ESAT Biology Mock 5

Statement 1 is correct: The nematicide acts as an environmental selective pressure. Individuals of *Species X* that carry alleles conferring resistance are more likely to survive and reproduce, passing on these alleles to offspring and thereby increasing the resistance allele frequency in the population.

Statement 2 is incorrect: Mutations arise spontaneously and at random. Environmental chemicals do not induce directed or purposeful mutations that specifically adapt an organism to the chemical challenge.

Statement 3 is correct: For a population to adapt via natural selection, the necessary beneficial alleles must either be pre-existing in the gene pool or arise randomly. *Species Y* became extinct because no individuals possessed alleles conferring sufficient tolerance to survive the lethal selective pressure of the nematicide.

Therefore, statements 1 and 3 only are correct.

To identify each box, trace the biological processes represented by the directed arrows:

1. **Box W (Atmospheric nitrogen gas, $\displaystyle \ce{N2}$ ):**
- Nitrogen is fixed from the atmosphere into soil ammonium by free-living nitrogen-fixing bacteria (arrow $\displaystyle \ce{W -> X}$ ).
- Nitrogen is fixed directly into legume plant biomass by mutualistic nitrogen-fixing bacteria in root nodules (arrow $\displaystyle \ce{W -> Z}$ ).
- Denitrifying bacteria return nitrogen gas to the atmosphere from nitrates (arrow $\displaystyle \ce{Y -> W}$ ).
Therefore, $\displaystyle \text{W} = \text{Atmospheric nitrogen gas}$ .

2. **Box X (Ammonium ions in soil, $\displaystyle \ce{NH4+}$ ):**
- Receives nitrogen from nitrogen fixation ( $\displaystyle \ce{W -> X}$ ) and from decomposition/ammonification of dead plant material by saprobionts (arrow $\displaystyle \ce{Z -> X}$ ).
- Is converted into nitrates by nitrifying bacteria (arrow $\displaystyle \ce{X -> Y}$ ).
Therefore, $\displaystyle \text{X} = \text{Ammonium ions in soil}$ .

3. **Box Y (Nitrate ions in soil, $\displaystyle \ce{NO3-}$ ):**
- Formed by nitrification of ammonium ( $\displaystyle \ce{X -> Y}$ ).
- Taken up/absorbed by plant roots for assimilation into amino acids and proteins (arrow $\displaystyle \ce{Y -> Z}$ ).
- Converted to $\displaystyle \ce{N2}$ gas by denitrifying bacteria under anaerobic conditions ( $\displaystyle \ce{Y -> W}$ ).
Therefore, $\displaystyle \text{Y} = \text{Nitrate ions in soil}$ .

4. Box Z (Nitrogen in plant biomass):
- Obtains nitrogen through mutualistic fixation ( $\displaystyle \ce{W -> Z}$ ) and absorption of nitrates ( $\displaystyle \ce{Y -> Z}$ ).
- Supplies dead organic matter to decomposers to form ammonium ( $\displaystyle \ce{Z -> X}$ ).
Therefore, $\displaystyle \text{Z} = \text{Nitrogen in plant biomass}$ .

This corresponds to row B.

To determine which species is most vulnerable to local extinction under worsening conditions and habitat fragmentation, we evaluate the effect of each trait:

1. Dietary niche: A specialist relies on specific food sources that may decline or disappear during environmental degradation, whereas a generalist can switch to alternative food sources. Therefore, a specialist niche increases vulnerability to extinction.
2. Dispersal ability: Low dispersal ability prevents individuals from moving between isolated pools or migrating to refugia as conditions deteriorate, increasing vulnerability.
3. Generation time: A long generation time means slower reproductive turnover, lower intrinsic rate of population increase ( $\displaystyle r$ ), and slower recovery from population bottlenecks, increasing extinction risk.

Species D possesses all three high-risk traits (specialist dietary niche, low dispersal ability, and long generation time), making it the most vulnerable to extinction.

Statement 1 is correct: Genes contain the genetic code to direct the synthesis of polypeptides (proteins) or functional RNA molecules. ATP (adenosine triphosphate) is a phosphorylated nucleotide molecule synthesized by metabolic enzymes/ATP synthase, not a polypeptide directly encoded by a gene.

Statement 2 is correct: Cyanobacteria are prokaryotes and therefore lack a nuclear membrane. Because there is no membrane barrier between the DNA and the ribosomes, ribosomes can attach to mRNA and translate it while it is still being transcribed by RNA polymerase (coupled transcription and translation).

