2023 Edexcel IGCSE Human Biology Practice Paper with Answers

To complete the table comparing blood vessels:

* **(A)** Veins have walls that are much thinner than those of arteries because they carry blood under much lower pressure. Therefore, the correct term is **Thin** (or thinner than an artery).
* **(B)** Capillaries are extremely small vessels designed to allow the exchange of substances. Their lumen is extremely narrow, typically just wide enough to allow red blood cells to pass through in single file. Therefore, the correct description is **Very narrow** (or microscopic).
* **(C)** Arteries carry blood away from the heart under high pressure and do not have valves along their length (with the exception of the semi-lunar valves at the base of the aorta and pulmonary artery). Therefore, the answer is **Absent** (or No).
* **(D)** Veins carry blood under low pressure and contain valves to prevent the backflow of blood. Therefore, the answer is **Present** (or Yes).
* **(E)** Capillaries do not have valves. Therefore, the answer is **Absent** (or No).

Award 1 mark for each correct identification (maximum of 5 marks):

* **(A)**: Thin / thinner / relatively thin [1 mark] (Reject: thick)
* **(B)**: Very narrow / microscopic / width of a single red blood cell [1 mark] (Reject: 'narrow' on its own, as this is already used for arteries)
* **(C)**: Absent / No / None [1 mark] (Accept: None, except at start of aorta/pulmonary artery)
* **(D)**: Present / Yes [1 mark]
* **(E)**: Absent / No / None [1 mark]

(a) To plot the graph accurately:
1. Assign running speed (independent variable) to the x-axis with unit (km/h) and lactic acid concentration (dependent variable) to the y-axis with unit (mmol/dm3).
2. Choose linear scales for both axes such that the plotted data covers more than half of the grid.
3. Plot each point precisely with a small cross (x) or dot-in-circle and join the points with straight, ruled lines or a smooth curve of best fit.

(b) From 8 km/h to 20 km/h, the lactic acid concentration increases rapidly from 1.5 to 11.5 mmol/dm3. This happens because the energy demand of the contracting muscles exceeds the rate at which aerobic respiration can supply ATP. Consequently, muscle cells carry out anaerobic respiration to meet this demand, which produces lactic acid as a byproduct.

(c) Lactic acid at 12 km/h = 2.8 mmol/dm3, and at 16 km/h = 6.2 mmol/dm3.
Increase = 6.2 - 2.8 = 3.4 mmol/dm3.
Percentage increase = \((3.4 / 2.8) \times 100 = 121.4\%\) (or 121% to 3 significant figures).

Part (a) [3 marks]:
- 1 mark for correct axes orientation with units (x-axis: running speed in km/h, y-axis: lactic acid concentration in mmol/dm3).
- 1 mark for suitable linear scales that utilize more than 50% of the graph paper.
- 1 mark for accurate plotting of all points and joining them with a smooth line or straight lines.

Part (b) [3 marks]:
- 1 mark for stating that lactic acid concentration increases exponentially/rapidly at higher speeds.
- 1 mark for explaining that aerobic respiration alone cannot supply enough energy/ATP for high-intensity exercise.
- 1 mark for stating that anaerobic respiration occurs, which produces lactic acid.

Part (c) [2 marks]:
- 1 mark for calculating the increase in concentration: 6.2 - 2.8 = 3.4 (mmol/dm3).
- 1 mark for correct percentage calculation: \((3.4 / 2.8) \times 100 = 121\%\) or 121.4% (accept 121.43%).

(a) To plot the graph accurately:
1. Assign environmental temperature to the horizontal x-axis with unit (°C) and rate of sweat production to the vertical y-axis with unit (g/min).
2. Choose linear, regular scales for both axes so that the data points span more than half of the grid space.
3. Plot each point precisely and connect them with a smooth curve of best fit.

(b) As environmental temperature increases, sweat gland activity increases. Sweat consists mainly of water. This water evaporates from the skin surface, which requires heat energy (latent heat of vaporization). This heat energy is drawn from the body, thereby cooling the skin and maintaining a stable core body temperature.

(c) Rate at 45 °C = 1.70 g/min, and rate at 20 °C = 0.05 g/min.
Ratio = 1.70 / 0.05 = 34.
Therefore, the ratio is 34 : 1.

Part (a) [3 marks]:
- 1 mark for correct axes orientation with units (x-axis: environmental temperature in °C, y-axis: rate of sweat production in g/min).
- 1 mark for choosing scales that are linear and cover more than half of the grid.
- 1 mark for accurately plotting all 6 data points and drawing a smooth curve.

