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The hepatic portal vein transports blood from the gastrointestinal tract (small intestine) directly to the liver. Because this blood has already delivered oxygen to the tissues of the digestive system, it is deoxygenated (having a low oxygen concentration). However, shortly after a meal rich in carbohydrates, glucose is actively absorbed from the ileum into the bloodstream, resulting in a high concentration of glucose in the hepatic portal vein before the liver regulates and stores the excess as glycogen.

1 mark for selecting B.
- Correctly identifies that hepatic portal vein blood has a low oxygen concentration (deoxygenated blood from the gut) and a high glucose concentration (newly absorbed nutrients from the digestion of carbohydrates).

Sweating without water replenishment leads to a decrease in the water potential of the blood (hemoconcentration). Osmoreceptors in the hypothalamus detect this change and signal the pituitary gland to release more antidiuretic hormone (ADH). An increase in the concentration of ADH in the blood increases the permeability of the walls of the collecting ducts in the kidney nephrons to water, allowing more water to be reabsorbed back into the blood, resulting in a low volume of concentrated urine.

1 mark for selecting C.
- Correctly links dehydration to an increase in blood ADH concentration and a corresponding increase in the permeability of the collecting duct walls to water.

(a) Bile salts are responsible for the physical emulsification of large lipid droplets into smaller droplets. This physical change significantly increases the surface area-to-volume ratio of the lipid substrate, allowing the lipase enzymes to bind to and digest lipids at a much faster rate.

(b) Chemical digestion of lipids by lipase yields fatty acids and glycerol. The accumulation of fatty acids increases the concentration of hydrogen ions (\(\text{H}^+\)) in the solution, which increases the acidity and consequently lowers the pH of the test tube mixture.

(c) In Tube C, the lipase enzyme was boiled before being added to the mixture. Boiling denatures the enzyme, permanently changing the shape of its active site. As a result, the active site is no longer complementary to the lipid substrate, meaning no enzyme-substrate complexes can form, no chemical digestion occurs, and the pH remains neutral/unchanged.

Part (a) [3 marks total]:
- 1 mark: Emulsification of lipids / breaking down large droplets into smaller droplets (Do NOT accept chemical breakdown).
- 1 mark: Increases the surface area (for lipase action).
- 1 mark: Increases the rate of enzyme-substrate collisions / faster rate of chemical digestion.

Part (b) [3 marks total]:
- 1 mark: Lipase chemically digests lipids into fatty acids and glycerol.
- 1 mark: Fatty acids are acidic / release \(\text{H}^+\) ions.
- 1 mark: Acidic products lower the pH of the solution.

Part (c) [2 marks total]:
- 1 mark: No digestion occurs / pH remains unchanged / no fatty acids produced.
- 1 mark: Boiling denatured the lipase / altered the active site shape so it can no longer bind with lipids.

(a) The graph shows that in the secondary immune response, antibody production begins almost immediately with a shorter lag phase. Additionally, the peak concentration of antibodies is significantly higher than in the primary response, and the antibody levels remain elevated in the blood for a much longer period of time.

(b) During the primary response, memory B cells and memory T cells are produced and remain in the lymphatic system. Upon secondary exposure to the same antigen, these memory B cells immediately recognise the antigen. They quickly divide by mitosis (clonal expansion) to differentiate into antibody-producing plasma cells and more memory cells, bypassing the slow step of initial antigen presentation.

(c) Active immunity is achieved when a person's own white blood cells (lymphocytes) actively synthesize antibodies after being exposed to a pathogen or vaccine. Passive immunity involves the direct transfer of pre-formed antibodies into the body from an external source (e.g., across the placenta or via breast milk), meaning the individual does not produce their own antibodies or memory cells.

Part (a) [3 marks total]:
- 1 mark: Secondary response has a shorter lag phase / starts faster.
- 1 mark: Secondary response reaches a higher maximum concentration of antibodies.
- 1 mark: Secondary response antibody levels remain high for longer / decrease slower.

Part (b) [3 marks total]:
- 1 mark: Memory cells persist in the blood/lymphatic system after the primary infection/vaccine.
- 1 mark: Memory cells rapidly recognise the same antigen upon re-exposure.
- 1 mark: Rapid cell division (mitosis) / clonal expansion occurs to produce plasma cells that secrete antibodies.

Part (c) [2 marks total]:
- 1 mark: Active immunity involves the body's own lymphocytes producing antibodies / development of memory cells.
- 1 mark: Passive immunity involves receiving pre-made antibodies / no memory cells produced.

