An original Thinka practice paper modelled on the structure and difficulty of the Jun 2025 (V4) Cambridge International A Level Biology (9700) paper. Not affiliated with or reproduced from Cambridge.
Paper 1 Multiple Choice
Forty multiple-choice questions based on AS syllabus. Choose the single best option out of A, B, C, D.
32 Question · 32 marks
Question 1 · multiple_choice
1 marks
An experiment was carried out to investigate the effect of two different inhibitors, X and Y, on the activity of an enzyme. In the presence of inhibitor X, the Michaelis-Menten constant (\(K_m\)) increased, but the maximum velocity (\(V_{max}\)) remained unchanged at high substrate concentrations. In the presence of inhibitor Y, the \(K_m\) remained unchanged, but the \(V_{max}\) decreased. Which statement correctly identifies the type of inhibition and the binding site of each inhibitor?
A.X is a competitive inhibitor that binds to the active site; Y is a non-competitive inhibitor that binds to an allosteric site.
B.X is a non-competitive inhibitor that binds to an allosteric site; Y is a competitive inhibitor that binds to the active site.
C.X is a competitive inhibitor that binds to an allosteric site; Y is a non-competitive inhibitor that binds to the active site.
D.X is a non-competitive inhibitor that binds to the active site; Y is a competitive inhibitor that binds to an allosteric site.
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Worked solution
A competitive inhibitor (X) competes with the substrate for the active site, increasing the value of \(K_m\) without changing \(V_{max}\). A non-competitive inhibitor (Y) binds to a site other than the active site (an allosteric site), reducing the overall rate of reaction (reducing \(V_{max}\)) but not affecting substrate affinity, meaning \(K_m\) remains unchanged.
Marking scheme
[1 mark] Correctly identifies inhibitor X as competitive (binding to the active site) and inhibitor Y as non-competitive (binding to an allosteric site).
Question 2 · multiple_choice
1 marks
In a mammalian cell, the concentration of sodium ions is 15 au inside and 145 au outside; potassium ions is 140 au inside and 5 au outside; glucose is 8 au inside and 5 au outside; and oxygen is 20 au inside and 100 au outside. Which transport mechanism is correctly matched to the direction of net movement of the substance under normal physiological conditions?
A.Sodium ions move out of the cell by facilitated diffusion.
B.Potassium ions move into the cell by active transport.
C.Glucose moves out of the cell by simple diffusion.
D.Oxygen moves out of the cell by active transport.
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Worked solution
Potassium ions are highly concentrated inside the cell (140 au) compared to outside (5 au). Moving potassium ions into the cell is against their concentration gradient, which requires active transport. Sodium ions moving out is also active transport (against gradient). Glucose is polar and cannot cross by simple diffusion. Oxygen is non-polar and moves in by simple diffusion.
Marking scheme
[1 mark] Correctly identifies that moving potassium into the cell requires active transport against its concentration gradient.
Question 3 · multiple_choice
1 marks
A segment of double-stranded DNA is shown: 5'-ATG GCC TTA CGA-3' (Coding strand) and 3'-TAC CGG AAT GCT-5' (Template strand). During transcription, this template strand is used to produce mRNA, which is then translated. Which of the following represents the correct mRNA sequence and the anticodon of the tRNA that will pair with the third codon of this mRNA?
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Worked solution
The mRNA is complementary to the template strand (3'-TAC CGG AAT GCT-5') and synthesized 5' to 3', yielding 5'-AUG GCC UUA CGA-3'. The third codon is 5'-UUA-3'. The complementary tRNA anticodon must be antiparallel, which is 3'-AAU-5'.
Marking scheme
[1 mark] Correctly identifies the 5'-to-3' mRNA sequence and the complementary, antiparallel 3'-to-5' tRNA anticodon.
Question 4 · multiple_choice
1 marks
A diploid animal cell contains 12 chromosomes (2n = 12). During the mitotic cell cycle, the quantity of DNA and the number of chromosomes/chromatids change. Which option correctly identifies the number of chromosomes, the number of chromatids, and the relative mass of DNA per cell during the specified stages?
A.G1 phase: Chromosomes = 12, Chromatids = 12, DNA mass = 2C; Metaphase: Chromosomes = 12, Chromatids = 24, DNA mass = 4C
B.G2 phase: Chromosomes = 24, Chromatids = 24, DNA mass = 4C; Anaphase: Chromosomes = 24, Chromatids = 24, DNA mass = 4C
C.Prophase: Chromosomes = 12, Chromatids = 24, DNA mass = 4C; Telophase (before cytokinesis): Chromosomes = 12, Chromatids = 12, DNA mass = 2C
D.Metaphase: Chromosomes = 12, Chromatids = 12, DNA mass = 4C; Anaphase: Chromosomes = 24, Chromatids = 0, DNA mass = 2C
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Worked solution
In G1, DNA is unreplicated so there are 12 chromosomes (12 chromatids/DNA molecules) and DNA mass is 2C. After replication in S phase and up to metaphase, there are 12 chromosomes, each with 2 sister chromatids (total 24 chromatids) and a DNA mass of 4C.
Marking scheme
[1 mark] Correctly identifies chromosomes, chromatids, and DNA mass in G1 and Metaphase.
Question 5 · multiple_choice
1 marks
Three biological molecules are described: Molecule 1 is a polysaccharide containing beta-1,4-glycosidic bonds forming straight, unbranched chains. Molecule 2 is a globular protein containing disulfide, ionic, and hydrogen bonds, and hydrophobic interactions between R-groups of a single polypeptide chain. Molecule 3 is a triglyceride containing three saturated fatty acid tails esterified to glycerol. Which option correctly identifies the description or properties of these molecules?
A.Molecule 1 is amylose; Molecule 2 is a quaternary structure protein; Molecule 3 has no double bonds in its hydrocarbon chains.
B.Molecule 1 is cellulose, providing high tensile strength to cell walls; Molecule 2 has a tertiary structure; Molecule 3 has no C=C double bonds in its fatty acid tails.
C.Molecule 1 is cellulose, easily hydrolysed for energy storage; Molecule 2 has a secondary structure; Molecule 3 is hydrophilic.
D.Molecule 1 is glycogen; Molecule 2 has a tertiary structure; Molecule 3 is a major component of cell membranes.
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Worked solution
Molecule 1 represents cellulose (beta-1,4-glycosidic bonds, unbranched), which provides structural support. Molecule 2 represents the tertiary structure of a single polypeptide chain with various R-group interactions. Molecule 3 is a saturated triglyceride, meaning its fatty acid tails contain no C=C double bonds.
Marking scheme
[1 mark] Correctly matches Molecule 1 to cellulose, Molecule 2 to tertiary structure, and Molecule 3 to absence of double bonds in saturated fatty acid tails.
Question 6 · multiple_choice
1 marks
Active loading of sucrose into phloem sieve tube elements involves several steps. Which sequence of events correctly describes the active loading of sucrose at a source?
A.Hydrogen ions are actively pumped out of the companion cell into the cell wall, then diffuse back in down their electrochemical gradient via a co-transporter protein, bringing sucrose with them, and sucrose then diffuses into the sieve tube element through plasmodesmata.
B.Sucrose is actively pumped out of the companion cell into the cell wall, then hydrogen ions diffuse into the companion cell through co-transporter proteins, and sucrose is actively transported into the sieve tube element.
C.Hydrogen ions are actively pumped out of the sieve tube element into the companion cell, then diffuse back down a concentration gradient, and sucrose moves by facilitated diffusion into the sieve tube element.
D.Sucrose is actively transported into the companion cell using ATP, then hydrogen ions diffuse out of the companion cell through plasmodesmata, and sucrose is actively pumped into the sieve tube element.
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Worked solution
Active loading begins with companion cells actively pumping hydrogen ions (H+) out into the cell wall space. H+ then diffuses back into the companion cell down its concentration gradient via co-transporter proteins, carrying sucrose along. Sucrose then diffuses into the sieve tube element via plasmodesmata.
Marking scheme
[1 mark] Correctly identifies the complete sequence of events for active phloem loading.
Question 7 · multiple_choice
1 marks
Consider four scenarios of immunity: Scenario 1 - a child receives a vaccine containing weakened pathogens; Scenario 2 - a newborn receives IgA antibodies from breast milk; Scenario 3 - a patient is injected with antitoxin antibodies after a snake bite; Scenario 4 - a person recovers from an infection of influenza and develops immunity. Which option correctly classifies the type of immunity for each scenario?
A.Scenario 1: Active artificial; Scenario 2: Passive natural; Scenario 3: Passive artificial; Scenario 4: Active natural
B.Scenario 1: Active natural; Scenario 2: Passive artificial; Scenario 3: Passive natural; Scenario 4: Active artificial
C.Scenario 1: Passive artificial; Scenario 2: Active natural; Scenario 3: Active artificial; Scenario 4: Passive natural
D.Scenario 1: Active artificial; Scenario 2: Passive artificial; Scenario 3: Passive natural; Scenario 4: Active natural
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Worked solution
Scenario 1 is active artificial because a vaccine induces an active immune response artificially. Scenario 2 is passive natural as antibodies are naturally transferred from mother to infant without triggering an immune response. Scenario 3 is passive artificial because pre-formed antibodies are injected. Scenario 4 is active natural because infection leads to a natural active immune response.
Marking scheme
[1 mark] Correctly classifies all four scenarios into active/passive and natural/artificial immunity.
Question 8 · multiple_choice
1 marks
Which tissues are present or absent in the human trachea, bronchi, and alveoli? (Key: present = yes, absent = no)
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Worked solution
Both trachea and bronchi contain cartilage, goblet cells, smooth muscle, and ciliated epithelium. Alveoli do not contain any of these tissues, being composed only of a single layer of squamous epithelial cells and elastic fibres to facilitate gas exchange.
Marking scheme
[1 mark] Correctly identifies the presence or absence of cartilage, goblet cells, smooth muscle, and ciliated epithelium in the trachea, bronchi, and alveoli.
Question 9 · multiple-choice
1 marks
An enzyme-catalysed reaction has a Michaelis-Menten constant (\(K_m\)) of \(2.0 \times 10^{-4}\text{ mol dm}^{-3}\). In the presence of a certain inhibitor, the rate of reaction at a substrate concentration of \(2.0 \times 10^{-4}\text{ mol dm}^{-3}\) is halved. However, when the substrate concentration is increased to \(2.0 \times 10^{-1}\text{ mol dm}^{-3}\), the rate of reaction is found to be the same as the maximum rate (\(V_{max}\)) achieved without the inhibitor. Which statement correctly identifies the type of inhibitor and the reason for this effect?
A.It is a competitive inhibitor because the maximum rate of reaction (\(V_{max}\)) can still be achieved at high substrate concentrations.
B.It is a competitive inhibitor because the value of the Michaelis-Menten constant (\(K_m\)) is decreased in its presence.
C.It is a non-competitive inhibitor because the rate of reaction at the original \(K_m\) substrate concentration is halved.
D.It is a non-competitive inhibitor because the inhibitor binds reversibly to the enzyme's active site.
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Worked solution
Since the maximum rate of reaction (\(V_{max}\)) is unaffected and can still be achieved at very high substrate concentrations, the inhibitor must be competitive. Competitive inhibitors compete with the substrate for the active site, and their inhibitory effect can be overcome by increasing the substrate concentration. A competitive inhibitor increases the apparent \(K_m\) (it does not decrease it).
Marking scheme
[1 mark] - Correctly identifies that a competitive inhibitor allows \(V_{max}\) to be reached at high substrate concentrations, selecting option A.
Question 10 · multiple-choice
1 marks
A double-stranded DNA molecule is analysed and found to contain 34% cytosine. Which of the following can be deduced with certainty about the base composition of a single strand of this DNA molecule?
A.It contains 16% adenine.
B.It contains 34% guanine.
C.The combined percentage of adenine and thymine is 32%.
D.The combined percentage of cytosine and guanine is 34%.
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Worked solution
In a double-stranded DNA molecule, if cytosine (C) is 34%, then guanine (G) must also be 34% because of complementary base pairing. Together, C + G make up 68% of the DNA. The remaining 32% must consist of adenine (A) and thymine (T). In any single strand, for every A in that strand, there is a T in the opposite strand, and vice versa. This means that the total count of A + T in one strand is exactly equal to the total count of A + T in the other strand. Thus, the percentage of (A + T) in any single strand is always equal to the overall percentage of (A + T) in the double-stranded molecule, which is 32%. Individual percentages of A, T, C, or G in a single strand cannot be determined with certainty from double-stranded data.
Marking scheme
[1 mark] - Explains that since double-stranded DNA has 68% C+G, it has 32% A+T, meaning each single strand must have exactly 32% A+T. Selects option C.
Question 11 · multiple-choice
1 marks
Plant cells are placed in a hypertonic sucrose solution. As water leaves the cells by osmosis, the protoplast shrinks away from the cell wall. At the exact point of incipient plasmolysis, which statement correctly describes the relationship between water potential (\(\Psi\)), solute potential (\(\Psi_s\)), and pressure potential (\(\Psi_p\)) of these cells?
A.\(\Psi_p\) is zero, so \(\Psi = \Psi_s\)
B.\(\Psi_s\) is zero, so \(\Psi = \Psi_p\)
C.\(\Psi\) is zero, so \(\Psi_s = -\Psi_p\)
D.\(\Psi_p\) is at its maximum positive value, so \(\Psi\) is greater than \(\Psi_s\)
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Worked solution
At the point of incipient plasmolysis, the protoplast has shrunk to the extent that it exerts no pressure against the cell wall. Consequently, the turgor pressure or pressure potential (\(\Psi_p\)) is zero. Since the overall water potential equation is \(\Psi = \Psi_s + \Psi_p\), when \(\Psi_p = 0\), the water potential (\(\Psi\)) is equal to the solute potential (\(\Psi_s\)).