Part (b) [3 marks]:
- 1 mark for mentioning that sweat contains water that evaporates from the skin surface.
- 1 mark for explaining that evaporation requires heat energy (or latent heat of vaporization).
- 1 mark for stating that this heat energy is removed from the body/blood, which cools the body down.

Part (c) [2 marks]:
- 1 mark for the correct calculation step: 1.70 / 0.05.
- 1 mark for the correct final ratio: 34 : 1 (or 34 to 1).

(a) Structure A is the left ventricle. Structure B is the aorta. Structure C is the semi-lunar (or aortic) valve. (b) The left ventricle must contract with high force to generate high hydrostatic pressure because it pumps blood around the entire systemic circulation (to the rest of the body), which has high resistance. The right ventricle only pumps blood to the lungs (pulmonary circulation), which is close to the heart and has low resistance, so a thinner wall is sufficient. (c) The aorta has a thick tunica media containing a high proportion of elastic fibres. These stretch when blood is forced out of the ventricle during systole and recoil when the ventricle relaxes during diastole, smoothing the blood flow and maintaining diastolic pressure. It also has a thick outer layer of collagen to prevent bursting under high pressure. (d) Coronary arteries supply the heart muscle (myocardium) with oxygenated blood containing glucose. If blocked (e.g., by atheroma), cardiac muscle cells are deprived of oxygen and cannot carry out aerobic respiration. They switch to anaerobic respiration, producing lactic acid, which lowers pH and damages enzymes, ultimately causing muscle cell death and a myocardial infarction.

Part (a) [3 marks total]: 1 mark for Left ventricle; 1 mark for Aorta; 1 mark for Semi-lunar valve (accept aortic valve). Part (b) [3.5 marks total]: 1 mark for identifying that the left ventricle pumps blood to the whole body / systemic circulation; 1 mark for identifying that the right ventricle pumps blood only to the lungs / pulmonary circulation; 1 mark for explaining that systemic circulation has higher resistance / requires higher pressure; 0.5 marks for stating this prevents damage to delicate capillaries in lungs. Part (c) [3 marks total]: 1 mark for elastic fibres stretching to accommodate blood surge; 1 mark for elastic fibres recoiling to maintain diastolic pressure; 1 mark for thick collagen layer/wall to withstand high pressure. Part (d) [3 marks total]: 1 mark for stating coronary arteries supply oxygen/glucose to heart muscle; 1 mark for stating blockage leads to lack of aerobic respiration; 1 mark for stating this causes cell death / heart attack.

(a) X = Glomerulus; Y = Bowman's capsule; Z = Collecting duct. (b) Ultrafiltration is driven by high hydrostatic pressure in the glomerulus (X). This high pressure is generated because the afferent arteriole supplying the glomerulus has a wider lumen than the efferent arteriole leaving it. This pressure forces water and small solutes (glucose, amino acids, urea, mineral ions) out of the blood. They pass through three filtration barriers: the fenestrations in the capillary endothelium, the basement membrane (which acts as the main selective filter), and the filtration slits of the podocytes on the wall of the Bowman's capsule (Y). Large proteins and blood cells are too big to pass through and remain in the capillaries. (c) When a person is dehydrated, the water potential of the blood plasma decreases (it becomes more concentrated). Osmoreceptors in the hypothalamus detect this change and stimulate the posterior pituitary gland to secrete more antidiuretic hormone (ADH) into the bloodstream. ADH travels to the kidneys and binds to receptors on the cells lining the collecting ducts (Z). This increases the permeability of the collecting duct walls to water by causing aquaporins (water channels) to insert into the membrane. Consequently, more water is reabsorbed by osmosis out of the collecting duct and back into the surrounding vasa recta capillaries. This results in a smaller volume of highly concentrated urine and restores the blood water potential to its set point.

Part (a) [3 marks total]: 1 mark for Glomerulus; 1 mark for Bowman's capsule; 1 mark for Collecting duct. Part (b) [4.5 marks total]: 1 mark for mentioning the afferent arteriole is wider than the efferent arteriole, creating high hydrostatic pressure; 1 mark for stating that the basement membrane acts as a molecular sieve/filter; 1 mark for listing substances that pass through (water, glucose, urea, salts); 1 mark for stating large plasma proteins and blood cells are too large to pass; 0.5 marks for mentioning podocytes / capillary fenestrations. Part (c) [5 marks total]: 1 mark for osmoreceptors in the hypothalamus detecting low blood water potential; 1 mark for pituitary gland releasing more ADH; 1 mark for ADH increasing the permeability of the collecting duct to water; 1 mark for more water being reabsorbed into the blood by osmosis; 1 mark for producing a low volume of concentrated urine / restoring normal blood water potential.

