Cambridge IAS-Level · Thinka-original Practice Paper

2023 Cambridge IAS-Level Biology (9700) Practice Paper with Answers

Thinka Nov 2023 (V2) Cambridge International A Level-Style Mock — Biology (9700)

100 marks150 mins2023
An original Thinka practice paper modelled on the structure and difficulty of the Nov 2023 (V2) Cambridge International A Level Biology (9700) paper. Not affiliated with or reproduced from Cambridge.

Paper 1 (Multiple Choice)

Answer all 40 multiple choice questions. For each question, choose the one you consider correct.
40 Question · 40 marks
Question 1 · Multiple Choice
1 marks
An enzyme-catalysed reaction was studied at various substrate concentrations. The Michaelis-Menten constant (\(K_m\)) was determined for two different substrates, X and Y. The \(K_m\) for substrate X is \(2.0 \times 10^{-3} \text{ mol dm}^{-3}\) and the \(K_m\) for substrate Y is \(5.0 \times 10^{-5} \text{ mol dm}^{-3}\). Which statement correctly interprets these results?
  1. A.The enzyme has a higher affinity for substrate X than for substrate Y.
  2. B.The enzyme has a higher affinity for substrate Y than for substrate X.
  3. C.The maximum velocity (\(V_{\max}\)) of the reaction with substrate Y must be higher than with substrate X.
  4. D.A higher concentration of substrate Y than substrate X is needed to reach half the maximum rate of reaction (\(\frac{1}{2} V_{\max}\)).
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Worked solution

The Michaelis-Menten constant (\(K_m\)) is the substrate concentration at which the reaction rate is half of its maximum velocity (\(V_{\max}\)). A lower \(K_m\) value indicates a higher affinity of the enzyme for that substrate, because less substrate is needed to occupy half of the active sites. Since \(5.0 \times 10^{-5} \text{ mol dm}^{-3}\) (for Y) is smaller than \(2.0 \times 10^{-3} \text{ mol dm}^{-3}\) (for X), the enzyme has a higher affinity for substrate Y than for substrate X.

Marking scheme

1 mark for identifying that a lower \(K_m\) corresponds to a higher affinity, making B the correct choice.
Question 2 · Multiple Choice
1 marks
Which of the following structures are found in both typical photosynthetic eukaryotic plant cells and typical prokaryotic cells?
  1. A.linear DNA, 70S ribosomes and a cell wall
  2. B.circular DNA, 70S ribosomes and cell surface membrane
  3. C.circular DNA, 80S ribosomes and cytoplasm
  4. D.linear DNA, 80S ribosomes and mitochondria
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Worked solution

Eukaryotic plant cells contain chloroplasts and mitochondria, both of which contain circular DNA and 70S ribosomes. Prokaryotic cells also contain circular DNA and 70S ribosomes. Both cell types possess a cell surface membrane. Linear DNA and 80S ribosomes are characteristic of the eukaryotic cytoplasm, but are not found in prokaryotes.

Marking scheme

1 mark for recognizing that circular DNA, 70S ribosomes, and a cell surface membrane are features shared by both prokaryotes and organelles within eukaryotic plant cells.
Question 3 · Multiple Choice
1 marks
A student observes two adjacent plant cells undergoing mitosis. In Cell 1, the chromosomes are aligned individually along the equator of the spindle. In Cell 2, sister chromatids have separated and are being pulled to opposite poles. Which stages of mitosis are observed in Cell 1 and Cell 2?
  1. A.Cell 1: metaphase; Cell 2: prophase
  2. B.Cell 1: metaphase; Cell 2: anaphase
  3. C.Cell 1: prophase; Cell 2: anaphase
  4. D.Cell 1: anaphase; Cell 2: telophase
Show answer & marking scheme

Worked solution

In metaphase (Cell 1), chromosomes align individually along the metaphase plate (equator). In anaphase (Cell 2), the centromeres divide and the sister chromatids are pulled apart toward opposite poles by the spindle fibres.

Marking scheme

1 mark for identifying Cell 1 as metaphase and Cell 2 as anaphase.
Question 4 · Multiple Choice
1 marks
Which feature of a capillary is essential for its function but is not found in an arteriole?
  1. A.A wall that is only a single layer of squamous endothelial cells thick
  2. B.The presence of a wide lumen to reduce blood pressure
  3. C.A thick layer of elastic tissue to withstand high pressure
  4. D.The presence of semilunar valves to prevent the backflow of blood
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Worked solution

Capillaries have walls that are extremely thin—consisting of only a single layer of endothelial cells (tunica intima) on a basement membrane. This minimizes diffusion distance. Arterioles have a tunica media containing smooth muscle and elastic fibres, making their walls much thicker. Capillaries do not have valves, and their lumen is very narrow (only wide enough for red blood cells to pass through in single file).

Marking scheme

1 mark for identifying the single-layered endothelial wall as the unique functional capillary feature.
Question 5 · Multiple Choice
1 marks
A codon on a strand of mRNA has the sequence 5'-GUC-3'. Which row correctly identifies the complementary anticodon on the tRNA and the template DNA triplet that coded for this codon?
  1. A.tRNA anticodon: 5'-CAG-3'; DNA template triplet: 3'-CAG-5'
  2. B.tRNA anticodon: 3'-CAG-5'; DNA template triplet: 3'-CAG-5'
  3. C.tRNA anticodon: 3'-CAG-5'; DNA template triplet: 3'-GAC-5'
  4. D.tRNA anticodon: 5'-GAC-3'; DNA template triplet: 3'-GAC-5'
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Worked solution

The mRNA codon is 5'-GUC-3'. The complementary DNA template strand must run antiparallel and have complementary bases, which is 3'-CAG-5'. The tRNA anticodon must also pair antiparallelly with the mRNA codon, meaning its 3' to 5' sequence is complementary to the mRNA's 5' to 3' sequence. Thus, the tRNA anticodon is 3'-CAG-5' (or 5'-GAC-3' if written 5' to 3'). Option B correctly lists 3'-CAG-5' for both.

Marking scheme

1 mark for correctly matching both the tRNA anticodon (3'-CAG-5') and the DNA template triplet (3'-CAG-5') based on complementary base pairing and antiparallel directionality.
Question 6 · Multiple Choice
1 marks
Which features are present in a mature, functional xylem vessel element but are absent in a mature, functional phloem sieve tube element?
  1. A.cellulose cell walls and plasmodesmata
  2. B.lignified cell walls and absence of cytoplasm
  3. C.companion cells and sieve plates
  4. D.peripheral cytoplasm and a cell surface membrane
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Worked solution

Xylem vessel elements are dead cells at maturity; they have no cytoplasm, no nucleus, and their walls are heavily thickened with waterproof lignin. Phloem sieve tube elements are living cells containing a thin layer of peripheral cytoplasm and a cell surface membrane, and their walls are made of cellulose (not lignified).

Marking scheme

1 mark for identifying that lignified walls and the complete absence of cytoplasm are features of mature xylem but not mature phloem sieve tubes.
Question 7 · Multiple Choice
1 marks
Which row correctly describes the distribution of cartilage, goblet cells, and smooth muscle in the trachea and the smaller bronchioles?
  1. A.Trachea: cartilage present, goblet cells present, smooth muscle present; Bronchioles: cartilage absent, goblet cells absent, smooth muscle present
  2. B.Trachea: cartilage present, goblet cells present, smooth muscle absent; Bronchioles: cartilage present, goblet cells present, smooth muscle present
  3. C.Trachea: cartilage present, goblet cells absent, smooth muscle present; Bronchioles: cartilage absent, goblet cells present, smooth muscle absent
  4. D.Trachea: cartilage absent, goblet cells present, smooth muscle present; Bronchioles: cartilage absent, goblet cells absent, smooth muscle absent
Show answer & marking scheme

Worked solution

The trachea contains C-shaped rings of cartilage, goblet cells in the ciliated epithelium, and smooth muscle. Bronchioles lack cartilage and goblet cells, but still contain smooth muscle to control the diameter of the airway.

Marking scheme

1 mark for identifying the correct distribution of these three tissue types in the trachea and bronchioles.
Question 8 · Multiple Choice
1 marks
Which statement correctly describes a difference between T-lymphocytes and B-lymphocytes in the human immune response?
  1. A.T-lymphocytes mature in the bone marrow, whereas B-lymphocytes mature in the thymus.
  2. B.T-lymphocytes produce and secrete soluble antibody molecules, whereas B-lymphocytes do not.
  3. C.T-lymphocytes are involved in cell-mediated immunity, whereas B-lymphocytes are involved in humoral immunity.
  4. D.T-lymphocytes have specific antigen receptors on their surface, whereas B-lymphocytes have non-specific receptors.
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Worked solution

T-lymphocytes are responsible for cell-mediated immunity (destroying infected host cells or activating other immune cells), whereas B-lymphocytes differentiate into plasma cells that secrete antibodies, which is the humoral immune response. T-lymphocytes mature in the thymus (B-lymphocytes mature in the bone marrow), and both cell types have highly specific antigen receptors.

