2024 Edexcel IGCSE Human Biology 模擬試題連答案詳解

(a) Tuberculosis is caused by a bacterium (specifically Mycobacterium tuberculosis). It is transmitted via airborne droplets when an infected person coughs, sneezes, or speaks, and another person inhales these droplets.

(b) When a patient starts taking antibiotics, the most susceptible bacteria are killed first. This makes the patient feel better. However, some bacteria are more resistant and survive longer. If the treatment is stopped early, these surviving, more resistant bacteria will multiply, causing a relapse of the infection. Furthermore, this selection pressure contributes to the development of widespread antibiotic resistance (such as MDR-TB), rendering standard antibiotics ineffective.

(c) Methods to control the spread include: vaccination using the BCG vaccine; isolating infected individuals during the contagious phase; improving ventilation in crowded spaces; and trace-and-test programs for close contacts.

(d) Calculation:
Increase in cases = \(1830 - 1500 = 330\)
Percentage increase = \(\frac{330}{1500} \times 100 = 22\%\)

(a) [3 marks total]
- 1 mark for identifying the pathogen as a bacterium / bacteria.
- 1 mark for mentioning airborne transmission / respiratory droplets.
- 1 mark for detail (e.g., droplets produced by coughing, sneezing, or talking, which are then inhaled by others).

(b) [4 marks total]
- 1 mark for noting that the initial phase of antibiotics kills the least resistant / most susceptible bacteria first.
- 1 mark for explaining that stopping early allows more resistant bacteria to survive.
- 1 mark for explaining that these surviving bacteria will multiply / cause a relapse.
- 1 mark for linking early cessation to the development of antibiotic resistance / superbugs.

(c) [2 marks total]
- 1 mark each for any two sensible control measures, such as: BCG vaccination; isolation of infectious patients; trace-and-test of contacts; masks; improving ventilation / housing conditions.

(d) [1 mark total]
- 1 mark for correct calculation and answer: 22% (allow 1 mark for correct method if calculation error occurs, e.g. showing \(\frac{330}{1500} \times 100\)).

(a) The pathogen genus is Plasmodium, and the vector is the female Anopheles mosquito.

(b) Mosquitoes lay their eggs and complete their larval and pupal stages in stagnant water. Draining these pools removes their breeding grounds, preventing them from reproducing. This reduces the mosquito population, which in turn reduces the number of vectors available to transmit the Plasmodium pathogen to humans.

(c) Introducing certain species of fish (like Gambusia/mosquitofish) to water bodies acts as a biological control because the fish feed on the mosquito larvae and pupae. This reduces the number of larvae that survive to become adult mosquitoes, thus lowering vector numbers and malaria transmission rates without using chemical insecticides.

(d) A mutation is a change in the DNA sequence of the pathogen. This can lead to a change in the amino acid sequence, altering the structure of the target protein or enzyme that the anti-malarial drug normally binds to or inhibits. If the drug can no longer bind to or affect the mutated protein, the pathogen survives and reproduces, passing on the resistant mutation to its offspring.

(a) [2 marks total]
- 1 mark for Plasmodium (accept Plasmodium species).
- 1 mark for female Anopheles mosquito.

(b) [3 marks total]
- 1 mark for stating that mosquitoes lay eggs / breed in stagnant water.
- 1 mark for explaining that draining water removes breeding sites / prevents development of larvae/pupae.
- 1 mark for linking this to a smaller vector population, reducing transmission.

(c) [2 marks total]
- 1 mark for stating that fish eat mosquito larvae / pupae.
- 1 mark for explaining that this limits the number of adult mosquitoes emerging, reducing vector density / transmission.

(d) [3 marks total]
- 1 mark for stating that mutation is a change in the gene/DNA sequence of Plasmodium.
- 1 mark for explaining that this alters the structure of the target protein / enzyme / receptor.
- 1 mark for explaining that the drug can no longer bind / inhibit the target, allowing the pathogen to survive and reproduce.

(a) To focus on a near object, the ciliary muscles in the eye contract. This causes the suspensory ligaments to slacken (become loose). Consequently, the elastic lens becomes thicker and more rounded (more convex), which increases its refractive power to focus light rays from the near object onto the retina.

(b) In bright light, the circular muscles of the iris contract and the radial muscles relax, causing the pupil to constrict (become smaller) to protect the retina from damage. In dim light, the radial muscles contract and the circular muscles relax, causing the pupil to dilate (become larger) to allow more light to enter.

(c) In a typical reflex arc, the pathway starts with a receptor detecting a stimulus. The nervous impulse then travels along the: sensory neurone, across a synapse into the central nervous system (specifically the spinal cord or brain) via the relay neurone, and then out along the motor neurone to the effector (muscle or gland).