Statement 3 is incorrect: Comparing the 15-nucleotide sequences:
- Pea plant vs baker's yeast: differs at 2 positions (position 6: C vs T; position 12: T vs C).
- Pea plant vs cyanobacterium: differs at 7 positions (positions 4, 5, 6 in codon 2; position 7 in codon 3; positions 10, 11, 12 in codon 4).
Therefore, the pea plant sequence is substantially more similar to baker's yeast (13/15 matching bases) than to the cyanobacterium (8/15 matching bases).

Thus, only conclusions 1 and 2 are correct.

We analyse the impact of the two environmental changes on the selective pressures acting on the population.

First, the water becomes turbid, reducing visibility. Since the predator relies on vision, this change reduces the predation risk for all fish, but specifically relaxes the selective pressure against Morph B, which was previously disadvantaged by being conspicuous.

Second, the food source decreases by $\displaystyle 50\%$ . This scarcity increases the intensity of intraspecific competition. Morph B is a strong competitor, while Morph A is a weak competitor. Therefore, the selection pressure favouring the strong competitor (Morph B) increases.

Since the penalty for being conspicuous is reduced and the advantage of being a strong competitor is amplified, the net selective pressure shifts strongly in favour of Morph B. Consequently, the relative frequency of Morph B will increase.

Statement 1 is correct: At a light intensity of $\displaystyle 0\text{ arbitrary units}$ , there is no light to drive photosynthesis. The net $\displaystyle \ce{CO2}$ uptake is $\displaystyle -4\ \mu\text{mol}\,\text{m}^{-2}\,\text{s}^{-1}$ , meaning that $\displaystyle \ce{CO2}$ is being released by cellular respiration at a rate of $\displaystyle 4\ \mu\text{mol}\,\text{m}^{-2}\,\text{s}^{-1}$ .

Statement 2 is incorrect: Point P is the light compensation point, where the net rate of $\displaystyle \ce{CO2}$ uptake is zero. At this point, the rate of $\displaystyle \ce{CO2}$ consumption by photosynthesis equals the rate of $\displaystyle \ce{CO2}$ production by respiration. Photosynthesis is actively taking place, so the light-dependent reactions are actively running.

Statement 3 is correct: $\displaystyle \text{Gross rate of photosynthesis} = \text{Net } \ce{CO2} \text{ uptake} + \text{Respiration rate}$ . At $\displaystyle 80\text{ arbitrary units}$ , net $\displaystyle \ce{CO2}$ uptake is $\displaystyle 10\ \mu\text{mol}\,\text{m}^{-2}\,\text{s}^{-1}$ . Adding the respiration rate of $\displaystyle 4\ \mu\text{mol}\,\text{m}^{-2}\,\text{s}^{-1}$ gives a gross $\displaystyle \ce{CO2}$ fixation rate of $\displaystyle 10 + 4 = 14\ \mu\text{mol}\,\text{m}^{-2}\,\text{s}^{-1}$ .

Therefore, statements 1 and 3 only are correct.

Statement 1 is correct: Unaffected parents (1 and 2) produce an affected child (3), and unaffected parents (4 and 6) produce an affected child (7). For a dominant condition with complete penetrance and no new mutations, every affected individual must have at least one affected parent. Thus, the condition cannot be dominant.

Statement 2 is correct: If the condition is X-linked recessive, affected male 7 has genotype $\displaystyle X^a Y$ and must have inherited the $\displaystyle X^a$ allele from his mother (individual 4). Because individual 4 is unaffected, she must be a carrier ( $\displaystyle X^A X^a$ ). Individual 6 is an unaffected male ( $\displaystyle X^A Y$ ). Female offspring 8 receives $\displaystyle X^A$ from her father (individual 6) and has a 50% chance of receiving $\displaystyle X^a$ from her mother (individual 4), giving a 50% probability of being a carrier ( $\displaystyle X^A X^a$ ).

Statement 3 is incorrect: While an unaffected offspring of two carriers has a prior probability of $\displaystyle \dfrac{2}{3}$ of being a carrier, individual 4 has an affected child (individual 7 with genotype $\displaystyle aa$ ). For individual 7 to be homozygous recessive, individual 4 must have contributed an ' $\displaystyle a$ ' allele. As individual 4 is phenotypically unaffected, her genotype is definitively $\displaystyle Aa$ (probability = 1, or 100%), not $\displaystyle \dfrac{2}{3}$ .

Therefore, only statements 1 and 2 are correct.

