An original Thinka practice paper modelled on the structure and difficulty of the Jun 2025 (V3) 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 best answer from the four options A, B, C or D.
40 PastPaper.question · 40 PastPaper.marks
PastPaper.question 1 · multiple-choice
1 PastPaper.marks
A plant stem is stained with phloroglucinol, which stains lignified cell walls red-brown. Which tissues or cell structures will show a red-brown color when viewed under a light microscope? 1. Sclerenchyma fibers, 2. Companion cells, 3. Xylem vessels, 4. Sieve tube elements.
A.1 and 3 only
B.2 and 4 only
C.1, 2 and 3 only
D.1, 3 and 4 only
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Sclerenchyma fibers and xylem vessels have highly lignified cell walls to provide mechanical support and withstand tension, respectively. Therefore, they will be stained red-brown by phloroglucinol. Companion cells and sieve tube elements are components of phloem and possess non-lignified cellulose cell walls, so they will not show a red-brown color.
PastPaper.markingScheme
1 mark for identifying that sclerenchyma fibers and xylem vessels are lignified, while companion cells and sieve tube elements are not, leading to the selection of option A.
PastPaper.question 2 · multiple-choice
1 PastPaper.marks
How will the maximum rate of reaction (\(V_{max}\)) and the Michaelis-Menten constant (\(K_m\)) of an enzyme-controlled reaction change when a competitive inhibitor is added?
A competitive inhibitor competes with the substrate for the active site. At very high substrate concentrations, the substrate outcompetes the inhibitor, allowing the enzyme to still reach its maximum rate of reaction (\(V_{max}\)), so \(V_{max}\) remains unchanged. However, a higher concentration of substrate is required to reach half of \(V_{max}\), which means the Michaelis-Menten constant (\(K_m\)) increases.
PastPaper.markingScheme
1 mark for identifying that competitive inhibition does not change the \(V_{max}\) of the enzyme but increases the \(K_m\) value.
PastPaper.question 3 · multiple-choice
1 PastPaper.marks
An antibody molecule is a quaternary protein with a specific Y-shaped structure. Which regions of an IgG antibody molecule form the antigen-binding sites, and which region determines the specific class of the antibody?
A.Antigen-binding sites: Variable regions of both heavy and light chains. Class determination: Constant region of heavy chains.
B.Antigen-binding sites: Variable regions of heavy chains only. Class determination: Constant region of light chains.
C.Antigen-binding sites: Constant regions of both heavy and light chains. Class determination: Variable region of heavy chains.
D.Antigen-binding sites: Variable regions of light chains only. Class determination: Constant region of both heavy and light chains.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
The antigen-binding sites are formed by the variable regions (V regions) at the amino-terminal ends of both the heavy and light polypeptide chains. The constant region of the heavy chain determines the class of the antibody (such as IgG, IgA, or IgM) and dictates its effector function.
PastPaper.markingScheme
1 mark for correctly matching the variable regions of both heavy and light chains to the antigen-binding sites, and the constant region of the heavy chain to the class determination.
PastPaper.question 4 · multiple-choice
1 PastPaper.marks
Which statement correctly describes the structural features of the fibrous protein collagen?
A.It is composed of three separate polypeptide chains wound around each other, with every third amino acid being glycine to allow close packing.
B.It is composed of a single polypeptide chain folded into a spherical shape with a high proportion of hydrophilic amino acid residues on its outer surface.
C.It is composed of four polypeptide chains held together by hydrophobic interactions and disulfide bonds, containing a non-protein prosthetic group.
D.It consists of three identical alpha-helices held together by hydrogen bonds, forming a highly soluble structure.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Collagen is a fibrous protein consisting of three helical polypeptide chains wound together to form a triple helix. The amino acid sequence consists of a repeating pattern where every third residue is glycine. Because glycine is the smallest amino acid (having only a hydrogen atom as its R-group), it allows the three chains to pack tightly together, forming a very strong structure. It is insoluble in water and is not globular.
PastPaper.markingScheme
1 mark for identifying the correct triple-helix structure of collagen containing glycine at every third position to facilitate close packing.
PastPaper.question 5 · multiple-choice
1 PastPaper.marks
A template strand of DNA has the sequence 3'- TAC GGC TTA CTA ACT -5'. During transcription and translation, what is the sequence (written from 5' to 3') of the anticodon on the tRNA molecule that base-pairs with the third codon of the transcribed mRNA?
A.5'- AAU -3'
B.5'- AUU -3'
C.5'- TTA -3'
D.5'- UUA -3'
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
First, transcribe the template DNA strand to mRNA: Template DNA: 3'- TAC GGC TTA CTA ACT -5'. Complementary mRNA strand: 5'- AUG CCG AAU GAU UGA -3'. The third codon on this mRNA is 5'- AAU -3'. The tRNA anticodon that base-pairs with 5'- AAU -3' is antiparallel and complementary, which is 3'- UUA -5'. Written in the standard 5' to 3' direction, this is 5'- AUU -3'.
PastPaper.markingScheme
1 mark for correctly transcribing the DNA template to mRNA, identifying the third codon (5'- AAU -3'), determining the complementary anticodon (3'- UUA -5'), and correctly writing it in the 5' to 3' direction (5'- AUU -3').
PastPaper.question 6 · multiple-choice
1 PastPaper.marks
Which statement describes the correct mode of action of penicillin on bacteria?
A.It binds to the 70S ribosome, preventing the translation of essential enzymes.
B.It inhibits the enzyme transpeptidase, preventing the formation of cross-links between peptidoglycan chains in the bacterial cell wall.
C.It binds directly to the bacterial cell membrane, causing leakage of ions and subsequent cell lysis.
D.It acts as a competitive inhibitor of DNA polymerase, halting replication of the bacterial chromosome.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Penicillin acts by inhibiting the enzyme transpeptidase, which is responsible for catalyzing the formation of peptide cross-links between peptidoglycan polymer chains in the bacterial cell wall. Without these cross-links, the cell wall is weakened. When water enters the bacterial cell by osmosis, the weakened cell wall cannot withstand the turgor pressure, leading to osmotic lysis.
PastPaper.markingScheme
1 mark for identifying that penicillin inhibits transpeptidase and prevents the cross-linking of peptidoglycan in bacterial cell walls.
PastPaper.question 7 · multiple-choice
1 PastPaper.marks
A student calibrates a light microscope using an eyepiece graticule and a stage micrometer. At a magnification of x100, each division on the stage micrometer is equal to 10 \(\mu\text{m}\). The student notes that 100 divisions of the eyepiece graticule line up exactly with 10 divisions of the stage micrometer. What is the actual length of a plant cell that spans 25 eyepiece graticule divisions under this same magnification?
A.2.5 \(\mu\text{m}\)
B.25.0 \(\mu\text{m}\)
C.250 \(\mu\text{m}\)
D.2500 \(\mu\text{m}\)
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
First, calculate the actual distance represented by the stage micrometer divisions: 10 divisions x 10 \(\mu\text{m}\) per division = 100 \(\mu\text{m}\). Since 100 eyepiece graticule (EPG) divisions line up with these 10 stage micrometer divisions, 100 EPG divisions = 100 \(\mu\text{m}\). Therefore, 1 EPG division = 1 \(\mu\text{m}\). A plant cell spanning 25 EPG divisions will have an actual length of: 25 divisions x 1 \(\mu\text{m}\) per division = 25.0 \(\mu\text{m}\).
PastPaper.markingScheme
1 mark for calculating that 1 eyepiece graticule division equals 1 \(\mu\text{m}\), and multiplying this by 25 to obtain the correct actual length of 25.0 \(\mu\text{m}\).
PastPaper.question 8 · multiple-choice
1 PastPaper.marks
Plant cells are placed in a hypertonic solution of sucrose. What changes will occur to the protoplast volume, the water potential of the cell, and the turgor pressure of these cells?
A.Protoplast volume: decreases; Water potential: decreases; Turgor pressure: decreases.
B.Protoplast volume: increases; Water potential: increases; Turgor pressure: increases.
C.Protoplast volume: decreases; Water potential: increases; Turgor pressure: decreases.
D.Protoplast volume: increases; Water potential: decreases; Turgor pressure: increases.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
A hypertonic solution has a lower water potential than the cytoplasm of the plant cells. Water will therefore leave the cells down a water potential gradient by osmosis. As water leaves, the protoplast shrinks away from the cell wall (decreasing in volume). The loss of water increases the concentration of solutes inside the cell, which decreases (makes more negative) the water potential of the cell. The turgor pressure also decreases as the protoplast no longer exerts pressure against the rigid cell wall.
