2024 OCR GCSE Twenty First Century Science - Biology B - J257 模擬試題連答案詳解

According to Fick's Law, the rate of diffusion is proportional to: \(\frac{\text{Surface Area} \times \text{Concentration Difference}}{\text{Thickness of Membrane}}\). If the concentration difference is doubled, the rate of diffusion doubles (factor of 2). If the membrane thickness is halved, the rate of diffusion doubles again (factor of 2). Therefore, the overall rate increases by a factor of 4: \(5.0\text{ units} \times 4 = 20.0\text{ units}\).

[1 mark] for selecting correct option C.

Using the same restriction endonuclease ensures that both the plasmid vector and the human insulin gene are cut at identical recognition sequences. This leaves complementary single-stranded overhanging bases (sticky ends) that can easily base-pair and seal together.

[1 mark] for identifying option A as correct.

During the follicular phase, estrogen levels increase and initially inhibit FSH release via negative feedback to prevent multiple follicles from maturing. When estrogen levels reach a high threshold, it triggers a positive feedback mechanism resulting in a rapid surge of LH, which triggers ovulation.

[1 mark] for selecting option A.

Using the Lincoln Index formula: \(N = \frac{M \times C}{R}\), where \(M = 80\) (initially marked), \(C = 60\) (total captured on Day 2), and \(R = 15\) (recaptured marked beetles). Substituting these values: \(N = \frac{80 \times 60}{15} = \frac{4800}{15} = 320\).

[1 mark] for the correct calculation leading to option C (320).

First, calculate the volume of water taken up: \(\text{Volume} = \text{Cross-sectional area} \times \text{Distance} = 0.5\text{ mm}^2 \times 40\text{ mm} = 20\text{ mm}^3\). Next, calculate the rate per minute: \(\text{Rate} = \frac{\text{Volume}}{\text{Time}} = \frac{20\text{ mm}^3}{10\text{ minutes}} = 2.0\text{ mm}^3\text{ minute}^{-1}\).

[1 mark] for selecting option B.

To produce monoclonal antibodies, a B-lymphocyte (which produces the desired antibody but cannot divide indefinitely) is fused with a myeloma (tumor) cell (which divides rapidly and indefinitely but does not make antibodies) to form a hybridoma cell. This combined cell is able to produce antibodies and divide rapidly.

[1 mark] for selecting option B.

Sweating causes water loss, which increases the solute concentration of the blood. This is detected by osmoreceptors in the hypothalamus, which stimulates the pituitary gland to release MORE ADH (antidiuretic hormone). ADH travels to the kidneys and increases the permeability of the collecting ducts, allowing more water to be reabsorbed back into the blood, resulting in concentrated urine.

[1 mark] for selecting the correct pathway in option B.

A continuous leak in the gas syringe causes all measured volumes of gas to be consistently lower than the true volumes. Because this error is consistent and predictable across all measurements, it is classified as a systematic error, which directly reduces the accuracy of the experiment.

[1 mark] for identifying option B as the correct classification and effect of the error.

When external humidity is high, the concentration of water vapour in the air surrounding the leaf is higher. This reduces the concentration gradient of water vapour between the air spaces inside the leaf (which are saturated with water) and the outside atmosphere. As a result, the rate of diffusion of water vapour out of the stomata decreases, thus decreasing the rate of transpiration.

[1 mark] - B: It decreases the concentration gradient of water vapour between the inside of the leaf and the external air. Reject any explanations stating that humidity increases the gradient or causes stomata to open wider to reduce transpiration.

The surface area to volume ratio (\(\text{SA:V}\)) of a sphere simplifies to \(\frac{3}{r}\). Therefore, the \(\text{SA:V}\) is inversely proportional to the radius \(r\). Since the radius of the larger organism is 10 times larger than that of the amoeba (\(500\text{ \mu m} / 50\text{ \mu m} = 10\)), its \(\text{SA:V}\) must be 10 times smaller.

[1 mark] - C: 10 times smaller. Award 1 mark for the correct calculation or understanding that \(\text{SA:V}\) is inversely proportional to the radius.

To ensure the complementary sticky ends of the human insulin gene and the bacterial plasmid can join together, they must be cut using the same restriction enzyme. After mixing the cut gene and plasmid, the enzyme DNA ligase is added to join the sugar-phosphate backbones of the DNA molecules together, forming recombinant DNA.

[1 mark] - B: Cut insulin gene with restriction enzyme \(\rightarrow\) Cut plasmid with same restriction enzyme \(\rightarrow\) Mix gene and plasmid \(\rightarrow\) Add ligase. Do not accept answers using different restriction enzymes or incorrect enzymes like protease or ligase for cutting.

LH (Luteinising Hormone) is released by the pituitary gland. A sudden surge in LH levels around day 14 of the menstrual cycle triggers ovulation (the release of a mature egg from the ovary).

[1 mark] - C: Luteinising Hormone (LH). Do not accept FSH (which stimulates follicle growth), Estrogen (which builds up the uterine lining), or Progesterone (which maintains the lining).

A double-blind trial is designed to eliminate psychological bias. Neither the trial participants nor the doctors/researchers administering the trial and collecting data know who is in the active drug group and who is in the placebo group.

[1 mark] - B: To prevent bias in reporting and analyzing results by ensuring neither the volunteers nor the researchers know who receives the drug.

