In an experiment investigating cellular respiration, yeast cells were incubated with glucose solution. Under which of the following conditions would the production of carbon dioxide be the highest per unit mass of glucose consumed?
- A.Aerobic conditions
- B.Anaerobic conditions
- C.Aerobic conditions with the addition of a mitochondrial respiratory inhibitor
- D.Anaerobic conditions with the addition of a glycolytic inhibitor
Worked solution
In aerobic respiration, one molecule of glucose is completely oxidized to yield 6 molecules of carbon dioxide. In anaerobic respiration (alcoholic fermentation) of yeast, one molecule of glucose is only partially broken down to yield 2 molecules of carbon dioxide. Therefore, aerobic conditions yield the highest amount of carbon dioxide per unit mass of glucose consumed.
Marking scheme
Correct option is A (1 mark). Complete oxidation yields 6 CO2 molecules per glucose, whereas anaerobic respiration yields only 2 CO2 molecules.
A double-stranded DNA molecule contains 22% adenine. What is the percentage of cytosine in the mRNA transcribed from a template strand of this DNA?
- A.28%
- B.22%
- C.44%
- D.Cannot be determined from the given information
Worked solution
Although we can determine that the total percentage of guanine and cytosine (G + C) in the double-stranded DNA is 56% (since A + T = 44%), the distribution of bases on the individual template strand is highly variable and cannot be determined. Therefore, the exact percentage of cytosine in the mRNA transcribed from the template strand cannot be determined.
Marking scheme
Correct option is D (1 mark). Base composition of a single template strand cannot be deduced solely from the total base composition of double-stranded DNA.
Which of the following comparisons between the blood in the hepatic portal vein and the hepatic vein of a healthy human is correct several hours after a protein-rich meal?
- A.The concentration of amino acids is higher in the hepatic portal vein than in the hepatic vein.
- B.The concentration of urea is higher in the hepatic portal vein than in the hepatic vein.
- C.The concentration of oxygen is higher in the hepatic vein than in the hepatic portal vein.
- D.The concentration of carbon dioxide is lower in the hepatic vein than in the hepatic portal vein.
Worked solution
After a protein-rich meal, proteins are digested into amino acids and absorbed into the capillaries of the small intestine, which drain into the hepatic portal vein. Thus, the hepatic portal vein has a very high amino acid concentration. As blood passes through the liver, excess amino acids are deaminated or used for protein synthesis, so the concentration of amino acids in the hepatic vein is lower. The liver also produces urea from deamination, making the urea concentration higher in the hepatic vein.
Marking scheme
Correct option is A (1 mark). Digested amino acids enter the liver via the hepatic portal vein, where they are processed, resulting in a lower concentration in the hepatic vein leaving the liver.
Four similar leafy shoots from a terrestrial dicotyledonous plant are set up in identical potometers under the same environmental conditions. Each shoot receives one of the following treatments: Shoot P: No treatment; Shoot Q: Upper surface of all leaves coated with vaseline; Shoot R: Lower surface of all leaves coated with vaseline; Shoot S: Both surfaces of all leaves coated with vaseline. Which of the following shows the expected order of the rate of movement of the air bubble in the potometers, from fastest to slowest?
- A.P > Q > R > S
- B.P > R > Q > S
- C.Q > P > S > R
- D.R > S > P > Q
Worked solution
In terrestrial dicotyledonous plants, stomata are mainly located on the lower epidermis of the leaves. Therefore, sealing the lower surface (R) reduces the transpiration rate much more than sealing the upper surface (Q). Sealing both surfaces (S) almost completely stops transpiration, while the untreated control (P) has the highest transpiration rate. Thus, the expected order from fastest to slowest is P > Q > R > S.
Marking scheme
Correct option is A (1 mark). Knowing that stomatal density is higher on the lower epidermis of dicot leaves leads to the order of P (highest) > Q > R > S (lowest).
Red onion epidermal cells were placed in a 10% sucrose solution. After 15 minutes, the cells were observed under a light microscope. What would be observed, and what substance(s) would occupy the space between the cell wall and the shrunken cell membrane?
- A.Turgid cells; water
- B.Turgid cells; 10% sucrose solution
- C.Plasmolyzed cells; 10% sucrose solution
- D.Plasmolyzed cells; air
Worked solution
Since the 10% sucrose solution is hypertonic to the cell sap of the red onion cells, water leaves the vacuole by osmosis, causing the protoplast to shrink away from the cell wall (plasmolysis). Because the cell wall is fully permeable to sucrose molecules while the cell membrane is selectively impermeable to sucrose, the 10% sucrose solution passes through the cell wall and occupies the space between the cell wall and the shrunken cell membrane.
Marking scheme
Correct option is C (1 mark). Plasmolysis occurs, and the cell wall's full permeability allows the external sucrose solution to occupy the gap.
During the process of eutrophication in a freshwater lake, which of the following is the primary cause of the rapid depletion of dissolved oxygen?
