An ecological investigation was carried out to study the distribution and abundance of three plant species (X, Y, and Z) along a coastal sand dune.
A continuous 50 m belt transect was established from the edge of the beach inland, using contiguous 1 m×1 m quadrats ( 50 quadrats in total). The percentage cover of each plant species was recorded in each quadrat. The results are shown in the graph below.
Which of the following statements is/are correct?
1 The frequency of occurrence of species Y across the entire 50 m transect is 60% .
2 At the 8 m position along the transect, the population density of species X is definitely higher than at the 18 m position.
3 At the 35 m position along the transect, species Y and species Z together cover the entire ground area within the quadrat.
A.
none of them
B.
1 only
C.
2 only
D.
3 only
E.
1 and 2 only
F.
1 and 3 only
G.
2 and 3 only
H.
1, 2 and 3
Answer and solution
Answer: B
1 is correct: Frequency is the percentage of sampled quadrats in which a species is present. Species Y is present continuously between 10 m and 40 m , which corresponds to 40−10=30 quadrats out of the 50 sampled quadrats. Therefore, the frequency is 5030×100%=60% .
2 is incorrect: Percentage cover measures the proportion of ground area covered by the aerial parts of the plant, whereas population density is the number of individual plants per unit area. A high percentage cover can be produced by a single large individual, while a lower percentage cover can contain many small seedlings. Thus, percentage cover alone does not definitively prove a higher population density.
3 is incorrect: At the 35 m position, species Y has a percentage cover of 30% and species Z has a percentage cover of 20% . The total cover is 30%+20%=50% , meaning 50% of the quadrat area consists of bare ground or other unrecorded species.
▸Question 2
Which of the following statements about the human circulatory system is/are correct?
1. The walls of capillaries consist of a single layer of cells, providing a short diffusion distance for substances.
2. The pulmonary artery transports oxygenated blood from the heart to the lungs.
3. During a cardiac cycle, the blood pressure in the aorta is significantly higher than the pressure in the pulmonary artery.
A.
none of them
B.
1 only
C.
2 only
D.
3 only
E.
1 and 2 only
F.
1 and 3 only
G.
2 and 3 only
H.
1, 2 and 3
Answer and solution
Answer: F
Statement 1 is correct. Capillaries are the primary sites of exchange between the blood and the surrounding tissues. Their walls are composed of a single layer of endothelial cells, which creates a very short diffusion pathway, facilitating rapid and efficient exchange of gases (oxygen, carbon dioxide), nutrients, and waste products.
Statement 2 is incorrect. While most arteries carry oxygenated blood, the pulmonary artery is a critical exception. It carries deoxygenated blood from the right ventricle of the heart to the lungs for oxygenation.
Statement 3 is correct. The aorta is the main artery leaving the left ventricle, supplying blood to the entire body (systemic circulation). The left ventricle must generate high pressure to pump blood over this long distance. The pulmonary artery leaves the right ventricle, supplying blood only to the lungs (pulmonary circulation), which is a much shorter, lower-resistance circuit. Consequently, the pressure in the aorta is substantially higher than in the pulmonary artery.
Since statements 1 and 3 are correct, and statement 2 is incorrect, the correct option is F.
▸Question 3
An investigation was carried out to study the effect of salinity on the germination of a crop species.
Three identical Petri dishes were prepared, each containing 20 viable seeds placed on filter paper moistened with a different concentration of sodium chloride solution: 0 mM \ceNaCl (control), 50 mM \ceNaCl , and 100 mM \ceNaCl .
All other environmental variables, including temperature and light conditions, were kept constant.
The cumulative number of germinated seeds in each dish was recorded once every day for 10 days.
The results are shown in the graph.
Which of the following statements is/are correct?
1 Over the first 5 days, the average rate of germination was highest in the 0 mM \ceNaCl solution.
2 At day 4, 30% of the seeds in the 50 mM \ceNaCl solution remained ungerminated.
3 The reciprocal of the time taken for 50% of the seeds in a treatment to germinate can be used as a measure of the relative rate of germination.
A.
none of them
B.
1 only
C.
2 only
D.
3 only
E.
1 and 2 only
F.
1 and 3 only
G.
2 and 3 only
H.
1, 2 and 3
Answer and solution
Answer: F
Statement 1 is correct: Over the first 5 days, 20 seeds germinated in the 0 mM \ceNaCl treatment (an average rate of 4.0 seeds day−1 ), compared with 10 seeds in 50 mM \ceNaCl ( 2.0 seeds day−1 ) and 1 seed in 100 mM \ceNaCl ( 0.2 seeds day−1 ).
Statement 2 is incorrect: At day 4, the graph shows that 6 seeds had germinated in the 50 mM \ceNaCl treatment out of the 20 seeds originally present. This means 20−6=14 seeds remained ungerminated. The percentage remaining ungerminated is therefore 2014×100%=70% , not 30% (30% is the proportion that had already germinated).
Statement 3 is correct: The rate of a process is inversely proportional to the time required to reach a specified endpoint ( rate∝t1 ). Using the reciprocal of the time taken for half the population to germinate ( 1/t50 ) provides a valid comparative measure of the germination rate across treatments.
▸Question 4
The graph shows the effect of different concentrations of auxin (IAA) on cell elongation in two different plant tissues, P and Q, measured as a percentage change relative to untreated controls ( 0% ).
Which of the following statements is/are correct?
1 Curve P represents the response of root tissue, and Curve Q represents the response of shoot tissue.
2 In a horizontal seedling where auxin accumulates on the lower side to 1.0 arbitrary units while the upper side remains at 10−4 arbitrary units , differential elongation will cause the shoot to bend downwards.
3 In a vertical root with a uniform auxin concentration of 1.0 arbitrary units , applying an inhibitor that reduces the auxin concentration throughout the root to 10−4 arbitrary units will increase the root elongation rate.
A.
1 only
B.
2 only
C.
3 only
D.
1 and 2 only
E.
1 and 3 only
F.
2 and 3 only
G.
1, 2 and 3
H.
none of the statements
Answer and solution
Answer: E
Statement 1 is correct: Root cells are much more sensitive to auxin than shoot cells. Roots are stimulated by very low concentrations of auxin (peaking at 10−4 arbitrary units ) and inhibited by higher concentrations, which corresponds to Curve P. Shoot cells require higher auxin concentrations for stimulation (peaking at 102 arbitrary units ), which corresponds to Curve Q.