PastPaper.markingScheme
1 mark for identifying that placement in a hypertonic solution leads to a decrease in protoplast volume, a decrease (more negative) in water potential, and a decrease in turgor pressure.
PastPaper.question 9 · multiple-choice
1 PastPaper.marks
Which row correctly identifies the cellular characteristics of a sieve tube element, a companion cell, and a xylem vessel element?
A.Sieve tube element: peripheral cytoplasm; Companion cell: high density of mitochondria; Xylem vessel element: lignified walls with pits
B.Sieve tube element: dead at maturity; Companion cell: lacks a nucleus; Xylem vessel element: possesses plasmodesmata
D.Sieve tube element: lacks a plasma membrane; Companion cell: active proton pumps in the tonoplast; Xylem vessel element: no end walls
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Sieve tube elements are living cells containing a thin layer of peripheral cytoplasm but no nucleus. Companion cells are highly metabolically active with many mitochondria. Xylem vessel elements are dead, hollow tubes with lignified walls containing pits to allow lateral movement of water.
PastPaper.markingScheme
1 mark for selecting the correct row containing all three correct cellular characteristics.
PastPaper.question 10 · multiple-choice
1 PastPaper.marks
An enzyme-catalysed reaction was monitored in both the presence and absence of an inhibitor. The maximum rate of reaction \(V_{\text{max}}\) was halved in the presence of the inhibitor, but the Michaelis-Menten constant \(K_{\text{m}}\) remained unchanged. Which statement is correct?
A.The inhibitor is competitive and its effect can be overcome by significantly increasing the substrate concentration.
B.The inhibitor binds reversibly to the active site, preventing substrate binding.
C.The inhibitor is non-competitive and does not affect the affinity of the enzyme for its substrate.
D.The inhibitor is non-competitive and decreases the affinity of the enzyme for its substrate.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
A non-competitive inhibitor reduces the overall rate of reaction (halving \(V_{\text{max}}\)) by binding to an allosteric site. Because it does not compete with the substrate for the active site, the affinity of the active enzymes for the substrate (measured by \(K_{\text{m}}\)) remains unchanged.
PastPaper.markingScheme
1 mark for identifying that a constant \(K_{\text{m}}\) with a decreased \(V_{\text{max}}\) indicates non-competitive inhibition without changing substrate affinity.
PastPaper.question 11 · multiple-choice
1 PastPaper.marks
Which statement correctly describes the relationship between the structure and function of an IgG antibody molecule?
A.The variable region has a specific tertiary structure that binds to a complementary antigen, while the constant region binds to receptors on phagocytes.
B.The hinge region is composed of hydrophobic amino acids to allow the antibody to span the hydrophobic core of the cell surface membrane.
C.The variable region is identical in all IgG antibodies, allowing them to bind to a wide variety of different pathogen surfaces.
D.Disulfide bonds only link the heavy polypeptide chains together and are absent between the heavy and light polypeptide chains.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
The variable regions of an antibody form the antigen-binding sites, which have a specific tertiary structure complementary to a specific antigen. The constant region is recognized by receptors on phagocytes, facilitating opsonisation and phagocytosis.
PastPaper.markingScheme
1 mark for identifying the correct relationship between antibody structure (variable and constant regions) and function.
PastPaper.question 12 · multiple-choice
1 PastPaper.marks
Which row correctly describes and compares the structures of collagen and haemoglobin?
A.Collagen: fibrous, three polypeptide chains forming a triple helix; Haemoglobin: globular, four polypeptide chains, each with a prosthetic haem group
B.Collagen: globular, high tensile strength due to covalent cross-links; Haemoglobin: fibrous, soluble due to hydrophilic R-groups on the outside
C.Collagen: fibrous, single polypeptide chain containing repeating amino acid sequences; Haemoglobin: globular, hydrophobic R-groups on the outside
D.Collagen: fibrous, soluble in water; Haemoglobin: globular, insoluble in water due to hydrophobic interactions between subunits
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Collagen is a fibrous, insoluble protein made of three polypeptide chains wound in a triple helix. Haemoglobin is a soluble, globular protein made of four polypeptide chains (two alpha, two beta subunits), each containing a prosthetic haem group.
PastPaper.markingScheme
1 mark for identifying the correct structural features of both collagen and haemoglobin.
PastPaper.question 13 · multiple-choice
1 PastPaper.marks
Which group of structures is present in eukaryotic plant cells but absent from both eukaryotic animal cells and prokaryotic cells?
A.plasmodesmata, chloroplasts, large permanent vacuole
Plasmodesmata, chloroplasts, and a large permanent vacuole are characteristic features of eukaryotic plant cells. Animal cells lack all three. Prokaryotic cells also lack chloroplasts (which are membrane-bound organelles), plasmodesmata, and a large permanent vacuole.
PastPaper.markingScheme
1 mark for correctly identifying the three structures unique to plant cells among the options.
PastPaper.question 14 · multiple-choice
1 PastPaper.marks
Plant cells with a water potential of \(-600\text{ kPa}\) are placed in four different sucrose solutions (W, X, Y, and Z) with water potentials of \(-200\text{ kPa}\), \(-400\text{ kPa}\), \(-600\text{ kPa}\), and \(-800\text{ kPa}\) respectively. Which statement correctly describes the net movement of water and the state of the cells?
A.In solution W, water enters the cells by osmosis until the cell membrane bursts.
B.In solution X, there is a net movement of water out of the cells, causing them to become plasmolysed.
C.In solution Y, there is no net movement of water because the water potential of the solution equals the water potential of the cells.
D.In solution Z, water enters the cells, causing them to become fully turgid.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
In solution Y, the water potential of the solution is \(-600\text{ kPa}\), which is identical to the water potential of the plant cells. Thus, the system is in dynamic equilibrium and there is no net movement of water. Plant cells do not burst (eliminating A) due to their cell walls; in solution X, water moves in because the solution has a higher water potential than the cells (eliminating B); in solution Z, water moves out because the solution has a lower water potential (eliminating D).
PastPaper.markingScheme
1 mark for identifying that no net movement of water occurs when water potentials are equal.
PastPaper.question 15 · multiple-choice
1 PastPaper.marks
Why is penicillin effective against growing bacterial cells but completely ineffective against viruses?
A.Penicillin inhibits transpeptidase, preventing the formation of cross-links in peptidoglycan cell walls, whereas viruses lack a cell wall.
B.Penicillin breaks existing peptide bonds in peptidoglycan, whereas viruses are surrounded by a protective protein capsid.
C.Penicillin prevents transcription by binding to bacterial RNA polymerase, whereas viruses use host cell enzymes for transcription.
D.Penicillin targets the 70S ribosomes of bacteria, whereas viruses do not possess ribosomes.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Penicillin is an antibiotic that inhibits the enzyme transpeptidase, preventing the cross-linking of peptidoglycan chains during bacterial cell wall synthesis. Since viruses do not have cell walls (they have a protein capsid instead) and do not undergo cell wall synthesis, penicillin has no effect on them.
PastPaper.markingScheme
1 mark for identifying penicillin's target (transpeptidase/cell wall synthesis) and explaining why it doesn't affect viruses.
PastPaper.question 16 · multiple-choice
1 PastPaper.marks
A segment of a template strand of DNA has the base sequence: \(3'\text{-TAC GGT CAT ATT-}5'\). During protein synthesis, what is the sequence of the anticodons on the four tRNA molecules, written in the \(5' \rightarrow 3'\) direction, in the order they bind to the mRNA transcribed from this DNA?
1. The DNA template sequence is \(3'\text{-TAC GGT CAT ATT-}5'\). 2. During transcription, complementary base pairing produces mRNA in the \(5' \rightarrow 3'\) direction: \(5'\text{-AUG CCA GUA UAA-}3'\). 3. During translation, tRNA anticodons bind antiparallel and complementary to the mRNA codons. For the first codon \(5'\text{-AUG-}3'\), the anticodon is \(3'\text{-UAC-}5'\), which is written \(5'\text{-CAU-}3'\) in the \(5' \rightarrow 3'\) direction. For the second codon \(5'\text{-CCA-}3'\), the anticodon is \(3'\text{-GGU-}5'\), written \(5'\text{-UGG-}3'\). For the third codon \(5'\text{-GUA-}3'\), the anticodon is \(3'\text{-CAU-}5'\), written \(5'\text{-UAC-}3'\). For the fourth codon \(5'\text{-UAA-}3'\), the anticodon is \(3'\text{-AUU-}5'\), written \(5'\text{-UUA-}3'\). Thus, the correct sequence is option A.
