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The correct terms to complete the passage are:
(i) **hypothalamus**: contains the osmoreceptors that detect osmotic changes in the blood plasma.
(ii) **pituitary**: the endocrine gland that stores and releases ADH under instructions from the hypothalamus.
(iii) **ADH** (or **antidiuretic hormone**): the specific hormone regulating water reabsorption.
(iv) **collecting duct**: the specific part of the nephron whose permeability to water is altered by ADH.
(v) **osmosis**: the passive movement of water molecules down a water potential gradient.
(vi) **decrease** (or **reduce**): because more water is reabsorbed into the bloodstream, less water is excreted, leading to a smaller volume of highly concentrated urine.

One mark is awarded for each correct term:
- (i) hypothalamus [1 mark] (accept: brain)
- (ii) pituitary [1 mark]
- (iii) ADH / antidiuretic hormone [1 mark]
- (iv) collecting duct / collecting ducts [1 mark] (accept: distal convoluted tubule / DCT)
- (v) osmosis [1 mark] (reject: diffusion / active transport)
- (vi) decrease / reduce / lower / fall [1 mark] (reject: increase)

To complete the passage:
(i) The hypothalamus contains osmoreceptors that monitor the concentration of blood plasma.
(ii) The hypothalamus stimulates the pituitary gland to release ADH.
(iii) The hormone is ADH (antidiuretic hormone).
(iv) ADH increases the permeability of the collecting duct of the nephron to water.
(v) Water is reabsorbed by osmosis.
(vi) Since more water is reabsorbed, the volume of urine produced decreases.

1 mark for each correct response:
(i) hypothalamus
(ii) pituitary
(iii) ADH / antidiuretic hormone
(iv) collecting duct / collecting tubules
(v) osmosis
(vi) decrease / decline / drop / reduce

(a) Temperature rise \(\Delta T = 46.5\,^{\circ}\text{C} - 18.5\,^{\circ}\text{C} = 28.0\,^{\circ}\text{C}\).
(b) Thermal energy transferred \(Q = m \times c \times \Delta T = 20.0\text{ g} \times 4.2\text{ J/g}^{\circ}\text{C} \times 28.0\,^{\circ}\text{C} = 2352\text{ J}\).
(c) Energy per gram in J/g = \(2352\text{ J} / 0.50\text{ g} = 4704\text{ J/g}\). To convert to kJ/g, divide by 1000: \(4704 / 1000 = 4.704\text{ kJ/g}\) (accept 4.7 or 4.70 kJ/g).
(d) Significant heat energy is lost to the surrounding air and the glass test tube rather than heating the water. Also, the peanut may not have burnt completely (incomplete combustion).
(e) Improvements: 1. Use a metal (copper) container instead of a glass test tube as metal is a better conductor of heat. 2. Place a lid on the container to prevent heat loss from the surface of the water. 3. Use draught shields around the apparatus. 4. Use a bomb calorimeter with a constant oxygen supply to ensure complete combustion.
(f) Bile is produced in the liver and stored in the gall bladder. It emulsifies large lipid droplets into many tiny droplets, increasing the surface area for lipase enzymes to digest lipids into fatty acids and glycerol. It is also alkaline, so it neutralises the acidic contents entering from the stomach to provide the optimum pH for pancreatic lipase.

(a) [1 mark]
- 28.0 (°C) (1)

(b) [2 marks]
- Correct substitution: \(20.0 \times 4.2 \times 28.0\) (1)
- 2352 (J) (1)

(c) [3 marks]
- Divide energy by mass: \(2352 / 0.50 = 4704\,(\text{J/g})\) (1)
- Divide by 1000 to convert to kJ: \(4704 / 1000\) (1)
- Correct final answer: 4.704 or 4.7 (kJ/g) (1)

(d) [2 marks]
- Heat/thermal energy is lost to the surrounding air / glass boiling tube (1)
- Incomplete combustion of the peanut / some unburnt peanut remaining (1)

(e) [3 marks]
- Any three from: Use a metal can/copper calorimeter (1); Add a lid to the container (1); Use draft shields/insulation (1); Stir the water regularly (1); Burn the sample in pure oxygen / use a bomb calorimeter (1)

(f) [3 marks]
- Emulsification of lipids / breaking large droplets into smaller droplets (1)
- Increases surface area for lipase action (1)
- Neutralises stomach acid / provides alkaline pH for pancreatic lipase (1)