Statement 2 is incorrect: For shoot tissue (Curve Q), an auxin concentration of 10−4 arbitrary units produces 0% change in elongation, whereas 1.0 arbitrary units produces approximately +70% stimulation. Because cells on the lower side elongate faster than cells on the upper side, the shoot bends upwards (away from gravity), not downwards.
Statement 3 is correct: For root tissue (Curve P), an auxin concentration of 1.0 arbitrary units causes elongation inhibition ( −60% ), whereas 10−4 arbitrary units causes maximal stimulation ( +40% ). Reducing the auxin concentration from 1.0 to 10−4 arbitrary units relieves the inhibition and stimulates elongation, thereby increasing the elongation rate.
Therefore, only statements 1 and 3 are correct.
▸Question 5
Genetically identical individuals of the water flea *Daphnia magna* were divided into four equal groups of twenty individuals. All four groups were raised in identical volumes of water, at the same temperature, with the same light cycle, and fed equal quantities of algae. The only difference between the groups was the concentration of a chemical cue (kairomone) released by a predator added to the water:
• Group 1: 0 a.u. of predator cue • Group 2: 10 a.u. of predator cue • Group 3: 50 a.u. of predator cue • Group 4: 100 a.u. of predator cue
Upon reaching maturity, the mean relative helmet length (the length of the defensive helmet as a percentage of body length) was measured for each group. The results are shown in the graph.
Which of the following statements could explain these results?
1 The difference in mean relative helmet length between Group 1 and Group 2 is caused by genetic differences between the individuals in the two groups.
2 The identical mean relative helmet length in Group 3 and Group 4 could be due to an environmental factor, such as food availability, becoming limiting.
3 The identical mean relative helmet length in Group 3 and Group 4 could be due to reaching the maximum phenotypic limit determined by the genotype of the clone.
A.
none of them
B.
1 only
C.
2 only
D.
3 only
E.
1 and 2 only
F.
1 and 3 only
G.
2 and 3 only
H.
1, 2 and 3
Answer and solution
Answer: G
Statement 1 is incorrect: the *Daphnia* used are genetically identical clones. Therefore, any phenotypic variation in helmet length between Group 1 and Group 2 must be caused by environmental factors (the difference in predator cue concentration triggering differential gene expression / phenotypic plasticity), not genetic differences.
Statement 2 is correct: helmet development requires metabolic resources. Because the food provided was kept constant across all groups, nutrient or energy availability could become a limiting factor preventing further growth in helmet size beyond 50 a.u. of cue.
Statement 3 is correct: the genotype defines the norm of reaction and sets an upper physiological/developmental limit on how large the helmet structure can grow. Even with higher concentrations of chemical cue, the maximum genetic potential of the trait may have already been reached in Group 3.
Therefore, only statements 2 and 3 can explain the results (option G).
▸Question 6
A short section of mRNA has the following sequence:
5′ - AUG CUC CAC UAA - 3′
The genetic code for selected amino acids is provided in the table below:
The mutation occurs at position 6, which is the third base of the second codon. The original codon is CUC, which codes for Leucine (Leu).
The mutation substitutes cytosine (C) with guanine (G), changing the codon from CUC to CUG.
According to the table, CUG also codes for Leucine (Leu). Therefore, there is no change in the amino acid sequence. This is an example of the degeneracy of the genetic code.
▸Question 7
Epidermolysis bullosa is an inherited condition in humans caused by a mutation in a gene that codes for a structural protein needed to anchor the outer layer of skin (epidermis) to underlying tissues.
In an experimental gene therapy treatment, skin stem cells were taken from an adult patient with this condition. A functional copy of the gene was inserted into the nuclear DNA of these stem cells using a viral vector. The genetically modified stem cells were then grown in the laboratory to form sheets of epidermal tissue, which were grafted onto the patient's skin to heal open wounds.
Which of the following statements about this gene therapy treatment is/are correct?
1 Transcription and translation of the inserted gene in the cells of the grafted epidermis can produce the functional structural protein.
2 When the genetically modified stem cells divide by mitosis, the inserted gene is replicated so that daughter cells retain the inserted gene.
3 The patient's biological children will inherit the inserted functional gene from this patient.
A.
1 only
B.
2 only
C.
3 only
D.
1 and 2 only
E.
1 and 3 only
F.
2 and 3 only
G.
1, 2 and 3
Answer and solution
Answer: D
Statement 1 is correct: the purpose of gene therapy is to introduce a functional gene that can be transcribed into mRNA and translated into the functional protein in the host cells.
Statement 2 is correct: before a stem cell divides by mitosis, all of its nuclear DNA—including the stably integrated functional gene—undergoes semi-conservative replication during interphase, ensuring that both daughter cells receive an identical copy of the gene.
Statement 3 is incorrect: this is an example of somatic gene therapy. Skin stem cells are somatic cells and do not differentiate into gametes (sperm or egg cells). Therefore, the genetic modification will not be passed on to the patient's offspring.
▸Question 8
Two cellular structures, structure X and structure Y, were imaged using electron microscopy. The micrographs are not shown.
Some of the data collected is shown in the table.
| | structure X | structure Y | | :--- | :---: | :---: | | actual length / nm | 2500 | 25 | | image length on micrograph / mm | 50 | 1.5 |
Which of the following statements is/are correct?
1 Structure X has been magnified 20 000 times.
2 Structure X could be a mitochondrion and structure Y could be a ribosome.
3 Structure Y can be resolved using a standard light microscope.
A.
none of them
B.
1 only
C.
2 only
D.
3 only
E.
1 and 2 only
F.
1 and 3 only
G.
2 and 3 only
H.
1, 2 and 3
Answer and solution
Answer: E
1 is correct: Magnification=actual sizeimage size . For structure X, the image size is 50 mm=50×106 nm . The actual size is 2500 nm . Therefore, magnification=2500 nm50×106 nm=20000 .
2 is correct: Structure X has an actual length of 2500 nm=2.5μm , which is within the typical size range of a mitochondrion ( 1–5μm ). Structure Y has an actual diameter/length of 25 nm , which matches the typical diameter of a ribosome ( 20–30 nm ).