(a) [4 marks total]
- 1 mark for ciliary muscles contracting.
- 1 mark for suspensory ligaments slackening / loosening.
- 1 mark for lens becoming thicker / more convex / rounder.
- 1 mark for light rays being bent (refracted) more to focus on the retina.

(b) [3 marks total]
- 1 mark for stating that circular muscles contract and radial muscles relax in bright light (or vice versa for dim light).
- 1 mark for describing the opposite muscular action in the other light condition (dim light: radial contract, circular relax).
- 1 mark for stating pupil constricts in bright light and dilates in dim light.

(c) [3 marks total]
- 1 mark for sensory neurone.
- 1 mark for relay neurone (located in CNS / spinal cord).
- 1 mark for motor neurone.

(a) Cystic fibrosis is a recessive disorder. Since the parents do not have cystic fibrosis but have an affected child (ff), they must both be carriers. Therefore, the genotype of both parents is heterozygous (Ff).
Gametes: F and f from both parents.
Crossing Ff x Ff gives the offspring genotypes:
- FF (normal, 25%)
- Ff (carrier, 50%)
- ff (cystic fibrosis, 25%)
Thus, the probability of their next child having cystic fibrosis is 25% (or 0.25, or 1 in 4).

(b) A mutation in the CFTR gene changes the sequence of DNA bases. During protein synthesis, transcription produces an mRNA molecule with a altered sequence of codons. During translation, this altered codon sequence results in a different sequence of amino acids (primary structure) being joined together. This changes the folding and tertiary structure of the protein, rendering the final CFTR channel protein non-functional or preventing it from inserting into the cell membrane.

(c) Two structural differences between DNA and RNA are:
1. DNA is double-stranded (forming a double helix), whereas RNA is typically single-stranded.
2. DNA contains the pentose sugar deoxyribose, whereas RNA contains ribose.
3. DNA has the nitrogenous base thymine (T), whereas RNA has uracil (U).

(a) [4 marks total]
- 1 mark for identifying both parents as heterozygous (Ff / carriers).
- 1 mark for showing correct gametes (F and f) for both parents.
- 1 mark for correct offspring genotypes (FF, Ff, ff) in a Punnett square or list.
- 1 mark for identifying the correct probability of having an affected child (25%, 0.25, or 1/4).

(b) [4 marks total]
- 1 mark for stating that a mutation is a change in the DNA base sequence.
- 1 mark for explaining that this changes the sequence of mRNA codons.
- 1 mark for explaining that a different sequence of amino acids is incorporated during translation.
- 1 mark for linking the different amino acid sequence to altered folding / tertiary structure, making the protein non-functional.

(c) [2 marks total]
- 1 mark each for any two correct differences (up to 2 marks):
* DNA is double-stranded vs. RNA is single-stranded.
* DNA contains deoxyribose vs. RNA contains ribose.
* DNA has thymine (T) vs. RNA has uracil (U).

(a) Alveoli are adapted for rapid gas exchange because:
1. They provide an extremely large total surface area across which diffusion can occur.
2. They have very thin walls (only one cell thick, made of squamous epithelium) which minimizes the diffusion distance.
3. They are surrounded by an extensive network of blood capillaries, maintaining a steep concentration gradient for oxygen and carbon dioxide.

(b) Emphysema is caused by long-term irritation from smoke, which leads to the breakdown of the elastic fibers and walls of the alveoli. The alveoli merge into larger, irregular air spaces. This drastically reduces the surface area available for gas exchange. Consequently, the rate of oxygen diffusion into the blood is significantly reduced. The loss of elasticity also makes it harder for the patient to exhale completely, leading to stale air remaining trapped in the lungs.

(c) Carbon monoxide has a much higher affinity for haemoglobin than oxygen does. It binds preferentially and irreversibly to haemoglobin in red blood cells to form carboxyhaemoglobin. This reduces the capacity of the blood to carry oxygen, meaning less oxygen is transported to tissues and organs, which can cause oxygen starvation, fatigue, and damage to aerobic tissues.

(a) [3 marks total]
- 1 mark for mentioning large surface area (due to millions of alveoli).
- 1 mark for thin walls / one cell thick / short diffusion path.
- 1 mark for excellent blood supply / dense capillary network (maintaining a steep concentration gradient) or moist lining (allowing gases to dissolve).

(b) [4 marks total]
- 1 mark for stating that emphysema causes walls of alveoli to break down / rupture.
- 1 mark for explaining that this creates larger air spaces with reduced surface area.
- 1 mark for stating this reduces the rate of diffusion of oxygen / carbon dioxide.
- 1 mark for explaining that loss of elastic tissue makes exhalation difficult, leaving stale air trapped in lungs.