- A.The rapid consumption of oxygen by the proliferating algae during daytime photosynthesis.
- B.The decrease in total oxygen production due to the death of submerged plants.
- C.The rapid multiplication of aerobic decomposers breaking down dead organic matter.
- D.The barrier formed by the dense algal bloom preventing oxygen from dissolving into the water.
Worked solution
When the algal bloom dies, a large amount of organic matter becomes available. Aerobic bacteria (decomposers) multiply rapidly and consume vast amounts of dissolved oxygen during cellular respiration as they decompose the dead algae. This rapid microbial respiration is the primary cause of oxygen depletion.
Marking scheme
Correct option is C (1 mark). Decomposition of dead algae by aerobic bacteria consumes the bulk of dissolved oxygen.
A student accidentally steps on a sharp pin, triggering a withdrawal reflex. If the dorsal root of the spinal nerve connected to the injured leg is completely severed, which of the following would be the outcome?
- A.The student feels the pain but cannot withdraw the leg.
- B.The student cannot feel the pain and cannot withdraw the leg.
- C.The student cannot feel the pain but can still withdraw the leg automatically.
- D.The student feels the pain and withdraws the leg normally.
Worked solution
The dorsal root of the spinal nerve contains sensory neurons. If the dorsal root is completely severed, sensory impulses from the receptors in the skin cannot enter the spinal cord. Consequently, no impulse can reach the brain to produce the sensation of pain, and no impulse can pass through the reflex arc to the motor neurons to trigger the withdrawal of the leg.
Marking scheme
Correct option is B (1 mark). Severing the dorsal root blocks all incoming sensory input, eliminating both sensation and the reflex response.
An aquatic food chain is represented below: Phytoplankton -> Zooplankton -> Small fish -> Osprey. If a non-biodegradable toxic heavy metal is released into the water body, which of the following statements about biomagnification is correct?
- A.The concentration of the toxic metal is highest in phytoplankton because they absorb it directly from water.
- B.The concentration of the toxic metal is highest in the osprey because it occupies the highest trophic level.
- C.The total amount of toxic metal in the osprey population is greater than that in the phytoplankton population.
- D.Organisms at higher trophic levels accumulate less toxic metal because energy is lost along the food chain.
Worked solution
Since the toxic heavy metal is non-biodegradable and cannot be excreted, it accumulates in the tissues of organisms. As energy is lost at each trophic level, consumers must ingest a large biomass of organisms from lower trophic levels. This leads to the biomagnification of the toxin, resulting in the highest concentration in the top predator, the osprey.
Marking scheme
Correct option is B (1 mark). Biomagnification results in the highest concentration of non-biodegradable toxins at the highest trophic level.
Which of the following correctly compares the primary and secondary immune responses in humans upon exposure to the same pathogen?
- A.Lag phase: Primary is Shorter; Secondary is Longer
- B.Cells involved in initial recognition: Primary is Memory cells; Secondary is Naive B cells
- C.Peak antibody concentration: Primary is Lower; Secondary is Higher
- D.Rate of antibody production: Primary is Faster; Secondary is Slower
Worked solution
In a primary immune response, naive B cells must be activated, clonal selection takes time, leading to a longer lag phase, a slower rate of antibody production, and a lower peak antibody concentration. In a secondary immune response, memory cells quickly recognize the antigen, resulting in a shorter lag phase, faster antibody production, and a much higher peak antibody concentration.
Marking scheme
Correct option is C (1 mark). Secondary immune response is faster and produces a much higher concentration of antibodies due to the presence of memory cells.
A woman with blood group A and a man with blood group B have a child with blood group O. What is the probability that their next child will be a boy with blood group AB?
- A.12.5%
- B.25%
- C.50%
- D.0%
Worked solution
Since the couple has an O-group child (\(ii\)), the parents must be heterozygous: the woman is \(I^A i\) and the man is \(I^B i\). The probability of their child having blood group AB (\(I^A I^B\)) is \(1/4\) (or 25%). The probability of having a boy is \(1/2\) (or 50%). Since these two events are independent, the overall probability of having a boy with blood group AB is \(1/4 \times 1/2 = 1/8\), which is 12.5%.
Marking scheme
Correct option is A (1 mark). Determination of parental genotypes as \(I^A i\) and \(I^B i\) gives a 1/4 chance for AB group, multiplied by 1/2 chance of being a boy, yielding 12.5%.
A leafy shoot is mounted in a potometer. Under a stable environment, the weight of the setup decreases by 4.2 g over 2 hours. At the same time, the volume of water absorbed by the shoot from the capillary tube is 4.5 cm³. (Assume 1 cm³ of water weighs 1 g). Which of the following can explain why the volume of water absorbed is greater than the weight loss?\n\n(1) Some water is retained in the cells to maintain turgidity.\n(2) Some water is consumed in photosynthesis.\n(3) Some water is produced in respiration.