(a) Substances moving from maternal to fetal blood: Oxygen, Glucose, Amino acids (accept: lipids, vitamins, mineral ions, water, antibodies). Waste substance moving from fetal to maternal blood: Carbon dioxide (accept: Urea). (b) Adaptations: 1. It has highly folded chorionic villi, which provides an extremely large surface area for diffusion. 2. The barrier between maternal and fetal blood is very thin (only two cell layers thick: the capillary endothelium and the trophoblast layer), which ensures a short diffusion pathway. 3. There is a counter-current or continuous flow of blood (maternal blood pool replenished constantly, fetal blood pumped by fetal heart) which maintains steep concentration gradients for efficient diffusion. (c) After the first trimester, the corpus luteum degenerates and the placenta takes over the production of hormones. It secretes progesterone, which maintains the thickness and vascularity of the endometrium (uterine lining), preventing miscarriage, and inhibits uterine contractions (prevents premature labor). It also secretes estrogen, which stimulates the growth of the uterus to accommodate the growing fetus and stimulates the development of the mammary glands in preparation for lactation.

Part (a) [4 marks total]: 1 mark each for any three valid substances from mother to fetus [Max 3]; 1 mark for one valid waste substance from fetus to mother [Max 1]. Part (b) [4.5 marks total]: 1.5 marks for adaptation 1 (1 mark for villi/large surface area, 0.5 mark for explaining how this speeds up rate of diffusion); 1.5 marks for adaptation 2 (1 mark for thin membrane, 0.5 mark for explaining it reduces diffusion distance); 1.5 marks for adaptation 3 (1 mark for blood flow/ventilation of area, 0.5 mark for explaining it maintains steep concentration gradient). Part (c) [4 marks total]: 1 mark for stating that the placenta takes over hormone production from the corpus luteum; 1 mark for stating it secretes progesterone; 1 mark for explaining progesterone maintains the endometrium / prevents uterine contractions; 1 mark for stating it secretes estrogen to stimulate uterine/mammary growth.

(a) Region P is the antigen-binding site (variable region), which has a specific 3D shape complementary to a specific antigen, allowing it to bind to the pathogen. Region Q is the constant region, which is identical in all antibodies of the same class and binds to receptors on phagocytes to facilitate phagocytosis. (b) 1. Agglutination: Antibodies have at least two antigen-binding sites, allowing them to bind to multiple pathogens simultaneously, clumping them together. This prevents them from spreading and makes it easier for phagocytes to engulf them. 2. Opsonisation: Antibodies coat the pathogen, acting as chemical markers that are easily recognized by phagocytes, stimulating phagocytosis. 3. Neutralisation: Antibodies bind to toxins released by pathogens (antitoxins) or to viral surface proteins, preventing them from entering host cells or causing damage. (c) Active immunity occurs when the body's own immune system is stimulated to produce antibodies and memory cells in response to exposure to an antigen. This provides long-term immunity. Examples include natural infection by a pathogen or artificial introduction via a vaccine. Passive immunity occurs when an individual receives pre-made antibodies from an external source. Because the individual's own B lymphocytes are not activated, no memory cells are produced, meaning the immunity is temporary (short-term) as the antibodies are eventually broken down. Examples include natural passage of maternal IgG antibodies across the placenta or IgA in breast milk, or artificial injection of monoclonal antibodies/antivenom.

Part (a) [2 marks total]: 1 mark for identifying P as the antigen-binding / variable region and stating it binds to a specific antigen; 1 mark for identifying Q as the constant region and stating it binds to phagocytes. Part (b) [4.5 marks total]: 1.5 marks for each described method (max 3 methods): 1 mark for naming/describing the mechanism (e.g., agglutination, opsonisation, neutralization) and 0.5 mark for explaining how this leads to the pathogen's destruction. Part (c) [6 marks total]: 1 mark for defining active immunity as the body producing its own antibodies/memory cells; 1 mark for stating active immunity provides long-term protection; 1 mark for a valid active immunity example (e.g., vaccination or catching a disease); 1 mark for defining passive immunity as receiving pre-formed antibodies from another source; 1 mark for stating passive immunity provides immediate but short-term protection (no memory cells); 1 mark for a valid passive immunity example (e.g., breast milk, antibody injection).

