2024 Cambridge IAL Biology (9700) 模擬試題連答案詳解

In the loop of Henle, the descending limb is highly permeable to water but impermeable to sodium and chloride ions. Water moves out of the descending limb by osmosis into the highly concentrated tissue fluid of the medulla, concentrating the filtrate inside the tubule. Conversely, the thick ascending limb is completely impermeable to water but allows the movement of sodium and chloride ions. In the thick region of the ascending limb, sodium and chloride ions are actively transported out into the medullary tissue fluid to maintain the high solute concentration of the medulla.

1 mark: Correctly identifies the water permeability, ion permeability, and active transport properties of both limbs as described in option A.

White-flowered plants are homozygous recessive (\(rr\)). When they are eliminated, all the remaining plants have red flowers and have either the genotype \(RR\) or \(Rr\). Because the heterozygotes (\(Rr\)) survive, they still carry the recessive \(r\) allele. When they reproduce, some of their offspring will receive the \(r\) allele. Therefore, while the overall frequency of the \(r\) allele in the population will decrease, it will not drop to zero.

1 mark: Correctly identifies that the allele frequency of the recessive allele decreases but remains above zero due to heterozygous carriers (B).

DCPIP is a blue dye that acts as an electron acceptor in the Hill reaction. When it is reduced by electrons originating from the photolysis of water during the light-dependent stage of photosynthesis, it becomes colorless, resulting in a decrease in light absorbance at 600 nm. Tube 1 has no light, so no photolysis occurs and DCPIP is not reduced. Tube 2 has chloroplasts and light, so photolysis occurs, electrons are transferred to DCPIP, reducing it, and absorbance decreases. Tube 3 has an inhibitor of DCPIP reduction, so no reduction occurs. Tube 4 has no chloroplasts, so no photolysis occurs. Therefore, only Tube 2 will show a significant decrease in absorbance.

1 mark: Correctly identifies that only Tube 2 will show a decrease in absorbance because it has all necessary components for the light-dependent reaction to proceed (A).

Using two different restriction enzymes that produce different ends (one sticky, one blunt) is known as directional cloning. It prevents the digested plasmid vector from self-ligating (recircularising) because the sticky end created by *Bam*HI cannot base-pair with the blunt end created by *Sma*I. It also ensures the insert is integrated in the correct orientation because the *Bam*HI-digested end of the insert can only ligate to the *Bam*HI-digested end of the plasmid, and the *Sma*I-digested end can only ligate to the *Sma*I-digested end of the plasmid.

1 mark: Correctly identifies the dual benefits of preventing plasmid self-ligation and ensuring correct gene orientation (B).

In cyclic photophosphorylation, electrons from PSI are passed back to the electron transport chain and return to the reaction center of PSI. Thus, no external source of electrons (water) is needed, meaning photolysis of water does not occur. In non-cyclic photophosphorylation, electrons from PSII are passed to PSI and eventually to NADP. The electrons lost from PSII must be replaced by electrons from the photolysis of water, which also produces oxygen as a waste product.

1 mark: Correctly identifies that cyclic photophosphorylation does not involve water photolysis while non-cyclic does (C).

When glucagon binds to its receptor on the hepatocyte membrane, it induces a conformational change that activates a G-protein. The active G-protein then activates the enzyme adenylyl cyclase. Adenylyl cyclase catalyses the conversion of ATP to cyclic AMP (cAMP), which acts as a second messenger. cAMP binds to and activates protein kinase A (PKA). PKA triggers an enzyme cascade that ultimately activates glycogen phosphorylase, the enzyme responsible for glycogenolysis (the breakdown of glycogen to glucose).

1 mark: Correctly identifies the step-by-step intracellular signaling cascade triggered by glucagon (A).

Because the two populations live in the same geographical area, speciation is sympatric (not allopatric, which requires geographical isolation). Because the two populations reproduce at different times of the year (early spring vs. mid-summer), the reproductive barrier is temporal isolation.

1 mark: Correctly identifies sympatric speciation and temporal reproductive isolation (B).

Individual X has three bands (1.3 kb, 1.1 kb, and 0.2 kb), which indicates they have both the normal allele (\(Hb^A\)) and the mutant allele (\(Hb^S\)). Thus, X is heterozygous (\(Hb^A Hb^S\)) and is a carrier of the sickle cell trait. Individual Y has only the 1.3 kb band, which indicates they only have the mutant allele. Thus, Y is homozygous recessive (\(Hb^S Hb^S\)) and has sickle cell anaemia. Individual Z has only the 1.1 kb and 0.2 kb bands, indicating they only have the normal allele. Thus, Z is homozygous normal (\(Hb^A Hb^A\)) and is unaffected.

1 mark: Correctly interprets the genotypes and phenotypes of individuals X, Y, and Z based on the restriction fragment lengths (B).

1. Initially, the population is in Hardy-Weinberg equilibrium. The frequency of the white phenotype (\(C^W C^W\)) is \(q^2 = 0.09\), so \(q (C^W) = 0.3\) and \(p (C^R) = 0.7\).
2. The genotype frequencies in the initial population are:
- \(C^R C^R = p^2 = 0.49\) (Red)
- \(C^R C^W = 2pq = 2 \times 0.7 \times 0.3 = 0.42\) (Pink)
- \(C^W C^W = q^2 = 0.09\) (White)
3. Selective herbivory removes all red-flowered plants (\(C^R C^R\)), leaving only \(C^R C^W\) and \(C^W C^W\) to reproduce.
4. The total surviving population frequency is \(0.42 + 0.09 = 0.51\).
5. The new frequency of the \(C^W\) allele (\(q'\)) among the survivors is calculated as:
\[q' = \frac{\text{frequency of } C^W C^W + 0.5 \times \text{frequency of } C^R C^W}{\text{total surviving frequency}}\]
\[q' = \frac{0.09 + 0.5 \times 0.42}{0.51} = \frac{0.30}{0.51} \approx 0.59\]

Award 1 mark for the correct option C.
- Correct identification of starting allele frequencies (p=0.7, q=0.3): 0.5 marks
- Correct recalculation of allele frequency after removing the red genotype: 0.5 marks

When blood glucose levels rise, glucose enters the pancreatic \(\beta\)-cells via GLUT2 transporter proteins by facilitated diffusion. The glucose is metabolised via glycolysis and respiration, increasing the concentration of ATP inside the cell. The high ATP-to-ADP ratio causes ATP-sensitive potassium (\(\text{K}^+\)) channels to close. As potassium ions can no longer leave the cell, the membrane depolarises. This depolarisation triggers the opening of voltage-gated calcium (\(\text{Ca}^{2+}\)) channels, allowing an influx of calcium ions into the cell. The increased intracellular calcium concentration stimulates the exocytosis of insulin-containing vesicles.

Award 1 mark for option A.
- Reject B because GLUT4 is insulin-dependent and not the primary entry mechanism in pancreatic beta-cells, and potassium channels must close, not open.
- Reject C and D due to incorrect directions of channel activities and potential changes.

At high light intensities (distances less than 10 cm), light is no longer the factor limiting the rate of photosynthesis (Statement 1 is correct). Instead, the process is limited by other environmental factors such as carbon dioxide concentration or temperature, which limit the rate of enzyme-controlled reactions in the light-independent stage (Statement 2 is correct). Extremely high light intensity in standard laboratory experiments does not denature enzymes, which is a process caused by heat, not light energy directly (Statement 3 is incorrect).

Award 1 mark for option A (1 and 2 only).