- A.(1) and (2) only
- B.(1) and (3) only
- C.(2) and (3) only
- D.(1), (2) and (3)
Worked solution
The weight loss of the entire setup represents the amount of water lost by transpiration (4.2 g). The water absorbed by the shoot is 4.5 g. The difference (0.3 g) represents water that remains inside the plant. This retained water can be used to maintain cell turgidity (1) and as a reactant in photosynthesis (2). Water produced in respiration (3) is an internal product and would reduce the net requirement for external water absorption, so it does not explain why the absorbed water is greater than the transpired water.
Marking scheme
Award 1 mark for selecting A. Explain that water absorbed = water transpired + water retained (for turgidity and photosynthesis).
The graph of the net \(\text{CO}_2\) uptake of a plant leaf at different light intensities at \(25\text{ }^\circ\text{C}\) starts at a negative \(\text{CO}_2\) uptake at zero light intensity, crosses the x-axis (compensation point) at light intensity X, and plateaus at high light intensity. If a chemical that specifically inhibits the electron transport chain of aerobic respiration is applied to the leaf, which of the following changes would be observed?
- A.The y-intercept (net \(\text{CO}_2\) uptake at zero light intensity) moves closer to zero.
- B.The light compensation point (X) shifts to a higher light intensity.
- C.The maximum rate of net \(\text{CO}_2\) uptake at high light intensity decreases significantly.
- D.The rate of photosynthesis at light intensity X becomes zero.
Worked solution
At zero light intensity, there is no photosynthesis, so the net \(\text{CO}_2\) release (negative uptake) is entirely due to respiration. Inhibiting aerobic respiration reduces the rate of respiration, making the rate of \(\text{CO}_2\) release smaller (i.e., closer to zero). Thus, the y-intercept moves upwards closer to zero. This also means less photosynthesis is needed to balance respiration, so the light compensation point (X) shifts to the left (lower light intensity), not the right.
Marking scheme
Award 1 mark for A. Correctly identify that inhibiting respiration reduces carbon dioxide release in the dark, shifting the y-intercept towards zero.
An experiment was conducted to study the absorption of potassium ions (\(\text{K}^+\)) by plant roots. The rate of \(\text{K}^+\) uptake by the roots was measured under two different conditions: aeration with air and aeration with nitrogen gas. The results are shown in the table below:\n\n| Aeration gas | Rate of \(\text{K}^+\) uptake (arbitrary units) |\n|---|---|\n| Air | 45 |\n| Nitrogen gas | 8 |\n\nWhich of the following can be concluded from these results?
- A.The uptake of \(\text{K}^+\) ions is mainly an active process that relies on aerobic respiration.
- B.In the absence of oxygen, root cells completely lose their ability to absorb \(\text{K}^+\) ions.
- C.Nitrogen gas acts as a competitive inhibitor of the carrier proteins for \(\text{K}^+\) transport.
- D.Passive transport of \(\text{K}^+\) ions requires ATP produced from anaerobic respiration.
Worked solution
Aeration with air provides oxygen for aerobic respiration, which generates abundant ATP for active transport, resulting in a high uptake rate (45 units). When air is replaced by nitrogen gas (depriving the roots of oxygen), the uptake rate drops significantly to 8 units. This indicates that the majority of \(\text{K}^+\) uptake is an active process dependent on aerobic respiration. However, the uptake is not zero, suggesting either passive diffusion or active transport fueled by anaerobic respiration still occurs, ruling out B and C. Passive transport does not require ATP, ruling out D.
Marking scheme
Award 1 mark for A. Candidates should recognize that a major decrease in active ion uptake under anaerobic conditions indicates dependence on aerobic respiration.
Which of the following substances absorbed from the small intestine are transported to the liver via the hepatic portal vein before entering the general circulation?\n\n(1) Glucose\n(2) Amino acids\n(3) Fatty acids\n(4) Vitamin A
- A.(1) and (2) only
- B.(1) and (3) only
- C.(2) and (4) only
- D.(3) and (4) only
Worked solution
Glucose (1) and amino acids (2) are water-soluble nutrients. They are absorbed into the blood capillaries of the villi and transported to the liver via the hepatic portal vein. Fatty acids (3) and Vitamin A (4, a fat-soluble vitamin) are absorbed into the lacteals of the villi and transported via the lymphatic system, bypassing the liver and entering the venous system near the heart directly.
Marking scheme
Award 1 mark for selecting A. Correctly distinguish between water-soluble substances (enter blood system and go to liver) and fat-soluble substances (enter lymphatic system).
A certain drug X is known to inhibit the action of acetylcholinesterase, an enzyme that breaks down the neurotransmitter acetylcholine at the neuromuscular junction. Which of the following would be the most likely effect of drug X on skeletal muscles?
- A.Muscle cells cannot be stimulated, leading to flaccid paralysis.