3 is incorrect: The theoretical resolution limit of a standard light microscope is approximately 200 nm ( 0.2μm ), determined by the wavelength of visible light. Structure Y ( 25 nm ) is well below this limit and cannot be resolved as an individual structure using a standard light microscope.
Hence, statements 1 and 2 only are correct.
▸Question 9
A sealed flask containing a suspension of yeast (*Saccharomyces cerevisiae*) in a glucose nutrient solution was incubated at 25∘C . The rate of oxygen uptake, the rate of carbon dioxide release, and the rate of ethanol production were measured after 1 hour and after 6 hours.
| | Time = 1 hour | Time = 6 hours | | :--- | :---: | :---: | | rate of \ceO2 uptake / arbitrary units | 8.4 | 0.2 | | rate of \ceCO2 release / arbitrary units | 8.5 | 5.6 | | rate of ethanol production / arbitrary units | 0.1 | 5.5 |
Which of the following statements is/are correct?
1 A single molecule of glucose metabolised by the yeast yielded fewer molecules of ATP at 6 hours than at 1 hour.
2 At 6 hours, the oxidation of \ceNADH to \ceNAD+ was coupled to the reduction of oxygen at the inner mitochondrial membrane.
3 At 6 hours, some of the energy originally stored in glucose remained in an organic waste product released into the medium.
A.
none of them
B.
1 only
C.
2 only
D.
3 only
E.
1 and 2 only
F.
1 and 3 only
G.
2 and 3 only
H.
1, 2 and 3
Answer and solution
Answer: F
At time = 1 hour, the high rate of \ceO2 uptake (8.4) paired with equal \ceCO2 release (8.5) and negligible ethanol production (0.1) indicates that aerobic cellular respiration is the dominant metabolic pathway.
At time = 6 hours, \ceO2 uptake has dropped near zero (0.2), while \ceCO2 release (5.6) and ethanol production (5.5) occur at a roughly 1:1 ratio, showing that the yeast have switched almost entirely to anaerobic alcoholic fermentation ( \ceC6H12O6−>2C2H5OH+2CO2 ).
- Statement 1 is correct: Aerobic respiration yields roughly 30–32 ATP molecules per glucose molecule, whereas alcoholic fermentation yields only 2 ATP per glucose (via substrate-level phosphorylation in glycolysis). Therefore, fewer ATP molecules are produced per glucose molecule at 6 hours. - Statement 2 is incorrect: Because oxygen is depleted at 6 hours, oxidative phosphorylation via the mitochondrial electron transport chain cannot occur. Instead, \ceNADH is re-oxidised to \ceNAD+ in the cytosol by reducing ethanal (acetaldehyde) to ethanol. - Statement 3 is correct: Ethanol is a two-carbon organic compound that is not fully oxidised; it retains a significant proportion of the chemical potential energy originally present in glucose.
Thus, statements 1 and 3 only are correct (option F).
▸Question 10
In a population of rats, resistance to the anticoagulant poison warfarin is determined by a single gene with two alleles, R and S .
The table provides information about the phenotypes associated with each genotype:
| Genotype | Effect of standard dose of warfarin | Health in the absence of warfarin | | :--- | :--- | :--- | | SS | fatal internal bleeding | normal health and survival | | RS | survives | normal health and survival | | RR | survives | severe vitamin K deficiency (often fatal) |
Which of the following statements is/are correct?
1 In habitats where warfarin is never applied, natural selection is likely to maintain a low frequency of the R allele in the population.
2 The application of warfarin causes the S allele to mutate into the R allele in surviving rats.
3 If all SS individuals in a population are killed by warfarin before they reproduce, the S allele will be completely eliminated from the next generation.
A.
none of them
B.
1 only
C.
2 only
D.
3 only
E.
1 and 2 only
F.
1 and 3 only
G.
2 and 3 only
H.
1, 2 and 3
Answer and solution
Answer: B
Statement 1 is correct: In the absence of warfarin, SS and RS rats show normal health and survival, whereas RR rats suffer high mortality from severe vitamin K deficiency. Natural selection will therefore act against the R allele in homozygous individuals, keeping the frequency of the R allele low in the population.
Statement 2 is incorrect: Mutations arise spontaneously and randomly. Warfarin acts as a selective agent that selects for pre-existing resistant genotypes, rather than directing or causing the mutation from S to R .
Statement 3 is incorrect: Heterozygous ( RS ) individuals survive exposure to warfarin and carry the S allele. When these surviving RS individuals mate with each other, they transmit the S allele to their offspring (producing SS , RS , and RR genotypes in a 1:2:1 expected ratio). Thus, the S allele is not eliminated from the subsequent generation.
Therefore, only statement 1 is correct.
▸Question 11
An agricultural investigation was carried out using two varieties of a crop plant, Variety X and Variety Y. For each variety, all plants used were genetically identical clones.
Groups of plants of each variety were grown across a range of soil phosphorus concentrations under strictly controlled environmental conditions: • equal volumes of soil containing identical baseline nutrients • identical temperature, light intensity, and photoperiod • watered with equal volumes of water • identical atmospheric \ceCO2 concentration
After 16 weeks of growth, the mean grain yield per plant was determined for each group. The results are shown in the graph below.
Which of the following statements is/are correct?
1 At a soil phosphorus concentration of 10 mg kg−1 , the difference in mean grain yield between Variety X and Variety Y is due to differences in their genotype.
2 For Variety X, the difference in mean grain yield between plants grown at 10 mg kg−1 and plants grown at 25 mg kg−1 is due to differences in their genotype.
3 Above 30 mg kg−1 , the plateau in mean grain yield for Variety X could be because another abiotic factor has become limiting.
A.
none of them
B.
1 only
C.
2 only
D.
3 only
E.
1 and 2 only
F.
1 and 3 only
G.
2 and 3 only
H.
1, 2 and 3
Answer and solution
Answer: F
Statement 1 is correct: At a soil phosphorus concentration of 10 mg kg−1 , all environmental variables (light, temperature, water, \ceCO2 , and soil nutrients) are identical for both Variety X and Variety Y. Any difference in phenotype (grain yield) observed between the two varieties in the same environment must be caused by differences in their genotype.