PCR primers must bind to the \(3'\) ends of the target DNA strands so that DNA polymerase can synthesise new strands in the \(5'\) to \(3'\) direction.
- The first strand is: \(5'\)- A T G C G T ... A A G C T A -\(3'\). The primer for this strand must be complementary to the \(3'\) end (\(5'\)- A A G C T A -\(3'\)). Its complementary sequence is \(3'\)- T T C G A T -\(5'\), which written from \(5'\) to \(3'\) is \(5'\)- T A G C T T -\(3'\).
- The complementary strand runs antiparallel: \(3'\)- T A C G C A ... T T C G A T -\(5'\). Its \(3'\) end is \(3'\)- T A C G C A -\(5'\). The primer for this strand must be complementary to this \(3'\) end, which is \(5'\)- A T G C G T -\(3'\).
Therefore, the correct primers are \(5'\)- A T G C G T -\(3'\) and \(5'\)- T A G C T T -\(3'\).

Award 1 mark for option A.
- Primers must bind antiparallel to the template strands and extend in the 5' to 3' direction.

In non-cyclic photophosphorylation, photolysis of water acts as the ultimate electron source, producing protons, electrons, and oxygen gas as a waste product. The electrons flow through photosystem II (PSII) and photosystem I (PSI) along an electron transport chain, which generates a proton gradient to drive ATP synthesis. The final electron acceptor at the end of this pathway is \(\text{NADP}^+\), which is reduced to form reduced NADP.

Award 1 mark for option A.
- B is incorrect because PSI is an intermediate electron carrier, not the final acceptor.
- C is incorrect because PSII is not the ultimate donor (water is).
- D is incorrect because reduced NADP is a product, not a donor.

Both extreme phenotypes (the largest/most active and the smallest/slowest lizards) are selected against by the two predators. The intermediate phenotype (medium-sized lizards) has the highest fitness and is selected for. This is a classic description of stabilising selection. Over time, the mean value of the trait (lizard size) remains constant, but the variation around the mean decreases, resulting in a narrower distribution with a decreased standard deviation.

Award 1 mark for option A.
- Disruptive selection (B) would select for both extremes and against the mean.
- Directional selection (C and D) occurs when one extreme phenotype is favoured over another.

Glucose is fully filtered at the glomerulus (concentration in filtrate is equal to that in blood plasma: 1.0 \(\text{g dm}^{-3}\)). It is completely absent from urine (0.0 \(\text{g dm}^{-3}\)), which indicates that 100% of the glucose is reabsorbed back into the blood. This selective reabsorption occurs in the proximal convoluted tubule via secondary active transport (co-transport with sodium ions) and facilitated diffusion. Therefore, option A is correct.

Award 1 mark for option A.
- B is incorrect because large proteins are not freely filtered (filtrate concentration is almost zero).
- C is incorrect because urea is concentrated due to the passive removal of water from the tubules, not active transport.
- D is incorrect because sodium ions are highly reabsorbed (the concentration rises because water is reabsorbed at a higher proportional rate).

To analyze gene expression (which genes are active and transcribed), mRNA is extracted from both types of tissue. Since mRNA is unstable and single-stranded, it is converted into complementary DNA (cDNA) using the enzyme reverse transcriptase. The cDNA from the cancer cells is labelled with a fluorescent dye of one colour (e.g., red) and the healthy cell cDNA with a different colour (e.g., green). Both samples are mixed and hybridised to the microarray slide. The fluorescence pattern indicates which genes are expressed in which tissue.

Award 1 mark for option A.
- B is incorrect because mRNA is not directly hybridised to microarrays in standard expression analysis, nor is reverse transcriptase used on the slide.
- C is incorrect because PCR of mRNA requires RT-PCR, and restriction enzymes and gel electrophoresis are not part of microarray hybridisation preparation.
- D is incorrect because it describes gene cloning, not gene expression profiling.

Glucose is filtered out of the blood in the glomerulus and enters the nephron. Under normal physiological conditions, 100% of the glucose is reabsorbed back into the blood from the proximal convoluted tubule (PCT) via sodium-glucose co-transporter proteins (SGLTs). However, these carrier proteins have a maximum transport capacity (\(T_m\)). When the blood glucose concentration is extremely high, the concentration of glucose in the filtrate also increases drastically, exceeding the capacity of these co-transporter proteins. The proteins become fully saturated, meaning they cannot transport any more glucose, and the excess unabsorbed glucose remains in the filtrate and is excreted in the urine.

1 mark for identifying that co-transporter proteins in the proximal convoluted tubule become saturated (Option B).

During photophosphorylation, the energy from electron transport is used to pump protons (\(H^+\)) across the thylakoid membrane from the stroma into the thylakoid space (lumen), creating a high concentration of protons inside. Protons then diffuse back down their electrochemical gradient from the thylakoid space to the stroma through the channel within ATP synthase. This movement of protons (chemiosmosis) drives the phosphorylation of ADP to ATP.

1 mark for identifying the correct direction of proton pumping, diffusion, and site of ATP synthesis (Option A).

At low light intensity, light is the limiting factor for all curves because the rate of photosynthesis is directly proportional to light intensity. At high light intensity, the rate of photosynthesis levels off because some other factor becomes limiting. For Curve 2 (0.04% \(\text{CO}_2\) at \(25^\circ\text{C}\)), the rate is lower than Curve 3 (0.15% \(\text{CO}_2\) at \(25^\circ\text{C}\)). Since the temperature is the same but the concentration of carbon dioxide is lower, increasing the carbon dioxide concentration increases the rate. Therefore, carbon dioxide concentration is the limiting factor for Curve 2 at high light intensity.

1 mark for correctly identifying carbon dioxide concentration as the limiting factor for Curve 2 under high light intensity (Option B).

During PCR: (1) Heating to \(95^\circ\text{C}\) denatures the double-stranded DNA template by breaking hydrogen bonds between complementary base pairs, separating it into single strands. (2) Cooling to \(55^\circ\text{C}\) allows short DNA primers to anneal (form hydrogen bonds) to complementary target sequences on the single strands. (3) Heating to \(72^\circ\text{C}\) provides the optimum temperature for the thermostable Taq polymerase to synthesise a new complementary DNA strand by extending from the primers.

1 mark for assigning the correct functions to all three temperatures in the PCR cycle (Option A).

Disruptive selection occurs when environmental conditions favour individuals at both phenotypic extremes (very light shells and very dark shells) over individuals with intermediate phenotypes. This leads to a bimodal distribution and an overall increase in the phenotypic variance within the population.

1 mark for identifying disruptive selection and an increase in phenotypic variance (Option B).

ADH is a peptide hormone that cannot cross the phospholipid bilayer of the target cell's membrane. It binds to specific receptors (GPCRs) on the basolateral cell-surface membrane of the collecting duct cells. This activates a G-protein, which in turn activates the enzyme adenylyl cyclase. Adenylyl cyclase converts ATP to cyclic AMP (cAMP), a second messenger. cAMP activates a protein kinase cascade that causes vesicles containing aquaporin water channels to move towards and fuse with the luminal (apical) membrane of the cells, increasing their permeability to water.

1 mark for outlining the correct signal transduction sequence from GPCR to aquaporin insertion at the luminal membrane (Option A).

Speciation has occurred because the \(4n\) and \(2n\) plants are reproductively isolated: they can no longer interbreed to produce fertile, viable offspring (their triploid offspring are sterile, which is a post-zygotic barrier). Since this reproductive isolation developed in the same physical locality (the meadow) without geographic separation, it is an example of sympatric speciation.

1 mark for identifying that sympatric speciation has occurred and that the sterile triploids represent a post-zygotic barrier (Option A).

