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Thinka Jun 2024 (V1) Cambridge International A Level-Style Mock — Biology (9700)

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An original Thinka practice paper modelled on the structure and difficulty of the Jun 2024 (V1) Cambridge International A Level Biology (9700) paper. Not affiliated with or reproduced from Cambridge.

Paper 11 (Multiple Choice)

Answer all 40 multiple-choice questions. For each question, there are four possible answers: A, B, C, and D.
33 PastPaper.question · 33 PastPaper.marks
PastPaper.question 1 · multiple-choice
1 PastPaper.marks
A student calibrates an eyepiece graticule using a stage micrometer. The stage micrometer scale is 1 mm long and is divided into 100 divisions. At \(\times 100\) magnification, 40 divisions of the eyepiece graticule align with 10 divisions of the stage micrometer. An animal cell is measured using the same microscope setup and is found to be 6 eyepiece graticule units wide. What is the actual width of the animal cell?
  1. A.1.5 \(\mu\text{m}\)
  2. B.15 \(\mu\text{m}\)
  3. C.150 \(\mu\text{m}\)
  4. D.1500 \(\mu\text{m}\)
PastPaper.showAnswers

PastPaper.workedSolution

1 stage micrometer division = 1 mm / 100 = 0.01 mm = 10 \(\mu\text{m}\). 10 stage micrometer divisions = 100 \(\mu\text{m}\). Since 40 eyepiece graticule units align with 10 stage micrometer divisions: 1 eyepiece graticule unit = 100 \(\mu\text{m}\) / 40 = 2.5 \(\mu\text{m}\). The actual width of the animal cell is 6 units \(\times\) 2.5 \(\mu\text{m}\) = 15 \(\mu\text{m}\).

PastPaper.markingScheme

Award 1 mark for the correct answer (B). [Method: 1 stage division = 10 \(\mu\text{m}\); 1 eyepiece unit = 100 \(\mu\text{m}\) / 40 = 2.5 \(\mu\text{m}\); cell size = 6 \(\times\) 2.5 \(\mu\text{m}\) = 15 \(\mu\text{m}\)]
PastPaper.question 2 · multiple-choice
1 PastPaper.marks
Which statement correctly describes the role of cholesterol in mammalian cell membranes?
  1. A.It acts as a barrier to the entry of all polar, water-soluble molecules.
  2. B.It regulates membrane fluidity by stabilizing the membrane at high temperatures and preventing crystallisation at low temperatures.
  3. C.It forms hydrophilic channels through which large macromolecular nutrients are actively transported.
  4. D.It binds specifically to cell-signalling molecules like hormones, acting as a cell surface receptor.
PastPaper.showAnswers

PastPaper.workedSolution

Cholesterol molecules regulate membrane fluidity. At high temperatures, they stabilize the membrane by pulling phospholipid fatty acid tails together, reducing fluidity. At low temperatures, they prevent the phospholipid tails from packing too closely together, preventing crystallisation and maintaining fluidity.

PastPaper.markingScheme

Award 1 mark for the correct statement about cholesterol's role in membrane fluidity (B).
PastPaper.question 3 · multiple-choice
1 PastPaper.marks
The primary structure of a polypeptide is determined by the sequence of amino acids. Which statement about the bonds that maintain the secondary, tertiary, and quaternary structures of proteins is correct?
  1. A.Secondary structure is maintained exclusively by covalent peptide bonds between adjacent R-groups.
  2. B.Tertiary structure is stabilized by hydrogen bonds, ionic bonds, disulfide bonds, and hydrophobic interactions involving R-groups.
  3. C.Quaternary structure is only present in proteins that consist of a single polypeptide chain folded into multiple domains.
  4. D.Disulfide bonds are easily disrupted by slight increases in temperature because they are weak non-covalent interactions.
PastPaper.showAnswers

PastPaper.workedSolution

The tertiary structure represents the complex 3D shape of a single polypeptide chain and is held together by four main types of bonds/interactions between R-groups: disulfide bonds (covalent), ionic bonds, hydrogen bonds, and hydrophobic interactions. Option A is incorrect because secondary structure is held by hydrogen bonds of the polypeptide backbone. Option C is incorrect because quaternary structure requires multiple polypeptide chains. Option D is incorrect because disulfide bonds are strong covalent bonds.

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Award 1 mark for the correct identification of the bonds stabilizing tertiary protein structure (B).
PastPaper.question 4 · multiple-choice
1 PastPaper.marks
An experiment was carried out to investigate the effect of an inhibitor on an enzyme-catalysed reaction. The results showed that the maximum rate of reaction (\(V_{\max}\)) remained unchanged, but a higher substrate concentration was required to reach half of the \(V_{\max}\) (\(K_{\text{m}}\) increased). Which statement correctly describes the type of inhibition and the mechanism involved?
  1. A.Non-competitive inhibition, because the inhibitor binds to the active site and prevents substrate binding.
  2. B.Non-competitive inhibition, because the inhibitor binds to an allosteric site and alters the shape of the active site.
  3. C.Competitive inhibition, because the inhibitor has a complementary shape to the active site and can be overcome by increasing substrate concentration.
  4. D.Competitive inhibition, because the inhibitor binds covalently to the active site and permanently deactivates the enzyme.
PastPaper.showAnswers

PastPaper.workedSolution

In competitive inhibition, the inhibitor binds reversibly to the active site of the enzyme because it is structurally similar to the substrate. This can be overcome by high substrate concentrations, so \(V_{\max}\) can still be achieved, although a higher concentration of substrate is needed to reach half \(V_{\max}\) (hence \(K_{\text{m}}\) is increased).

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Award 1 mark for identifying competitive inhibition and its mechanism (C).
PastPaper.question 5 · multiple-choice
1 PastPaper.marks
Which factors are essential for the movement of water up xylem vessels in a continuous column from roots to leaves? 1. Active transport of water molecules into xylem vessels at the root. 2. Cohesion between water molecules due to hydrogen bonding. 3. Adhesion between water molecules and the hydrophilic cellulose walls of xylem vessels. 4. High hydrostatic pressure at the top of the xylem column.
  1. A.1 and 2 only
  2. B.2 and 3 only
  3. C.1, 3 and 4 only
  4. D.2, 3 and 4 only
PastPaper.showAnswers

PastPaper.workedSolution

Water moves up the xylem by cohesion-tension. Cohesion (hydrogen bonds between water molecules) keeps the column intact, while adhesion (attraction between water and the cellulose wall) prevents the column from falling due to gravity. Water enters the xylem down a water potential gradient passively, and there is low (negative) hydrostatic pressure at the top of the xylem due to transpiration.

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Award 1 mark for selecting option B (statements 2 and 3 only).
PastPaper.question 6 · multiple-choice
1 PastPaper.marks
Which row correctly matches the infectious disease with both its causative pathogen type and its primary method of transmission?
  1. A.Tuberculosis | Bacterium | Inhalation of airborne droplets containing the pathogen
  2. B.Malaria | Virus | Female Anopheles mosquito vector during a blood meal
  3. C.Cholera | Bacterium | Vector transmission by common houseflies onto food
  4. D.HIV/AIDS | Protoctist | Direct exchange of bodily fluids or sharing of needles
PastPaper.showAnswers

PastPaper.workedSolution

Tuberculosis is caused by Mycobacterium tuberculosis (a bacterium) and is transmitted via airborne droplets. Malaria is caused by Plasmodium (a protoctist, not a virus). Cholera is caused by Vibrio cholerae (a bacterium) primarily transmitted via contaminated water or food (waterborne), not fly vectors. HIV/AIDS is caused by the Human Immunodeficiency Virus (a virus, not a protoctist).

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Award 1 mark for the correct disease, pathogen, and transmission mode combination (A).
PastPaper.question 7 · multiple-choice
1 PastPaper.marks
A diploid animal cell has 12 chromosomes (\(2n = 12\)). What is the number of chromosomes and the number of chromatids present in this cell during metaphase and during anaphase of mitosis?
  1. A.Metaphase: 12 chromosomes, 24 chromatids. Anaphase: 24 chromosomes, 0 chromatids.
  2. B.Metaphase: 24 chromosomes, 24 chromatids. Anaphase: 12 chromosomes, 12 chromatids.
  3. C.Metaphase: 12 chromosomes, 12 chromatids. Anaphase: 24 chromosomes, 24 chromatids.
  4. D.Metaphase: 24 chromosomes, 48 chromatids. Anaphase: 24 chromosomes, 0 chromatids.
PastPaper.showAnswers

PastPaper.workedSolution

In metaphase, there are 12 sister chromatid pairs aligned at the spindle equator, which are counted as 12 chromosomes containing 24 chromatids. During anaphase, the centromeres split and sister chromatids separate to become 24 single-stranded individual chromosomes. They are no longer referred to as chromatids because they are not joined at a centromere.