Marking scheme
[1 mark] - Recalls that pressure potential is zero at incipient plasmolysis and deduces that water potential equals solute potential. Selects option A.
Question 12 · multiple-choice
1 marks
Which of the following cellular components are present in both a healthy eukaryotic leaf palisade mesophyll cell and a cell of the bacterium *Vibrio cholerae*?
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Worked solution
Eukaryotic plant cells contain chloroplasts and mitochondria, both of which possess 70S ribosomes and circular DNA (reflecting their endosymbiotic evolutionary origins). Therefore, a mesophyll cell has 70S ribosomes and circular DNA. The bacterium *Vibrio cholerae* (a prokaryote) also has 70S ribosomes and circular DNA. Both cells require RNA polymerase for transcription. However, only the bacterium has a peptidoglycan cell wall, while the mesophyll cell has a cellulose cell wall. Therefore, components 1, 2, and 4 are found in both, but 3 is not.
Marking scheme
[1 mark] - Recognizes that chloroplasts/mitochondria contain 70S ribosomes and circular DNA, and both cells contain RNA polymerase but have different cell wall compositions. Selects option A.
Question 13 · multiple-choice
1 marks
During the process of active loading of sucrose into the phloem companion cells of a plant, which of the following events occur?
1. Active transport of hydrogen ions (\(\text{H}^+\)) out of the companion cell cytoplasm into the cell wall apoplast. 2. Co-transport of sucrose and hydrogen ions (\(\text{H}^+\)) back into the companion cell down an electrochemical gradient. 3. Diffusion of sucrose from the companion cell into the sieve tube element via plasmodesmata.
A.1, 2 and 3
B.1 and 2 only
C.2 and 3 only
D.1 and 3 only
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Worked solution
All three processes occur during phloem loading. First, proton pumps actively transport hydrogen ions (\(\text{H}^+\)) out of the companion cell into the apoplast, establishing a high proton concentration gradient outside. Second, these hydrogen ions diffuse back into the companion cell through co-transporter proteins, carrying sucrose molecules against their concentration gradient (co-transport). Third, once inside the companion cell, sucrose diffuses down its concentration gradient into the adjacent sieve tube element through the plasmodesmata.
Marking scheme
[1 mark] - Identifies all three listed steps as correct components of the phloem loading mechanism. Selects option A.
Question 14 · multiple-choice
1 marks
Which of the following changes shifts the oxygen dissociation curve of adult hemoglobin to the right, and what is the physiological advantage of this shift?
A.An increase in carbon dioxide concentration; it decreases hemoglobin's affinity for oxygen, allowing more oxygen to be released to respiring tissues.
B.A decrease in hydrogen ion concentration; it increases hemoglobin's affinity for oxygen, allowing more oxygen to be loaded at the lungs.
C.An increase in temperature; it increases hemoglobin's affinity for oxygen, preventing premature oxygen loss during exercise.
D.A decrease in carbon dioxide concentration; it decreases hemoglobin's affinity for oxygen, promoting rapid delivery of oxygen to muscles.
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Worked solution
The Bohr effect states that an increase in carbon dioxide concentration (which leads to an increase in hydrogen ions and a decrease in pH) shifts the oxygen dissociation curve of hemoglobin to the right. A rightward shift means that, at any given partial pressure of oxygen (\(p\text{O}_2\)), hemoglobin has a lower affinity for oxygen. This is physiologically advantageous because it facilitates the unloading of oxygen to tissues that are actively respiring and producing carbon dioxide.
Marking scheme
[1 mark] - Correctly identifies that increased carbon dioxide shifts the curve to the right, decreasing affinity to release oxygen to tissues. Selects option A.
Question 15 · multiple-choice
1 marks
A diploid eukaryotic cell has a diploid chromosome number of \(2n = 8\). How many chromosomes and how many DNA molecules are present in this cell during prophase of mitosis and during anaphase of mitosis?
A.Prophase: 8 chromosomes, 16 DNA molecules; Anaphase: 16 chromosomes, 16 DNA molecules
B.Prophase: 8 chromosomes, 8 DNA molecules; Anaphase: 8 chromosomes, 16 DNA molecules
C.Prophase: 16 chromosomes, 16 DNA molecules; Anaphase: 8 chromosomes, 8 DNA molecules
D.Prophase: 8 chromosomes, 16 DNA molecules; Anaphase: 16 chromosomes, 8 DNA molecules
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Worked solution
During prophase of mitosis, the DNA has already replicated in interphase. There are 8 chromosomes, each consisting of 2 sister chromatids. Since each chromatid contains one double-stranded DNA molecule, there are 16 DNA molecules. During anaphase, the sister chromatids separate at the centromere and are pulled to opposite poles. Once separated, each chromatid is classified as an individual chromosome, so the chromosome number temporarily doubles to 16. The total number of DNA molecules in the cell remains 16.
Marking scheme
[1 mark] - Correctly calculates chromosome and DNA molecule counts for both prophase and anaphase. Selects option A.
Question 16 · multiple-choice
1 marks
Which of the following correctly pairs a medical scenario with the specific type of immunity it confers to the individual?
A.Injection of anti-venom containing antibodies after a venomous snake bite — Artificial passive immunity
B.Infection with the measles virus followed by full recovery — Artificial active immunity
C.Transfer of protective maternal antibodies across the placenta to a fetus — Natural active immunity
D.Receiving a booster dose of a tetanus toxoid vaccine — Natural passive immunity
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Worked solution
Injection of pre-formed antibodies (such as snake anti-venom or tetanus immunoglobulin) provides instant protection but does not trigger the recipient's immune system to make memory cells. Thus, it is artificial (medical intervention) and passive (ready-made antibodies). Natural active immunity comes from natural infection (measles). Natural passive immunity comes from placenta/colostrum transfer. Artificial active immunity comes from vaccination.
Marking scheme
[1 mark] - Identifies anti-venom injection as artificial passive immunity and recognizes the other options as incorrect pairings. Selects option A.
Question 17 · multiple-choice
1 marks
A stage micrometer has scale divisions of \(0.1\text{ mm}\). At magnification \(\times 100\), 40 eyepiece graticule units align perfectly with 8 stage micrometer divisions. When a plant cell is viewed under the same magnification, its width spans exactly 15 eyepiece graticule units. What is the actual width of the plant cell in micrometres (\(\mu\text{m}\))?
A.30
B.150
C.300
D.600
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Worked solution
One stage micrometer division equals \(0.1\text{ mm}\) which is \(100\ \mu\text{m}\). Therefore, 8 divisions equal \(800\ \mu\text{m}\). Since 40 eyepiece graticule units (epu) align with these 8 divisions, 1 epu is equal to \(800\ \mu\text{m} / 40 = 20\ \mu\text{m}\). The plant cell is 15 epu wide, so its actual width is \(15 \times 20\ \mu\text{m} = 300\ \mu\text{m}\).
Marking scheme
Award 1 mark for the correct calculation of 300 micrometres, corresponding to Option C.
Question 18 · multiple-choice
1 marks
Glycogen has a higher frequency of \(\alpha\)-1,6-glycosidic branches than amylopectin. Which statement correctly explains why glycogen is more suitable as an energy storage molecule in animals than amylopectin is in plants?
A.It allows glycogen to form rigid, unbranched fibrils that strengthen the animal cell membrane.
B.It reduces the number of terminal ends, slowing down hydrolysis to maintain stable blood glucose levels.
C.It increases the overall solubility of the polysaccharide so that it can be actively transported through the blood plasma.
D.It provides a greater number of terminal ends per unit volume, allowing rapid mobilization of glucose to meet higher metabolic demands.
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Worked solution
The higher degree of branching in glycogen compared to amylopectin means there are far more terminal ends per unit volume. This allows glycogen-hydrolysing enzymes to act on multiple branches simultaneously, rapidly mobilizing glucose molecules to meet the high and sudden energy demands of active animal cells. Plants have a lower metabolic rate and do not require such rapid glucose release.
Marking scheme
Award 1 mark for identifying that increased branching provides more terminal ends for rapid enzymatic hydrolysis to meet higher metabolic demands (Option D).
Question 19 · multiple-choice
1 marks
An enzyme-controlled reaction is carried out in the presence of a reversible competitive inhibitor. How do the maximum rate of reaction (\(V_{\max}\)) and the Michaelis-Menten constant (\(K_m\)) of this reaction compare to the values obtained without the inhibitor?
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Worked solution
A competitive inhibitor competes with the substrate for the active site of the enzyme. At infinitely high substrate concentrations, the substrate outcompetes the inhibitor completely, meaning \(V_{\max}\) remains unchanged. However, because a higher concentration of substrate is required to reach half of this maximum velocity, the apparent affinity of the enzyme for its substrate decreases, which is reflected as an increase in \(K_m\).
Marking scheme
Award 1 mark for correctly stating that \(V_{\max}\) remains unchanged and \(K_m\) increases (Option B).
Question 20 · multiple-choice
1 marks
Groups of uniform sweet potato cylinders were placed in sucrose solutions of concentrations \(0.0\), \(0.2\), \(0.4\), \(0.6\), and \(0.8\text{ mol dm}^{-3}\). After two hours, the percentage changes in mass of the cylinders were calculated as \(+12\%\), \(+5\%\), \(-2\%\), \(-8\%\), and \(-12\%\) respectively. Which of the following statements is a correct deduction from this experiment?
A.The water potential of the sweet potato cells is equal to the water potential of a sucrose solution between \(0.2\text{ mol dm}^{-3}\) and \(0.4\text{ mol dm}^{-3}\).
B.In the \(0.0\text{ mol dm}^{-3}\) solution, the sweet potato cells have become fully plasmolysed.
C.In the \(0.4\text{ mol dm}^{-3}\) solution, water entered the sweet potato cells by osmosis down a water potential gradient.
D.In the \(0.8\text{ mol dm}^{-3}\) solution, the cellulose cell walls have actively contracted to reduce the volume of the tissue.
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Worked solution
The point of no net movement of water (where percentage change in mass is \(0\%\)) corresponds to the concentration of sucrose where the water potential of the sucrose solution matches the internal water potential of the potato cells. Since the mass change goes from \(+5\%\) (net water entry at \(0.2\text{ mol dm}^{-3}\)) to \(-2\%\) (net water exit at \(0.4\text{ mol dm}^{-3}\)), the point of zero mass change must lie between these two concentrations.
Marking scheme
Award 1 mark for the deduction that the water potential of the sweet potato cells is equivalent to a sucrose solution between \(0.2\) and \(0.4\text{ mol dm}^{-3}\) (Option A).
Question 21 · multiple-choice
1 marks
A template strand of DNA has the base sequence: `3'- T A C G G T C A T T C A -5'`. What is the sequence of the tRNA anticodons that align with the mRNA transcribed from this sequence?
A.`3'- U A C - 5'`, `3'- G G U - 5'`, `3'- C A U - 5'`, `3'- U C A - 5'`
B.`5'- U A C - 3'`, `5'- G G U - 3'`, `5'- C A U - 3'`, `5'- U C A - 3'`
C.`3'- A U G - 5'`, `3'- C C A - 5'`, `3'- G U A - 5'`, `3'- A G U - 5'`
D.`5'- A U G - 3'`, `5'- C C A - 3'`, `5'- G U A - 3'`, `5'- A G U - 3'`
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Worked solution
1. Transcription of the template DNA sequence `3'- T A C G G T C A T T C A -5'` produces the complementary mRNA sequence `5'- A U G C C A G U A A G U -3'`. 2. During translation, tRNA molecules pair antiparallelly with the mRNA codons. The codon `5'-AUG-3'` pairs with the anticodon `3'-UAC-5'`. The codon `5'-CCA-3'` pairs with the anticodon `3'-GGU-5'`. The codon `5'-GUA-3'` pairs with the anticodon `3'-CAU-5'`. The codon `5'-AGU-3'` pairs with the anticodon `3'-UCA-5'`. Thus, the correct sequence of anticodons is shown in Option A.
Marking scheme
Award 1 mark for correctly transcribing the mRNA sequence and aligning it with the antiparallel tRNA anticodons written 3'-to-5' (Option A).
Question 22 · multiple-choice
1 marks
Which statement correctly describes the mechanism of active loading of sucrose into phloem companion cells?
A.Protons (hydrogen ions) are actively pumped out of the companion cells, creating an electrochemical gradient that drives sucrose in via a cotransporter protein.
B.Sucrose molecules are actively pumped directly into the companion cells via carrier proteins using energy directly released from ATP hydrolysis.
C.Protons and sucrose molecules diffuse down their respective concentration gradients from the companion cells into sieve tube elements.
D.Sucrose is actively loaded via simple diffusion through large plasmodesmata in the companion cell wall.
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Worked solution
Active phloem loading relies on an electrochemical gradient of protons. Proton pumps actively transport hydrogen ions (protons) out of the companion cells and into the cell wall space. Protons then diffuse back into the companion cells down their concentration and electrochemical gradient through co-transporter proteins, which bring sucrose molecules into the cell against the sucrose concentration gradient.
Marking scheme
Award 1 mark for identifying that active sucrose loading is driven by a proton electrochemical gradient via a cotransporter (Option A).
Question 23 · multiple-choice
1 marks
During which phase of the cardiac cycle do the semi-lunar valves at the base of the aorta open?
A.When the pressure in the left atrium becomes greater than the pressure in the left ventricle
B.When the pressure in the left ventricle becomes greater than the pressure in the aorta
C.When the pressure in the aorta becomes lower than the pressure in the left atrium
D.When the left atrium begins to contract during atrial systole
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Worked solution
The semi-lunar valves open when ventricular pressure exceeds the pressure in the arteries. For the aortic valve, this occurs during ventricular systole when the pressure within the contracting left ventricle rises above the diastolic pressure within the aorta, allowing blood to enter the systemic circulation.