(a) Cardiac output is the total volume of blood pumped by the left ventricle of the heart per minute.
(b) \(\text{Cardiac output} = \text{stroke volume} \times \text{heart rate}\).
(c) Resting cardiac output = \(72\text{ bpm} \times 75\text{ cm}^3 = 5400\text{ cm}^3\text{/min}\). To convert to \(\text{dm}^3\text{/min}\), divide by 1000: \(5400 / 1000 = 5.4\text{ dm}^3\text{/min}\).
(d) Given exercise cardiac output = \(21.45\text{ dm}^3\text{/min} = 21450\text{ cm}^3\text{/min}\). \(\text{Stroke volume} = \text{Cardiac output} / \text{Heart rate} = 21450 / 165 = 130\text{ cm}^3\).
(e) During exercise, blood flow is redirected. Arterioles supplying active skeletal muscles undergo vasodilation (widen), increasing blood flow to them. Conversely, arterioles supplying non-essential organs (such as the digestive tract and kidneys) undergo vasoconstriction (narrow), reducing blood flow. This ensures that more oxygen and glucose are delivered to the contracting muscle cells for aerobic respiration, and carbon dioxide is removed more rapidly.
(f) The left ventricle pumps blood into the systemic circulation. Its muscular wall is much thicker than that of the right ventricle because it needs to generate high enough pressure to pump blood all around the body, whereas the right ventricle only pumps blood a short distance to the lungs under lower pressure.

(a) [1 mark]
- Volume of blood pumped (by the left ventricle / heart) per minute (1)

(b) [1 mark]
- Cardiac output = stroke volume × heart rate (or CO = SV × HR) (1)

(c) [3 marks]
- Substitution: \(72 \times 75\) (1)
- Correct calculation in cm³: 5400 (cm³/min) (1)
- Correct conversion and final answer: 5.4 (dm³/min) (1)

(d) [3 marks]
- Convert cardiac output to cm³: \(21.45 \times 1000 = 21450\,(\text{cm}^3\text{/min})\) (1)
- Rearrangement and substitution: \(21450 / 165\) (1)
- Correct final answer: 130 (cm³) (1)

(e) [4 marks]
- Vasodilation of arterioles / blood vessels supplying active skeletal muscles (1)
- Vasoconstriction of arterioles / blood vessels supplying digestive organs / kidneys / skin (1)
- This diverts more blood (carrying oxygen and glucose) to the working muscles (1)
- To support increased rate of aerobic respiration / release more energy (1)

(f) [2 marks]
- Left ventricle (1)
- Thicker muscular wall (than the right ventricle) (1)

(a) Resting minute ventilation = \(\text{Breathing rate} \times \text{Tidal volume} = 15\text{ breaths/min} \times 0.48\text{ dm}^3\text{/breath} = 7.2\text{ dm}^3\text{/min}\).
(b) After exercise: \(\text{Minute ventilation} = \text{Breathing rate} \times \text{Tidal volume}\).
Rearranging gives: \(\text{Tidal volume} = \text{Minute ventilation} / \text{Breathing rate} = 36.4\text{ dm}^3\text{/min} / 28\text{ breaths/min} = 1.3\text{ dm}^3\).
(c) Resting tidal volume = 0.48 dm³; Exercise tidal volume = 1.30 dm³.
Increase in tidal volume = \(1.30 - 0.48 = 0.82\text{ dm}^3\).
Percentage increase = \((0.82 / 0.48) \times 100 = 170.833...\%\). To 1 decimal place, this is 170.8%.
(d) The alveoli are adapted in several ways: 1. They provide a very large total surface area across the lungs. 2. Their walls are only one cell thick (made of flattened squamous epithelium), which minimizes the diffusion distance for gases. 3. They are surrounded by an extensive network of capillaries which maintains a steep concentration gradient by continually removing oxygen and delivering carbon dioxide. 4. They have a moist lining, allowing oxygen to dissolve before diffusing across the membrane.
(e) During inspiration: 1. The external intercostal muscles contract, pulling the ribcage upwards and outwards. 2. The diaphragm contracts and flattens, increasing the volume of the thoracic cavity (thereby decreasing air pressure inside the lungs, drawing air in).