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Award 1 mark for the correct combination of chromosome and chromatid numbers (A).
PastPaper.question 8 · multiple-choice
1 PastPaper.marks
Which statement correctly distinguishes between active artificial immunity and passive natural immunity?
  1. A.Active artificial immunity involves the injection of pre-formed antibodies, whereas passive natural immunity involves the injection of a weakened pathogen.
  2. B.Active artificial immunity leads to the production of memory cells by the recipient's immune system, whereas passive natural immunity involves the transfer of antibodies across the placenta and does not produce memory cells.
  3. C.Active artificial immunity provides immediate short-term protection, whereas passive natural immunity provides delayed long-term protection.
  4. D.Active artificial immunity relies on T-lymphocyte activation only, whereas passive natural immunity relies on B-lymphocyte activation only.
PastPaper.showAnswers

PastPaper.workedSolution

Active artificial immunity (e.g., vaccination) triggers an immune response in the recipient, resulting in clonal selection, expansion, and the generation of long-lived memory cells. Passive natural immunity involves the natural transfer of maternal antibodies (e.g., across the placenta or via colostrum), which provides immediate but temporary protection without stimulating the recipient's immune system to produce memory cells.

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Award 1 mark for the correct immunological distinction (B).
PastPaper.question 9 · Multiple Choice
1 PastPaper.marks
Under a light microscope equipped with a stage micrometer and an eyepiece graticule, a student calibrates the eyepiece scale. The stage micrometer has scale divisions of \( 0.01 \) mm. Using a \( \times 10 \) objective lens, the student finds that \( 40 \) eyepiece graticule units align exactly with \( 16 \) divisions of the stage micrometer. What is the value of each eyepiece graticule unit under this objective lens?
  1. A.\( 0.25 \) micrometres
  2. B.\( 4.0 \) micrometres
  3. C.\( 25.0 \) micrometres
  4. D.\( 40.0 \) micrometres Gold value for stage alignment.
PastPaper.showAnswers

PastPaper.workedSolution

One stage micrometer division equals \( 0.01 \) mm, which is equivalent to \( 10 \) micrometres. Therefore, \( 16 \) stage micrometer divisions equal \( 16 \times 10 = 160 \) micrometres. Since \( 40 \) eyepiece graticule units align with these \( 16 \) divisions, each eyepiece graticule unit represents \( 160 / 40 = 4.0 \) micrometres.

PastPaper.markingScheme

1 mark for the correct option (B).
PastPaper.question 10 · multiple_choice
1 PastPaper.marks
Which row in the table correctly identifies the characteristics of a mitochondrion, a ribosome, and a lysosome?
  1. A.Mitochondrion: double membrane present, DNA present, hydrolytic enzymes absent; Ribosome: double membrane absent, DNA absent, hydrolytic enzymes absent; Lysosome: double membrane absent, DNA absent, hydrolytic enzymes present
  2. B.Mitochondrion: double membrane present, DNA absent, hydrolytic enzymes absent; Ribosome: double membrane absent, DNA absent, hydrolytic enzymes absent; Lysosome: double membrane present, DNA absent, hydrolytic enzymes present
  3. C.Mitochondrion: double membrane present, DNA present, hydrolytic enzymes present; Ribosome: double membrane absent, DNA present, hydrolytic enzymes absent; Lysosome: double membrane absent, DNA absent, hydrolytic enzymes present
  4. D.Mitochondrion: double membrane absent, DNA present, hydrolytic enzymes absent; Ribosome: double membrane absent, DNA absent, hydrolytic enzymes absent; Lysosome: double membrane present, DNA absent, hydrolytic enzymes present
PastPaper.showAnswers

PastPaper.workedSolution

A mitochondrion has a double membrane, contains its own DNA, and does not contain hydrolytic enzymes (which are characteristic of lysosomes). A ribosome is a non-membrane-bound organelle (no membrane), does not contain DNA, and has no hydrolytic enzymes. A lysosome is surrounded by a single membrane (double membrane is absent), contains no DNA, and contains hydrolytic enzymes. Therefore, option A is correct.

PastPaper.markingScheme

1 mark for the correct option A.
PastPaper.question 11 · multiple_choice
1 PastPaper.marks
Which statement correctly explains why a phospholipid molecule has different properties compared to a triglyceride molecule?
  1. A.One of the three fatty acids in a triglyceride is replaced by a hydrophilic phosphate-containing group in a phospholipid.
  2. B.Phospholipids contain three ester bonds per molecule, whereas triglycerides contain only two.
  3. C.Phospholipids contain a hydrophobic phosphate group, whereas triglycerides contain only hydrophilic fatty acid chains.
  4. D.Triglycerides contain a glycerol backbone, whereas phospholipids contain a different three-carbon alcohol.
PastPaper.showAnswers

PastPaper.workedSolution

In a phospholipid, one of the three fatty acids found in a triglyceride is replaced by a hydrophilic phosphate-containing group. This gives the phospholipid a polar, hydrophilic head and two non-polar, hydrophobic tails, whereas a triglyceride has three hydrophobic fatty acid tails and is entirely hydrophobic.

PastPaper.markingScheme

1 mark for the correct option A.
PastPaper.question 12 · multiple_choice
1 PastPaper.marks
An enzyme-controlled reaction was carried out at different substrate concentrations. The Michaelis-Menten constant (Km) was determined. The experiment was then repeated in the presence of a competitive inhibitor. What is the effect of the competitive inhibitor on the maximum rate of reaction (Vmax) and the Michaelis-Menten constant (Km)?
  1. A.Vmax is decreased and Km is increased
  2. B.Vmax remains unchanged and Km is increased
  3. C.Vmax is decreased and Km remains unchanged
  4. D.Vmax remains unchanged and Km is decreased
PastPaper.showAnswers

PastPaper.workedSolution

A competitive inhibitor competes with the substrate for the active site of the enzyme. At infinitely high substrate concentrations, the substrate outcompetes the inhibitor, meaning the maximum rate of reaction (Vmax) remains unchanged. However, a higher concentration of substrate is required to reach half of Vmax, which means the Michaelis-Menten constant (Km) increases (indicating a lower affinity of the enzyme for its substrate in the presence of the inhibitor).

PastPaper.markingScheme

1 mark for the correct option B.
PastPaper.question 13 · multiple_choice
1 PastPaper.marks
Plant cells with a solute potential of -1200 kPa and a pressure potential of 400 kPa are placed in a sucrose solution with a water potential of -600 kPa. Which statement correctly describes the net movement of water and the final state of the cells?
  1. A.Water enters the cells because the water potential of the surrounding solution is higher than that of the cells, making the cells more turgid.
  2. B.Water leaves the cells because the water potential of the surrounding solution is lower than that of the cells, making the cells plasmolysed.
  3. C.Water enters the cells because the water potential of the surrounding solution is lower than that of the cells, causing the cells to eventually burst.
  4. D.Water leaves the cells because the water potential of the cells is higher than that of the surrounding solution, making the cells flaccid.
PastPaper.showAnswers

PastPaper.workedSolution

The initial water potential of the plant cells is calculated using the formula: water potential = solute potential + pressure potential = -1200 kPa + 400 kPa = -800 kPa. Since the surrounding sucrose solution has a water potential of -600 kPa (which is higher/less negative than -800 kPa), water moves into the cells down a water potential gradient by osmosis. This net entry of water increases the pressure potential of the cells, making them more turgid.

PastPaper.markingScheme

1 mark for the correct option A.
PastPaper.question 14 · multiple_choice
1 PastPaper.marks
A cell from an organism with a diploid number of 2n = 8 undergoes mitosis. Which row correctly describes the number of chromosomes and the number of DNA molecules present in the cell at metaphase and at anaphase?
  1. A.Metaphase: 8 chromosomes, 16 DNA molecules; Anaphase: 16 chromosomes, 16 DNA molecules
  2. B.Metaphase: 8 chromosomes, 8 DNA molecules; Anaphase: 8 chromosomes, 8 DNA molecules
  3. C.Metaphase: 16 chromosomes, 16 DNA molecules; Anaphase: 8 chromosomes, 16 DNA molecules
  4. D.Metaphase: 8 chromosomes, 16 DNA molecules; Anaphase: 8 chromosomes, 16 DNA molecules
PastPaper.showAnswers

PastPaper.workedSolution

At metaphase of mitosis, there are 8 chromosomes, each consisting of two sister chromatids. Since each chromatid contains one double-stranded DNA molecule, there are 16 DNA molecules in total. At anaphase, the sister chromatids separate and are pulled to opposite poles. Once separated, each chromatid is counted as an individual chromosome. Therefore, there are now 16 chromosomes in the cell, but the total number of DNA molecules remains 16. Hence, option A is correct.