Statement 2 is incorrect: Within Variety X, all individual plants are genetically identical clones and share the exact same genotype. The difference in grain yield between plants grown at 10 mg kg−1 and 25 mg kg−1 is due solely to the difference in their environment (the concentration of available phosphorus).
Statement 3 is correct: For Variety X, grain yield increases with phosphorus concentration up to 30 mg kg−1 , after which further increases in phosphorus yield no additional growth. This indicates that phosphorus is no longer the factor limiting the rate of photosynthesis and growth; another abiotic factor (such as light intensity, temperature, or \ceCO2 concentration) has become the limiting factor.
Therefore, only statements 1 and 3 are correct.
▸Question 12
A botanist crosses a homozygous purple-flowered plant (genotype BB ) with a homozygous white-flowered plant (genotype bb ). Flower colour is controlled by a single autosomal gene, where the purple allele ( B ) is dominant to the white allele ( b ).
One of the offspring produced from this cross has entirely white flowers. Genetic analysis confirms that both parent plants are the true biological parents, and that all somatic cells and germ cells of this white-flowered offspring have the same genotype.
Which of the following statements could explain this result?
1 A loss-of-function mutation in the B allele occurred during meiosis in the BB parent.
2 A loss-of-function mutation in the B allele occurred during the first mitotic division of the embryo.
3 Chromosome non-disjunction during meiosis in the BB parent produced a gamete lacking the chromosome carrying the B gene.
A.
none of them
B.
1 only
C.
2 only
D.
3 only
E.
1 and 2 only
F.
1 and 3 only
G.
2 and 3 only
H.
1, 2 and 3
Answer and solution
Answer: F
Statement 1 could explain the result: If a mutation occurs in the B allele during meiosis in the BB parent, the resulting gamete will carry a non-functional mutant allele. When fertilised by a normal b gamete from the bb parent, the zygote will carry only non-functional/recessive alleles. All somatic and germ cells derived from this zygote by mitosis will have the identical genotype, giving white flowers.
Statement 2 cannot explain the result: If a mutation occurs during the first mitotic division of the diploid zygote, it will affect only one daughter cell or lineage. The other lineage will retain the functional B allele, resulting in genetic mosaicism. The stem states that all somatic cells and germ cells have the same genotype, so this is ruled out.
Statement 3 could explain the result: Non-disjunction in the BB parent during meiosis can produce an (n−1) gamete lacking the chromosome that carries the B gene. Fertilisation by a normal (n) gamete carrying allele b from the bb parent produces a monosomic zygote carrying only the b allele. All subsequent mitotic divisions propagate this identical genotype to all cells, resulting in white flowers.
Therefore, statements 1 and 3 only could explain this result (option F).
▸Question 13
The table shows some mRNA codons and the corresponding amino acids or translation stop signals:
| Codon | Amino acid / function | |---|---| | AUG | Met | | CCU | Pro | | CCG | Pro | | CGA | Arg | | CGU | Arg | | GAA | Glu | | GAG | Glu | | GAU | Asp | | UAA | Stop | | UAG | Stop | | UGC | Cys | | UGG | Trp | | UUA | Leu | | UUG | Leu | | UUC | Phe | | UUU | Phe |
A synthetic mRNA molecule has the following sequence of 30 nucleotides, beginning with a start codon at the 5′ end:
AUG UUC CGU GAA UGG CCU GAU UGC UUA GAG
Single-base substitution mutations took place at both the 8th and 14th nucleotide positions in this sequence.
Using only the information provided and translating the mRNA sequence from left to right, which of the following statements could be correct for the resulting polypeptide?
1 This polypeptide could be only four amino acids long.
2 The third amino acid in the polypeptide could be unaffected by the mutation.
3 If translation does not stop prematurely, the resulting polypeptide could contain seven different amino acids.
A.
none of them
B.
1 only
C.
2 only
D.
3 only
E.
1 and 2 only
F.
1 and 3 only
G.
2 and 3 only
H.
1, 2 and 3
Answer and solution
Answer: F
Let us break the sequence into codons and locate the mutated positions:
- Codon 1 (bases 1–3): AUG (Met) - Codon 2 (bases 4–6): UUC (Phe) - Codon 3 (bases 7–9): CGU (Arg) — the 8th base is the middle base, G . - Codon 4 (bases 10–12): GAA (Glu) - Codon 5 (bases 13–15): UGG (Trp) — the 14th base is the middle base, G . - Codon 6 (bases 16–18): CCU (Pro) - Codon 7 (bases 19–21): GAU (Asp) - Codon 8 (bases 22–24): UGC (Cys) - Codon 9 (bases 25–27): UUA (Leu) - Codon 10 (bases 28–30): GAG (Glu)
Now consider the effects of the mutations using only the table provided:
1. Codon 5 (UGG) with base 14 mutated: The middle base changes from G to A , C , or U . The matching entries in the table with pattern U_G are UAG (Stop) and UUG (Leu). If the mutation is G→A , codon 5 becomes UAG (Stop). Translation halts after codon 4, producing a polypeptide that is only 4 amino acids long (Met–Phe–(mutated amino acid)–Glu). Thus, statement 1 is correct.
2. Codon 3 (CGU) with base 8 mutated: The middle base changes from G to A , C , or U . In the table, the only codons for Arg are CGU and CGA , both of which require G as the second nucleotide. Therefore, any substitution of the second base cannot code for Arg. The only listed codon matching C_U is CCU (Pro). Hence, the third amino acid must change to Pro and cannot remain unaffected. Thus, statement 2 is incorrect.
3. Full translation without premature termination: Codon 5 must become UUG (Leu). Codon 3 becomes CCU (Pro). The resulting sequence of 10 amino acids is: Met (1), Phe (2), Pro (3), Glu (4), Leu (5), Pro (6), Asp (7), Cys (8), Leu (9), Glu (10) The distinct amino acids present in this polypeptide are {Met, Phe, Pro, Glu, Leu, Asp, Cys}, which is exactly 7 different amino acids. Thus, statement 3 is correct.
Hence, statements 1 and 3 only could be correct (Option F).
▸Question 14
Cells of a unicellular eukaryote were placed into a solution containing three different uncharged solutes: X, Y, and Z. The concentration of each solute in the external solution was maintained at a constant 10mmoldm−3 throughout the experiment.