(a) Arteries carry blood under high pressure directly from the heart. To withstand this pressure without bursting, they have thick walls containing a substantial outer layer of collagen. They also contain a thick middle layer of smooth muscle and elastic fibres. The elastic fibres stretch as high-pressure blood surges through, and recoil during diastole to maintain a continuous high pressure and smooth out the pulsatile blood flow.

(b) Capillaries have walls that are extremely thin—only one cell thick (made of squamous endothelial cells). This structure provides a very short diffusion pathway, allowing rapid exchange of oxygen, carbon dioxide, and nutrients between blood and surrounding tissues.

(c) First, calculate flow rate per second:
\(\text{Rate} = \frac{180\text{ cm}^3}{45\text{ s}} = 4\text{ cm}^3\text{/s}\).
Next, convert to minutes by multiplying by 60:
\(4\text{ cm}^3\text{/s} \times 60\text{ s/min} = 240\text{ cm}^3\text{/min}\).

Part (a) [4 marks total]:
- 1 mark: Thick muscular/elastic walls (to prevent bursting/withstand high pressure).
- 1 mark: Elastic fibres stretch to accommodate the surge of blood.
- 1 mark: Elastic fibres recoil to maintain pressure / push blood along.
- 1 mark: Narrow lumen to help maintain high blood pressure.

Part (b) [2 marks total]:
- 1 mark: Walls are only one cell thick / made of a single layer of endothelial cells.
- 1 mark: Shortens the diffusion distance / allows rapid diffusion of gases.

Part (c) [2 marks total]:
- 1 mark: Correct method of calculation shown (e.g., \(180 / 45 = 4\) and then \(4 \times 60\) OR \((180 / 45) \times 60\)).
- 1 mark: Correct answer of \(240\) with units \(\text{cm}^3\text{/min}\).

(a) When blood becomes more concentrated (low water potential), osmoreceptors in the hypothalamus detect this change. The hypothalamus signals the posterior pituitary gland to secrete more antidiuretic hormone (ADH) into the bloodstream. ADH travels to the kidneys, where it increases the permeability of the collecting duct walls to water. As a result, more water is reabsorbed from the filtrate back into the blood capillaries by osmosis, leaving a smaller, more concentrated volume of urine.

(b) ADH acts specifically on the kidney, targeting the epithelial cells of the collecting ducts (and distal convoluted tubules) to increase their permeability to water.

(c) When a person is dehydrated, the kidneys reabsorb maximum water, leading to a small volume of urine that is highly concentrated and thus has a dark yellow/amber appearance.

Part (a) [5 marks total]:
- 1 mark: Osmoreceptors in the hypothalamus detect the increase in blood concentration / low water potential.
- 1 mark: Pituitary gland is stimulated to release MORE ADH into the blood.
- 1 mark: ADH travels in the blood to the kidneys.
- 1 mark: ADH increases the permeability of the collecting ducts (to water).
- 1 mark: More water is reabsorbed back into the blood, leading to a small volume of concentrated urine.

Part (b) [2 marks total]:
- 1 mark: Kidney (target organ).
- 1 mark: Collecting duct (part of nephron) [accept distal convoluted tubule].

Part (c) [1 mark total]:
- 1 mark: Dark yellow color / concentrated appearance / low volume of urine.

(a) Goblet cells, which are interspersed among ciliated cells, synthesize and secrete sticky mucus. This mucus acts as a barrier, trapping inhaled dust, dirt particles, and pathogens. The ciliated epithelial cells have tiny hair-like projections called cilia that beat in a coordinated, rhythmic fashion to sweep the trapped mucus upwards toward the throat (pharynx), where it can be swallowed or coughed out.

(b) The constant, rhythmic beating motion of cilia requires a continuous supply of chemical energy in the form of ATP. Mitochondria are the sites of aerobic respiration, which generates the ATP needed to power the motor proteins responsible for movement of the cilia.

(c) To find the magnification, first convert the image size from millimetres to micrometres to ensure both values use the same units:
\(6.0\text{ mm} = 6.0 \times 1000 = 6000\ \mu\text{m}\).
Next, apply the magnification formula:
\(\text{Magnification} = \frac{\text{Image Size}}{\text{Actual Size}}\)
\(\text{Magnification} = \frac{6000\ \mu\text{m}}{15\ \mu\text{m}} = 400\).
Therefore, the magnification is \(\times 400\).