EcoRI cuts the plasmid within the promoter region of the antibiotic resistance gene. Insertion of the human gene at this site will disrupt the sequence of the promoter. Because the promoter is disrupted, RNA polymerase cannot bind to initiate transcription of the antibiotic resistance gene, and therefore the gene will not be expressed. The transformed bacteria containing recombinant plasmids will be sensitive to the antibiotic, not resistant. Option A is incorrect because EcoRI produces sticky ends, not blunt ends. Option C is incorrect because transcription cannot occur without a functional promoter. Option D is incorrect because restriction with a single enzyme (EcoRI) allows the gene to insert in either orientation.

1 mark for identifying that insertion at the EcoRI site disrupts the promoter, preventing expression of the resistance gene (Option B).

To find the allele frequency of \(W\) in the surviving population:
1. First, find the total number of surviving individuals: \(80\text{ (red)} + 200\text{ (pink)} + 40\text{ (white)} = 320\text{ individuals}\).
2. Since each individual is diploid, the total number of alleles in the gene pool is \(320 \times 2 = 640\).
3. Next, calculate the number of \(W\) alleles in the survivors:
- Pink-flowered survivors (\(RW\)) each have 1 \(W\) allele: \(200 \times 1 = 200\).
- White-flowered survivors (\(WW\)) each have 2 \(W\) alleles: \(40 \times 2 = 80\).
- Total \(W\) alleles = \(200 + 80 = 280\).
4. Finally, calculate the frequency of the \(W\) allele: \(\frac{280}{640} = 0.4375 \approx 0.44\).

Award 1 mark for the correct selection of option B.
- Reject other options due to calculation errors (e.g., option A is a miscalculation, option D is the frequency of the \(R\) allele).

When ADH secretion increases, ADH binds to receptor proteins on the cell surface membrane of collecting duct cells. This activates a signaling cascade that causes vesicles containing aquaporins to fuse with the luminal (apical) membrane, inserting aquaporins. This increases the permeability of the collecting duct to water, allowing more water to be reabsorbed by osmosis into the hypertonic medulla and back into the blood. Consequently, blood water potential increases (becomes less negative / returns to normal). Since more water is reabsorbed, a smaller volume of more concentrated urine is produced, which increases the concentration of urea in the urine.

Award 1 mark for identifying that ADH causes insertion of aquaporins, increases blood water potential, and increases urine urea concentration (Option B).

At 0.3% \(\text{NaHCO}_3\) under low light intensity, the rate of photosynthesis has already levelled off. Increasing the carbon dioxide source (\(\text{NaHCO}_3\)) further does not increase the rate. However, when the light intensity is increased to 'high', the rate of photosynthesis increases significantly. This indicates that under low light conditions, light intensity is the factor limiting the rate of photosynthesis at this concentration.

Award 1 mark for identifying Option C as the correct interpretation of limiting factors from the experimental results.

Restriction endonucleases cut DNA at specific recognition sites, often leaving sticky ends (unpaired nucleotides) that can pair with complementary sticky ends on another DNA molecule. DNA ligase catalyses the formation of covalent phosphodiester bonds between adjacent nucleotides, linking the sugar-phosphate backbones. A marker gene (such as an antibiotic resistance gene or a fluorescent protein gene) is used to identify host bacterial cells that have successfully taken up the plasmid during transformation.

Award 1 mark for identifying Row A as containing the correct roles for all three components.

The photolysis of water takes place on the inner side of the thylakoid membrane, releasing protons into the thylakoid lumen. The electrons lost from the reaction centre of PSI (P700) during photoactivation are replaced by electrons originating from Photosystem II (PSII), which are passed along an electron transport chain. The final electron acceptor at the end of this non-cyclic pathway is \(\text{NADP}^+\), which is reduced to NADPH (reduced NADP) in the stroma.

Award 1 mark for identifying the correct locations and pathways corresponding to Row A.

Natural selection operates on pre-existing genetic variation. Mutations arise randomly and spontaneously, independent of the presence of selective agents such as antibiotics. When an antibiotic is introduced, it acts as a selective agent: individuals that already possess a mutation conferring resistance have a selective advantage, enabling them to survive, reproduce, and pass on the resistance allele to their offspring, thereby increasing the allele frequency over generations.

Award 1 mark for selecting Option B.
- Reject A and D as antibiotics do not induce specific directed mutations.
- Reject C because antibiotics do not increase mutation rates to cause adaptation.

When blood glucose concentration falls, the change is detected by \(\alpha\) cells in the islets of Langerhans of the pancreas, which respond by secreting glucagon. Glucagon is a peptide hormone, so it cannot cross cell membranes. It binds to specific G-protein-coupled receptors on the cell surface membrane of liver cells (hepatocytes). This binding activates a G-protein, which in turn activates the enzyme adenylyl cyclase. Adenylyl cyclase catalyses the conversion of ATP to cyclic AMP (cAMP), which acts as a second messenger to initiate a cascade that activates glycogen phosphorylase, promoting glycogenolysis.

Award 1 mark for identifying Option C.
- Reject A: glucagon is secreted by \(\alpha\) cells, not \(\beta\) cells.
- Reject B: glucagon binds to external cell-surface receptors, it does not enter the cytoplasm of the hepatocytes.
- Reject D: insulin secretion decreases, not increases.

The first step of each cycle of PCR is denaturation, which occurs at about \(90-95^\circ\text{C}\). This high temperature breaks the hydrogen bonds between the complementary base pairs of the double-stranded DNA template, separating it into single strands. Annealing occurs at around \(50-65^\circ\text{C}\) (not \(72^\circ\text{C}\)), extension/elongation occurs at \(72^\circ\text{C}\) with Taq polymerase adding nucleotides to the 3' end of DNA primers (not 5'), and primers are short single-stranded DNA molecules, not RNA.

Award 1 mark for choosing Option A.
- Reject B: \(72^\circ\text{C}\) is the temperature for elongation, not annealing.
- Reject C: \(55^\circ\text{C}\) is the annealing temperature, and nucleotides are added to the 3' end, not the 5' end.
- Reject D: Primers used in PCR are DNA, not RNA molecules.

(a) A selective agent is an environmental factor (biotic or abiotic) that influences the survival and reproductive success of individuals within a population. In this scenario, the primary selective agent is the low partial pressure of oxygen (\(p\text{O}_2\)) or hypoxia.

(b) Random mutations occurred in the \(\alpha\)- and \(\beta\)-globin genes, producing new alleles that code for high-affinity hemoglobin. In high-altitude environments, the low partial pressure of oxygen acts as a powerful selection pressure. Individuals possessing the high-affinity alleles are better able to load oxygen at low partial pressures, reducing hypoxia and maintaining metabolic efficiency. Consequently, these individuals have a survival advantage over those with low-affinity alleles, making them more likely to survive to reproductive age, reproduce, and pass on the advantageous alleles to their offspring. Over many generations, this differential reproductive success causes the frequency of the high-affinity hemoglobin alleles to increase within the population.

(c) The type of selection is directional selection. This is because the environmental conditions shifted (from low altitude to high altitude, which is a new selection pressure), favoring one extreme phenotype (highly increased oxygen affinity) over the ancestral mean. This causes the average phenotype of the population to shift in a single direction over time.