PastPaper.markingScheme

1 mark for the correct option A.
PastPaper.question 15 · multiple_choice
1 PastPaper.marks
A double-stranded DNA molecule contains 22% adenine. What is the correct percentage of cytosine and the total percentage of purine bases in this DNA molecule?
  1. A.Cytosine = 28%, Total purines = 50%
  2. B.Cytosine = 28%, Total purines = 22%
  3. C.Cytosine = 22%, Total purines = 50%
  4. D.Cytosine = 28%, Total purines = 72%
PastPaper.showAnswers

PastPaper.workedSolution

By complementary base pairing, if Adenine (A) = 22%, then Thymine (T) must also be 22%. Together, A + T = 44%. The remaining bases (Guanine (G) and Cytosine (C)) make up 100% - 44% = 56%. Since G = C, the percentage of Cytosine is 56% / 2 = 28%. Purines are double-ring bases (Adenine and Guanine). In any double-stranded DNA molecule, total purines (A + G) always equal 50% of the total bases because A = T and G = C. Checking this: 22% (A) + 28% (G) = 50%.

PastPaper.markingScheme

1 mark for the correct option A.
PastPaper.question 16 · multiple_choice
1 PastPaper.marks
Which processes are involved in the active loading of sucrose into phloem sieve tube elements at a source plant cell? 1. Active transport of protons (H+) out of the companion cells into the cell wall. 2. Co-transport of protons (H+) and sucrose back into the companion cells down a proton gradient. 3. Diffusion of sucrose from companion cells into sieve tube elements through plasmodesmata.
  1. A.1, 2 and 3
  2. B.1 and 2 only
  3. C.2 and 3 only
  4. D.1 and 3 only
PastPaper.showAnswers

PastPaper.workedSolution

All three statements are correct. First, protons (H+) are actively pumped out of the companion cells into the apoplast (cell wall pathway). Second, protons diffuse back into the companion cells down their concentration gradient through co-transporter proteins, carrying sucrose along with them. Third, once inside the companion cells, sucrose diffuses into the sieve tube elements via the plasmodesmata. Thus, 1, 2, and 3 are all involved.

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1 mark for the correct option A.
PastPaper.question 17 · multiple_choice
1 PastPaper.marks
During which phase of the cardiac cycle is the blood pressure in the left ventricle higher than the blood pressure in the aorta?
  1. A.When the atrioventricular (bicuspid) valve is open.
  2. B.When the semi-lunar valve of the aorta is open.
  3. C.When the left ventricle is undergoing diastole.
  4. D.When the left atrium is contracting.
PastPaper.showAnswers

PastPaper.workedSolution

Blood flows down a pressure gradient. In order for the semi-lunar valve of the aorta to open and allow blood to be ejected from the left ventricle into the systemic circulation, the pressure in the left ventricle must exceed the pressure in the aorta. Thus, when the semi-lunar valve of the aorta is open, the ventricular pressure is higher than the aortic pressure.

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1 mark for the correct option B.
PastPaper.question 18 · multiple-choice
1 PastPaper.marks
Which of the following statements about membrane components and their effect on cell membrane fluidity is correct?
  1. A.An increase in cholesterol content always increases membrane fluidity at all temperatures.
  2. B.Phospholipids with longer fatty acid chains increase membrane fluidity because they have more hydrophobic interactions.
  3. C.High proportions of unsaturated fatty acid tails increase membrane fluidity because their 'kinks' prevent close packing of lipids.
  4. D.Transmembrane proteins decrease fluidity because they are covalently bonded to the polar head groups of the phospholipids.
PastPaper.showAnswers

PastPaper.workedSolution

Unsaturated fatty acids contain one or more carbon-carbon double bonds, which introduce a rigid 'kink' in the hydrocarbon tail. These kinks prevent the phospholipid molecules from packing closely together, thereby maintaining membrane fluidity even at lower temperatures. Cholesterol acts as a buffer for membrane fluidity, decreasing it at high temperatures and preventing solidifying at low temperatures. Longer fatty acid chains increase the van der Waals interactions between tails, which actually decreases membrane fluidity. Transmembrane proteins are held in place by hydrophobic and hydrophilic interactions, not covalent bonds.

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[1 mark] - Correct letter C chosen.
PastPaper.question 19 · multiple-choice
1 PastPaper.marks
An exocrine pancreatic cell synthesizes and secretes the digestive enzyme amylase. What is the correct sequence of organelles and structures through which this newly synthesized protein travels before it is released from the cell?
  1. A.Rough endoplasmic reticulum \(\rightarrow\) transport vesicle \(\rightarrow\) Golgi body \(\rightarrow\) secretory vesicle \(\rightarrow\) cell surface membrane
  2. B.Nucleolus \(\rightarrow\) rough endoplasmic reticulum \(\rightarrow\) Golgi body \(\rightarrow\) secretory vesicle \(\rightarrow\) cell surface membrane
  3. C.Smooth endoplasmic reticulum \(\rightarrow\) transport vesicle \(\rightarrow\) Golgi body \(\rightarrow\) secretory vesicle \(\rightarrow\) cell surface membrane
  4. D.Rough endoplasmic reticulum \(\rightarrow\) secretory vesicle \(\rightarrow\) lysosome \(\rightarrow\) transport vesicle \(\rightarrow\) cell surface membrane
PastPaper.showAnswers

PastPaper.workedSolution

Proteins destined for secretion are synthesized by ribosomes bound to the rough endoplasmic reticulum (RER). After entering the lumen of the RER, they are packaged into transport vesicles and moved to the Golgi body for post-translational modification (e.g., glycosylation). From the Golgi body, the modified proteins are packaged into secretory vesicles, which then move to and fuse with the cell surface membrane, releasing the protein via exocytosis.

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[1 mark] - Correct letter A chosen.
PastPaper.question 20 · multiple-choice
1 PastPaper.marks
A student performs four biochemical tests on an unknown liquid food sample. The results are shown below:

* **Iodine test:** yellow-brown
* **Benedict's test on original sample (heated):** blue
* **Benedict's test on sample after acid hydrolysis and neutralisation (heated):** brick-red precipitate
* **Biuret test:** purple
* **Emulsion test:** cloudy/milky emulsion

Which biological molecules are present in this food sample?
  1. A.reducing sugar, non-reducing sugar, and protein
  2. B.non-reducing sugar, protein, and lipid
  3. C.starch, reducing sugar, and lipid
  4. D.starch, non-reducing sugar, and protein
PastPaper.showAnswers

PastPaper.workedSolution

1. The Iodine test remaining yellow-brown indicates that starch is absent.
2. The original Benedict's test remaining blue indicates that reducing sugars are absent.
3. The Benedict's test turning brick-red after acid hydrolysis (which breaks glycosidic bonds) and neutralisation indicates that non-reducing sugars (e.g., sucrose) are present.
4. The Biuret test turning purple indicates that protein is present.
5. The Emulsion test forming a cloudy emulsion indicates that lipid is present.
Therefore, the molecules present are non-reducing sugar, protein, and lipid.

PastPaper.markingScheme

[1 mark] - Correct letter B chosen.
PastPaper.question 21 · multiple-choice
1 PastPaper.marks
Which row correctly matches the infectious disease with its pathogen type and its primary method of transmission?
  1. A.Cholera | Bacterium | Vector-borne
  2. B.Malaria | Protoctist | Water-borne
  3. C.Tuberculosis | Bacterium | Airborne droplets
  4. D.HIV/AIDS | Virus | Vector-borne
PastPaper.showAnswers

PastPaper.workedSolution

A is incorrect because Cholera is caused by a bacterium (Vibrio cholerae) but is transmitted primarily via water-borne or food-borne routes, not vector-borne. B is incorrect because Malaria is caused by a protoctist (Plasmodium spp.) and is vector-borne, not water-borne. C is correct because Tuberculosis is caused by the bacterium Mycobacterium tuberculosis and is transmitted via airborne droplets. D is incorrect because HIV/AIDS is caused by a virus (HIV) and is transmitted by exchange of body fluids (not vectors).