Statement 1 is correct: Between 0 and 6 hours ( $\displaystyle \text{G}_1$ phase), the DNA has not yet replicated, so each chromosome consists of a single chromatid (nuclear DNA mass is $\displaystyle 6.0\text{ pg}$ ). Between 10 and 14 hours ( $\displaystyle \text{G}_2$ phase), DNA replication is complete and the DNA mass has doubled to $\displaystyle 12.0\text{ pg}$ . Each chromosome now consists of two sister chromatids, meaning the total number of chromatids in the cell has doubled.

Statement 2 is incorrect: DNA synthesis (S phase) is represented by the rise in nuclear DNA mass, which occurs between 6 and 10 hours (duration = $\displaystyle 4\text{ hours}$ ). The $\displaystyle \text{G}_1$ phase lasts from 0 to 6 hours (duration = $\displaystyle 6\text{ hours}$ ). Thus, the S phase is shorter than the $\displaystyle \text{G}_1$ phase.

Statement 3 is correct: Adding an inhibitor of mitotic spindle formation at 8 hours allows DNA synthesis to finish normally at 10 hours, reaching $\displaystyle 12.0\text{ pg}$ . However, the inhibitor prevents chromosome separation and cytokinesis at 14–15 hours. The cell is arrested before cell division can occur, so its nuclear DNA mass remains at $\displaystyle 12.0\text{ pg}$ at 18 hours.

Therefore, only statements 1 and 3 are correct.

Statement 1 is correct: Between $\displaystyle t = 0$ and $\displaystyle t = 30\text{ min}$ , the graph is linear. The gradient represents the initial rate: $\displaystyle \text{rate} = \dfrac{6.0\text{ cm}^3}{30\text{ min}} = 0.20\text{ cm}^3\text{ min}^{-1}$ . Converting this into hours gives $\displaystyle 0.20\text{ cm}^3\text{ min}^{-1} \times 60\text{ min h}^{-1} = 12.0\text{ cm}^3\text{ h}^{-1}$ .

Statement 2 is incorrect: The respiratory quotient is defined as $\displaystyle \text{RQ} = \dfrac{\text{volume of }\ce{CO2}\text{ produced}}{\text{volume of }\ce{O2}\text{ consumed}}$ . For an $\displaystyle \text{RQ}$ of $\displaystyle 0.7$ , for every $\displaystyle 1.0\text{ cm}^3$ of $\displaystyle \ce{O2}$ consumed, only $\displaystyle 0.7\text{ cm}^3$ of $\displaystyle \ce{CO2}$ is produced. Without $\displaystyle \ce{KOH}$ to absorb the $\displaystyle \ce{CO2}$ , the net change in gas volume is $\displaystyle 0.7 - 1.0 = -0.3\text{ cm}^3$ , resulting in a net decrease in chamber gas volume.

Statement 3 is correct: In a closed respirometer, as $\displaystyle \ce{O2}$ is consumed and $\displaystyle \ce{CO2}$ is absorbed by $\displaystyle \ce{KOH}$ , the partial pressure of oxygen steadily drops. At lower oxygen concentrations, the rate of oxidative phosphorylation slows because oxygen is the terminal electron acceptor, making $\displaystyle \ce{O2}$ concentration limiting.

In the ZW sex-determination system, males are $\displaystyle \text{ZZ}$ and females are $\displaystyle \text{ZW}$ . Female offspring always inherit their single $\displaystyle \text{Z}$ chromosome from their male parent and their $\displaystyle \text{W}$ chromosome from their female parent.

Cross 1: $\displaystyle \text{Z}^B\text{Z}^b \times \text{Z}^b\text{W}$ * The male parent produces gametes with $\displaystyle \text{Z}^B$ (probability $\displaystyle 0.5$ ) and $\displaystyle \text{Z}^b$ (probability $\displaystyle 0.5$ ).
* The female parent produces gametes with $\displaystyle \text{Z}^b$ (probability $\displaystyle 0.5$ ) and $\displaystyle \text{W}$ (probability $\displaystyle 0.5$ ).
* A female offspring receives the $\displaystyle \text{W}$ chromosome (probability $\displaystyle 0.5$ ). For her to be non-barred ( $\displaystyle \text{Z}^b\text{W}$ ), she must inherit $\displaystyle \text{Z}^b$ from the father (probability $\displaystyle 0.5$ ).
* Expected proportion of all offspring that are non-barred females = $\displaystyle 0.5 \times 0.5 = 0.25$ .