(a) Max 2 marks:
- Definition of selective agent: environmental factor / pressure that influences survival/reproduction of organisms. [1]
- Identification of agent: low partial pressure of oxygen / low oxygen level / hypoxia (Reject: "altitude" alone). [1]

(b) Max 5 marks:
- Spontaneous/random mutation in globin genes (\(\alpha\) or \(\beta\)) occurs to produce new alleles. [1]
- Low oxygen level acts as the selection pressure. [1]
- Individuals with high-affinity hemoglobin can bind oxygen more effectively in hypoxic conditions. [1]
- These individuals have a survival advantage / are more likely to survive. [1]
- Successful individuals reproduce and pass on high-affinity alleles to offspring. [1]
- Frequency of high-affinity alleles increases in the gene pool over generations. [1]

(c) Max 3 marks:
- Directional selection. [1]
- Explanation: Environment changed / selection pressure acts against one extreme and favors the other extreme. [1]
- Result: Mean phenotype of the population shifts towards high-affinity hemoglobin. [1]

(a) Sympatric speciation occurs within the same geographical location without physical barriers. Ecological niche differentiation takes place as different sub-populations exploit different food resources (benthic vs limnetic). Disruptive selection acts on the population, favoring the extreme phenotypes (well-adapted bottom feeders and well-adapted open-water feeders) over intermediate phenotypes. Over time, assortative mating occurs where benthic fish prefer to mate with other benthic fish, and limnetic with limnetic. This reduces gene flow between the groups. Distinct genetic variations accumulate due to natural selection and genetic drift, eventually leading to the formation of two reproductively isolated species.

(b) Reproductive isolation prevents gene flow and genetic mixing between the two diverging populations. This allows each population's gene pool to evolve independently of the other. Consequently, distinct mutations, allele frequency changes, and adaptations accumulate in response to different selection pressures, ensuring that differences are maintained and speciation is completed.

(c) Pre-zygotic isolating mechanisms prevent fertilization or the formation of a zygote (e.g., habitat, behavioral, or temporal barriers), whereas post-zygotic isolating mechanisms occur after fertilization and reduce the viability or fertility of the hybrid offspring (e.g., hybrid inviability or hybrid sterility). An example of a pre-zygotic mechanism is behavioral isolation, where different courtship displays prevent interbreeding.

(a) Max 5 marks:
- Sympatric speciation occurs in the same geographic area / without geographic barrier. [1]
- Ecological separation / niche differentiation (benthic vs. limnetic). [1]
- Disruptive selection acts against intermediate phenotypes. [1]
- Assortative mating / behavioral barriers decrease gene flow between the sub-populations. [1]
- Independent accumulation of mutations / different selection pressures act on each group. [1]
- Development of reproductive isolation. [1]

(b) Max 3 marks:
- Prevents gene flow / genetic recombination between populations. [1]
- Keeps gene pools separate / allows independent evolution of gene pools. [1]
- Allows different alleles/mutations to accumulate. [1]
- Solidifies species boundaries even if populations overlap. [1]

(c) Max 2 marks:
- Difference: Pre-zygotic prevents mating/fertilization, whereas post-zygotic occurs after fertilization (affects hybrid viability/fertility). [1]
- Example: Behavioral isolation (courtship) / ecological isolation (different habitats within lake) / temporal isolation (different breeding times). [1]

(a) A decrease in blood glucose concentration is detected by the \(\alpha\) (alpha) cells located in the islets of Langerhans in the pancreas. When blood glucose is low, less glucose enters the \(\alpha\) cells, which triggers cell depolarization. This depolarization stimulates the exocytosis of glucagon from vesicles within the \(\alpha\) cells directly into the bloodstream.

(b) Glucagon binds to specific, complementary G-protein-coupled receptors (GPCRs) on the external surface of the hepatocyte cell membrane. This binding changes the conformation of the receptor, activating an associated G-protein. The active G-protein stimulates the membrane-bound enzyme adenylyl cyclase. Adenylyl cyclase converts ATP into cyclic AMP (cAMP), which acts as a second messenger. cAMP binds to and activates protein kinase A (PKA), which then initiates a phosphorylation cascade that activates glycogen phosphorylase kinase, which in turn phosphorylates and activates glycogen phosphorylase.

(c) Active glycogen phosphorylase catalyzes glycogenolysis, which is the breakdown of stored intracellular glycogen into glucose-1-phosphate (subsequently converted to glucose). This glucose is transported out of hepatocytes into the bloodstream down its concentration gradient. Another response of hepatocytes is gluconeogenesis (the synthesis of glucose from non-carbohydrate precursors like amino acids or glycerol) or the inhibition of glycogenesis (glycogen synthesis).

(a) Max 3 marks:
- Detected by \(\alpha\) (alpha) cells. [1]
- Located in islets of Langerhans / pancreas. [1]
- Endocrine response: secretion/exocytosis of glucagon into the blood. [1]

(b) Max 5 marks:
- Glucagon binds to specific GPCR / receptor on hepatocyte membrane. [1]
- Receptor activates G-protein. [1]
- G-protein stimulates adenylyl cyclase. [1]
- Adenylyl cyclase converts ATP to cyclic AMP / cAMP (second messenger). [1]
- cAMP activates protein kinase A. [1]
- Phosphorylation cascade occurs, leading to activation of glycogen phosphorylase. [1]

(c) Max 2 marks:
- Role of glycogen phosphorylase: Catalyzes glycogenolysis / breakdown of glycogen to glucose. [1]
- Other response: Gluconeogenesis / conversion of amino acids or glycerol to glucose OR inhibition of glycogenesis (glycogen synthesis) OR release of glucose via GLUT proteins. [1]

(a) Osmoreceptors are located in the hypothalamus of the brain. When the water potential of the blood decreases, water moves out of the osmoreceptor cells by osmosis, down a water potential gradient. This loss of water causes the osmoreceptor cells to shrink (lose volume), which triggers nerve impulses that travel to the posterior pituitary gland, stimulating the release of ADH into the bloodstream.

(b) ADH travels in the blood and binds to specific receptors on the basolateral membranes of the collecting duct epithelial cells. This binding activates a G-protein, which stimulates adenylyl cyclase to produce cAMP (a second messenger). cAMP activates a protein kinase enzyme, initiating a signaling cascade. This cascade causes intracellular vesicles containing aquaporin channel proteins to migrate towards the luminal (apical) membrane of the cell. The vesicles fuse with this membrane via exocytosis, inserting the aquaporins and greatly increasing its permeability to water.

(c) Since the interstitial fluid of the renal medulla has a very low water potential (is highly hypertonic) due to the countercurrent multiplier system of the loop of Henle, water is drawn rapidly out of the collecting duct lumen by osmosis through the newly inserted aquaporins. This reabsorption of water back into the blood capillaries leaves behind a highly concentrated, low volume of urine.

(a) Max 3 marks:
- Location: Hypothalamus. [1]
- Response: Water leaves osmoreceptor cells by osmosis down water potential gradient. [1]
- Consequence: Cells shrink, triggering nerve impulses to the posterior pituitary (to release ADH). [1]

(b) Max 5 marks:
- ADH binds to receptors on basolateral membrane of collecting duct cells. [1]
- Activation of G-protein and adenylyl cyclase to produce cAMP. [1]
- Activation of protein kinase / intracellular cell-signalling cascade. [1]
- Vesicles with aquaporins move towards the luminal / apical membrane. [1]
- Vesicles fuse with membrane via exocytosis. [1]
- Aquaporins are inserted, increasing permeability to water. [1]

(c) Max 2 marks:
- Medullary interstitial fluid has a low water potential / is hypertonic. [1]
- Water moves out of the filtrate by osmosis down water potential gradient into the interstitium/blood, leaving concentrated/low-volume urine. [1]

(a) The water plant releases oxygen bubbles during photosynthesis. The gas is collected in a capillary tube of known diameter within the photosynthesiometer. After a set period of time (e.g., 10 minutes), the syringe of the apparatus is used to draw the bubble of collected gas along the tube next to a millimeter scale. The length of the bubble is measured, and the volume of oxygen produced is calculated (using \(V = \pi r^2 l\)) to give a rate of oxygen production per unit time.