- B.Continuous stimulation of muscle cells, leading to repeated contractions or spasms.
- C.Failure of action potential propagation along the motor neurone.
- D.Rapid depletion of neurotransmitters in the synaptic vesicles of the motor neurone.
Worked solution
Acetylcholinesterase breaks down acetylcholine in the synaptic cleft to terminate the stimulus. Inhibiting this enzyme means acetylcholine remains in the cleft, continuously binding to receptors on the muscle cell membrane. This results in continuous stimulation, causing repeated muscle contractions or spasms. It does not block stimulation (A), nor does it affect action potential propagation (C) or vesicle storage (D).
Marking scheme
Award 1 mark for B. Correctly identify the physiological consequence of preventing neurotransmitter breakdown.
The pedigree below shows the inheritance of a genetic disorder in a family:\n\n* Parents 1 (normal male) and 2 (normal female) have three children: 3 (affected male), 4 (normal female), and 5 (normal male).\n* Individual 4 marries individual 6 (affected male), and they have an affected daughter (7).\n\nBased on this pedigree, which of the following statements is/are correct?\n\n(1) The disorder is caused by a recessive allele.\n(2) The gene for this disorder must be located on an autosome.\n(3) The genotype of individual 4 must be heterozygous.
- A.(1) and (2) only
- B.(1) and (3) only
- C.(2) and (3) only
- D.(1), (2) and (3)
Worked solution
1. Since unaffected parents 1 and 2 produced an affected child (3), the disease allele must be recessive. Statement (1) is correct.\n2. Both autosomal recessive (parents: Dd x Dd -> 3: dd; 4: Dd; 6: dd -> 7: dd) and X-linked recessive (parents: X^D Y x X^D X^d -> 3: X^d Y; 4: X^D X^d; 6: X^d Y -> 7: X^d X^d) inheritances are fully compatible with the pedigree. Thus, the gene does not have to be autosomal; it can be X-linked. Statement (2) is incorrect.\n3. For daughter 7 to be affected (dd or X^d X^d), she must receive one disease allele from her father (6) and one from her mother (4). Since mother 4 is phenotypically normal, she must carry one normal allele and one disease allele, meaning she must be heterozygous (Dd or X^D X^d). Statement (3) is correct.
Marking scheme
Award 1 mark for B. Correctly identify the recessive nature of the inheritance and show that individual 4 must be a carrier under both autosomal and sex-linked modes, while recognizing that the mode cannot be definitively determined between autosomal and sex-linked.
The table below shows some mRNA codons and the amino acids they code for:\n\n| Codon | Amino acid |\n|---|---|\n| GAC | Aspartic acid |\n| CUG | Leucine |\n| CAG | Glutamine |\n| GUC | Valine |\n\nA segment of a template strand of DNA has the sequence `3'-CAG-GAC-GTC-5'`. What is the sequence of the polypeptide synthesized from this template?
- A.Valine - Leucine - Glutamine
- B.Glutamine - Aspartic acid - Valine
- C.Aspartic acid - Leucine - Glutamine
- D.Leucine - Valine - Aspartic acid
Worked solution
1. Transcription produces an mRNA strand complementary and antiparallel to the DNA template strand.\n2. DNA template: `3'-CAG-GAC-GTC-5'`\n3. Complementary mRNA (5' to 3'): `5'-GUC-CUG-CAG-3'`\n4. Read the codons from 5' to 3':\n * First codon: `GUC` -> Valine\n * Second codon: `CUG` -> Leucine\n * Third codon: `CAG` -> Glutamine\nTherefore, the polypeptide sequence is Valine - Leucine - Glutamine.
Marking scheme
Award 1 mark for A. Must correctly perform complementary transcription with correct directionality (antiparallel) and translate using the codon table.
Which of the following descriptions correctly compares primary succession and secondary succession?
- A.Primary succession starts in an area where soil is already present, whereas secondary succession starts on bare rock.
- B.The pioneer community of primary succession usually consists of woody shrubs, whereas that of secondary succession consists of lichens.
- C.Secondary succession generally reaches a climax community faster than primary succession because soil and seeds are already present.
- D.Biodiversity decreases during primary succession but increases during secondary succession.
Worked solution
Primary succession occurs on newly formed or exposed land with no soil (e.g., bare rock), while secondary succession occurs in areas where an existing community has been cleared (e.g., by fire or farming) but soil remains intact. Because soil, nutrients, and seeds/spores are already present in secondary succession, it reaches the climax community much faster than primary succession. In both successions, biodiversity increases over time.
Marking scheme
Award 1 mark for C. Recognize the role of pre-existing soil and seed banks in speeding up secondary succession.
The concentration of dissolved oxygen (DO) and biochemical oxygen demand (BOD) were measured in a river downstream from a discharge point of untreated domestic sewage. Which of the following statements is correct regarding the river section immediately downstream from the discharge point?