The intracellular concentration of each solute was measured over 60minutes . At 30minutes , a respiratory inhibitor was added to the solution. The results are shown in the graph.
Which of the following statements is/are correct?
1 The accumulation of solute Y inside the cell between 0 and 30 minutes requires metabolic energy (ATP).
2 Solute X is transported into the cell against its concentration gradient.
3 After the addition of the respiratory inhibitor at 30 minutes, solute Y leaves the cell by moving down its concentration gradient.
A.
none of them
B.
1 only
C.
2 only
D.
3 only
E.
1 and 2 only
F.
1 and 3 only
G.
2 and 3 only
H.
1, 2 and 3
Answer and solution
Answer: F
Statement 1 is correct: Between 0 and 30minutes , the intracellular concentration of solute Y increases to 45mmoldm−3 , which is significantly higher than the external concentration of 10mmoldm−3 . Accumulating a solute against its concentration gradient requires active transport, which relies on metabolic energy (ATP).
Statement 2 is incorrect: The intracellular concentration of solute X increases from 0 up to 10mmoldm−3 , matching the extracellular concentration. Thus, solute X moves into the cell down its concentration gradient until equilibrium is reached, not against it.
Statement 3 is correct: When the respiratory inhibitor is added at 30minutes , ATP synthesis is blocked and active transport ceases. Because the intracellular concentration of solute Y ( 45mmoldm−3 ) is higher than the extracellular concentration ( 10mmoldm−3 ), solute Y moves passively out of the cell down its concentration gradient, leading to the observed decrease in intracellular concentration.
▸Question 15
The allele frequencies in the gene pool of a sexually reproducing population can change over generations as a result of various biological processes and environmental events.
Which of the following would be expected to cause a change in allele frequencies in the gene pool of a population?
1 widespread application of a chemical herbicide across an agricultural weed population
2 immigration of fertile individuals from a genetically distinct neighbouring population
3 a severe population bottleneck caused by a natural disaster that randomly kills the majority of individuals
4 random mating occurring within a very large population in the absence of mutation, migration, and differential selection
A.
1 and 2 only
B.
1 and 4 only
C.
2 and 3 only
D.
3 and 4 only
E.
1, 2 and 3 only
F.
1, 2 and 4 only
G.
2, 3 and 4 only
H.
1, 2, 3 and 4
Answer and solution
Answer: E
Allele frequencies in a gene pool change through evolutionary mechanisms such as natural selection, gene flow, mutation, and genetic drift:
- Statement 1 describes directional natural selection: applying an herbicide exerts selection pressure, increasing the frequency of alleles conferring resistance. - Statement 2 describes gene flow: immigration of individuals from a genetically distinct population introduces new alleles or changes the proportion of existing alleles. - Statement 3 describes genetic drift via a population bottleneck: a catastrophic event randomly reduces population size, causing non-representative sampling of alleles that alters allele frequencies. - Statement 4 describes the conditions required for Hardy-Weinberg equilibrium (large population, random mating, no mutation, no migration, no selection), under which allele frequencies remain constant from generation to generation.
Therefore, statements 1, 2, and 3 only cause a change in allele frequencies.
▸Question 16
An experiment was conducted to investigate the breakdown of hydrogen peroxide catalysed by catalase extracted from human liver cells.
In a control experiment, 10 cm3 of 1.0 mol dm−3\ceH2O2 was mixed with 1.0 cm3 of human catalase solution at 25 °C and pH 7.0 . The reaction was allowed to proceed to completion, and the total volume of oxygen gas produced was recorded over time, producing curve X on the graph.
The experiment was repeated under three separate modifications: - Modification 1: Carried out at 37 °C instead of 25 °C . - Modification 2: Carried out using 10 cm3 of 0.5 mol dm−3\ceH2O2 at 25 °C . - Modification 3: Carried out at 25 °C after boiling the catalase solution for 10 minutes and then cooling it.
Which row in the table correctly identifies the curve obtained for each modification?
| | Modification 1 | Modification 2 | Modification 3 | | :---: | :---: | :---: | :---: | | A | P | S | T | | B | P | R | T | | C | P | S | R | | D | Q | S | T | | E | Q | R | T | | F | Q | S | R | | G | R | S | T | | H | R | P | T |
**Modification 1 ( 37 °C ):** Human catalase has an optimum temperature close to normal body temperature ( 37 °C ). Increasing the temperature from 25 °C to 37 °C increases the kinetic energy of the molecules, resulting in more frequent collisions exceeding the activation energy and a faster initial rate of reaction. Because the amount of substrate is unchanged, the final volume of oxygen produced remains 40 cm3 . This corresponds to curve P.
Modification 2 (halved substrate concentration): Using 10 cm3 of 0.5 mol dm−3\ceH2O2 halves the total amount of substrate available, so the plateau volume of \ceO2 is halved to 20 cm3 . The lower substrate concentration also reduces the initial rate of enzyme-substrate collisions. This corresponds to curve S.
Modification 3 (boiled enzyme): Boiling at 100 °C irreversibly denatures human catalase by breaking the hydrogen and ionic bonds maintaining its tertiary structure. The active site is lost, so no catalysed reaction occurs, resulting in 0 cm3 of oxygen produced. This corresponds to curve T.
Therefore, the correct row is A (P, S, T).
▸Question 17
The graph shows the blood lactic acid concentration of two athletes, Athlete 1 and Athlete 2, during an exercise test where running speed was increased incrementally in stages.
Which of the following statements correctly describe(s) the athletes?
1 At a running speed of 11 km h−1 , Athlete 1 relies more on anaerobic respiration to meet their energy demands than Athlete 2. 2 At a running speed of 5 km h−1 , the rate of lactic acid production in both athletes is zero. 3 If both athletes run at 11 km h−1 for 15 minutes, Athlete 1 will incur a larger oxygen debt than Athlete 2.
A.
none of them
B.
1 only
C.
2 only
D.
3 only
E.
1 and 2 only
F.
1 and 3 only
G.
2 and 3 only
H.
1, 2 and 3
Answer and solution
Answer: F
Statement 1 is correct: At 11 km h−1 , Athlete 1's blood lactic acid concentration has risen sharply above baseline (to over 5 mmol dm−3 ), indicating substantial anaerobic glycolysis, whereas Athlete 2's concentration remains at resting baseline (around 1 mmol dm−3 ), showing that their energy demand is met almost entirely by aerobic respiration.