PastPaper.markingScheme
1 mark for correctly transcribing the DNA template into mRNA and then determining the complementary antiparallel tRNA anticodon sequence in the correct \(5' \rightarrow 3'\) orientation.
PastPaper.question 17 · multiple-choice
1 PastPaper.marks
Which statements correctly describe the movement of water from the soil into the xylem of a root? 1. Water entering the apoplast pathway moves through cell walls and intercellular spaces. 2. Water in the symplast pathway moves through plasmodesmata. 3. The Casparian strip blocks the symplast pathway, forcing water into the apoplast. 4. Active transport of mineral ions into the xylem lowers its water potential, encouraging osmosis.
A.1, 2 and 3 only
B.1, 2 and 4 only
C.2 and 4 only
D.1 and 3 only
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Statement 1 is correct because the apoplast pathway consists of cell walls and intercellular spaces. Statement 2 is correct because the symplast pathway is the living path through cytoplasm connected by plasmodesmata. Statement 3 is incorrect because the suberised Casparian strip in the endodermis blocks the apoplast pathway, forcing water into the symplast pathway. Statement 4 is correct because active transport of mineral ions into the xylem vessels lowers the solute potential and water potential, drawing water into the xylem by osmosis down a water potential gradient. Therefore, statements 1, 2, and 4 are correct.
PastPaper.markingScheme
Award 1 mark for selecting B. Reject other options as they contain incorrect statement combinations.
PastPaper.question 18 · multiple-choice
1 PastPaper.marks
Which row correctly matches a structural feature of a transport cell to its functional significance?
A.Companion cell - Lignified walls - Prevents collapse under high tension
B.Xylem vessel element - Plates with sieve pores - Allows bidirectional flow of organic solutes
C.Companion cell - Numerous mitochondria - Provides ATP for active loading of sucrose
D.Xylem vessel element - Cytoplasm containing large vacuoles - Facilitates rapid water movement
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Companion cells have high metabolic activity and numerous mitochondria to produce ATP required for the active loading of sucrose into sieve tube elements. Xylem vessel elements are dead cells with no cytoplasm or organelles, so they do not have vacuoles, and they do not have sieve pores (which are characteristic of phloem sieve tube elements). Companion cells do not have lignified walls; xylem vessel elements have lignified walls to prevent collapse under high tension.
PastPaper.markingScheme
Award 1 mark for selecting C. Reject A because companion cells do not have lignified walls. Reject B because xylem vessels do not contain sieve pores. Reject D because xylem vessels do not contain cytoplasm or vacuoles.
PastPaper.question 19 · multiple-choice
1 PastPaper.marks
A person is bitten by a venomous snake and is immediately injected with an antivenom containing specific antibodies. Two months later, the person is bitten by the same species of snake but does not have immunity and requires another injection of antivenom. Which row correctly describes the type of immunity provided by the first injection and the reason why long-term immunity was not established?
A.Active artificial | No memory cells were produced because no antigen was injected.
B.Passive artificial | No memory cells were produced because no antigen was injected.
C.Passive natural | The antibodies were rapidly broken down before they could trigger an immune response.
D.Active natural | The dose of venom was too low to stimulate B-lymphocyte cloning.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
The injection of antibodies (antivenom) provides artificial passive immunity. It is artificial because the antibodies are introduced from an external source, and passive because the recipient's own immune system did not produce them. No long-term immunity is established because no antigen was injected, meaning no B-lymphocytes or T-lymphocytes were stimulated to undergo clonal selection, clonal expansion, and differentiation into memory cells.
PastPaper.markingScheme
Award 1 mark for selecting B. Reject other options because they incorrectly classify the type of immunity or state an incorrect reason for the lack of long-term immunity.
PastPaper.question 20 · multiple-choice
1 PastPaper.marks
An enzyme-controlled reaction is carried out in the presence of a non-competitive inhibitor. How are the maximum rate of reaction (\(V_{max}\)) and the Michaelis-Menten constant (\(K_m\)) affected compared to the reaction without the inhibitor?
Non-competitive inhibitors bind to an allosteric site rather than the active site. This alters the tertiary structure and shape of the active site, reducing the rate of reaction. Since increasing the substrate concentration cannot overcome this type of inhibition, the maximum possible rate of reaction (\(V_{max}\)) decreases. However, because the affinity of the unaffected enzyme molecules for the substrate remains unchanged, the Michaelis-Menten constant (\(K_m\)), which is a measure of substrate affinity, remains unchanged.
PastPaper.markingScheme
Award 1 mark for selecting B. Reject A because \(K_m\) does not increase. Reject C because \(V_{max}\) is not unchanged. Reject D because \(V_{max}\) is decreased.
PastPaper.question 21 · multiple-choice
1 PastPaper.marks
Which statements correctly describe differences between a molecule of collagen and a molecule of haemoglobin? 1. Collagen consists of three polypeptide chains, whereas haemoglobin consists of four polypeptide chains. 2. Collagen contains a high proportion of hydrophobic amino acids on its outer surface, whereas haemoglobin has hydrophilic amino acids on its outer surface. 3. Collagen has a structural role in the body, whereas haemoglobin has a transport function. 4. Collagen contains prosthetic groups, whereas haemoglobin does not.
A.1, 2 and 3 only
B.1, 3 and 4 only
C.2 and 3 only
D.1 and 4 only
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Statement 1 is correct because collagen is a fibrous triple helix of three polypeptide chains, while haemoglobin is a globular quaternary protein of four polypeptide chains (two alpha and two beta). Statement 2 is correct because collagen is insoluble and has hydrophobic R-groups on its outside, whereas haemoglobin is soluble and has hydrophilic R-groups on its outer surface. Statement 3 is correct because collagen forms structural fibers in tendons and bone, while haemoglobin transports oxygen in red blood cells. Statement 4 is incorrect because haemoglobin contains haem prosthetic groups, whereas collagen does not have prosthetic groups. Therefore, statements 1, 2, and 3 are correct.
PastPaper.markingScheme
Award 1 mark for selecting A. Reject other options as they omit correct statements or include statement 4.
PastPaper.question 22 · multiple-choice
1 PastPaper.marks
Which sequence correctly represents the pathway of a newly synthesised membrane-bound glycoprotein from its translation to its final destination in the cell surface membrane?
A.Free ribosome -> Golgi body -> transport vesicle -> cell surface membrane
B.Rough endoplasmic reticulum -> transport vesicle -> Golgi body -> secretory vesicle -> cell surface membrane
Translation of membrane-bound and secretory proteins begins on ribosomes bound to the rough endoplasmic reticulum (RER). After folding within the RER, the proteins are packaged into transport vesicles that travel to and fuse with the Golgi body. In the Golgi body, modification (such as glycosylation to form a glycoprotein) occurs. Modified proteins then bud off the Golgi in secretory vesicles which fuse with the cell surface membrane.
PastPaper.markingScheme
Award 1 mark for selecting B. Reject A because translation of membrane-bound glycoproteins occurs on bound ribosomes (RER), not free ribosomes. Reject C because lysosomes are not part of the standard pathway to the cell surface membrane. Reject D because translation does not occur in the nucleus.
PastPaper.question 23 · multiple-choice
1 PastPaper.marks
The table describes three processes, X, Y, and Z, by which substances move across cell membranes. Process X: Movement against a concentration gradient; requires ATP; requires a carrier protein. Process Y: Movement down a concentration gradient; does not require ATP; requires a channel or carrier protein. Process Z: Movement down a concentration gradient; does not require ATP; does not require a membrane protein. Which row correctly identifies these three processes?
A.Process X: Active transport | Process Y: Facilitated diffusion | Process Z: Simple diffusion
B.Process X: Facilitated diffusion | Process Y: Active transport | Process Z: Osmosis
C.Process X: Active transport | Process Y: Simple diffusion | Process Z: Facilitated diffusion
D.Process X: Endocytosis | Process Y: Facilitated diffusion | Process Z: Simple diffusion
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Process X is active transport because it moves substances against their concentration gradient, requires metabolic energy (ATP), and uses carrier proteins. Process Y is facilitated diffusion because it moves substances down a concentration gradient without ATP, using transport proteins (channels or carriers). Process Z is simple diffusion because substances move down a concentration gradient directly through the phospholipid bilayer without the aid of membrane proteins.