Marking scheme

1 mark for identifying the difference in the type of immunity (cell-mediated vs humoral) between T-lymphocytes and B-lymphocytes.
Question 9 · multiple_choice
1 marks
The table shows some features of eukaryotic cell structures. Which row correctly describes a cell structure?
  1. A.Structure: Lysosome | Membrane: Single | Function: Contains hydrolytic enzymes | Presence: Animal cells
  2. B.Structure: Nucleolus | Membrane: Single | Function: Synthesises ribosomal RNA | Presence: Plant and animal cells
  3. C.Structure: Centriole | Membrane: Double | Function: Organises microtubules of the spindle | Presence: Animal cells only
  4. D.Structure: Rough endoplasmic reticulum | Membrane: None | Function: Translates mRNA into proteins | Presence: Plant and animal cells
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Worked solution

Lysosomes are bounded by a single membrane and contain hydrolytic enzymes (such as proteases, nucleases, and lipases) used to break down biological molecules. They are present in animal cells. The nucleolus is not membrane-bound (no membrane). Centrioles are not membrane-bound. The rough endoplasmic reticulum is a single membrane-bound organelle.

Marking scheme

Award 1 mark for selecting option A, which correctly identifies that a lysosome is single-membrane bound, contains hydrolytic enzymes, and is present in animal cells.
Question 10 · multiple_choice
1 marks
Which statement correctly compares the structures and properties of collagen and haemoglobin?
  1. A.Collagen consists of three polypeptide chains wound into a triple helix, whereas haemoglobin consists of four polypeptide chains, each associated with a haem group.
  2. B.Collagen is a soluble globular protein with hydrophobic R-groups on its outer surface, whereas haemoglobin is an insoluble fibrous protein.
  3. C.Collagen contains a high proportion of glycine residues to allow tight packing, whereas haemoglobin consists entirely of beta-pleated sheets held by disulfide bonds.
  4. D.Both collagen and haemoglobin are quaternary proteins that contain prosthetic groups containing iron ions (\(\text{Fe}^{2+}\)).
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Worked solution

Collagen is a fibrous protein made of three polypeptide chains wound together to form a tight triple helix. Haemoglobin is a quaternary globular protein composed of four polypeptide chains (two alpha and two beta chains), each containing a prosthetic haem group with an iron ion. Collagen does not contain a prosthetic group and is insoluble, while haemoglobin is soluble.

Marking scheme

Award 1 mark for identifying option A as the correct comparison between collagen and haemoglobin.
Question 11 · multiple_choice
1 marks
Which row correctly matches a component of the cell surface membrane with its role in maintaining the structure or function of the membrane?
  1. A.Component: Cholesterol | Role: Regulates membrane fluidity by preventing fatty acid tails of phospholipids from packing too closely at low temperatures
  2. B.Component: Glycoprotein | Role: Acts as a primary barrier to prevent the passage of polar molecules and ions
  3. C.Component: Phospholipid bilayer | Role: Acts as a specific cell-surface receptor for water-soluble hormones
  4. D.Component: Peripheral protein | Role: Forms hydrogen bonds with water molecules outside the cell to stabilise the membrane structure
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Worked solution

Cholesterol acts as a fluidity buffer in cell membranes. At low temperatures, it prevents the fatty acid tails of phospholipids from packing too closely together, which prevents the membrane from becoming too rigid. The phospholipid bilayer (not glycoprotein) acts as a barrier to polar molecules. Glycoproteins/glycolipids (not phospholipids) act as receptors. Carbohydrate chains of glycoproteins/glycolipids (not peripheral proteins) form hydrogen bonds with water to stabilise the membrane.

Marking scheme

Award 1 mark for selecting option A, which correctly matches cholesterol with its role in regulating membrane fluidity.
Question 12 · multiple_choice
1 marks
An experiment was carried out to investigate the effect of two different inhibitors, X and Y, on the rate of an enzyme-controlled reaction. The concentration of the enzyme was kept constant. At very high substrate concentrations, the rate of reaction with inhibitor X reached the same maximum rate (\(V_{\max}\)) as the control reaction without inhibitor. However, the rate of reaction with inhibitor Y remained significantly lower than the control reaction, even at the highest substrate concentrations. Which statement correctly explains these results?
  1. A.Inhibitor X is a competitive inhibitor that binds reversibly to the active site, whereas inhibitor Y is a non-competitive inhibitor that binds to an allosteric site.
  2. B.Inhibitor X is a non-competitive inhibitor that changes the shape of the active site, whereas inhibitor Y is a competitive inhibitor that competes with the substrate.
  3. C.Inhibitor X binds permanently to an allosteric site, whereas inhibitor Y binds reversibly to the active site.
  4. D.Inhibitor X increases the activation energy of the reaction, whereas inhibitor Y decreases the affinity of the enzyme for its substrate.
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Worked solution

A competitive inhibitor (X) competes with substrate molecules for the active site. At very high substrate concentrations, the substrate molecules outcompete the inhibitor, so the maximum rate of reaction (\(V_{\max}\)) is achieved. A non-competitive inhibitor (Y) binds to an allosteric site, altering the shape of the active site so that the substrate can no longer bind. This effectively reduces the concentration of active enzyme, meaning \(V_{\max}\) is decreased and cannot be restored by adding more substrate.

Marking scheme

Award 1 mark for option A, which correctly identifies inhibitor X as a competitive inhibitor and inhibitor Y as a non-competitive inhibitor based on their effects on \(V_{\max}\).
Question 13 · multiple_choice
1 marks
During which stages of the mitotic cell cycle do the following events occur?

- Stage 1: Chromatin condenses to form visible chromosomes, and the nucleolus disappears.
- Stage 2: Sister chromatids are pulled apart and move to opposite poles of the spindle.
- Stage 3: Nuclear envelopes reform around the two groups of chromosomes.
  1. A.Stage 1: prophase | Stage 2: anaphase | Stage 3: telophase
  2. B.Stage 1: interphase | Stage 2: metaphase | Stage 3: telophase
  3. C.Stage 1: prophase | Stage 2: metaphase | Stage 3: anaphase
  4. D.Stage 1: metaphase | Stage 2: anaphase | Stage 3: cytokinesis
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Worked solution

Chromatin condensation and disappearance of the nucleolus occur during prophase (Stage 1). The separation of sister chromatids and their movement to opposite poles occurs during anaphase (Stage 2). The reformation of the nuclear envelopes occurs during telophase (Stage 3).

Marking scheme

Award 1 mark for selecting option A, which correctly identifies the mitotic stages for all three stages described.
Question 14 · multiple_choice
1 marks
A DNA template strand contains 18% adenine and 26% thymine. What is the percentage of uracil in the complementary mRNA molecule transcribed from this template strand?
  1. A.18%
  2. B.26%
  3. C.44%
  4. D.56%
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Worked solution

During transcription, complementary base pairing occurs between the DNA template strand and the forming mRNA molecule. Adenine (A) on the DNA template strand pairs with Uracil (U) on the RNA strand. Therefore, the percentage of uracil in the mRNA will be exactly equal to the percentage of adenine in the template strand, which is 18%.

Marking scheme

Award 1 mark for option A, since uracil in mRNA is complementary to adenine in the DNA template strand (18%).
Question 15 · multiple_choice
1 marks
Which row correctly describes the presence (\(\checkmark\)) or absence (\(\times\)) of cartilage, smooth muscle, and elastic fibres in different parts of the human gas exchange system?
  1. A.Trachea: Cartilage \(\checkmark\), Smooth muscle \(\checkmark\), Elastic fibres \(\checkmark\) | Bronchioles: Cartilage \(\times\), Smooth muscle \(\checkmark\), Elastic fibres \(\checkmark\) | Alveoli: Cartilage \(\times\), Smooth muscle \(\times\), Elastic fibres \(\checkmark\)
  2. B.Trachea: Cartilage \(\checkmark\), Smooth muscle \(\times\), Elastic fibres \(\checkmark\) | Bronchioles: Cartilage \(\checkmark\), Smooth muscle \(\checkmark\), Elastic fibres \(\times\) | Alveoli: Cartilage \(\times\), Smooth muscle \(\checkmark\), Elastic fibres \(\checkmark\)
  3. C.Trachea: Cartilage \(\checkmark\), Smooth muscle \(\checkmark\), Elastic fibres \(\times\) | Bronchioles: Cartilage \(\times\), Smooth muscle \(\times\), Elastic fibres \(\checkmark\) | Alveoli: Cartilage \(\times\), Smooth muscle \(\times\), Elastic fibres \(\times\)
  4. D.Trachea: Cartilage \(\times\), Smooth muscle \(\checkmark\), Elastic fibres \(\checkmark\) | Bronchioles: Cartilage \(\checkmark\), Smooth muscle \(\times\), Elastic fibres \(\checkmark\) | Alveoli: Cartilage \(\times\), Smooth muscle \(\times\), Elastic fibres \(\checkmark\)
Show answer & marking scheme

Worked solution

The trachea contains C-shaped rings of cartilage, smooth muscle, and elastic fibres in its walls. Bronchioles lack cartilage entirely but contain smooth muscle to control airway diameter and elastic fibres. Alveoli lack both cartilage and smooth muscle but contain abundant elastic fibres to allow for expansion and elastic recoil during ventilation.

