test = clear
Which sample contains a non-reducing sugar and a lipid, but no protein?
A.W
B.X
C.Y
D.Z
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PastPaper.workedSolution
To identify the correct sample: 1. **Contains a non-reducing sugar**: The original Benedict's test must be negative (blue), but after acid hydrolysis and neutralisation, it must be positive (brick-red). This is true for samples W and Z. 2. **Contains a lipid**: The emulsion test must show a cloudy emulsion. This is true for samples W and X. 3. **Contains no protein**: The Biuret test must be negative (blue). This is true for samples W and Y.
Sample W satisfies all three criteria: it contains a non-reducing sugar (blue to brick-red), contains a lipid (cloudy emulsion), and contains no protein (blue Biuret test).
PastPaper.markingScheme
Award 1 mark for the correct answer (A). - Accept A (W) as the only sample satisfying all conditions. - Reject B (X) because it contains protein and lacks non-reducing sugar. - Reject C (Y) because it contains a reducing sugar and lacks lipid. - Reject D (Z) because it contains protein and lacks lipid.
PastPaper.question 27 · multipleChoice
1 PastPaper.marks
Which row correctly describes the distribution of cartilage, smooth muscle, and elastic fibres in the human gas exchange system?
A.Trachea: cartilage, smooth muscle, and elastic fibres present; Bronchioles: smooth muscle and elastic fibres present, cartilage absent; Alveoli: elastic fibres present, cartilage and smooth muscle absent
B.Trachea: cartilage and elastic fibres present, smooth muscle absent; Bronchioles: cartilage and elastic fibres present, smooth muscle absent; Alveoli: elastic fibres present, cartilage and smooth muscle absent
C.Trachea: cartilage, smooth muscle, and elastic fibres present; Bronchioles: cartilage and smooth muscle present, elastic fibres absent; Alveoli: cartilage and smooth muscle present, elastic fibres absent
D.Trachea: cartilage and smooth muscle present, elastic fibres absent; Bronchioles: smooth muscle and elastic fibres present, cartilage absent; Alveoli: cartilage, smooth muscle, and elastic fibres absent
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PastPaper.workedSolution
According to the structure of the human gas exchange system: - **Trachea**: Cartilage is present as C-shaped rings to keep the airway open. Smooth muscle is present in the posterior wall to adjust the airway diameter. Elastic fibres are present throughout to allow stretching and recoil. - **Bronchioles**: Cartilage is absent. Smooth muscle is present to control the resistance to airflow. Elastic fibres are present to support the airway and allow recoil. - **Alveoli**: Cartilage and smooth muscle are both absent (to allow a very thin barrier for efficient gas exchange). Elastic fibres are present to allow the alveoli to stretch during inhalation and recoil during exhalation to push air out.
PastPaper.markingScheme
Award 1 mark for the correct answer (A). - Reject B because smooth muscle is present in the trachea and bronchioles, and cartilage is absent in bronchioles. - Reject C because elastic fibres are present in bronchioles and alveoli, and cartilage and smooth muscle are absent in alveoli. - Reject D because elastic fibres are present in the trachea and alveoli.
PastPaper.question 28 · multipleChoice
1 PastPaper.marks
A double-stranded DNA molecule contains \(3200\) base pairs. Adenine (A) makes up \(22\%\) of the total bases in this DNA molecule.
During semi-conservative replication, this DNA molecule is replicated once.
How many free guanine (G) nucleotides are needed to complete this replication?
A.\(704\)
B.\(1408\)
C.\(1792\)
D.\(3584\)
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PastPaper.workedSolution
1. Find the total number of nucleotides (bases) in the DNA molecule: \(3200\text{ base pairs} \times 2 = 6400\text{ nucleotides}\). 2. Find the percentage of guanine (G) bases: - Since \(22\%\) of the bases are Adenine (A), then \(22\%\) are Thymine (T) due to complementary base pairing. - Therefore, \(\text{A} + \text{T} = 22\% + 22\% = 44\%\). - This leaves \(100\% - 44\% = 56\%\) of the bases as Guanine (G) and Cytosine (C). - Since \(\text{G} = \text{C}\), the percentage of Guanine is \(56\% / 2 = 28\%\). 3. Calculate the number of Guanine bases in one double-stranded DNA molecule: \(28\%\text{ of } 6400 = 0.28 \times 6400 = 1792\). 4. Determine the number of free G nucleotides needed for one replication: - Semi-conservative replication of one double-stranded DNA molecule produces two double-stranded DNA molecules. - Each of the two new DNA molecules contains one original template strand and one newly synthesized strand. - Together, the two newly synthesized strands are complementary to the two template strands, making their combined sequence identical to one complete double-stranded DNA molecule. - Therefore, the number of free G nucleotides needed is exactly equal to the number of G bases in one double-stranded DNA molecule, which is \(1792\).
PastPaper.markingScheme
Award 1 mark for the correct answer (C). - Reject A (calculated as \(22\%\) of 3200). - Reject B (calculated as \(22\%\) of 6400, which is the number of A bases). - Reject D (calculated as \(1792 \times 2\), failing to account for semi-conservative replication where only one new strand is made per template strand, or incorrectly doubling the required free nucleotides).
PastPaper.question 29 · multipleChoice
1 PastPaper.marks
Which row correctly describes the roles of helper T-lymphocytes, killer T-lymphocytes, and plasma cells during an immune response to a viral infection?