(b) Sodium hydrogencarbonate solution dissociates to release dissolved carbon dioxide (\(\text{CO}_2\)) into the water. Distilled water contains very little dissolved carbon dioxide. This ensures that carbon dioxide concentration is abundant and not a limiting factor for photosynthesis, allowing the effect of light intensity to be investigated without interference.

(c) At low light intensities, both curves will show a linear, proportional increase because light intensity is the limiting factor. As light intensity continues to increase, both curves will flatten out (plateau) because light intensity is no longer limiting. However, the curve for the higher concentration (\(0.10\text{ mol dm}^{-3}\)) will plateau at a significantly higher maximum rate of photosynthesis than the \(0.01\text{ mol dm}^{-3}\) curve. This is because at \(0.01\text{ mol dm}^{-3}\), carbon dioxide concentration becomes the limiting factor at a lower light intensity, restricting the rate of the light-independent reaction.

(a) Max 3 marks:
- Oxygen gas bubble is collected in the capillary tube. [1]
- Syringe is used to draw the bubble along a scale. [1]
- Length of the bubble is measured for a fixed/known duration of time to calculate the volume of gas (rate = volume/time). [1]

(b) Max 2 marks:
- Dissolves to release carbon dioxide (\(\text{CO}_2\)). [1]
- Prevents carbon dioxide from acting as a limiting factor / ensures constant supply of carbon dioxide. [1]

(c) Max 5 marks:
- At low light intensity: both curves show linear/proportional increase (light intensity is limiting). [1]
- At high light intensity: both curves plateau/level off (light is no longer limiting). [1]
- High conc (\(0.10\text{ mol dm}^{-3}\)) plateaus at a higher rate than low conc (\(0.01\text{ mol dm}^{-3}\)). [1]
- Explanation for plateau: another factor (carbon dioxide/temperature) is limiting. [1]
- Explanation for difference: low conc curve is limited by carbon dioxide availability at lower light intensities. [1]

(a)
(i) DNA primers: They are short, single-stranded DNA sequences that are complementary to the start and end of the target DNA region. They bind (anneal) to the template DNA to provide a free 3' hydroxyl (-OH) group, which is required for DNA polymerase to initiate synthesis, and they define the specific region of DNA to be amplified.
(ii) Taq polymerase: This is a heat-stable DNA polymerase that synthesizes the complementary DNA strands by adding free deoxynucleoside triphosphates (dNTPs) to the primers. Its heat stability prevents it from denaturing during the high-temperature denaturation phase of PCR.

(b)
1. Denaturation (at \(94\text{--}96^\circ\text{C}\)): The high temperature breaks the hydrogen bonds between complementary base pairs of the double-stranded DNA, separating it into two single strands.
2. Annealing (at \(50\text{--}65^\circ\text{C}\)): The temperature is lowered to allow the primers to form hydrogen bonds (anneal) with their complementary sequences on the single-stranded template DNA.
3. Extension (at \(72^\circ\text{C}\)): This is the optimum temperature for Taq polymerase, which synthesizes the complementary strands by moving along the templates and adding free nucleotides.

(c) Gel electrophoresis separates DNA fragments primarily by size (length). An electric current is applied across an agarose gel matrix. Since DNA molecules have a net negative charge due to their phosphate groups, they migrate towards the positive electrode (anode). The gel acts as a molecular sieve; smaller DNA fragments can navigate through the pores of the gel more quickly and thus travel further than larger, heavier fragments.

(a) Max 4 marks (2 marks per sub-part):
(i) Primers:
- Bind/anneal to complementary target sequences. [1]
- Provide starting point / free 3'-OH group for DNA polymerase. [1]
(ii) Taq polymerase:
- Synthesizes complementary strands by joining free nucleotides (dNTPs). [1]
- Is thermostable / does not denature at high temperatures. [1]

(b) Max 4 marks:
- Denaturation at \(94\text{--}96^\circ\text{C}\): hydrogen bonds break to separate double-stranded DNA into single strands. [1]
- Annealing at \(50\text{--}65^\circ\text{C}\): primers bind to single-stranded templates. [1]
- Extension at \(72^\circ\text{C}\): Taq polymerase synthesizes new strands from the primers. [1]
- All three names paired with correct temperature ranges. [1]

(c) Max 2 marks:
- DNA is negatively charged (due to phosphate groups) and moves towards the positive electrode / anode. [1]
- Gel acts as a molecular sieve / mesh, so smaller fragments move faster / further than larger fragments. [1]

The pathway begins with osmoreceptor cells in the hypothalamus shrinking as water leaves them by osmosis due to low blood water potential. This initiates action potentials that travel down the pituitary stalk to the posterior pituitary, where ADH is released via exocytosis into surrounding capillaries. ADH circulates in the bloodstream and binds to complementary G-protein coupled receptors on the basolateral membranes of the collecting duct cells. The resulting intracellular cascade involves the activation of adenylyl cyclase, which increases cAMP levels. cAMP activates protein kinase A, which promotes the translocation of aquaporin-containing vesicles to the apical membrane facing the lumen. Fusion of these vesicles inserts water channels, greatly increasing the membrane's water permeability. Water leaves the lumen down the steep osmotic gradient created by the hypertonic medullary interstitium.

Award 1 mark for each point up to a maximum of 10 marks: 1. Osmoreceptors in the hypothalamus detect low water potential of blood. 2. Nerve impulses sent along axons to posterior pituitary gland. 3. ADH released from posterior pituitary into the blood. 4. ADH transported to the kidneys. 5. ADH binds to complementary receptors on the cell surface membrane of collecting duct epithelial cells. 6. Activation of G-protein. 7. G-protein activates adenylyl cyclase. 8. Adenylyl cyclase converts ATP to cyclic AMP (cAMP). 9. cAMP acts as second messenger and activates protein kinase A / phosphorylation cascade. 10. Vesicles containing aquaporins move towards and fuse with the apical / luminal membrane. 11. Permeability of the collecting duct to water increases. 12. Water moves out of the collecting duct by osmosis down a water potential gradient.

The scenario describes sympatric speciation driven by ecological and temporal isolation. The key factor is that both species exist on the same small island (no geographical barrier). Genetic variation within the ancestral population allowed colonization of different soil types (calcareous vs volcanic). This created divergent selection pressures. The soil chemistry affected flowering physiology, causing a split in flowering times (temporal isolation). Because they flower at different times, they cannot fertilize one another. This absence of gene flow allows natural selection and genetic drift to alter the allele frequencies of each subpopulation independently. Over time, morphological and genetic differences accumulate to the point where they are reproductively isolated biological species.

Award 1 mark for each point up to a maximum of 10 marks: 1. Sympatric speciation defined as speciation occurring in the same geographical area / without physical barriers. 2. Genetic variation existed in the ancestral population due to mutation. 3. Different soil types (calcareous vs volcanic) acted as different selection pressures. 4. Disruptive selection occurred favoring phenotypes adapted to each soil type. 5. Palms on calcareous soil flowered early, palms on volcanic soil flowered late. 6. Temporal isolation established / different flowering times prevented cross-pollination. 7. Gene flow between the two groups was prevented / reduced. 8. Assortative mating / mating only with individuals of the same group. 9. Distinct mutations and allele combinations accumulated in each group. 10. Natural selection led to changes in allele frequencies in each subpopulation. 11. Reproductive isolation arose (no longer able to interbreed to produce fertile offspring).