(a) [2 marks]
- Substitution: \(15 \times 0.48\) (1)
- Correct answer: 7.2 (dm³/min) (1)

(b) [3 marks]
- Rearrangement: \(\text{Tidal volume} = \text{Minute ventilation} / \text{Breathing rate}\) (1)
- Substitution: \(36.4 / 28\) (1)
- Correct answer: 1.3 or 1.30 (dm³) (1)

(c) [3 marks]
- Calculate difference: \(1.30 - 0.48 = 0.82\) (1)
- Division by original value and multiplication by 100: \((0.82 / 0.48) \times 100\) (1)
- Correct final value to 1 d.p.: 170.8 (%) (1)

(d) [4 marks]
- Large surface area (due to millions of alveoli) (1)
- Short diffusion pathway / thin wall / only one cell thick (1)
- Excellent blood supply / dense network of capillaries to maintain steep concentration gradient (1)
- Moist lining allows gases to dissolve to assist diffusion (1)

(e) [2 marks]
- External intercostal muscles contract to pull ribcage up and out (1)
- Diaphragm contracts and flattens (1)

(a) To plot the graph:
- Label the x-axis as 'Environmental Temperature (°C)' and the y-axis as 'Mean Sweat Production (g/min)'.
- Choose a scale that uses at least half of the grid (e.g., x-axis from 20 to 40 with 2 cm representing 5 °C, and y-axis from 0 to 3.5 with 2 cm representing 0.5 g/min).
- Plot all 6 points accurately: (20, 0.2), (24, 0.4), (28, 0.8), (32, 1.4), (36, 2.2), (40, 3.2).
- Draw a smooth curve or use a ruler to connect the points sequentially.

(b) The trend is that as environmental temperature increases, the mean sweat production also increases. The increase is not constant (non-linear); the rate of sweating increases more rapidly at higher temperatures (e.g., an increase of 0.2 g/min between 20 °C and 24 °C compared to 1.0 g/min between 36 °C and 40 °C).

(c) An increase in external temperature is detected by thermoreceptors in the skin and internal temperature changes are detected by receptors in the hypothalamus. The hypothalamus (thermoregulatory center) coordinates the response, sending nerve impulses to the sweat glands in the dermis of the skin. The sweat glands secrete sweat onto the skin surface. As the water in sweat evaporates, it takes latent heat of vaporization away from the blood vessels in the dermis, lowering body temperature back to the set point.

(d) Hypothalamus.

(e) Calculation:
- Sweat production at 28 °C = 0.8 g/min
- Sweat production at 36 °C = 2.2 g/min
- Increase = \(2.2 - 0.8 = 1.4\text{ g/min}\)
- Percentage increase = \(\frac{1.4}{0.8} \times 100 = 175\%\)

(a) Graph [5 marks]:
- Axes labeled correctly with units: 'Environmental Temperature (°C)' on horizontal axis, 'Mean Sweat Production (g/min)' on vertical axis [1]
- Suitable linear scale chosen so that points occupy at least half of the grid [1]
- Points plotted accurately to within +/- half a small square [1]
- Clean, smooth curve or straight lines connecting points (no thick lines or double lines) [1]
- Graph is correctly oriented (independent variable on x-axis, dependent on y-axis) [1]

(b) Description [2 marks]:
- Sweat production increases as temperature increases [1]
- The rate of increase becomes greater/steeper at higher temperatures (non-linear relationship) [1]

(c) Explanation [4 marks]:
- Rise in temperature detected by thermoreceptors (in skin/hypothalamus) [1]
- Nerve impulses sent from the hypothalamus to sweat glands [1]
- Sweat glands secrete sweat onto the surface of the skin [1]
- Evaporation of water in sweat transfers/absorbs heat energy (latent heat) from the body/skin, resulting in cooling [1]

(d) Brain structure [1 mark]:
- Hypothalamus [1]
*(Reject: Pituitary gland, Cerebrum)*

(e) Calculation [1 mark]:
- Correct answer of 175% [1]
*(Accept correct working showing \(\frac{2.2 - 0.8}{0.8} \times 100\))*

(a) To plot the graph:
- Label the x-axis as 'Time (minutes)' and the y-axis as 'Heart Rate (beats per minute / bpm)'.
- Select a scale that fills at least half the page (e.g., x-axis: 2 cm = 2 minutes; y-axis: 2 cm = 20 bpm, starting from 60 or 0).
- Plot all 10 coordinates carefully.
- Connect the points with straight, ruled lines or a smooth curve.