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[1 mark] - Correct letter C chosen.
PastPaper.question 22 · multiple-choice
1 PastPaper.marks
Which of the following explains the shift of the oxyhaemoglobin dissociation curve to the right (the Bohr effect) in actively respiring tissues?
  1. A.An increase in carbon dioxide concentration decreases the blood pH, reducing the affinity of haemoglobin for oxygen.
  2. B.A decrease in carbon dioxide concentration increases the blood pH, reducing the affinity of haemoglobin for oxygen.
  3. C.An increase in oxygen concentration increases the affinity of haemoglobin for carbon dioxide.
  4. D.A decrease in temperature increases the affinity of haemoglobin for oxygen, shifting the curve to the right.
PastPaper.showAnswers

PastPaper.workedSolution

In actively respiring tissues, carbon dioxide is produced. This CO2 diffuses into red blood cells and reacts with water to form carbonic acid (catalysed by carbonic anhydrase), which dissociates into hydrogen ions (H+) and hydrogencarbonate ions. The increase in hydrogen ions decreases the blood pH (increases acidity). These hydrogen ions bind to haemoglobin, altering its tertiary structure and reducing its affinity for oxygen. Consequently, haemoglobin releases oxygen more readily, shifting the oxyhaemoglobin dissociation curve to the right.

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[1 mark] - Correct letter A chosen.
PastPaper.question 23 · multiple-choice
1 PastPaper.marks
During the movement of water across the root cortex of a plant, what characterizes the apoplast pathway?
  1. A.Water moves from the cytoplasm of one cell to another via plasmodesmata.
  2. B.Water moves through the non-living cell walls and intercellular spaces until it reaches the Casparian strip.
  3. C.Water moves through the cytoplasm and vacuoles of cells, crossing the tonoplast membrane.
  4. D.Water moves against a water potential gradient using active transport proteins in the cell wall.
PastPaper.showAnswers

PastPaper.workedSolution

The apoplast pathway involves the movement of water and dissolved mineral ions through the non-living parts of the plant, specifically the cellulose cell walls and intercellular spaces. This movement is rapid and driven by cohesion-tension. However, upon reaching the endodermis, the apoplast pathway is blocked by the suberin-rich, impermeable Casparian strip, which forces water to pass across the cell surface membrane and enter the symplast pathway.

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[1 mark] - Correct letter B chosen.
PastPaper.question 24 · multiple-choice
1 PastPaper.marks
Four events that occur during the mitotic cell cycle are described below:

1. Centromeres split and sister chromatids are pulled to opposite poles of the spindle.
2. Chromosomes condense and become visible; spindle fibres begin to form.
3. Chromosomes align individually along the equator of the spindle.
4. Nuclear envelopes reform around each group of newly separated chromosomes.

Which row correctly identifies the stages of mitosis in which these events occur?
  1. A.1 = Anaphase, 2 = Prophase, 3 = Metaphase, 4 = Telophase
  2. B.1 = Metaphase, 2 = Prophase, 3 = Anaphase, 4 = Telophase
  3. C.1 = Anaphase, 2 = Interphase, 3 = Metaphase, 4 = Telophase
  4. D.1 = Telophase, 2 = Prophase, 3 = Metaphase, 4 = Anaphase
PastPaper.showAnswers

PastPaper.workedSolution

Event 1 corresponds to Anaphase, where centromeres divide and chromatids (now individual chromosomes) are pulled apart to opposite poles. Event 2 corresponds to Prophase, which involves chromatin condensation and the start of spindle formation. Event 3 corresponds to Metaphase, where chromosomes line up individually at the equator. Event 4 corresponds to Telophase, where the nuclear envelopes reform around the two new nuclei. This corresponds to the sequence: 1 = Anaphase, 2 = Prophase, 3 = Metaphase, 4 = Telophase.

PastPaper.markingScheme

[1 mark] - Correct letter A chosen.
PastPaper.question 25 · multiple-choice
1 PastPaper.marks
Which statement correctly describes the difference between a competitive inhibitor and a non-competitive inhibitor of an enzyme?
  1. A.Competitive inhibitors bind to the active site and can be overcome by increasing substrate concentration; non-competitive inhibitors bind to an allosteric site and cannot be overcome by increasing substrate concentration.
  2. B.Competitive inhibitors bind to an allosteric site and decrease \(V_{\text{max}}\); non-competitive inhibitors bind to the active site and leave \(V_{\text{max}}\) unchanged.
  3. C.Competitive inhibitors increase the activation energy of the reaction; non-competitive inhibitors have no effect on the activation energy.
  4. D.Competitive inhibitors form strong covalent bonds with the active site; non-competitive inhibitors always form temporary hydrogen bonds with the substrate.
PastPaper.showAnswers

PastPaper.workedSolution

Competitive inhibitors have a similar shape to the substrate and bind to the active site. Because they compete with the substrate, increasing the substrate concentration can overcome the inhibition and allow the reaction to reach the maximum rate (\(V_{\text{max}}\)). Non-competitive inhibitors bind to an allosteric site (a site other than the active site), changing the tertiary structure and shape of the active site so the substrate can no longer bind. Since they do not compete for the active site, increasing substrate concentration cannot overcome non-competitive inhibition, and \(V_{\text{max}}\) is reduced.

PastPaper.markingScheme

[1 mark] - Correct letter A chosen.
PastPaper.question 26 · multiple-choice
1 PastPaper.marks
Which row correctly matches the membrane component with its description and function?
  1. A.Glycoprotein | Has carbohydrate chains attached to a protein | Involved in cell-to-cell recognition
  2. B.Cholesterol | A hydrophobic lipid molecule within the bilayer | Increases membrane fluidity at high temperatures
  3. C.Phospholipid | A molecule with a hydrophobic phosphate head | Forms a barrier to water-soluble substances
  4. D.Channel protein | A peripheral membrane protein | Allows active transport of ions against a concentration gradient
PastPaper.showAnswers

PastPaper.workedSolution

Glycoproteins consist of carbohydrate chains covalently bonded to proteins. They project into the extracellular fluid and act as receptor molecules in cell-to-cell recognition. Cholesterol stabilizes membrane fluidity and actually decreases membrane fluidity at high temperatures. Phospholipids have hydrophilic (water-attracting) phosphate heads and hydrophobic fatty acid tails. Channel proteins are integral membrane proteins involved in passive facilitated diffusion, not active transport.

PastPaper.markingScheme

1 mark for identifying the correct matching row. Reject options B, C, and D due to incorrect properties of cholesterol, phospholipids, or channel proteins.
PastPaper.question 27 · multiple-choice
1 PastPaper.marks
Which features are found in both mitochondria and chloroplasts, but are NOT found in the nucleus of a eukaryotic cell?

1. Circular DNA
2. 70S ribosomes
3. Double outer membrane
  1. A.1 and 2 only
  2. B.1 and 3 only
  3. C.2 and 3 only
  4. D.1, 2 and 3
PastPaper.showAnswers

PastPaper.workedSolution

Both mitochondria and chloroplasts contain circular DNA and 70S ribosomes, reflecting their evolutionary origin from endosymbiotic prokaryotes. In contrast, the nucleus of a eukaryotic cell contains linear DNA associated with histone proteins, and it does not contain 70S ribosomes. However, all three structures (mitochondria, chloroplasts, and the nucleus) are surrounded by a double outer membrane (the nuclear envelope in the case of the nucleus). Therefore, features 1 and 2 are correct, while feature 3 is incorrect because a double membrane is present in the nucleus.

PastPaper.markingScheme

1 mark for identifying that circular DNA and 70S ribosomes are absent in the nucleus but present in both mitochondria and chloroplasts, while the double membrane is common to all three.
PastPaper.question 28 · multiple-choice
1 PastPaper.marks
An unknown solution was subjected to four different food tests. The results were as follows:
- Test 1: Iodine in potassium iodide solution was added: the solution remained yellow-brown.
- Test 2: Biuret reagent was added: the solution turned purple.
- Test 3: Benedict's solution was added and heated to \(80\ ^\circ\text{C}\): the solution remained blue.
- Test 4: A second sample of the solution was boiled with dilute hydrochloric acid, neutralized with sodium hydrogencarbonate, and then heated with Benedict's solution to \(80\ ^\circ\text{C}\): the solution turned red.