1. Plasma (E) is the liquid component of blood that transports key solutes including glucose, amino acids, mineral ions, carbon dioxide, urea, and hormones.
2. Red blood cells (A) have a biconcave disc shape to increase the surface-area-to-volume ratio and lack a nucleus to allow more space for haemoglobin.
3. Lymphocytes (B) are white blood cells that produce specific proteins called antibodies to destroy targeted pathogens.
4. Phagocytes (C) are phagocytic white blood cells that recognize, engulf, and break down pathogens inside a vacuole using enzymes.
5. Platelets (D) are small fragments of cells that release chemicals to trigger blood clotting when tissues are damaged.
6. Arteries (F) carry blood under high pressure away from the heart, which requires thick muscular and elastic walls to stretch and recoil.
7. Veins (G) carry blood at low pressure back to the heart, containing valves to prevent blood from flowing backward.
8. Capillaries (H) form extensive networks within tissues and are only one cell thick to ensure a short diffusion pathway.
9. Valves (J) are physical flaps present in both veins and lymph vessels to maintain unidirectional fluid flow.
10. Lymphatic vessels (I) collect excess tissue fluid that did not re-enter the blood capillaries and return it (as lymph) back to the subclavian veins.

Award 1 mark for each correct match up to a maximum of 10 marks:
- 1: E
- 2: A
- 3: B
- 4: C
- 5: D
- 6: F
- 7: G
- 8: H
- 9: J
- 10: I

1. Follicle Stimulating Hormone (D) stimulates primary follicles in the ovaries to mature into Graafian follicles.
2. Estrogen (A) repairs and starts building up the endometrium after menstruation ends.
3. Luteinising Hormone (C) reaches a peak concentration on day 14, stimulating ovulation and causing the follicle wall to convert into the corpus luteum.
4. Progesterone (B) is produced by the corpus luteum to maintain the highly vascularised endometrium during the secretory phase.
5. The oviduct (E) carries the secondary oocyte where it can be fertilised by a sperm cell.
6. The uterus (F) (specifically the endometrium layer) is where the blastocyst implants and the placenta develops.
7. The testis (G) contains seminiferous tubules for spermatogenesis and interstitial cells that secrete testosterone.
8. The epididymis (H) is a tightly coiled tube situated on the outer surface of each testis where sperm mature.
9. The vas deferens (I) is the muscular duct that transports mature sperm to the urethra.
10. The prostate gland (J) adds alkaline secretions to sperm to neutralise the acidic environment of the vagina.

Award 1 mark for each correct match up to a maximum of 10 marks:
- 1: D
- 2: A
- 3: C
- 4: B
- 5: E
- 6: F
- 7: G
- 8: H
- 9: I
- 10: J

(a) Drinking 1000 cm³ of water dilutes the blood plasma, lowering its solute concentration and raising its water potential. This change is detected by osmoreceptors in the hypothalamus of the brain. The hypothalamus subsequently signals the posterior pituitary gland to secrete less antidiuretic hormone (ADH) into the blood. Due to lower levels of ADH, the cells lining the collecting ducts of the nephrons in the kidneys become less permeable to water. As a result, less water is reabsorbed by osmosis back into the surrounding capillary networks, leaving a larger volume of unabsorbed water in the collecting ducts to be excreted as dilute urine.

(b) (i) Rate of urine production = \(\frac{\text{Total urine volume}}{\text{Time}}\) = \(\frac{900 \text{ cm}^3}{150 \text{ min}} = 6.0 \text{ cm}^3\text{/min}\).
(ii) Percentage of water excreted = \(\frac{\text{Volume of urine}}{\text{Volume of water ingested}} \times 100\) = \(\frac{900}{1000} \times 100 = 90\%\).

(c) To ensure a valid comparison, variables that must be controlled include: the physical activity levels of the subjects during the testing window (both kept at rest), the ambient room temperature (21 °C), the type of liquid consumed (pure water), and the initial hydration state or diet of the subjects prior to the test. A larger sample size increases reliability because it reduces the impact of anomalous individual biological differences (such as varying metabolic rates or kidney efficiencies) and allows for the calculation of an accurate, representative mean.