In this experiment, vacuum infiltration is first used to draw liquid into the air spaces of the leaf discs, making them sink. During photosynthesis, the light-dependent reactions split water, producing oxygen gas. As oxygen accumulates in the intercellular spaces, it drives out the liquid, making the tissue buoyant again. Because oxygen production is directly related to the rate of light-dependent reactions, the time taken for discs to rise (ET50) serves as an inverse proxy for photosynthetic rate. When adapting this to study temperature, temperature becomes the independent variable. This must be controlled precisely using water baths, and light intensity (a potential confounding variable) must be held strictly constant using a water jacket to block the infrared heat emitted by the lamp.

Award 1 mark for each point up to a maximum of 10 marks: Principles (max 6 marks): 1. Sodium hydrogencarbonate provides carbon dioxide. 2. Light energy drives photolysis of water. 3. Photolysis produces oxygen gas. 4. Oxygen gas fills the intercellular / spongy mesophyll air spaces. 5. Increased buoyancy / decreased density causes the leaf discs to rise. 6. Rate of flotation (1/ET50) is a proxy for the rate of photosynthesis. Adaptation (max 5 marks): 7. Keep light intensity constant by maintaining a fixed distance to the lamp. 8. Use a water jacket / heat shield to prevent the lamp from heating the experiment. 9. Vary temperature using water baths at 5 different temperatures (e.g., 10 to 50 degrees Celsius). 10. Equilibrate leaf discs and solution at each temperature before light exposure. 11. Keep other variables constant (e.g., concentration of hydrogencarbonate, disc diameter, plant species).

The process begins with mRNA extraction because beta cells express high levels of insulin mRNA, which already has introns spliced out. Reverse transcriptase copies this into single-stranded cDNA. DNA polymerase converts this to double-stranded DNA. Plasmids and cDNA are cleaved with the same restriction endonuclease so that their sticky ends are complementary. DNA ligase seals the nicks in the sugar-phosphate backbone. Bacteria are transformed by heat-shocking them in the presence of calcium ions, making their membranes transiently permeable to DNA. Marker genes, such as those encoding resistance to ampicillin, are present on the plasmid. Plating the bacteria on ampicillin-containing agar ensures only transformed cells grow, as non-transformed cells lack the resistance gene.

Award 1 mark for each point up to a maximum of 10 marks: 1. Extract mRNA from human pancreatic beta cells. 2. Reverse transcriptase synthesizes single-stranded cDNA. 3. DNA polymerase produces double-stranded cDNA. 4. cDNA has no introns (suitable for bacterial expression). 5. Restiction enzymes produce sticky ends on cDNA and plasmid. 6. Complementary base-pairing occurs between sticky ends of cDNA and plasmid. 7. DNA ligase joins the DNA fragments by forming phosphodiester bonds. 8. Plasmids are introduced to E. coli via transformation (heat shock / electroporation / calcium chloride). 9. Plasmids contain marker genes (e.g., antibiotic resistance or GFP). 10. Bacteria are plated on agar containing the antibiotic. 11. Only transformed bacteria survive / express the marker.

The light-dependent stage occurs in the thylakoid membranes. Non-cyclic flow involves a Z-scheme where electrons flow from water to NADP+, generating both ATP (via the cytochrome complex/proton gradient) and NADPH. Photolysis of water at the oxygen-evolving complex of PSII supplies the replacing electrons. Cyclic flow occurs when NADP+ is in short supply or more ATP is needed. Electrons from PSI are diverted back to the plastoquinone pool / cytochrome complex instead of going to NADP+ reductase. This cycle continues to pump protons and synthesize ATP via chemiosmosis but yields no reduced NADP and requires no photolysis of water.

Award 1 mark for each point up to a maximum of 10 marks: Non-cyclic photophosphorylation (max 6 marks): 1. Absorption of light energy by photosynthetic pigments causes photoactivation of PSII and PSI. 2. High-energy electrons emitted from reaction centre chlorophyll a. 3. Electrons from PSII pass down an electron transport chain to PSI. 4. Energy from electron transport chain pumps protons into the thylakoid lumen. 5. Proton / electrochemical gradient drives ATP synthesis via ATP synthase (chemiosmosis). 6. Photolysis of water replaces electrons lost from PSII: 2H2O -> 4H+ + 4e- + O2. 7. Electrons from PSI and protons reduce NADP to reduced NADP. Cyclic photophosphorylation (max 4 marks): 8. Involves Photosystem I (PSI) only. 9. Excited electrons are recycled back to PSI via an electron transport chain. 10. Drives proton pumping and ATP synthesis. 11. No reduced NADP is produced. 12. No photolysis of water occurs / no oxygen is produced.

The binding of glucagon to its receptor is the signal reception phase. Transduction occurs via G-proteins and adenylyl cyclase, which amplifies the signal by generating many cAMP molecules. Each cAMP molecule activates protein kinase A. The kinase cascade represents a significant amplification mechanism: a single glucagon molecule can lead to the release of tens of thousands of glucose molecules. The final target enzyme is glycogen phosphorylase, which cleaves glucose residues from glycogen as glucose-1-phosphate, which is then isomerised to glucose. Glucose accumulation within the cell creates a concentration gradient, driving its efflux into the blood plasma via facilitated diffusion transporters (GLUT2).

Award 1 mark for each point up to a maximum of 10 marks: 1. Glucagon acts as the first messenger and binds to a specific G-protein coupled receptor. 2. Receptor undergoes a conformational change. 3. G-protein is activated. 4. Activated G-protein activates adenylyl cyclase. 5. Adenylyl cyclase converts ATP to cyclic AMP (cAMP). 6. cAMP acts as the second messenger. 7. cAMP activates protein kinase A (PKA). 8. PKA activates phosphorylase kinase. 9. Phosphorylase kinase activates glycogen phosphorylase. 10. Enzyme cascade amplifies the signal. 11. Glycogen phosphorylase catalyses glycogenolysis (glycogen to glucose). 12. Glucose leaves hepatocyte via facilitated diffusion through GLUT proteins.

Both processes operate on the principle of differential reproductive success based on phenotypic traits. The core difference lies in the source of the selection pressure (the environment versus human choice) and the ultimate biological outcome. Natural selection selects for survival and reproductive capability under natural conditions, optimizing fitness. Artificial selection selects for specific economic or aesthetic traits, sometimes selecting for alleles that would be deleterious in nature (e.g., heavily muscled cattle that require Caesarean sections to give birth, or crops that lack chemical defenses and require human pesticide application).

Award 1 mark for each point up to a maximum of 10 marks: Similarities (max 4 marks): 1. Both require genetic variation (arising from mutation). 2. Both involve differential survival / reproduction of individuals with favored alleles. 3. Both change allele frequencies in the gene pool over generations. 4. Both can lead to distinct phenotypic changes. Differences (max 7 marks): 5. Selection pressure: Environmental factors in natural selection vs humans in artificial selection. 6. Purpose: No predetermined goal in natural selection vs defined goals in artificial selection. 7. Fitness: Natural selection increases fitness for wild survival vs artificial selection often decreases wild fitness. 8. Rate: Natural selection is relatively slow vs artificial selection which is very rapid. 9. Inbreeding: Artificial selection often leads to inbreeding depression / homozygous recessive defects, rare in natural selection. 10. Examples: Natural selection (e.g., antibiotic resistance in bacteria) AND Artificial selection (e.g., high-yielding wheat or domestic dog breeds).