Which biological molecules are present in the unknown solution?
  1. A.starch, protein, reducing sugar
  2. B.starch, non-reducing sugar, protein
  3. C.protein and non-reducing sugar only
  4. D.reducing sugar and non-reducing sugar only
PastPaper.showAnswers

PastPaper.workedSolution

The negative iodine test (yellow-brown) indicates that starch is absent. The positive Biuret test (purple) indicates that protein is present. The negative direct Benedict's test (remained blue) indicates that reducing sugars are absent. The positive Benedict's test after acid hydrolysis (turned red) indicates that a non-reducing sugar was present and was hydrolyzed into reducing sugars. Therefore, the solution contains protein and non-reducing sugar only.

PastPaper.markingScheme

1 mark for analyzing all four qualitative food tests and identifying that only protein and non-reducing sugar are present in the solution.
PastPaper.question 29 · multiple-choice
1 PastPaper.marks
The relative amounts of elastic fibers, smooth muscle, and collagen in the walls of four different blood vessels, P, Q, R, and S, are described below:
- Vessel P: high elastic fibers, medium smooth muscle, high collagen
- Vessel Q: low elastic fibers, high smooth muscle, low collagen
- Vessel R: absent elastic fibers, absent smooth muscle, absent collagen
- Vessel S: low elastic fibers, low smooth muscle, medium collagen

Which row correctly identifies blood vessels P, Q, R, and S?
  1. A.P = arteriole, Q = large artery, R = capillary, S = vein
  2. B.P = large artery, Q = arteriole, R = capillary, S = vein
  3. C.P = large artery, Q = vein, R = capillary, S = arteriole
  4. D.P = vein, Q = arteriole, R = capillary, S = large artery
PastPaper.showAnswers

PastPaper.workedSolution

Vessel P must be a large artery (such as the aorta) because it requires high amounts of elastic fibers to stretch and recoil under high pressure. Vessel Q must be an arteriole because it has a relatively thick wall of smooth muscle to control blood flow via vasoconstriction and vasodilation, but has low elastic fibers. Vessel R must be a capillary because its wall is only one cell thick (endothelium) and entirely lacks elastic fibers, smooth muscle, and collagen. Vessel S must be a vein, which has thin walls with low elastic fibers and smooth muscle, and medium collagen.

PastPaper.markingScheme

1 mark for correctly matching the wall composition of P, Q, R, and S to a large artery, arteriole, capillary, and vein, respectively.
PastPaper.question 30 · multiple-choice
1 PastPaper.marks
Which row correctly describes transport in the xylem and transport in the phloem of a flowering plant?
  1. A.Main mechanism of bulk flow: Xylem = Transpiration pull created by negative pressure (tension) | Phloem = Hydrostatic pressure gradient created by active loading of sucrose
  2. B.State of transport cells: Xylem = Living cells containing cytoplasm but no nucleus | Phloem = Dead cells with no cell contents
  3. C.Direction of transport: Xylem = Bidirectional depending on the season | Phloem = Unidirectional from roots to leaves
  4. D.Energy requirement for bulk flow: Xylem = Direct active transport using ATP from companion cells | Phloem = Passive movement requiring no metabolic energy at any stage
PastPaper.showAnswers

PastPaper.workedSolution

Bulk flow in the xylem is driven by transpiration pull (tension) created by negative pressure as water evaporates from leaves. Bulk flow in the phloem is driven by a hydrostatic pressure gradient created by the active loading of sucrose at source tissues. Xylem vessels consist of dead cells with no cytoplasm, whereas phloem sieve tube elements consist of living cells containing cytoplasm. Xylem transport is unidirectional (upwards), while phloem transport is bidirectional (from source to sink). Phloem transport requires metabolic energy (ATP) for loading, whereas xylem transport is entirely passive.

PastPaper.markingScheme

1 mark for identifying that transpiration pull drives xylem bulk flow while a hydrostatic pressure gradient drives phloem bulk flow.
PastPaper.question 31 · multiple-choice
1 PastPaper.marks
Which option correctly describes the tissues present in the wall of a bronchus and the wall of an alveolus?
  1. A.In a bronchus, cartilage, ciliated epithelium, smooth muscle, and elastic fibers are all present. In an alveolus, cartilage, ciliated epithelium, and smooth muscle are absent, but elastic fibers are present.
  2. B.In a bronchus, cartilage is present, but ciliated epithelium, smooth muscle, and elastic fibers are absent. In an alveolus, ciliated epithelium is present, but cartilage and smooth muscle are absent.
  3. C.In a bronchus, cartilage and smooth muscle are absent, but ciliated epithelium and elastic fibers are present. In an alveolus, smooth muscle and elastic fibers are present, but cartilage is absent.
  4. D.In a bronchus, cartilage, ciliated epithelium, and elastic fibers are present, but smooth muscle is absent. In an alveolus, cartilage is present, but elastic fibers are absent.
PastPaper.showAnswers

PastPaper.workedSolution

The wall of a bronchus contains cartilage plates to keep the airway open, ciliated epithelium with goblet cells to trap and sweep away pathogens, smooth muscle to regulate airway diameter, and elastic fibers. Thus, all four tissues are present. The wall of an alveolus consists of a single layer of squamous epithelium (non-ciliated) and elastic fibers to allow stretch and recoil during breathing, but entirely lacks cartilage and smooth muscle.

PastPaper.markingScheme

1 mark for correctly identifying the presence/absence of all four tissue types in both the bronchus and alveolus.
PastPaper.question 32 · multiple-choice
1 PastPaper.marks
In the production of monoclonal antibodies, B-lymphocytes (plasma cells) are fused with cancer cells (myeloma cells) to form hybridoma cells. What is the reason for this fusion?
  1. A.Plasma cells produce the specific antibody but do not divide in culture, while myeloma cells divide continuously but do not produce the specific antibody.
  2. B.Plasma cells divide continuously but do not produce antibodies, while myeloma cells produce the specific antibody but do not divide.
  3. C.Fusion combines the antigen-binding sites of the myeloma cells with the constant regions of the plasma cell antibodies.
  4. D.Fusion allows the myeloma cells to undergo meiosis to produce genetically diverse antibodies.
PastPaper.showAnswers

PastPaper.workedSolution

Plasma cells are specialized, terminally differentiated cells that produce a single type of specific antibody, but they are short-lived and cannot divide in culture. Myeloma cells are cancerous plasma cells that can divide continuously and indefinitely in culture, but they do not produce the specific antibody of interest. Fusing them produces hybridoma cells, which combine the ability to divide indefinitely with the ability to synthesize and secrete the specific antibody.

PastPaper.markingScheme

1 mark for explaining that fusion combines the specific antibody production of plasma cells with the continuous division of myeloma cells.
PastPaper.question 33 · multiple-choice
1 PastPaper.marks
A double-stranded DNA molecule contains \(20\%\) adenine.

What is the percentage of cytosine in this DNA molecule, and how many hydrogen bonds would be found in a section of this DNA that is 100 base pairs long?
  1. A.cytosine = \(30\%\); hydrogen bonds = 260
  2. B.cytosine = \(30\%\); hydrogen bonds = 240
  3. C.cytosine = \(20\%\); hydrogen bonds = 260
  4. D.cytosine = \(20\%\); hydrogen bonds = 240
PastPaper.showAnswers

PastPaper.workedSolution

1. Calculate the percentage of cytosine: Since DNA is double-stranded, adenine (A) is equal to thymine (T), so \(A = 20\%\) and \(T = 20\%\). This leaves \(100\% - (20\% + 20\%) = 60\%\) of the DNA for guanine (G) and cytosine (C). Since \(G = C\), the percentage of cytosine is \(60\% \div 2 = 30\%\).
2. Calculate the number of hydrogen bonds in a 100 base pair (bp) long section: 100 bp means there are 200 total bases. Since \(20\%\) of the bases are adenine, there are \(20\%\) of 200 = 40 adenine bases. Therefore, there are 40 A-T base pairs. Since \(30\%\) of the bases are cytosine, there are \(30\%\) of 200 = 60 cytosine bases, meaning there are 60 G-C base pairs. Each A-T base pair contains 2 hydrogen bonds, and each G-C base pair contains 3 hydrogen bonds. Total hydrogen bonds = \((40 \times 2) + (60 \times 3) = 80 + 180 = 260\).

PastPaper.markingScheme

1 mark for correctly calculating the percentage of cytosine as \(30\%\) and the number of hydrogen bonds in a 100 bp DNA section as 260.