Part (a) [3 marks total]:
- 1 mark: Goblet cells secrete mucus.
- 1 mark: Mucus traps pathogens / bacteria / dust.
- 1 mark: Cilia beat / sweep mucus upwards/away from the lungs (towards the mouth/throat).

Part (b) [2 marks total]:
- 1 mark: Cilia movement requires energy / ATP.
- 1 mark: Mitochondria carry out aerobic respiration to produce ATP / release energy.

Part (c) [3 marks total]:
- 1 mark: Convert millimetres to micrometres (\(6.0\text{ mm} = 6000\ \mu\text{m}\)).
- 1 mark: Set up calculation using \(\text{Magnification} = \text{Image} / \text{Actual}\) (i.e., \(6000 / 15\)).
- 1 mark: Correct evaluation of \(\times 400\) or \(400\) (must indicate multiplication factor).

(a) The parents are heterozygous, so both have the genotype \(\text{Ff}\).
Each parent produces gametes containing either \(\text{F}\) or \(\text{f}\).
Crossing the gametes in a Punnett square yields:
- \(\text{FF}\) (homozygous dominant, healthy)
- \(\text{Ff}\) (heterozygous, healthy carrier)
- \(\text{Ff}\) (heterozygous, healthy carrier)
- \(\text{ff}\) (homozygous recessive, has cystic fibrosis)

The ratio of healthy offspring (including carriers) to offspring with cystic fibrosis is \(3:1\).

(b) Looking at the four possible offspring combinations, two out of the four are heterozygous carriers (\(\text{Ff}\)). Therefore, the probability is \(\frac{2}{4} = 0.5\) or \(50\%\).

(c) In individuals with cystic fibrosis, a mutated membrane channel protein prevents the normal movement of salt and water out of cells, causing mucus secretions to become extremely thick and sticky. This viscous mucus physically blocks the pancreatic duct. As a result, digestive enzymes (such as amylase, trypsin, and lipase) synthesized in the pancreas cannot enter the duodenum. This severely reduces chemical digestion, meaning nutrients cannot be absorbed, leading to malnutrition and weight loss.

Part (a) [4 marks total]:
- 1 mark: Correct parental genotypes shown (\(\text{Ff} \times \text{Ff}\)) and gametes (\(\text{F}\) and \(\text{f}\)).
- 1 mark: Punnett square grid drawn correctly with offspring genotypes (\(\text{FF}\), \(\text{Ff}\), \(\text{Ff}\), \(\text{ff}\)).
- 1 mark: Link genotypes to phenotypes correctly (\(\text{FF}\) and \(\text{Ff}\) are healthy/carriers; \(\text{ff}\) has cystic fibrosis).
- 1 mark: Correct phenotypic ratio stated as \(3:1\) (healthy to cystic fibrosis).

Part (b) [1 mark total]:
- 1 mark: Correct probability of \(0.5\) / \(50\%\) / \(1/2\) / \(2\) in \(4\).

Part (c) [3 marks total]:
- 1 mark: Mucus is thick and sticky.
- 1 mark: Mucus blocks the pancreatic duct.
- 1 mark: Digestive enzymes cannot reach the small intestine / duodenum, leading to incomplete digestion or absorption of food.

(a) Phenolphthalein is a pH indicator that turns pink in alkaline conditions. The sodium hydroxide present in the agar makes the cube alkaline, causing it to appear pink.

(b) Surface Area of Cube B = \(6 \times (2\text{ cm} \times 2\text{ cm}) = 24\text{ cm}^2\).
Volume of Cube B = \(2\text{ cm} \times 2\text{ cm} \times 2\text{ cm} = 8\text{ cm}^3\).
SA:V Ratio = \(24 : 8 = 3 : 1\).

(c) (i) The rate of diffusion depends on factors such as temperature and the concentration gradient of the acid, both of which were kept constant. The overall size or surface area of the cube does not affect the speed of movement of individual acid molecules.
(ii) Convert the side length of Cube B to mm: \(2\text{ cm} = 20\text{ mm}\).
Total volume of Cube B = \(20\text{ mm} \times 20\text{ mm} \times 20\text{ mm} = 8000\text{ mm}^3\).
The acid diffuses \(3\text{ mm}\) from each of the 6 faces. The unpenetrated central core is a cube with side length: \(20\text{ mm} - (2 \times 3\text{ mm}) = 14\text{ mm}\).
Volume of unpenetrated core = \(14\text{ mm} \times 14\text{ mm} \times 14\text{ mm} = 2744\text{ mm}^3\).
Volume penetrated by acid = \(8000\text{ mm}^3 - 2744\text{ mm}^3 = 5256\text{ mm}^3\).
Percentage of volume penetrated = \(\frac{5256}{8000} \times 100\% = 65.7\%\).