Marking scheme
Award 1 mark for identifying that the aortic valve opens when left ventricular pressure exceeds aortic pressure (Option B).
Question 24 · multiple-choice
1 marks
Which row correctly identifies the presence (\(\checkmark\)) or absence (\(\times\)) of tissues in the wall of a healthy human bronchiole?
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Worked solution
A bronchiole is a narrow airway that lacks cartilage (unlike the trachea and bronchi) and lacks goblet cells (to prevent thick mucus secretions from blocking the small passageways). However, its walls still contain elastic fibres (and smooth muscle) to support structures and allow elastic recoil during ventilation.
Marking scheme
Award 1 mark for selecting the row indicating that bronchioles lack cartilage, contain elastic fibres, and lack goblet cells (Option B).
Question 25 · multiple choice
1 marks
An experiment was carried out to investigate the effect of two different inhibitors, X and Y, on an enzyme-controlled reaction. Under normal conditions, the Michaelis-Menten constant \(K_m\) is 2.5 arbitrary units (au) and the maximum velocity \(V_{max}\) is 100 au. In the presence of inhibitor X, \(K_m\) increases to 6.0 au and \(V_{max}\) remains 100 au. In the presence of inhibitor Y, \(K_m\) remains 2.5 au and \(V_{max}\) decreases to 45 au. Which statement correctly identifies the types of inhibition shown by X and Y?
A.X is a competitive inhibitor and Y is a non-competitive inhibitor.
B.X is a non-competitive inhibitor and Y is a competitive inhibitor.
C.Both X and Y are competitive inhibitors.
D.Both X and Y are non-competitive inhibitors.
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Worked solution
Competitive inhibitors bind to the active site and compete with the substrate. This increases the Michaelis-Menten constant \(K_m\) (representing lower affinity), but the maximum rate of reaction \(V_{max}\) remains unchanged because a high substrate concentration can overcome the inhibition. Non-competitive inhibitors bind to an allosteric site, decreasing \(V_{max}\) because they reduce the total concentration of active enzyme molecules, but they do not affect the binding affinity of the substrate at the active site, leaving \(K_m\) unchanged.
Marking scheme
1 mark for identifying X as a competitive inhibitor and Y as a non-competitive inhibitor based on their effects on \(K_m\) and \(V_{max}\).
Question 26 · multiple choice
1 marks
A radioactive amino acid is introduced into a culture of mucus-secreting goblet cells from the mammalian trachea. In which sequence of organelles will the radioactivity be detected over time as the glycoprotein mucin is synthesized, modified, and secreted?
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Worked solution
Protein synthesis occurs on ribosomes. The newly synthesized polypeptide enters the cisternae of the rough endoplasmic reticulum (RER) for folding. It is then transported via transport vesicles to the Golgi body, where it undergoes modification (such as glycosylation) to form mucin. Finally, it is packaged into secretory vesicles that migrate to and fuse with the cell surface membrane to release the product.
Marking scheme
1 mark for identifying the correct sequential pathway: ribosome -> rough endoplasmic reticulum -> Golgi body -> secretory vesicle.
Question 27 · multiple choice
1 marks
The table describes how the rate of uptake of three substances (1, 2, and 3) into a plant cell is affected by a respiratory inhibitor and by a membrane protein-blocking agent. Substance 1: unaffected by respiratory inhibitor, unaffected by protein-blocking agent. Substance 2: unaffected by respiratory inhibitor, decreased by protein-blocking agent. Substance 3: decreased by respiratory inhibitor, decreased by protein-blocking agent. Which transport mechanisms are used by substances 1, 2, and 3?
A.1 = simple diffusion, 2 = facilitated diffusion, 3 = active transport
B.1 = facilitated diffusion, 2 = simple diffusion, 3 = active transport
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Worked solution
Substance 1 is unaffected by both inhibitors, meaning it does not require cellular energy (ATP) or membrane proteins, which matches simple diffusion. Substance 2 requires a protein carrier or channel (it is decreased by the protein-blocking agent) but does not require metabolic energy (it is unaffected by the respiratory inhibitor), which matches facilitated diffusion. Substance 3 requires both ATP (it is decreased by the respiratory inhibitor) and a carrier protein (it is decreased by the protein-blocking agent), which matches active transport.
Marking scheme
1 mark for correctly matching all three transport mechanisms based on their response to the inhibitors.
Question 28 · multiple choice
1 marks
Which level of protein structure and associated bonding is correctly described?
A.Primary structure is stabilized by hydrogen bonds between the R-groups of adjacent amino acids.
B.Secondary structure is stabilized by peptide bonds forming alpha-helices and beta-pleated sheets.
C.Tertiary structure is stabilized by hydrophobic interactions, ionic bonds, hydrogen bonds, and disulfide bonds between R-groups.
D.Quaternary structure is only present in proteins consisting of identical polypeptide chains held by covalent peptide bonds.
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Worked solution
Tertiary structure refers to the overall 3D shape of a single polypeptide chain, which is stabilized by various interactions between the amino acid R-groups, including hydrophobic interactions, ionic bonds, hydrogen bonds, and covalent disulfide bonds.
Marking scheme
1 mark for selecting the correct definition of tertiary protein structure and its stabilizing bonds.
Question 29 · multiple choice
1 marks
A diploid animal cell in the G1 phase of the cell cycle contains \(y\) picograms of DNA. The cell undergoes a complete mitotic division. What is the mass of DNA in a single nucleus of this cell during the G2 phase and during the prophase of mitosis?
A.G2 phase = \(y\); Prophase = \(y\)
B.G2 phase = \(2y\); Prophase = \(2y\)
C.G2 phase = \(2y\); Prophase = \(y\)
D.G2 phase = \(y\); Prophase = \(2y\)
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Worked solution
In the S phase of interphase, DNA replication occurs, doubling the DNA mass from \(y\) to \(2y\). Thus, during the G2 phase, the cell contains \(2y\) picograms of DNA. Prophase is the first stage of mitosis, before any nuclear or cellular division has occurred, so the nucleus still contains \(2y\) picograms of DNA.
Marking scheme
1 mark for identifying that both G2 phase and prophase nuclei contain \(2y\) picograms of DNA.
Question 30 · multiple choice
1 marks
Which sequence of events correctly describes the active loading of sucrose into a companion cell and its subsequent entry into a sieve tube element at a source?
A.Protons are pumped out of the companion cell by active transport -> protons diffuse back into the companion cell down their gradient via a co-transporter protein, bringing sucrose with them -> sucrose diffuses into the sieve tube element via plasmodesmata.
B.Protons are pumped into the companion cell by active transport -> protons diffuse out via a co-transporter, carrying sucrose into the sieve tube element -> water potential of the sieve tube element increases.
C.Sucrose is actively pumped out of the companion cell -> protons follow sucrose by facilitated diffusion into the sieve tube element -> hydrostatic pressure decreases.
D.Companion cells actively transport sucrose directly into the sieve tube element using ATP -> water moves out of the sieve tube element by osmosis.
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Worked solution
The companion cell actively pumps protons out of its cytoplasm into the cell wall space using ATP. This creates a proton gradient. Protons then diffuse back into the companion cell down their concentration gradient through a co-transporter protein, bringing sucrose along with them against its concentration gradient. Once accumulated in high concentration, sucrose diffuses into the sieve tube element down its concentration gradient through plasmodesmata.
Marking scheme
1 mark for identifying the correct order: proton pumping, co-transport of sucrose with protons, and diffusion into sieve tubes via plasmodesmata.
Question 31 · multiple choice
1 marks
The following reactions occur inside a red blood cell as it passes through actively respiring tissues: (1) \(\text{CO}_2 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{CO}_3\) (catalyzed by Enzyme X); (2) \(\text{H}_2\text{CO}_3 \rightarrow \text{H}^+ + \text{HCO}_3^-\); (3) \(\text{H}^+ + \text{HbO}_8 \rightarrow \text{HHb} + 4\text{O}_2\). Which row correctly identifies Enzyme X, the role of hemoglobin in reaction (3), and the movement of \(\text{HCO}_3^-\) to maintain electrical neutrality?
A.Enzyme X is carbonic anhydrase; role of Hb is buffer; \(\text{HCO}_3^-\) diffuses out of the cell as \(\text{Cl}^-\) ions diffuse in.
B.Enzyme X is carbonic anhydrase; role of Hb is catalyst; \(\text{HCO}_3^-\) diffuses into the cell as \(\text{Cl}^-\) ions diffuse out.
C.Enzyme X is carboxypeptidase; role of Hb is buffer; \(\text{HCO}_3^-\) diffuses out of the cell as \(\text{Na}^+\) ions diffuse in.
D.Enzyme X is catalase; role of Hb is oxygen carrier only; \(\text{HCO}_3^-\) is actively transported out of the cell.
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Worked solution
Enzyme X is carbonic anhydrase, which catalyzes the hydration of carbon dioxide. In reaction 3, hemoglobin binds to free hydrogen ions (H+), acting as a buffer to prevent a drop in cellular pH. To maintain electrical neutrality, the hydrogencarbonate ions (HCO3-) diffuse out of the red blood cell into the plasma down a concentration gradient, while chloride ions (Cl-) diffuse in from the plasma (known as the chloride shift).
Marking scheme
1 mark for identifying carbonic anhydrase as Enzyme X, hemoglobin acting as a buffer, and the exit of hydrogencarbonate ions compensated by chloride ions entering.
Question 32 · multiple choice
1 marks
Which row correctly identifies the presence (yes) or absence (no) of tissues in the wall of a healthy human terminal bronchiole?
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Worked solution
Terminal bronchioles do not contain cartilage, as support structures are not needed at this fine level of the airway. They also lack goblet cells (which would obstruct small airways with mucus). However, they contain smooth muscle (to regulate airflow) and are lined with ciliated epithelium to clear remaining fine debris.
Marking scheme
1 mark for identifying that terminal bronchioles lack cartilage and goblet cells, but contain smooth muscle and ciliated epithelium.
Paper 2 AS Level Structured
Answer all six structured questions in the spaces provided on the exam paper.
6 Question · 60 marks
Question 1 · Structured
10 marks
Alkaline phosphatase (ALP) is an enzyme that catalyses the hydrolysis of phosphate groups from organic molecules, such as p-nitrophenyl phosphate (pNPP), to produce a yellow-coloured product, p-nitrophenol, under alkaline conditions.
(a) Explain the term **activation energy** of an enzyme-catalysed reaction. [2]
(b) A student investigated the effect of substrate concentration on the rate of reaction of ALP. Describe how the student could use a colorimeter to obtain a quantitative measure of the initial rate of reaction. [4]
(c) The student repeated the investigation in the presence of a competitive inhibitor. Explain the effect of a competitive inhibitor on the rate of reaction of ALP, and how this effect can be overcome at very high substrate concentrations. [4]
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Worked solution
(a) Activation energy is the minimum amount of energy that reactant molecules must possess before they can undergo a chemical reaction. Enzymes act as catalysts by lowering this activation energy barrier. They do this by holding substrate molecules in a way that puts physical stress on their chemical bonds or by creating a highly favorable microenvironment (e.g., local pH or charge) within the active site, facilitating transition state formation.
(b) To obtain a quantitative measure of the initial rate of reaction using a colorimeter: 1. Prepare a series of reaction mixtures with different concentrations of the substrate pNPP, keeping the volume and concentration of alkaline phosphatase (ALP) constant. 2. Set the colorimeter to a wavelength of approximately 405 nm (or use a blue filter), which is the wavelength absorbed by the yellow product, p-nitrophenol. 3. Calibrate the colorimeter using a blank (containing buffer and enzyme or substrate only) to set the absorbance to zero. 4. Mix the enzyme and substrate, quickly transfer the mixture to a cuvette, place it in the colorimeter, and record the absorbance at regular, short time intervals (e.g., every 10 or 15 seconds) for the first minute. 5. Plot a graph of absorbance (y-axis) against time (x-axis) and draw a tangent to the steepest, linear portion of the curve at the start ($t = 0$) to calculate the gradient, which represents the initial rate of reaction.
(c) A competitive inhibitor has a molecular structure that is complementary/similar in shape to the substrate (pNPP). It binds reversibly to the active site of ALP, preventing substrate molecules from entering and forming enzyme-substrate complexes, thereby reducing the rate of reaction. However, because the binding is reversible, if the substrate concentration is increased to a very high level, the substrate molecules vastly outnumber the inhibitor molecules. This greatly increases the probability of a substrate molecule colliding with and binding to an active site rather than an inhibitor molecule. Consequently, at very high substrate concentrations, the effect of the competitive inhibitor is negligible, and the maximum rate of reaction ($V_{max}$) can still be reached.
Marking scheme
(a) [Max 2 marks] - Definition: minimum energy required to start a chemical reaction / transition state energy barrier. (1) - Role of enzyme: enzymes lower the activation energy / provide alternative pathway. (1)
(b) [Max 4 marks] - Use of filter/wavelength: use blue filter / wavelength of 405 nm. (1) - Standardization: calibrate with a blank / maintain constant temperature or enzyme volume. (1) - Data collection: record absorbance at regular time intervals (e.g., 10s / 15s). (1) - Determination of initial rate: plot absorbance against time and calculate the gradient of the initial linear region (tangent at t=0). (1)
(c) [Max 4 marks] - Inhibitor structure: competitive inhibitor has similar shape to substrate / is complementary to active site. (1) - Mode of action: binds to active site / blocks substrate entry / prevents enzyme-substrate (ES) complex formation. (1) - Effect of substrate concentration: increasing substrate concentration increases probability of substrate binding / outcompeting inhibitor. (1) - Vmax achievement: maximum velocity (Vmax) is unchanged / can still be achieved at very high substrate concentrations. (1)
Question 2 · Structured
10 marks
Eukaryotic and prokaryotic cells share some structural features but also possess distinct differences.