PastPaper.markingScheme
Award 1 mark for selecting A. Reject other options because they misidentify the membrane transport processes based on the provided descriptions.
PastPaper.question 24 · multiple-choice
1 PastPaper.marks
During translocation in the phloem of a plant, what causes the high hydrostatic pressure at the source?
A.Active transport of sucrose out of the sieve tube elements, raising the water potential.
B.Active transport of hydrogen ions out of the sieve tube elements, lowering the water potential.
C.Diffusion of sucrose into the sieve tube elements, followed by the active uptake of water.
D.Active loading of sucrose into the sieve tube elements, which lowers the water potential and causes water to enter by osmosis.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Translocation in the phloem is driven by a hydrostatic pressure gradient. At the source, companion cells actively load sucrose into the sieve tube elements. This high concentration of sucrose lowers the solute potential and water potential inside the sieve tube elements. As a result, water moves from the surrounding tissues (such as the xylem) into the sieve tube elements by osmosis down a water potential gradient. The entry of water increases the hydrostatic pressure inside the sieve tube elements at the source, driving mass flow.
PastPaper.markingScheme
Award 1 mark for selecting D. Reject A because active loading of sucrose lowers (makes more negative) the water potential, not raises it. Reject B because it is the loading of sucrose, not hydrogen ions, that directly lowers water potential. Reject C because sucrose enters by active loading (co-transport), and water uptake is passive (osmosis), not active.
PastPaper.question 25 · multiple-choice
1 PastPaper.marks
Which row correctly identifies structural features of mature xylem vessel elements and mature sieve tube elements?
C.Xylem vessel elements: Non-lignified walls, cytoplasm absent, sieve plates present; Sieve tube elements: Lignified walls, cytoplasm present, end walls open
D.Xylem vessel elements: Lignified walls, cytoplasm absent, sieve plates present; Sieve tube elements: Non-lignified walls, cytoplasm present, end walls open
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Mature xylem vessel elements are dead, hollow tubes with no cytoplasm, lignified cell walls, and completely open end walls to allow uninterrupted water flow. In contrast, mature sieve tube elements are living cells that contain a thin layer of peripheral cytoplasm (though lacking a nucleus and many organelles) and non-lignified cellulose cell walls, with end walls modified into perforated sieve plates.
PastPaper.markingScheme
B is correct because xylem vessels are lignified, dead (no cytoplasm) and have open end walls, whereas sieve tube elements are non-lignified, living (contain cytoplasm) and have sieve plates.
PastPaper.question 26 · multiple-choice
1 PastPaper.marks
An experiment was carried out to investigate the effect of substrate concentration on the rate of an enzyme-controlled reaction. The experiment was performed under three conditions: (1) Enzyme alone, (2) Enzyme + competitive inhibitor X, (3) Enzyme + non-competitive inhibitor Y. Which statement correctly describes the curves obtained when the rate of reaction is plotted against substrate concentration?
A.Curve 2 reaches the same maximum rate (Vmax) as Curve 1, but at a higher substrate concentration. Curve 3 never reaches the same Vmax as Curve 1.
B.Curve 2 and Curve 3 both reach the same maximum rate (Vmax) as Curve 1, but at higher substrate concentrations.
C.Curve 2 never reaches the same maximum rate (Vmax) as Curve 1. Curve 3 reaches the same Vmax as Curve 1, but at a higher substrate concentration.
D.Neither Curve 2 nor Curve 3 can reach the same maximum rate (Vmax) as Curve 1, regardless of the substrate concentration.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Competitive inhibitors bind to the active site and can be outcompeted by increasing the substrate concentration. Therefore, with competitive inhibitor X (Curve 2), the reaction will eventually reach the same Vmax as the uninhibited enzyme (Curve 1) if the substrate concentration is sufficiently high. Non-competitive inhibitors bind to an allosteric site and permanently reduce the maximum potential rate of reaction regardless of substrate concentration, meaning Curve 3 (with inhibitor Y) will have a lower Vmax than Curve 1.
PastPaper.markingScheme
A is correct because competitive inhibition can be overcome by high substrate concentrations (reaching Vmax), whereas non-competitive inhibition cannot (Vmax is reduced).
PastPaper.question 27 · multiple-choice
1 PastPaper.marks
Which functional role is correctly matched with the specific region of an IgG antibody molecule?
A.Constant region of the light chain — binds to specific antigens to form an antigen-antibody complex.
B.Variable regions of both heavy and light chains — form the highly specific antigen-binding sites.
C.Hinge region — contains the only disulfide bonds present in the entire antibody molecule.
D.Variable region of the heavy chain — binds to receptors on the surface of phagocytes during opsonisation.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
The variable regions (V domains) of both the heavy and light chains at the tips of the 'Y' shape combine to form the specific antigen-binding sites. The constant region of the light chain does not bind antigens. Disulfide bonds are found throughout the antibody molecule, not just in the hinge. The constant region of the heavy chain (specifically the Fc region), not the variable region, binds to phagocyte receptors.
PastPaper.markingScheme
B is correct because the antigen-binding sites are formed by the variable regions of both heavy and light chains.
PastPaper.question 28 · multiple-choice
1 PastPaper.marks
Which row correctly compares the structural features of a collagen molecule and a haemoglobin molecule?
A.Collagen: Three polypeptide chains, fibrous, soluble in water; Haemoglobin: Four polypeptide chains, globular, insoluble in water
B.Collagen: Three polypeptide chains, fibrous, insoluble in water; Haemoglobin: Four polypeptide chains, globular, soluble in water
C.Collagen: Four polypeptide chains, globular, insoluble in water; Haemoglobin: Three polypeptide chains, fibrous, soluble in water
D.Collagen: Three polypeptide chains, globular, soluble in water; Haemoglobin: Four polypeptide chains, fibrous, insoluble in water
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
A collagen molecule is a fibrous, structural protein consisting of three polypeptide chains wound together into a tight triple helix, and is completely insoluble in water. Haemoglobin is a globular transport protein consisting of four polypeptide chains (two alpha and two beta chains), each with a prosthetic haem group, and is soluble in water due to its outer hydrophilic R-groups.
PastPaper.markingScheme
B is correct because collagen is fibrous, insolubly constructed of three chains, while haemoglobin is a soluble globular protein made of four chains.
PastPaper.question 29 · multiple-choice
1 PastPaper.marks
Radioactively labelled amino acids are supplied to a eukaryotic cell actively secreting glycoproteins. In which order would the radioactivity be detected in the cell organelles?
Amino acids are first incorporated into polypeptides at the ribosomes on the rough endoplasmic reticulum (RER). The proteins are packaged into transport vesicles and sent to the Golgi body, where carbohydrate chains are added to form glycoproteins. The mature glycoproteins are then packaged into secretory vesicles which move to and fuse with the cell surface membrane, releasing the proteins via exocytosis.
PastPaper.markingScheme
A is correct because the secretory pathway starts with protein synthesis on the RER, followed by modification in the Golgi body, transport via secretory vesicles, and fusion with the cell surface membrane.
PastPaper.question 30 · multiple-choice
1 PastPaper.marks
Plant cell vacuoles contain a cell sap with an initial water potential of -600 kPa. Four identical plant cells are placed in different solutions and allowed to reach equilibrium: Cell 1 in a solution with a water potential of -200 kPa; Cell 2 in a solution with a water potential of -600 kPa; Cell 3 in a solution with a water potential of -1000 kPa; Cell 4 in pure water. Which of the following statements about these cells is correct?
A.Cell 1 will become plasmolysed because water leaves the cell down a water potential gradient.
B.Cell 2 will have a higher solute concentration inside its vacuole at equilibrium than it did initially.
C.Cell 3 will undergo plasmolysis as the protoplast shrinks and the cell membrane pulls away from the cell wall.
D.Cell 4 will swell and burst because the cell wall is unable to resist the intake of water.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
In Cell 3, the external solution has a lower water potential (-1000 kPa) than the cell sap (-600 kPa), so water moves out of the cell by osmosis down a water potential gradient. The vacuole and protoplast shrink, causing the plasma membrane to pull away from the cell wall, which is the definition of plasmolysis. In Cell 1, water enters the cell, making it turgid. In Cell 2, there is no net movement of water, so solute concentration remains unchanged. In Cell 4, the cell becomes fully turgid but does not burst due to the mechanical strength of the cellulose cell wall.