Paper 21 (AS Level Structured)

Answer all six structured questions. Write answers in the spaces provided on the question paper.
6 PastPaper.question · 60 PastPaper.marks
PastPaper.question 1 · Structured
10 PastPaper.marks
(a) Insulin is a protein hormone secreted by pancreatic beta cells. Describe the roles of the rough endoplasmic reticulum (RER) and the Golgi apparatus in the synthesis, folding, and packaging of insulin. [4] (b) Microtubules form part of the cytoskeleton. Explain the role of microtubules in the pathway of insulin secretion from these cells. [2] (c) The secretion of insulin is an active process requiring ATP. Describe how the structure of a mitochondrion is adapted to maximize ATP production. [4]
PastPaper.showAnswers

PastPaper.workedSolution

(a) Ribosomes on the RER translate the mRNA coding for insulin into a polypeptide chain, which enters the RER lumen where it folds into its tertiary conformation. Transport vesicles containing the folded protein bud off the RER and fuse with the Golgi apparatus. In the Golgi, the protein undergoes post-translational modification (cleavage into active insulin) and is packaged into secretory vesicles. (b) Microtubules act as physical tracks along which transport and secretory vesicles are guided, using motor proteins, from the Golgi apparatus to the cell surface membrane for exocytosis. (c) The double membrane structure creates an intermembrane space for high proton concentration. The highly folded inner membrane (cristae) increases surface area for ATP synthase complexes and electron transport chain proteins. The matrix contains soluble enzymes, coenzymes, and circular DNA necessary for the Krebs cycle and Link reaction.

PastPaper.markingScheme

(a) 1 mark for mentioning translation on RER ribosomes; 1 mark for folding of the polypeptide inside the RER lumen; 1 mark for vesicle transport from RER to Golgi; 1 mark for Golgi modification and packaging into secretory vesicles. (Max 4 marks) (b) 1 mark for microtubules acting as tracks for vesicle movement; 1 mark for specifying movement from Golgi to the cell surface membrane/exocytosis. (Max 2 marks) (c) 1 mark for cristae increasing surface area; 1 mark for ATP synthase/electron carriers on the inner membrane; 1 mark for compartmentalization of intermembrane space for proton gradient; 1 mark for fluid matrix containing Krebs cycle enzymes. (Max 4 marks)
PastPaper.question 2 · Structured
10 PastPaper.marks
(a) In the mammalian small intestine, glucose is absorbed into epithelial cells via cotransport. Describe how the active transport of sodium ions out of epithelial cells into the blood establishes a concentration gradient that drives this cotransport. [3] (b) Explain how glucose is transported from the lumen of the intestine into the epithelial cells against its concentration gradient. [4] (c) Contrast facilitated diffusion with active transport. [3]
PastPaper.showAnswers

PastPaper.workedSolution

(a) Sodium-potassium pump proteins in the basolateral membrane of the epithelial cells actively transport \( \text{Na}^+ \) ions out of the cell into the blood, using energy from ATP hydrolysis. This lowers the concentration of \( \text{Na}^+ \) inside the cytoplasm relative to the lumen of the intestine, establishing a steep electrochemical/concentration gradient for \( \text{Na}^+ \). (b) \( \text{Na}^+ \) ions in the lumen bind to sodium-glucose cotransporter proteins (symporters) on the microvilli/apical membrane. Glucose also binds to the same cotransporter protein. \( \text{Na}^+ \) moves down its concentration gradient into the cell, which provides the energy to transport glucose against its concentration gradient into the cell. (c) Facilitated diffusion is passive (no ATP required) and moves substances down their concentration gradient, utilizing both channel and carrier proteins. Active transport requires metabolic energy (ATP) to move substances against their concentration gradient, utilizing only carrier proteins/pumps.

PastPaper.markingScheme

(a) 1 mark for active transport of \( \text{Na}^+ \) out of cells via sodium-potassium pump; 1 mark for energy source being ATP hydrolysis; 1 mark for lowering internal sodium concentration to create a steep gradient from lumen to cell. (Max 3 marks) (b) 1 mark for sodium-glucose cotransporter/symporter proteins on microvilli; 1 mark for binding of both sodium and glucose; 1 mark for sodium moving down its gradient; 1 mark for this movement driving the transport of glucose against its gradient. (Max 4 marks) (c) 1 mark for active transport requiring ATP vs facilitated diffusion being passive; 1 mark for active transport moving against gradient vs facilitated diffusion moving down gradient; 1 mark for facilitated diffusion using channels or carrier proteins vs active transport using carrier proteins/pumps only. (Max 3 marks)
PastPaper.question 3 · Structured
10 PastPaper.marks
(a) Cholera is an infectious disease. State the genus and species of the pathogen that causes cholera, and explain how it is transmitted between hosts. [3] (b) Describe how the toxin produced by this pathogen interacts with epithelial cells in the small intestine to cause severe watery diarrhoea. [4] (c) Explain why oral rehydration therapy (ORT) is an effective treatment for cholera, even though it does not destroy the cholera pathogen. [3]
PastPaper.showAnswers

PastPaper.workedSolution

(a) The disease is caused by the bacterium Vibrio cholerae. Transmission occurs via the faecal-oral route, typically when an uninfected person drinks water or eats food contaminated with the faeces of an infected individual containing the pathogen. (b) The bacterium releases cholera toxin (choleragen). The toxin binds to GM1 ganglioside receptors on the surface membrane of intestinal epithelial cells. This triggers the activation of adenylate cyclase, raising intracellular cyclic AMP (cAMP) levels. Elevated cAMP opens CFTR chloride channels, causing active secretion of chloride ions (\( \text{Cl}^- \)) into the intestinal lumen, which lowers the water potential of the lumen, causing water to move out of the epithelial cells and blood into the lumen by osmosis. (c) ORT contains water, glucose, and sodium salts in balanced concentrations. Sodium ions and glucose are cotransported into epithelial cells, which lowers the water potential of the epithelial cells and tissue fluid. Water follows by osmosis from the intestinal lumen back into the body tissues, rehydrating the patient and preventing circulatory shock while the immune system clears the bacterial infection.

PastPaper.markingScheme

(a) 1 mark for Vibrio cholerae (must be correctly spelled, genus capitalized); 1 mark for transmission via contaminated water/food; 1 mark for faecal-oral route explanation. (Max 3 marks) (b) 1 mark for toxin binding to receptors (GM1) on epithelial cells; 1 mark for activation of adenylate cyclase and increase in cAMP; 1 mark for opening of chloride channels and efflux of chloride ions into the lumen; 1 mark for water moving down its water potential gradient into the lumen by osmosis. (Max 4 marks) (c) 1 mark for ORT containing sodium ions and glucose; 1 mark for cotransport of sodium and glucose into epithelial cells; 1 mark for water moving back into the tissues/blood by osmosis to prevent dehydration. (Max 3 marks)
PastPaper.question 4 · Structured
10 PastPaper.marks
(a) The cohesion-tension theory explains the movement of water up xylem vessels. Describe how the chemical properties of water molecules contribute to this movement. [4] (b) A student used a potometer to investigate the rate of water uptake in a woody stem. Explain why the rate of water uptake measured by the potometer may not be exactly equal to the rate of transpiration. [2] (c) Describe how three structural adaptations of xerophytic leaves reduce the rate of transpiration in arid environments. [4]
PastPaper.showAnswers

PastPaper.workedSolution

(a) Water molecules are polar because oxygen is more electronegative than hydrogen, creating partial charges. This allows hydrogen bonds to form between adjacent water molecules, resulting in cohesion, which maintains a continuous, unbroken column of water in the xylem. Water molecules also form hydrogen bonds with the hydrophilic cellulose of the xylem vessel walls, causing adhesion, which prevents the column from falling back down under gravity. Tension is established as water evaporates from mesophyll walls into air spaces. (b) The potometer measures the rate of water uptake, which is not identical to transpiration because some water is used by the plant cells for photosynthesis, some is used to maintain cell turgidity (causing growth/cell expansion), and some is produced internally during respiration. (c) Adaptations include: 1. Sunken stomata in pits, which trap moist air to reduce the water potential gradient. 2. Rolled leaves, which shield stomata from air currents and trap humid air. 3. Thick waxy cuticle, which acts as a hydrophobic barrier to reduce cuticular water loss. 4. Reduced leaf surface area (or spine-like leaves), which minimizes the area available for evaporation.