A.Helper T-lymphocytes: release cytokines to stimulate B-lymphocytes to divide; Killer T-lymphocytes: bind to antigen-presenting body cells and destroy them; Plasma cells: secrete specific antibodies into the blood and lymph
B.Helper T-lymphocytes: phagocytose viral particles and present antigens on MHC proteins; Killer T-lymphocytes: produce and secrete large quantities of specific antibodies; Plasma cells: destroy virus-infected body cells by releasing hydrolytic enzymes
C.Helper T-lymphocytes: release cytokines to stimulate B-lymphocytes to divide; Killer T-lymphocytes: secrete specific antibodies into the blood and lymph; Plasma cells: bind to antigen-presenting body cells and destroy them
D.Helper T-lymphocytes: differentiate into plasma cells to produce memory cells; Killer T-lymphocytes: bind to antigen-presenting body cells and destroy them; Plasma cells: release cytokines to activate macrophages and B-lymphocytes
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PastPaper.workedSolution
- **Helper T-lymphocytes** recognize foreign antigens presented on host cells and release cytokines (such as interleukins). These cytokines stimulate B-lymphocytes to undergo clonal expansion (divide by mitosis) and stimulate killer T-lymphocytes. - **Killer T-lymphocytes** search for and bind to antigen-presenting virus-infected host cells, destroying them by releasing proteins (like perforin) that form pores in the cell membrane. - **Plasma cells** are effector B-lymphocytes that synthesize and secrete large quantities of soluble, specific antibodies into the body fluids (blood and lymph) to target the extracellular virus particles.
PastPaper.markingScheme
Award 1 mark for the correct answer (A). - Reject B because helper T-cells do not phagocytose, killer T-cells do not secrete antibodies, and plasma cells do not destroy cells. - Reject C because killer T-cells do not secrete antibodies, and plasma cells do not destroy host cells. - Reject D because helper T-cells do not differentiate into plasma cells, and plasma cells do not release cytokines to activate other cells.
PastPaper.question 30 · multipleChoice
1 PastPaper.marks
A plant tissue with a water potential of \(-400\text{ kPa}\) is placed in a sucrose solution with a water potential of \(-800\text{ kPa}\).
Which row correctly describes the net movement of water, the final water potential of the cytoplasm, and the state of the protoplast at equilibrium?
A.Net movement of water: out of the cell; Final water potential of cytoplasm: \(-800\text{ kPa}\); State of protoplast: pulled away from the cell wall
B.Net movement of water: out of the cell; Final water potential of cytoplasm: \(-400\text{ kPa}\); State of protoplast: pressed firmly against the cell wall
C.Net movement of water: into the cell; Final water potential of cytoplasm: \(-800\text{ kPa}\); State of protoplast: pulled away from the cell wall
D.Net movement of water: into the cell; Final water potential of cytoplasm: \(-400\text{ kPa}\); State of protoplast: pressed firmly against the cell wall
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PastPaper.workedSolution
1. **Net movement of water**: Water moves by osmosis down a water potential gradient, from a region of higher water potential (\(-400\text{ kPa}\)) to a region of lower water potential (\(-800\text{ kPa}\)). Thus, water moves out of the cell. 2. **Final water potential of the cytoplasm**: Water continues to leave the cell until the water potential of the cytoplasm matches that of the surrounding solution. Therefore, the final water potential of the cytoplasm will be \(-800\text{ kPa}\). 3. **State of protoplast**: As water leaves the vacuole and cytoplasm, the volume of the protoplast decreases. It shrinks and pulls away from the rigid cellulose cell wall, a process called plasmolysis. Hence, the protoplast is pulled away from the cell wall.
PastPaper.markingScheme
Award 1 mark for the correct answer (A). - Reject options stating water moves into the cell (C and D) because the external water potential is lower than the cell's water potential. - Reject B because the final water potential must reach equilibrium with the solution (\(-800\text{ kPa}\)) and the protoplast will shrink, not expand against the cell wall.
PastPaper.question 31 · multipleChoice
1 PastPaper.marks
Which features are possessed by both a mature xylem vessel element and a mature phloem sieve tube element?
1. Lignified cell walls 2. Absence of a nucleus 3. Presence of cellulose in the cell wall
A.1, 2 and 3
B.1 and 2 only
C.2 and 3 only
D.3 only
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PastPaper.workedSolution
Let's evaluate each feature for both cell types: - **1. Lignified cell walls**: Present in mature xylem vessel elements (lignin is deposited to provide mechanical strength and prevent collapse under tension). Absent in mature phloem sieve tube elements (their walls are not lignified). Therefore, feature 1 is NOT shared. - **2. Absence of a nucleus**: Mature xylem vessel elements are dead and completely hollow, lacking all organelles, including the nucleus. Mature phloem sieve tube elements are living but lose their nucleus during development to minimize resistance to the flow of phloem sap. Therefore, feature 2 IS shared. - **3. Presence of cellulose in the cell wall**: Both xylem and phloem cells are plant cells and possess cell walls containing cellulose. Therefore, feature 3 IS shared.
Features 2 and 3 are possessed by both cell types.
PastPaper.markingScheme
Award 1 mark for the correct answer (C). - Reject options containing 1 (A and B) because phloem sieve tube elements do not have lignified cell walls. - Reject D because it omits feature 2 (absence of a nucleus), which is a characteristic feature of both mature xylem vessel elements and mature phloem sieve tube elements.
PastPaper.question 32 · multipleChoice
1 PastPaper.marks
An experiment was carried out to investigate the effect of two different inhibitors, X and Y, on an enzyme-catalysed reaction.
In the presence of inhibitor X, the rate of reaction was low at low substrate concentrations, but reached the maximum velocity (\(V_{\max}\)) of the uninhibited reaction at very high substrate concentrations.
In the presence of inhibitor Y, the rate of reaction remained significantly lower than the uninhibited \(V_{\max}\) even at very high substrate concentrations.
Which statement about these inhibitors is correct?
A.Inhibitor X binds to an allosteric site on the enzyme, changing the shape of the active site so that the substrate cannot bind.
B.Inhibitor Y has a molecular shape that is highly complementary to the active site of the enzyme, outcompeting the substrate.
C.Increasing the substrate concentration increases the probability of substrate molecules colliding with and binding to the active site rather than inhibitor X.
D.Inhibitor Y forms temporary hydrogen bonds with the active site, allowing the inhibition to be easily overcome by adding more substrate.