PastPaper.markingScheme
C is correct because water moves from the cell (-600 kPa) to the hypertonic external environment (-1000 kPa), resulting in plasmolysis.
PastPaper.question 31 · multiple-choice
1 PastPaper.marks
Which of the following describes the mechanism of action of penicillin on bacteria?
A.It binds to ribosomal subunits, inhibiting translation and preventing protein synthesis.
B.It inhibits the enzyme transpeptidase, preventing the cross-linking of peptidoglycan chains in the cell wall.
C.It acts as a competitive inhibitor of DNA polymerase, halting bacterial DNA replication.
D.It damages the phospholipid bilayer of the cell membrane, leading to the leakage of cellular contents.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Penicillin acts by inhibiting the enzyme glycopeptide transpeptidase, which is responsible for making the cross-links between peptidoglycan chains in the cell wall of growing bacteria. Without these cross-links, the cell wall is weakened and cannot withstand the internal osmotic pressure, causing the bacterium to burst (lyse) when water enters.
PastPaper.markingScheme
B is correct because penicillin specifically targets the cell wall synthesis of bacteria by inhibiting transpeptidase.
PastPaper.question 32 · multiple-choice
1 PastPaper.marks
A segment of DNA has the template strand sequence: 3'- T A C G G C T T A C T G -5'. Which row correctly identifies the codons on the mRNA transcribed from this DNA template, and the anticodons on the tRNA molecules that would pair with these codons during translation?
Transcription of the DNA template strand 3'- T A C G G C T T A C T G -5' produces mRNA with complementary RNA bases in the 5' to 3' direction: 5'- A U G C C G A A U G A C -3'. This gives the codons: AUG, CCG, AAU, GAC. During translation, the tRNA anticodons that pair complementary with these codons must be RNA sequences containing uracil instead of thymine, resulting in: UAC, GGC, UUA, CUG respectively.
PastPaper.markingScheme
A is correct because transcription of the DNA template yields mRNA codons (with U instead of T), and complementary tRNA anticodons (also containing U) pair with them.
PastPaper.question 33 · multiple-choice
1 PastPaper.marks
Which row correctly describes the features of a mature phloem sieve tube element?
A mature sieve tube element lacks a nucleus and a tonoplast (the membrane surrounding the vacuole) to reduce resistance to the flow of phloem sap. However, it still contains a thin peripheral layer of cytoplasm and is connected to adjacent companion cells via plasmodesmata.
PastPaper.markingScheme
1 mark for the correct option B. Sieve tube elements contain cytoplasm and plasmodesmata but lack a nucleus and tonoplast.
PastPaper.question 34 · multiple-choice
1 PastPaper.marks
An experiment investigates the effect of a competitive inhibitor on an enzyme-controlled reaction. Which statement is correct about the effects of increasing the substrate concentration in the presence of a fixed concentration of a competitive inhibitor?
A.The value of \(K_m\) decreases because the enzyme has a higher affinity for the substrate.
B.The maximum rate of reaction (\(V_{max}\)) can still be reached because the inhibitor can be displaced from the active site.
C.The rate of reaction is always lower than without the inhibitor, regardless of how high the substrate concentration is.
D.The inhibitor binds covalently to the active site, permanently preventing the substrate from binding.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
A competitive inhibitor has a similar shape to the substrate and competes for the active site. Because the binding is temporary and reversible, increasing the substrate concentration increases the probability of a substrate molecule binding to the active site rather than an inhibitor molecule. At very high substrate concentrations, the inhibitor is effectively displaced, and the maximum rate of reaction (\(V_{max}\)) can still be achieved.
PastPaper.markingScheme
1 mark for the correct option B. Reject options suggesting that the maximum rate cannot be reached or that Km decreases.
PastPaper.question 35 · multiple-choice
1 PastPaper.marks
Which row correctly matches each medical scenario with the type of immunity achieved? Scenario 1: A child is given an injection of antitoxins after being bitten by a venomous snake. Scenario 2: A breastfed infant receives IgA antibodies through colostrum. Scenario 3: A person produces antibodies and memory cells after recovering from measles. Scenario 4: A teenager receives a tetanus toxoid injection.
Scenario 1 involves injecting pre-formed antibodies (antitoxins), which is artificial passive immunity. Scenario 2 involves transferring maternal antibodies naturally via milk, which is natural passive immunity. Scenario 3 involves the body's own immune response to a pathogen, which is natural active immunity. Scenario 4 involves injecting an antigen (toxoid) to trigger an active immune response, which is artificial active immunity.
PastPaper.markingScheme
1 mark for the correct option A. All four scenarios must be correctly matched to their corresponding types of immunity.
PastPaper.question 36 · multiple-choice
1 PastPaper.marks
Which statements about the structure of a collagen molecule are correct? 1. It is a polymer of amino acids consisting of three polypeptide chains wound around each other to form a triple helix. 2. Every third amino acid in each polypeptide chain is glycine, which allows the chains to lie close together. 3. Hydrogen bonds and covalent cross-links are the main forces holding adjacent triple helices together to form fibrils. 4. It is a soluble protein that performs a structural function in the extracellular matrix.
A.1, 2, 3 and 4
B.1, 2 and 3 only
C.1 and 2 only
D.2, 3 and 4 only
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Statements 1, 2, and 3 are correct. Collagen consists of three polypeptide chains in a triple helix (statement 1). Glycine is the smallest amino acid (having only a hydrogen atom as its R-group) and occurs at every third position, allowing the chains to pack tightly together (statement 2). Adjacent triple helices are held together by covalent cross-links and hydrogen bonds to form tough fibrils (statement 3). Statement 4 is incorrect because collagen is an insoluble structural protein.
PastPaper.markingScheme
1 mark for the correct option B. Reject options containing statement 4 since collagen is insoluble.
PastPaper.question 37 · multiple-choice
1 PastPaper.marks
Which of the following cellular structures can be resolved only by using a transmission electron microscope (TEM) and not by a high-quality light microscope? 1. Microtubules of the spindle fibres (diameter 25 nm). 2. The cell surface membrane (thickness 7 nm). 3. A mitochondrion (length 3 micrometers).
A.1, 2 and 3
B.1 and 2 only
C.2 and 3 only
D.1 only
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
The maximum resolution of a light microscope is approximately 200 nm. Any structures smaller than 200 nm cannot be resolved. Microtubules (25 nm) and the cell surface membrane (7 nm) are far below this limit, so they can only be resolved using an electron microscope. Mitochondria (length ~3 micrometers) are larger than 200 nm and can easily be resolved using a light microscope, even if their inner structure cannot be resolved in detail.
PastPaper.markingScheme
1 mark for the correct option B. Reject choices including statement 3 as mitochondria are within the resolution limit of light microscopes.
PastPaper.question 38 · multiple-choice
1 PastPaper.marks
Plant cells with a water potential of \(-600\text{ kPa}\) are placed in a sucrose solution with a water potential of \(-300\text{ kPa}\). Which row correctly describes the net movement of water and the final state of the cells after reaching equilibrium?
A.Net movement of water: into the cells | Final state of the cells: turgid
B.Net movement of water: into the cells | Final state of the cells: plasmolysed
C.Net movement of water: out of the cells | Final state of the cells: flaccid
D.Net movement of water: out of the cells | Final state of the cells: plasmolysed
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Water moves by osmosis from a region of higher water potential (less negative, \(-300\text{ kPa}\)) to a region of lower water potential (more negative, \(-600\text{ kPa}\)). Therefore, the net movement of water is into the plant cells. As water enters, the protoplast pushes against the cell wall, causing the cells to become turgid.
PastPaper.markingScheme
1 mark for the correct option A. Water moves into the cells from higher to lower water potential, resulting in turgidity.
PastPaper.question 39 · multiple-choice
1 PastPaper.marks
Which statement correctly describes the action of penicillin on growing bacterial cells?
A.It binds to 70S ribosomes, stopping the translation of crucial bacterial proteins.
B.It inhibits the enzyme transpeptidase, preventing the cross-linking of peptidoglycan chains in the cell wall.
C.It disrupts the cell surface membrane, making it highly permeable and leading to cell leakage.
D.It inhibits DNA helicase, preventing the replication of the bacterial chromosome.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
Penicillin functions by inhibiting the enzyme transpeptidase, which is responsible for forming the cross-links between peptidoglycan chains in the bacterial cell wall. This weakens the wall of growing bacteria, making them susceptible to osmotic lysis (bursting) due to internal turgor pressure.