(b) Between 2 and 10 minutes, the student is performing vigorous exercise:
- Active muscle tissues contract more rapidly, which requires more energy in the form of ATP.
- This energy is released via aerobic respiration, demanding a higher supply of oxygen and glucose.
- Respiration also produces carbon dioxide, which must be removed to prevent cellular pH from dropping.
- The heart rate increases to pump blood more quickly, ensuring faster delivery of oxygen and glucose, and rapid removal of carbon dioxide from the muscles to the lungs.

(c) After high-intensity exercise ceases at 10 minutes, the heart rate remains elevated because:
- Muscle cells carried out anaerobic respiration during intense exercise due to insufficient oxygen supply.
- This led to the accumulation of lactic acid, creating an 'oxygen debt'.
- The heart rate stays elevated to transport oxygenated blood to the liver and muscles to break down/oxidize the toxic lactic acid into carbon dioxide and water.

(d) Calculation:
- \(\text{Cardiac Output} = \text{Heart Rate} \times \text{Stroke Volume}\)
- At 10 minutes, Heart Rate = 150 bpm
- Stroke Volume = 80 cm³
- \(\text{Cardiac Output} = 150 \text{ bpm} \times 80 \text{ cm}^3 = 12,000\text{ cm}^3\text{/min}\) (or \(12\text{ dm}^3\text{/min}\) / \(12\text{ liters/min}\)).

(a) Graph [5 marks]:
- Correct labels with units on both axes: 'Time (minutes)' on x-axis and 'Heart Rate (beats per minute / bpm)' on y-axis [1]
- Scale is linear, appropriate, and uses more than half of the grid [1]
- All 10 points plotted correctly to within half a small square [1]
- Points connected with a neat line (either smooth curve or straight point-to-point lines) [1]
- No extrapolation beyond the plotted data points (0 to 18 min) [1]

(b) Explanation of heart rate increase [4 marks]:
- Muscular contraction during exercise requires more energy / ATP [1]
- Increased rate of aerobic respiration is needed [1]
- More oxygen and glucose must be delivered to working muscle cells [1]
- Carbon dioxide (waste product) must be removed more rapidly [1]
- Heart rate increases to pump blood faster to meet these cellular demands [1]
*(Max 4 marks)*

(c) Post-exercise recovery [3 marks]:
- Anaerobic respiration occurred during exercise [1]
- Lactic acid accumulated in the muscle tissue, building up an oxygen debt [1]
- High heart rate is maintained to transport oxygen to the muscles/liver to oxidize/break down lactic acid [1]

(d) Calculation [1 mark]:
- Correct answer of 12,000 cm³/min (or 12 dm³/min or 12 L/min) with appropriate unit [1]
*(Accept 12,000 with any correct variation of unit: cm³/min, mL/min, dm³/min, L/min. Deduct mark if unit is missing or incorrect)*

**(a)**
An experimental design can be structured using the CORMS approach:
- **C** (Change): Change the independent variable, which is the temperature. This can be achieved by placing the yeast and glucose mixtures in water baths set at different temperatures (e.g., 20 °C, 30 °C, 40 °C, 50 °C, and 60 °C).
- **O** (Organism): Use the same species and concentration of yeast throughout the investigation.
- **R** (Repeat): Repeat the experiment at least three times at each temperature to calculate a reliable mean and identify anomalies.
- **M** (Measure): Measure the volume of carbon dioxide gas produced using a gas syringe (or count the number of bubbles produced) over a set time period.
- **S** (Same): Control variables such as the volume and concentration of the glucose solution, the volume of yeast suspension, and ensure anaerobic conditions by adding a layer of liquid paraffin/oil on top of the mixture to exclude oxygen.

**(b)**
Three key control variables include:
1. Concentration of the glucose solution.
2. Volume of the yeast suspension.
3. pH of the mixture (controlled using a buffer solution).

**(c)**
1. Calculate the mean volume of carbon dioxide produced at 40 °C:
\(\text{Mean Volume} = \frac{12.4 + 13.1 + 11.7}{3} = \frac{37.2}{3} = 12.4\text{ cm}^3\)
2. Calculate the rate per minute by dividing the mean volume by the total time (10 minutes):
\(\text{Rate} = \frac{12.4\text{ cm}^3}{10\text{ min}} = 1.24\text{ cm}^3\text{/min}\)

**(d)**
When temperature increases to 60 °C, the excess thermal energy breaks the bonds holding the tertiary structure of the respiratory enzymes together. This causes the enzymes to denature, changing the specific shape of their active sites. Consequently, the substrate (glucose) can no longer fit into the active site, preventing the formation of enzyme-substrate complexes and stopping anaerobic respiration.