Cross 2: $\displaystyle \text{Z}^b\text{Z}^b \times \text{Z}^B\text{W}$ * The male parent produces only $\displaystyle \text{Z}^b$ gametes (probability $\displaystyle 1.0$ ).
* The female parent produces $\displaystyle \text{Z}^B$ gametes (probability $\displaystyle 0.5$ ) and $\displaystyle \text{W}$ gametes (probability $\displaystyle 0.5$ ).
* All female offspring receive $\displaystyle \text{Z}^b$ from the father and $\displaystyle \text{W}$ from the mother, giving genotype $\displaystyle \text{Z}^b\text{W}$ (non-barred female).
* Expected proportion of all offspring that are non-barred females = $\displaystyle 1.0 \times 0.5 = 0.50$ .

Therefore, the correct row is C.

Statement 1 is correct: Cell type P possesses a plasma membrane, nucleus, mitochondria, ribosomes, and a cell wall. This combination is characteristic of walled eukaryotes such as plant cells (with cellulose cell walls) and fungal cells (with chitin cell walls).

Statement 2 is incorrect: Cell type Q possesses a cell wall, plasma membrane, and ribosomes, but lacks a nucleus and mitochondria. This is characteristic of a prokaryotic cell (e.g. a bacterium). Bacteria possess 70S ribosomes and actively synthesise proteins. Many bacteria also carry out aerobic cellular respiration using electron transport chains embedded in their plasma membrane, or anaerobic respiration/fermentation, despite lacking mitochondria.

Statement 3 is correct: Cell type S possesses a plasma membrane but lacks a nucleus, mitochondria, ribosomes, and a cell wall. During maturation, mammalian erythrocytes (red blood cells) lose their nucleus, mitochondria, and ribosomes to maximise space for haemoglobin, leaving only a plasma membrane enclosing the cytoplasm.

Therefore, only statements 1 and 3 are correct.

Statement 1 is correct: In successful gene therapy, the introduced functional gene is expressed by the target cells (liver cells) through transcription into mRNA and translation into functional Factor VIII protein.

Statement 2 is incorrect: This is a form of somatic gene therapy. The genetic modification affects only somatic cells (liver cells) and does not alter the germline cells (spermatocytes) in the testes. Because the male patient still carries the mutant Factor VIII allele on the X chromosome of his germline cells, all of his X-bearing sperm will carry the mutant allele, which will be passed to his biological daughters.

Statement 3 is correct: Delivering a functional Factor VIII gene into the liver cells introduces additional nucleotide sequences into the DNA content of those cells, altering the total base sequence present.

Therefore, only statements 1 and 3 are correct.

1. Calculate the mean number of stomata per field of view: $\displaystyle \text{Total count} = 28 + 34 + 31 + 29 + 35 + 35 = 192$ $\displaystyle \text{Mean count per field} = \dfrac{192}{6} = 32$ 2. Calculate the mean stomatal density per $\displaystyle \text{mm}^2$ : $\displaystyle \text{Density} = \dfrac{32}{0.20\text{ mm}^2} = 160\text{ stomata mm}^{-2}$ 3. Convert the total leaf area from $\displaystyle \text{cm}^2$ to $\displaystyle \text{mm}^2$ : $\displaystyle 1\text{ cm} = 10\text{ mm} \implies 1\text{ cm}^2 = (10\text{ mm})^2 = 100\text{ mm}^2$ $\displaystyle \text{Total leaf area} = 15.0\text{ cm}^2 \times 100\text{ mm}^2\text{ cm}^{-2} = 1500\text{ mm}^2$ 4. Estimate the total number of stomata: $\displaystyle \text{Total stomata} = 160\text{ stomata mm}^{-2} \times 1500\text{ mm}^2 = 240\,000 = 2.4 \times 10^5$

From the graph, we extract the rate of reaction at both temperatures:
- At $\displaystyle 40 \, ^{\circ}\text{C}$ , the rate of reaction is $\displaystyle 50 \, \text{arbitrary units}$ .
- At $\displaystyle 60 \, ^{\circ}\text{C}$ , the rate of reaction is $\displaystyle 10 \, \text{arbitrary units}$ .