PastPaper.markingScheme
1 mark for the correct option B. Reject options that describe protein synthesis inhibition (A) or membrane disruption (C).
PastPaper.question 40 · multiple-choice
1 PastPaper.marks
A segment of double-stranded DNA has a total of 600 bases. 180 of these bases are adenine. How many hydrogen bonds are present between the complementary base pairs in this segment of DNA?
A.660
B.720
C.780
D.900
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
In double-stranded DNA, Adenine (A) pairs with Thymine (T) via 2 hydrogen bonds, and Guanine (G) pairs with Cytosine (C) via 3 hydrogen bonds. 1. Since there are 180 adenine bases, there must also be 180 thymine bases, making 180 A-T pairs. 2. Total A-T bases = 180 + 180 = 360 bases. 3. The remaining bases represent G-C base pairs: 600 - 360 = 240 bases. 4. Therefore, there are 120 G-C pairs (120 guanine bases and 120 cytosine bases). 5. Calculate the hydrogen bonds: Hydrogen bonds from A-T pairs = 180 pairs \(\times\) 2 = 360. Hydrogen bonds from G-C pairs = 120 pairs \(\times\) 3 = 360. Total hydrogen bonds = 360 + 360 = 720.
PastPaper.markingScheme
1 mark for the correct calculation leading to option B. Identify the correct number of G-C pairs (120) and A-T pairs (180), then multiply by their respective hydrogen bonds (3 and 2) to get 720.
Paper 2 (AS Level Structured Questions)
Answer all six structured questions. Write your answers in the spaces provided on the question paper.
6 PastPaper.question · 60 PastPaper.marks
PastPaper.question 1 · structured-short-answer
10 PastPaper.marks
Answer all parts of the question. (a) Distinguish between the apoplast and symplast pathways of water movement in the root of a plant. [3 marks] (b) Explain how the Casparian strip forces water into the symplast pathway and state the significance of this. [3 marks] (c) Xerophytic plants have adaptations to reduce water loss by transpiration. Explain how two morphological features of xerophytic leaves reduce water loss. [4 marks]
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a) The apoplast pathway involves water movement through cell walls and intercellular spaces, whereas the symplast pathway involves water movement through the cytoplasm and plasmodesmata. The apoplast pathway is passive and non-living, while the symplast pathway requires water to cross the selectively permeable cell surface membrane. (b) The Casparian strip is made of suberin, which is a waterproof substance. It blocks the cell wall route at the endodermis, forcing water to pass across the cell surface membrane into the cytoplasm. This allows the plant cell membranes to select and regulate which ions and solutes enter the xylem. (c) Two morphological adaptations are: 1. Sunken stomata in pits trap moist air, which reduces the water potential gradient between the inside of the leaf and the atmosphere, thereby reducing transpiration. 2. A thick waxy cuticle on the leaf surface is impermeable to water, which increases the diffusion distance and reduces cuticular transpiration.
PastPaper.markingScheme
(a) 1 mark for apoplast through cell walls / intercellular spaces; 1 mark for symplast through cytoplasm / plasmodesmata; 1 mark for apoplast being non-living / passive while symplast crosses selectively permeable membrane. Max 3 marks. (b) 1 mark for Casparian strip contains suberin; 1 mark for blocking apoplast pathway at endodermis; 1 mark for forcing water into symplast / cytoplasm to regulate ions. Max 3 marks. (c) 1 mark for naming adaptation (e.g., sunken stomata, rolled leaves, thick cuticle) and 1 mark for explaining its mechanism. Repeat for second adaptation. Max 4 marks.
PastPaper.question 2 · structured-short-answer
10 PastPaper.marks
An experiment was carried out to investigate the effect of substrate concentration on the rate of an enzyme-controlled reaction. (a) State two variables, other than substrate concentration, that must be controlled in this investigation, and explain how each is controlled. [4 marks] (b) Explain the shape of the curve expected when the rate of reaction is plotted against substrate concentration at: (i) low substrate concentrations [3 marks] (ii) high substrate concentrations [3 marks]
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a) 1. Temperature: must be controlled because it affects kinetic energy and molecular collisions. It is controlled using a thermostatically-controlled water bath. 2. pH: must be controlled because pH changes can denature the enzyme by altering charges on the active site R-groups. It is controlled using a buffer solution. (b) (i) At low substrate concentrations, the rate of reaction increases proportionally with substrate concentration because substrate concentration is the limiting factor. Many active sites are empty, so increasing substrate concentration increases collision frequency. (ii) At high substrate concentrations, the rate of reaction plateaus and reaches its maximum rate \(V_{\max}\) because all active sites are saturated with substrate. Substrate is no longer the limiting factor; instead, enzyme concentration is the limiting factor.
PastPaper.markingScheme
(a) 1 mark for naming variable (temperature, pH, enzyme concentration) and 1 mark for description of control (water bath, buffer, fixed volume/concentration respectively). Repeat for second variable. Max 4 marks. (b)(i) 1 mark for rate is proportional to substrate concentration; 1 mark for substrate concentration being the limiting factor; 1 mark for active sites being unoccupied. Max 3 marks. (b)(ii) 1 mark for rate plateaus / reaches \(V_{\max}\); 1 mark for active sites saturated / enzyme-substrate complexes formed at maximum rate; 1 mark for enzyme concentration being the limiting factor. Max 3 marks.
PastPaper.question 3 · structured-short-answer
10 PastPaper.marks
(a) Describe the structure of an antibody molecule, referencing its variable regions, constant regions, and disulfide bonds. [4 marks] (b) Contrast active immunity and passive immunity. [3 marks] (c) Distinguish between a primary immune response and a secondary immune response after a vaccination. [3 marks]
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a) An antibody molecule is a Y-shaped protein composed of four polypeptide chains, comprising two identical heavy chains and two identical light chains. It features highly specific variable regions at the tips of the Y-arms that act as antigen-binding sites, while the rest of the molecule consists of constant regions. Disulfide bonds hold the heavy and light chains together. (b) Active immunity involves the production of antibodies by the person's own immune system, resulting in memory cells and long-term protection. Passive immunity involves introducing external antibodies, providing immediate but short-term protection without memory cells. (c) A primary response occurs on first exposure, has a lag phase, and produces a low concentration of antibodies. A secondary response is much faster, produces a far higher concentration of antibodies, and is sustained longer due to the presence of memory cells.
PastPaper.markingScheme
(a) 1 mark for Y-shape / four chains; 1 mark for variable regions forming antigen-binding sites; 1 mark for constant regions forming the stem/base; 1 mark for disulfide bonds holding chains together. Max 4 marks. (b) 1 mark for active producing own antibodies while passive receives them; 1 mark for active producing memory cells while passive does not; 1 mark for active being long-term while passive is temporary. Max 3 marks. (c) 1 mark for primary being slower / secondary being faster; 1 mark for secondary producing a higher antibody concentration; 1 mark for secondary being due to memory cells. Max 3 marks.
PastPaper.question 4 · structured-short-answer
10 PastPaper.marks
(a) Describe the structure of a collagen molecule and explain how its structure is related to its function in the body. [6 marks] (b) Contrast the structure of collagen with that of haemoglobin. [4 marks]
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a) Collagen is a fibrous structural protein consisting of three polypeptide chains wound together in a tight triple helix. Every third amino acid is glycine, which is small enough to fit in the close-packed center of the helix. Hydrogen bonds stabilize the triple helix. Adjacent collagen molecules form covalent cross-links with one another to produce fibrils, which aggregate to form thick, strong fibres. This arrangement provides immense tensile strength and non-elastic support to structures like tendons, bone, and skin. (b) Collagen is a fibrous protein, whereas haemoglobin is a globular protein. Collagen is composed of three polypeptide chains in a triple helix, whereas haemoglobin is composed of four polypeptide chains in a spherical quaternary structure. Collagen has hydrophobic amino acids on its surface (making it insoluble), while haemoglobin has hydrophilic residues on its surface (making it soluble). Finally, haemoglobin contains prosthetic haem groups with iron ions, whereas collagen contains no prosthetic groups.
PastPaper.markingScheme
(a) 1 mark for being a fibrous protein; 1 mark for three polypeptide chains in a triple helix; 1 mark for glycine as every third amino acid; 1 mark for hydrogen bonds stabilizing helix; 1 mark for covalent cross-links forming fibrils/fibres; 1 mark for function related to tensile strength / tendons / skin / non-elastic. Max 6 marks. (b) 1 mark for fibrous vs globular; 1 mark for three chains vs four chains; 1 mark for insoluble vs soluble; 1 mark for no prosthetic group vs haem group present. Max 4 marks.