When body temperature rises, the thermoregulatory centre in the brain triggers vasodilation. This is the widening of arterioles supplying the skin surface capillaries, allowing more warm blood to flow close to the skin surface, increasing heat loss by radiation. Note that capillaries themselves do not have muscle walls to constrict or dilate; it is the arterioles supplying them that dilate.

[1 mark] - B: Vasodilation of arterioles near the skin surface to increase blood flow to capillaries. Reject option C because capillaries cannot vasoconstrict. Reject option A because vasoconstriction occurs in cold conditions.

In human muscle cells, anaerobic respiration breaks down glucose into lactic acid only (no carbon dioxide is produced). In yeast cells, anaerobic respiration (fermentation) breaks down glucose into ethanol and carbon dioxide.

[1 mark] - B: Human muscle cells produce lactic acid only, while yeast cells produce ethanol and carbon dioxide. Accept correct differentiation of products.

Root hair cells have a large surface area that allows them to efficiently absorb water and mineral ions from the surrounding soil.

Award 1 mark for 'root hair cells' or 'root hair cell'.
Do not accept 'root cells' or 'xylem'.

Veins carry blood back to the heart under low pressure, so they contain valves to ensure blood flows in one direction only.

Award 1 mark for 'vein' or 'veins'.
Do not accept 'artery' or 'capillary'.

Restriction enzymes (or restriction endonucleases) cut DNA at specific recognition sites, leaving 'sticky ends' or 'blunt ends'.

Award 1 mark for 'restriction enzymes' or 'restriction endonucleases' or 'restriction enzyme'.

Follicle stimulating hormone (FSH) is released by the pituitary gland and travels in the blood to the ovaries, where it triggers egg follicle maturation.

Award 1 mark for 'Follicle stimulating hormone' or 'FSH'.
Accept minor spelling variations.

An individual with two different alleles (e.g., Bb) is described as heterozygous.

Award 1 mark for 'heterozygous'.
Do not accept 'heterozygote'.

During vasodilation, the diameter of blood vessels supplying the skin capillaries increases, allowing more blood to flow close to the skin surface so heat can be radiated away.

Award 1 mark for 'vasodilation'.
Do not accept 'vasoconstriction'.

Antibiotics are substances that kill bacteria or inhibit their growth. They do not work against viruses because viruses replicate inside host cells and do not have the same cellular machinery.

Award 1 mark for 'antibiotics' or 'antibiotic'.
Do not accept 'antibodies' or 'antiseptics'.

In yeast, anaerobic respiration (fermentation) breaks down glucose into ethanol and carbon dioxide.

Award 1 mark for 'ethanol' or 'alcohol'.
Do not accept 'lactic acid' or 'lactate'.

Xylem is the specialized transport tissue in plants that is responsible for the upward translocation of water and dissolved minerals from the roots to the leaves.

1 mark for 'xylem' (accept 'xylem vessels' or 'xylem tissue'). Reject 'phloem'.

A synapse is the junction between two neurones where the electrical impulse triggers the release of neurotransmitters, which diffuse across the gap.

1 mark for 'synapse' (accept 'synaptic cleft' or 'synaptic gap').

During anaerobic respiration (fermentation) in yeast, glucose is broken down to produce carbon dioxide and ethanol.

1 mark for 'ethanol' (accept 'alcohol'). Reject 'lactic acid'.

Follicle-stimulating hormone (FSH) is released by the pituitary gland and travels in the blood to the ovaries, where it stimulates an egg follicle to mature.

1 mark for 'FSH' or 'follicle-stimulating hormone'.

Restriction enzymes (or restriction endonucleases) are used in genetic engineering to cut DNA molecules at precise, specific base sequences.

1 mark for 'restriction enzyme' (accept 'restriction endonuclease'). Reject 'ligase'.

Haemoglobin is the iron-rich protein molecule found in red blood cells that reversibly binds to oxygen to transport it around the body.

1 mark for 'haemoglobin' (accept 'hemoglobin').

The hypothalamus contains receptors that monitor blood temperature and coordinates the body's thermoregulatory responses.

1 mark for 'hypothalamus'. Reject 'pituitary gland'.

Hybridoma cells are created by fusing a tumor cell (myeloma) with a specific antibody-producing B-lymphocyte.

1 mark for 'B-lymphocyte' (accept 'B cell' or 'lymphocyte'). Reject 'T cell' or general 'white blood cell'.

Active transport is an active process requiring energy in the form of ATP. This energy is produced through cellular respiration (specifically aerobic respiration) in the mitochondria of root hair cells.

Award 1 mark for 'Respiration' (or 'Aerobic respiration' or 'Cellular respiration'). Do not accept 'Anaerobic respiration' on its own without qualification as it does not typically sustain healthy active transport in roots over long periods.

The right ventricle pumps deoxygenated blood into the pulmonary artery, which branches to carry the blood to both lungs where gas exchange occurs.

Award 1 mark for 'Pulmonary artery'. Reject 'pulmonary vein' or 'artery' on its own.

Restriction enzymes (or restriction endonucleases) are used to recognize specific sequences of DNA and cut them, leaving either sticky or blunt ends. This allows the target gene to be isolated.

Award 1 mark for 'Restriction enzyme' or 'Restriction endonuclease'.

Follicle-stimulating hormone (FSH) is released by the pituitary gland. It travels in the blood to the ovaries where it causes an egg to mature in its follicle.

Award 1 mark for 'Follicle-stimulating hormone' or 'FSH'.