Statement 2 is incorrect: At 5 km h−1 , both curves show a constant baseline concentration of approximately 1 mmol dm−3 . Even at low intensities and rest, lactic acid is continuously produced (e.g. by erythrocytes) and cleared by the liver and heart at an equal rate, so the rate of production is not zero.
Statement 3 is correct: Running at 11 km h−1 causes significant lactic acid accumulation in Athlete 1, whereas Athlete 2 remains below their lactate threshold. Consequently, Athlete 1 will require a greater volume of extra oxygen post-exercise to metabolise the accumulated lactic acid, meaning they incur a larger oxygen debt.
▸Question 18
Researchers investigated the role of Protein Q in the synthesis of a stress-response metabolite, M. Three bacterial strains were compared: a wild type, a mutant lacking the gene for Protein Q, and a strain engineered to contain high cellular levels of Protein Q. The concentration of metabolite M (arbitrary units) was measured in each strain after incubation at either 20°C or 40°C. The results are shown in the table below: | Strain | Level of Protein Q | Concentration of M at 20°C | Concentration of M at 40°C | | :--- | :--- | :--- | :--- | | Wild Type | Normal | 5 | 100 | | Mutant | None | 5 | 25 | | Engineered | High | 5 | 250 | Which of the following conclusions is best supported by the data?
A.
Protein Q is essential for the initiation of metabolite M synthesis.
B.
Protein Q acts as a temperature sensor that triggers the production of metabolite M.
C.
The production of metabolite M is proportional to the concentration of Protein Q at both temperatures.
D.
Protein Q enhances the production of metabolite M but only in the presence of the thermal stimulus.
E.
The presence of Protein Q suppresses the basal production of metabolite M at 20°C.
F.
Metabolite M synthesis is regulated solely by the concentration of Protein Q.
Answer and solution
Answer: D
First, let's examine the role of the thermal stimulus. The mutant strain, which lacks Protein Q, still shows an increase in metabolite M from 5 to 25 units when the temperature is raised to 40°C. This shows that M synthesis can be triggered by temperature alone, meaning Protein Q is not essential for the process to start. This eliminates option A.
Next, consider the effect of Protein Q concentration at the lower temperature. At 20°C, all three strains produce the same low level of M (5 units), regardless of the level of Protein Q. The engineered strain, with high levels of Protein Q, does not produce more M than the wild type or the mutant. This demonstrates that Protein Q by itself is not sufficient to increase M production; the thermal stimulus is also required. This rules out options C and F.
Finally, let's look at the results at 40°C. Here, there is a clear correlation between the level of Protein Q and the concentration of M. The mutant (no Q) produces 25 units, the wild type (normal Q) produces 100 units, and the engineered strain (high Q) produces 250 units. This shows that Protein Q enhances or amplifies the production of M, but only when the thermal stimulus is present.
This combination of observations—that Protein Q is neither necessary nor sufficient on its own, but enhances the response to a thermal stimulus—best supports conclusion D.
▸Question 19
In a species of plant, the allele for tall stems ( T ) is dominant to the allele for dwarf stems ( t ).
A tall plant is crossed with a dwarf plant. The resulting offspring show a 1:1 ratio of tall plants to dwarf plants.
Two tall plants, each with the same genotype as the original tall parent, are then crossed with each other.
What is the expected proportion of offspring from this second cross that will be dwarf?
A.
0
B.
1/4
C.
1/2
D.
3/4
Answer and solution
Answer: B
The solution requires two steps of reasoning.
First, we must deduce the genotype of the original tall parent plant. It is crossed with a dwarf plant. Since dwarf stems is a recessive trait, the dwarf plant must have the genotype tt . The tall parent has the dominant phenotype, so its genotype is either TT (homozygous dominant) or Tt (heterozygous).
The cross is a test cross. The offspring show a 1:1 ratio of tall to dwarf. - If the tall parent were TT , the cross would be TT×tt . All offspring would be Tt and therefore tall. This contradicts the observed 1:1 ratio. - If the tall parent were Tt , the cross would be Tt×tt . The expected offspring genotypes are Tt and tt in a 1:1 ratio. This corresponds to a 1:1 phenotypic ratio of tall ( Tt ) to dwarf ( tt ), which matches the observation. Therefore, the original tall parent plant must be heterozygous ( Tt ).
Second, we determine the outcome of the new cross. Two plants with this genotype ( Tt ) are crossed with each other: Tt×Tt .
We can use a Punnett square to find the expected proportions of genotypes in the offspring: - Parent gametes: T and t from each parent. - Offspring genotypes: TT , Tt , Tt , and tt . - This gives a genotypic ratio of 1 TT : 2 Tt : 1 tt .
The question asks for the proportion of offspring that will be dwarf. The dwarf phenotype corresponds to the homozygous recessive genotype ( tt ).
From the Punnett square, 1 out of the 4 equally likely outcomes is tt . Therefore, the expected proportion of dwarf offspring is 1/4.
▸Question 20
In mice, the allele for black fur, B , is dominant to the allele for brown fur, b .
Two heterozygous black mice are crossed. A black offspring from this cross is selected at random and crossed with a heterozygous black mouse.
What is the probability that their first offspring has brown fur?
A.
0
B.
121
C.
61
D.
41
E.
31
F.
21
Answer and solution
Answer: C
The first cross is Bb×Bb , which gives genotype probabilities
BB:Bb:bb=41:21:41.
The selected offspring is known to have black fur, so it is either BB or Bb . Conditional on being black,
P(Bb∣black)=3/41/2=32.
If the selected mouse is BB , crossing it with Bb cannot produce a brown offspring. If it is Bb , the cross Bb×Bb produces a brown bb offspring with probability 1/4 . Therefore,
P(brown offspring)=32×41=61.
The correct option is C.
▸Question 21
A diploid cell contains 12 chromosomes.
Which row correctly gives the number of chromosomes in each daughter cell after mitosis and in each gamete after meiosis?
A.
Mitosis 6; meiosis 6
B.
Mitosis 6; meiosis 12
C.
Mitosis 12; meiosis 6
D.
Mitosis 12; meiosis 12
E.