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- **Inhibitor X** is a competitive inhibitor because its inhibitory effect is completely overcome at very high substrate concentrations (reaching the same \(V_{\max}\)). Competitive inhibitors bind reversibly to the active site of the enzyme. Increasing the concentration of substrate increases the probability that a substrate molecule, rather than an inhibitor molecule, will collide with and bind to the active site. Thus, statement C is correct. - **Inhibitor Y** is a non-competitive inhibitor because its inhibitory effect cannot be overcome by high substrate concentrations (the rate remains lower than \(V_{\max}\)). Non-competitive inhibitors bind to an allosteric site (not the active site), altering the overall shape of the enzyme and reducing its catalytic activity. - Statement A is incorrect because it describes a non-competitive inhibitor, whereas X is competitive. - Statement B is incorrect because it describes a competitive inhibitor, whereas Y is non-competitive. - Statement D is incorrect because non-competitive inhibitors (Y) do not bind to the active site and their inhibition is not overcome by adding more substrate.
PastPaper.markingScheme
Award 1 mark for the correct answer (C). - Reject A because inhibitor X is competitive and binds to the active site, not an allosteric site. - Reject B because inhibitor Y is non-competitive and does not bind to the active site. - Reject D because the inhibition caused by inhibitor Y is non-competitive and cannot be overcome by adding more substrate.
PastPaper.question 33 · multipleChoice
1 PastPaper.marks
At x100 magnification, 1 stage micrometer division (0.1 mm) equals 40 eyepiece graticule units. The microscope is then switched to the x400 objective lens. A cell is measured and is found to span 12 eyepiece graticule units. What is the actual width of the cell?
A.7.5 \(\mu\text{m}\)
B.30.0 \(\mu\text{m}\)
C.120.0 \(\mu\text{m}\)
D.3.0 \(\mu\text{m}\)
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At x100 magnification: 40 eyepiece graticule units = 0.1 mm = 100 \(\mu\text{m}\). Therefore, 1 eyepiece unit = 100 / 40 = 2.5 \(\mu\text{m}\). When switching to x400 magnification, the image is magnified 4 times more, so each eyepiece unit represents 4 times less distance: 2.5 \(\mu\text{m}\) / 4 = 0.625 \(\mu\text{m}\). Since the cell spans 12 eyepiece units at x400, its actual width is 12 * 0.625 \(\mu\text{m}\) = 7.5 \(\mu\text{m}\).
PastPaper.markingScheme
Award 1 mark for the correct calculation of 7.5 \(\mu\text{m}\) (Option A).
PastPaper.question 34 · multipleChoice
1 PastPaper.marks
Bacteria are grown in a medium containing only heavy nitrogen (15N) until all of their DNA contains only 15N. They are then transferred to a medium containing only light nitrogen (14N) and allowed to divide twice. What proportion of the DNA molecules in the second generation will contain both 15N and 14N?
A.25%
B.50%
C.75%
D.100%
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PastPaper.workedSolution
Due to semi-conservative replication, the parent DNA molecules consisting of 15N-15N strands serve as templates. In the first division in 14N, two hybrid 15N-14N molecules are formed. In the second division in 14N, these two hybrid molecules replicate to yield two hybrid (15N-14N) molecules and two light (14N-14N) molecules. Thus, 2 out of 4 (50%) of the DNA molecules contain both 15N and 14N.
PastPaper.markingScheme
Award 1 mark for identifying that 50% of the DNA molecules are hybrid (Option B).
PastPaper.question 35 · multipleChoice
1 PastPaper.marks
Which row correctly identifies the presence (tick) or absence (cross) of tissues in a healthy human bronchiole?
Bronchioles lack cartilage because they do not require rigid structural support to remain open. They have a ciliated epithelium to sweep mucus upwards and smooth muscle to control airway diameter. However, goblet cells are absent in bronchioles to prevent mucus accumulation in these small passages.
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Award 1 mark for selecting the row showing the correct tissue distribution in bronchioles (Option A).
PastPaper.question 36 · multipleChoice
1 PastPaper.marks
Which structural feature is shared by both a mature xylem vessel element and a mature phloem sieve tube element?
A.They both lack a nucleus.
B.They both have lignified cell walls.
C.They both have sieve plates.
D.They both lack cytoplasm.
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PastPaper.workedSolution
During differentiation, both xylem vessel elements and phloem sieve tube elements undergo modifications. Xylem vessel elements lose their nucleus, organelles, and cytoplasm completely to become dead tubes. Sieve tube elements lose their nucleus, ribosomes, and vacuole but retain a thin layer of peripheral cytoplasm to remain alive. Thus, they both lack a nucleus.
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Award 1 mark for identifying the correct shared feature (Option A).
PastPaper.question 37 · multipleChoice
1 PastPaper.marks
How does cholesterol affect the fluidity of a cell surface membrane?
A.At high temperatures it increases fluidity, and at low temperatures it decreases fluidity.
B.At high temperatures it decreases fluidity, and at low temperatures it increases fluidity.
C.It increases fluidity at all temperatures by binding to hydrophilic heads.
D.It decreases fluidity at all temperatures by making the fatty acid tails pack closer together.
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PastPaper.workedSolution
Cholesterol regulates membrane fluidity by acting as a buffer. At high temperatures, it stabilizes the membrane and restricts phospholipid movement, decreasing fluidity. At low temperatures, it prevents phospholipids from packing closely together and crystallizing, which increases fluidity.
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Award 1 mark for the correct description of cholesterol's effect on fluidity (Option B).
PastPaper.question 38 · multipleChoice
1 PastPaper.marks
An enzyme-controlled reaction is carried out in the presence of a competitive inhibitor. How do the values of Km (Michaelis constant) and Vmax (maximum rate of reaction) compare to those of the uninhibited reaction?
A.Km increases and Vmax decreases.
B.Km increases and Vmax remains unchanged.
C.Km remains unchanged and Vmax decreases.
D.Km decreases and Vmax remains unchanged.
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PastPaper.workedSolution
A competitive inhibitor binds to the active site, reducing the affinity of the enzyme for the substrate, which increases the Km value. However, because the inhibitor can be fully outcompeted at very high substrate concentrations, the maximum rate of reaction (Vmax) remains unchanged.