(d) Variables to control include:
1. Temperature of the acid solution.
2. Concentration of the hydrochloric acid.
3. Concentration of sodium hydroxide/indicator in the agar cubes.
4. Time the cubes are left in the acid (5 minutes).

(e) Keep the size of the agar cube constant (e.g., only use \(2\text{ cm}\) cubes). Place the acid in water baths at different set temperatures (e.g., \(20\ ^\circ\text{C}\), \(30\ ^\circ\text{C}\), \(40\ ^\circ\text{C}\), and \(50\ ^\circ\text{C}\)) and allow them to equilibrate before introducing the agar cubes.

(a)
- Phenolphthalein is an indicator (1)
- It turns pink in alkaline/basic conditions / in the presence of sodium hydroxide (1)

(b)
- Calculation of surface area: \(24\text{ cm}^2\) (1)
- Calculation of volume: \(8\text{ cm}^3\) (1)
- Correct simplified ratio: \(3:1\) (allow 3) (1)

(c)(i)
- Temperature / concentration gradient is the same (1)
- Speed/kinetic energy of diffusing acid molecules is not affected by cube size (1)

(c)(ii)
- Calculates total volume as \(8000\text{ mm}^3\) or \(8\text{ cm}^3\) (1)
- Calculates unpenetrated core volume as \(2744\text{ mm}^3\) or \(2.744\text{ cm}^3\) (1)
- Calculates correct percentage: \(65.7\%\) (1)

(d)
- Concentration of acid (1)
- Temperature of the solution (1)
- Concentration of sodium hydroxide in agar (1)
- Time spent in acid (1)
(Accept any three for 1 mark each, max 3)

(e)
- Keep cube size constant (1)
- Use different temperatures using water baths and allow to equilibrate before starting (1)

(a) Safety precaution: Wear safety goggles to protect the eyes from hot fat spitting or sparks from the burning food. Alternatively, use a heatproof mat or clamp to avoid burning hands, or verify no nut allergies are present before handling peanuts.

(b) (i) Mass of water = \(20\text{ g}\) (since \(20\text{ cm}^3\) was used and \(1\text{ cm}^3 = 1\text{ g}\)).
Temperature rise = \(42.0\ ^\circ\text{C} - 18.5\ ^\circ\text{C} = 23.5\ ^\circ\text{C}\).
\(\text{Energy (J)} = 20\text{ g} \times 4.2 \times 23.5\ ^\circ\text{C} = 1974\text{ J}\).

(ii) \(\text{Energy per gram (J/g)} = \frac{\text{Energy (J)}}{\text{Mass of food (g)}} = \frac{1974\text{ J}}{0.50\text{ g}} = 3948\text{ J/g}\).

(c) (i) Reasons for a lower calculated value:
1. Heat is lost directly to the surrounding air instead of being transferred to the water.
2. Heat is absorbed by the glass of the boiling tube rather than the water.
3. Incomplete combustion of the food sample occurs, leaving unburnt carbon (soot) or residue on the needle.
4. Heat is lost while transferring the burning peanut from the Bunsen burner flame to the position beneath the boiling tube.

(ii) Modifications to improve accuracy:
1. Use a copper can/calorimeter rather than glass, as copper conducts heat much more efficiently than glass.
2. Put a draught shield around the apparatus to prevent air currents from blowing heat away.
3. Use a bomb calorimeter where food is burnt in an enclosed, oxygen-filled chamber surrounded by water.
4. Stir the water continuously with the thermometer to ensure heat is evenly distributed throughout.

(d) Independent variable: The type of food sample used.
Dependent variable: The temperature rise of the water (or energy released).