(a) Copy and complete the table below by placing a tick (✓) if the feature is present or a cross (✗) if the feature is absent.
(b) Explain how the structure of the inner membrane of a mitochondrion is adapted to its function. [3]
(c) Transmission electron microscopes (TEM) are used to study cell organelles. Explain the difference between magnification and resolution, and explain why a TEM has a much higher resolution than a light microscope. [4]
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Worked solution
(a) Table Completion: - Circular DNA: Present in Chloroplast (✓), Present in Mitochondrion (✓), Present in Typical Bacterial Cell (✓) [Both mitochondria and chloroplasts have circular DNA due to their endosymbiotic origin]. - 70S Ribosomes: Present in Chloroplast (✓), Present in Mitochondrion (✓), Present in Typical Bacterial Cell (✓) [Eukaryotic organelles have 70S ribosomes inside them]. - Peptidoglycan cell wall: Absent in Chloroplast (✗), Absent in Mitochondrion (✗), Present in Typical Bacterial Cell (✓).
(b) The inner membrane of a mitochondrion is adapted to its function of aerobic respiration in the following ways: 1. It is highly folded to form structures called cristae, which vastly increases the surface area of the membrane available for chemical reactions. 2. It contains a high density of electron transport chain proteins (cytochromes/carriers) embedded within the lipid bilayer. 3. It contains ATP synthase enzymes (stalked particles) which utilize the proton gradient across the inner membrane (from the intermembrane space to the matrix) to generate ATP during chemiosmosis.
(c) Differences between magnification and resolution, and TEM resolution: - Magnification is defined as the number of times larger an image is compared to the actual size of the object (calculated as Image size / Actual size). - Resolution is defined as the minimum distance between two objects at which they can still be distinguished as separate, distinct entities (i.e., the capacity to show fine detail). - The resolution of a microscope is limited by the wavelength of the radiation used to view the specimen. A TEM uses a beam of electrons instead of light waves. Electrons have a much shorter wavelength than visible light (approx. 100,000 times shorter), allowing the electron microscope to resolve structures that are much closer together (down to 0.5 nm, compared to the 200 nm limit of light microscopes).
Marking scheme
(a) [3 marks] - 1 mark for Circular DNA row completely correct (all three ticks). - 1 mark for 70S ribosomes row completely correct (all three ticks). - 1 mark for Peptidoglycan cell wall row completely correct (cross, cross, tick).
(b) [Max 3 marks] - Cristae / Folding: inner membrane is folded into cristae to provide a large surface area. (1) - Proteins / ETC: provides space for electron transport chain proteins / proton pumps. (1) - ATP Synthase: contains ATP synthase / stalked particles for ATP synthesis / oxidative phosphorylation. (1) - Barrier function: impermeable to protons to maintain a proton gradient. (1)
(c) [Max 4 marks] - Magnification definition: number of times larger the image is relative to the actual specimen size. (1) - Resolution definition: ability to distinguish between two close points as separate / detail shown. (1) - Wavelength concept: resolution depends on the wavelength of radiation used (shorter wavelength = higher resolution). (1) - TEM explanation: electrons have a much shorter wavelength than light, allowing higher resolution. (1)
Question 3 · Structured
10 marks
Plants must continuously absorb water from the soil to replace water lost by transpiration.
(a) Describe the pathways taken by water as it moves from the root hair cells, through the cortex, to the xylem of a root. In your answer, refer to the apoplastic and symplastic pathways. [4]
(b) Plants that live in dry environments (xerophytes) possess anatomical adaptations to reduce transpiration. Describe three anatomical adaptations of a xerophytic leaf, and explain how each adaptation reduces the rate of water loss. [6]
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Worked solution
(a) Water pathways through the root: 1. Water is absorbed from the soil into root hair cells by osmosis down a water potential gradient. 2. In the apoplast pathway, water moves through the non-living parts of the tissue—specifically through the cellulose cell walls and intercellular spaces. This pathway is fast and passive, driven by tension. 3. In the symplast pathway, water moves through the living contents of cells. It enters the cytoplasm of a cell via the plasma membrane and then passes from cell to cell through plasmodesmata (cytoplasmic connections between adjacent cells). 4. When water reaching the endodermis via the apoplast pathway encounters the Casparian strip (composed of impermeable suberin), the apoplast pathway is completely blocked. Water is forced to cross the selectively permeable cell membrane of the endodermal cells and enter the symplast pathway, allowing the plant to regulate the entry of solutes into the xylem.
(b) Xerophytic leaf adaptations and mechanisms: 1. **Thick waxy cuticle**: The cuticle on the epidermis is composed of hydrophobic waxy cutin. A thick cuticle increases the diffusion distance for water molecules and acts as a highly effective physical barrier, significantly reducing cuticular transpiration. 2. **Sunken stomata / stomata in pits**: Stomata are located in deep depressions (pits) rather than on the flat epidermal surface. This shelters the stomata from air currents and traps a microclimate of highly humid/moist air right outside the pore. This decreases the water vapour potential gradient between the intercellular spaces inside the leaf and the external air, reducing the rate of diffusion/transpiration. 3. **Rolled leaves (e.g., in Marram grass)**: The leaf rolls inwards, placing the stomata on the inner surface. This creates a confined, sheltered environment where water vapour accumulates and is trapped inside the roll. The humidity remains extremely high inside the roll, which decreases the water vapour potential gradient and greatly slows down transpiration.
Marking scheme
(a) [Max 4 marks] - Apoplast pathway: water moves through cell walls / intercellular spaces. (1) - Symplast pathway: water moves through cytoplasm / plasmodesmata. (1) - Endodermis / Casparian strip: Casparian strip is made of suberin / is waterproof. (1) - Function of strip: blocks the apoplast pathway, forcing water into the symplast / through cell membranes (allowing selective absorption of ions). (1)
(b) [Max 6 marks - 1 mark for adaptation, 1 mark for explanation for each of three adaptations] - Thick waxy cuticle (1) -> reduces non-stomatal water loss / increases diffusion distance / waterproof barrier. (1) - Sunken stomata / stomata in pits (1) -> traps a layer of moist air / decreases water vapour potential gradient. (1) - Rolled leaves (1) -> traps moist air inside / shelters stomata from wind / decreases water vapour potential gradient. (1) - Hairs on leaf surface / trichomes (1) -> traps moist air / reduces air flow near stomata / decreases water vapour potential gradient. (1) - Reduced leaves / spines (1) -> decreases surface area for transpiration. (1)
[Accept any three appropriate pairs up to 6 marks]
Question 4 · Structured
10 marks
The cell surface membrane acts as a barrier that controls the movement of substances into and out of cells.
(a) Describe the fluid mosaic model of membrane structure, explaining what is meant by the terms 'fluid' and 'mosaic'. [3]
(b) Explain how the properties of phospholipids cause them to form a bilayer in an aqueous environment. [3]
(c) Distinguish between the processes of facilitated diffusion and active transport. [4]
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Worked solution
(a) Under the fluid mosaic model: - **Fluid**: The phospholipid molecules can move laterally (from side to side) within their own monolayer, and some membrane proteins are also free to move within the lipid bilayer, giving the membrane flexibility and dynamic properties. - **Mosaic**: The membrane is composed of various components, mainly phospholipids and proteins, where protein molecules of different sizes and shapes are scattered/embedded within the phospholipid bilayer in an asymmetrical, mosaic-like pattern.
(b) Phospholipid molecules have an amphipathic nature: 1. They possess a polar, hydrophilic phosphate-containing head that is attracted to water molecules. 2. They possess two non-polar, hydrophobic fatty acid tails that are repelled by water. 3. In an aqueous environment (such as the extracellular fluid and intracellular cytoplasm), the phospholipids spontaneously arrange themselves so that their hydrophilic heads point outwards to face the water on both sides of the membrane, while the hydrophobic tails face inwards, pointing away from the water to form a hydrophobic core. This hydrophobic core is held together by weak hydrophobic interactions, stabilizing the bilayer structure.
(c) Distinctions between facilitated diffusion and active transport: 1. **Energy requirement**: Facilitated diffusion is a passive process that does not require metabolic energy (ATP), whereas active transport is an active process that requires energy input in the form of ATP. 2. **Concentration gradient**: Facilitated diffusion moves substances down a concentration gradient (from a region of higher concentration to a region of lower concentration), whereas active transport moves substances against a concentration gradient (from a region of lower concentration to a region of higher concentration). 3. **Proteins involved**: Facilitated diffusion can occur through either channel proteins (which can be gated) or carrier proteins. Active transport is highly specific and always involves carrier proteins that act as ATP-powered pumps (such as the sodium-potassium pump).
Marking scheme
(a) [Max 3 marks] - Fluid concept: phospholipid (and protein) molecules can move laterally within the monolayer. (1) - Mosaic concept: protein molecules are embedded / scattered throughout the bilayer in an irregular / mosaic pattern. (1) - Composition: mentions both phospholipids and proteins. (1)
(b) [Max 3 marks] - Amphipathic nature: heads are hydrophilic / polar, and tails are hydrophobic / non-polar. (1) - Alignment: heads face outwards towards aqueous environment (extracellular and cytoplasm). (1) - Core structure: tails face inwards / point towards each other, creating a hydrophobic core. (1)
(c) [Max 4 marks] - Energy difference: facilitated diffusion does not require ATP / is passive, active transport requires ATP / is active. (1) - Direction of movement: facilitated diffusion is down a concentration gradient, active transport is against a concentration gradient. (1) - Protein types: facilitated diffusion uses channel and carrier proteins; active transport uses carrier proteins / pumps only. (1) - Specificity/directionality: active transport is highly directional (unidirectional), facilitated diffusion can occur in both directions depending on the gradient. (1)
Question 5 · Structured
10 marks
The human gas exchange system is highly specialized to facilitate the rapid exchange of gases while protecting the body from damage.
(a) Describe the structure and function of cartilage in the trachea and bronchi. [3]
(b) Explain how goblet cells and ciliated epithelial cells work together to keep the airways clean and free from pathogens. [3]
(c) Describe the changes that occur in the thorax during expiration (breathing out) at rest. [4]
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Worked solution
(a) Cartilage is a tough, flexible connective tissue. In the trachea, it is structured as incomplete C-shaped rings, which prevent the trachea from collapsing when the pressure inside falls during inhalation, while allowing the esophagus behind it to expand during swallowing. In the bronchi, the cartilage is present as smaller, irregular plates that keep the airways open, maintaining a low-resistance pathway for airflow.
(b) Goblet cells and ciliated epithelial cells form a highly effective defense system: 1. Goblet cells synthesize and secrete sticky mucus onto the inner lining of the airways. This mucus is sticky and traps inhaled dust, pollen, and pathogens (such as bacteria and viruses). 2. Ciliated epithelial cells possess numerous hair-like projections called cilia on their apical membranes. These cilia beat in a coordinated, rhythmic, wave-like fashion to sweep the trapped-particle-containing mucus upwards, away from the alveoli and lungs towards the throat (pharynx), where it can be swallowed (and pathogens killed by stomach acid) or coughed out.
(c) During quiet, passive expiration at rest: 1. The external intercostal muscles relax, causing the ribcage to move downwards and inwards due to gravity. 2. The diaphragm muscle relaxes, allowing it to return to its dome-shaped position, pushing up into the thoracic cavity. 3. These relaxation movements, coupled with the natural elastic recoil of the lungs (alveolar elastic fibers), cause a decrease in the volume of the thoracic cavity. 4. According to Boyle's law, this decrease in volume increases the air pressure inside the lungs (alveolar pressure) above atmospheric pressure. 5. Air is consequently forced out of the lungs down a pressure gradient until the pressures equalize.
Marking scheme
(a) [Max 3 marks] - Structure: C-shaped rings in the trachea, irregular plates in the bronchi. (1) - Mechanical function: keeps the trachea/bronchi open / prevents collapse during inhalation / low-pressure. (1) - Swallowing link: C-shape allows the esophagus to expand during swallowing of food. (1)
(b) [Max 3 marks] - Goblet cell function: secrete / produce sticky mucus. (1) - Trapping: mucus traps dust, dirt, and pathogens. (1) - Ciliated cell function: cilia beat rhythmically to move / sweep the mucus away from the lungs / up to the pharynx/mouth. (1)
(c) [Max 4 marks] - Intercostals: external intercostal muscles relax. (1) - Ribcage movement: ribs move downwards and inwards. (1) - Diaphragm: diaphragm muscle relaxes and domes upwards. (1) - Volume/Pressure change: thoracic volume decreases, which increases pressure inside the lungs above atmospheric pressure, forcing air out. (1)
Question 6 · Structured
10 marks
The synthesis of proteins requires the precise transcription of DNA into mRNA followed by translation at the ribosomes.
(a) State three differences between the processes of transcription and DNA replication. [3]
(b) A portion of a DNA template strand has the base sequence: 3'- T A C G G T G A A A C T -5'.