The percentage decrease between $\displaystyle 40 \, ^{\circ}\text{C}$ and $\displaystyle 60 \, ^{\circ}\text{C}$ is calculated using the initial rate at $\displaystyle 40 \, ^{\circ}\text{C}$ as the reference: $\displaystyle \text{Percentage decrease} = \dfrac{50 - 10}{50} \times 100\% = \dfrac{40}{50} \times 100\% = 80\%$ Biologically, enzymes are non-living protein molecules, so they do not "die". Instead, at temperatures above the optimum, the increased kinetic energy causes intramolecular vibrations that break hydrogen bonds and ionic interactions holding the tertiary structure together. This denatures the enzyme by permanently altering the conformation of the active site so that the substrate can no longer bind.

The question requires us to interpret ecological data, distinguishing between frequency, abundance, and distribution. In each zone, 20 quadrats were surveyed.

A: In Zone 1, the frequency of Species X is $\displaystyle \dfrac{16}{20} = 80\%$ , with a total abundance of 32. For Species Y, the frequency is $\displaystyle \dfrac{5}{20} = 25\%$ , with a total abundance of 120. Species X indeed has a higher frequency of occurrence ( $\displaystyle 80\% > 25\%$ ) despite a lower total abundance ( $\displaystyle 32 < 120$ ). This statement is correct.

To confirm, let's examine the other options.
B: In Zone 2, the frequency of Species X is $\displaystyle \dfrac{8}{20} = 40\%$ , while for Species Y it is $\displaystyle \dfrac{18}{20} = 90\%$ . So Species X has a lower frequency, making this statement incorrect.
C: The frequency of Species X in Zone 1 is $\displaystyle \dfrac{16}{20} = 80\%$ , not $\displaystyle 40\%$ .
D: A clumped distribution is characterised by high numbers of individuals in a few locations. Species Y is more clumped in Zone 1 (120 individuals across 5 quadrats) than in Zone 2 (45 individuals across 18 quadrats). The statement claims the reverse.
E: The mean density of Species Y in Zone 1 is the total individuals divided by the total quadrats: $\displaystyle \dfrac{120}{20} = 6.0$ . The value $\displaystyle 0.25$ is its frequency of occurrence, $\displaystyle \dfrac{5}{20}$ .

Statement 1 is incorrect: Mutations occur spontaneously and at random with respect to an organism's needs. The environment acts as a selective agent on pre-existing or randomly arising genetic variation; it does not induce specific, beneficial mutations to solve an environmental challenge.

Statement 2 is correct: Directional selection occurs when natural selection favors individuals at one extreme of a phenotypic distribution (here, larger beak depths better suited for cracking large, tough seeds), shifting the population mean over time.

Statement 3 is correct: For natural selection to cause evolutionary change across generations, the phenotypic variation must be heritable (have a genetic basis) so that individuals with higher fitness pass their advantageous alleles on to offspring.

Therefore, statements 2 and 3 only are correct.

1 is correct: As the root curves downwards, the upper outer surface ( $\displaystyle \text{X}$ ) forms the convex side with a larger arc length than the lower inner surface ( $\displaystyle \text{Y}$ ). For this curvature to occur, cells on the upper side ( $\displaystyle \text{X}$ ) must have elongated more than cells on the lower side ( $\displaystyle \text{Y}$ ), so cells at $\displaystyle \text{X}$ are longer.

2 is correct: When a root is oriented horizontally, gravity causes auxin transport to be directed towards the lower side of the root tip. As a result, auxin accumulates at $\displaystyle \text{Y}$ , making its concentration higher at $\displaystyle \text{Y}$ than at $\displaystyle \text{X}$ .

3 is correct: Unlike in shoots where high auxin concentrations promote elongation, root elongation cells are sensitive to auxin such that high concentrations inhibit elongation. The high concentration of auxin at $\displaystyle \text{Y}$ inhibits elongation of cells on the lower side, while cells at $\displaystyle \text{X}$ continue to elongate, resulting in positive gravitropism (downward curvature).

Therefore, statements 1, 2, and 3 are all correct.

The net rate of CO₂ uptake is the difference between the rate of photosynthesis (which consumes CO₂) and the rate of respiration (which produces CO₂).

Net CO₂ Uptake = Rate of Photosynthesis - Rate of Respiration

At a light intensity of 0, there is no photosynthesis. The table shows a net uptake of -2.0 mg dm⁻² h⁻¹, which means CO₂ is being released. This value represents the rate of respiration.

Rate of Respiration = 2.0 mg dm⁻² h⁻¹

As light intensity increases from 0 to 90 arbitrary units, the net CO₂ uptake increases. This indicates that in this range, light intensity is the limiting factor for the rate of photosynthesis.