PastPaper.question 5 · structured-short-answer
10 PastPaper.marks
(a) Describe the pathway taken by a protein from its synthesis to its secretion from a eukaryotic cell. Reference the organelles involved in this process. [5 marks] (b) Distinguish between the structure of a eukaryotic cell and a prokaryotic cell. [5 marks]
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a) Protein synthesis begins on ribosomes bound to the rough endoplasmic reticulum (RER). The synthesized protein enters the lumen of the RER, where it is folded into its native shape. Transport vesicles containing the protein bud from the RER and fuse with the Golgi apparatus. Inside the Golgi body, the protein is chemically modified (e.g., glycosylated) and sorted. Secretory vesicles containing the final protein then bud from the Golgi, migrate along microtubule tracts to the cell surface membrane, and fuse with it to release the protein via exocytosis. (b) Eukaryotic cells have a true nucleus with linear DNA associated with histone proteins, whereas prokaryotic cells lack a nucleus and have circular naked DNA. Eukaryotic cells contain membrane-bound organelles (such as mitochondria, chloroplasts, and ER), which are absent in prokaryotic cells. Eukaryotic ribosomes are larger (80S), whereas prokaryotic ribosomes are smaller (70S). Eukaryotic cell walls, if present, are made of cellulose or chitin, while prokaryotic cell walls are composed of peptidoglycan.
PastPaper.markingScheme
(a) 1 mark for synthesis on RER ribosomes; 1 mark for folding inside RER lumen; 1 mark for transport vesicles moving from RER to Golgi; 1 mark for modification / sorting in Golgi body; 1 mark for secretory vesicles fusing with cell membrane for exocytosis. Max 5 marks. (b) 1 mark for nucleus with linear DNA/histones vs no nucleus/naked circular DNA; 1 mark for presence vs absence of membrane-bound organelles; 1 mark for 80S vs 70S ribosomes; 1 mark for cellulose/chitin vs peptidoglycan cell walls. Max 5 marks.
PastPaper.question 6 · structured-short-answer
10 PastPaper.marks
(a) Distinguish between active transport and facilitated diffusion. [4 marks] (b) Plant cells were placed in a solution with a water potential of -600 kPa. The initial water potential of the cytoplasm was -300 kPa. (i) Explain the direction of water movement and the resulting state of the plant cells. [3 marks] (ii) Describe what would happen if red blood cells were placed in a solution with a much higher water potential (e.g., pure water), and explain why this outcome differs from that of plant cells. [3 marks]
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a) Active transport moves substances against their concentration gradient (from a region of low concentration to high concentration), which requires energy from ATP hydrolysis. Facilitated diffusion moves substances down their concentration gradient and is passive (requiring no ATP). Active transport relies exclusively on specific carrier proteins, whereas facilitated diffusion can utilize both channel and carrier proteins. (b) (i) Water will move out of the plant cells because the water potential of the external solution (-600 kPa) is lower than that of the cytoplasm (-300 kPa). Water moves down a water potential gradient by osmosis, causing the protoplast to shrink away from the cell wall, leaving the cells plasmolysed. (ii) Red blood cells will swell and burst (undergo haemolysis) because water enters down a water potential gradient. This differs from plant cells because red blood cells lack a cell wall to resist internal pressure, whereas plant cells have a rigid cellulose cell wall that prevents bursting and instead makes the cell turgid.
PastPaper.markingScheme
(a) 1 mark for active transport being against gradient vs facilitated diffusion being down gradient; 1 mark for active transport requiring ATP vs passive; 1 mark for active transport using carrier proteins only vs channel/carrier proteins; 1 mark for active transport only in living systems. Max 4 marks. (b)(i) 1 mark for water moving out of plant cell; 1 mark for osmosis down water potential gradient; 1 mark for protoplast shrinking / plasmolysis. Max 3 marks. (b)(ii) 1 mark for red blood cells swelling and bursting / haemolysis; 1 mark for lack of cell wall in red blood cells; 1 mark for plant cells having a rigid cell wall to resist pressure / prevent bursting. Max 3 marks.
Paper 3 (Advanced Practical Skills 1)
Answer all questions. Show your working and use appropriate units where necessary.
2 PastPaper.question · 40 PastPaper.marks
PastPaper.question 1 · practical-investigation
20 PastPaper.marks
Section instructions: Answer all questions. Show your working and use appropriate units where necessary.
Question 1 Osmosis involves the net movement of water molecules down a water potential gradient across a partially permeable membrane. You are required to investigate the effect of different concentrations of sucrose solution on the length of potato cylinders, and to use your results to estimate the concentration of an unknown sucrose solution, U.
You are provided with: * \(1.0\text{ mol dm}^{-3}\) sucrose solution, labelled S * Distilled water, labelled W * Sucrose solution of unknown concentration, labelled U * Six potato cylinders, cut to at least \(50\text{ mm}\) in length
(a) (i) You are required to prepare a range of sucrose concentrations from S. Complete the dilution plan to show how you would prepare \(20.0\text{ cm}^3\) of each of the following concentrations: \(0.8\text{ mol dm}^{-3}\), \(0.6\text{ mol dm}^{-3}\), \(0.4\text{ mol dm}^{-3}\), and \(0.2\text{ mol dm}^{-3}\). (ii) State how you would measure the length of the potato cylinders accurately to ensure consistency.
(b) A student performed this investigation using potato cylinders cut precisely to \(40.0\text{ mm}\) in length. The cylinders were immersed in the prepared sucrose solutions and solution U for 30 minutes. After 30 minutes, the final length of each potato cylinder was measured. The results obtained are shown below: * In \(1.0\text{ mol dm}^{-3}\) sucrose: final length = \(37.0\text{ mm}\) * In \(0.8\text{ mol dm}^{-3}\) sucrose: final length = \(37.5\text{ mm}\) * In \(0.6\text{ mol dm}^{-3}\) sucrose: final length = \(38.5\text{ mm}\) * In \(0.4\text{ mol dm}^{-3}\) sucrose: final length = \(40.0\text{ mm}\) * In \(0.2\text{ mol dm}^{-3}\) sucrose: final length = \(41.5\text{ mm}\) * In \(0.0\text{ mol dm}^{-3}\) sucrose (distilled water): final length = \(43.0\text{ mm}\) * In solution U: final length = \(39.0\text{ mm}\)
(i) Prepare a single table to record all of these results. Your table must show: the concentration of sucrose solution, the initial length of the potato cylinders, the final length of the potato cylinders, the change in length of the potato cylinders (final length \(-\) initial length), and the percentage change in length of the potato cylinders. (ii) Describe the key features of the line graph plotted using these results, specifying the axes, scale, plotting rules, and line shape. (iii) Use the provided experimental results to estimate the concentration of sucrose solution U. Show your working. (iv) Explain the results obtained for the potato cylinder immersed in the \(0.2\text{ mol dm}^{-3}\) sucrose solution. Use the term water potential in your answer. (v) Suggest two ways to modify this procedure to obtain a more reliable estimate of the concentration of sucrose solution U.
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a) (i) Dilution Table: - For \(0.8\text{ mol dm}^{-3}\): \(16.0\text{ cm}^3\) of S + \(4.0\text{ cm}^3\) of W - For \(0.6\text{ mol dm}^{-3}\): \(12.0\text{ cm}^3\) of S + \(8.0\text{ cm}^3\) of W - For \(0.4\text{ mol dm}^{-3}\): \(8.0\text{ cm}^3\) of S + \(12.0\text{ cm}^3\) of W - For \(0.2\text{ mol dm}^{-3}\): \(4.0\text{ cm}^3\) of S + \(16.0\text{ cm}^3\) of W
(a) (ii) Use a digital calliper or a sharp ruler with millimetre subdivisions. Align the cylinder flat against the scale and view at eye level (perpendicularly) to avoid parallax error.
(b) (ii) Graph Description: - x-axis: Sucrose Concentration / \(\text{mol dm}^{-3}\), scaled linearly from 0.0 to 1.0. - y-axis: Percentage Change in Length / \(\\%\), scaled linearly from -8.0 to +8.0. - Points plotted using small, neat crosses (x). - Points connected by a smooth, best-fit curve or straight lines from point to point with no extrapolation beyond the plotted values.