(c) [3 marks total]
- 1 mark for stating that carbon monoxide binds to haemoglobin in red blood cells.
- 1 mark for mentioning that it forms carboxyhaemoglobin / binds irreversibly / has a higher affinity than oxygen.
- 1 mark for explaining that this reduces the oxygen-carrying capacity of the blood / less oxygen is delivered to body tissues.

(a) Arteries transport blood away from the heart under high pressure. They have thick walls containing a large amount of muscle and elastic tissue to withstand and maintain this high pressure. The elastic fibers allow the artery wall to stretch and recoil as blood surges from the ventricles, helping to smooth out blood flow. Veins transport blood back to the heart under low pressure. They have thinner walls and a wider lumen to reduce resistance to blood flow. They also contain pocket-style semi-lunar valves to prevent the backflow of blood, as blood pressure is low and blood must often travel against gravity.

(b) Valves ensure that blood flows in only one direction through the heart. The atrioventricular (AV) valves prevent blood from flowing back into the atria when the ventricles contract (systole). The pressure of the contracting ventricle forces these valves closed. The semi-lunar valves at the entrances to the aorta and pulmonary artery open during ventricular contraction but close when the ventricles relax (diastole), preventing blood from flowing back into the ventricles from the arteries.

(c) Calculation:
Cardiac Output = Heart Rate \(\times\) Stroke Volume
Cardiac Output = \(75 \text{ beats/min} \times 80 \text{ mL}\) = \(6000 \text{ mL/min}\)
To convert to \(\text{dm}^3\) (or liters) per minute:
\(6000 \text{ mL} / 1000 = 6.0 \text{ dm}^3/\text{min}\)

(a) [5 marks total]
- 1 mark for noting arteries have thick walls of muscle/elastic tissue to withstand high pressure.
- 1 mark for describing the role of elastic fibers in stretching and recoiling to maintain blood pressure.
- 1 mark for noting veins have a wide lumen to minimize resistance to blood flow.
- 1 mark for noting veins have valves because blood is at a lower pressure.
- 1 mark for stating valves prevent backflow of blood in veins.

(b) [3 marks total]
- 1 mark for stating that valves ensure one-way flow of blood.
- 1 mark for explaining that high pressure in contracting ventricles shuts the atrioventricular (AV) valves to prevent backflow into the atria.
- 1 mark for explaining that when ventricles relax, the back pressure of blood in the arteries (aorta/pulmonary artery) shuts the semi-lunar valves to prevent backflow into ventricles.

(c) [2 marks total]
- 1 mark for correct calculation of 6000 mL/min (or showing the correct formula / working).
- 1 mark for converting to 6.0 with correct unit (\(\text{dm}^3/\text{min}\) or \(\text{L/min}\)).

(a) In the 'lock and key' model, the enzyme is like a lock and the substrate is like a complementary key. The enzyme has a specifically shaped region called the active site. Only a substrate with a complementary shape can fit into this active site to form an enzyme-substrate complex. Once bound, the reaction takes place, products are formed, and they are released from the active site.

(b) As temperature increases towards the optimum, the enzyme and substrate molecules gain more kinetic energy. This causes them to move faster, increasing the frequency of successful collisions between the substrate and the active site, thereby increasing the rate of reaction. Above the optimum temperature, the excessive thermal energy causes the atoms within the enzyme molecule to vibrate violently. This breaks the hydrogen bonds and other weak bonds holding the enzyme's tertiary structure together. The active site loses its specific shape (denaturation), meaning the substrate can no longer fit, and the reaction stops.

(c) Proteins are composed of carbon, hydrogen, oxygen, nitrogen, and usually sulfur. Simple carbohydrates (like glucose) only contain carbon, hydrogen, and oxygen. Therefore, the elements present in all proteins but not in simple carbohydrates are nitrogen (always present in the amine group of amino acids) and sulfur (present in some amino acids like cysteine).

(a) [3 marks total]
- 1 mark for mentioning the specific / complementary shape of the active site and substrate.
- 1 mark for describing the substrate binding to the active site (forming an enzyme-substrate complex).
- 1 mark for stating that the substrate is converted to products, which then leave the active site unchanged.

(b) [5 marks total]
- 1 mark for explaining that increasing temperature gives molecules more kinetic energy.
- 1 mark for stating that molecules move faster, leading to more frequent successful collisions.
- 1 mark for stating that above the optimum temperature, bonds (hydrogen/ionic) holding the enzyme's tertiary structure break.
- 1 mark for explaining that the active site changes shape / denatures.
- 1 mark for explaining that the substrate can no longer fit into the denatured active site.