- A.The rapid decrease in DO is mainly due to the intensive respiration of aquatic plants.
- B.The high BOD is caused by a large amount of inorganic nutrients in the sewage.
- C.The population of aerobic decomposers increases rapidly, leading to the drop in DO.
- D.Fish populations increase because of the abundance of organic matter in the sewage.
Worked solution
Untreated domestic sewage contains abundant organic matter. This provides a food source for aerobic bacteria (decomposers), causing their population to increase rapidly. Their intensive aerobic respiration consumes dissolved oxygen, leading to a sharp decline in DO immediately downstream. BOD is high because of the high concentration of organic matter, not inorganic nutrients. Fish populations decrease due to suffocation from lack of oxygen.
Marking scheme
Award 1 mark for C. Correctly identify the link between organic matter input, decomposer population growth, aerobic respiration, and DO drop.
A person received two injections of the same vaccine. The concentration of antibodies in the blood was measured over time. Which of the following cellular events explains why the secondary immune response (after the second injection) is much faster and stronger than the primary response?
- A.During the second injection, phagocytes recognize the antigen faster and produce more antibodies.
- B.Memory B cells produced after the first injection rapidly proliferate and differentiate into plasma cells upon the second exposure.
- C.T-killer cells directly transform into antibody-producing plasma cells during the secondary immune response.
- D.The antigen in the second injection stimulates helper T cells to secrete antibodies directly.
Worked solution
During the primary immune response (first injection), some B cells differentiate into memory B cells. Upon the second exposure to the same antigen (second injection), these memory B cells recognize the antigen quickly, rapidly proliferate, and differentiate into plasma cells, which produce massive amounts of antibodies in a short time. Phagocytes do not make antibodies (A). T-killer cells and helper T cells are involved in cell-mediated immunity or activation, but they do not make antibodies (C, D).
Marking scheme
Award 1 mark for B. Correctly identify the role of memory B cells in rapid differentiation into plasma cells during the secondary response.
Question 21 · multiple_choice
1 marksIn an investigation of photosynthesis, DCPIP (a blue dye) is used to detect the light-dependent reactions. Which of the following setups will show the fastest rate of decolourisation of DCPIP?
- A.Isolated chloroplasts suspended in an isotonic sucrose solution under blue light
- B.Isolated chloroplasts suspended in an isotonic sucrose solution under green light
- C.Boiled isolated chloroplasts suspended in an isotonic sucrose solution under red light
- D.Isolated chloroplasts suspended in distilled water in the dark
Worked solution
DCPIP is a blue dye that acts as an electron acceptor. During the light-dependent reactions of photosynthesis, photolysis of water releases electrons, which reduce DCPIP and cause it to turn from blue to colourless. Isotonic sucrose solution maintains the integrity of the chloroplasts. Chlorophyll absorbs blue light most efficiently compared to green light, resulting in the highest rate of light reaction and thus the fastest decolourisation. Boiled chloroplasts have denatured proteins and destroyed thylakoid membranes, preventing the light reaction. No light reaction occurs in the dark.
Marking scheme
Award 1 mark for selecting option A. Candidates must identify that intact chloroplasts in isotonic solution under highly absorbed light (blue light) will perform light reactions most rapidly to reduce DCPIP.
Question 22 · multiple_choice
1 marksA student uses a potometer to measure the rate of transpiration of a leafy shoot. Which of the following changes will decrease the rate of movement of the air bubble in the potometer?\n\nI. Removing half of the leaves from the shoot.\nII. Increasing the relative humidity of the surrounding air.\nIII. Applying vaseline to the lower surface of all leaves.
- A.I and II only
- B.I and III only
- C.II and III only
- D.I, II and III
Worked solution
The movement of the air bubble in a potometer represents the rate of water absorption, which closely approximates the rate of transpiration. \n- I: Removing leaves reduces the total surface area for transpiration, thus decreasing the transpiration rate.\n- II: Increasing relative humidity decreases the water vapor concentration gradient between the air spaces inside the leaves and the external atmosphere, decreasing the transpiration rate.\n- III: In most land plants (especially dicots), stomata are primarily located on the lower epidermis. Applying vaseline to the lower surface blocks the stomata, significantly reducing water loss via transpiration.\nTherefore, all three changes will decrease the rate of bubble movement.
Marking scheme
Award 1 mark for option D. Students need to understand how leaf surface area, humidity, and stomatal blockage affect transpiration rate and thus water uptake measured by the potometer.
Question 23 · multiple_choice
1 marksPotato cylinders of equal mass are placed in sucrose solutions of different concentrations. The percentage change in mass of the potato cylinders is plotted against the sucrose concentration. The curve crosses the x-axis (0% mass change) at 0.3 M. If potato cylinders are placed in a 0.1 M sucrose solution, which of the following statements is correct?
- A.The water potential of the potato cells is lower than that of the 0.1 M sucrose solution.