The hypothalamus contains receptors sensitive to the temperature of the blood flowing through the brain. It coordinates the body's response to changes in temperature to maintain homeostasis.

Award 1 mark for 'Hypothalamus'. Do not accept 'brain' or 'cerebrum'.

During anaerobic respiration (fermentation), yeast breaks down glucose in the absence of oxygen to produce ethanol and carbon dioxide.

Award 1 mark for 'Ethanol' or 'Alcohol'.

A hybridoma cell is formed by fusing a B-lymphocyte (which produces the specific desired antibody) with a cancer cell (myeloma, which divides rapidly). This allows the hybridoma to divide repeatedly to produce many clones of the antibody.

Award 1 mark for 'Hybridoma' (or 'Hybridoma cell').

Alfred Russel Wallace independently formulated a theory of evolution by natural selection and sent his ideas to Darwin in 1858, which prompted their joint publication.

Award 1 mark for 'Alfred Russel Wallace' or 'Alfred Wallace' or 'Wallace'.

Active transport is required because the concentration of nitrate ions in the soil is lower than the concentration inside the root hair cell. Because passive diffusion cannot occur against a concentration gradient, the cell must use energy released from cellular respiration to actively pump the ions into the cell.

1 mark: For identifying that nitrate ions are at a lower concentration in the soil than in the root hair cell (or that they move against/up the concentration gradient). 1 mark: For stating that this process requires energy from respiration.

In a double circulatory system, blood passes through the heart twice in one complete circuit. After blood passes through the lungs, its pressure drops significantly. Returning to the heart allows it to be pumped a second time, raising the blood pressure so that oxygenated blood can travel rapidly to respiring tissues to support a high metabolic rate.

1 mark: Recognises that blood pressure drops after passing through the lungs (or that the heart repressurises blood before it goes to the body). 1 mark: Explains that this allows rapid delivery of oxygen/glucose to body cells to support a high rate of respiration.

Not all bacterial cells will successfully take up the modified plasmid during the transformation stage. By including an antibiotic resistance gene as a marker, the bacteria can be grown on agar containing the antibiotic. Only the cells that have taken up the plasmid will survive, allowing scientists to easily identify and select the transformed bacteria.

1 mark: To identify or select which host bacterial cells have successfully taken up the plasmid. 1 mark: By showing survival on a medium containing the antibiotic (or killing off non-transformed bacteria).

High levels of progesterone exert negative feedback on the pituitary gland. This inhibits the secretion of Follicle-Stimulating Hormone (FSH), which prevents new follicles from maturing, and Luteinising Hormone (LH), which prevents ovulation. Consequently, no new eggs are released during pregnancy.

1 mark: State that progesterone inhibits the release of FSH and/or LH from the pituitary gland. 1 mark: State that without FSH/LH, no further follicles mature or are ovulated.

Sweat consists mostly of water. As sweat evaporates from the surface of the skin, it requires heat energy (latent heat of vaporisation). This thermal energy is absorbed from the skin and blood flowing near the skin surface, thereby cooling the body down as the heat is transferred to the environment.

1 mark: State that water in sweat evaporates from the skin surface. 1 mark: State that this evaporation transfers thermal energy away from the skin/body to the surroundings.

By day five, the most susceptible bacteria have been killed, which reduces the patient's symptoms, but the more resistant bacteria may still survive. If the patient stops taking the antibiotics early, these remaining resistant bacteria will survive and multiply. This can cause a relapse of the infection and contributes to the spread of antibiotic resistance.

1 mark: State that the most resistant bacteria survive if the course is not finished. 1 mark: State that these surviving resistant bacteria will reproduce, leading to increased antibiotic resistance or a relapse.

The rapid growth of algae blocks sunlight from reaching plants deeper in the river, preventing them from photosynthesising and causing them to die. Decomposers (bacteria) multiply rapidly as they feed on the dead plant matter. These decomposers use up the dissolved oxygen in the water through aerobic respiration, leaving insufficient oxygen for the fish to survive.

1 mark: For explaining that plants die due to lack of light, or that decomposers/bacteria break down dead organic matter. 1 mark: For stating that decomposers respire aerobically, depleting dissolved oxygen levels so fish suffocate.

In anaerobic respiration, oxygen is absent. Without oxygen, glucose is only partially broken down (incomplete oxidation) to form lactic acid. Much of the chemical energy remains locked up within the chemical bonds of the lactic acid molecule, resulting in a much lower yield of ATP (only \( 2 \) molecules per glucose compared to around \( 32 \) to \( 38 \) in aerobic respiration).

1 mark: State that glucose is only partially/incompletely broken down (or oxidised). 1 mark: State that significant chemical energy remains trapped in the bonds of lactic acid (or that only \( 2 \) ATP molecules are produced compared to a much higher amount in aerobic respiration).

An amoeba has a high surface area to volume ratio, so diffusion is fast enough to supply all parts of the cell. A human has a much smaller surface area to volume ratio and a larger diffusion distance, requiring specialized exchange surfaces (like lungs) and transport systems to deliver oxygen to internal cells.

One mark for stating that single-celled organisms have a large surface area to volume ratio (or short diffusion distance) [1]. One mark for explaining that multicellular organisms have a small surface area to volume ratio (or large diffusion distance) and need specialized systems to overcome this [1].

A vector (e.g., a plasmid or a virus) is used to carry the isolated gene. It then inserts and integrates this foreign gene into the genome of the target cell/host organism, allowing the cell to express the new characteristic.