PastPaper.markingScheme

(a) 1 mark for dipolar nature of water/hydrogen bonding; 1 mark for cohesion holding water molecules together in a continuous column; 1 mark for adhesion of water to xylem cell walls (cellulose); 1 mark for tension created by evaporation pulling the water column up. (Max 4 marks) (b) 1 mark for noting water used in photosynthesis / maintaining cell turgidity; 1 mark for noting water is produced by aerobic respiration. (Max 2 marks) (c) Award 1 mark per adaptation and explanation pair: e.g., sunken stomata / rolled leaves trap a layer of humid air, reducing the water potential gradient (1 mark); thick waxy cuticle reduces cuticular transpiration (1 mark); hairs on leaves trap moist air/reduce air movement (1 mark); reduced surface area/needles minimize evaporative surface area (1 mark). (Max 4 marks)
PastPaper.question 5 · Structured
10 PastPaper.marks
(a) Describe the behaviour of chromosomes during metaphase and anaphase of mitosis. [4] (b) Explain the distinct roles of the centromere and the spindle microtubules during the mitotic cell cycle. [3] (c) Explain why DNA replication must occur during the S phase of interphase before mitosis can proceed. [3]
PastPaper.showAnswers

PastPaper.workedSolution

(a) During metaphase, chromosomes (each made of two sister chromatids) condense and align individually along the equator (metaphase plate) of the cell. Spindle microtubules attach to the kinetochores of the centromeres. During anaphase, the centromeres split. Spindle microtubules shorten/contract, pulling the sister chromatids (now individual chromosomes) apart, centromere first, to opposite poles of the spindle. (b) The centromere holds two sister chromatids together until they are ready to separate and provides the site of assembly for the kinetochore where spindle microtubules attach. The spindle microtubules polymerize and depolymerize to move chromosomes during metaphase and contract to pull chromatids apart during anaphase, ensuring accurate segregation. (c) DNA replication is essential to synthesize an identical copy of each chromosome (forming sister chromatids). This ensures that when the cell divides during mitosis, each of the two daughter cells receives a complete, identical diploid set of genetic information, maintaining genetic stability across generations of cells.

PastPaper.markingScheme

(a) 1 mark for chromosomes aligning along the cell equator in metaphase; 1 mark for spindle attachment to centromeres; 1 mark for splitting of centromeres in anaphase; 1 mark for sister chromatids being pulled to opposite poles. (Max 4 marks) (b) 1 mark for centromere holding sister chromatids together; 1 mark for centromere/kinetochore acting as attachment site for spindle; 1 mark for spindle microtubules contracting/shortening to pull chromatids apart. (Max 3 marks) (c) 1 mark for doubling the quantity of DNA/producing sister chromatids; 1 mark for ensuring both daughter cells receive an exact/identical copy of the genome; 1 mark for maintaining genetic stability / diploid chromosome number. (Max 3 marks)
PastPaper.question 6 · Structured
10 PastPaper.marks
(a) State the biochemical nature of antibodies and describe how their molecular structure is related to their specific functions. [5] (b) Distinguish between active immunity and passive immunity, providing one example of how each type can be acquired. [5]
PastPaper.showAnswers

PastPaper.workedSolution

(a) Antibodies are globular glycoproteins. They consist of four polypeptide chains: two identical heavy chains and two identical light chains held together by covalent disulfide bonds. Each antibody has: 1. A variable region (Fab) with a unique tertiary structure that forms antigen-binding sites complementary to a specific antigen, allowing selective binding. 2. A constant region (Fc) that binds to receptors on phagocytic cells (such as macrophages), facilitating opsonisation and phagocytosis. 3. A hinge region that provides flexibility, enabling the antibody to bind to antigens spaced at varying distances or to cross-link pathogens (agglutination). (b) Active immunity occurs when an individual's own immune system is stimulated to produce antibodies and memory cells in response to an antigen. This requires exposure to the antigen, activation of B/T lymphocytes, and takes time to develop, but provides long-term protection. An example is acquiring immunity after recovering from an infection (natural) or via vaccination (artificial). Passive immunity occurs when an individual receives pre-formed antibodies produced by another organism. It does not involve lymphocyte activation or memory cell production, providing immediate but short-term protection as the antibodies are metabolized. An example is antibodies passing across the placenta/colostrum to a fetus (natural) or an injection of monoclonal antibodies/antitoxin (artificial).

PastPaper.markingScheme

(a) 1 mark for describing the four-chain structure (two heavy, two light) held by disulfide bonds; 1 mark for variable regions forming antigen-binding sites complementary to specific antigens; 1 mark for constant region binding to receptors on phagocytes; 1 mark for hinge region allowing flexibility/agglutination; 1 mark for identifying antibodies as glycoproteins. (Max 5 marks) (b) 1 mark for active immunity involving antigen exposure and lymphocyte activation vs passive involving direct receipt of antibodies; 1 mark for active immunity producing memory cells vs passive producing no memory cells; 1 mark for active being long-term vs passive being short-term/temporary; 1 mark for active example (natural infection or vaccination); 1 mark for passive example (breast milk, placenta, or injection of antibodies/monoclonal antibodies). (Max 5 marks)

Paper 31 (Advanced Practical)

Answer both experimental questions. Complete practical setups, document observations in tables, plot graphs, and answer analytical sub-questions.
2 PastPaper.question · 40 PastPaper.marks
PastPaper.question 1 · practical
20 PastPaper.marks
You are required to investigate the effect of concentration of acid on the rate of diffusion into model cells. You are provided with:
- **H**: 1.0 mol dm^-3 hydrochloric acid solution
- **W**: Distilled water
- **A**: Three blocks of agar (approx. 2.0 cm x 2.0 cm x 2.0 cm) containing phenolphthalein indicator and sodium hydroxide (colored pink)
- White tile, scalpel, ruler, stopclock, boiling tubes, and beakers.

**Procedure**:
1. Prepare a serial dilution of the 1.0 mol dm^-3 hydrochloric acid solution, **H**, to obtain the following concentrations: 0.50 mol dm^-3, 0.25 mol dm^-3, and 0.125 mol dm^-3. Use **W** to make up the volume to 20 cm^3 for each concentration.

(a)(i) Complete Table 1.1 to show how you will prepare these three concentrations of hydrochloric acid.

Table 1.1
| Concentration of hydrochloric acid / mol dm^-3 | Volume of stock solution H / cm^3 | Volume of distilled water W / cm^3 |
|---|---|---|
| 0.50 | | |
| 0.25 | | |
| 0.125 | | |

2. Cut four blocks of agar from **A**, each measuring exactly 10 mm x 10 mm x 10 mm.
3. Put 15 cm^3 of the stock solution **H** (1.0 mol dm^-3) into a boiling tube. Place one agar block into this tube and start the stopclock immediately.
4. Measure the time taken (t), in seconds, for the pink color to completely disappear from the block. Record this time.
5. Repeat steps 3 and 4 using 15 cm^3 of each of the other three concentrations of hydrochloric acid prepared in Table 1.1.

(a)(ii) Prepare a single, comprehensive table (Table 1.2) to record all of your results, including the concentration of hydrochloric acid, the time taken (t) in seconds, and the calculated rate of diffusion (1/t) in s^-1.

(a)(iii) Identify the main source of experimental error in determining the exact point at which the pink color disappears, and suggest one improvement to minimize this error.

(a)(iv) State the independent variable and the dependent variable in this investigation.

(b) Plot a graph of the rate of diffusion (1/t) on the y-axis against the concentration of hydrochloric acid on the x-axis.

(c) Explain the relationship shown on your graph using the kinetic theory of diffusion.
PastPaper.showAnswers

PastPaper.workedSolution

**(a)(i) Table 1.1 Serial Dilution Scheme:**
- For 0.50 mol dm^-3: 10 cm^3 of **H** (1.0 mol dm^-3) + 10 cm^3 of **W**.
- For 0.25 mol dm^-3: 10 cm^3 of the prepared 0.50 mol dm^-3 solution + 10 cm^3 of **W**.
- For 0.125 mol dm^-3: 10 cm^3 of the prepared 0.25 mol dm^-3 solution + 10 cm^3 of **W**.

**(a)(ii) Expected Experimental Results Table (Table 1.2):**
| Concentration of HCl / mol dm^-3 | Time taken for pink color to disappear, t / s | Rate of diffusion (1/t) / s^-1 |
|---|---|---|
| 1.00 | 85 | 0.0118 |
| 0.50 | 170 | 0.0059 |
| 0.25 | 345 | 0.0029 |
| 0.125 | 695 | 0.0014 |
*(Note: Raw times should be recorded to the nearest whole second. Calculated rates should be formatted to a consistent number of significant figures, e.g., 3 s.f.)*

**(a)(iii) Sources of Error & Improvements:**
- **Source of Error**: Subjective judgment of when the pink color has completely disappeared, especially in the center of the block.
- **Improvement**: Cut the agar blocks in half at a set time interval to check for a remaining pink core, or use a color standard / colorimeter to standardize the endpoint.