- B.There is a net movement of water out of the potato cells.
- C.The potato cells will absorb water, expand, and eventually burst.
- D.The solute potential of the 0.1 M sucrose solution is lower than that of the potato cells.
Worked solution
Since the mass change is 0% at 0.3 M, the water potential of the potato cells is equal to the water potential of a 0.3 M sucrose solution. A 0.1 M sucrose solution has a higher water potential (less negative) than 0.3 M sucrose solution, and thus has a higher water potential than the potato cells. Therefore, the water potential of the potato cells is lower than that of the 0.1 M solution (Option A is correct). Consequently, water will move net into the potato cells by osmosis (making Option B incorrect). Since plant cells have rigid cell walls, they will become turgid but will not burst (making Option C incorrect). Solute potential is inversely related to solute concentration; thus, the 0.1 M solution (lower solute concentration) has a higher solute potential than the potato cells (making Option D incorrect).
Marking scheme
Award 1 mark for option A. Candidates must understand the relationship between solute concentration, water potential, and osmotic movement of water in plant cells.
Question 24 · multiple_choice
1 marksA student accidentally touches a hot object and withdraws their hand. If the sensory neuron in the dorsal root supplying the hand is completely severed, what will happen when the student's hand touches the hot object again?
- A.They can feel the pain but cannot withdraw their hand.
- B.They can withdraw their hand but cannot feel the pain.
- C.They can neither feel the pain nor withdraw their hand.
- D.They can feel the pain and withdraw their hand normally.
Worked solution
The sensory neuron in the dorsal root is responsible for transmitting nerve impulses from receptors (detecting heat/pain) to the spinal cord. If the sensory neuron is severed, nerve impulses cannot enter the spinal cord. As a result, impulses cannot be transmitted to the relay neuron and motor neuron to trigger the reflex action (no withdrawal). Additionally, impulses cannot be transmitted up the spinal cord to the brain, meaning the person will not feel any pain. Therefore, they can neither feel the pain nor withdraw their hand.
Marking scheme
Award 1 mark for option C. Candidates must understand that the dorsal root contains sensory fibers, and damage to it prevents both reflex response and sensory perception.
Question 25 · multiple_choice
1 marksWhich of the following processes in the human digestive system does NOT involve the action of enzymes?
- A.Emulsification of lipids in the duodenum
- B.Hydrolysis of starch in the mouth
- C.Digestion of peptides into amino acids in the small intestine
- D.Conversion of proteins to polypeptides in the stomach
Worked solution
Emulsification of lipids is a physical process where large lipid droplets are broken down into smaller droplets to increase the surface area for lipase action. This is carried out by bile salts present in bile, which is not an enzyme. Hydrolysis of starch in the mouth involves salivary amylase (an enzyme). Digestion of peptides into amino acids involves peptidases (enzymes). Conversion of proteins to polypeptides in the stomach involves pepsin (an enzyme).
Marking scheme
Award 1 mark for option A. Candidates must distinguish between chemical digestion (catalyzed by enzymes) and physical digestion (like emulsification by bile salts).
Question 26 · multiple_choice
1 marksA cell of an animal species with a diploid chromosome number of 4 (\(2n = 4\)) is undergoing cell division. At a certain stage, there are exactly 4 chromosomes, each consisting of two sister chromatids, aligned individually along the equatorial plane of the cell. Which stage of cell division is described?
- A.Metaphase of mitosis
- B.Metaphase I of meiosis
- C.Metaphase II of meiosis
- D.Anaphase of mitosis
Worked solution
In mitosis metaphase of a \(2n = 4\) organism, all 4 chromosomes (each consisting of two sister chromatids) align individually along the equator. In meiosis Metaphase I, homologous chromosomes pair up as bivalents (2 pairs), rather than aligning individually. In meiosis Metaphase II, the cell is haploid (\(n = 2\)), so only 2 chromosomes would align at the equator. In mitosis anaphase, sister chromatids have already separated into single-chromatid chromosomes moving to opposite poles, so there would be 8 chromosomes in total instead of 4 aligned at the equator.
Marking scheme
Award 1 mark for option A. Candidates must be able to recognize the alignment pattern and chromosome count of a diploid cell (\(2n = 4\)) in different cell division stages.
Question 27 · multiple_choice
1 marksIn humans, red-green color blindness is an X-linked recessive trait. A woman with normal vision, whose father was color-blind, marries a man with normal vision. What is the probability that their first child will be a color-blind son?
- A.0%
- B.25%
- C.50%
- D.75%
Worked solution
Let \(X^B\) be the dominant allele for normal vision and \(X^b\) be the recessive allele for color blindness. The woman has normal vision but her father was color-blind (\(X^b Y\)), meaning she must have inherited the \(X^b\) allele from her father, making her a carrier with genotype \(X^B X^b\). The man has normal vision, so his genotype is \(X^B Y\). \nWhen they cross (\(X^B X^b \times X^B Y\)), the possible genotypes of their offspring are:\n- \(X^B X^B\) (normal daughter, 25%)\n- \(X^B X^b\) (normal carrier daughter, 25%)\n- \(X^B Y\) (normal son, 25%)\n- \(X^b Y\) (color-blind son, 25%)\nTherefore, the probability of having a color-blind son is 25% (or 1/4).