One mark for identifying that the vector carries/transports the gene into the target cell/host DNA [1]. One mark for stating that this allows the gene to be integrated or expressed by the host organism [1].

Water evaporates and diffuses out of the leaves through the stomata. This loss of water creates a tension or pulling force (transpiration pull) that draws water upwards through the xylem, maintained by the cohesion between water molecules.

One mark for explaining that evaporation/transpiration creates a tension or pulling force (transpiration pull) at the leaf end [1]. One mark for mentioning that cohesion (water molecules sticking together) allows a continuous column of water to be pulled up the xylem [1].

Overuse kills non-resistant bacteria, removing competition. Resistant bacteria survive, reproduce, and pass on the resistance genes/alleles to their offspring, increasing their proportion in the population.

One mark for stating that antibiotics kill non-resistant bacteria, leaving resistant ones to survive (natural selection/selective pressure) [1]. One mark for explaining that these survivors reproduce and pass on their resistance genes/alleles to offspring [1].

Arterioles widen (dilate) near the skin surface to increase blood flow through the skin capillaries. This allows more heat energy to be transferred/radiated from the blood to the external environment, lowering core body temperature.

One mark for explaining that arterioles widen (dilate), increasing blood flow to skin capillaries [1]. (Do NOT accept 'capillaries widen'). One mark for explaining that this increases heat loss/transfer from the blood to the surroundings via radiation/conduction [1].

FSH stimulates the growth of follicles in the ovary, which causes the ovaries to secrete estrogen. High levels of estrogen then inhibit (prevent) the further secretion of FSH from the pituitary gland.

One mark for stating that FSH stimulates the ovaries to produce/release estrogen [1]. One mark for explaining that estrogen inhibits FSH production (negative feedback) to prevent more eggs/follicles from developing [1].

Anaerobic respiration in yeast (fermentation) produces carbon dioxide gas. This carbon dioxide gas forms bubbles that expand and cause the bread dough to rise.

One mark for identifying that carbon dioxide is produced during anaerobic respiration in yeast [1]. One mark for explaining that the carbon dioxide gas bubbles cause the dough to expand/rise [1].

Step 1: Calculate the energy stored in the biomass of the producers: \(1.2 \times 10^6\text{ kJ} \times 0.015 = 18,000\text{ kJ}\) (or \(1.8 \times 10^4\text{ kJ}\)). Step 2: Calculate the efficiency of energy transfer from the producers to the primary consumers: \(\frac{1,800\text{ kJ}}{18,000\text{ kJ}} \times 100 = 10\%\).

Award 1 mark for calculating the energy in producer biomass: \(18,000\text{ kJ}\) (or \(1.8 \times 10^4\text{ kJ}\)). Award 1 mark for setting up the correct fraction for efficiency: \(\frac{1,800}{18,000} \times 100\) (or equivalent). Award 1 mark for the correct final answer: 10 (accept 10%).

Step 1: Calculate the initial stroke volume: \(\text{Stroke Volume} = \frac{\text{Cardiac Output}}{\text{Heart Rate}} = \frac{4.8\text{ dm}^3/\text{min}}{60\text{ beats/min}} = 0.08\text{ dm}^3\text{/beat}\). Step 2: Calculate the stroke volume during exercise: \(0.08\text{ dm}^3 \times 1.25 = 0.10\text{ dm}^3\text{/beat}\). Step 3: Calculate the new cardiac output: \(155\text{ beats/min} \times 0.10\text{ dm}^3\text{/beat} = 15.5\text{ dm}^3/\text{min}\).

Award 1 mark for calculating the initial stroke volume: \(0.08\text{ dm}^3\) (or \(80\text{ cm}^3\)). Award 1 mark for calculating the stroke volume during exercise: \(0.10\text{ dm}^3\) (or \(100\text{ cm}^3\)). Award 1 mark for the correct final cardiac output during exercise: 15.5 (accept 15.5 dm^3/min).

Step 1: Rearrange the magnification formula to solve for actual size: \(\text{Actual Size} = \frac{\text{Image Size}}{\text{Magnification}}\). Step 2: Convert the image length from millimetres to micrometres: \(48\text{ mm} \times 1,000 = 48,000\ \mu\text{m}\). Step 3: Calculate the actual length: \(\frac{48,000}{15,000} = 3.2\ \mu\text{m}\).

Award 1 mark for converting units correctly (either \(48\text{ mm}\) to \(48,000\ \mu\text{m}\) or converting a final value in millimetres to micrometres). Award 1 mark for rearranging the magnification formula correctly: \(\text{Actual} = \frac{\text{Image}}{\text{Magnification}}\). Award 1 mark for the correct final answer to 2 significant figures: 3.2.

Step 1: Calculate the volume of water taken up using the cylinder volume formula (area x distance): \(0.80\text{ mm}^2 \times 45\text{ mm} = 36\text{ mm}^3\). Step 2: Convert the time into hours: \(15\text{ minutes} = 0.25\text{ hours}\). Step 3: Calculate the rate of water uptake: \(\frac{36\text{ mm}^3}{0.25\text{ hours}} = 144\text{ mm}^3/\text{hour}\) (or \(36 \times 4 = 144\)).

Award 1 mark for calculating the volume of water absorbed: \(36\text{ mm}^3\). Award 1 mark for a correct time conversion or multiplication factor to scale up to 1 hour (e.g., dividing by 0.25 or multiplying by 4). Award 1 mark for the correct final rate: 144.