**(a) Experimental Method [Max 6 marks]**
- **C**: State how to vary the temperature using water baths (at least 3 different temperatures) (1)
- **O**: Use the same concentration/source of yeast (1)
- **R**: Repeat each temperature at least 3 times and calculate a mean (1)
- **M1**: Measure volume of gas produced using a gas syringe / measuring cylinder over water (or count bubbles) (1)
- **M2**: State a specific, realistic time frame for gas measurement (e.g., 5 to 10 minutes) (1)
- **S1**: Keep volume/concentration of glucose solution constant (1)
- **S2**: Keep yeast suspension volume constant (1)
- **S3**: Use a layer of paraffin/oil to ensure anaerobic conditions (1)

**(b) Control Variables [Max 3 marks]**
- Concentration of glucose solution (1)
- Volume of yeast suspension (1)
- pH of the mixture (1)
*(Accept: mass/concentration of yeast)*

**(c) Calculation [3 marks]**
- Summing up values and finding correct mean volume of 12.4 cm³ (1)
- Dividing mean volume by 10 minutes (1)
- Correct final answer of 1.24 with units of cm³/min (or cm³ min⁻¹) (1)
*(Allow error carried forward (ECF) from step 1 to step 2)*

**(d) Explanation [3 marks]**
- Enzymes are denatured at high temperatures / 60 °C (1)
- The active site of the enzymes changes shape (1)
- Substrate can no longer bind to the active site / no enzyme-substrate complexes are formed (1)

The correct completed passage is: When the water potential of the blood plasma is too low, specialized osmoreceptor cells in the (i) **hypothalamus** detect this change. These cells stimulate the (ii) **pituitary** gland to secrete more of the hormone (iii) **ADH** (or antidiuretic hormone) into the bloodstream. This hormone is transported to the kidneys, where it binds to receptors on the cells of the (iv) **collecting duct**, increasing their permeability to water. Consequently, water moves out of the filtrate and back into the blood capillaries by the process of (v) **osmosis**. This homeostatic response results in the excretion of a small volume of (vi) **concentrated** (or hypertonic) urine.

1 mark for each correct term:
- (i) hypothalamus (accept: osmoregulatory center / brain)
- (ii) pituitary / posterior pituitary (reject: other glands)
- (iii) ADH / antidiuretic hormone
- (iv) collecting duct (accept: collecting ducts / distal convoluted tubule / DCT)
- (v) osmosis (reject: active transport / diffusion)
- (vi) concentrated / hypertonic (accept: low volume / high concentration / yellow)

(a) When a person is dehydrated, the water potential of the blood decreases. This change is detected by osmoreceptors in the hypothalamus, which stimulates the posterior pituitary gland to secrete more antidiuretic hormone (ADH) into the bloodstream. ADH travels to the kidneys, where it binds to receptors on the cells of the collecting ducts. This increases the permeability of the collecting duct walls to water by causing the insertion of aquaporins (water channels) into the cell membranes. Consequently, more water is reabsorbed from the filtrate back into the blood capillaries of the medulla by osmosis. This produces a low volume of highly concentrated urine and helps restore the blood water potential back to normal.

(b) (i) First, calculate the difference in urine volume: \(6.2 \text{ dm}^3 - 0.8 \text{ dm}^3 = 5.4 \text{ dm}^3\).
Next, calculate the percentage decrease relative to the original volume (Period A):
\(\frac{5.4}{6.2} \times 100 = 87.096...\%\)
Rounding to 3 significant figures gives \(87.1\%\) (or 87%).

(ii) For Period A: \(6.2 \text{ dm}^3 \times 80 \text{ mmol/dm}^3 = 496 \text{ mmol}\).
For Period B: \(0.8 \text{ dm}^3 \times 600 \text{ mmol/dm}^3 = 480 \text{ mmol}\).
Comparison and significance: The total amount of solute excreted is very similar (496 mmol vs 480 mmol). This indicates that ADH specifically regulates the reabsorption of water (volume control) rather than the excretion of solutes. The kidney still successfully excretes similar amounts of waste solutes under both conditions to maintain homeostatic solute balance, but alters the water volume in which they are dissolved.