Mitosis 24; meiosis 6
F.
Mitosis 24; meiosis 12
Answer and solution
Answer: C
Mitosis preserves chromosome number, so each daughter cell has 12 chromosomes. Meiosis halves the diploid number to form gametes, so each gamete has 6. The correct option is C.
▸Question 22
The pedigree shows the inheritance of an X-linked recessive condition in a family.
Individual 5 has been diagnosed with Klinefelter syndrome (karyotype 47, XXY) and expresses the recessive condition.
Assuming no new mutations occurred and no crossing over took place between the gene locus and the centromere, in which parent did the non-disjunction event occur, and at which stage of cell division?
A.
mother; meiosis I
B.
mother; meiosis II
C.
mother; mitosis after fertilisation
D.
father; meiosis I
E.
father; meiosis II
F.
father; mitosis after fertilisation
G.
either parent; meiosis I
H.
either parent; meiosis II
Answer and solution
Answer: B
1. Let the dominant normal allele be XR and the recessive mutant allele causing the condition be Xr .
2. Individual 1 is an affected male, so his genotype is XrY . He must pass his Xr chromosome to his daughter, Individual 3.
3. Individual 3 is unaffected, so she must be a carrier with genotype XRXr .
4. Individual 4 is an unaffected male, so his genotype is XRY .
5. Individual 5 has Klinefelter syndrome (47, XXY) and expresses the recessive condition. To express an X-linked recessive phenotype, he must not possess the dominant allele XR . Therefore, his genotype must be XrXrY .
6. The Y chromosome must have been inherited from the father (Individual 4) via normal segregation in sperm formation.
7. Both Xr chromosomes must therefore come from the mother (Individual 3).
8. The mother has genotype XRXr . In meiosis I, homologous chromosomes ( XR and Xr ) separate into different daughter cells. In meiosis II, sister chromatids separate. For an ovum to receive two identical copies of Xr , the sister chromatids of the replicated Xr chromosome must have failed to separate during meiosis II in the mother.
9. If non-disjunction had occurred in the mother during meiosis I, the egg would have received both homologous chromosomes ( XRXr ), which upon fertilisation would yield an unaffected male of genotype XRXrY .
Therefore, non-disjunction occurred in the mother during meiosis II (option B).
▸Question 23
A culture of secretory cells has a total tissue mass of 10 mg . The average mass of a single cell in the culture is 1 ng .
Each cell synthesizes a specific protein at a rate of 105 molecules per second. The mass of one molecule of this protein is 10−18 g .
What is the total mass of the protein synthesized by the culture in 100 minutes?
A.
0.1 mg
B.
1.0 mg
C.
6.0 mg
D.
10 mg
E.
60 mg
F.
6.0 g
Answer and solution
Answer: C
We calculate the total mass in three steps: finding the number of cells, the total number of molecules produced, and the final mass.
1. Calculate the number of cells:
Number of cells=Mass per cellTotal tissue mass=1×10−9 g10×10−3 g=107 cells
2. Calculate total molecules produced: Convert time to seconds: 100 minutes=100×60=6000 s .
Total molecules=(Cells)×(Rate)×(Time)
=107×105×6000=6×1015 molecules
3. Convert to mass:
Total mass=(6×1015)×(10−18 g/molecule)=6×10−3 g
Converting to milligrams: 6×10−3 g=6.0 mg .
▸Question 24
A suspension contains a mixture of two strains of yeast: a wild-type strain that secretes the enzyme invertase, and a mutant strain that does not. The total concentration of yeast cells in the suspension is 8.0×106 cells mL −1 and the invertase activity is 300 arbitrary units (AU) mL −1 .
A separate control sample containing only the wild-type strain at a concentration of 1.0×106 cells mL −1 has an invertase activity of 50 AU mL −1 .
Assuming the mutant strain has zero invertase activity, what is the concentration of the mutant strain in the mixed suspension?
A.
1.0×106 cells mL −1
B.
2.0×106 cells mL −1
C.
4.0×106 cells mL −1
D.
5.0×106 cells mL −1
E.
6.0×106 cells mL −1
Answer and solution
Answer: B
The control sample establishes the relationship between cell concentration and enzyme activity for the wild-type strain. A concentration of 1.0×106 cells mL −1 gives an activity of 50 AU mL −1 .
The mixed suspension has a total activity of 300 AU mL −1 . Since the mutant strain contributes no activity, this entire amount must come from the wild-type cells. The activity is 6 times higher than in the control ( 300/50=6 ), so the concentration of wild-type cells must also be 6 times higher.
The diagram shows a circular plasmid vector containing an origin of replication ( ori ), an ampicillin resistance gene ( AmpR ), and a β -galactosidase gene ( lacZ ).
The enzyme β -galactosidase hydrolyses the colourless substrate X-gal into a blue product in the presence of the inducer IPTG.
Two recombinant plasmids were produced by inserting a target gene into one of the unique restriction sites: - Plasmid A has the target gene inserted into the Eco RI site located within the coding sequence of lacZ . - Plasmid B has the target gene inserted into the Pst I site located within the coding sequence of AmpR .
Samples of ampicillin-sensitive, β -galactosidase-deficient bacteria were separately transformed with either Plasmid A or Plasmid B. The transformed bacteria were spread onto two sets of agar plates: - Plate 1: Nutrient agar containing ampicillin, X-gal, and IPTG - Plate 2: Nutrient agar containing X-gal and IPTG (without ampicillin)
Which row in the table correctly predicts the growth and appearance of each bacterial culture on the two plates?
| | Plasmid A on Plate 1 (with ampicillin) | Plasmid A on Plate 2 (without ampicillin) | Plasmid B on Plate 1 (with ampicillin) | Plasmid B on Plate 2 (without ampicillin) | | :--- | :--- | :--- | :--- | :--- | | A | blue colonies | blue colonies | white colonies | white colonies | | B | white colonies | white colonies | no growth | blue colonies | | C | white colonies | blue colonies | no growth | white colonies | | D | no growth | white colonies | blue colonies | blue colonies | | E | blue colonies | white colonies | no growth | blue colonies | | F | white colonies | white colonies | blue colonies | no growth | | G | no growth | blue colonies | white colonies | white colonies | | H | white colonies | blue colonies | blue colonies | no growth |
Plasmid A on Plate 1: blue colonies | Plasmid A on Plate 2: blue colonies | Plasmid B on Plate 1: white colonies | Plasmid B on Plate 2: white colonies
B.