Using the same restriction enzyme ensures that both the target gene (human insulin gene) and the vector (bacterial plasmid) are cut at the same specific recognition sequence. This produces complementary single-stranded overhangs called 'sticky ends'. These sticky ends can then pair through hydrogen bonding between complementary bases, allowing DNA ligase to permanently join the backbones.

1 mark for identifying that using the same restriction enzyme produces complementary sticky ends that can pair up.

According to Fick's Law: \(\text{Rate of Diffusion} \propto \frac{\text{Surface Area} \times \text{Concentration Difference}}{\text{Thickness}}\). Let the original Rate = \(\frac{1 \times 1}{1} = 1\). With the new changes: New Surface Area = \(0.5\), New Thickness = \(2\). New Rate = \(\frac{0.5 \times 1}{2} = 0.25\). Therefore, the rate of diffusion is reduced to one-quarter (\(25\%\)) of its original value.

1 mark for applying Fick's Law mathematically to show that halving surface area and doubling thickness results in a rate of \(0.25\) times the original (one-quarter).

1. Calculate the volume of water taken up (volume of a cylinder represented by bubble movement): \(\text{Volume} = \text{Cross-sectional Area} \times \text{Distance} = 0.8\text{ mm}^2 \times 45\text{ mm} = 36\text{ mm}^3\). 2. Calculate the rate of transpiration per minute: \(\text{Rate} = \frac{\text{Volume}}{\text{Time}} = \frac{36\text{ mm}^3}{15\text{ minutes}} = 2.4\text{ mm}^3\text{/minute}\).

1 mark for calculating volume as \(36\text{ mm}^3\) and dividing by \(15\text{ minutes}\) to get \(2.4\text{ mm}^3\text{/minute}\).

ADH (anti-diuretic hormone) increases the permeability of the kidney's collecting ducts to water. This causes more water to be reabsorbed from the filtrate back into the blood. Consequently, the volume of urine produced is small and highly concentrated. Since excess water is reabsorbed into the blood, the concentration of solutes in the blood decreases (diluting the blood).

1 mark for identifying that high ADH causes increased water reabsorption, leading to concentrated urine and dilute blood (low solute concentration).

During the follicular phase of the menstrual cycle, the developing follicle secretes increasing amounts of estrogen. Once estrogen levels reach a high threshold, they exert positive feedback on the pituitary gland, triggering a rapid and large release (surge) of Luteinising Hormone (LH). This LH surge causes the follicle to rupture and release the mature egg (ovulation).

1 mark for stating that high estrogen levels cause the LH surge which triggers ovulation.

Natural selection relies on pre-existing genetic variation arising from random mutations. Some bacteria in the population already possessed resistance alleles before exposure to penicillin. When penicillin was applied, it acted as a selective pressure, killing the non-resistant bacteria. The resistant bacteria survived, reproduced, and passed on the resistance alleles to their offspring.

1 mark for identifying that random mutations occurred before exposure, providing a selective advantage that allowed survival and reproduction.

Fertilizer runoff causes rapid algal growth (algal bloom). The algae block sunlight from reaching deeper water, causing submerged plants to die because they cannot photosynthesize. Bacterial decomposers multiply rapidly as they feed on the dead plants. These bacteria respire aerobically, using up the dissolved oxygen in the water, which leads to anoxia and causes fish to suffocate and die.

1 mark for identifying the correct chronological order of events during eutrophication: algal bloom, light blockage, plant death, bacterial decomposition leading to oxygen depletion, and fish suffocation.

In a lateral flow test, the sample pad contains mobile monoclonal antibodies conjugated to a colored label (e.g., gold nanoparticles). If the virus is present, it binds to these mobile antibodies. As the fluid flows along the strip, it reaches the test line (T-line), where fixed (immobilized) monoclonal antibodies are bound to the membrane. These fixed antibodies bind to another site on the virus-antibody complex, trapping it and concentrating the colored labels to form a visible line.

1 mark for explaining that the test line has immobilized antibodies that capture the migrating virus-antibody complex, resulting in a colored band.

According to Fick's Law: \(\text{Rate of diffusion} \propto \frac{\text{Surface Area} \times \text{Concentration Difference}}{\text{Thickness of membrane}}\). Let us evaluate each option: Option A: Doubling both surface area (\(2\times\)) and thickness (\(2\times\)) results in a rate change of \(\frac{2}{2} = 1\) (no change). Option B: Doubling concentration difference (\(2\times\)) and thickness (\(2\times\)) results in a rate change of \(\frac{2}{2} = 1\) (no change). Option C: Doubling surface area (\(2\times\)) and halving thickness (\(0.5\times\)) results in a rate change of \(\frac{2}{0.5} = 4\) times faster. Option D: Halving both concentration difference (\(0.5\times\)) and thickness (\(0.5\times\)) results in a rate change of \(\frac{0.5}{0.5} = 1\) (no change). Therefore, Option C results in the greatest increase.

Award 1 mark for selecting the correct option (C). No partial marks are awarded for this single-mark multiple-choice question.

Using the same restriction enzyme ensures that both the target human DNA and the bacterial plasmid are cut at the same specific recognition sequence, leaving complementary single-stranded 'sticky ends'. These complementary bases can then pair up via hydrogen bonding, allowing the DNA ligase enzyme to join the backbones together. If different enzymes were used, the sticky ends would not be complementary and would not bind together.