**(a)(iv) Variables:**
- **Independent variable**: Concentration of hydrochloric acid / mol dm^-3.
- **Dependent variable**: Rate of diffusion / s^-1 (or time taken for pink color to disappear / s).

**(b) Graph Details:**
- **x-axis**: Concentration of hydrochloric acid / mol dm^-3 (linear scale from 0 to 1.0).
- **y-axis**: Rate of diffusion (1/t) / s^-1 (linear scale).
- **Plotting**: All points plotted accurately with a small 'x' or encircled dot. Straight line of best fit drawn through the origin.

**(c) Explanation of Results:**
- As the concentration of hydrochloric acid increases, the rate of diffusion increases.
- This is because a higher external concentration of hydrogen ions creates a steeper concentration gradient between the outside and the inside of the agar block.
- According to kinetic theory, molecules possess random kinetic energy; a steeper gradient results in a greater net movement of hydrogen ions into the block per unit time, leading to more rapid neutralization of the sodium hydroxide.

PastPaper.markingScheme

**(a)(i) Dilution Scheme (3 marks):**
- M1: Correct volumes of stock H and water for 0.50 mol dm^-3 (10 and 10).
- M2: Correct carry-over of 10 cm^3 from previous beaker to make subsequent concentrations.
- M3: Final volumes in all tubes are equal (20 cm^3 before removal for next dilution or 10 cm^3 final).

**(a)(ii) Table 1.2 Presentation (6 marks):**
- M1: Table drawn with clear cells and borders using a ruler.
- M2: Headings include appropriate units separated by a solidus (/ mol dm^-3, / s, / s^-1).
- M3: All raw time measurements are recorded in whole seconds only.
- M4: Correct trend shown (time increases as concentration decreases).
- M5: Rates calculated correctly to 3 significant figures.
- M6: At least four concentrations are recorded.

**(a)(iii) Error and Improvement (2 marks):**
- M1: Identifies difficulty in judging the precise endpoint due to subjectivity of color disappearance.
- M2: Suggests using a white card background for comparison / cutting a control block to show 'clear' stage / using a colorimeter.

**(a)(iv) Variables (2 marks):**
- M1: Identifies concentration of acid as the independent variable.
- M2: Identifies time or 1/time as the dependent variable.

**(b) Graph Plotting (4 marks):**
- M1: Axes labeled with variable name and correct units (x: concentration / mol dm^-3; y: rate / s^-1).
- M2: Scale chosen so that points occupy more than half of the grid in both dimensions; scales are linear and easy to read.
- M3: Points plotted accurately within half a small square.
- M4: A clean, thin, straight line of best fit drawn with a ruler, not passing through anomalous points.

**(c) Conclusion (3 marks):**
- M1: Describes relationship (rate is directly proportional to / increases with concentration).
- M2: Explains that higher concentration increases the steepness of the concentration gradient.
- M3: References random kinetic movement of ions causing faster net diffusion/neutralization.
PastPaper.question 2 · practical
20 PastPaper.marks
You are provided with a microscope and a prepared slide, **K1**, which is a transverse section of a xerophytic leaf (*Ammophila arenaria*).

(a)(i) Draw a large, low-power plan diagram of the entire leaf section to show the distribution of tissues. Your drawing must show the correct proportions of the structures. Do not draw any individual cells. Label the thick cuticle and a vascular bundle on your diagram.

(a)(ii) Identify three structural adaptations visible on slide **K1** that reduce water loss by transpiration. Explain how each adaptation achieves this.

(b)(i) Focus on one vascular bundle on slide **K1** under high power. Draw a small group of cells containing three adjacent bundle sheath cells and two adjacent xylem vessel elements. Label the cell wall of one cell and the lumen of one vessel element.

(b)(ii) Use an eyepiece graticule to measure the total thickness of the leaf blade at its thickest point (in eyepiece graticule units) and the diameter of the vascular bundle you drew (in eyepiece graticule units). Calculate the ratio of leaf thickness to vascular bundle diameter. Show your working.

(c) Figure 2.1 is a photomicrograph of a transverse section of a mesophytic leaf (*Ligustrum*).

Complete Table 2.1 to state four observable differences between the leaf section shown in **K1** and the leaf section shown in Figure 2.1.

Table 2.1
| Feature | Slide K1 (Xerophytic leaf) | Figure 2.1 (Mesophytic leaf) |
|---|---|---|
| 1. | | |
| 2. | | |
| 3. | | |
| 4. | | |
PastPaper.showAnswers

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**(a)(i) Low-power Plan Diagram:**
- Sharp, continuous lines drawn with a hard pencil. No shading.
- Large size (occupies at least half of the page space provided).
- Rolled morphology of the *Ammophila* leaf shown correctly, with invaginations (hinge cells at the bottom of the grooves) on the upper (inner) epidermis.
- Labels: Cuticle must point clearly to the outer, thick waxy layer on the outer/abaxial epidermis. Vascular bundle must point to one of the circular arrangements of tissues inside the mesophyll ridges.

**(a)(ii) Structural Adaptations & Explanations:**
1. *Rolled leaf shape*: Traps moist air inside the central cavity, reducing the water vapor potential gradient between the leaf interior and the outside air.
2. *Sunken stomata (in pits/grooves)*: Traps a layer of moist, humid air outside the stomata, increasing local humidity and lowering the rate of diffusion of water vapor.
3. *Hairs/trichomes on upper epidermis*: Restricts air movement (wind) across the stomata, trapping a layer of humid air and reducing transpiration.

**(b)(i) High-power Drawing:**
- Shows only five cells in total (3 bundle sheath cells, 2 xylem vessels) showing realistic shapes.
- Xylem vessels should be drawn with thicker walls compared to bundle sheath cells, and angular/lumen-containing shapes.
- Clean lines with no overlapping or sketchiness.
- Labels: 'Cell wall' points to the boundary of a bundle sheath cell; 'Lumen' points to the empty center of a xylem vessel.

**(b)(ii) Ratio Calculation:**
- Leaf thickness measurement = 48 graticule units.
- Vascular bundle diameter = 12 graticule units.
- Ratio = 48 / 12 = 4.0 (or written as 4:1).

**(c) Table 2.1 Comparison of K1 and Figure 2.1:**
1. *Leaf Shape*: Rolled or highly curved in **K1** vs. flat/planar in **Figure 2.1**.
2. *Stomata Position*: Located only in pits on the inner epidermis in **K1** vs. evenly distributed on the lower epidermis in **Figure 2.1**.
3. *Cuticle*: Extremely thick outer cuticle in **K1** vs. thin cuticle in **Figure 2.1**.
4. *Hairs / Trichomes*: Present inside the rolls/inner epidermis in **K1** vs. absent/very few in **Figure 2.1**.

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**(a)(i) Low-power Plan Diagram (5 marks):**
- M1: Clear, sharp, single lines with no sketching or shading.
- M2: Drawing is large enough (occupies at least 50% of the space).
- M3: Shows the leaf correctly as rolled/curved with ridges/grooves on the inner side.
- M4: Correct proportion: vascular bundles shown in the ridges and not touching the outer boundary.
- M5: Correctly labels thick cuticle on the outer boundary and a vascular bundle.

**(a)(ii) Adaptations (3 marks):**
- M1: Rolled leaf / hinge cells AND traps moisture / reduces water vapor potential gradient.
- M2: Stomata in pits / grooves AND reduces wind movement / increases local humidity.
- M3: Thick cuticle on outer surface AND reduces cuticular transpiration.

**(b)(i) High-power Drawing (5 marks):**
- M1: Draw only the requested number of cells (exactly 3 bundle sheath and 2 xylem elements) adjacent to each other.
- M2: Cell walls of xylem vessels drawn as double lines, thicker than bundle sheath walls.
- M3: Xylem vessels shown with correct polygonal/angular shapes, larger than bundle sheath cells.
- M4: Cell walls meet cleanly with no gaps or triple-line overlaps.
- M5: Correct labels for cell wall and lumen.

**(b)(ii) Ratio Calculation (3 marks):**
- M1: Records both measurements with appropriate graticule units.
- M2: Shows correct math division of leaf thickness by vascular bundle diameter.
- M3: Gives ratio simplified to a single decimal place or a whole-number ratio (e.g., 4:1).

**(c) Table 2.1 Differences (4 marks):**
- M1: Table contrasts the same feature for both specimens on the same row.
- M2: Correctly contrasts leaf shape (rolled vs. flat).
- M3: Correctly contrasts stomatal location (sunken in pits vs. superficial/lower epidermis).
- M4: Correctly contrasts epidermal hairs (present vs. absent).

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