(i) State the base sequence of the mRNA transcript produced from this template, indicating the 5' and 3' ends. [2]
(ii) A mutation occurs where the ninth base (A) in the DNA template strand is substituted with G. Explain why this mutation does not alter the amino acid sequence of the resulting polypeptide. [2]
(c) Describe the role of transfer RNA (tRNA) in the translation of mRNA. [3]
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Worked solution
(a) Three key differences between transcription and DNA replication are: 1. **Template length**: DNA replication copies the entire genome (all chromosomes) to prepare for cell division, whereas transcription only copies a specific section of DNA (a single gene or group of genes). 2. **Enzymes involved**: DNA replication requires DNA polymerase (along with helicase and ligase), while transcription requires RNA polymerase. 3. **Base pairing & nucleotides**: DNA replication uses deoxyribonucleotides (A, T, C, G) and pairs A with T; transcription uses ribonucleotides (A, U, C, G) and pairs A with U (Uracil). 4. **Product**: Replication produces a double-stranded DNA molecule, whereas transcription produces a single-stranded mRNA molecule.
(b) (i) The complementary base sequence of mRNA is transcribed in the 5' to 3' direction from the 3' to 5' DNA template. Therefore: - Template: 3'- T A C G G T G A A A C T -5' - mRNA: 5'- A U G C C A C U U U G A -3'
(ii) The original template triplet is G A A (bases 7, 8, 9), which transcribes to the mRNA codon C U U. This codon codes for the amino acid Leucine. When the ninth base (A) is substituted with G, the template triplet becomes G A G. This triplet transcribes to the mRNA codon C U C. Because the genetic code is **degenerate** (or redundant), multiple codons can code for the same amino acid; both C U U and C U C code for Leucine. Therefore, the same amino acid is incorporated into the polypeptide, and its primary structure is unaltered.
(c) The role of tRNA in translation: 1. tRNA acts as an adapter molecule that carries a specific amino acid to the ribosome. Each tRNA molecule is activated by binding to its corresponding amino acid at its 3' CCA terminal attachment site. 2. Each tRNA has a specific, exposed triplet of bases called an anticodon. During translation, the tRNA anticodon binds to its complementary codon on the mRNA molecule via hydrogen bonds. 3. This complementary base pairing (codon-anticodon interaction) ensures that amino acids are brought to the ribosome and aligned in the precise sequence specified by the genetic code on the mRNA, allowing peptide bonds to be formed between adjacent amino acids to form a polypeptide.
(b) [4 marks] (i) [2 marks] - Correct sequence: A U G C C A C U U U G A. (1) - Correct directionality: 5' at the left and 3' at the right. (1)
(ii) [2 marks] - Original vs Mutated codon: Original mRNA codon is C U U, mutated codon is C U C. (1) - Degeneracy of code: both codons code for the same amino acid / Leucine (due to the degenerate nature of the genetic code). (1)
(c) [Max 3 marks] - Specific amino acid transport: tRNA carries a specific amino acid to the ribosome. (1) - Anticodon-codon pairing: anticodon on tRNA binds to the complementary codon on mRNA by hydrogen bonding. (1) - Translation accuracy: ensures the correct sequence of amino acids in the growing polypeptide chain. (1)
Paper 3 Advanced Practical Skills
Conduct specified experimental work, present findings in tables/graphs, and perform microscope observations/drawings.
2 Question · 40 marks
Question 1 · Practical
20 marks
Question 1 [20 marks]
You are required to investigate the effect of different concentrations of copper sulfate solution, I, on the activity of the enzyme amylase. Copper sulfate is known to act as an inhibitor of amylase.
You are provided with the following materials:
E: 1.0% amylase solution
S: 1.0% starch solution
I: 1.0% copper sulfate solution
W: distilled water
Iodine in potassium iodide solution, in a dropper bottle
Spotting tiles, pipettes, test-tubes, and a stop-watch
(a) (i) Describe how you will perform a 2-fold serial dilution of the 1.0% copper sulfate solution, I, to produce four further concentrations of inhibitor: 0.5%, 0.25%, 0.125%, and 0.0625%, using distilled water, W. Your plan should specify the volumes of I and W required to make 10 cm3 of each concentration. [3]
(ii) Identify the independent variable and the dependent variable in this investigation. [2]
(iii) Carry out the investigation using the following instructions:
Label five test-tubes with the concentrations of copper sulfate solution prepared: 1.0%, 0.5%, 0.25%, 0.125%, and 0.0625%. Label a sixth tube '0.0%' (using distilled water only).
Add 2 cm3 of the corresponding inhibitor concentration (or distilled water for 0.0%) to each tube.
Add 2 cm3 of amylase solution, E, to each tube. Shake gently and allow the mixture to incubate at room temperature for 5 minutes to let the inhibitor interact with the enzyme.
During the incubation, place one drop of iodine solution into each of several wells on a spotting tile.
Add 2 cm3 of starch solution, S, to the first tube, mix thoroughly, and immediately start the stop-watch.
Every 30 seconds, remove a drop of the mixture and add it to a well containing iodine solution. Note the time when the starch is completely broken down (i.e. when the iodine solution remains yellow-brown instead of turning blue-black). If the reaction does not complete within 10 minutes, stop and record the time as "> 600 s".
Repeat steps 5 and 6 for each of the remaining mixtures.
Prepare a single table in which to record all your results. [5]
(iv) State the effect of copper sulfate concentration on the time taken for starch hydrolysis and explain this effect in terms of enzyme-inhibitor interactions. [3]
(v) Identify two potential sources of error in this investigation, and for each suggest an improvement to reduce its impact. [4]
(vi) Suggest how you would modify this investigation to find the minimum concentration of copper sulfate that completely inhibits amylase activity under these conditions. [3]
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Worked solution
(a) (i) Serial Dilution Steps: 1. Label four tubes: 0.5%, 0.25%, 0.125%, and 0.0625%. 2. Add 5 cm3 of distilled water (W) to each of these four tubes. 3. Take 5 cm3 of 1.0% copper sulfate (I) and transfer it to the 0.5% tube, mixing thoroughly (total volume = 10 cm3). 4. Take 5 cm3 of the 0.5% solution and transfer it to the 0.25% tube, mixing thoroughly. 5. Repeat this 2-fold transfer sequentially to the 0.125% and 0.0625% tubes. Discard 5 cm3 from the final 0.0625% tube to leave 5 cm3 in each tube.
(a) (ii) Variables: - Independent variable: Concentration of copper sulfate solution / %. - Dependent variable: Time taken for complete starch hydrolysis / s.
(a) (iii) Table of Results: A typical table should have headers with units, e.g., 'Concentration of copper sulfate solution / %' and 'Time taken for complete starch hydrolysis / s'. The data must show that as the concentration of copper sulfate increases, the time taken for starch hydrolysis increases (e.g., 0.0% takes 60 s, 0.0625% takes 120 s, 0.125% takes 240 s, 0.25% takes 450 s, and higher concentrations take >600 s).
(a) (iv) Conclusion and Explanation: As copper sulfate concentration increases, the time taken for complete hydrolysis of starch increases. This is because copper ions (Cu2+) act as enzyme inhibitors that bind to amylase, altering its tertiary structure and the shape of its active site. This prevents starch from binding, reducing the rate of enzyme-substrate complex formation.
(a) (v) Sources of Error and Improvements: 1. Error: Judging the endpoint of starch breakdown by eye is subjective and difficult. Improvement: Use a colorimeter to measure light absorbance at regular intervals to determine a precise, objective end-point. 2. Error: Temperature fluctuations during the experiment. Improvement: Keep all tubes in a thermostatically controlled water bath throughout the experiment.
(a) (vi) Modification: 1. Prepare a narrower, closer range of concentrations of copper sulfate (e.g., between 0.0% and 0.125% with increments of 0.02%). 2. Keep all other variables constant (temperature, starch concentration, enzyme concentration, pH). 3. Determine the lowest concentration in this range where blue-black color persists indefinitely (showing complete inhibition).
Marking scheme
(a) (i) [3 marks] - M1: Describes adding an equal volume (5 cm3) of distilled water to tubes. - M2: Describes sequential transfer of 5 cm3 of copper sulfate solution from one tube to the next (2-fold dilution). - M3: States that 10 cm3 is prepared at each step before transfer / 5 cm3 final volume remains.
(a) (ii) [2 marks] - M1: Independent variable: concentration of copper sulfate / inhibitor. - M2: Dependent variable: time taken for starch hydrolysis / time for iodine to remain yellow-brown.
(a) (iii) [5 marks] - M1: Table with clear columns, borders, and complete headings with units: 'Concentration of copper sulfate / %' and 'Time / s' (no units in the table body). - M2: Table contains all 6 specified concentrations (1.0, 0.5, 0.25, 0.125, 0.0625, 0.0). - M3: Correct trend shown (time increases as concentration increases). - M4: Times recorded consistently in seconds (e.g., multiples of 30 s). - M5: Space or column included for replicates or a highly detailed recording of colors over time.
(a) (iv) [3 marks] - M1: States that higher concentration of copper sulfate increases the time taken for starch hydrolysis (decreases rate). - M2: Explains that copper ions act as inhibitors and bind to amylase, changing the tertiary structure of the enzyme / active site. - M3: States that this prevents the formation of enzyme-substrate complexes.
(a) (v) [4 marks] - M1: Source of error 1: Endpoint is subjective / difficult to determine when yellow-brown is reached. - M2: Improvement 1: Use a colorimeter / color standard card. - M3: Source of error 2: Fluctuations in room temperature. - M4: Improvement 2: Carry out incubation and reaction in a thermostatically controlled water bath.
(a) (vi) [3 marks] - M1: Test a narrower/smaller range of copper sulfate concentrations. - M2: Choose concentrations below the lowest concentration that caused complete inhibition (e.g., between 0% and 0.125%). - M3: Keep all other environmental variables (temperature, pH, volumes) constant.
Question 2 · Practical
20 marks
Question 2 [20 marks]
You are provided with slide K1, which is a transverse section through the stem of a herbaceous dicotyledonous plant.
(a) (i) Draw a large, low-power plan diagram of a quarter sector of the stem shown on slide K1 to show the distribution of tissues. Your drawing should show the correct proportions of the different tissues. No individual cells should be drawn. Use a sharp pencil and make clear, continuous lines with no shading. [5]
(ii) Using the high-power lens of your microscope, locate a single vascular bundle on slide K1. Identify the large xylem vessel elements and the smaller, adjacent phloem sieve tube elements.
Make a high-power drawing of one xylem vessel element and two adjacent phloem sieve tube elements. The cell walls of all cells must be shown as double lines. Label your drawing to show a xylem vessel and a phloem sieve tube. [4]
(b) (i) You are required to estimate the actual diameter of a xylem vessel element on slide K1.
An eyepiece graticule has been calibrated under the high-power objective (×400) such that:
1 eyepiece graticule unit (epu) = 2.5 μm
The diameter of the chosen xylem vessel element spans exactly 18 epu under this magnification.
Calculate the actual diameter of the xylem vessel element in micrometres (μm). Show your working and state the answer to an appropriate number of significant figures. [2]
(ii)Figure 2.1 is a photomicrograph of a transverse section through the stem of a xerophytic plant of a different species.
[Figure 2.1: A transverse section of a xerophytic stem showing a thick cuticle, heavily lignified sub-epidermal sclerenchyma, sunken stomata, and vascular bundles arranged in a ring with large sclerenchyma caps.]
Prepare a table to compare the observable anatomical features of the stem on slide K1 with the stem in Figure 2.1. Your comparison should include both similarities and differences. [5]
(c) Some xerophytic plants have adaptations in their stem anatomy to survive in environments where water is scarce. Explain how two anatomical features visible in the stem in Figure 2.1 adapt the plant to dry environments. [4]
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Worked solution
(a) (i) Low-power Plan Diagram: - The drawing must show a quarter sector of a circle (representing the stem section). - Outer layer (epidermis) drawn with a single thin, clean line. - Vascular bundles should be drawn in a ring, showing correct relative proportions (occupying a middle layer between the cortex and pith). - Inside each bundle, tissues (xylem and phloem) should be delimited by light lines, but no individual cells must be drawn. - Central pith must be correctly proportioned relative to the cortex width.
(a) (ii) High-power Drawing: - Exactly one large xylem vessel element and two adjacent, smaller phloem sieve tube elements must be drawn. - All cells must have double-line walls to represent cell wall thickness. - The xylem vessel should be drawn with significantly thicker walls than the phloem sieve tubes. - Label lines must be straight, drawn with a ruler, and point directly to a 'xylem vessel' and a 'phloem sieve tube'.
(b) (i) Calculation: Actual diameter = number of epu × calibration value Actual diameter = 18 × 2.5 μm = 45 μm The final answer is 45 μm.
(b) (ii) Comparison Table: Features compared should include: 1. Cuticle: Slide K1 has a thin cuticle; Figure 2.1 has a thick cuticle. 2. Stomata: Slide K1 has stomata level with the epidermis (not sunken); Figure 2.1 has sunken stomata / stomata in pits. 3. Sclerenchyma/Support tissue: Slide K1 has little/no sub-epidermal sclerenchyma; Figure 2.1 has a prominent layer of sub-epidermal sclerenchyma and large caps over the vascular bundles. 4. Vascular bundle arrangement (Similarity): Both stems have vascular bundles arranged in a ring. 5. Pith (Similarity): Both stems possess a central region of parenchyma/pith.
(c) Adaptations: 1. Thick cuticle: Forms a waterproof barrier that reduces cuticular water loss (transpiration). 2. Sunken stomata: Traps a layer of moist air outside the stomatal pore, reducing the water vapour potential gradient between the inside of the stem and the atmosphere, thereby reducing transpiration rate. 3. Sub-epidermal sclerenchyma: Provides mechanical strength and structural support to keep the stem upright and prevent collapse when cells lose turgor pressure during drought stress.
Marking scheme
(a) (i) [5 marks] - M1: Clear, sharp, continuous lines drawn with a sharp pencil, with no shading. - M2: Drawing is large, occupying at least half of the available space. - M3: Correct quarter-sector shape representing the stem section. - M4: Correct proportion of tissues (e.g., width of cortex relative to vascular bundle layer and central pith). - M5: No individual cells drawn, and tissues in the vascular bundles are correctly delimited.