Marking scheme

Award 1 mark for option A, which correctly identifies the distribution of cartilage, smooth muscle, and elastic fibres.
Question 16 · multiple_choice
1 marks
Penicillin is an antibiotic used to treat bacterial infections. It works by inhibiting the enzyme glycopeptide transpeptidase. Which process is directly prevented by the action of penicillin?
  1. A.The synthesis of peptidoglycan monomers within the bacterial cytoplasm
  2. B.The formation of cross-links between peptidoglycan chains in the cell wall
  3. C.The replication of circular DNA prior to binary fission
  4. D.The secretion of autolysins that break down the bacterial cell membrane
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Worked solution

Penicillin inhibits the enzyme transpeptidase, which is responsible for forming covalent cross-links between peptidoglycan chains in the bacterial cell wall. This inhibition prevents the cell wall from being reinforced, leading to osmotic lysis when water enters the cell.

Marking scheme

Award 1 mark for option B, which correctly states that penicillin prevents the formation of cross-links between peptidoglycan chains.
Question 17 · multiple_choice
1 marks
Which of the following structures contain both 70S ribosomes and circular DNA? 1. Chloroplast; 2. Mitochondrion; 3. Nucleoplasm of a eukaryotic cell; 4. Cytoplasm of a prokaryotic cell
  1. A.1, 2 and 4 only
  2. B.1 and 2 only
  3. C.2 and 4 only
  4. D.1, 2, 3 and 4
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Worked solution

Chloroplasts, mitochondria, and prokaryotic cells all contain 70S ribosomes and circular DNA as part of their evolutionary origin (endosymbiotic theory for the organelles). The nucleoplasm of eukaryotic cells contains linear chromosomes and does not contain ribosomes, which are assembled in the nucleolus as subunits and function as 80S ribosomes in the cytoplasm/RER. Thus, 1, 2, and 4 are correct.

Marking scheme

1 mark for identifying that chloroplasts, mitochondria, and prokaryotic cytoplasm contain both 70S ribosomes and circular DNA, while eukaryotic nucleoplasm does not.
Question 18 · multiple_choice
1 marks
An enzyme-catalysed reaction was investigated to determine its Michaelis-Menten constant (\(K_m\)). The maximum rate of reaction (\(V_{max}\)) was found to be \(80\text{ }\mu\text{mol dm}^{-3}\text{ s}^{-1}\). At what substrate concentration will the rate of reaction be \(40\text{ }\mu\text{mol dm}^{-3}\text{ s}^{-1}\)?
  1. A.\(0.5 \times K_m\)
  2. B.\(K_m\)
  3. C.\(2 \times K_m\)
  4. D.\(V_{max} - K_m\)
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Worked solution

The Michaelis-Menten constant (\(K_m\)) is defined as the substrate concentration at which the rate of the enzyme-catalysed reaction is half of the maximum rate (\(V_{max}\)). Since the maximum rate is \(80\text{ }\mu\text{mol dm}^{-3}\text{ s}^{-1}\), half of this rate is \(40\text{ }\mu\text{mol dm}^{-3}\text{ s}^{-1}\). Therefore, the substrate concentration at this rate is equal to \(K_m\).

Marking scheme

1 mark for identifying that half of the maximum rate of reaction corresponds to a substrate concentration equal to the Michaelis-Menten constant (\(K_m\)).
Question 19 · multiple_choice
1 marks
Four identical cylinders of potato tissue were placed into four different sucrose solutions (\(P\), \(Q\), \(R\), and \(S\)) of different concentrations. After two hours, the percentage change in mass of each cylinder was calculated: Solution \(P\): \(-4.5\%\), Solution \(Q\): \(+2.1\%\), Solution \(R\): \(0.0\%\), Solution \(S\): \(-1.8\%\). Which list ranks the sucrose solutions from the highest (least negative) water potential to the lowest (most negative) water potential?
  1. A.\(P \rightarrow S \rightarrow R \rightarrow Q\)
  2. B.\(Q \rightarrow R \rightarrow S \rightarrow P\)
  3. C.\(Q \rightarrow S \rightarrow R \rightarrow P\)
  4. D.\(P \rightarrow R \rightarrow S \rightarrow Q\)
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Worked solution

A higher (least negative, closest to zero) water potential in the external solution means water will move down the water potential gradient into the potato cells by osmosis, causing an increase in mass. Therefore, solution \(Q\) (mass change of \(+2.1\%\)) has the highest water potential. At solution \(R\) (\(0.0\%\) mass change), the water potential is equal to the potato tissue. Solutions \(S\) and \(P\) have lower water potentials than the potato tissue, causing water to leave the cells and resulting in a loss of mass, with \(P\) (mass change of \(-4.5\%\)) having the lowest (most negative) water potential. The correct order is \(Q \rightarrow R \rightarrow S \rightarrow P\).

Marking scheme

1 mark for determining the correct order of water potentials from highest (least negative / mass gain) to lowest (most negative / mass loss).
Question 20 · multiple_choice
1 marks
A diploid cell of a specific organism has 12 chromosomes (\(2n = 12\)). How many chromatids and how many telomeres are present in this cell during metaphase of mitosis?
  1. A.12 chromatids and 24 telomeres
  2. B.24 chromatids and 24 telomeres
  3. C.24 chromatids and 48 telomeres
  4. D.48 chromatids and 48 telomeres
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Worked solution

During metaphase of mitosis, the DNA has replicated, so each of the 12 chromosomes consists of two sister chromatids. This gives \(12 \times 2 = 24\) chromatids. Each chromatid is a single linear double-stranded DNA molecule, which has 2 telomeres (one at each end). Therefore, 24 chromatids will have \(24 \times 2 = 48\) telomeres.

Marking scheme

1 mark for calculating both the correct number of chromatids (24) and telomeres (48) in metaphase.
Question 21 · multiple_choice
1 marks
Which feature is present in mature xylem vessel elements but absent in mature phloem sieve tube elements?
  1. A.Cytoplasm
  2. B.Lignified cell walls
  3. C.Plasmodesmata
  4. D.Sieve plates
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Worked solution

Mature xylem vessel elements are dead cells that are heavily reinforced with lignin in their cell walls to withstand tension. In contrast, mature phloem sieve tube elements are living cells with non-lignified cellulose walls. Cytoplasm, plasmodesmata, and sieve plates are characteristic of sieve tube elements but absent in xylem vessels.

Marking scheme

1 mark for identifying that lignified cell walls are present in mature xylem but completely absent in phloem sieve tube elements.
Question 22 · multiple_choice
1 marks
Which row correctly identifies the presence or absence of tissues/structures in the walls of the trachea, bronchioles, and alveoli of a healthy human?
  1. A.Trachea: cartilage, goblet cells, smooth muscle all present; Bronchioles: cartilage and goblet cells absent, smooth muscle present; Alveoli: cartilage, goblet cells, smooth muscle all absent
  2. B.Trachea: cartilage, goblet cells, smooth muscle all present; Bronchioles: cartilage present, goblet cells and smooth muscle absent; Alveoli: cartilage and goblet cells absent, smooth muscle present
  3. C.Trachea: cartilage absent, goblet cells and smooth muscle present; Bronchioles: cartilage and goblet cells present, smooth muscle present; Alveoli: cartilage, goblet cells, smooth muscle all absent
  4. D.Trachea: cartilage and smooth muscle present, goblet cells absent; Bronchioles: cartilage, goblet cells, smooth muscle all absent; Alveoli: cartilage absent, goblet cells present, smooth muscle absent
Show answer & marking scheme

Worked solution

In a healthy human, the trachea wall contains cartilage rings, goblet cells, and smooth muscle. Bronchioles lack cartilage and goblet cells (to prevent clogging of small airways) but retain smooth muscle to regulate airflow. Alveoli lack cartilage, goblet cells, and smooth muscle, consisting of simple squamous epithelium and elastic fibres to facilitate rapid gas exchange.