However, between 90 and 120 arbitrary units, the net CO₂ uptake plateaus at 10.5 mg dm⁻² h⁻¹. Increasing the light intensity further does not increase the net uptake. This means that at 120 arbitrary units, light is no longer the limiting factor. The rate of photosynthesis is now limited by another factor that was held constant, such as CO₂ concentration or temperature. This directly supports conclusion D.

Let's examine the other options:

A: At a light intensity of 10, the net CO₂ uptake is 0. This is the light compensation point, where the rate of photosynthesis is equal to the rate of respiration (both are 2.0 mg dm⁻² h⁻¹). The rate of photosynthesis is not zero.

B: Between 30 and 60 units, the rate of uptake is clearly increasing as light intensity increases. Therefore, light is the limiting factor in this range, not CO₂ concentration.

C: The rate of respiration is assumed to be constant. The increase in net uptake is due to the rate of photosynthesis increasing with light intensity.

E: The maximum *net* rate of CO₂ uptake is 10.5 mg dm⁻² h⁻¹. The maximum *gross* rate of photosynthesis is the sum of the maximum net uptake and the rate of respiration: $\displaystyle 10.5 + 2.0 = 12.5$ mg dm⁻² h⁻¹.

Statement 1 is correct: Blood enters the capillary at position $\displaystyle \text{Y}$ (deoxygenated, lowest oxygen partial pressure) and flows towards position $\displaystyle \text{X}$ , picking up oxygen along the way so that it exits at $\displaystyle \text{X}$ with its highest oxygen partial pressure. Moving along the spatial axis from $\displaystyle \text{X}$ to $\displaystyle \text{Y}$ , the partial pressure of oxygen in the blood decreases.

Statement 2 is correct: A key feature of counter-current exchange is that blood leaving at $\displaystyle \text{X}$ is in diffusion contact with freshly arriving, oxygen-rich water at $\displaystyle \text{X}$ , allowing blood oxygen partial pressure to reach high values (e.g. ~80%). In contrast, water leaving at $\displaystyle \text{Y}$ has transferred most of its oxygen along the lamella (e.g. ~15-20%). Therefore, the oxygen partial pressure in blood leaving at $\displaystyle \text{X}$ can exceed that in water leaving at $\displaystyle \text{Y}$ .

Statement 3 is incorrect: At position $\displaystyle \text{Y}$ , the incoming deoxygenated blood has a lower oxygen partial pressure than the leaving water. Oxygen always diffuses down its partial pressure gradient, so net diffusion at $\displaystyle \text{Y}$ is from the water into the blood.

The condition is X-linked recessive. The mother does not have the condition, so she can be an unaffected carrier ( $\displaystyle X^A X^a$ ). The father does not have the condition, so his genotype is $\displaystyle X^A Y$ .

The child has only one copy of the gene in every somatic cell, and this single copy is the mutant allele ( $\displaystyle X^a$ ).

- Statement 1 is correct: If non-disjunction of the sex chromosomes occurs during meiosis in the father, a sperm may carry no sex chromosome (nullisomic). If this sperm fertilises an ovum carrying the maternal $\displaystyle X^a$ allele, the zygote will have a single X chromosome ( $\displaystyle X^a O$ ). Because every somatic cell arises from this zygote by mitosis, all cells will have only one copy of the gene (the mutant allele $\displaystyle X^a$ ), leading to expression of the condition.

- Statement 2 is correct: If a chromosomal deletion removes this gene locus from the father's X chromosome during spermatogenesis, the sperm will carry an X chromosome lacking this gene. When this sperm fertilises an ovum carrying $\displaystyle X^a$ , the child will have one functional mutant allele and one deleted copy ( $\displaystyle X^a X^{\Delta}$ ). Thus, all cells will contain only one copy of the gene, which is the mutant allele.

- Statement 3 is incorrect: A male inherits his X chromosome from his mother (the paternal grandmother) and his Y chromosome from his father (the paternal grandfather). Therefore, a father cannot inherit an X-linked allele from his father, nor can he pass it on.

Therefore, statements 1 and 2 only could explain the observation.

In Environment 1, the allele *a* is maintained at a stable frequency (balanced polymorphism) because the heterozygote *Aa* has a significant survival advantage (0.95) over the homozygote *AA* (0.60), which offsets the lethal loss of *aa* individuals.

In Environment 2, the survival of *AA* increases to 0.95, becoming equal to that of *Aa*. Consequently, there is no longer a heterozygote advantage to preserve the *a* allele. However, the *aa* genotype remains lethal (0.00).