(a)
- Correct precaution (e.g., wear safety goggles / do not touch hot glassware) (1)
- Correct linked explanation (e.g., protects eyes from hot oil spitting / prevents skin burns) (1)

(b)(i)
- Correct temperature rise calculation: \(23.5\ ^\circ\text{C}\) (1)
- Correct substitution of values: \(20 \times 4.2 \times 23.5\) (1)
- Correct final calculation: \(1974\text{ J}\) (1)

(b)(ii)
- Dividing energy by the mass of the food: \(\frac{1974}{0.50}\) (1)
- Correct calculation: \(3948\text{ J/g}\) (1)

(c)(i)
- Heat lost to the surrounding air (1)
- Heat absorbed by the glass tube (1)
- Incomplete combustion of the peanut (1)
- Heat lost during transfer from Bunsen burner (1)
(Accept any three for 1 mark each, max 3)

(c)(ii)
- Use a copper can/metal container instead of glass (1)
- Use a draught shield to minimize convection currents (1)
- Use a bomb calorimeter / burn in pure oxygen (1)
- Stir water to distribute heat evenly (1)
(Accept any three for 1 mark each, max 3)

(d)
- Independent variable: type of food sample (1)
- Dependent variable: temperature rise of water (accept energy released) (1)

Ciliated epithelial cells have tiny hair-like projections called cilia that beat rhythmically to sweep mucus and trapped dust/pathogens out of the airways. This movement requires a significant amount of energy in the form of ATP. Mitochondria are the organelles responsible for aerobic respiration, which generates ATP. Therefore, these cells contain a high density of mitochondria.

Award 1 mark for the correct option (B).
- Reject other options: A (Ribosomes are for protein synthesis), C (Lysosomes contain digestive enzymes), D (Golgi apparatus is for modifying and packaging proteins).

Benedict's reagent is used to test for the presence of reducing sugars. When heated with a reducing sugar, the copper(II) ions in Benedict's solution are reduced, causing a colour change from blue to green, yellow, orange, or brick-red (depending on concentration). Lactose is a reducing sugar present in milk and will yield a brick-red precipitate. Sucrose is a non-reducing sugar and would not react directly. Lipids and proteins are tested using different reagents (emulsion test and Biuret test respectively).

Award 1 mark for the correct option (B).
- Reject A: Sucrose is a non-reducing sugar and does not give a positive Benedict's test.
- Reject C and D: Lipids and proteins do not react with Benedict's reagent.

When a person is dehydrated, the hypothalamus detects a low water potential in the blood and stimulates the pituitary gland to release more antidiuretic hormone (ADH). ADH increases the permeability of the collecting duct walls to water. As a result, more water is reabsorbed by osmosis back into the blood capillaries, leading to the production of a low volume of highly concentrated urine.

Award 1 mark for the correct option (A).
- Reject B: Increased permeability leads to a low volume of concentrated urine, not dilute.
- Reject C and D: Dehydration increases permeability via ADH, it does not decrease it.

(a)
(i) Pulmonary artery
(ii) Left atrium
(iii) Bicuspid / mitral valve

(b)
- The left ventricle has to pump blood to the rest of the body (systemic circulation), which is a much longer distance/has higher resistance.
- This requires a much higher pressure to be generated by thicker cardiac muscle.
- The right ventricle only pumps blood to the lungs (pulmonary circulation), which is close to the heart and requires lower pressure to prevent damage to delicate capillaries.

(c)
- Cardiac output formula: \(\text{Cardiac Output} = \text{Heart Rate} \times \text{Stroke Volume}\)
- Calculation of output in cm³: \(55 \text{ beats/min} \times 80 \text{ cm}^3 = 4400 \text{ cm}^3\text{/min}\)
- Conversion to dm³: \(4400 / 1000 = 4.4 \text{ dm}^3\text{/min}\)

Part (a) [3 marks]:
- (i) 1 mark for pulmonary artery (reject pulmonary vein)
- (ii) 1 mark for left atrium / auricle
- (iii) 1 mark for bicuspid valve / mitral valve (reject tricuspid)

Part (b) [3 marks]:
- 1 mark for stating the left ventricle pumps blood to the body / systemic circulation, whereas the right ventricle pumps to the lungs / pulmonary circulation.
- 1 mark for identifying that pumping to the body requires higher pressure / overcomes greater resistance.
- 1 mark for stating that too high pressure in pulmonary circulation would damage alveoli / capillaries in the lungs.

Part (c) [4 marks]:
- 1 mark for stating correct formula: \(\text{Cardiac Output} = \text{Heart Rate} \times \text{Stroke Volume}\)
- 1 mark for correct substituted calculation: \(55 \times 80\)
- 1 mark for calculating 4400 (cm³/min)
- 1 mark for correct final value in dm³: 4.4 (dm³/min)

(a)
- A pathogen is any microorganism (or agent) that causes infectious disease.

(b)
- The secondary immune response is much faster than the primary response (shorter lag phase).
- It produces a significantly higher concentration of antibodies.
- The concentration of antibodies remains high in the blood for a longer duration.
- This is because memory B-cells and memory T-cells remain in the blood after the first infection/vaccination.
- Upon second exposure, these memory cells recognise the antigen and rapidly divide by mitosis to produce plasma cells that secrete antibodies.