PCR relies on thermal cycling to copy DNA. The choice of temperatures is critical: 95C for breaking the strong hydrogen bonds between GC and AT pairs; 55C to allow primers to form hydrogen bonds with the template without the two genomic strands reannealing; and 72C to maximize the catalytic rate of Taq polymerase. Taq is sourced from Thermus aquaticus, ensuring it survives the 95C denaturing step. In electrophoresis, the agarose matrix creates physical resistance. Because all DNA molecules have a constant charge-to-mass ratio (due to the regular spacing of negative phosphate groups), their movement is determined entirely by physical size, with smaller molecules experiencing less resistance and moving further.

Award 1 mark for each point up to a maximum of 10 marks: PCR (max 6 marks): 1. DNA is mixed with Taq polymerase, primers, and free nucleotides (dNTPs). 2. Heated to 95 degrees Celsius to separate DNA strands / break hydrogen bonds. 3. Cooled to 55 degrees Celsius to allow primers to anneal / bind to complementary DNA sequences. 4. Heated to 72 degrees Celsius to allow Taq polymerase to extend / synthesize new strands. 5. Taq polymerase is thermostable so does not need to be added in each cycle. 6. Cycle repeated to exponentially duplicate / amplify target DNA. Gel Electrophoresis (max 5 marks): 7. DNA samples loaded into wells in agarose gel. 8. Voltage / electric current applied. 9. DNA fragments move toward positive electrode / anode because phosphate groups are negatively charged. 10. Smaller fragments travel faster / further through the pores of the gel (separation by size). 11. DNA bands visualized using fluorescent stain / UV light / radioactive probes.

This question tests the understanding of mammalian osmoregulation.

Part (a) requires a clear description of the events in the descending and ascending limbs of the loop of Henle. The essential concept is that active transport of ions in the ascending limb sets up an osmotic gradient that draws water out of the descending limb, which in turn concentrates the filtrate entering the ascending limb. This positive feedback process, combined with the opposite directions of flow (countercurrent), multiplies the concentration gradient.

Part (b) requires a step-by-step description of the cell-signalling pathway of ADH. It is important to mention the specific membranes (basolateral vs. apical/luminal), the role of the G-protein, adenylyl cyclase, cAMP as a second messenger, and the translocation and fusion of aquaporin-containing vesicles with the luminal membrane.

Part (a) [Max 5 marks]
1. Sodium (\(\text{Na}^+\)) and chloride (\(\text{Cl}^-\)) ions are actively transported / pumped out of the thick ascending limb. [1]
2. These ions enter the tissue fluid of the renal medulla. [1]
3. The ascending limb is impermeable to water, so water cannot follow by osmosis. [1]
4. The descending limb is permeable to water but impermeable to ions / solutes. [1]
5. Water leaves the descending limb by osmosis into the medullary tissue fluid (and is removed by the vasa recta). [1]
6. This concentrates the filtrate in the descending limb, reaching a maximum concentration at the tip of the loop. [1]
7. Countercurrent flow (filtrate flowing in opposite directions in parallel limbs) allows the osmotic gradient to be multiplied / established from the cortex down into the deep medulla. [1]

Part (b) [Max 5 marks]
8. ADH binds to specific receptors on the basolateral membrane / cell surface membrane of collecting duct cells. [1]
9. This binding activates a G-protein. [1]
10. The active G-protein activates adenylyl cyclase. [1]
11. Adenylyl cyclase catalyses the production of cyclic AMP (cAMP) from ATP (cAMP acts as a second messenger). [1]
12. cAMP activates a cascade of protein kinases. [1]
13. This causes vesicles containing aquaporins (water channel proteins) to move towards and fuse with the luminal / apical membrane. [1]
14. This increases the permeability of the luminal membrane, allowing water to move down a water potential gradient out of the filtrate by osmosis. [1]

This question tests knowledge of genetic technology, including the production of recombinant human proteins, the function of promoters in controlling gene expression, and the practical details of the polymerase chain reaction (PCR).

Part (a) evaluates the candidate's understanding of the differences between prokaryotic and eukaryotic expression systems. Eukaryotic proteins that undergo post-translational modifications like glycosylation (glycoproteins) cannot be produced in their active forms in bacteria because bacteria lack a Golgi body and endoplasmic reticulum.

Part (b) tests the concept of promoters as control sequences. A tissue-specific promoter (like the casein promoter) is active only where the correct transcription factors are present, ensuring local expression (e.g., in milk) and avoiding systemic side effects.

Part (c) tests the precise temperatures and physiological actions during a single PCR cycle: denaturation (92–95 °C), annealing (50–65 °C), and extension (70–75 °C), along with the roles of primers, hydrogen bond cleavage, and Taq polymerase.

Part (a) [Max 3 marks]
1. Prokaryotes / E. coli lack membrane-bound organelles / rough endoplasmic reticulum / Golgi apparatus. [1]
2. Prokaryotes cannot perform post-translational modifications / glycosylation (addition of sugar chains). [1]
3. Factor IX is a glycoprotein and requires glycosylation to be fully functional / active. [1]
4. Eukaryotes have appropriate chaperones for correct 3D folding / disulfide bond formation of complex proteins (which bacteria lack). [1]

Part (b) [Max 3 marks]
5. A promoter is a non-coding sequence of DNA where RNA polymerase binds to initiate transcription. [1]
6. A tissue-specific / mammary-specific promoter (e.g., casein promoter) is ligated / linked to the human Factor IX gene sequence. [1]
7. The transcription factors required to activate this promoter are only synthesized / active in mammary gland cells. [1]
8. This restricts transcription of the Factor IX gene to the mammary glands, preventing systemic expression (which could harm the host sheep) and allowing collection from milk. [1]

Part (c) [Max 4 marks]
9. Denaturation: heated to 92–95 °C to break hydrogen bonds between complementary bases, separating double-stranded DNA into single strands. [1]
10. Annealing: cooled to 50–65 °C to allow primers to bind / anneal to complementary sequences at the 3' ends of the target DNA. [1]
11. Extension: heated to 70–75 °C (accept 72 °C) which is the optimum temperature for Taq polymerase. [1]
12. Taq polymerase adds free nucleotides / dNTPs to extend the complementary strands starting from the primers. [1]