Since *a* alleles are constantly removed from the gene pool via the death of *aa* individuals, and there is no longer a reproductive advantage for *Aa* carriers to compensate for this loss, the frequency of allele *a* will steadily decrease over time (directional selection).

1. Total number of hydrogen bonds in the DNA segment:
- The DNA segment is 300 base pairs long, meaning it contains $\displaystyle 300 \times 2 = 600$ total bases.
- Thymine (T) accounts for $\displaystyle 36\%$ of the bases: $\displaystyle 0.36 \times 600 = 216\text{ T bases}$ .
- By complementary base pairing in double-stranded DNA, the number of adenine (A) bases equals the number of thymine bases, so there are $\displaystyle 216\text{ A-T base pairs}$ .
- The remaining base pairs are guanine-cytosine (G-C) pairs: $\displaystyle 300 - 216 = 84\text{ G-C base pairs}$ .
- Each A-T pair has 2 hydrogen bonds, and each G-C pair has 3 hydrogen bonds.
- Total hydrogen bonds $\displaystyle = (216 \times 2) + (84 \times 3) = 432 + 252 = 684$ .

2. Number of peptide bonds in the polypeptide:
- The mRNA has 300 nucleotides, which corresponds to $\displaystyle 300 / 3 = 100\text{ codons}$ .
- The final codon is a stop codon, which signals the termination of translation and does not code for an amino acid.
- Therefore, the polypeptide consists of $\displaystyle 100 - 1 = 99\text{ amino acids}$ .
- In a linear polypeptide containing $\displaystyle n$ amino acids, there are $\displaystyle n - 1$ peptide bonds: $\displaystyle 99 - 1 = 98\text{ peptide bonds}$ .

Thus, row C is correct.

Let us break the sequence into triplets (codons):
- Codon 1 (bases 1–3): $\displaystyle \text{AUG}$ (Met)
- Codon 2 (bases 4–6): $\displaystyle \text{UUA}$ (Leu)
- Codon 3 (bases 7–9): $\displaystyle \text{GAA}$ (Glu) — base 8 is mutated
- Codon 4 (bases 10–12): $\displaystyle \text{GGA}$ (Gly)
- Codon 5 (bases 13–15): $\displaystyle \text{UCG}$ (Ser)
- Codon 6 (bases 16–18): $\displaystyle \text{UGG}$ (Trp) — base 17 is mutated
- Codon 7 (bases 19–21): $\displaystyle \text{GAG}$ (Glu)
- Codon 8 (bases 22–24): $\displaystyle \text{CAA}$ (Gln)
- Codon 9 (bases 25–27): $\displaystyle \text{UAA}$ (Stop)

**Mutation at base 8 (Codon 3: $\displaystyle \text{G\_A}$ ):**
From the table, the only alternative codon starting with $\displaystyle \text{G}$ and ending with $\displaystyle \text{A}$ is $\displaystyle \text{GGA}$ (Gly). Thus, codon 3 must now code for Gly.

**Mutation at base 17 (Codon 6: $\displaystyle \text{U\_G}$ ):**
From the table, the possible alternative codons starting with $\displaystyle \text{U}$ and ending with $\displaystyle \text{G}$ are:
- $\displaystyle \text{UAG}$ (Stop)
- $\displaystyle \text{UCG}$ (Ser)
- $\displaystyle \text{UUG}$ (Leu)

Evaluating statement 1:
If base 17 mutates to $\displaystyle \text{A}$ , codon 6 becomes $\displaystyle \text{UAG}$ (Stop codon). Translation terminates, producing a polypeptide with 5 amino acids (Met-Leu-Gly-Gly-Ser). Statement 1 is correct.

Evaluating statement 2:
In this truncated 5-amino-acid polypeptide (Met-Leu-Gly-Gly-Ser), there are 4 distinct types of amino acids (Met, Leu, Gly, Ser). Statement 2 is correct.

Evaluating statement 3:
If base 17 mutates to $\displaystyle \text{C}$ (Ser) or $\displaystyle \text{U}$ (Leu), translation proceeds to codon 9 (Stop), producing an 8-amino-acid polypeptide:
- With Ser at position 6: Met, Leu, Gly, Gly, Ser, Ser, Glu, Gln $\displaystyle \rightarrow 6$ unique types (Met, Leu, Gly, Ser, Glu, Gln).
- With Leu at position 6: Met, Leu, Gly, Gly, Ser, Leu, Glu, Gln $\displaystyle \rightarrow 6$ unique types (Met, Leu, Gly, Ser, Glu, Gln).
It is impossible to produce 7 different types of amino acids. Statement 3 is incorrect.