Award 1 mark for the correct option (B). Reject all other options.

Root hair cells have a large surface area to increase the rate of absorption of water and mineral ions from the soil.

1 mark: Root hair cell(s). Do not accept 'root cell' or 'hair cell'.

The vena cava is the main vein that returns deoxygenated blood from the upper and lower body to the right atrium.

1 mark: Vena cava. Accept 'superior vena cava' or 'inferior vena cava'.

Plasmids are small, circular double-stranded DNA molecules distinct from a cell's chromosomal DNA, commonly used as vectors in genetic engineering.

1 mark: Plasmid(s).

Follicle Stimulating Hormone (FSH) is secreted by the pituitary gland and causes an egg to mature in its follicle in the ovary.

1 mark: FSH or Follicle Stimulating Hormone.

Osmoregulation is the process of maintaining salt and water balance (osmotic pressure) across membranes within the body.

1 mark: Osmoregulation. Reject homeostasis (too broad, as homeostasis refers to all internal conditions).

Anaerobic respiration in animal cells glucose is converted to lactic acid (and releases a small amount of energy).

1 mark: Lactic acid. Accept 'lactate'. Do not accept 'carbon dioxide' or 'ethanol'.

An organism is heterozygous for a gene if it has two different alleles of that gene.

1 mark: Heterozygous. Reject heterozygote.

Charles Darwin proposed the theory of evolution by natural selection, where organisms better adapted to their environment tend to survive and produce more offspring.

1 mark: Natural selection. Accept 'survival of the fittest'.

Active transport is the process used to move substances from an area of lower concentration to an area of higher concentration (against a concentration gradient), requiring energy from respiration.

1 mark: Active transport. Reject: diffusion, osmosis, passive transport.

The pulmonary artery carries deoxygenated blood from the right ventricle of the heart to the lungs so that carbon dioxide can be released and oxygen absorbed.

1 mark: Pulmonary artery. Accept: pulmonary trunk. Reject: pulmonary vein, artery (on its own).

Luteinising hormone (LH) is produced by the pituitary gland. A sudden surge in LH levels midway through the cycle triggers ovulation.

1 mark: Luteinising hormone / LH. Reject: FSH, oestrogen, progesterone.

Negative feedback loops detect deviations from a norm and trigger mechanisms to bring the level back down or up to the set point, maintaining a constant internal environment.

1 mark: Negative feedback (mechanism / loop). Reject: positive feedback, homeostasis (on its own).

Anaerobic respiration in yeast converts glucose into ethanol and carbon dioxide: \(\text{glucose} \rightarrow \text{ethanol} + \text{carbon dioxide}\).

1 mark: Carbon dioxide. Accept: \(CO_2\). Reject: lactic acid.

Plasmids are small, circular rings of DNA found in bacteria that exist independently of the bacterial chromosome, and are commonly used as vectors in recombinant DNA technology.

1 mark: Plasmid(s). Reject: chromosomes, nuclei.

An individual is described as heterozygous for a particular gene if they possess two different alleles (such as one dominant and one recessive allele) at a given locus.

1 mark: Heterozygous. Reject: homozygous, genotype.

Antibiotics are medicines used to treat bacterial infections. They do not work on viral infections because viruses replicate inside host cells, making them difficult to target without damaging host tissues, and viruses do not have the cellular machinery targeted by antibiotics.

1 mark: Antibiotic(s). Reject: antibody / antibodies, antivirals, antiseptics, painkillers.

Water and dissolved mineral ions are absorbed by the root hair cells and are transported upwards through the plant stem to the leaves via the xylem tissue.

1 mark for xylem (accept xylem tissue, xylem vessels, or xylem vessel). Reject phloem.

The pulmonary artery is the vessel that transports deoxygenated blood away from the right ventricle of the heart and into the lungs to receive oxygen.

1 mark for pulmonary artery (accept pulmonary arteries). Reject pulmonary vein.

Following ovulation, the remains of the follicle develop into the corpus luteum, which secretes progesterone to maintain the thickness of the uterus lining in preparation for a potential fertilised egg.

1 mark for progesterone (accept phonetically correct spelling).

During anaerobic respiration in yeast, glucose is broken down into ethanol and carbon dioxide.

1 mark for ethanol (accept alcohol). Reject lactic acid.

Root hair cells absorb mineral ions from the soil against a concentration gradient using active transport. Active transport requires energy, which is released by aerobic respiration occurring inside the mitochondria of the cell.

1 mark: Identify that mineral ions are absorbed by active transport (against a concentration gradient). 1 mark: Explain that active transport requires energy released from respiration in the mitochondria.

The walls of both the alveoli and the capillaries are only one cell thick. This extremely thin boundary minimizes the diffusion distance that oxygen and carbon dioxide must travel, which significantly increases the overall rate of gas exchange.

1 mark: For stating that the thin walls reduce/minimize the diffusion distance. 1 mark: For linking the shorter diffusion distance to an increased rate of diffusion/gas exchange.

Restriction enzymes cut DNA at specific base sequences. Using the same enzyme on both the donor DNA and the plasmid ensures they both end up with complementary single-stranded 'sticky ends', allowing them to easily anneal and join together using DNA ligase.

1 mark: For stating that it produces matching/complementary sticky ends. 1 mark: For explaining that this allows the gene and plasmid to join/pair up/anneal.