Marking scheme

1 mark for identifying the correct distribution of cartilage, goblet cells, and smooth muscle in all three regions.
Question 23 · multiple_choice
1 marks
Why is penicillin effective against bacteria but not against viruses?
  1. A.Penicillin inhibits bacterial ribosomes, whereas viruses use host cell 80S ribosomes.
  2. B.Penicillin prevents peptidoglycan cross-linking in bacterial cell walls, whereas viruses do not possess a cell wall.
  3. C.Penicillin binds to specific cell membrane receptors on bacteria, which are absent on viral envelopes.
  4. D.Penicillin degrades bacterial DNA, whereas viral nucleic acids are protected by a protein capsid.
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Worked solution

Penicillin is an antibiotic that inhibits the enzyme transpeptidase, which is responsible for forming cross-links in the peptidoglycan cell wall of bacteria. This weakens the cell wall, causing osmotic lysis of growing bacteria. Since viruses do not have a cell wall (nor do they have peptidoglycan), penicillin has no effect on them.

Marking scheme

1 mark for identifying that penicillin's mechanism of action targets peptidoglycan cross-linking in bacterial cell walls, which viruses lack.
Question 24 · multiple_choice
1 marks
A segment of double-stranded DNA contains 1500 base pairs. This segment codes for a polypeptide. If 150 base pairs consist of introns and non-coding regulatory sequences that are spliced out, what is the maximum number of amino acids in the synthesized polypeptide?
  1. A.450
  2. B.500
  3. C.900
  4. D.1350
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Worked solution

The total coding segment is \(1500 - 150 = 1350\) base pairs of DNA. When transcribed, this yields an mRNA molecule with \(1350\) coding nucleotides. Since each codon consists of 3 nucleotides (a triplet codon), the maximum number of codons is \(1350 / 3 = 450\). Thus, the maximum number of amino acids in the polypeptide chain is 450.

Marking scheme

1 mark for calculating the correct number of amino acids by subtracting the introns (150 bp) from total base pairs (1500 bp) and dividing the remaining coding sequence (1350 bp) by 3.
Question 25 · multiple_choice
1 marks
An experiment was conducted using radioactive amino acids to track the pathway of protein synthesis and secretion in goblet cells of the trachea. Which sequence shows the correct pathway of radioactive label detection in these cells?
  1. A.Nucleus \(\rightarrow\) rough endoplasmic reticulum \(\rightarrow\) Golgi body \(\rightarrow\) secretory vesicles
  2. B.Rough endoplasmic reticulum \(\rightarrow\) Golgi body \(\rightarrow\) secretory vesicles \(\rightarrow\) cell surface membrane
  3. C.Rough endoplasmic reticulum \(\rightarrow\) smooth endoplasmic reticulum \(\rightarrow\) Golgi body \(\rightarrow\) lysosome
  4. D.Ribosome \(\rightarrow\) smooth endoplasmic reticulum \(\rightarrow\) secretory vesicles \(\rightarrow\) cell surface membrane
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Worked solution

In secretory cells like goblet cells, proteins (such as mucin) are synthesized on the ribosomes of the rough endoplasmic reticulum (RER). They are then transported in vesicles to the Golgi body, where they are modified (e.g., glycosylated to form glycoproteins). From the Golgi body, they are packaged into secretory vesicles which move towards the apical membrane and fuse with the cell surface membrane, releasing the proteins via exocytosis.

Marking scheme

Award 1 mark for identifying the correct physiological pathway of a secreted protein (RER -> Golgi body -> secretory vesicles -> cell surface membrane).
Question 26 · multiple_choice
1 marks
The rate of an enzyme-controlled reaction was measured at different substrate concentrations. The experiment was repeated in the presence of an inhibitor. At high substrate concentrations, the rate of the reaction containing the inhibitor approached the maximum rate (\(V_{\max}\)) of the control reaction. Which statement correctly describes the inhibitor and its effect on the Michaelis-Menten constant (\(K_{\m}\)) of the enzyme?
  1. A.Competitive inhibitor, which decreases the \(K_{\m}\) value of the enzyme
  2. B.Competitive inhibitor, which increases the \(K_{\m}\) value of the enzyme
  3. C.Non-competitive inhibitor, which decreases the \(K_{\m}\) value of the enzyme
  4. D.Non-competitive inhibitor, which increases the \(K_{\m}\) value of the enzyme
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Worked solution

The inhibitor is competitive because its effect can be overcome at high substrate concentrations, allowing the reaction to reach the original \(V_{\max}\). A competitive inhibitor decreases the affinity of the enzyme for its substrate, which increases the value of the Michaelis-Menten constant (\(K_{\m}\)) because a higher concentration of substrate is required to reach half of the \(V_{\max}\).

Marking scheme

Award 1 mark for identifying that the inhibitor is competitive and that it increases the \(K_{\m}\) value.
Question 27 · multiple_choice
1 marks
A cell from an animal species with a diploid number of \(2n = 16\) undergoes mitosis. What are the correct numbers of chromosomes and chromatids present in this cell during metaphase and during anaphase?
  1. A.Metaphase: 16 chromosomes, 32 chromatids; Anaphase: 32 chromosomes, 0 chromatids
  2. B.Metaphase: 16 chromosomes, 32 chromatids; Anaphase: 16 chromosomes, 16 chromatids
  3. C.Metaphase: 32 chromosomes, 32 chromatids; Anaphase: 32 chromosomes, 0 chromatids
  4. D.Metaphase: 16 chromosomes, 16 chromatids; Anaphase: 32 chromosomes, 32 chromatids
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Worked solution

During metaphase, there are 16 chromosomes aligned at the equator, each consisting of two sister chromatids joined at a centromere, giving a total of 32 chromatids. During anaphase, the centromeres split and the sister chromatids separate to opposite poles. Once separated, each chromatid is defined as an individual chromosome. Therefore, during anaphase, there are 32 chromosomes and 0 chromatids (as they are no longer paired).

Marking scheme

Award 1 mark for selecting the row that correctly identifies 16 chromosomes and 32 chromatids in metaphase, and 32 chromosomes and 0 chromatids in anaphase.
Question 28 · multiple_choice
1 marks
A section of a template DNA strand has the sequence: \(3'\text{- T A C G G G C A T A T T -}5'\). Which row correctly identifies the corresponding mRNA codon sequence and the complementary tRNA anticodon sequence?
  1. A.mRNA: \(5'\text{- U A C G G G C A U A U U -}3'\); tRNA anticodons: \(A U G\), \(C C C\), \(G U A\), \(U A A\)
  2. B.mRNA: \(5'\text{- A U G C C C G U A U A A -}3'\); tRNA anticodons: \(U A C\), \(G G G\), \(C A U\), \(A U U\)
  3. C.mRNA: \(5'\text{- A U G C C C G U A U A A -}3'\); tRNA anticodons: \(T A C\), \(G G G\), \(C A T\), \(A T T\)
  4. D.mRNA: \(5'\text{- U A C G G G C A U A U U -}3'\); tRNA anticodons: \(U A C\), \(G G G\), \(C A U\), \(A U U\)
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Worked solution

Transcription of the template DNA strand \(3'\text{- T A C G G G C A T A T T -}5'\) produces complementary mRNA in the \(5'\) to \(3'\) direction: \(5'\text{- A U G C C C G U A U A A -}3'\). During translation, the tRNA anticodons align with these mRNA codons. The anticodons (which contain uracil instead of thymine) will be complementary to the codons, giving: \(U A C\), \(G G G\), \(C A U\), \(A U U\).

Marking scheme

Award 1 mark for identifying the correct complementary mRNA strand containing uracil (not thymine) and the corresponding tRNA anticodons.
Question 29 · multiple_choice
1 marks
The pressures operating at the arterial and venous ends of a capillary bed are given below:

- Hydrostatic pressure of blood: arterial end = \(4.6\text{ kPa}\); venous end = \(2.3\text{ kPa}\)
- Hydrostatic pressure of tissue fluid: arterial end = \(1.2\text{ kPa}\); venous end = \(1.2\text{ kPa}\)
- Oncotic (osmotic) pressure of blood: arterial end = \(3.3\text{ kPa}\); venous end = \(3.3\text{ kPa}\)
- Oncotic (osmotic) pressure of tissue fluid: arterial end = \(1.0\text{ kPa}\); venous end = \(1.0\text{ kPa}\)

What is the net filtration pressure at the arterial end and the net pressure at the venous end of this capillary bed?
  1. A.Arterial end: \(+1.1\text{ kPa}\) (outwards); Venous end: \(-1.2\text{ kPa}\) (inwards)
  2. B.Arterial end: \(+2.5\text{ kPa}\) (outwards); Venous end: \(-0.2\text{ kPa}\) (inwards)
  3. C.Arterial end: \(+1.1\text{ kPa}\) (inwards); Venous end: \(-1.2\text{ kPa}\) (outwards)
  4. D.Arterial end: \(+1.3\text{ kPa}\) (outwards); Venous end: \(-1.0\text{ kPa}\) (inwards)
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Worked solution

The formula for net filtration pressure is: \(\text{Net Pressure} = (\text{Blood Hydrostatic Pressure} - \text{Tissue Hydrostatic Pressure}) - (\text{Blood Oncotic Pressure} - \text{Tissue Oncotic Pressure})\).