(c) [2 marks total]
- 1 mark for Nitrogen (N).
- 1 mark for Sulfur (S) (accept Phosphorus if referring to some conjugated proteins, but accept Sulfur as standard amino acid component difference; reject other elements).

(a) When the water potential of the blood decreases (e.g., due to sweating or dehydration), osmoreceptors in the hypothalamus of the brain detect this change. The hypothalamus stimulates the posterior pituitary gland to release more antidiuretic hormone (ADH) into the bloodstream. ADH travels to the kidneys, where it increases the permeability of the walls of the collecting ducts and distal convoluted tubules to water. As a result, more water is reabsorbed by osmosis out of the filtrate and back into the blood capillaries. This produces a small volume of concentrated urine and restores the blood's water potential to normal.

(b) Ultrafiltration occurs in the Bowman's capsule of the nephron. High blood pressure is created in the glomerulus because the afferent arteriole entering the glomerulus is wider than the efferent arteriole leaving it. This high pressure forces water and small soluble molecules (such as glucose, urea, amino acids, and mineral ions) out of the blood capillaries through the basement membrane and into the Bowman's capsule, forming the glomerular filtrate. Large molecules like proteins and red blood cells are too big to pass through and remain in the blood.

(c) In a healthy person, glucose is filtered out of the blood during ultrafiltration. However, as the filtrate passes through the proximal convoluted tubule (PCT), all of the glucose is actively reabsorbed back into the surrounding blood capillaries. Since 100% of the glucose is reabsorbed under normal conditions, none remains in the fluid to be excreted in the urine.

(a) [5 marks total]
- 1 mark for stating that osmoreceptors in the hypothalamus detect the decrease in water potential.
- 1 mark for stating that the pituitary gland releases more ADH.
- 1 mark for explaining that ADH increases the permeability of the collecting duct (and/or distal convoluted tubule) to water.
- 1 mark for explaining that more water is reabsorbed back into the blood (by osmosis).
- 1 mark for stating that this results in a low volume of highly concentrated urine.

(b) [3 marks total]
- 1 mark for identifying that ultrafiltration occurs between the glomerulus and Bowman's capsule.
- 1 mark for explaining that high pressure is generated because the afferent arteriole is wider than the efferent arteriole.
- 1 mark for explaining that small molecules (water, glucose, ions, urea) are forced through the filter/basement membrane while large proteins/cells stay in the blood.

(c) [2 marks total]
- 1 mark for stating that glucose is selectively reabsorbed in the proximal convoluted tubule (PCT).
- 1 mark for explaining that this occurs via active transport, so all of it is returned to the blood and none is left in the urine.

Part (a): Cardiac output is the volume of blood pumped by one ventricle of the heart per minute. To calculate resting cardiac output: \(72\text{ bpm} \times 70\text{ mL} = 5040\text{ mL min}^{-1}\). Convert to \(\text{dm}^3\text{ min}^{-1}\) by dividing by 1000: \(5040 / 1000 = 5.04\text{ dm}^3\text{ min}^{-1}\). To calculate exercise cardiac output: \(160\text{ bpm} \times 120\text{ mL} = 19200\text{ mL min}^{-1}\). Convert to \(\text{dm}^3\text{ min}^{-1}\) by dividing by 1000: \(19200 / 1000 = 19.2\text{ dm}^3\text{ min}^{-1}\). Part (b): During exercise, cellular respiration in muscles increases, producing more carbon dioxide. This dissolves in the blood, lowering the blood pH. Chemoreceptors in the aorta and carotid arteries detect this decrease in pH and send nerve impulses to the medulla oblongata (cardiac center) in the brain. The medulla sends nerve impulses down the sympathetic nerve to the sinoatrial node (SAN / pacemaker) to increase the heart rate. Simultaneously, the endocrine system coordinates a response where adrenaline is secreted by the adrenal glands into the blood, which directly stimulates the SAN to increase the rate of contraction. Part (c): An increased cardiac output increases the rate of blood flow to actively contracting skeletal muscles. This delivers more oxygen and glucose required for a higher rate of aerobic respiration, while also rapidly removing waste products such as carbon dioxide and lactic acid.