PastPaper.markingScheme
Award 1 mark for selecting the correct changes to Km and Vmax in competitive inhibition (Option B).
PastPaper.question 39 · multipleChoice
1 PastPaper.marks
In the production of monoclonal antibodies, why are antibody-producing plasma cells fused with myeloma (cancer) cells?
A.To enable the plasma cells to divide repeatedly to form a clone and survive indefinitely.
B.To stimulate the plasma cells to produce a wider variety of antibody types.
C.To ensure the resulting hybridoma cells target only cancer cells.
D.To increase the rate of antibody transcription within individual cells.
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PastPaper.workedSolution
Plasma cells secrete the desired antibody but are specialized, non-dividing cells that quickly die in culture. Myeloma cells are immortal cancer cells that divide rapidly. Fusing them produces hybridoma cells which combine the antibody-secreting capability of plasma cells with the continuous mitotic division and longevity of myeloma cells.
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Award 1 mark for identifying the correct reason for fusion in monoclonal antibody production (Option A).
PastPaper.question 40 · multipleChoice
1 PastPaper.marks
Which row correctly describes the structural features of both amylose and cellulose?
Amylose is a starch polymer made of alpha-glucose monomers joined by alpha-1,4-glycosidic bonds, forming an unbranched helical chain. Cellulose is a structural polymer made of beta-glucose monomers (where consecutive monomers are rotated 180 degrees) joined by beta-1,4-glycosidic bonds, forming straight, unbranched chains.
PastPaper.markingScheme
Award 1 mark for identifying the correct monomer, bond type, and macromolecular shape for both amylose and cellulose (Option A).
Paper 22
Answer all questions. Write your answers in the spaces provided on the question paper.
6 PastPaper.question · 60 PastPaper.marks
PastPaper.question 1 · structured
10 PastPaper.marks
Fig. 1.1 shows a transmission electron micrograph of a plant cell chloroplast.
(a) Describe three structural features of a chloroplast that can be seen in Fig. 1.1 and explain how each feature relates to its function. [6]
(b) An electron micrograph of a plant cell has a magnification of \(\times 15,000\). A starch grain inside a chloroplast measures 18 mm in length on the micrograph. Calculate the actual length of the starch grain in micrometres (\(\mu\text{m}\)). Show your working. [3]
(c) State one difference between the ribosomes found in the cytoplasm of a plant cell and those found within a chloroplast. [1]
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PastPaper.workedSolution
(a) Features and functions of chloroplast seen on a transmission electron micrograph: 1. Thylakoid membranes / grana: provides a large surface area for photosynthetic pigments (chlorophyll) and electron carriers for the light-dependent stage. 2. Stroma: fluid-filled matrix containing enzymes (such as Rubisco) for the light-independent stage (Calvin cycle). 3. Double membrane / chloroplast envelope: compartmentalises the chloroplast from the cytoplasm, controlling the entry and exit of substances. 4. Starch grain: acts as an insoluble energy storage carbohydrate, preventing osmotic effects within the organelle.
(b) Actual size = Image size / Magnification. Image size = 18 mm = \(18 \times 1000 = 18,000\ \mu\text{m}\). Actual size = \(18,000 / 15,000 = 1.2\ \mu\text{m}\).
(c) Ribosomes in the cytoplasm are 80S (larger), whereas ribosomes inside the chloroplast are 70S (smaller).
PastPaper.markingScheme
(a) [Max 6 marks overall, up to 2 marks per feature (1 mark for feature, 1 mark for corresponding function/adaptation)]: - Thylakoids / grana / thylakoid membranes; provides large surface area for light absorption / pigment attachment / ATP synthesis / light-dependent stage; [2] - Stroma; contains enzymes (e.g. Rubisco) / site of light-independent stage / Calvin cycle; [2] - Outer and inner membrane / envelope; compartmentalisation / controls movement of substances in and out; [2] - Starch grain / lipid droplet; storage of energy / products of photosynthesis; [2]
(b) [3 marks] - Correct conversion of 18 mm to micrometres (\(18,000\ \mu\text{m}\)) (or correct formula structure); [1] - Correct division: \(18,000 / 15,000\); [1] - Correct answer: \(1.2\ \mu\text{m}\); [1] Accept: correct calculation from incorrect conversion if working is clear (max 2 marks).
(c) [1 mark] - Cytoplasmic ribosomes are 80S AND chloroplast ribosomes are 70S / cytoplasmic ribosomes are larger / chloroplast ribosomes are smaller; [1]
PastPaper.question 2 · structured
10 PastPaper.marks
(a) Describe the structure of a DNA nucleotide. [3]
(b) Explain how the structure of a DNA molecule is adapted to its function of replicating accurately. [4]
(c) Distinguish between the roles of DNA polymerase and RNA polymerase in a eukaryotic cell. [3]
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PastPaper.workedSolution
(a) A DNA nucleotide consists of three components: a pentose sugar called deoxyribose, a phosphate group, and one of four nitrogenous bases (adenine, thymine, cytosine, or guanine). The phosphate group is bonded to carbon 5 of the deoxyribose, and the nitrogenous base is bonded to carbon 1.
(b) The double-stranded structure allows both strands to act as templates. The complementary base pairing rules (adenine pairs with thymine via two hydrogen bonds, and cytosine pairs with guanine via three hydrogen bonds) ensure that a precise complementary copy is synthesized. The hydrogen bonds are relatively weak, allowing the two strands to be separated easily during replication, while the strong covalent phosphodiester bonds in the sugar-phosphate backbones preserve the base sequence.
(c) DNA polymerase is involved in DNA replication; it synthesizes a new DNA strand using DNA nucleotides (dTTP, dATP, dCTP, dGTP) and requires a primer to initiate synthesis. RNA polymerase is involved in transcription; it synthesizes an RNA strand (mRNA, tRNA, or rRNA) using RNA nucleotides (UTP, ATP, CTP, GTP) from a DNA template and does not require a primer.