1. Preparation of the Independent Variable: Prepare 5 different concentrations of \(\text{NaHCO}_3\) by simple dilution from a stock solution of 2.0\\%. To make 20 \(\text{cm}^3\) of each concentration: for 2.0\\%, use 20 \(\text{cm}^3\) of stock; for 1.5\\%, mix 15 \(\text{cm}^3\) of stock with 5 \(\text{cm}^3\) of distilled water; for 1.0\\%, mix 10 \(\text{cm}^3\) of stock with 10 \(\text{cm}^3\) of distilled water; for 0.5\\%, mix 5 \(\text{cm}^3\) of stock with 15 \(\text{cm}^3\) of distilled water; for 0.0\\%, use 20 \(\text{cm}^3\) of distilled water. Measure all volumes using accurate graduated pipettes. 2. Measurement of the Dependent Variable: Place a freshly cut 5 \(\text{cm}\) stem of Elodea canadensis upside down in a boiling tube containing the respective concentration of \(\text{NaHCO}_3\). Ensure the stem is cut obliquely under water to prevent air locks in the xylem. Set up a photosynthesiometer (Audus micro-apparatus) consisting of a capillary tube attached to a syringe and a millimeter scale. Allow the plant to equilibrate for 10 minutes under the light source before beginning measurements. Draw the oxygen bubble produced into the capillary tube using the syringe and measure its length (L) in mm using the scale after a fixed time of 5 minutes. Calculate the rate of photosynthesis as length of gas bubble produced per unit time (\(\text{mm min}^{-1}\) ) or calculate volume as \(V = \pi r^2 L\) and express rate in \(\text{mm}^3\text{ min}^{-1}\) if the capillary radius (r) is known. 3. Control of Variables: (a) Light Intensity: Position a high-intensity LED light source at a fixed distance (e.g., 15 cm) from the boiling tube, using a ruler to measure. LED lights minimize heat generation. Light intensity must be controlled as it is a limiting factor for light-dependent reactions. (b) Temperature: Immerse the boiling tube in a water bath kept constant at 20 degrees Celsius (monitored with a thermometer). Temperature affects the kinetic energy of enzymes in the Calvin cycle (such as RuBisCO). (c) Plant Material: Use Elodea pieces of equal length (5 cm) and mass from the same parent plant to ensure consistent chlorophyll concentration. 4. Control Experiment: Set up an identical apparatus but replace the living Elodea with a dead, boiled Elodea plant or glass beads of equivalent volume to prove gas production is due to active photosynthesis. 5. Reliability: Repeat the procedure at least three times at each concentration of \(\text{NaHCO}_3\), calculating the mean rate and standard deviation to identify anomalies. 6. Safety and Risk Assessment: Wear safety goggles and gloves to prevent skin/eye irritation from \(\text{NaHCO}_3\). Use a scalpel to cut Elodea on a cutting tile, cutting away from the body. Keep electrical light sources away from water and ensure hands are dry.

Independent Variable (IV) Manipulation [max 3 marks]: MP1 - Selects and states at least 5 different concentrations of \(\text{NaHCO}_3\) in a logical range between 0.0\\% and 2.0\\%. MP2 - Describes a correct simple or serial dilution method using named apparatus (graduated pipettes/syringes) to prepare concentrations. MP3 - Shows explicit calculation details/volumes of stock solution (2.0\\%) and distilled water required to make at least one intermediate concentration. Dependent Variable (DV) Measurement [max 4 marks]: MP4 - Mentions cutting the stem of Elodea obliquely/under water and placing it upside down in the tube. MP5 - Explains the use of a photosynthesiometer/capillary tube and syringe to capture the gas bubble. MP6 - Describes a specific equilibration time (e.g., 5 to 10 minutes) before recording starts. MP7 - States that the length of the gas bubble is measured over a specified constant time interval (e.g., 5 minutes) and explains how rate is calculated. Control of Variables [max 4 marks]: MP8 - Identifies temperature as a variable to be controlled, describes using a water bath, and explains that temperature affects enzyme-controlled reactions (Calvin cycle / RuBisCO). MP9 - Identifies light intensity as a variable to be controlled, describes using a ruler to set a fixed distance from an LED light, and explains that light intensity affects the light-dependent stage. MP10 - Identifies plant physiological state, describes using pieces of the same length/mass/from the same parent plant, and explains this ensures equal photosynthetic capacity. Control Experiment [max 1 mark]: MP11 - Describes a control setup using boiled/dead plant or glass beads to show gas volume changes are due to active photosynthesis. Reliability [max 1 mark]: MP12 - States that the experiment should be repeated at each concentration at least 3 times and a mean calculated. Safety [max 2 marks]: MP13 - Identifies sharp scalpel hazard and states mitigation (cut away from self on a cutting tile). MP14 - Identifies electrical hazard with water near light source and states mitigation (dry hands / keep electrical items separated from liquids) or wear goggles for chemical safety.

[Part a] Independent Variable: ADH concentration (in \(\text{pg cm}^{-3}\)). Dependent Variable: Rate of water movement / trans-epithelial water flux (in \(\mu\text{m}^3\text{ s}^{-1}\) ). [Part b] 1. Temperature: Keep the organ bath at mammalian body temperature (e.g., 37 degrees Celsius) using a thermostatically controlled heating jacket. Reason: Mammalian enzymes and transport proteins (aquaporins) function optimally at this temperature; low temperature decreases kinetic energy and flux rates, while extremely high temperature denatures proteins. 2. pH: Use a physiological buffer (e.g., bicarbonate buffer) to keep the pH constant at 7.4. Reason: Fluctuations in pH change the ionic charge of amino acid R-groups, causing conformational changes in ADH receptors or aquaporin channels, altering their binding affinity. [Part c] Experimental Setup: Group 1 (Control) consists of isolated collecting ducts treated with a range of ADH concentrations (e.g., 0, 10, 20, 30, 40, 50 \(\text{pg cm}^{-3}\)) without 'Aquasnap'. Group 2 (Inhibitor Group) consists of isolated collecting ducts treated with the exact same range of ADH concentrations in the presence of a constant, fixed concentration of 'Aquasnap'. Measure the mean trans-epithelial water flux for both groups. Plot rate of water flux (y-axis) against ADH concentration (x-axis) for both datasets. If 'Aquasnap' is a competitive inhibitor of the ADH pathway, increasing the ADH concentration will eventually overcome the inhibition. The maximum rate of water flux (\(V_{\text{max}}\)) will remain the same as the control group, but a higher ADH concentration (higher \(K_{\text{m}}\)) is needed to reach half-\(V_{\text{max}}\). If 'Aquasnap' is a non-competitive inhibitor (or directly blocks the channels downstream), the maximum rate (\(V_{\text{max}}\)) will be significantly lower in the 'Aquasnap' group compared to the control group, even at high, saturating ADH concentrations. [Part d] Spearman's rank is suitable because: (1) The data consists of two continuous variables (ADH concentration and water flux); (2) The relationship is monotonic (water flux increases continuously with ADH concentration in this range); (3) The sample size is small. Null Hypothesis: There is no significant correlation between the concentration of ADH and the rate of water movement (trans-epithelial water flux) in isolated collecting ducts.

Part (a) [2 marks]: MP1 - Identifies independent variable as ADH concentration. MP2 - Identifies dependent variable as rate of water movement / trans-epithelial water flux. Part (b) [4 marks]: MP3 - Identifies first factor to control (e.g., Temperature at 37 degrees Celsius using heating jacket/water bath). MP4 - Gives correct biological reason for first factor (ensures normal kinetic energy / prevents denaturation / maintains mammalian physiological rate). MP5 - Identifies second factor to control (e.g., pH at 7.4 using a buffer OR solute concentration using isotonic saline). MP6 - Gives correct biological reason for second factor (pH prevents denaturation/charge changes of receptors; isotonicity prevents osmotic swelling/shrinking of cells). Part (c) [6 marks]: MP7 - States the use of two groups: one with ADH only and one with ADH + constant concentration of 'Aquasnap'. MP8 - Specifies using a wide range of ADH concentrations (at least 5 concentrations) for both groups. MP9 - Describes plotting water flux rate against ADH concentration for both datasets. MP10 - Explains that if competitive, the maximum rate (\(V_{\text{max}}\)) of water flux will be the same in both setups at high ADH concentration. MP11 - Explains that if competitive, the curve for the inhibitor setup will be shifted to the right / require higher ADH concentration to reach half-maximum. MP12 - Explains that if non-competitive (blocking aquaporins downstream), the maximum rate (\(V_{\text{max}}\)) of water flux will be lower in the 'Aquasnap' group than the control. Part (d) [3 marks]: MP13 - Explains that Spearman's rank is used to test for a correlation / association between two continuous variables. MP14 - Explains that the data displays a monotonic trend / the sample size is small. MP15 - Correctly states the null hypothesis: There is no significant correlation between ADH concentration and the rate of water movement / trans-epithelial water flux.