(c)
- When a high percentage of the population is vaccinated, they become immune to the disease.
- This limits the spread of the pathogen because there are fewer susceptible hosts to transmit it.
- As a result, unvaccinated individuals (e.g., newborn babies, immunocompromised people) are highly unlikely to come into contact with an infected person.
- The transmission cycle of the pathogen is broken.

Part (a) [1 mark]:
- 1 mark for definition: disease-causing microorganism / organism.

Part (b) [5 marks]:
- 1 mark for stating the response is faster / has a shorter lag phase.
- 1 mark for stating a higher concentration of antibodies is produced.
- 1 mark for stating antibody levels stay elevated for longer.
- 1 mark for mentioning the presence/role of memory cells (B or T cells).
- 1 mark for explaining that memory cells quickly differentiate/divide into antibody-producing plasma cells.

Part (c) [4 marks]:
- 1 mark for stating that a high proportion of the population is immunised.
- 1 mark for stating that vaccinated individuals cannot catch/transmit the disease.
- 1 mark for explaining that the chain of transmission is broken / pathogen cannot spread.
- 1 mark for linking this to protection of unvaccinated / vulnerable individuals.

(a)
- Receptors in the skin detect the high temperature (stimulus).
- An electrical impulse is generated and travels along the sensory neurone to the central nervous system (spinal cord).
- The impulse passes across a synapse to a relay neurone in the grey matter of the spinal cord.
- The impulse passes across another synapse to a motor neurone, which carries it to the effector (bicep muscle), causing it to contract and pull the hand away.

(b)
- An electrical impulse arrives at the presynaptic membrane.
- This causes vesicles containing neurotransmitters to fuse with the presynaptic membrane.
- Neurotransmitters are released into the synaptic cleft by exocytosis.
- They diffuse across the short gap (synaptic cleft) down a concentration gradient.
- They bind to complementary receptors on the postsynaptic membrane, generating a new electrical impulse.

(c)
- Nervous: faster speed of response; Endocrine: slower speed of response.
- Nervous: impulses travel via neurones; Endocrine: hormones travel through the bloodstream.
- Nervous: short-lasting effect; Endocrine: long-lasting effect.
- Nervous: localized response; Endocrine: widespread response.

Part (a) [4 marks]:
- 1 mark for receptor detecting stimulus and generating an impulse in sensory neurone.
- 1 mark for impulse traveling to CNS / spinal cord / coordinator.
- 1 mark for transmission through relay neurone to motor neurone.
- 1 mark for motor neurone carrying impulse to effector / muscle causing contraction.

Part (b) [4 marks]:
- 1 mark for arrival of impulse triggering release of neurotransmitter from vesicles.
- 1 mark for neurotransmitter diffusing across the synaptic cleft.
- 1 mark for binding to specific receptors on the postsynaptic membrane.
- 1 mark for triggering a new electrical impulse in the postsynaptic neurone.

Part (c) [2 marks]:
- Any two correct differences (1 mark each):
- Speed: nervous is faster vs endocrine is slower
- Pathway: nervous travels along neurones vs endocrine travels in blood
- Duration: nervous is short-lived vs endocrine is longer-lasting
- Target: nervous is highly localized vs endocrine is widespread/systemic

(a)
- Vasodilation: Arterioles near the skin surface dilate, allowing more blood to flow through capillaries close to the skin surface, increasing heat loss by radiation.
- Sweating: Sweat glands secrete sweat onto the skin surface. Heat energy from the body is used to evaporate the water in sweat, cooling the body down.
- Hair flattening: Erector muscles relax, causing hairs to lie flat against the skin, which prevents a layer of warm, insulating air from being trapped.

(b)
- Osmoreceptors in the hypothalamus detect the high solute concentration (low water potential) of the blood plasma.
- The pituitary gland is stimulated to release more ADH into the bloodstream.
- ADH travels to the kidneys, where it increases the permeability of the collecting ducts of the nephrons.
- More water is reabsorbed from the filtrate back into the blood capillaries, resulting in a small volume of highly concentrated urine.