Part (a) (4 marks total): 1 mark for defining cardiac output as the volume of blood pumped by the heart/ventricle per minute. 1 mark for correct resting calculation: \(5.04\text{ dm}^3\text{ min}^{-1}\). 1 mark for correct exercise calculation: \(19.2\text{ dm}^3\text{ min}^{-1}\). 1 mark for providing correct units (\(\text{dm}^3\text{ min}^{-1}\)) for both values. Part (b) (4 marks total): Max 4 marks from: 1 mark for identifying that increased CO2 / reduced pH is detected by chemoreceptors. 1 mark for stating that impulses are sent to the medulla oblongata. 1 mark for stating that sympathetic nerves transmit impulses to the SAN / pacemaker to increase heart rate. 1 mark for stating that the hormone adrenaline is released from the adrenal glands. 1 mark for stating that adrenaline travels in the blood and stimulates the SAN. Part (c) (2 marks total): Max 2 marks from: 1 mark for stating that it increases the delivery of oxygen / glucose to working muscles. 1 mark for stating that it allows for a faster rate of aerobic respiration to produce energy / ATP. 1 mark for stating that it speeds up the removal of carbon dioxide / lactic acid / metabolic waste.

(a) At \(37^\circ\text{C}\), there is a much greater risk of cultivating human pathogens which are adapted to human body temperature. Incubating at \(30^\circ\text{C}\) reduces this danger while still allowing rapid bacterial growth.

(b) For Antiseptic B, the diameter is 20 mm, so the radius \(r = \frac{20}{2} = 10\text{ mm}\).
\(\text{Area} = \pi r^2 = 3.14 \times (10)^2 = 3.14 \times 100 = 314.0\text{ mm}^2\).

(c) For Antiseptic C, the diameter is 26 mm, so the radius \(r = 13\text{ mm}\).
\(\text{Area for C} = 3.14 \times (13)^2 = 3.14 \times 169 = 530.66\text{ mm}^2\).
Percentage increase = \(\frac{\text{Area of C} - \text{Area of B}}{\text{Area of B}} \times 100 = \frac{530.66 - 314.0}{314.0} \times 100 = \frac{216.66}{314.0} \times 100 \approx 69.00\%\).
To the nearest whole number, this is 69%.

(d) Control variables: Same volume of antiseptic solution added to the discs, identical size/material of filter paper discs, same depth/type of nutrient agar, and same starting concentration/volume of the bacterial culture spread on the plates.

Part (a) [2 marks total]:
- 1 mark for stating that \(37^\circ\text{C}\) promotes growth of human pathogens / harmful bacteria.
- 1 mark for stating that \(30^\circ\text{C}\) is safer for school laboratory conditions while still allowing growth.

Part (b) [3 marks total]:
- 1 mark for finding correct radius (10 mm).
- 1 mark for correct substitution into formula: \(3.14 \times 10^2\).
- 1 mark for correct final answer with units (314.0 mm² / accept 314 mm²).

Part (c) [4 marks total]:
- 1 mark for calculating area of C (530.66 mm²).
- 1 mark for calculating absolute increase: \(530.66 - 314 = 216.66\).
- 1 mark for setting up percentage increase calculation: \(\frac{216.66}{314} \times 100\).
- 1 mark for correct final percentage rounded to the nearest whole number (69%).

Part (d) [2.25 marks total]:
- 1.125 marks for each valid control variable (up to 2), e.g., disc size, volume of antiseptic, concentration of bacteria inoculant, agar composition.

(a) Mean distance before = \(\frac{18 + 22 + 19 + 21 + 20}{5} = \frac{100}{5} = 20\text{ cm} = 0.20\text{ m}\).

(b) Using the formula \(t = \sqrt{\frac{2d}{g}}\) with \(d = 0.20\text{ m}\) and \(g = 9.8\text{ m/s}^2\):
\(t = \sqrt{\frac{2 \times 0.20}{9.8}} = \sqrt{\frac{0.40}{9.8}} = \sqrt{0.0408163} \approx 0.202\text{ s}\).

(c) Using the formula with \(d = 0.13\text{ m}\):
\(t = \sqrt{\frac{2 \times 0.13}{9.8}} = \sqrt{\frac{0.26}{9.8}} = \sqrt{0.0265306} \approx 0.163\text{ s}\).

(d) When the ruler drops, photoreceptors in the retina of the eye detect the movement (stimulus). Impulses travel along the sensory neurone to the central nervous system (brain/spinal cord). Synapses pass the impulse to relay neurones and then to motor neurones. Motor neurones conduct impulses to the muscles of the hand (effectors), which contract to grasp the ruler.

Part (a) [2 marks total]:
- 1 mark for showing correct working: \(\frac{100}{5} = 20\text{ cm}\).
- 1 mark for converting to meters (0.20 m).

Part (b) [3 marks total]:
- 1 mark for correct substitution into formula: \(\sqrt{\frac{0.4}{9.8}}\).
- 1 mark for correct square root division: \(\sqrt{0.0408}\).
- 1 mark for correct final answer of 0.202 s (accept 0.202).