(c) In a patient with diabetes mellitus, blood glucose levels are extremely high and exceed the renal threshold for reabsorption in the proximal convoluted tubule. As a result, not all glucose is reabsorbed, and some remains in the nephron filtrate. This glucose lowers the water potential of the filtrate, reducing the osmotic gradient between the filtrate and the surrounding medulla tissue. Therefore, less water is reabsorbed by osmosis out of the tubule, resulting in a large volume of urine containing glucose. In contrast, a diabetes insipidus patient has normal blood glucose levels and normal proximal convoluted tubule function, so all glucose is reabsorbed, meaning no glucose is present in their urine.

(a) Up to 5 marks:
- Blood water potential decreases / solute concentration increases [1]
- Detected by osmoreceptors in hypothalamus [1]
- Pituitary gland releases more ADH [1]
- ADH increases permeability of collecting duct (to water) [1]
- By inserting aquaporins / water channels [1]
- Water reabsorbed by osmosis (back into blood) [1]
- Low volume / concentrated urine produced [1]

(b) (i) Up to 2 marks:
- Correct working showing change divided by original: \((6.2 - 0.8) / 6.2\) [1]
- Correct calculation: \(87.1\%\) or \(87\%\) [1] (Accept \(87.096...\%\))

(b) (ii) Up to 3 marks:
- Correct calculation of solutes: Period A = \(496 \text{ mmol}\) AND Period B = \(480 \text{ mmol}\) [1]
- Solute amounts are very similar / almost unchanged [1]
- Explanation: ADH regulates water reabsorption/volume, not solute excretion / waste solutes must still be eliminated regardless of hydration state [1]

(c) Up to 3 marks:
- In diabetes mellitus, high blood glucose exceeds the capacity of proximal convoluted tubule to reabsorb it / glucose remains in filtrate [1]
- Glucose lowers water potential of filtrate, reducing water reabsorption by osmosis [1]
- In diabetes insipidus, blood glucose is normal so all glucose is reabsorbed, resulting in glucose-free urine [1]

(a) At the start of exercise, muscle activity and cellular respiration increase, producing more carbon dioxide. This increases the concentration of hydrogen ions, lowering the pH of the blood. This change is detected by chemoreceptors located in the carotid arteries and the arch of the aorta. These receptors send nerve impulses to the cardiovascular center in the medulla oblongata of the brain. The medulla then sends impulses along the sympathetic nerve to the sinoatrial node (SAN / pacemaker) of the heart, causing it to increase the rate of electrical impulses, thereby increasing the heart rate.

(b) (i) For Individual X:
Peak \(CO = 185 \text{ bpm} \times 110 \text{ cm}^3 = 20350 \text{ cm}^3\text{/min}\)
To convert to \(\text{dm}^3\text{/min}\): \(20350 / 1000 = 20.35 \text{ dm}^3\text{/min}\) (or \(20.4\))

For Individual Y:
Peak \(CO = 165 \text{ bpm} \times 160 \text{ cm}^3 = 26400 \text{ cm}^3\text{/min}\)
To convert to \(\text{dm}^3\text{/min}\): \(26400 / 1000 = 26.40 \text{ dm}^3\text{/min}\) (or \(26.4\))

(ii) Individual Y (the athlete) achieves a significantly higher peak cardiac output (\(26.40 \text{ dm}^3\text{/min}\) compared to \(20.35 \text{ dm}^3\text{/min}\) for Individual X) despite having a lower maximum heart rate (\(165 \text{ bpm}\) vs \(185 \text{ bpm}\)). This superior performance is due to Individual Y's much larger stroke volume (\(160 \text{ cm}^3\) vs \(110 \text{ cm}^3\)). The physiological adaptation responsible is hypertrophy of the heart muscle (stronger, larger ventricles), which allows more blood to be pumped per contraction, facilitating a greater volume of oxygen-carrying red blood cells to reach active tissues per minute with less cardiac strain.

(c) During exercise, active muscles require a continuous and rapid supply of ATP for contraction. Aerobic respiration is highly efficient but requires oxygen as the terminal electron acceptor. If cardiac output does not increase sufficiently, the delivery of oxygenated blood cannot keep pace with demand. The muscles are forced to switch to anaerobic respiration to produce ATP. Anaerobic respiration is inefficient (producing only 2 ATP per glucose molecule compared to 36-38 in aerobic) and produces lactic acid. Lactic acid buildup lowers pH, causing muscle fatigue, pain, and eventual failure of contraction. Therefore, a high cardiac output is necessary to maintain fully aerobic conditions.