(a) (ii) [4 marks] - M1: Drawing shows exactly one large xylem vessel element and two smaller phloem sieve tube elements. - M2: Cell walls of all three cells drawn as double lines to represent thickness. - M3: Xylem vessel is drawn with thicker walls than the phloem cells. - M4: Labels 'xylem vessel' and 'phloem sieve tube' correctly point to the respective cells with straight ruler-drawn lines.
(b) (i) [2 marks] - M1: Shows correct working: 18 × 2.5. - M2: Correct final answer: 45 μm (with correct unit of micrometres).
(b) (ii) [5 marks] - M1: Table format with clear column and row headers. - M2: States at least one similarity (e.g., vascular bundles arranged in a ring / presence of central pith). - M3: Difference 1: Cuticle is thin in Slide K1 but thick in Figure 2.1. - M4: Difference 2: Stomata are superficial/not sunken in Slide K1 but sunken in pits in Figure 2.1. - M5: Difference 3: Sub-epidermal sclerenchyma is absent/reduced in Slide K1 but present as a thick, continuous layer in Figure 2.1.
(c) [4 marks] - M1: Identifies thick cuticle. - M2: Explains that thick cuticle reduces cuticular water loss / transpiration. - M3: Identifies sunken stomata. - M4: Explains that sunken stomata trap moist air, reducing the water vapour potential gradient and lowering transpiration. - (Accept: Identifies sub-epidermal sclerenchyma [1] and explains that it provides support to prevent stem collapse during wilting / turgor loss [1]).
Paper 4 A Level Structured
Answer all ten structured questions in the spaces provided on the exam paper.
10 Question · 100 marks
Question 1 · Structured
10 marks
(a) Describe how an increase in blood water potential is detected and how this leads to a decrease in the secretion of antidiuretic hormone (ADH) from the pituitary gland. [3 marks] (b) Explain the effect of a low concentration of ADH on the cells of the collecting duct in the kidneys. [4 marks] (c) Diabetes insipidus is a condition where either ADH is not produced (central) or the kidneys do not respond to ADH (nephrogenic). Explain how a mutation in the gene coding for the ADH receptor could lead to nephrogenic diabetes insipidus, and predict the effect on the volume and concentration of urine produced. [3 marks]
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Worked solution
(a) Osmoreceptors in the hypothalamus detect the increase in blood water potential. The osmoreceptors absorb water by osmosis and swell. This prevents nerve impulses (action potentials) from being sent along the axons of neurosecretory cells to the posterior pituitary gland, resulting in less ADH being released into the blood. (b) With low ADH, ADH does not bind to receptors on the cell surface membrane of collecting duct epithelial cells. This decreases the concentration of active cyclic AMP (second messenger) inside the cells. Vesicles containing aquaporins (water channel proteins) are not fused with the apical membrane (luminal membrane). Instead, aquaporins are removed from the membrane by endocytosis. The membrane remains impermeable to water, so water cannot move out of the collecting duct down the water potential gradient. (c) A mutation alters the DNA base sequence, resulting in a change in the primary structure of the ADH receptor protein, which alters its tertiary structure. The ADH receptor's active site / binding site changes shape so ADH can no longer bind. This prevents the intracellular signaling cascade (no activation of G-proteins/adenylyl cyclase). Consequently, a large volume of dilute urine is produced.
Marking scheme
(a) 1. Osmoreceptors in the hypothalamus detect the increase in blood water potential; 2. Osmoreceptors gain water by osmosis and swell / expand; 3. Fewer / no nerve impulses / action potentials are sent along neurosecretory cells; 4. To the posterior pituitary gland, reducing the release of ADH into the blood; [Max 3 marks]
(b) 1. Less ADH binds to receptors on the cell surface membrane of collecting duct cells; 2. Inactive / less G-protein activation, leading to decreased adenylyl cyclase activity / less cAMP produced; 3. Vesicles containing aquaporins / water channel proteins do not fuse / insert into the cell membrane / luminal membrane; 4. Aquaporins are retrieved from the membrane via endocytosis; 5. The luminal membrane of the collecting duct remains impermeable to water; 6. Less water is reabsorbed by osmosis into the tissue fluid / blood; [Max 4 marks]
(c) 1. Mutation changes the primary structure / amino acid sequence of the receptor; 2. Changes tertiary structure / 3D shape of the receptor, so ADH cannot bind / fit; 3. No intracellular signaling / no aquaporins inserted; 4. Resulting in a large volume of urine / dilute urine / low concentration urine; [Max 3 marks]
Question 2 · Structured
10 marks
(a) Distinguish between cyclic and non-cyclic photophosphorylation in the light-dependent stage of photosynthesis. [4 marks] (b) The herbicide DCMU blocks the flow of electrons from photosystem II (PSII) to the electron transport chain. (i) Explain why the addition of DCMU stops the production of oxygen. [2 marks] (ii) Predict and explain the effect of DCMU on the rate of the Calvin cycle (light-independent stage). [4 marks]
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Worked solution
(a) Cyclic photophosphorylation involves only Photosystem I (PSI), whereas non-cyclic photophosphorylation involves both Photosystem I (PSI) and Photosystem II (PSII). Cyclic photophosphorylation does not involve the photolysis of water, whereas non-cyclic photophosphorylation does. In cyclic, the excited electrons return to the same photosystem (PSI), whereas in non-cyclic, electrons from PSII replace those lost in PSI, and electrons from water photolysis replace those lost in PSII. Cyclic produces only ATP, whereas non-cyclic produces both ATP and reduced NADP. Oxygen is produced only in non-cyclic. (b) (i) Photolysis of water occurs at PSII to replace the electrons lost when chlorophyll a absorbs light. If electron flow from PSII is blocked, the oxidized chlorophyll a cannot accept further electrons, preventing photolysis of water, and therefore stopping the release of oxygen. (ii) If electron flow is blocked, non-cyclic photophosphorylation stops, so no reduced NADP and less ATP are produced. The Calvin cycle requires ATP and reduced NADP. Specifically, reduced NADP and ATP are needed to reduce glycerate 3-phosphate (GP) to triose phosphate (TP). ATP is also needed to regenerate ribulose bisphosphate (RuBP) from TP. Therefore, without these, the concentration of GP increases initially, TP decreases, and the rate of the Calvin cycle decreases or stops.
Marking scheme
(a) 1. Cyclic involves PSI only, non-cyclic involves PSI and PSII; 2. Cyclic does not involve photolysis of water, non-cyclic does; 3. Cyclic produces ATP only, non-cyclic produces ATP and reduced NADP; 4. Cyclic does not produce oxygen, non-cyclic produces oxygen; 5. In cyclic, electrons return to the reaction center / chlorophyll a, in non-cyclic, they are accepted by NADP; [Max 4 marks]
(b)(i) 1. Water photolysis is linked to PSII / occurs to replace electrons lost from PSII; 2. If electron flow is blocked, PSII remains oxidized / cannot receive electrons, so photolysis of water ceases / oxygen is not released; [Max 2 marks]
(b)(ii) 1. No reduced NADP and less ATP are produced from the light-dependent stage; 2. ATP and reduced NADP are required to reduce GP to TP; 3. ATP is required to regenerate RuBP from TP; 4. Lack of these products stops the Calvin cycle / prevents synthesis of organic molecules; [Max 4 marks]
Question 3 · Structured
10 marks
(a) Outline the role of NAD and FAD in the Krebs cycle. [3 marks] (b) In an investigation, the rate of respiration of yeast cells was measured using different respiratory substrates: glucose, maltose, and arachidonic acid (a highly saturated fatty acid). (i) State why the respiratory quotient (RQ) of arachidonic acid is lower than the RQ of glucose. [2 marks] (ii) Under anaerobic conditions, yeast can respire glucose but cannot respire fatty acids. Explain why anaerobic respiration in yeast cannot use fatty acids as a substrate. [3 marks] (iii) State the pathways and products of anaerobic respiration in yeast. [2 marks]
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Worked solution
(a) NAD and FAD act as coenzymes that accept hydrogen atoms (protons and electrons) during the oxidation reactions of the Krebs cycle. They become reduced to NADH and FADH2, which transfer these hydrogen atoms to the electron transport chain on the inner mitochondrial membrane (cristae) for ATP synthesis via oxidative phosphorylation. (b) (i) Lipids (like arachidonic acid) contain a higher proportion of hydrogen atoms and fewer oxygen atoms per molecule compared to carbohydrates (like glucose). Therefore, more oxygen is required to oxidize the hydrogen to water, resulting in a lower ratio of CO2 produced to O2 consumed. (ii) Fatty acids must be broken down by beta-oxidation in the mitochondrial matrix to produce acetyl CoA, which enters the Krebs cycle. The Krebs cycle and oxidative phosphorylation require oxygen as the terminal electron acceptor. Under anaerobic conditions, there is no oxygen, so the link reaction, Krebs cycle, and oxidative phosphorylation cease. Yeast can only perform glycolysis under anaerobic conditions, which uses glucose but not fatty acids. (iii) The pathway is glycolysis followed by ethanol fermentation (ethanal pathway). The products are ethanol, carbon dioxide, and ATP (net 2 molecules per glucose).
Marking scheme
(a) 1. Act as hydrogen carriers / coenzymes; 2. Become reduced (to reduced NAD / reduced FAD) by accepting protons and electrons from Krebs cycle intermediates; 3. Transport hydrogens / electrons to the electron transport chain / cristae; [Max 3 marks]
(b)(i) 1. Fatty acids have a higher ratio of hydrogen to oxygen / are more reduced than glucose; 2. More oxygen is required for their complete oxidation (per carbon atom) / more water is produced; [Max 2 marks]
(b)(ii) 1. Fatty acids enter respiration via link reaction / Krebs cycle / beta-oxidation; 2. These pathways only operate under aerobic conditions / require oxygen (indirectly) to regenerate oxidized NAD/FAD; 3. Under anaerobic conditions, only glycolysis can occur / only glucose can be metabolized; [Max 3 marks]
(b)(iii) 1. Glycolysis and ethanol pathway / fermentation; 2. Products: ethanol, carbon dioxide, and ATP; [Max 2 marks]
Question 4 · Structured
10 marks
(a) Explain how environmental factors act as stabilizing selection forces in a constant environment. [3 marks] (b) A population of land snails, *Cepaea nemoralis*, lives in a woodland habitat. These snails have shells that are either yellow, brown, or pink, and may have dark bands. Song thrushes (*Turdus philomelos*) are predators of these snails. They find them by sight and break their shells on stones ("anvils") to eat them. (i) Describe how natural selection operates on the snail population if the woodland floor becomes covered in dark brown leaf litter during autumn. [4 marks] (ii) State the type of selection that would occur if both the yellow and brown shell phenotypes are favored over the intermediate pink phenotype, and describe its effect on the genetic diversity of the snail population. [3 marks]
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Worked solution
(a) In a constant environment, the existing phenotype of a population is already well-adapted. Extreme phenotypes (at both ends of the distribution) are selected against because they are less fit or suffer higher mortality. Individuals with the intermediate / modal phenotype are selected for and are more likely to survive, reproduce, and pass on their alleles. This maintains the phenotypic characteristics and reduces phenotypic variation. (b) (i) Snails with brown shells have a selective advantage as they are better camouflaged against the dark brown leaf litter. Yellow and pink snails are more visible to the song thrushes. Therefore, song thrushes will preferentially prey on yellow and pink snails. Brown-shelled snails have a higher survival rate, are more likely to reproduce, and pass on the alleles for brown shells to their offspring. Over generations, the frequency of the allele for brown shells increases in the population. (ii) This is disruptive selection. Disruptive selection favors both extreme phenotypes (yellow and brown) and selects against the intermediate phenotype (pink). This leads to a bimodal distribution of phenotypes and maintains or increases genetic diversity / genetic polymorphism within the population.
Marking scheme
(a) 1. Extreme phenotypes are selected against / have lower survival; 2. Intermediate / mean / modal phenotype is selected for / has selective advantage; 3. Alleles for extreme phenotypes decrease in frequency; 4. Phenotypic variation is reduced / characteristics of the population remain stable; [Max 3 marks]
(b)(i) 1. Brown-shelled snails are camouflaged / yellow and pink snails are visible; 2. Song thrushes act as a selection pressure, preying more on yellow and pink snails; 3. Brown snails survive and reproduce (differential survival); 4. Alleles for brown shell color are passed on to offspring; 5. Frequency of the brown shell allele increases; [Max 4 marks]
(b)(ii) 1. Disruptive selection; 2. Selects against intermediate phenotype / favors both extremes; 3. Maintains or increases genetic diversity / polymorphism / results in two distinct sub-populations; [Max 3 marks]
Question 5 · Structured
10 marks
(a) In genetic engineering, vectors are used to transfer genes into host cells. Plasmids are commonly used as vectors. (i) State two features of a plasmid that make it suitable for use as a vector. [2 marks] (ii) Explain the role of restriction endonucleases and DNA ligase in the production of a recombinant plasmid. [4 marks] (b) Severe combined immunodeficiency (SCID) is a genetic disorder that can be treated using gene therapy. Explain why gene therapy using retroviruses to insert a functional gene into a patient's T-lymphocytes provides only a temporary cure, whereas using hematopoietic stem cells can provide a longer-lasting treatment. [4 marks]
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Worked solution
(a) (i) Plasmids have a small size, circular double-stranded DNA structure, an origin of replication (to replicate independently in host cells), unique restriction sites (polylinker), and marker genes (like antibiotic resistance genes) to identify transformed cells. (ii) Restriction endonucleases recognize specific palindromic target sequences and cut the sugar-phosphate backbone of DNA, producing either sticky or blunt ends. The same restriction enzyme is used to cut both the plasmid DNA and the gene of interest to ensure complementary sticky ends. DNA ligase then seals the nicks in the sugar-phosphate backbone by forming phosphodiester bonds between the complementary base-paired sticky ends of the gene and the plasmid. (b) T-lymphocytes have a limited lifespan and die after a certain period, so they must be replaced. When the genetically modified T-lymphocytes die, the functional gene is lost, and symptoms return. Hematopoietic stem cells are undifferentiated cells that can divide by mitosis to produce a continuous supply of functional T-lymphocytes. Because stem cells can self-renew indefinitely, the functional gene remains in the body long-term, providing a continuous supply of healthy T-lymphocytes.