Part (c) [2.25 marks total]:
- 1.125 marks for substitution: \(\sqrt{\frac{0.26}{9.8}}\).
- 1.125 marks for correct final answer of 0.163 s.

Part (d) [4 marks total]:
- 1 mark for receptor detects stimulus (photoreceptors in retina detect falling ruler).
- 1 mark for transmission of electrical impulse along sensory neurone to coordinator (CNS/brain).
- 1 mark for transmission across synapses via neurotransmitters to motor neurone.
- 1 mark for impulse reaching effector (muscles in hand) causing contraction to catch ruler.

(a) The parental genotypes are both \(Ff\).
Crossing \(Ff \times Ff\) gives a Punnett square:
- Gametes: \(F\) and \(f\) from each parent.
- Offspring genotypes: \(FF\) (Normal), \(Ff\) (Carrier), \(Ff\) (Carrier), \(ff\) (Cystic Fibrosis).
- Offspring phenotypes: 3 healthy (including 2 carriers) : 1 with cystic fibrosis.

(b) The carriers are the heterozygotes (\(Ff\)). Two out of the four possible offspring combinations are carriers, so the probability is \(\frac{2}{4} = 50\%\).

(c) The probability of one child having cystic fibrosis (\(ff\)) is \(\frac{1}{4} = 0.25\).
Since each birth is an independent event, the probability that all three children have cystic fibrosis is:
\(\frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{64} \approx 0.0156\) (or \(1.56\%\)).

(d) Fetal cells can be obtained via amniocentesis (sampling amniotic fluid) or chorionic villus sampling (sampling placental tissue). DNA is extracted from these cells and analysed for the recessive \(f\) allele. An ethical issue is that these procedures carry a small risk of miscarriage of a healthy fetus, or they may present parents with a difficult decision regarding pregnancy termination.

Part (a) [3 marks total]:
- 1 mark for correct parental gametes labeled (F and f).
- 1 mark for correct offspring genotypes in the grid (FF, Ff, Ff, ff).
- 1 mark for linking genotypes to phenotypes correctly (FF/Ff = unaffected/normal, ff = has cystic fibrosis).

Part (b) [2 marks total]:
- 1 mark for identification of carrier genotypes (Ff).
- 1 mark for correct probability as a percentage (50%).

Part (c) [3 marks total]:
- 1 mark for identifying the single-birth probability as 1/4 (or 0.25).
- 1 mark for multiplying the probabilities of independent events: \((1/4) \times (1/4) \times (1/4)\).
- 1 mark for correct final answer: 1/64, 0.0156, or 1.56%.

Part (d) [3.25 marks total]:
- 1 mark for naming a screening method (CVS / Amniocentesis).
- 1 mark for explaining that fetal DNA/chromosomes are tested for the CF allele.
- 1.25 marks for discussing a clear ethical issue (e.g., risk of miscarriage, right to life, potential termination of pregnancy).

(a) Ventilation rate calculation:
- At rest: \(12 \text{ breaths/min} \times 0.5 \text{ dm}^3 = 6.0\text{ dm}^3/\text{min}\).
- During vigorous exercise: \(28 \text{ breaths/min} \times 2.5 \text{ dm}^3 = 70.0\text{ dm}^3/\text{min}\).

(b) Percentage increase calculation:
\(\text{Percentage increase} = \frac{\text{Vigorous} - \text{Rest}}{\text{Rest}} \times 100 = \frac{70.0 - 6.0}{6.0} \times 100 = \frac{64.0}{6.0} \times 100 \approx 1066.67\%\) (accept 1067% or 1066.7%).

(c) During vigorous exercise, respiration in muscles increases, producing more carbon dioxide. This decreases blood pH, which is detected by chemoreceptors. The respiratory center in the medulla sends more frequent nerve impulses to the external intercostal muscles and the diaphragm. The intercostal muscles contract harder and faster, increasing chest cavity volume change, resulting in deeper and faster breathing.

Part (a) [4 marks total]:
- 1 mark for correct rest calculation (6.0).
- 1 mark for correct vigorous exercise calculation (70.0).
- 1 mark for correct units (dm³/min or dm^3 min^-1).
- 1 mark for showing working.

Part (b) [3.25 marks total]:
- 1 mark for calculating the difference (64.0 dm³/min).
- 1 mark for dividing by rest value (64 / 6).
- 1.25 marks for correct percentage conversion and calculation (1066.7% or 1067%).