(c)
- Gland: Pituitary gland (posterior pituitary)
- Nephron part: Collecting duct (accept distal convoluted tubule)

Part (a) [4 marks]:
- 1 mark for identifying vasodilation of arterioles near skin surface.
- 1 mark for explaining that vasodilation increases blood flow to surface capillaries, increasing heat loss via radiation.
- 1 mark for describing sweat production and cooling via evaporation.
- 1 mark for hair erector muscles relaxing / hairs lying flat to prevent trapping of insulating air.

Part (b) [4 marks]:
- 1 mark for hypothalamus / osmoreceptors detecting low water potential / high concentration of blood.
- 1 mark for increased secretion of ADH by the pituitary gland.
- 1 mark for ADH increasing the permeability of the collecting ducts.
- 1 mark for more water being reabsorbed back into the blood / producing a lower volume of concentrated urine.

Part (c) [2 marks]:
- 1 mark for pituitary gland.
- 1 mark for collecting duct (accept distal convoluted tubule; reject loop of Henle / Bowman's capsule).

(a)
- Highly folded surface with villi and microvilli, which vastly increases the surface area for diffusion/active transport.
- The wall of each villus is only one cell thick (single layer of epithelial cells), which provides a short diffusion pathway.
- Excellent capillary network (blood supply) within each villus, constantly carrying away absorbed nutrients to maintain a steep concentration gradient.
- Presence of lacteals to absorb fatty acids and glycerol.

(b)
- In the stomach, the enzyme pepsin (an endopeptidase working at low pH) begins the hydrolysis of proteins into shorter polypeptide chains.
- In the duodenum (small intestine), trypsin (secreted by the pancreas) continues breaking down proteins/polypeptides.
- Peptidases (exopeptidases) located in the membranes of the small intestine epithelial cells break down the small peptides into amino acids.
- Amino acids are the final soluble products ready for absorption.

(c)
- Temperature change (\(\Delta T\)) = \(33.0 - 18.0 = 15.0\) °C
- Energy absorbed by water (\(Q\)) = \(m \times c \times \Delta T = 20.0 \text{ g} \times 4.2 \text{ J/g/°C} \times 15.0\text{ °C} = 1260 \text{ J}\)
- Energy released per gram of food = \(1260 \text{ J} / 0.5 \text{ g} = 2520 \text{ J/g}\)

Part (a) [4 marks]:
- 1 mark for mentioning villi / microvilli to increase surface area.
- 1 mark for thin epithelium / wall of villus is one cell thick for a short diffusion distance.
- 1 mark for rich blood capillary network to maintain a steep concentration gradient.
- 1 mark for lacteals inside villi to absorb lipid/fat products.

Part (b) [4 marks]:
- 1 mark for protein digestion starting in stomach with pepsin.
- 1 mark for digestion continuing in duodenum/small intestine with trypsin (from pancreas).
- 1 mark for action of peptidases to break down polypeptides/peptides into amino acids.
- 1 mark for identifying amino acids as the final absorbed end-product.

Part (c) [2 marks]:
- 1 mark for correct working showing calculation of energy absorbed by water: \(20 \times 4.2 \times 15 = 1260\) (J).
- 1 mark for correct final answer of 2520 (J/g) (allow error carried forward from minor arithmetic slips in step 1, provided the division by 0.5 is correctly applied).

(a)
1. Set up five water baths at different temperatures (e.g., \(10^\circ\text{C}\), \(20^\circ\text{C}\), \(30^\circ\text{C}\), \(40^\circ\text{C}\), and \(50^\circ\text{C}\)).
2. Put a fixed volume (e.g., \(10\text{ cm}^3\)) of glucose solution of a known concentration into five separate, identical pieces of Visking tubing.
3. Tie both ends of each piece of Visking tubing securely to seal them.
4. Rinse the outside of each Visking tubing with distilled water to remove any traces of spilled glucose.
5. Submerge each Visking tubing in a test tube containing a fixed volume (e.g., \(20\text{ cm}^3\)) of distilled water.
6. Place each test tube into its designated water bath and leave them for a fixed time duration (e.g., 20 minutes).
7. Remove the tubing, mix the surrounding water, and measure the concentration of glucose using glucose test strips or a colorimeter after performing a Benedict's test.

(b)
(i) Independent variable: Temperature.
(ii) Dependent variable: Concentration of glucose in the surrounding distilled water / rate of diffusion of glucose.
(iii) Control variable: Any one from: surface area/length of Visking tubing, volume of glucose solution inside, concentration of glucose solution inside, volume of distilled water outside, or time duration.