Marking scheme
(a)(i) 1. Small size / circular DNA (easy to isolate / introduce into cells); 2. Multiple cloning sites / unique restriction sites; 3. Marker genes (e.g., antibiotic resistance) to identify transformed cells; 4. Origin of replication (allows independent replication); [Max 2 marks]
(a)(ii) 1. Restriction endonuclease cuts DNA at specific recognition / palindromic sites; 2. Produces sticky ends (or blunt ends); 3. Same restriction enzyme cuts both plasmid and donor DNA to produce complementary sticky ends; 4. DNA ligase joins plasmid and donor DNA; 5. By forming phosphodiester bonds between adjacent nucleotides; [Max 4 marks]
(b) 1. T-lymphocytes have a short/limited lifespan / are not self-renewing; 2. Genetically modified T-lymphocytes eventually die, so the treatment must be repeated / is temporary; 3. Hematopoietic stem cells can divide indefinitely / self-renew; 4. Stem cells differentiate to produce a continuous supply of functional T-lymphocytes containing the active gene; [Max 4 marks]
Question 6 · Structured
10 marks
(a) In sweet peas, flower color is controlled by two genes, \(A/a\) and \(B/b\), which are located on different chromosomes. A dominant allele of gene \(A\) is required for the synthesis of a colorless precursor. A dominant allele of gene \(B\) is required to convert this precursor into a purple pigment. If either dominant allele is absent, the flowers are white. (i) State the name given to this type of gene interaction. [1 mark] (ii) A cross was carried out between two double heterozygous purple-flowered plants (\(AaBb \times AaBb\)). Determine the expected phenotypic ratio of purple-flowered to white-flowered offspring. Show your working by drawing a genetic diagram or Punnett square. [5 marks] (b) Explain how a single base substitution mutation in the gene for an enzyme involved in pigment synthesis can result in a completely non-functional enzyme. [4 marks]
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Worked solution
(a) (i) Epistasis (specifically, duplicate recessive epistasis / complementary epistasis). (ii) Gametes of both parents: AB, Ab, aB, ab. The offspring genotypes from the Punnett square: - Purple flowers require at least one dominant allele at both loci (A_B_): - AABB (1), AABb (2), AaBB (2), AaBb (4) -> Total = 9. - White flowers occur when either gene has homozygous recessive alleles (A_bb, aaB_, or aabb): - AAbb (1), Aabb (2), aaBB (1), aaBb (2), aabb (1) -> Total = 7. Phenotypic ratio: 9 Purple : 7 White. (b) A base substitution mutation replaces one nucleotide with another. This can change a single codon in the mRNA. This may result in a different amino acid being incorporated into the polypeptide chain during translation (missense mutation). If this amino acid is located in the active site or is crucial for folding, it can disrupt ionic, hydrogen, or disulfide bonds, changing the tertiary structure of the enzyme. The shape of the active site is altered, meaning the substrate can no longer fit to form an enzyme-substrate complex. Alternatively, the mutation could introduce a premature stop codon (nonsense mutation), producing a truncated, non-functional polypeptide.
(a)(ii) 1. Parent genotypes: \(AaBb \times AaBb\); 2. Correct gametes: AB, Ab, aB, ab; 3. Correct Punnett square or list showing 16 possible combinations; 4. Correctly identifies purple genotypes (A_B_) and white genotypes (A_bb, aaB_, aabb); 5. Correct phenotypic ratio: 9 purple : 7 white; [Max 5 marks]
(b) 1. Base substitution changes one codon in the mRNA; 2. Changes the primary structure / amino acid sequence of the protein (missense mutation); 3. Alters hydrogen, ionic, or disulfide bonds; 4. Changes the tertiary structure / 3D shape of the enzyme; 5. Active site shape changes, so substrate is no longer complementary / cannot bind / no ESCs formed; 6. OR: introduction of a premature stop codon (nonsense mutation) leads to a truncated polypeptide; [Max 4 marks]
Question 7 · Structured
10 marks
(a) Describe how an action potential arriving at the presynaptic membrane of a cholinergic synapse leads to the release of acetylcholine. [4 marks] (b) Explain how acetylcholine stimulates the postsynaptic membrane, and outline how its action is rapidly terminated to prevent continuous stimulation. [4 marks] (c) Organophosphates are chemicals used as insecticides that inhibit the enzyme acetylcholinesterase. Predict and explain the effect of organophosphate exposure on the transmission of nerve impulses across a cholinergic synapse. [2 marks]
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Worked solution
(a) When an action potential arrives, it depolarizes the presynaptic membrane. This depolarization causes voltage-gated calcium ion channels in the presynaptic membrane to open. Calcium ions (\(Ca^{2+}\)) diffuse rapidly down their electrochemical gradient into the presynaptic neurone cytoplasm. The influx of calcium ions stimulates synaptic vesicles containing acetylcholine to move towards and fuse with the presynaptic membrane, releasing acetylcholine into the synaptic cleft by exocytosis. (b) Acetylcholine diffuses across the synaptic cleft and binds to specific receptor proteins on the ligand-gated sodium ion channels in the postsynaptic membrane. This binding causes the channels to open, allowing sodium ions (\(Na^+\)) to diffuse into the postsynaptic neurone, depolarizing the postsynaptic membrane and generating an action potential if the threshold potential is reached. To terminate stimulation, acetylcholinesterase hydrolyzes acetylcholine into acetate and choline, which dissociate from the receptors and close the sodium channels. Choline is reabsorbed into the presynaptic bulb. (c) If acetylcholinesterase is inhibited, acetylcholine is not broken down and remains bound to receptors in the postsynaptic membrane. This keeps ligand-gated sodium channels open, causing continuous influx of sodium ions and persistent depolarization, leading to continuous firing of action potentials / muscle spasms.
Marking scheme
(a) 1. Depolarization of presynaptic membrane; 2. Opens voltage-gated calcium channels; 3. Calcium ions (\(Ca^{2+}\)) diffuse into the presynaptic neurone / synaptic knob; 4. Causes synaptic vesicles containing acetylcholine to move to and fuse with the presynaptic membrane; 5. Acetylcholine is released into the synaptic cleft by exocytosis; [Max 4 marks]
(b) 1. Acetylcholine diffuses across the synaptic cleft; 2. Binds to receptors / ligand-gated sodium channels on the postsynaptic membrane; 3. Sodium channels open, sodium ions (\(Na^+\)) diffuse into the postsynaptic cytoplasm, causing depolarization / EPSP / action potential; 4. Acetylcholinesterase hydrolyzes acetylcholine into choline and acetate; 5. This prevents continuous depolarization / allows channels to close; [Max 4 marks]
(c) 1. Acetylcholine remains in the synaptic cleft / bound to receptors; 2. Continuous depolarization of the postsynaptic membrane / continuous action potentials; [Max 2 marks]
Question 8 · Structured
10 marks
(a) Captive breeding programmes are often used in conservation to prevent the extinction of endangered species. (i) Discuss the problems associated with captive breeding programmes. [4 marks] (ii) Explain the role of assisted reproduction techniques, such as artificial insemination (AI) and *in vitro* fertilization (IVF), in maintaining genetic diversity in captive populations. [3 marks] (b) National parks are areas of land protected by law to conserve ecosystems and biodiversity. Explain why conserving a whole ecosystem is more effective than focusing conservation efforts on a single endangered species. [3 marks]
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(a) (i) Captive breeding programmes face several challenges: captive populations are small, which can lead to inbreeding depression and a loss of genetic diversity. Animals may experience behavioral changes and lose natural behaviors, such as hunting or avoiding predators, making reintroduction difficult. There is an increased risk of disease transmission in confined conditions. Some species simply refuse to breed in captivity due to stress or lack of suitable environmental cues. (ii) Assisted reproduction techniques allow genetic material (semen/eggs) to be transported easily across the world without moving the physical animals, which is expensive and stressful. Artificial insemination (AI) allows sperm from a male in one zoo to fertilize a female in another, preventing inbreeding. In vitro fertilization (IVF) and embryo transfer can maximize the number of offspring produced by genetically valuable females, and frozen gametes/embryos preserve gene pools for the future. (b) Conserving an entire ecosystem is more effective because it protects all species (including producers, decomposers, and predators) and their complex food webs and interactions (e.g., pollination, seed dispersal). It preserves the abiotic conditions and habitats required for survival, which is essential because a single species cannot survive without its ecological niche and supporting habitat. It is also more cost-effective and prevents other non-target species in the area from becoming endangered.
Marking scheme
(a)(i) 1. Small gene pool leads to inbreeding depression / reduced genetic variation; 2. Behavioral changes / loss of natural skills (e.g., hunting, predator avoidance, social structures); 3. Difficulty in reintroducing animals to the wild / high mortality rate on release; 4. Increased susceptibility to diseases in closed environments; 5. Some species do not breed well in captivity (due to stress / lack of environmental cues); [Max 4 marks]
(a)(ii) 1. Allows transfer of genetic material (semen / eggs) between zoos without moving the animals; 2. Reduces inbreeding / allows controlled mating between genetically distinct individuals; 3. Frozen gametes / embryos can be stored long-term (gene banks) to preserve genetic variation; [Max 3 marks]
(b) 1. Protects whole food webs / trophic levels / ecological niches / interdependent species; 2. Preserves the physical habitat and abiotic factors necessary for survival; 3. Is more sustainable / cost-effective than protecting multiple individual species; 4. Prevents other species in the community from becoming endangered; [Max 3 marks]
Question 9 · Structured
10 marks
Sickle cell anemia is a genetic disorder caused by a mutation in the gene coding for the \(\beta\)-globin polypeptide of hemoglobin. In adults, fetal hemoglobin (\(\text{HbF}\), structure \(\alpha_2\gamma_2\)) is normally switched off, and adult hemoglobin (\(\text{HbA}\), structure \(\alpha_2\beta_2\)) is produced. The transcription factor BCL11A represses the transcription of the \(\gamma\)-globin gene.
A modern gene-editing approach uses the CRISPR-Cas9 system *ex vivo* to disrupt the erythroid-specific enhancer region of the *BCL11A* gene in hematopoietic stem cells (HSCs), which are then returned to the patient.
(a) Describe how the single guide RNA (sgRNA) directs the Cas9 endonuclease to the target DNA sequence in the *BCL11A* enhancer. [3]
(b) Explain how disrupting the *BCL11A* gene enhancer in hematopoietic stem cells leads to a reduction in the symptoms of sickle cell anemia. [4]
(c) State three advantages of using gene editing (such as CRISPR-Cas9) to treat genetic disorders compared to using traditional gene therapy with viral vectors. [3]
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Worked solution
(a) The single guide RNA (sgRNA) contains a sequence of approximately 20 nucleotides that is complementary to the target DNA sequence in the BCL11A enhancer. The sgRNA guides the Cas9 endonuclease to this specific locus, where it binds via complementary base-pairing. Cas9 also recognizes a protospacer adjacent motif (PAM) sequence on the non-target DNA strand to initiate binding and make a double-stranded break.
(b) In normal adult erythroid cells, the BCL11A protein is expressed and acts as a repressor, binding to the promoter of the \(\gamma\)-globin gene and silencing its transcription. By using CRISPR-Cas9 to disrupt the erythroid-specific enhancer of the *BCL11A* gene, the gene is no longer transcribed, meaning BCL11A protein is not produced. This removes the repression on the \(\gamma\)-globin gene, allowing its transcription and translation. The resulting \(\gamma\)-globin chains combine with \(\alpha\)-globin chains to form functional fetal hemoglobin (\(\text{HbF}\)). \(\text{HbF}\) prevents the polymerization of sickle-type hemoglobin (\(\text{HbS}\)), reducing red blood cell sickling, hemolysis, and vaso-occlusive crises.
(c) 1. Precision: Gene editing targets a specific sequence in the genome, whereas viral vectors insert therapeutic genes randomly. 2. Safety: It avoids the risk of insertional mutagenesis, which occurs when a viral vector inserts a gene into an active oncogene or tumor suppressor gene, potentially causing cancer. 3. Reduced Immune Response: It avoids the use of modified viral vectors, which can trigger severe immune reactions in the patient.