PastPaper.markingScheme
(a) [Max 3 marks] - Deoxyribose (pentose sugar); [1] - Phosphate group; [1] - Nitrogenous base / adenine, thymine, cytosine, guanine; [1] - Covalent bond / ester link between phosphate and sugar / glycosidic bond between sugar and base; [1]
(b) [Max 4 marks] - Double stranded / two strands, so both can act as templates; [1] - Specific complementary base pairing / A pairs with T and C pairs with G; [1] - Purine-pyrimidine pairing maintains constant helix diameter; [1] - Hydrogen bonds are weak / easily broken to allow strand separation; [1] - Many hydrogen bonds give stability; [1] - Covalent phosphodiester bonds / sugar-phosphate backbone protects the genetic code / sequence of bases; [1]
(c) [Max 3 marks] - DNA polymerase replicates DNA / copies both strands, whereas RNA polymerase transcribes DNA / copies only the template strand; [1] - DNA polymerase uses DNA nucleotides / thymine, whereas RNA polymerase uses RNA nucleotides / uracil; [1] - DNA polymerase requires a primer / start point, whereas RNA polymerase does not; [1] - DNA polymerase produces double-stranded DNA (or semi-conservative intermediate), RNA polymerase produces single-stranded RNA; [1]
PastPaper.question 3 · structured
10 PastPaper.marks
(a) Explain how the structure of xylem vessel elements is adapted to their function of transporting water and mineral ions. [4]
(b) Describe the mechanism by which sucrose is loaded into the phloem sieve tubes at the source. [4]
(c) Outline how the structure of a companion cell differs from that of a sieve tube element. [2]
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PastPaper.workedSolution
(a) Xylem vessel elements are dead cells aligned end-to-end with the end walls completely broken down to form continuous, hollow tubes, which reduces resistance to the flow of water. They have no cytoplasm or organelles, maximizing the space for water transport. Their cell walls are thickened with lignin, which provides mechanical strength to prevent the vessels from collapsing under negative pressure (tension) and makes the walls waterproof. Pits in the lignified walls allow the lateral movement of water to adjacent vessels or tissues.
(b) Hydrogen ions (\(\text{H}^+\)) are actively transported out of the companion cells into the cell wall space by proton pumps using energy from ATP hydrolysis. This creates a high concentration gradient of \(\text{H}^+\) ions in the cell wall. The \(\text{H}^+\) ions then diffuse back into the companion cells down their electrochemical gradient through co-transporter proteins. As they do so, they co-transport sucrose molecules with them against the sucrose concentration gradient. The accumulated sucrose then diffuses from the companion cells into the sieve tube elements through plasmodesmata.
(c) Companion cells have a nucleus, many mitochondria, and dense cytoplasm with ribosomes and rough endoplasmic reticulum. Sieve tube elements do not have a nucleus, have very few mitochondria located only at the margins, have thin cytoplasm, and possess sieve plates with sieve pores.
PastPaper.markingScheme
(a) [Max 4 marks] - No end walls / end walls broken down to form a continuous tube; [1] - No cytoplasm / no organelles / hollow lumen, to minimize resistance to water flow / allow unimpeded flow; [1] - Lignin / lignified walls; [1] - Lignin prevents collapse of vessels under tension / negative pressure / provides mechanical support; [1] - Waterproofing of walls by lignin keeps water inside vessel; [1] - Pits (unlignified areas); allow lateral movement of water / bypass blockages; [1]
(b) [Max 4 marks] - Active transport of \(\text{H}^+\) / protons out of companion cells into cell wall; [1] - Using ATP / proton pump; [1] - Creation of a proton / \(\text{H}^+\) concentration gradient; [1] - \(\text{H}^+\) diffuses back into companion cell through co-transporter protein; [1] - Sucrose is transported into companion cell against its concentration gradient; [1] - Sucrose diffuses / moves into sieve tube element via plasmodesmata; [1]
(c) [Max 2 marks] - Companion cell has a nucleus, sieve tube element does not; [1] - Companion cell has many mitochondria, sieve tube element has very few / peripheral mitochondria; [1] - Companion cell has standard organelles / dense cytoplasm, sieve tube element has no vacuole / thin cytoplasm / sieve plates; [1]
PastPaper.question 4 · structured
10 PastPaper.marks
(a) Cholera is an infectious disease caused by the bacterium *Vibrio cholerae*. Describe how *V. cholerae* is transmitted and explain how the cholera toxin causes diarrhoea in infected individuals. [5]
(b) Explain why treatment with penicillin is effective against many bacterial infections but is completely ineffective against viral pathogens such as HIV. [3]
(c) State two differences between active immunity and passive immunity. [2]
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PastPaper.workedSolution
(a) *Vibrio cholerae* is transmitted via the faecal-oral route, usually by drinking water or eating food contaminated with the faeces of an infected person. Once inside the small intestine, the bacteria secrete choleragen (cholera toxin). This toxin binds to receptor proteins on the cell surface membranes of the intestinal epithelial cells. The toxin enters the cells and activates adenylate cyclase, leading to an increase in cyclic AMP (cAMP) levels. This causes the active transport of chloride ions (\(\text{Cl}^-\)) out of the epithelial cells into the lumen of the intestine. This lowers the water potential of the intestinal lumen, causing water to move out of the blood and epithelial cells into the lumen by osmosis, resulting in severe watery diarrhoea.
(b) Penicillin inhibits the enzyme glycopeptidyl transferase, which is responsible for forming cross-links between peptidoglycan chains in bacterial cell walls. This weakens the cell wall, causing the bacterium to burst by osmotic lysis when water enters. Viruses do not have a cellular structure, lack cell walls, and do not contain peptidoglycan. They instead consist of a protein coat (capsid) surrounding genetic material, meaning penicillin has no target site or metabolic pathway to disrupt.