Marking scheme
Award 1 mark for option B. Candidates must identify the maternal carrier genotype and correctly calculate the offspring probability for an X-linked cross.
Question 28 · multiple_choice
1 marksWhich of the following processes in the nitrogen cycle is carried out by anaerobic bacteria?
- A.Nitrification
- B.Nitrogen fixation
- C.Denitrification
- D.Decomposition of organic nitrogen to ammonium
Worked solution
Denitrification is the process of converting nitrates (\(\text{NO}_3^-\)) into nitrogen gas (\(\text{N}_2\)). This process is carried out by denitrifying bacteria (such as Pseudomonas), which are facultative anaerobes. They perform this conversion under anaerobic (oxygen-poor) conditions, such as in waterlogged soils, using nitrate as an alternative electron acceptor to oxygen. Nitrification requires oxygen and is carried out by aerobic nitrifying bacteria.
Marking scheme
Award 1 mark for option C. Candidates must know the specific conditions and bacterial roles in the different steps of the nitrogen cycle.
Question 29 · multiple_choice
1 marksWhen untreated domestic sewage is discharged into a river, it causes a series of ecological changes. Which of the following shows the correct sequence of events immediately following the discharge?
- A.Rapid growth of algae -> Death of fish -> Decrease in dissolved oxygen -> Increase in aerobic decomposers
- B.Increase in dissolved oxygen -> Rapid growth of algae -> Increase in aerobic decomposers -> Death of fish
- C.Increase in organic matter -> Increase in aerobic decomposers -> Decrease in dissolved oxygen -> Death of fish
- D.Death of fish -> Increase in organic matter -> Decrease in dissolved oxygen -> Increase in aerobic decomposers
Worked solution
Untreated domestic sewage contains a high concentration of organic matter. Immediately after discharge, the abundant organic matter provides nutrients for aerobic decomposers (bacteria and fungi), causing their population to increase rapidly. As these decomposers multiply, they consume a large amount of dissolved oxygen through aerobic respiration, leading to a sharp decrease in dissolved oxygen levels. The resulting oxygen depletion causes fish and other aquatic organisms to die from suffocation.
Marking scheme
Award 1 mark for option C. Candidates must understand the mechanism of biochemical oxygen demand (BOD) increase and oxygen depletion caused by organic discharge.
Question 30 · multiple_choice
1 marksA segment of double-stranded DNA contains 30% adenine (A). What is the percentage of cytosine (C) in this DNA segment?
- A.20%
- B.30%
- C.40%
- D.70%
Worked solution
According to Chargaff's rules for double-stranded DNA, base pairing is complementary: Adenine (A) pairs with Thymine (T), and Guanine (G) pairs with Cytosine (C). Therefore, the percentage of A equals the percentage of T, and the percentage of G equals the percentage of C. \nGiven A = 30%, then T must also be 30%. \nTogether, A + T = 30% + 30% = 60%. \nRemaining percentage for G + C is 100% - 60% = 40%. \nSince G = C, the percentage of Cytosine (C) is 40% / 2 = 20%.
Marking scheme
Award 1 mark for option A. Candidates must apply Chargaff's rules of base pairing to calculate base composition in double-stranded DNA.
Question 31 · Multiple Choice
1 marksIn a study of cellular respiration, isolated mitochondria are suspended in an oxygenated buffer solution. Which of the following substrates, when added to the suspension, would lead to the most rapid increase in oxygen consumption?
- A.Glucose
- B.Pyruvate
- C.Glycogen
- D.Starch
Worked solution
Glycolysis occurs in the cytosol, where glucose is broken down into pyruvate. Isolated mitochondria lack the enzymes required for glycolysis and cannot directly utilize glucose, glycogen, or starch. Pyruvate, however, can directly enter the mitochondria to undergo the link reaction, Krebs cycle, and oxidative phosphorylation, which consumes oxygen. Therefore, adding pyruvate leads to a rapid increase in oxygen consumption.
Marking scheme
Award 1 mark for the correct answer B. No marks are given for incorrect options. (Method: recall that glycolysis occurs in cytosol and link reaction/Krebs cycle occurs in mitochondria).
Question 32 · Multiple Choice
1 marksWhich of the following comparisons between the blood in the hepatic portal vein and the hepatic vein of a healthy human 2 hours after a meal rich in carbohydrates and proteins is/are correct?
(1) The concentration of glucose in the hepatic portal vein is higher than that in the hepatic vein.