Marking scheme
(a) 1. sgRNA contains a sequence complementary to the target DNA in the *BCL11A* enhancer; [1] 2. sgRNA binds to the target DNA strand via complementary base-pairing; [1] 3. sgRNA guides / delivers the Cas9 endonuclease to the specific genomic locus; [1] 4. Reference to Cas9 recognizing the PAM (Protospacer Adjacent Motif) sequence; [1] [Max 3]
(b) 1. Disrupting the enhancer prevents transcription / expression of the *BCL11A* gene; [1] 2. BCL11A protein (repressor) is not produced / concentration decreases; [1] 3. Repression of the \(\gamma\)-globin gene is removed / transcription of \(\gamma\)-globin gene is activated; [1] 4. \(\gamma\)-globin polypeptides are synthesized and combine with \(\alpha\)-globin to form fetal hemoglobin / \(\text{HbF}\); [1] 5. Presence of \(\text{HbF}\) prevents sickling of red blood cells / prevents polymerisation of HbS / improves oxygen transport; [1] [Max 4]
(c) 1. Precision / high specificity (as it targets a specific, known site in the host genome); [1] 2. Lowers risk of insertional mutagenesis / activation of oncogenes / disruption of essential host genes; [1] 3. Avoids the use of viral vectors which can cause an adverse immune response / inflammatory response; [1] 4. No risk of the viral vector regaining virulence / replicating; [1] [Max 3]
Question 10 · Structured
10 marks
Brown adipose tissue (BAT) plays an important role in non-shivering thermogenesis in newborn mammals. Mitochondria in BAT contain a specialized protein called thermogenin (UCP1) in their inner membranes. Thermogenin acts as a proton channel, allowing protons (\(H^+\)) to diffuse from the intermembrane space into the mitochondrial matrix down their electrochemical gradient, bypassing ATP synthase.
(a) Explain the effect of thermogenin on: (i) the synthesis of ATP by oxidative phosphorylation. [2] (ii) the production of heat energy. [2]
(b) Triolein (\(C_{57}H_{104}O_6\)) is a major lipid substrate stored in brown adipose tissue and used for thermogenesis. (i) Complete the balanced chemical equation for the aerobic respiration of triolein: \[ C_{57}H_{104}O_6 + \text{ } O_2 \rightarrow 57 CO_2 + 52 H_2O \] Calculate the respiratory quotient (RQ) for triolein. Show your working and express your final answer to two decimal places. [3]
(ii) Explain why lipids, such as triolein, have a lower RQ value and a higher energy density than carbohydrates, such as glucose. [3]
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Worked solution
(a)(i) ATP synthesis is significantly reduced or stopped. This is because thermogenin provides an alternative pathway for protons (\(H^+\)) to return to the matrix, which dissipates the proton gradient (proton motive force). As a result, fewer protons flow through ATP synthase, meaning there is less energy available to phosphorylate ADP to ATP.
(a)(ii) Heat production increases. The potential energy stored within the proton gradient (derived from the electron transport chain) is dissipated as kinetic/thermal energy (heat) instead of being conserved as chemical energy in ATP.
(b)(i) To balance the equation for oxygen: On the product side, we have \((57 \times 2) + 52 = 114 + 52 = 166\) oxygen atoms. On the reactant side, triolein contains 6 oxygen atoms. Therefore, the remaining oxygen atoms must come from the molecular oxygen: \(166 - 6 = 160\) oxygen atoms, which corresponds to \(80\) molecules of \(O_2\). Equation: \(C_{57}H_{104}O_6 + 80 O_2 \rightarrow 57 CO_2 + 52 H_2O\). \(RQ = \frac{\text{volume/moles of } CO_2 \text{ produced}}{\text{volume/moles of } O_2 \text{ consumed}} = \frac{57}{80} = 0.7125 \approx 0.71\).
(b)(ii) Lipids have a much higher ratio of hydrogen-to-oxygen atoms in their molecules compared to carbohydrates (which have a 2:1 H:O ratio and already contain significant oxygen). This means lipids are more highly reduced. During respiration, more hydrogens are transferred via NAD and FAD to the electron transport chain, resulting in more oxidative phosphorylation and a higher yield of ATP per gram (higher energy density). Because there is less internal oxygen in lipids, a larger amount of external oxygen is needed relative to carbon dioxide produced to act as the final electron acceptor and form water, resulting in a lower ratio of \(CO_2\) produced to \(O_2\) consumed (lower RQ).
Marking scheme
(a)(i) 1. ATP synthesis is reduced / stopped; [1] 2. Proton gradient / proton motive force is dissipated / reduced (as protons bypass ATP synthase); [1]
(a)(ii) 1. Heat production is increased; [1] 2. Potential energy of the proton gradient / energy from the electron transport chain is released as thermal/heat energy (rather than being converted to chemical energy in ATP); [1]
(b)(i) 1. 80 (molecules of \(O_2\)); [1] 2. Working showing: \(RQ = \frac{\text{moles of } CO_2}{\text{moles of } O_2} = \frac{57}{80}\); [1] 3. RQ = 0.71; [1] (Accept 0.71 only; do not accept 0.7 or 0.713 without showing rounding to 2 d.p.)
(b)(ii) 1. Lipids have a higher proportion of hydrogen atoms (relative to oxygen / carbon) than carbohydrates / are more reduced; [1] 2. More hydrogen is released to NAD/FAD to be transferred to the electron transport chain / more oxidative phosphorylation occurs, yielding more ATP per unit mass (higher energy density); [1] 3. More external oxygen is required to act as the terminal electron acceptor (to combine with protons to form water) relative to the amount of \(CO_2\) produced, which decreases the RQ; [1] [Max 3]
Paper 5 Planning, Analysis & Evaluation
Analyze data, plan investigations, evaluate experimental setups, and interpret statistical parameters.
2 Question · 30 marks
Question 1 · Written Analytical
15 marks
A student investigated the effect of substrate concentration on the activity of the enzyme invertase (sucrase) to determine its Michaelis-Menten constant (Km). They were provided with a stock solution of 1.0 mol dm-3 sucrose, a 1% invertase solution, Benedict's reagent, and a colorimeter.
(a) Describe a detailed method the student could use to: - Prepare five different concentrations of sucrose solution ranging from 0.05 to 0.50 mol dm-3 by simple dilution. - Carry out the investigation to obtain the data needed to calculate the initial rate of reaction at each concentration. Ensure your method is detailed enough for another person to follow. [8]
(b) (i) Describe how the student would use a graph of initial rate of reaction against substrate concentration to determine the Km of invertase. [2] (ii) Explain why a double-reciprocal (Lineweaver-Burk) plot is more accurate than a direct plot for determining Vmax and Km. [1]
(c) Describe how the values of Vmax and Km would change if a competitive inhibitor were added to the reactions. [2]
(d) Identify two key variables (other than substrate concentration) that must be kept constant, and for each, describe how it is controlled and why its control is necessary. [2]
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Worked solution
(a) Dilutions: To make 20 cm3 of each sucrose concentration, mix the 1.0 mol dm-3 stock with distilled water: 0.50 mol dm-3 (10 cm3 stock + 10 cm3 water), 0.40 mol dm-3 (8 cm3 stock + 12 cm3 water), 0.30 mol dm-3 (6 cm3 stock + 14 cm3 water), 0.20 mol dm-3 (4 cm3 stock + 16 cm3 water), 0.10 mol dm-3 (2 cm3 stock + 18 cm3 water). Method: Set up a thermostatically controlled water bath at 37 degrees Celsius. Pre-equilibrate 5 cm3 of each sucrose dilution and 5 cm3 of the 1% invertase solution in separate tubes. Mix the enzyme and substrate, starting a timer immediately. At regular intervals (e.g., 30, 60, 90, 120, 150 seconds), withdraw a sample and add it to Benedict's reagent to stop the reaction. Heat with Benedict's, then use a colorimeter to measure absorbance. Convert absorbance to concentration of reducing sugars using a calibration curve. Plot reducing sugar concentration against time, and calculate the gradient of the initial straight-line portion to find the initial rate of reaction. Repeat each concentration 3 times to calculate a mean. (b) (i) From the curve, identify the maximum rate (Vmax) and divide by 2 to find half-Vmax. Find the corresponding substrate concentration on the x-axis, which represents Km. (ii) Vmax is an asymptote on a direct plot and is hard to identify precisely, whereas a double-reciprocal plot forms a straight line allowing accurate extrapolation of intercepts. (c) Vmax remains unchanged; Km increases. (d) Temperature: controlled using a water bath; ensures constant kinetic energy and collision rate. pH: controlled using a buffer solution; prevents alteration of the shape of the enzyme's active site.
Marking scheme
(a) [Max 8 marks] - M1: Dilution scheme: Correct volumes of stock and water calculated to make a range of 5 concentrations (e.g. 0.5, 0.4, 0.3, 0.2, 0.1 mol dm-3) with a constant total volume. - M2: Use of a thermostatically controlled water bath (e.g. 35 to 40 degrees Celsius). - M3: Pre-equilibration of enzyme and substrate in separate tubes before mixing. - M4: Standardised volumes of enzyme and substrate in all trials. - M5: Sampling at regular timed intervals (at least 3 points within the first few minutes) to find the initial rate. - M6: Stopping the reaction at each interval (e.g. by heating with Benedict's or adding acid/alkali). - M7: Measurement of reducing sugars produced using a colorimeter and converting to concentration (using a calibration curve). - M8: Plotting product concentration against time and measuring the gradient of the initial linear phase. - M9: Replicates (at least 3) for each concentration and calculating a mean. - M10: Safety precaution: safety goggles when handling hot water and Benedict's reagent.
(b) [Max 3 marks] - (i) Determine the maximum rate (Vmax) from the curve plateau, find half-Vmax, and read the corresponding substrate concentration on the x-axis [2] - (ii) Direct plot has an asymptotic Vmax which is hard to estimate, while double-reciprocal plot yields a straight line for easy extrapolation [1]
(d) [Max 2 marks] - Award 1 mark per variable with correct control and explanation (max 2): 1. pH: controlled with buffer; keeps ionic charges/active site shape stable. 2. Temperature: controlled with water bath; keeps kinetic energy/rate of collisions stable. 3. Enzyme concentration: controlled by using identical volume/concentration of invertase; keeps total active sites constant.
Question 2 · Written Analytical
15 marks
An ecologist investigated the species diversity of ground-dwelling beetles in two woodland habitats: an ancient deciduous woodland and a commercial pine plantation, using pitfall traps.
(a) Describe a method the ecologist could use to sample the beetle populations in both woodlands to obtain reliable data to calculate Simpson's Index of Diversity. [7]
(b) The ecologist calculated Simpson's Index of Diversity (D) using the formula: D = 1 - Σ(n/N)2, where n is the number of individuals of a particular species and N is the total number of individuals of all species. The results for the deciduous woodland are shown below: - Species A: n = 42 - Species B: n = 18 - Species C: n = 12 - Species D: n = 28 - Species E: n = 5 Calculate Simpson's Index of Diversity (D) for this woodland. Show your working and give your answer to two decimal places. [3]
(c) (i) In the pine plantation, the Simpson's Index of Diversity was 0.38. Compare the biodiversity of the two woodlands and suggest ecological reasons for the difference. [2] (ii) State what a Simpson's Index of Diversity of 0.38 indicates about the stability of the pine plantation ecosystem in response to environmental change. [1]
(d) The ecologist wanted to determine if there is a significant difference in the population density of Species A between the two woodlands. State the statistical test they should use, and explain why this test is appropriate. [2]
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Worked solution
(a) Sampling method: Divide each woodland area into a grid and use a random number generator to select coordinates for placing the traps. Set up at least 10 to 15 identical pitfall traps in each wood. Ensure the rim of each trap is flush with the soil surface and cover the top with a slightly raised tile to protect captures from predators and rain. Leave traps for exactly 24 hours. Identify species using a key, count all individuals, and safely release them. Replicate the sampling on multiple days to ensure reliability. (b) Working: Total N = 42 + 18 + 12 + 28 + 5 = 105. Calculate (n/N)2 for each species: Species A: (42/105)2 = 0.1600; Species B: (18/105)2 = 0.0294; Species C: (12/105)2 = 0.0131; Species D: (28/105)2 = 0.0711; Species E: (5/105)2 = 0.0023. Sum of (n/N)2 = 0.1600 + 0.0294 + 0.0131 + 0.0711 + 0.0023 = 0.2759. D = 1 - 0.2759 = 0.72 (to 2 decimal places). (c) (i) Deciduous woodland has higher biodiversity (0.72) than the pine plantation (0.38). This is because deciduous woodland has a variety of tree species, multi-layered vegetation (canopy, shrub, herb layers), and diverse leaf litter, offering more niches and food sources, while the pine plantation is a monoculture with heavy shade and acidic pine litter, supporting fewer niches. (ii) The pine plantation has low stability; it is highly vulnerable to disruption, as a single pest or disease outbreak could collapse the simple food web. (d) Student's t-test (two-sample unpaired t-test). It is appropriate because it compares the means of two independent groups (the two woodlands) to see if the difference between them is statistically significant, and the population density data is continuous/interval.
Marking scheme
(a) [Max 7 marks] - M1: Grid-based random sampling: using a coordinate system and a random number generator to place traps. - M2: Traps: at least 10-15 pitfall traps per woodland to ensure reliability. - M3: Standardisation of traps: identical trap size, depth, and cover (e.g., raised slate). - M4: Standardised time: traps left for a consistent duration (e.g., 24 hours). - M5: Identification: use of an identification key to categorize beetle species. - M6: Quantification: counting the number of individuals of each species. - M7: Ethical practice: safe release of beetles after counting. - M8: Replicates: sampling repeated on multiple days under similar conditions to calculate a mean.
(b) [Max 3 marks] - M1: Correct total N = 105 [1] - M2: Correct calculation of sum of squares of (n/N) = 0.276 (or 0.28) [1] - M3: Correct final answer of D = 0.72 (accept 0.72 - 0.73 depending on rounding steps) [1]
(c) [Max 3 marks] - (i) Deciduous woodland has higher species diversity than pine plantation, due to more niches / varied food sources / more microhabitats in deciduous vs a pine monoculture [2] - (ii) Ecosystem has low stability / high vulnerability to environmental change [1]
(d) [Max 2 marks] - Student's t-test (two-sample unpaired t-test) [1] - Appropriate because it compares the means of two separate, independent groups of continuous data [1]
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