At the arterial end: \(\text{Net Pressure} = (4.6 - 1.2) - (3.3 - 1.0) = 3.4 - 2.3 = +1.1\text{ kPa}\) (pointing outwards, causing filtration).

At the venous end: \(\text{Net Pressure} = (2.3 - 1.2) - (3.3 - 1.0) = 1.1 - 2.3 = -1.2\text{ kPa}\) (pointing inwards, causing reabsorption).

Marking scheme

Award 1 mark for showing correct calculation steps for both the arterial (+1.1 kPa) and venous (-1.2 kPa) net pressures.
Question 30 · multiple_choice
1 marks
Which row correctly describes the distribution of tissues within the wall of a bronchus and the wall of a terminal bronchiole?
  1. A.Bronchus: cartilage present, smooth muscle present, ciliated epithelium present, goblet cells present. Terminal bronchiole: cartilage absent, smooth muscle present, ciliated epithelium present, goblet cells absent.
  2. B.Bronchus: cartilage absent, smooth muscle present, ciliated epithelium present, goblet cells present. Terminal bronchiole: cartilage absent, smooth muscle absent, ciliated epithelium absent, goblet cells present.
  3. C.Bronchus: cartilage present, smooth muscle present, ciliated epithelium present, goblet cells absent. Terminal bronchiole: cartilage present, smooth muscle present, ciliated epithelium absent, goblet cells absent.
  4. D.Bronchus: cartilage present, smooth muscle absent, ciliated epithelium present, goblet cells present. Terminal bronchiole: cartilage absent, smooth muscle present, ciliated epithelium absent, goblet cells absent.
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Worked solution

A bronchus contains cartilage (in plates), smooth muscle, ciliated epithelium, and goblet cells. A terminal bronchiole does not have cartilage or goblet cells (to prevent mucus blockage in very narrow passages), but it still possesses smooth muscle and a ciliated epithelium (often simple cuboidal and ciliated) to control airway diameter and sweep up any remaining particles.

Marking scheme

Award 1 mark for correctly matching the tissue composition of both structures (bronchus: cartilage, smooth muscle, ciliated epithelium, goblet cells present; terminal bronchiole: cartilage and goblet cells absent, smooth muscle and ciliated epithelium present).
Question 31 · multiple_choice
1 marks
Which statements correctly describe differences between T-lymphocytes and B-lymphocytes?

1. T-lymphocytes mature in the thymus gland, whereas B-lymphocytes mature in the bone marrow.
2. T-lymphocytes produce antibodies that circulate in blood plasma, whereas B-lymphocytes do not.
3. T-lymphocytes have specific receptors on their cell surface membranes, whereas B-lymphocytes do not.
4. T-helper cells release cytokines to stimulate B-lymphocytes, whereas B-lymphocytes can differentiate into antibody-secreting plasma cells.
  1. A.1 and 4 only
  2. B.1, 2 and 4 only
  3. C.2 and 3 only
  4. D.1, 3 and 4 only
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Worked solution

Statement 1 is correct: T-cells migrate to and mature in the thymus, while B-cells mature in the bone marrow. Statement 2 is incorrect: B-lymphocytes differentiate into plasma cells that produce antibodies, whereas T-lymphocytes do not produce antibodies. Statement 3 is incorrect: both cell types possess specific surface receptors (T-cell receptors and B-cell receptors/antibodies). Statement 4 is correct: activated T-helper cells release cytokines that trigger B-cell clonal expansion, and B-cells differentiate into antibody-secreting plasma cells.

Marking scheme

Award 1 mark for identifying statements 1 and 4 as the only correct options.
Question 32 · multiple_choice
1 marks
How does penicillin act on bacteria to treat infections?
  1. A.It binds to 70S ribosomes, inhibiting translation and preventing protein synthesis.
  2. B.It inhibits the enzyme transpeptidase, preventing the cross-linking of peptidoglycan chains in the cell wall.
  3. C.It disrupts the phospholipid bilayer of the cell surface membrane, causing cell lysis.
  4. D.It inhibits DNA gyrase, preventing DNA replication and transcription.
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Worked solution

Penicillin is a beta-lactam antibiotic. It inhibits the enzyme transpeptidase (glycoprotein peptidase), which catalyzes the cross-linking of peptidoglycan chains in the bacterial cell wall. This prevents the synthesis of a strong, structural cell wall. When the bacterium attempts to grow, the wall is too weak to withstand internal turgor pressure, leading to osmotic lysis (bursting) of the cell.

Marking scheme

Award 1 mark for identifying the correct mechanism of action of penicillin: inhibition of transpeptidase and prevention of peptidoglycan cross-linking.
Question 33 · Multiple Choice
1 marks
An experiment was carried out to investigate the effect of two different inhibitors, X and Y, on the rate of reaction of an enzyme. The concentrations of the enzyme and the substrate were kept constant, except that the substrate concentration was varied to determine \(K_m\) and \(V_{\text{max}}\). Inhibitor X increased the value of \(K_m\) but did not change \(V_{\text{max}}\). Inhibitor Y decreased \(V_{\text{max}}\) but did not change \(K_m\). Which statement correctly identifies the types of inhibition shown by X and Y?
  1. A.X is a competitive inhibitor and binds to the active site; Y is a non-competitive inhibitor and binds to an allosteric site.
  2. B.X is a non-competitive inhibitor and binds to an allosteric site; Y is a competitive inhibitor and binds to the active site.
  3. C.X is a competitive inhibitor and binds to an allosteric site; Y is a non-competitive inhibitor and binds to the active site.
  4. D.X is a non-competitive inhibitor and binds to the active site; Y is a competitive inhibitor and binds to an allosteric site.
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Worked solution

Competitive inhibitors compete with the substrate for the active site. They increase \(K_m\) (reducing substrate affinity) but do not affect \(V_{\text{max}}\) because high substrate concentrations can overcome the inhibition. Non-competitive inhibitors bind to an allosteric site, reducing the maximum rate of reaction (\(V_{\text{max}}\)) by rendering some enzymes inactive, but do not affect the affinity of the unaffected enzymes for the substrate, so \(K_m\) remains unchanged. Therefore, X is a competitive inhibitor and Y is a non-competitive inhibitor.

Marking scheme

1 mark for identifying that inhibitor X is competitive and binds to the active site, and inhibitor Y is non-competitive and binds to an allosteric site.
Question 34 · Multiple Choice
1 marks
The path taken by a newly synthesized membrane protein from its site of synthesis to its final destination in the cell surface membrane involves several organelles. Which sequence correctly identifies the order of organelles and structures through which the protein or its precursor passes?
  1. A.nucleolus \(\rightarrow\) rough endoplasmic reticulum \(\rightarrow\) transport vesicle \(\rightarrow\) Golgi body \(\rightarrow\) secretory vesicle
  2. B.rough endoplasmic reticulum \(\rightarrow\) transport vesicle \(\rightarrow\) Golgi body \(\rightarrow\) secretory vesicle \(\rightarrow\) cell surface membrane
  3. C.free ribosome \(\rightarrow\) Golgi body \(\rightarrow\) transport vesicle \(\rightarrow\) rough endoplasmic reticulum \(\rightarrow\) secretory vesicle
  4. D.rough endoplasmic reticulum \(\rightarrow\) secretory vesicle \(\rightarrow\) Golgi body \(\rightarrow\) transport vesicle \(\rightarrow\) cell surface membrane
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Worked solution

Proteins destined for insertion into the cell surface membrane are synthesized by ribosomes bound to the rough endoplasmic reticulum (RER). They are folded within the RER and then packaged into transport vesicles. These vesicles fuse with the Golgi body, where the proteins undergo post-translational modifications (e.g., glycosylation). Modified proteins are then packaged into secretory vesicles which move to and fuse with the cell surface membrane.