(a) Up to 4 marks:
- Increased carbon dioxide / lowered pH detected by chemoreceptors [1]
- In carotid arteries / aorta [1]
- Nerve impulses sent to cardiovascular center in medulla oblongata [1]
- Medulla sends impulses along sympathetic nerve [1]
- To sinoatrial node (SAN) / pacemaker to increase heart rate [1]

(b) (i) Up to 3 marks:
- Correct formula application / multiplication of HR and SV [1]
- Division by 1000 to convert to \(\text{dm}^3\text{/min}\) [1]
- Correct final values: Individual X = \(20.35 \text{ dm}^3\text{/min}\) AND Individual Y = \(26.40 \text{ dm}^3\text{/min}\) (allow 20.4 and 26.4) [1]

(b) (ii) Up to 3 marks:
- Athlete has higher peak cardiac output but lower peak heart rate [1]
- Athlete has a much larger stroke volume [1]
- Explanation: Heart muscle hypertrophy / stronger ventricular walls allows more blood per beat, providing more efficient oxygen delivery with less strain on the heart muscle [1]

(c) Up to 3 marks:
- Aerobic respiration requires a continuous oxygen supply to produce ATP [1]
- Insufficient oxygen delivery forces muscles to switch to anaerobic respiration [1]
- Anaerobic respiration produces lactic acid / causes fatigue / produces less ATP per glucose [1]

For resting cardiac output: \(\text{Cardiac Output} = \text{Heart Rate} \times \text{Stroke Volume} = 72 \times 70 = 5040\text{ cm}^3/\text{min}\). Converting to \(\text{dm}^3\): \(5040 / 1000 = 5.04\text{ dm}^3/\text{min}\). For exercise cardiac output: \(\text{Cardiac Output} = 150 \times 110 = 16500\text{ cm}^3/\text{min}\). Converting to \(\text{dm}^3\): \(16500 / 1000 = 16.50\text{ dm}^3/\text{min}\).

(a) Rest calculation: \(72 \times 70 = 5040\text{ cm}^3/\text{min}\) (1 mark); Conversion factor dividing by 1000 for both values (1 mark); Correct values of 5.04 dm3/min and 16.50 dm3/min (1 mark).

(b) Nervous coordination: High CO2 / low blood pH is detected by chemoreceptors in aorta/carotid arteries (1 mark); Nerve impulses are sent to the medulla oblongata (1 mark); Impulses travel down the sympathetic nerve to the sinoatrial node (SAN) to increase heart rate (1 mark). Endocrine coordination: Adrenal glands secret adrenaline (1 mark); Adrenaline is transported through the bloodstream to the heart (1 mark); Adrenaline binds to receptors on the sinoatrial node (SAN) to increase the rate and force of contraction (1 mark).

(c) Benefits: More oxygen and glucose delivered to active skeletal muscles (1 mark); increases the rate of aerobic respiration to meet higher energy demands (1 mark); produces more ATP for muscle contraction (1 mark); faster removal of metabolic waste products like carbon dioxide (1 mark) and lactic acid (1 mark); delays onset of muscle fatigue and minimizes oxygen debt (1 mark).

Temperature change (\(\Delta T\)) = \(39.0 - 21.0 = 18.0\text{ degrees C}\). Heat absorbed by water (\(Q\)) = \(m \times c \times \Delta T = 20\text{ g} \times 4.2\text{ J/g degrees C} \times 18\text{ degrees C} = 1512\text{ Joules}\). Convert to kilojoules (kJ): \(1512 / 1000 = 1.512\text{ kJ}\). Energy content per gram of food = \(\text{Total energy} / \text{mass of food} = 1.512\text{ kJ} / 1.5\text{ g} = 1.008\text{ kJ/g}\) (or 1.01 kJ/g).

(a) Temperature difference of 18 degrees C calculated (1 mark); Total heat energy calculated in Joules: \(20 \times 4.2 \times 18 = 1512\text{ J}\) (1 mark); Conversion to kJ: 1.512 kJ (1 mark); Dividing energy by mass of food to get 1.008 or 1.01 kJ/g (1 mark).

(b) Error 1: Heat lost to surroundings / boiling tube (1 mark); Minimize by using an insulated container / draft shield / bomb calorimeter (1 mark). Error 2: Incomplete combustion of food (1 mark); Minimize by ensuring food is completely burned by immediately relighting it until it no longer burns (1 mark).