Therefore, only statements 1 and 2 are correct (option E).

The data requires analysing two opposing control loops:

1. Glucose Tolerance Test (High Glucose stimulus): The mutant responds identically to the wild-type, returning high glucose to baseline normally. This indicates the system for lowering blood glucose (insulin secretion and receptor binding) is fully functional. This rules out defects in insulin synthesis (A) or insulin receptor function.
2. Fasting Test (Low Glucose stimulus): The mutant fails to maintain blood glucose, allowing it to drop dangerously low. This indicates a failure in the system for raising blood glucose (glucagon secretion and glycogenolysis).

Option C is incorrect because if insulin were constantly high, the clearance of the glucose load would likely be accelerated, or the baseline would be fundamentally unstable; the specific failure to *raise* glucose during fasting, combined with a normal ability to *lower* it after a meal, points directly to a defect in the glucagon pathway (Option D).

1. Variation 1 (doubling catalase volume / concentration): Doubling the enzyme concentration doubles the number of available active sites, thereby doubling the initial rate of reaction (steeper initial slope). Because the total amount of hydrogen peroxide is unchanged ( $\displaystyle 10\text{ cm}^3$ of $\displaystyle 2.0\%$ ), the final yield of $\displaystyle \ce{O2}$ remains $\displaystyle 40\text{ cm}^3$ . Thus, Variation 1 corresponds to curve P.

2. Variation 2 (doubling substrate volume at constant concentration): The substrate concentration is unchanged ( $\displaystyle 2.0\%$ ), so the initial rate is the same as the control. However, there are twice as many moles of $\displaystyle \ce{H2O2}$ available, yielding twice the total volume of oxygen ( $\displaystyle 80\text{ cm}^3$ ). Thus, Variation 2 corresponds to curve Q.

3. **Variation 3 ( $\displaystyle 75\text{ }^\circ\text{C}$ ):** At $\displaystyle 75\text{ }^\circ\text{C}$ , thermal energy disrupts hydrogen and ionic bonds stabilizing the tertiary structure of catalase, causing rapid denaturation of the active site. Catalytic activity ceases almost immediately before all substrate can react, causing the curve to level off prematurely at a very low yield ( $\displaystyle 10\text{ cm}^3$ ). Thus, Variation 3 corresponds to curve R.

4. **Variation 4 ( $\displaystyle 10\text{ }^\circ\text{C}$ ):** At $\displaystyle 10\text{ }^\circ\text{C}$ , molecules have lower kinetic energy, resulting in a lower frequency of successful collisions between enzyme active sites and substrate molecules. The rate is slower throughout (shallower initial gradient), but because the enzyme is not denatured, all substrate is eventually converted to reach the full $\displaystyle 40\text{ cm}^3$ . Thus, Variation 4 corresponds to curve S.

Hence, the correct row is A.

1. Identify the equilibrium point: The table shows that the tissue has $\displaystyle 0\%$ change in mass in the solution with a water potential of $\displaystyle -0.8 \text{ MPa}$ .
2. Determine cell water potential: At this equilibrium, the water potential of the tissue ( $\displaystyle \Psi_{\text{cell}}$ ) is equal to that of the external solution. Thus, $\displaystyle \Psi_{\text{cell}} = -0.8 \text{ MPa}$ .
3. Apply the water potential equation: The relationship is $\displaystyle \Psi_{\text{cell}} = \Psi_s + \Psi_p$ .
4. Solve for solute potential:

The listed features (circular DNA, no nuclear envelope, 70S ribosomes, lack of membrane-bound organelles, cell wall) identify the organism as a prokaryote (such as a bacterium).

- Statement 1 is incorrect: Although it lacks mitochondria, many prokaryotes are capable of aerobic respiration. They use electron transport chains and respiratory enzymes embedded in their cell surface membrane to carry out oxidative phosphorylation and synthesise ATP.
- Statement 2 is correct: Ribosomes consist of ribosomal RNA (rRNA) and proteins. In order to produce ribosomes, the organism's DNA genome must contain genes that code for rRNA molecules.
- Statement 3 is correct: Prokaryotes divide asexually by binary fission. Mitosis is a process of nuclear division unique to eukaryotes that involves chromosome condensation, spindle apparatus formation, and nuclear membrane breakdown.

Therefore, only statements 2 and 3 are correct (option G).