Marking scheme

1 mark for the correct chronological sequence of organelles and transport structures involved in membrane protein synthesis and routing.
Question 35 · Multiple Choice
1 marks
A diploid cell of an organism contains 12 pairs of chromosomes (\(2n = 24\)). During the mitotic cell cycle, the quantity of DNA and the number of chromosomes and chromatids change. Which row in the table correctly shows the number of chromosomes, chromatids, and DNA molecules present in this cell during prophase and telophase? (Assume telophase is measured before cytokinesis is complete, for the two reforming nuclei combined within the single cell).
  1. A.Prophase: 24 chromosomes, 48 chromatids, 48 DNA molecules; Telophase: 48 chromosomes, 0 chromatids, 48 DNA molecules
  2. B.Prophase: 24 chromosomes, 24 chromatids, 24 DNA molecules; Telophase: 24 chromosomes, 24 chromatids, 24 DNA molecules
  3. C.Prophase: 48 chromosomes, 48 chromatids, 96 DNA molecules; Telophase: 48 chromosomes, 0 chromatids, 48 DNA molecules
  4. D.Prophase: 24 chromosomes, 48 chromatids, 48 DNA molecules; Telophase: 24 chromosomes, 0 chromatids, 24 DNA molecules
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Worked solution

In prophase, DNA has replicated, so each of the 24 chromosomes consists of 2 sister chromatids joined at a centromere. This gives 24 chromosomes, 48 chromatids, and 48 DNA molecules. During anaphase, the sister chromatids separate to become individual chromosomes, doubling the chromosome count in the single cell to 48. In telophase, before cytokinesis separates the cytoplasm, there are 48 individual chromosomes, 0 chromatids (since they are no longer paired at centromeres), and 48 DNA molecules within the cell boundaries.

Marking scheme

1 mark for identifying the correct counts of chromosomes, chromatids, and DNA molecules in both stages.
Question 36 · Multiple Choice
1 marks
Which row correctly describes the structural features of an artery, a vein, and a capillary?
  1. A.Artery: thick tunica media with many elastic fibers and smooth muscle; Vein: thin tunica media, few elastic fibers, valves present; Capillary: wall consists of a single layer of endothelial cells only, no elastic fibers, no valves.
  2. B.Artery: thin tunica media with many elastic fibers and smooth muscle; Vein: thick tunica media, valves present; Capillary: wall consists of a single layer of endothelial cells, elastic fibers present.
  3. C.Artery: thick tunica media with no smooth muscle; Vein: thin tunica media, no valves; Capillary: wall consists of a single layer of endothelial cells and collagen fibers.
  4. D.Artery: thick tunica media with elastic fibers and valves; Vein: thin tunica media with no elastic fibers, valves present; Capillary: wall is several cells thick to withstand pressure.
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Worked solution

Arteries must withstand and maintain high hydrostatic pressure, thus possessing a thick tunica media with abundant elastic fibers and smooth muscle. Veins carry blood under low pressure and have a much thinner tunica media, fewer elastic fibers, and valves to prevent backflow. Capillaries consist solely of a single layer of endothelial cells (tunica intima) to allow rapid diffusion, completely lacking elastic fibers, smooth muscle, and valves.

Marking scheme

1 mark for correctly matching the histological features of arteries, veins, and capillaries.
Question 37 · Multiple Choice
1 marks
A section of a DNA template strand has the sequence: \(3'\text{-TACGCTAGGAAG-5'}\). During transcription, this template strand is used to synthesize a molecule of mRNA, which is then translated. What will be the sequence of anticodons on the tRNA molecules that bind to the codons of this mRNA sequence (written in the \(3' \rightarrow 5'\) direction from left to right)?
  1. A.\(3'\text{-UAC-5'}\), \(3'\text{-GCU-5'}\), \(3'\text{-AGG-5'}\), \(3'\text{-AAG-5'}\)
  2. B.\(5'\text{-UAC-3'}\), \(5'\text{-GCU-3'}\), \(5'\text{-AGG-3'}\), \(5'\text{-AAG-3'}\)
  3. C.\(3'\text{-AUG-5'}\), \(3'\text{-CGA-5'}\), \(3'\text{-UCC-5'}\), \(3'\text{-UUC-5'}\)
  4. D.\(3'\text{-ATG-5'}\), \(3'\text{-CGT-5'}\), \(3'\text{-TCC-5'}\), \(3'\text{-TTC-5'}\)
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Worked solution

The DNA template strand \(3'\text{-TACGCTAGGAAG-5'}\) is transcribed into mRNA by complementary base pairing, yielding the mRNA sequence: \(5'\text{-AUGCGAUCCUUC-3'}\). The mRNA codons are \(5'\text{-AUG-3'}\), \(5'\text{-CGA-3'}\), \(5'\text{-UCC-3'}\), and \(5'\text{-UUC-3'}\). The complementary tRNA anticodons bind antiparallel, so they are written in the \(3' \rightarrow 5'\) direction as: \(3'\text{-UAC-5'}\), \(3'\text{-GCU-5'}\), \(3'\text{-AGG-5'}\), and \(3'\text{-AAG-5'}\).

Marking scheme

1 mark for deriving the correct complementary mRNA codons and identifying the correct antiparallel tRNA anticodon sequences.
Question 38 · Multiple Choice
1 marks
Which statement correctly describes the function of T-helper lymphocytes in the mammalian immune system?
  1. A.They release cytokines that stimulate B-lymphocytes to divide and differentiate into plasma cells, and activate macrophages.
  2. B.They directly secrete large quantities of antigen-specific antibodies into the blood plasma to neutralize pathogens.
  3. C.They engulf pathogens by phagocytosis and present their antigens on major histocompatibility complex (MHC) proteins.
  4. D.They release perforins and granzymes to induce apoptosis in host cells infected with intracellular viruses.
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Worked solution

T-helper cells express T-cell receptors that bind to specific antigens presented on antigen-presenting cells (APCs). Once activated, they secrete signaling molecules called cytokines (such as interleukins). Cytokines stimulate clonal expansion and differentiation of antigen-specific B-lymphocytes into antibody-secreting plasma cells, and they also activate macrophages to perform phagocytosis more efficiently. Option B is incorrect because plasma cells secrete antibodies. Option C is incorrect because phagocytes present antigens. Option D describes the role of cytotoxic (T-killer) lymphocytes.

Marking scheme

1 mark for identifying the correct physiological role of cytokines released by activated T-helper lymphocytes.
Question 39 · Multiple Choice
1 marks
A tissue section from the human gas exchange system is examined under a light microscope. The specimen contains ciliated epithelial cells, goblet cells, smooth muscle, and elastic fibers, but completely lacks cartilage. Which part of the gas exchange system is being observed?
  1. A.Trachea
  2. B.Bronchus
  3. C.Bronchiole
  4. D.Alveolus
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Worked solution

Trachea and bronchi both possess cartilage to keep the airways open (C-shaped rings in trachea, irregular plates in bronchi). Alveoli lack cilia, goblet cells, and smooth muscle entirely, consisting of simple squamous epithelium. Bronchioles lack cartilage but still contain smooth muscle, elastic fibers, and ciliated epithelium (with goblet cells present in larger bronchioles, although fewer than in the bronchi). Thus, the specimen is a bronchiole.

Marking scheme

1 mark for identifying the bronchiole based on the presence of ciliated epithelium, smooth muscle, and the absence of cartilage.
Question 40 · Multiple Choice
1 marks
A student uses a light microscope fitted with an eyepiece graticule and a stage micrometer to measure the diameter of a plant cell. The stage micrometer has scale divisions that are \(0.01\text{ mm}\) apart. At a magnification of \(\times 400\), 10 divisions of the stage micrometer align exactly with 50 divisions of the eyepiece graticule. The student then observes the plant cell at the same magnification and finds that the cell has a diameter of 30 eyepiece graticule divisions. What is the actual diameter of the plant cell?
  1. A.\(6\ \mu\text{m}\)
  2. B.\(60\ \mu\text{m}\)
  3. C.\(150\ \mu\text{m}\)
  4. D.\(600\ \mu\text{m}\)
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Worked solution

1. Find the distance of 10 stage micrometer divisions: \(10 \times 0.01\text{ mm} = 0.1\text{ mm} = 100\ \mu\text{m}\).
2. Calculate the value of 1 eyepiece graticule division (epu): \(100\ \mu\text{m} / 50 = 2\ \mu\text{m}\) per epu.
3. Calculate the actual size of the plant cell: \(30\text{ epu} \times 2\ \mu\text{m}/\text{epu} = 60\ \mu\text{m}\).

Marking scheme

1 mark for correct calibration of the eyepiece graticule and calculation of the cell's actual diameter.

Paper 2 (AS Structured Questions)

Answer all six structured questions in the spaces provided.
6 Question · 60 marks
Question 1 · Structured
10 marks
Amylase is an enzyme that catalyses the hydrolysis of starch. An investigation was carried out into the effect of temperature on the rate of reaction of a bacterial amylase in the presence and absence of a competitive inhibitor, compound X. The substrate concentration was kept constant throughout the investigation.

At 40 °C, the rate of reaction without compound X is 2.8 \(\mu\text{mol dm}^{-3}\text{ min}^{-1}\) and with compound X is 1.2 \(\mu\text{mol dm}^{-3}\text{ min}^{-1}\).