(b) (iii) Determination of U: - Percentage change in length for U is \(-2.5\\%\). - Finding \(-2.5\\%\) on the y-axis, tracing horizontally to intercept the line/curve, and dropping down to the x-axis yields a sucrose concentration of approximately \(0.53\text{ mol dm}^{-3}\) (accept \(0.48 - 0.54\text{ mol dm}^{-3}\)).
(b) (iv) Explanation: - The water potential of the outer \(0.2\text{ mol dm}^{-3}\) sucrose solution is higher (less negative) than the water potential of the cell sap inside the potato cells. - Water moves down a water potential gradient into the cells by osmosis across the partially permeable cell membranes, increasing cell turgor and causing cell/tissue expansion (increase in length).
(b) (v) Improvements: 1. Perform 3 replicates for each concentration and calculate the mean percentage change to improve reliability. 2. Cover the tubes with caps or Parafilm during incubation to prevent evaporation, which would change the sucrose concentration.
PastPaper.markingScheme
(a) (i) [3 marks] - 1 mark for correct volumes of S (16.0, 12.0, 8.0, 4.0). - 1 mark for correct volumes of W (4.0, 8.0, 12.0, 16.0). - 1 mark for specifying units in the headings (/ cm3) and writing all volumes to 1 decimal place (e.g. 16.0, not 16).
(a) (ii) [1 mark] - 1 mark for describing the use of a ruler with mm divisions/callipers AND avoiding parallax error by viewing at 90-degree angle.
(b) (i) [6 marks] - 1 mark for correct column headers with units in slash format. - 1 mark for recording all initial lengths as 40.0 mm. - 1 mark for correct calculation of change in length (with positive/negative signs correctly assigned). - 2 marks for correct calculation of percentage change in length to 1 decimal place (1 mark if 1-2 calculations are incorrect or poorly rounded). - 1 mark for keeping units out of the body of the table.
(b) (ii) [4 marks] - 1 mark for correct axis labels with units: x-axis 'Concentration of sucrose / mol dm-3' and y-axis 'Percentage change in length / %'. - 1 mark for linear, appropriate scale (plotted points cover more than half the grid). - 1 mark for plotting all 6 standard points accurately with a neat cross/dot. - 1 mark for connecting points with a smooth curve or straight lines between points, with no extrapolation.
(b) (iii) [2 marks] - 1 mark for showing the coordinate tracing on the graph at -2.5% change in length. - 1 mark for the correct estimated value of U (0.48 to 0.54 mol dm-3).
(b) (iv) [2 marks] - 1 mark for stating that the 0.2 mol dm-3 solution has a higher water potential than the potato cell sap. - 1 mark for linking the water entry via osmosis across the partially permeable membrane to the increase in cylinder length/turgor.
(b) (v) [2 marks] - 1 mark for suggesting replicates (at least 3) for each concentration to calculate a mean. - 1 mark for suggesting preventing evaporation (e.g., using a lid/Parafilm) or using a narrower range of concentrations around the estimate (e.g., 0.4 to 0.6 mol dm-3 at 0.05 mol dm-3 intervals).
PastPaper.question 2 · practical-investigation
20 PastPaper.marks
Question 2 You are provided with a prepared slide of a transverse section of a stem, labelled J1.
(a) (i) Describe how you would draw a large, low-power plan diagram of a sector of the stem shown on slide J1 to show the distribution of tissues, including at least two vascular bundles. Explain what labels you would add to identify the xylem and the sclerenchyma. [5] (ii) Suggest one visible feature of the xylem vessels in J1 that supports their function in the transport of water. [1]
(b) (i) Describe the key details to include in a high-power drawing of a group of three adjacent parenchyma cells from the pith of slide J1. [4]
(c) A student calibrated the eyepiece graticule of a microscope using a stage micrometer. The stage micrometer scale is \(1.0\text{ mm}\) long and is divided into 100 equal subdivisions. * At a magnification of \(\times100\), 40 eyepiece graticule units (epu) aligned exactly with 8 subdivisions of the stage micrometer.
(i) Calculate the actual length of one eyepiece graticule unit (epu) at this magnification in micrometres (\(\mu\text{m}\)). Show your working. [2] (ii) The student used the same microscope and magnification (\(\times100\)) to measure the radial width of a vascular bundle on slide J1. The width of the vascular bundle was measured as \(18\text{ epu}\). Calculate the actual radial width of the vascular bundle in micrometres (\(\mu\text{m}\)). Show your working. [2]
(d) A student was also provided with Fig. 2.1, which is a photomicrograph of a transverse section through the stem of a different plant species, Zea mays (maize, a monocotyledonous plant). Prepare a table to compare three observable structural differences between the stem on slide J1 (a typical dicotyledonous stem) and the stem shown in Fig. 2.1. [6]
PastPaper.showAnswersPastPaper.hideAnswers
PastPaper.workedSolution
(a) (i) Low-power plan diagram description: - The drawing must be made using a sharp HB pencil with clean, continuous lines and no shading. - It should be a sector (wedge-shape) of the stem, not the entire circle, to ensure it is drawn large. - It must show at least two vascular bundles arranged in a peripheral ring, depicting tissue layers: epidermis, cortex, vascular bundles, and central pith. - No individual cells must be drawn. - Xylem should be labelled in the inner part of the vascular bundles, and sclerenchyma should be labelled as caps on the outer part of the vascular bundles.
(a) (ii) Xylem feature: - Thick/lignified cell walls (to withstand negative pressure/tension) OR hollow lumens/no end walls (to provide an unobstructed pathway for water flow).
(b) (i) High-power cell drawing details: - Draw exactly three cells, all of which must touch (be adjacent). - Cell walls must be represented by double lines to show cell wall thickness, with a middle lamella or intercellular spaces where the cells meet. - Cells should have irregular, polygonal or rounded shapes characteristic of parenchyma, with no shading or sketching.
(c) (ii) Measurement Calculation: - Width of vascular bundle = \(18\text{ epu}\). - Actual width = \(18 \times 2.0\\ \mu\text{m} = 36.0\\ \mu\text{m}\).
(d) Comparison Table: | Feature | Slide J1 (Typical Dicotyledonous Stem) | Fig. 2.1 (Monocotyledonous Stem, Zea mays) | | :--- | :--- | :--- | | 1. Arrangement of vascular bundles | Arranged in a ring near the outer edge | Scattered throughout the ground tissue | | 2. Differentiation of ground tissue | Distinctly separated into cortex and central pith | Undifferentiated ground tissue (no separate cortex/pith) | | 3. Sclerenchyma sheath | Sclerenchyma present only as caps over individual bundles | Sclerenchyma forms a complete sheath surrounding each bundle |
PastPaper.markingScheme
(a) (i) [5 marks] - 1 mark for drawing with clear, thin, continuous lines, no sketching or shading. - 1 mark for drawing a correct representative sector (wedge) containing at least two bundles. - 1 mark for drawing the correct layers of tissues in proportional thickness (epidermis, cortex, vascular bundles, pith). - 1 mark for drawing no individual cells inside the tissues. - 1 mark for correct labels to xylem (inner part of bundle) and sclerenchyma (outer cap/ring).
(a) (ii) [1 mark] - 1 mark for stating thick/lignified cell walls OR hollow lumens/absence of end walls.
(b) (i) [4 marks] - 1 mark for drawing exactly three adjacent cells in contact with each other. - 1 mark for drawing double lines representing cell walls in all three cells. - 1 mark for drawing parenchyma-like irregular/polygonal shapes of appropriate large size. - 1 mark for clean line quality with no shading.
(c) (i) [2 marks] - 1 mark for showing that 1 stage micrometer subdivision equals 10 micrometres, so 8 subdivisions equal 80 micrometres. - 1 mark for correct division (80 / 40) yielding 2.0 micrometres.
(c) (ii) [2 marks] - 1 mark for showing the multiplication of 18 epu by the calibrated value (2.0 micrometres). - 1 mark for the correct answer of 36.0 micrometres (must include unit \(\mu\text{m}\)).
(d) [6 marks] - 1 mark for drawing a table with clear headers comparing Slide J1 and Fig. 2.1. - 1 mark for comparing vascular bundle arrangement (ring vs scattered). - 1 mark for comparing ground tissue separation (distinct pith/cortex vs undifferentiated). - 1 mark for comparing sclerenchyma distribution (bundle caps vs complete bundle sheath). - 2 marks for clear, accurate, and contrasting statements across all three compared features (1 mark per complete row).