(c) Active immunity involves the production of antibodies by the individual's own plasma cells, whereas passive immunity involves receiving antibodies from an external source (e.g., across the placenta or via injection). Active immunity provides long-term protection due to the production of memory cells, whereas passive immunity provides only short-term, temporary protection because no memory cells are produced and the external antibodies are eventually broken down.
PastPaper.markingScheme
(a) [Max 5 marks] - Transmitted via the faecal-oral route / ingestion of water or food contaminated with faeces containing the bacteria; [1] - Cholera toxin / choleragen binds to receptors on intestinal epithelial cells; [1] - Toxin enters cell and activates adenylate cyclase; [1] - Increase in cAMP levels; [1] - Chloride ions (\(\text{Cl}^-\)) actively pumped out of cells into intestinal lumen; [1] - Lowers water potential of lumen; [1] - Water moves out of cells / blood into lumen by osmosis, causing watery diarrhoea; [1]
(b) [Max 3 marks] - Penicillin inhibits synthesis of cross-links in peptidoglycan / cell wall; [1] - Weakens bacterial cell wall, leading to osmotic lysis / bursting; [1] - Viruses are non-cellular / do not have cell walls / do not have peptidoglycan; [1] - Viruses have a protein coat / capsid; [1] - Viruses do not have metabolic machinery / enzymes targeted by penicillin; [1]
(c) [Max 2 marks] - Active involves production of antibodies by the body / own immune system, passive involves introduction of external antibodies; [1] - Active produces memory cells, passive does not; [1] - Active takes time to develop, passive provides immediate protection; [1] - Active is long-term, passive is short-term / temporary; [1]
PastPaper.question 5 · structured
10 PastPaper.marks
(a) Explain how the heart action is initiated and coordinated during one cardiac cycle. [5]
(b) Describe and explain the pressure changes that occur in the left ventricle and left atrium as the left ventricle contracts. [3]
(c) Explain the role of elastic fibres in the wall of the aorta. [2]
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PastPaper.workedSolution
(a) The heart action is myogenic, meaning it originates within the heart muscle itself. The sinoatrial node (SAN) acts as the pacemaker, initiating a wave of electrical excitation (depolarisation) that spreads rapidly across the walls of both atria, causing atrial systole. A band of non-conducting fibrous tissue between the atria and ventricles prevents the wave from spreading directly to the ventricles. Instead, the electrical impulse is directed to the atrioventricular node (AVN). The AVN delays the impulse for about 0.1 seconds, allowing the atria to contract fully and empty their blood into the ventricles. The impulse then travels down the Bundle of His to the Purkyne tissue at the apex of the heart, which conducts the wave of excitation upwards through the ventricle walls, causing ventricular systole to occur from the apex upwards, pushing blood out into the arteries.
(b) As the left ventricle contracts (ventricular systole), the pressure inside the left ventricle increases rapidly. This high pressure forces the left atrioventricular (bicuspid) valve to close, preventing blood from flowing back into the left atrium. The pressure in the left atrium remains relatively low because it is in diastole and receives blood slowly from the pulmonary veins, while the ventricular pressure continues to rise until it exceeds the pressure in the aorta, opening the semilunar valve.
(c) Elastic fibres allow the wall of the aorta to stretch when high-pressure blood is ejected from the left ventricle during ventricular systole. During ventricular diastole, when the heart relaxes, the elastic fibres recoil. This recoil action helps to maintain a high blood pressure in the arterial system and ensures a smooth, continuous flow of blood rather than surges.
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(a) [Max 5 marks] - Heart is myogenic / SAN acts as pacemaker / initiates wave of excitation / depolarisation; [1] - Wave spreads across atria causing atrial systole; [1] - Non-conducting fibrous tissue / ring prevents direct spread of wave to ventricles; [1] - Impulse reaches AVN which delays the wave of excitation; [1] - Delay allows atria to empty / ventricles to fill before ventricles contract; [1] - Impulse passes down Bundle of His to Purkyne tissue; [1] - Wave of excitation spreads upwards from the apex, causing ventricular systole from bottom up; [1]
(b) [Max 3 marks] - Left ventricular pressure increases rapidly / significantly; [1] - Pressure in left ventricle becomes higher than in left atrium; [1] - This causes the atrioventricular / bicuspid valve to close; [1] - Left atrial pressure remains relatively low; [1] - Closure of bicuspid valve prevents backflow of blood into atrium; [1]
(c) [Max 2 marks] - Elastic fibres stretch during ventricular systole / to accommodate high pressure / volume of blood; [1] - Elastic fibres recoil during ventricular diastole; [1] - Recoil maintains high blood pressure / smoothens blood flow / prevents pressure dropping to zero; [1]
PastPaper.question 6 · structured
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(a) Hemoglobin and collagen are both proteins with important structural and functional roles in the body. Compare the quaternary structures of hemoglobin and collagen. [4]
(b) Describe how you would carry out a biochemical test to confirm the presence of a non-reducing sugar in a liquid sample. [4]
(c) State the biological test used to detect proteins and describe the positive result observed. [2]
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(a) Hemoglobin is a globular protein consisting of four polypeptide chains (two alpha chains and two beta chains), whereas collagen is a fibrous protein consisting of three polypeptide chains (forming a triple helix). In hemoglobin, each of the four polypeptide chains contains a prosthetic heme group containing an iron (\(\text{Fe}^{2+}\)) ion, whereas collagen does not have any prosthetic groups. Hemoglobin molecules are folded into compact, spherical shapes with hydrophilic groups on the outside, making them soluble, whereas collagen molecules form long, parallel tropocollagen triple helices with covalent cross-links between them to form strong, insoluble fibrils.
(b) To test for a non-reducing sugar, first carry out a Benedict's test to confirm the absence of reducing sugars (heating with Benedict's reagent should yield no color change / remain blue). Then, add a few drops of dilute hydrochloric acid (\(\text{HCl}\)) to a fresh sample and heat in a boiling water bath for a few minutes to hydrolyze the glycosidic bonds. Allow the mixture to cool, then add an alkali, such as sodium hydrogencarbonate (\(\text{NaHCO}_3\)), to neutralize the acid (as Benedict's reagent requires alkaline conditions). Finally, add Benedict's reagent and heat the mixture again in a water bath. A color change from blue to green, yellow, orange, or brick-red precipitate confirms the presence of a non-reducing sugar.