(a) Explain how a non-competitive inhibitor decreases the rate of an enzyme-controlled reaction, and how this differs from the action of a competitive inhibitor such as compound X. [3]

(b) (i) Calculate the percentage decrease in the rate of reaction caused by compound X at 40 °C. Show your working and give your answer to one decimal place. [2]

(ii) With reference to the active site and molecular collisions, explain the effect of the competitive inhibitor compound X on the rate of reaction at 40 °C. [3]

(c) Explain why the rate of reaction of the amylase enzyme falls to zero at temperatures above 70 °C, even in the absence of compound X. [2]
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Worked solution

(a) A non-competitive inhibitor binds to an allosteric site of the enzyme, altering its tertiary structure and changing the shape of the active site so that the substrate is no longer complementary, preventing the formation of enzyme-substrate complexes. This differs from a competitive inhibitor, which binds directly to the active site and whose inhibitory effect can be overcome by high substrate concentrations.

(b) (i) Decrease in rate = 2.8 - 1.2 = 1.6 \(\mu\text{mol dm}^{-3}\text{ min}^{-1}\).
Percentage decrease = \(\frac{1.6}{2.8} \times 100 = 57.1\\%\).

(ii) Compound X has a shape complementary to the active site of the enzyme, enabling it to compete directly with substrate molecules for binding. This reduces the frequency of successful collisions between the substrate and the active site, leading to fewer enzyme-substrate complexes forming per unit time.

(c) At temperatures above 70 °C, high kinetic energy causes weak bonds (such as hydrogen and ionic bonds) holding the tertiary structure of the protein together to break. The enzyme denatures, causing a permanent change in the shape of its active site so that the substrate can no longer bind and no enzyme-substrate complexes can form.

Marking scheme

(a) Max 3 marks:
1. Non-competitive inhibitor binds to an allosteric site / site other than the active site;
2. Changes tertiary structure / 3D shape of enzyme;
3. Active site shape is altered / is no longer complementary, so substrate cannot bind / no enzyme-substrate complexes (ESCs) form;
4. Contrast: Competitive inhibitor binds to active site / its effect can be overcome by increasing substrate concentration (whereas non-competitive cannot);

(b)(i) Max 2 marks:
1. Correct calculation formula: \(\frac{2.8 - 1.2}{2.8} \times 100\) (or equivalent shown in working);
2. Correct final answer to one decimal place: 57.1% (Accept 57.1. Reject 57 or 57.14);

(b)(ii) Max 3 marks:
1. Compound X has a similar shape to the substrate / shape complementary to the active site;
2. Competes with the substrate to bind to the active site;
3. Reduces the frequency of successful collisions / fewer ESCs form per unit time;

(c) Max 2 marks:
1. High kinetic energy causes hydrogen / ionic / hydrophobic bonds to break;
2. Loss of tertiary structure / denaturation of enzyme;
3. Active site changes shape permanently, so substrate cannot bind / no ESCs form.
Question 2 · Structured
10 marks
Eukaryotic cells contain many specialized organelles that allow biological processes to occur efficiently. Protein synthesis and secretion involve the coordinated action of several membranous organelles.

(a) Describe the path taken by a newly synthesized protein from the site of its translation until it is secreted from the cell, detailing the role of each organelle involved. [4]

(b) Plant cells contain a large, permanent vacuole.
(i) Name the single membrane that surrounds the vacuole of a plant cell. [1]
(ii) State two functions of the permanent vacuole in plant cells. [2]

(c) Microtubules are structural components of the cytoskeleton.
Describe the structure of microtubules and explain how they contribute to the movement of organelles within the cell. [3]
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Worked solution

(a) Translation occurs on ribosomes bound to the rough endoplasmic reticulum (RER). The polypeptide enters the lumen of the RER, where it is folded into its tertiary structure. Transport vesicles bud off from the RER and travel to and fuse with the Golgi body. In the Golgi body, proteins are modified (such as by adding carbohydrate chains to form glycoproteins) and sorted. Secretory vesicles then bud off the trans face of the Golgi body, move along microtubules, and fuse with the cell surface membrane to release the protein by exocytosis.

(b) (i) Tonoplast
(ii) The permanent vacuole stores cell sap (water, inorganic ions, pigments, or organic solutes) and maintains cell turgor pressure / turgidity to support the plant structure.

(c) Microtubules are hollow, unbranched cylinders composed of polymerised alpha- and beta-tubulin dimers. They act as intracellular 'tracks' along which motor proteins (such as kinesin or dynein) attach to organelles and transport them across the cytoplasm using energy released from ATP hydrolysis.

Marking scheme

(a) Max 4 marks:
1. Translation occurs on ribosomes bound to the rough endoplasmic reticulum (RER);
2. Protein enters the RER lumen for folding into tertiary structure;
3. Transport vesicles bud off RER and move to / fuse with Golgi body;
4. In Golgi body, protein is modified / processed (e.g. addition of carbohydrate chain to form glycoprotein) / packaged into secretory vesicles;
5. Secretory vesicles move to cell surface membrane and fuse with it to release protein via exocytosis;

(b)(i) 1 mark:
1. Tonoplast;

(b)(ii) Max 2 marks:
1. Storage of water / inorganic ions / organic solutes / waste products / pigments;
2. Maintaining turgor pressure / keeping cell turgid (for support);
3. Hydrolysis of macromolecules (contains lysosomal-like enzymes);

(c) Max 3 marks:
1. Hollow cylinders / tubes;
2. Made of tubulin / protofilaments (composed of \(\alpha\)- and \(\beta\)-tubulin dimers);
3. Act as intracellular tracks / scaffolding;
4. Motor proteins (kinesin/dynein) attach to organelle and 'walk' along microtubules using ATP hydrolysis;
Question 3 · Structured
10 marks
Mitosis is a highly regulated process in the eukaryotic cell cycle that ensures genetic consistency across cell generations.

(a) During one of the stages of mitosis, sister chromatids are pulled apart to opposite poles of the spindle.
(i) Identify this stage of mitosis and describe the main events that occur during it. [3]
(ii) Explain the role of the centromere during mitosis. [2]

(b) (i) Distinguish between the terms *chromosome* and *chromatid*. [2]
(ii) State the name of the stage of the mitotic cell cycle during which DNA replication occurs and explain why DNA replication is essential before mitosis can take place. [3]
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Worked solution

(a) (i) The stage is anaphase. During anaphase, the centromeres divide and the spindle fibres shorten and contract. This pulls the sister chromatids (now individual chromosomes) apart, centromeres leading, towards opposite poles of the cell.
(ii) The centromere holds the two sister chromatids together until they are ready to separate in anaphase. It is also the site of attachment for spindle fibres (at the kinetochore), allowing proper alignment and equal distribution of chromosomes to each pole.

(b) (i) A chromosome is a single, double-stranded molecule of DNA (or a structure containing two identical sister chromatids joined at a centromere). A chromatid is one of the two identical copies of DNA that make up a duplicated chromosome prior to division.
(ii) DNA replication occurs during the S (synthesis) phase of interphase. It is essential because it duplicates the genetic material so that each daughter cell receives an identical, complete copy of the genome, maintaining genetic stability and ensuring no genes are lost.

Marking scheme

(a)(i) Max 3 marks:
1. Anaphase;
2. Centromeres split / divide;
3. Spindle fibres contract / shorten;
4. Chromatids pulled / separated to opposite poles (centromeres first / V-shape);

(a)(ii) Max 2 marks:
1. Holds sister chromatids together (until anaphase);
2. Attachment site for spindle fibres / microtubules (via kinetochore);
3. Ensures equal segregation / correct alignment of chromosomes;

(b)(i) Max 2 marks:
1. Chromosome: refers to the complete unit of condensed chromatin; can exist as a single chromatid (G1 phase) or two sister chromatids (after S phase);
2. Chromatid: is one of the two identical strands of a replicated chromosome joined by a centromere;

(b)(ii) Max 3 marks:
1. S phase / synthesis phase (of interphase);
2. Duplicates DNA / genetic material exactly;
3. Ensures both daughter cells receive a full / diploid set of chromosomes / genetically identical copy;
4. Prevents loss of genetic information / maintains genetic stability;
Question 4 · Structured
10 marks
Mammals possess a closed, double circulatory system to meet their high metabolic demands. Blood flows through different types of blood vessels with specialized walls.

(a) (i) Explain what is meant by a *double* circulatory system. [2]
(ii) Suggest two advantages of a double circulatory system compared to a single circulatory system. [2]

(b) Complete the table below by placing a tick (\(\checkmark\)) if the feature is present and a cross (\(\times\)) if it is absent in the wall or structure of each vessel.

| Feature | Artery | Vein | Capillary |
| :--- | :---: | :---: | :---: |
| Has a wall containing elastic fibres | ... | ... | ... |
| Has a wall containing smooth muscle | ... | ... | ... |
| Has a lumen that is about 7–8 \(\mu\text{m}\) wide | ... | ... | ... | [3]

(c) Describe the process of tissue fluid formation at the arteriole end of a capillary bed, and explain how most of this fluid is returned to the blood at the venule end. [3]
Show answer & marking scheme

Worked solution

(a) (i) A double circulatory system means that blood passes through the heart twice for each complete circuit of the body, involving two separate circuits: the pulmonary circulation (to the lungs) and the systemic circulation (to the body tissues).
(ii) Advantages include: maintaining a higher blood pressure in the systemic circulation to deliver oxygen and nutrients rapidly to metabolizing tissues, and keeping a lower blood pressure in the pulmonary circulation to protect delicate lung capillaries and allow slower flow for optimal gas exchange.