During anaerobic respiration (fermentation) in yeast, glucose is broken down to produce ethanol and carbon dioxide; the carbon dioxide gas bubbles cause the bread dough to rise. In contrast, anaerobic respiration in human muscle cells produces lactic acid, which builds up in muscles and causes fatigue and aching.

1 mark: For identifying that yeast produces carbon dioxide which makes dough rise. 1 mark: For identifying that muscle cells produce lactic acid which leads to muscle fatigue/pain.

When core body temperature increases, blood vessels (arterioles) supplying the skin capillaries dilate (widen). This process, called vasodilation, increases blood flow near the surface of the skin, allowing more thermal energy to be lost to the surrounding environment by radiation.

1 mark: For stating that blood vessels/arterioles dilate, increasing blood flow close to the skin surface. 1 mark: For explaining that this increases heat loss to the environment via radiation/convection/conduction.

If a patient stops taking antibiotics early, the most resistant bacteria survive because they have not been exposed to the drug for long enough to be killed. These surviving resistant bacteria then reproduce rapidly, passing on their resistance genes to their offspring and increasing the population of resistant strains.

1 mark: For explaining that finishing early allows the most resistant bacteria to survive. 1 mark: For stating that these survivors reproduce and pass on their resistance genes/alleles.

In IVF, FSH (Follicle Stimulating Hormone) is administered to stimulate the development and maturation of multiple egg follicles in the ovaries simultaneously. LH (Luteinising Hormone) is then given to trigger ovulation (the release of these mature eggs) so they can be collected for external fertilisation.

1 mark: For explaining that FSH stimulates multiple follicles/eggs to mature. 1 mark: For explaining that LH triggers the release/ovulation of these eggs for collection.

Decomposers play a vital role in recycling nutrients by breaking down dead organic matter and waste. A decline in decomposers would stop the release of essential mineral ions (such as nitrates and magnesium) back into the soil, meaning producers cannot absorb these nutrients to grow and photosynthesise, eventually decreasing their overall biomass.

1 mark: For explaining that decomposers break down dead matter/waste to release/recycle mineral ions (e.g., nitrates) into the soil. 1 mark: For linking the lack of these mineral ions to reduced plant growth/photosynthesis/biomass.

1. The promoter acts as the binding site for RNA polymerase. 2. This allows transcription of the human insulin gene to occur in the bacterial cell, leading to the translation and production of the insulin protein.

[1 mark] for identifying that the promoter is a DNA sequence where RNA polymerase binds or which initiates transcription. [1 mark] for explaining that this ensures the insulin gene is expressed or transcribed in the bacterial host.

1. The capillary walls are extremely thin, being composed of a single layer of endothelial cells, which minimizes the diffusion distance for nutrients and gases. 2. Capillaries are highly branched and form dense networks, which vastly increases the surface area available for diffusion.

[1 mark] for stating the walls are one cell thick or very thin to reduce diffusion distance. [1 mark] for stating that capillaries have a large surface area / form vast networks to maximize exchange rate OR that the narrow lumen slows blood flow to allow time for diffusion.

1. Bubbles are not of a uniform size, meaning counting them does not provide an accurate or reliable measurement of the total gas volume. 2. Measuring the volume directly using a gas syringe or graduated capillary tube provides precise quantitative data on the actual volume of oxygen produced.

[1 mark] for explaining that bubbles can vary in size or can be missed if released too fast. [1 mark] for explaining that measuring volume directly (e.g., using a gas syringe) gives a more precise, quantitative, and valid measurement of the gas produced.

1. Water moves into both cells by osmosis, down a water potential gradient from the dilute solution into the cells. 2. The red blood cell has only a flexible cell membrane and no cell wall, so it bursts under pressure, whereas the rigid cellulose cell wall of the plant cell prevents it from bursting by resisting the turgor pressure.

[1 mark] for identifying that water enters the cells by osmosis / down a water potential gradient. [1 mark] for explaining that the red blood cell lacks a cell wall and cannot withstand the pressure (and bursts), whereas the plant cell has a rigid cell wall that prevents lysis.

1. Photosynthesis is a chemical process regulated by enzymes. High temperatures break the bonds holding the enzyme's tertiary structure. 2. This denatures the enzyme, altering the shape of the active site so that substrates can no longer bind and form enzyme-substrate complexes.

[1 mark] for stating that high temperatures denature the enzymes involved in photosynthesis. [1 mark] for explaining that denaturation changes the shape of the active site, preventing the substrate from binding and preventing reactions from occurring.

1. Penicillin works by preventing bacteria from building functional peptidoglycan cell walls, causing them to burst. 2. Viruses do not have cell walls or their own metabolic machinery (they use host cells), so penicillin has no effect on them.

[1 mark] for explaining that penicillin specifically targets bacterial structures/processes, such as cell wall synthesis. [1 mark] for explaining that viruses lack these target structures, are non-cellular, or replicate inside host cells, making them immune to the antibiotic.

First, set up the percentage efficiency calculation: \(\text{Efficiency} = \frac{\text{Energy in sheep}}{\text{Energy from Sun}} \times 100\). Substitute the values: \(\text{Efficiency} = \frac{3.24 \times 10^3}{1.8 \times 10^6} \times 100 = 0.18\%\). Convert the decimal \(0.18\) to standard form with 2 significant figures: \(1.8 \times 10^{-1}\%\).