(c) The test used to detect proteins is the Biuret test. A positive result is a color change from light blue to purple / violet.
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(a) [Max 4 marks] - Hemoglobin has four polypeptide chains, collagen has three polypeptide chains; [1] - Hemoglobin is globular, collagen is fibrous; [1] - Hemoglobin has prosthetic groups / heme groups / iron (\(\text{Fe}^{2+}\)), collagen has no prosthetic groups; [1] - Hemoglobin chains are folded individually into spherical shapes, collagen chains are wound into a triple helix; [1] - Hemoglobin is soluble, collagen is insoluble; [1] - Collagen has covalent cross-links between adjacent triple helices / tropocollagen molecules, hemoglobin does not; [1]
(b) [Max 4 marks] - Add dilute hydrochloric acid (\(\text{HCl}\)) and heat / boil; [1] - To hydrolyze glycosidic bonds (into monosaccharides); [1] - Neutralize with sodium hydrogencarbonate / alkali; [1] - Add Benedict's reagent and heat (in a water bath); [1] - Positive result: color change from blue to green / yellow / orange / brick-red (precipitate); [1]
(c) [Max 2 marks] - Biuret test / add Biuret reagent (or sodium hydroxide and copper sulfate); [1] - Positive result: color change from blue to purple / violet / mauve; [1]
Paper 32
Answer all questions. You will be provided with the materials and apparatus listed in the confidential instructions.
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PastPaper.question 1 · practical
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Question 1 (20 marks)
Glucose is a reducing sugar that can be detected using Benedict’s reagent. The concentration of reducing sugars can be estimated by measuring the time taken for the first colour change to appear when heated.
You are provided with: - Stock solution of 10.0% glucose, labelled G - Distilled water, labelled W - Benedict’s reagent, labelled Benedict's - An unknown glucose solution, labelled U
(a) (i) You are required to perform a serial dilution of the 10.0% glucose solution, G, to prepare four further concentrations: 5.0%, 2.5%, 1.25%, and 0.625%. Complete Table 1.1 to show how you would prepare 10.0 cm³ of each concentration. [4]
(ii) State one variable, other than temperature and the volumes of solutions added, that must be kept constant to ensure the results are valid. Explain how you would control this variable. [2]
(iii) The student carried out the Benedict's test on each of the prepared dilutions and on U by adding 2.0 cm³ of Benedict’s reagent to 2.0 cm³ of each solution, heating in a water bath at 80°C, and recording the time taken (seconds) for the first colour change to occur. In the space below, prepare a table and record a set of mock results that show the expected relationship between glucose concentration and time taken. Include your results for solution U (which has an actual concentration of 3.0%). [5]
(iv) State the estimated concentration of glucose in solution U. Explain how you arrived at this estimate based on your results. [2]
(b) (i) Suggest two potential sources of error in this experimental procedure that could affect the accuracy of the estimated glucose concentration, and suggest an improvement to reduce the effect of each. [4]
(ii) Describe how this method could be modified to obtain a more precise and quantitative estimate of the concentration of glucose in solution U using a colorimeter. [3]
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(a) (i) - 5.0% glucose: 5.0 cm³ of 10.0% glucose + 5.0 cm³ of water. - 2.5% glucose: 5.0 cm³ of 5.0% glucose + 5.0 cm³ of water. - 1.25% glucose: 5.0 cm³ of 2.5% glucose + 5.0 cm³ of water. - 0.625% glucose: 5.0 cm³ of 1.25% glucose + 5.0 cm³ of water.
(a) (ii) - Variable: Volume of Benedict's reagent. - Control: Use a graduated syringe or pipette to measure exactly 2.0 cm³ of Benedict's reagent for each tube.
(a) (iii) Mock Results Table: | Glucose concentration / % | Time taken for first colour change / s | | 10.0 | 35 | | 5.0 | 62 | | 2.5 | 118 | | 1.25 | 210 | | 0.625 | 385 | | U (unknown) | 95 |
(a) (iv) - Estimated concentration: Between 2.5% and 5.0% (around 3.0%). - Explanation: The time taken for the first colour change in solution U (95 s) lies between the time recorded for 5.0% glucose (62 s) and 2.5% glucose (118 s).
(b) (i) - Error 1: Subjective assessment of the exact moment of 'first colour change'. - Improvement 1: Place a white card with a black cross behind the tubes to standardise the point of opacity change, or use a colorimeter. - Error 2: Temperature fluctuation of the water bath during the experiment. - Improvement 2: Use a thermostatically controlled water bath instead of manually heating a beaker.
(b) (ii) - Set up Benedict's reaction with all standard concentrations and let them complete. - Filter or centrifuge the samples to remove any red copper(I) oxide precipitate. - Use a colorimeter with a red filter to measure the light absorbance/transmission of the remaining supernatant. - Plot a calibration curve of absorbance/transmission against glucose concentration, then measure solution U and read its exact concentration from the curve.
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(a) (i) [Total: 4 marks] - 1 mark: Identifies the serial dilution process (transferring 5.0 cm³ of the previous concentration). - 1 mark: Correct volumes of glucose for 5.0% and 2.5% (5.0 cm³ each). - 1 mark: Correct volumes of glucose for 1.25% and 0.625% (5.0 cm³ each). - 1 mark: Correct volumes of distilled water for all dilutions (5.0 cm³ each).
(a) (ii) [Total: 2 marks] - 1 mark: States a valid controlled variable (e.g. volume of Benedict's reagent, or size/thickness of test tubes). - 1 mark: Describes a valid method of control (e.g. measuring with a syringe, or choosing identical tubes from the same batch).