(b)
| Feature | Artery | Vein | Capillary |
| :--- | :---: | :---: | :---: |
| Has a wall containing elastic fibres | \(\checkmark\) | \(\checkmark\) | \(\times\) |
| Has a wall containing smooth muscle | \(\checkmark\) | \(\checkmark\) | \(\times\) |
| Has a lumen that is about 7–8 \(\mu\text{m}\) wide | \(\times\) | \(\times\) | \(\checkmark\) |

(c) At the arteriole end, the hydrostatic pressure of the blood is higher than the osmotic (oncotic) pressure. This net outward filtration force drives water and small dissolved solutes out of the capillary through small gaps in the endothelium into the tissue spaces, forming tissue fluid. At the venule end, the hydrostatic pressure falls significantly due to friction and fluid loss, making it lower than the osmotic pressure (which remains high due to large plasma proteins remaining in the blood). This creates a net inward force, causing water to return to the capillary by osmosis down a water potential gradient.

Marking scheme

(a)(i) Max 2 marks:
1. Blood passes through the heart twice for each complete circuit;
2. Separate pulmonary (heart to lungs and back) and systemic (heart to body and back) circulations;

(a)(ii) Max 2 marks:
1. Allows high pressure in systemic circulation / rapid flow of blood to body tissues;
2. Low pressure in pulmonary circulation protects lung tissues / capillaries;
3. Oxygenated and deoxygenated blood are kept completely separate;

(b) 3 marks:
- All 9 entries correct = 3 marks
- 7 or 8 entries correct = 2 marks
- 5 or 6 entries correct = 1 mark
- Less than 5 correct = 0 marks

*Correct table entries:*
- Elastic fibres: Artery (\(\checkmark\)), Vein (\(\checkmark\)), Capillary (\(\times\))
- Smooth muscle: Artery (\(\checkmark\)), Vein (\(\checkmark\)), Capillary (\(\times\))
- Lumen 7-8 \(\mu\text{m}\): Artery (\(\times\)), Vein (\(\times\)), Capillary (\(\checkmark\))

(c) Max 3 marks:
1. At arteriole end, blood hydrostatic pressure exceeds solute / oncotic pressure;
2. Water and small solutes forced out of capillaries (through endothelial pores) into tissue spaces;
3. At venule end, hydrostatic pressure is lower than oncotic pressure (due to plasma proteins);
4. Water moves back into capillary by osmosis down water potential gradient;
Question 5 · Structured
10 marks
Protein synthesis is a multi-step process that transfers genetic information from DNA inside the nucleus to polypeptide chains assembled on ribosomes.

(a) Transcription is the first stage of protein synthesis.
(i) Describe the role of RNA polymerase during transcription. [3]
(ii) State two differences between the structure of a DNA molecule and an mRNA molecule. [2]

(b) Transfer RNA (tRNA) plays an essential role in translating the genetic code during protein synthesis.
(i) Describe how a tRNA molecule is adapted to its function. [3]
(ii) The codon on a molecule of mRNA is 5'-AUG-3'. State the sequence of the anticodon on the complementary tRNA molecule and state the name of the amino acid that this tRNA carries. [2]
Show answer & marking scheme

Worked solution

(a) (i) RNA polymerase binds to the promoter region of DNA, unwinding the double helix to expose the template strand. It then moves along the template strand, lining up free RNA nucleotides by complementary base pairing (A to U, T to A, C to G). RNA polymerase catalyzes the formation of phosphodiester bonds to join adjacent RNA nucleotides together, forming a primary mRNA transcript.
(ii) DNA is double-stranded, contains deoxyribose sugar, and contains thymine, whereas mRNA is single-stranded, contains ribose sugar, and contains uracil.

(b) (i) A tRNA molecule has a specific three-base anticodon sequence that is complementary to an mRNA codon. It also features a specific amino acid attachment site at its 3' end to carry a corresponding amino acid. The single-stranded RNA folds into a stable cloverleaf shape held by internal hydrogen bonding, which is recognized by specific enzymes.
(ii) The complementary anticodon sequence is 3'-UAC-5' (accept UAC). The amino acid carried by this tRNA is methionine.

Marking scheme

(a)(i) Max 3 marks:
1. Binds to promoter (on DNA) and unwinds DNA double helix / breaks hydrogen bonds;
2. Moves along template strand (in 3' to 5' direction);
3. Aligns free RNA nucleotides by complementary base pairing;
4. Catalyzes formation of phosphodiester bonds between RNA nucleotides;

(a)(ii) Max 2 marks (any two from):
1. DNA is double-stranded / double helix, whereas mRNA is single-stranded;
2. DNA contains deoxyribose sugar, whereas mRNA contains ribose sugar;
3. DNA contains thymine (T), whereas mRNA contains uracil (U);
4. DNA is long / has hydrogen bonds between strands, whereas mRNA is relatively short / linear;

(b)(i) Max 3 marks:
1. Has a specific anticodon loop (three bases) complementary to mRNA codon;
2. Has an amino acid binding site (at the 3' end) to carry a specific amino acid;
3. Cloverleaf/folded shape held by hydrogen bonds between bases, providing stability;
4. Has specific shapes/sites recognized by aminoacyl-tRNA synthetase enzymes;

(b)(ii) 2 marks:
1. Anticodon: UAC (accept 3'-UAC-5' or CAU if 5' to 3' specified, but standard is UAC / 3'-UAC-5') [1];
2. Amino acid: Methionine (accept Met) [1];
Question 6 · Structured
10 marks
Plants possess highly specialized transport tissues to move water, minerals, and organic compounds over long distances.

(a) Plant stems contain vascular bundles consisting of xylem and phloem.
(i) Distinguish between the transport functions of xylem tissue and phloem tissue. [2]
(ii) Explain how xylem vessels are adapted for the efficient transport of water and mineral ions. [4]

(b) Phloem tissue contains sieve tube elements and companion cells.
(i) Describe how the structure of a companion cell is adapted to support the function of a sieve tube element. [3]
(ii) Name the cytoplasmic junctions that connect companion cells directly to the adjacent sieve tube elements. [1]
Show answer & marking scheme

Worked solution

(a) (i) Xylem tissue transports water and dissolved mineral ions unidirectionally from the roots to the leaves. Phloem tissue transports organic solutes / assimilates (such as sucrose and amino acids) bidirectionally from sources (where they are produced) to sinks (where they are used or stored).
(ii) Xylem vessels are made of dead cells with no cytoplasm or organelles, leaving a completely hollow lumen that offers very low resistance to water flow. Their end walls are lost, creating continuous tubes for an uninterrupted column of water. The walls are thickened with lignin, which provides mechanical strength to prevent the vessel collapsing under the high tension created by transpiration pull. Pits in the lignified walls allow the lateral movement of water between adjacent vessels.

(b) (i) Companion cells have many mitochondria to generate ATP via aerobic respiration to power the active loading of sucrose. They contain many ribosomes and rough endoplasmic reticulum to synthesize the transport proteins, such as proton pumps and co-transporter proteins, required for transport. They also contain a prominent nucleus to coordinate metabolic activities for both themselves and the sieve tube elements.
(ii) Plasmodesmata (singular: plasmodesma).

Marking scheme

(a)(i) Max 2 marks:
1. Xylem: transports water and mineral ions upwards / from roots to leaves / unidirectionally;
2. Phloem: transports assimilates / sucrose / organic solutes from source to sink / bidirectionally;

(a)(ii) Max 4 marks:
1. Lignified walls resist negative pressure / tension, preventing collapse;
2. Lignin provides waterproofing, preventing water loss from the vessel;
3. Cells are dead / contain no cytoplasm / no organelles, offering minimal resistance to flow;
4. End walls are absent / perforated, allowing a continuous column of water;
5. Pits in walls allow lateral movement of water (to bypass blocks);

(b)(i) Max 3 marks:
1. Large number of mitochondria to provide ATP for active transport / loading of sucrose;
2. Presence of proton-pumping proteins / sucrose co-transporter proteins in the membrane;
3. Numerous ribosomes / rough ER to produce transport proteins;
4. Nucleus is present to direct protein synthesis and metabolic processes for the anucleate sieve tube element;

(b)(ii) 1 mark:
1. Plasmodesmata (accept plasmodesma);

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