Part (c) [4 marks total]:
- 1 mark for increased CO2 (or lower pH) detected by chemoreceptors / medulla.
- 1 mark for medulla sending impulses via motor nerves.
- 1 mark for intercostal muscles contracting faster/stronger.
- 1 mark for resulting increase in both breathing frequency and thoracic volume change.

(a) Cardiac Output rest:
\(52 \times 95 = 4940\text{ cm}^3/\text{min}\).
Converting to \\text{dm}^3/\\text{min}\: \(\frac{4940}{1000} = 4.94\text{ dm}^3/\text{min}\).

Cardiac Output peak exercise:
\(180 \times 155 = 27900\text{ cm}^3/\text{min}\).
Converting to \\text{dm}^3/\\text{min}\: \(\frac{27900}{1000} = 27.90\text{ dm}^3/\text{min}\).

(b) Ratio calculation:
\(\text{Ratio} = \frac{27.90}{4.94} \approx 5.6477\).
To 2 decimal places, this is 5.65.

(c) Unidirectional flow is maintained by valves: atrioventricular (bicuspid and tricuspid) valves prevent backflow from ventricles into atria, and semilunar valves prevent backflow from arteries (aorta/pulmonary artery) into ventricles. High pressure is generated by the thick muscular wall of the left ventricle, which contracts powerfully to force blood out into the aorta to the systemic circulation.

Part (a) [4 marks total]:
- 1 mark for rest calculation in cm³ (4940).
- 1 mark for peak exercise in cm³ (27900).
- 1 mark for correct division by 1000 for both values.
- 1 mark for correct final answers in dm³/min (4.94 and 27.90).

Part (b) [2.25 marks total]:
- 1.125 marks for division setup: \(27.90 / 4.94\).
- 1.125 marks for correct final ratio rounded to 2 decimal places (5.65).

Part (c) [5 marks total]:
- 1 mark for mentioning atrioventricular valves prevent backflow to atria.
- 1 mark for mentioning semilunar valves prevent backflow from arteries to ventricles.
- 1 mark for stating that blood flows from high to low pressure when ventricles contract.
- 1 mark for identifying the thick muscular wall of the left ventricle.
- 1 mark for linking ventricle wall thickness to generating high systemic blood pressure.

(a) Rate calculation:
- At \(20^\circ\text{C}\): \(\text{Rate} = \frac{1000}{360} \approx 2.777 \approx 2.78\text{ s}^{-1}\).
- At \(40^\circ\text{C}\): \(\text{Rate} = \frac{1000}{90} \approx 11.11\text{ s}^{-1}\).

(b) At \(40^\circ\text{C}\), the enzyme and substrate molecules have high kinetic energy, causing them to move faster. This results in more frequent successful collisions, forming more enzyme-substrate complexes per second. At \(60^\circ\text{C}\), the excessive heat disrupts the hydrogen bonds maintaining the specific 3D shape of the enzyme's active site. The active site is denatured, meaning the starch substrate is no longer complementary and cannot bind, so no reaction occurs.

(c) To find a more precise optimum temperature, the student should test temperatures at smaller, narrower intervals (e.g., every 2 degrees instead of 10 degrees) between \(30^\circ\text{C}\) and \(50^\circ\text{C}\). Additionally, they could take samples more frequently (e.g., every 10 seconds) to determine the exact end-point more accurately.

Part (a) [3 marks total]:
- 1 mark for correct calculation setup for both.
- 1 mark for \(20^\circ\text{C}\) rate: 2.78 (accept 2.8).
- 1 mark for \(40^\circ\text{C}\) rate: 11.11.

Part (b) [5 marks total]:
- 1 mark for stating that rising temperature increases kinetic energy of molecules.
- 1 mark for stating that this leads to more frequent successful collisions between amylase and starch.
- 1 mark for stating that \(60^\circ\text{C}\) causes denaturation of the amylase enzyme.
- 1 mark for explaining that denaturation changes the shape of the active site.
- 1 mark for stating that starch can no longer bind / is no longer complementary to the active site.

Part (c) [3.25 marks total]:
- 1.25 marks for suggesting a narrower temperature range surrounding the current optimum (e.g., between \(30^\circ\text{C}\) and \(50^\circ\text{C}\)).
- 1 mark for suggesting smaller intervals (e.g., testing at \(32, 34, 36, 38, 42, 44, 46, 48^\circ\text{C}\)).
- 1 mark for increasing sampling frequency (e.g., testing every 10 or 15 seconds).

(a) Rate of change calculation:
- Between \(25^\circ\text{C}\) and \(35^\circ\text{C}\):
\(\text{Rate of change} = \frac{65 - 12}{35 - 25} = \frac{53}{10} = 5.3\text{ g/hour per }^\circ\text{C}\).