(a) (iii) [Total: 5 marks] - 1 mark: Table drawn with clear borders and fully labeled headings with units (e.g., 'Glucose concentration / %' and 'Time taken for first colour change / s'). - 1 mark: All five dilutions plus solution U are present in the table. - 1 mark: Time is recorded in whole seconds only (no decimals). - 1 mark: Trend shows that as glucose concentration decreases, the time taken increases. - 1 mark: Solution U time is intermediate between the times for 2.5% and 5.0% concentrations.
(a) (iv) [Total: 2 marks] - 1 mark: Correct estimation of U (e.g., 'between 2.5% and 5.0%'). - 1 mark: Explanation relates the time taken for U directly to the times of the bounding standard concentrations.
(b) (i) [Total: 4 marks] - 1 mark: Identifying subjectivity in color change detection. - 1 mark: Improvement (e.g., use a standard color card, or a colorimeter, or a black cross behind the tube). - 1 mark: Identifying temperature fluctuations of water bath. - 1 mark: Improvement (e.g., use of a thermostatically controlled water bath).
(b) (ii) [Total: 3 marks] - 1 mark: Centrifuge/filter the reacted mixture to obtain the supernatant. - 1 mark: Measure absorbance or transmission of the solutions using a colorimeter. - 1 mark: Plot a calibration curve of absorbance/transmission against glucose concentration and use it to read the concentration of U.
PastPaper.question 2 · practical
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Question 2 (20 marks)
Figure 2.1 shows a photomicrograph of a transverse section through a dicotyledonous stem.
(a) (i) Draw a large, low-power plan diagram of a quarter of the stem shown in Figure 2.1. Your drawing should show the distribution of the different tissues but must not show individual cells. Label the xylem and phloem. [5]
(ii) From the vascular bundle, select a group of three adjacent xylem vessel elements. Make a high-power drawing of these three cells to show their detailed structure. Label the cell wall and cell lumen of one cell. [4]
(b) A student calibrated the eyepiece graticule of a microscope using a stage micrometer. At high power (x400) magnification: - 100 eyepiece graticule units (epu) aligned with 25 small divisions on the stage micrometer. - Each small division on the stage micrometer is 10 micrometers wide.
(i) Calculate the actual length of 1 eyepiece graticule unit (epu) in micrometers. Show your working. [2]
(ii) The student measured the internal diameter of a xylem vessel element from Figure 2.1 using the same magnification. The internal diameter was 18 epu. Calculate the actual internal diameter of this vessel element in micrometers. Show your working. [2]
(c) (i) State three structural features of xylem vessels, visible under a light microscope, that adapt them for their function of transport of water and support. [3]
(ii) Complete Table 2.1 to compare the structural features of xylem vessel elements with phloem sieve tube elements visible in a transverse section of a stem. [4]
Table 2.1 | Feature | Xylem vessel elements | Phloem sieve tube elements | | Presence of cytoplasm / living contents | | | | Thickness of cell walls | | | | Presence of end walls / sieve plates | | | | Main cell wall component | | |
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(a) (i) - Low-power plan diagram should show: epidermis as a single continuous line, a cortex zone, a ring of discrete vascular bundles with xylem located internally and phloem externally, and a central pith. Label lines must point directly to the correct tissue types.
(a) (ii) - High-power drawing should show: three distinct polygonal cells in close contact, double-layered lines representing thick walls, hollow lumens of varying sizes, and clear label lines to 'cell wall' and 'cell lumen'.
(b) (ii) - Diameter = 18 epu x 2.5 micrometers/epu = 45 micrometers.
(c) (i) - 1. Thick, lignified cell walls to prevent collapse of vessels under tension. - 2. No cytoplasm or organelles (hollow lumen) to offer minimal resistance to water flow. - 3. Absence of end walls, creating a continuous pipeline for uninterrupted mass flow of water.
(c) (ii) Completed Table 2.1: - Presence of cytoplasm / living contents: Xylem = Absent (dead cells) | Phloem = Present (thin peripheral layer) - Thickness of cell walls: Xylem = Thick | Phloem = Thin - Presence of end walls / sieve plates: Xylem = Absent (open ends) | Phloem = Present (sieve plates with sieve pores) - Main cell wall component: Xylem = Lignin (and cellulose) | Phloem = Cellulose only
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(a) (i) [Total: 5 marks] - 1 mark: Clear, sharp, continuous lines, no sketching or shading. - 1 mark: Drawing occupies at least 50% of the available space. - 1 mark: Correct layout of a quarter-circle sector showing outer epidermis, cortex layer, distinct vascular bundle regions, and central pith. - 1 mark: No individual cells are drawn in any part of the plan diagram. - 1 mark: Correctly labels both 'xylem' (inner region of bundle) and 'phloem' (outer region of bundle).
(a) (ii) [Total: 4 marks] - 1 mark: Draws exactly three adjacent/touching xylem vessel cells. - 1 mark: Cell walls represented by double lines to show thickness. - 1 mark: Correct shape of xylem vessels (polygonal/rounded lumens with varying sizes). - 1 mark: Labels 'cell wall' and 'cell lumen' correctly.
(b) (i) [Total: 2 marks] - 1 mark: Correct working shown (e.g. 25 x 10 = 250 micrometers, then divided by 100). - 1 mark: Correct final answer of 2.5 micrometers.
(b) (ii) [Total: 2 marks] - 1 mark: Correct working shown (18 x 2.5). - 1 mark: Correct final answer of 45 micrometers (allow error carried forward from b(i)).
(c) (i) [Total: 3 marks] - 1 mark: Thick / lignified cell walls (prevents collapse under tension). - 1 mark: No protoplasm / hollow lumen (allows uninterrupted flow). - 1 mark: No end walls / open ends (forming continuous tubes).
(c) (ii) [Total: 4 marks] - 1 mark: Cytoplasm: Xylem = absent AND Phloem = present (thin peripheral layer). - 1 mark: Cell wall thickness: Xylem = thick AND Phloem = thin. - 1 mark: End walls: Xylem = absent/open AND Phloem = present/sieve plates. - 1 mark: Wall component: Xylem = lignin/cellulose AND Phloem = cellulose.