(2) The concentration of urea in the hepatic portal vein is lower than that in the hepatic vein.
(3) The concentration of carbon dioxide in the hepatic portal vein is higher than that in the hepatic vein.
- A.(1) and (2) only
- B.(1) and (3) only
- C.(2) and (3) only
- D.(1), (2) and (3)
Worked solution
(1) is correct: 2 hours after a meal, digested carbohydrates are absorbed as glucose into the hepatic portal vein. The liver converts excess glucose into glycogen, resulting in a lower glucose concentration in the hepatic vein. (2) is correct: Excess amino acids are deaminated in the liver to produce urea, which is then released into the hepatic vein, so urea concentration is higher in the hepatic vein. (3) is incorrect: The liver is highly metabolic and undergoes active respiration, releasing carbon dioxide into the hepatic vein, making the carbon dioxide concentration in the hepatic vein higher than that in the hepatic portal vein.
Marking scheme
Award 1 mark for the correct answer A (statements 1 and 2 only). No marks for other options.
Question 33 · Multiple Choice
1 marksA pot of plant is placed horizontally in a completely dark chamber. After 24 hours, the shoot is observed to bend and grow upwards. Which of the following statements correctly explains this response?
- A.Auxin accumulates on the upper side of the shoot, inhibiting cell elongation.
- B.Auxin accumulates on the upper side of the shoot, promoting cell elongation.
- C.Auxin accumulates on the lower side of the shoot, promoting cell elongation.
- D.Auxin accumulates on the lower side of the shoot, inhibiting cell elongation.
Worked solution
Under the influence of gravity, auxin accumulates on the lower side of the horizontally placed shoot. In shoots, a higher concentration of auxin promotes cell elongation. Consequently, cells on the lower side elongate more rapidly than those on the upper side, causing the shoot to bend upwards (negative geotropism).
Marking scheme
Award 1 mark for selecting C. Distinguish between shoot response (promoted by auxin) and root response (inhibited by auxin).
Question 34 · Multiple Choice
1 marksA segment of double-stranded DNA contains 30% cytosine (C). What is the percentage of adenine (A) in the mRNA molecule transcribed from this DNA segment?
- A.20%
- B.30%
- C.40%
- D.Cannot be determined from the given information.
Worked solution
In double-stranded DNA, if C = 30%, then G = 30%. This leaves 40% for A + T, meaning across both strands, A = 20% and T = 20%. However, mRNA is transcribed from only one of the two strands (the template strand). The exact percentage of T on the template strand (which determines the percentage of A in the mRNA) cannot be deduced because the distribution of bases between the two individual strands is not necessarily symmetrical or equal. Therefore, the percentage of A in the mRNA cannot be determined.
Marking scheme
Award 1 mark for D. Deduce that single-strand base composition cannot be calculated from double-strand composition alone.
Question 35 · Multiple Choice
1 marksIn humans, red-green color blindness is a sex-linked recessive disorder. A woman with normal vision, whose father was red-green color blind, marries a man with normal vision. What is the probability that their first child will be a color-blind boy?
- A.12.5%
- B.25%
- C.50%
- D.100%
Worked solution
Let the normal vision allele be \(X^B\) and the color-blind allele be \(X^b\). The woman has normal vision but her father was color-blind (\(X^b Y\)), so she must have inherited her father's \(X^b\) chromosome, making her genotype \(X^B X^b\). The husband has normal vision, so his genotype is \(X^B Y\). Crossing \(X^B X^b \times X^B Y\) yields: \(X^B X^B\) (normal girl), \(X^B X^b\) (normal carrier girl), \(X^B Y\) (normal boy), and \(X^b Y\) (color-blind boy). The probability of having a color-blind boy as their first child is 1 out of 4, or 25%.
Marking scheme
Award 1 mark for B. Correctly identify parental genotypes and cross probabilities.
Question 36 · Multiple Choice
1 marksIn a marine ecosystem contaminated with a non-biodegradable pesticide, which of the following organisms in the food chain: "Phytoplankton -> Zooplankton -> Small fish -> Large fish -> Sea eagle" is expected to accumulate the highest concentration of the toxin, and why?
- A.Phytoplankton, because they directly absorb the toxin from the water over a large surface-area-to-volume ratio.
- B.Zooplankton, because they consume a massive quantity of phytoplankton.
- C.Sea eagle, because the toxin is non-biodegradable and accumulates progressively along the trophic levels.
- D.Small fish, because they have a high metabolic rate which accelerates toxin absorption.
Worked solution
The sea eagle, being at the highest trophic level, will accumulate the highest concentration of the non-biodegradable pesticide. This phenomenon is known as biomagnification. Because the pesticide cannot be broken down or excreted, it is passed along the food chain and becomes increasingly concentrated at each successive trophic level.
Marking scheme
Award 1 mark for C. The concept of biomagnification dictates that top predators accumulate the highest concentrations of non-biodegradable toxins.