- Between \(35^\circ\text{C}\) and \(45^\circ\text{C}\):
\(\text{Rate of change} = \frac{180 - 65}{45 - 35} = \frac{115}{10} = 11.5\text{ g/hour per }^\circ\text{C}\).

(b) Thermoreceptors in the skin and the hypothalamus detect an increase in core body temperature/blood temperature. The hypothalamus acts as the coordinating center and sends electrical impulses along motor nerves to effectors. These effectors carry out corrective mechanisms: sweat glands are stimulated to produce more sweat (evaporative cooling), and arteriole walls relax (vasodilation) to increase blood flow near the skin surface, increasing heat loss by radiation.

(c) Vasoconstriction occurs in cold conditions when the muscular walls of arterioles supplying the skin surface capillaries contract. This narrows the arteriole lumen, diverting blood flow through shunt vessels deeper in the skin. This reduces blood flow to the skin capillaries, minimizing heat loss by radiation.

Part (a) [4 marks total]:
- 1 mark for correct calculation working for 25–35°C.
- 1 mark for correct rate of 5.3 g/hour/°C.
- 1 mark for correct calculation working for 35–45°C.
- 1 mark for correct rate of 11.5 g/hour/°C.

Part (b) [4.25 marks total]:
- 1 mark for identifying the hypothalamus as the thermoregulatory center detecting blood temperature.
- 1 mark for impulses sent via autonomic/motor nervous system.
- 1 mark for explaining sweat glands releasing sweat to cool the skin via evaporation.
- 1.25 marks for detailing the vasodilation response (arterioles dilate, more blood to surface).

Part (c) [3 marks total]:
- 1 mark for explaining that muscles in arteriole walls contract/narrow.
- 1 mark for mentioning blood is diverted away from capillaries near the skin surface.
- 1 mark for stating this reduces heat loss by radiation to keep core body warm.

(a) To find where the mass change is \(0\%\):
The interval is between \(0.2\text{ mol/dm}^3\) (\(+5.0\%\)) and \(0.4\text{ mol/dm}^3\) (\(-2.5\%\)).
The total concentration change is \(0.4 - 0.2 = 0.2\text{ mol/dm}^3\).
The corresponding change in percentage mass is \(5.0 - (-2.5) = 7.5\%\).
We want to find the concentration \(C\) where the percentage change is \(0\%\).
The distance from \(+5.0\%\) to \(0\%\) is \(5.0\%\).
Using linear interpolation:
\(C = 0.2 + \left( \frac{5.0}{7.5} \times 0.2 \right)\)
\(C = 0.2 + \left( \frac{2}{3} \times 0.2 \right) = 0.2 + 0.133 = 0.33\text{ mol/dm}^3\).

(b) In the \(0.0\text{ mol/dm}^3\) solution, the external water potential is higher than inside the potato cells. Water enters the cells by osmosis down a water potential gradient, causing cells to expand and become turgid, increasing in mass. In the \(0.8\text{ mol/dm}^3\) solution, the external water potential is lower than inside the cells. Water leaves the cells by osmosis, causing the cell membrane to pull away from the cell wall (plasmolysed) and decrease in mass.

(c) The potato cylinders must be gently blotted with a paper towel before weighing to remove any excess surface liquid which would artificially increase the recorded mass. The balance should be tared to zero before each reading, and multiple cylinders should be used for each concentration to calculate a mean.

Part (a) [4 marks total]:
- 1 mark for identifying that the isotonic point lies between 0.2 and 0.4 mol/dm³.
- 1 mark for finding the total percentage change difference over the interval (7.5%).
- 1 mark for setting up the fractional interpolation: \(\frac{5.0}{7.5} \times 0.2\).
- 1 mark for correct final answer of 0.33 mol/dm³ (accept 0.33 or 0.333).

Part (b) [5 marks total]:
- 1 mark for stating osmosis is the movement of water down a water potential gradient.
- 1 mark for linking \(0.0\text{ mol/dm}^3\) to higher external water potential, causing water inflow.
- 1 mark for using the term 'turgid' to describe the cells in \(0.0\text{ mol/dm}^3\).
- 1 mark for linking \(0.8\text{ mol/dm}^3\) to lower external water potential, causing water outflow.
- 1 mark for using the term 'plasmolysed' or 'flaccid' to describe cells in \(0.8\text{ mol/dm}^3\).

Part (c) [2.25 marks total]:
- 1.125 marks for blotting potato cylinders to remove excess surface liquid.
- 1.125 marks for repeating the trials to calculate a mean and exclude anomalies.