(c) Digestion: Bile (produced in liver, stored in gallbladder) emulsifies lipids into smaller droplets to increase surface area (1 mark); Lipase enzyme from pancreas breaks down lipids into fatty acids and glycerol (1 mark). Absorption: Fatty acids and glycerol diffuse into the epithelial cells of the villi in the ileum (1 mark); they are reassembled into triglycerides (1 mark) and enter the lacteals / lymphatic capillaries (1 mark). Energy differences: Lipids have a higher proportion of hydrogen-carbon bonds relative to oxygen than carbohydrates (1 mark); therefore, oxidation of lipids yields significantly more ATP/energy per unit mass than carbohydrates (1 mark).

Increase in blood glucose for Person B = \(14.5 - 7.0 = 7.5\text{ mmol/dm}^3\). Percentage increase = \((7.5 / 7.0) \times 100 = 107.142...\%\). Rounded to one decimal place = 107.1%.

(a) Calculate actual glucose increase: \(14.5 - 7.0 = 7.5\text{ mmol/dm}^3\) (1 mark); Calculate percentage increase: \((7.5 / 7.0) \times 100\) (1 mark); Correct answer rounded to 1 decimal place: 107.1% (1 mark).

(b) Role of pancreas: High blood glucose levels detected by beta-cells in the Islets of Langerhans (1 mark); Pancreas responds by secreting insulin hormone into the blood (1 mark). Role of liver: Insulin travels in blood and binds to complementary receptors on hepatocytes/liver cells (1 mark); Insulin stimulates liver cells to absorb more glucose (1 mark); Liver cells convert excess glucose to glycogen / glycogenesis (1 mark); Blood glucose levels decrease back to the normal range / negative feedback loop completed (1 mark).

(c) Type 1 causes/treatment: Caused by autoimmune destruction of beta-cells in the pancreas, leading to absolute lack of insulin production (1 mark); treated with regular insulin injections / blood glucose monitoring (1 mark). Type 2 causes/treatment: Caused by body cells becoming resistant / unresponsive to insulin, often associated with obesity, age, or poor diet (1 mark); treated with lifestyle modifications like reduced carbohydrate diet, weight loss, and exercise (1 mark) or oral medication like metformin (1 mark). Age of onset difference: Type 1 typically has early onset in life / childhood, whereas Type 2 onset is usually later in life / adulthood (1 mark).

For healthy alveolus (r = 100 micrometers): \(\text{SA} = 4 \times \pi \times 100^2 = 40000\pi\). \(\text{V} = \frac{4}{3} \times \pi \times 100^3 = \frac{4000000}{3}\pi\). \(\text{SA:V} = \frac{40000\pi}{\frac{4000000}{3}\pi} = \frac{3}{100} = 0.03\mu\text{m}^{-1}\). For emphysematous alveolus (r = 300 micrometers): \(\text{SA} = 4 \times \pi \times 300^2 = 360000\pi\). \(\text{V} = \frac{4}{3} \times \pi \times 300^3 = \frac{108000000}{3}\pi = 36000000\pi\). \(\text{SA:V} = \frac{360000\pi}{36000000\pi} = \frac{3}{300} = 0.01\mu\text{m}^{-1}\).

(a) Correct calculation of Surface Area and Volume (or simplified expression \(SA:V = 3/r\)) for healthy alveolus (1 mark); Healthy SA:V ratio calculated as 0.03 (micrometers^-1) (1 mark); Correct calculation of Surface Area and Volume (or simplified expression \(SA:V = 3/r\)) for emphysematous alveolus (1 mark); Emphysematous SA:V ratio calculated as 0.01 (micrometers^-1) (1 mark).

(b) Adaptations: Alveolar wall is only one cell thick / squamous epithelium (1 mark); providing a very short diffusion pathway for gases (1 mark); Surrounded by a dense capillary network (1 mark); which maintains a steep concentration gradient for oxygen and carbon dioxide (1 mark); Moist lining allows gases to dissolve before diffusing (1 mark).

(c) Smoke effects: Cigarette smoke contains toxic chemicals and irritants that cause inflammation in the lungs (1 mark); This inflammation attracts white blood cells / phagocytes (1 mark); Phagocytes release elastase / protease enzymes (1 mark); These enzymes digest and break down elastin / elastic fibers in the alveolar walls (1 mark); Alveolar walls rupture and fuse, creating larger and fewer air spaces (1 mark); This severely reduces the total surface area and the surface area to volume ratio, significantly lowering the rate of gas exchange / oxygen absorption (1 mark).