1 mark: Correct substitution of values into the percentage efficiency formula: \(\frac{3.24 \times 10^3}{1.8 \times 10^6} \times 100\). 1 mark: Calculation of the percentage as \(0.18\%\). 1 mark: Correct standard form to 2 significant figures: \(1.8 \times 10^{-1}\) (accept with or without the \(\%\) sign).

Use the formula: \(\text{Cardiac output} = \text{stroke volume} \times \text{heart rate}\). Rearrange to find stroke volume: \(\text{Stroke volume} = \frac{\text{Cardiac output}}{\text{Heart rate}}\). First, convert cardiac output to \(cm^3\text{ min}^{-1}\): \(24.0 \times 1000 = 24000\text{ cm}^3\text{ min}^{-1}\). Divide by heart rate: \(\text{Stroke volume} = \frac{24000}{160} = 150\text{ cm}^3\text{ per beat}\).

1 mark: Correct conversion of \(24.0\text{ dm}^3\) to \(24000\text{ cm}^3\) OR division of final answer in \(dm^3\) by 1000. 1 mark: Correct rearrangement of formula to make stroke volume the subject: \(\text{Stroke volume} = \frac{24.0}{160}\) (or \(\frac{24000}{160}\)). 1 mark: Correct final answer: 150 (accept \(150\text{ cm}^3\)).

Calculate RQ using the formula: \(\text{RQ} = \frac{\text{Volume of } \text{CO}_2 \text{ produced}}{\text{Volume of } \text{O}_2 \text{ consumed}}\). Substitute the values: \(\text{RQ} = \frac{23.9}{34.2} = 0.6988...\). Round to 2 decimal places: \(0.70\). Comparing \(0.70\) to the reference table, the substrate is lipids.

1 mark: Correctly setting up the division: \(\frac{23.9}{34.2}\). 1 mark: Calculating the RQ to 2 decimal places as 0.70 (accept 0.7). 1 mark: Correctly identifying 'Lipids' as the primary substrate based on their calculated RQ value.

1. Population estimate: \(\frac{48 \times 40}{15} = 128\). 2. Percentage of the population marked in the first capture: \(\frac{48}{128} \times 100\% = 37.5\%\).

1 mark: Correctly calculating the estimated population size as 128. 1 mark: Showing a correct method to find the percentage of the estimated population that was marked: \(\frac{48}{\text{population estimate}} \times 100\). 1 mark: Correct calculation of 37.5% (accept error carried forward from incorrect population size).

1. Detection: Sweating on a hot day causes the blood plasma to become more concentrated (lower water potential). This is detected by receptor cells in the hypothalamus of the brain. 2. Hormone Release: The hypothalamus stimulates the pituitary gland to release more Anti-Diuretic Hormone (ADH) into the bloodstream. 3. Kidney Response: ADH travels to the kidneys where it increases the permeability of the collecting ducts to water. This allows more water to be reabsorbed back into the blood by osmosis, resulting in a smaller volume of highly concentrated urine. 4. Negative Feedback: The reabsorbed water dilutes the blood, returning its concentration to the normal level. As the blood concentration returns to normal, the hypothalamus detects this and signals the pituitary gland to reduce ADH release, completing the negative feedback loop.

Level 3 (5-6 marks): Comprehensive explanation covering all four aspects: detection by the hypothalamus, release of ADH from the pituitary gland, increased collecting duct permeability leading to more water reabsorption in the kidneys, and a clear explanation of negative feedback. Level 2 (3-4 marks): Clear explanation of most aspects, identifying ADH and its role in increasing water reabsorption in the kidney, with some reference to detection or negative feedback. Level 1 (1-2 marks): Simple points, such as identifying ADH or stating that kidneys reabsorb more water when the body is dehydrated. Indicative content: Sweating causes water loss from blood. Detected by hypothalamus. Pituitary gland releases ADH. ADH increases permeability of kidney collecting ducts. More water is reabsorbed. Less urine produced. Return to normal triggers reduction in ADH release (negative feedback).

1. Isolation: The human insulin gene is identified and cut out of human DNA using a specific restriction enzyme, leaving sticky ends. 2. Vector Preparation: A plasmid is extracted from a bacterium and cut open using the same restriction enzyme, ensuring matching sticky ends. 3. Insertion: The human insulin gene and plasmid are joined together by the enzyme DNA ligase to form a recombinant plasmid. 4. Transformation: This recombinant plasmid is inserted back into host bacterial cells. 5. Culturing: The genetically modified bacteria are grown in industrial fermenters where they multiply and produce human insulin, which is then extracted and purified. 6. Advantages: This method avoids ethical or religious issues associated with animal slaughter. The insulin produced is identical to human insulin, meaning there is less risk of allergic reactions. Production is reliable, scalable, and carries no risk of animal-transmitted pathogens.

Level 3 (5-6 marks): Detailed and accurate description of the genetic engineering process (using restriction enzymes, plasmids, ligase, and transformation) and a clear discussion of at least two advantages of using GM bacteria over animal extraction. Level 2 (3-4 marks): Good description of the genetic process using key terms, with at least one advantage, or a highly technical description of the process without advantages. Level 1 (1-2 marks): Simple statements about inserting human genes into bacteria, or simple points about advantages of GM insulin. Indicative content: Human insulin gene cut with restriction enzyme. Plasmid cut with same enzyme. Joined with DNA ligase. Recombinant plasmid inserted into bacteria. Bacteria grown in fermenter. Advantages: Ethical/religious acceptance, identical to human insulin (fewer allergies), high yield, no risk of animal diseases.