Plasmid A on Plate 1: white colonies | Plasmid A on Plate 2: white colonies | Plasmid B on Plate 1: no growth | Plasmid B on Plate 2: blue colonies
C.
Plasmid A on Plate 1: white colonies | Plasmid A on Plate 2: blue colonies | Plasmid B on Plate 1: no growth | Plasmid B on Plate 2: white colonies
D.
Plasmid A on Plate 1: no growth | Plasmid A on Plate 2: white colonies | Plasmid B on Plate 1: blue colonies | Plasmid B on Plate 2: blue colonies
E.
Plasmid A on Plate 1: blue colonies | Plasmid A on Plate 2: white colonies | Plasmid B on Plate 1: no growth | Plasmid B on Plate 2: blue colonies
F.
Plasmid A on Plate 1: white colonies | Plasmid A on Plate 2: white colonies | Plasmid B on Plate 1: blue colonies | Plasmid B on Plate 2: no growth
G.
Plasmid A on Plate 1: no growth | Plasmid A on Plate 2: blue colonies | Plasmid B on Plate 1: white colonies | Plasmid B on Plate 2: white colonies
H.
Plasmid A on Plate 1: white colonies | Plasmid A on Plate 2: blue colonies | Plasmid B on Plate 1: blue colonies | Plasmid B on Plate 2: no growth
Answer and solution
Answer: B
In Plasmid A, insertion into the Eco RI site disrupts the lacZ gene (insertional inactivation). The bacteria cannot produce functional β -galactosidase, so X-gal is not cleaved and the colonies remain white. The AmpR gene is intact, so the bacteria are resistant to ampicillin and will grow on Plate 1 as white colonies. On Plate 2 (no ampicillin), they will also grow as white colonies.
In Plasmid B, insertion into the Pst I site disrupts the AmpR gene. The bacteria are ampicillin-sensitive and therefore cannot grow on Plate 1 (containing ampicillin), resulting in no growth. On Plate 2 (lacking ampicillin), they are able to grow. Since their lacZ gene is intact, they express active β -galactosidase, which cleaves X-gal to form blue colonies.
Therefore, row B is correct.
▸Question 26
A functional polypeptide consists of a single chain of 150 amino acids. The complete coding region of the messenger RNA (mRNA) that translates this polypeptide (excluding the stop codon) has the following base composition:
Which row in the table correctly gives the percentage of thymine (T) in the double-stranded DNA coding region of this gene, and the total number of hydrogen bonds between the complementary base pairs in this DNA region?
| | percentage of thymine (T) | total number of hydrogen bonds | | --- | --- | --- | | A | 24% | 900 | | B | 24% | 1080 | | C | 30% | 540 | | D | 30% | 1080 | | E | 30% | 2160 | | F | 36% | 900 | | G | 36% | 1080 | | H | 60% | 1080 |
A.
percentage of thymine (T): 24%, total number of hydrogen bonds: 900
B.
percentage of thymine (T): 24%, total number of hydrogen bonds: 1080
C.
percentage of thymine (T): 30%, total number of hydrogen bonds: 540
D.
percentage of thymine (T): 30%, total number of hydrogen bonds: 1080
E.
percentage of thymine (T): 30%, total number of hydrogen bonds: 2160
F.
percentage of thymine (T): 36%, total number of hydrogen bonds: 900
G.
percentage of thymine (T): 36%, total number of hydrogen bonds: 1080
H.
percentage of thymine (T): 60%, total number of hydrogen bonds: 1080
Answer and solution
Answer: D
1. Length of the coding region: - A polypeptide of 150 amino acids is coded for by 150×3=450 nucleotides in the mRNA coding region. - Therefore, the corresponding double-stranded DNA coding region consists of 450 base pairs (a total of 900 nucleotides across both strands).
2. Base composition of the double-stranded DNA: - In transcription, mRNA adenine (A) is complementary to template thymine (T), and mRNA uracil (U) is complementary to template adenine (A). - The coding (non-template) strand will have A matching mRNA A and T matching mRNA U. - Across both DNA strands: Total A=Total T=(36%+24%)×450=60%×450=270 bases each - The percentage of thymine (T) in the double-stranded DNA (out of all 900 bases) is: 900270×100%=30% *(Or simply 236%+24%=30% )*.
3. Number of base pairs and hydrogen bonds: - There are 270 A-T base pairs (which have 2 hydrogen bonds each): 270×2=540 hydrogen bonds - The remaining base pairs are G-C pairs: 450−270=180 pairs (which have 3 hydrogen bonds each): 180×3=540 hydrogen bonds - Total number of hydrogen bonds: 540+540=1080 Therefore, row D is correct.
▸Question 27
DNA sequencing was used to analyse a specific gene locus in a child and both biological parents. Paternity and maternity were genetically confirmed. The table below shows the nucleotide base present at a specific position within this locus on both chromosomes for each individual. | Individual
| Chromosome 1 Base | Chromosome 2 Base | | :--- | :--- |
: | :---: | | Mother | T | T | | Father | T | T | | Child | T | A |
Which statement best explains the origin of the Adenine (A) base found in the child?
A.
It arose from a spontaneous mutation in the germline of one of the parents.
B.
It was generated by crossing over between homologous chromosomes during meiosis I.
C.
It resulted from the independent assortment of chromosomes during meiosis I.
D.
It is a recessive allele that was present in the genotype of both parents but masked in their phenotype.
E.
It originated from a somatic mutation in the mother that was transferred to the child during pregnancy.
Answer and solution
Answer: A
We examine the genotypes provided in the table:
Mother:(T,T)Father:(T,T)Child:(T,A)
Both parents are homozygous for T , meaning the A allele is completely absent from their genomes. Standard inheritance mechanisms—such as segregation, independent assortment, and crossing over—only rearrange existing alleles. They cannot generate new nucleotide bases.
Since the A allele is novel, it must have arisen via a change in the DNA sequence, which is the definition of a mutation. Furthermore, for this change to be present in every cell of the child (or at least the cells sequenced), the mutation must have occurred in the germline (gametes) of one of the parents.