(c)
Change in glucose concentration:
\(0.33\text{ mg/cm}^3 - 0.12\text{ mg/cm}^3 = 0.21\text{ mg/cm}^3\)

Percentage increase calculation:
\(\frac{0.21}{0.12} \times 100 = 175\%\)

(d)
At higher temperatures, the glucose and water molecules possess more kinetic energy. This causes them to move faster and results in more frequent collisions and a faster net rate of diffusion across the selectively permeable membrane.

(a) Max 6 marks:
- MP1: Mention of at least five different, controlled temperatures using water baths (1)
- MP2: Use of a fixed volume / concentration of glucose solution inside the Visking tubing (1)
- MP3: Tying/sealing both ends of the Visking tubing (1)
- MP4: Rinsing the outside of the Visking tubing with distilled water (1)
- MP5: Submerging the Visking tubing in a fixed volume of distilled water (1)
- MP6: Leaving the tubes for a set/controlled time duration (1)
- MP7: Method of detecting/measuring glucose concentration in the water (e.g., Benedict's reagent / glucose test strips) (1)

(b) 3 marks:
- (i) Temperature (1)
- (ii) Concentration of glucose in the water / rate of glucose diffusion (1)
- (iii) Volume of glucose inside / volume of water outside / length of Visking tubing / time duration (1)

(c) 2 marks:
- 1 mark for calculating the difference: \(0.33 - 0.12 = 0.21\) (1)
- 1 mark for the correct percentage: \(175\%\) (1) (Accept 175)

(d) 2 marks:
- MP1: Increased temperature increases the kinetic energy of molecules / molecules move faster (1)
- MP2: Leading to faster net movement / diffusion through the pores of the Visking tubing (1)

(a)
1. Use buffer solutions of at least five different pH values (e.g., pH 5, 6, 7, 8, 9).
2. Mix a fixed volume (e.g., \(5\text{ cm}^3\)) of starch solution with a fixed volume (e.g., \(2\text{ cm}^3\)) of a specific pH buffer solution in a test tube.
3. Add a fixed volume (e.g., \(2\text{ cm}^3\)) of amylase solution to the mixture and immediately start a stopwatch.
4. Prepare a spotting tile with a drop of iodine solution in each well.
5. At regular intervals (e.g., every 10 seconds), take a sample of the mixture using a pipette and add it to a drop of iodine on the tile.
6. Record the time when the iodine stops turning blue-black and remains orange-brown (showing all starch has been digested).
7. Repeat the experiment for each pH buffer, keeping the temperature constant (e.g., using a water bath).

(b)
(i) Independent variable: pH.
(ii) Dependent variable: Time taken for starch to be completely digested / rate of starch digestion.
(iii) Control variables (any two): Temperature, volume of starch solution, concentration of starch solution, volume of amylase solution, concentration of amylase solution.

(c)
(i) Calculation of rate:
\$$\text{rate} = \frac{1}{45\text{ s}} \approx 0.0222...\text{ s}^{-1}\$$
To 3 decimal places, the rate is \(0.022\text{ s}^{-1}\).

(ii) Explanation:
pH 7 is closer to the optimum pH of amylase than pH 5. At pH 5, the highly acidic conditions begin to denature the amylase molecules, changing the shape of their active sites. This prevents starch substrates from fitting into the active site, decreasing the formation of enzyme-substrate complexes and slowing down the reaction rate, which increases the time taken to digest the starch.

(a) Max 6 marks:
- MP1: Use of buffer solutions at different pH values (at least 5 values) (1)
- MP2: Constant/fixed volume of starch solution AND amylase solution (1)
- MP3: Use of iodine solution on a spotting tile to test for the presence of starch (1)
- MP4: Regular sampling intervals (e.g., every 10 or 20 seconds) (1)
- MP5: Accurate description of the endpoint (when iodine remains orange-brown / does not turn blue-black) (1)
- MP6: Use of a water bath to maintain a constant temperature (1)
- MP7: Repeats at each pH to calculate an average / ensure reliability (1)

(b) 4 marks:
- (i) pH (1)
- (ii) Time taken for starch to be digested / rate of digestion (1)
- (iii) Any two from: Temperature, concentration of starch, volume of starch, concentration of amylase, volume of amylase (2)

(c) 3 marks total:
- (i) \(0.022\) (1) (Accept 0.022 s^-1; reject other decimal places)
- (ii) MP1: pH 7 is closer to the optimum pH of the enzyme / pH 5 is further from the optimum (1)
- MP2: At pH 5, active sites are altered/partially denatured, resulting in fewer enzyme-substrate complexes / slower rate of reaction (1)