(a) [Max 6 marks]
- Blood dilution / lower blood concentration / increased water potential of blood identified (1)
- Change detected by osmoreceptors in the hypothalamus (1)
- Pituitary gland secretes less ADH into the blood (1)
- Collecting ducts of the kidneys become less permeable to water (1)
- Less water is reabsorbed by osmosis back into the blood capillaries (1)
- Resulting in a larger volume of dilute urine (1)

(b) (i) [1.5 marks]
- Correct working: 900 / 150 (0.5)
- Correct answer with units: 6.0 cm³/min (1)
(ii) [2 marks]
- Correct working: (900 / 1000) * 100 (1)
- Correct answer: 90% (1)

(c) [5 marks]
- Any three controlled variables listed: physical activity level, room temperature, age, body mass, pre-test diet/fluid intake (3)
- Explaining reliability: minimizes the effect of individual anomalies / allows calculation of a reliable mean (2)

(a) Both parents are carriers, meaning their genotypes are Ff. The gametes they produce are F and f. The genetic cross yields the following offspring genotypes in a 1:2:1 ratio: FF (normal), Ff (carrier), Ff (carrier), and ff (affected). The probability of producing an offspring with cystic fibrosis (genotype ff) is 25% or 1 in 4.

(b) The mutation in the CFTR gene causes the chloride channel protein to be absent or non-functional. Consequently, chloride ions cannot be transported out of epithelial cells into the mucus layer of the airways. As solute concentration outside the cells remains low, water does not move out of the cells by osmosis. The mucus lining the trachea and bronchi therefore remains exceptionally thick and sticky. Cilia on the ciliated epithelium cannot sweep this thick mucus out of the airways. Pathogenic bacteria and viruses become trapped in the stagnant mucus, leading to chronic respiratory infections.

(c) Support for newborn screening: Early diagnosis allows for immediate treatment, such as airway clearance therapy, specialized diet, and pancreatic enzyme supplements, which significantly slows lung damage and improves life expectancy. It also provides parents with genetic counseling for future family planning. Arguments against: Screening can yield false-positive results, causing unnecessary panic, stress, and anxiety for parents. Identifying carrier status (Ff) may cause psychological distress or future discrimination regarding employment or health insurance. Testing infants also raises ethical issues regarding lack of personal consent.

(a) [4.5 marks]
- Correct parental genotypes identified: Ff and Ff (1)
- Correct gametes shown: F and f (1)
- Punnett square with correct combinations: FF, Ff, Ff, ff (1)
- Genotype ff correctly identified as the affected genotype (0.5)
- Probability correctly stated: 25% / 1 in 4 / 0.25 (1)

(b) [5 marks]
- CFTR gene mutation leads to defective/absent chloride channels (1)
- Chloride ions are not actively transported out of cells (1)
- Water does not move out of cells by osmosis (1)
- Mucus remains extremely thick and sticky (1)
- Cilia are unable to sweep/move the mucus, trapping pathogens and causing infection (1)

(c) [5 marks]
- Support points (Max 3): Early diagnosis allows earlier treatment (1); improves prognosis/life expectancy (1); allows parents to access genetic counseling (1)
- Against points (Max 3): False positives cause anxiety (1); carrier identification can lead to stigma or insurance discrimination (1); ethical concerns over testing individuals without their consent (1)
- Note: Must include arguments from both sides for full marks.

(a) The Athlete's heart rate is consistently lower than the Control's at all measured recovery intervals. The Athlete's heart rate drops much more rapidly after exercise stops, reaching a stable resting baseline of 60 bpm by minute 4, whereas the Control's heart rate is still declining at minute 5 (72 bpm). This indicates that the Athlete's cardiovascular system is more efficient. Due to a larger stroke volume, the Athlete's heart delivers oxygen more efficiently, clearing the oxygen debt and flushing out lactic acid faster than the Control's.

(b) (i) For the Athlete: Heart rate decreased from 145 to 65 bpm. Decrease = \(145 - 65 = 80 \text{ bpm}\). Percentage decrease = \(\frac{80}{145} \times 100 = 55.17\%\) (accept 55.2% or 55%).
(ii) For the Control: Heart rate decreased from 160 to 95 bpm. Decrease = \(160 - 95 = 65 \text{ bpm}\). Percentage decrease = \(\frac{65}{160} \times 100 = 40.63\%\) (accept 40.6% or 41%).

(c) Regular aerobic training causes cardiac hypertrophy, where the muscular walls of the ventricles (primarily the left ventricle) become thicker and stronger, and the chamber volume increases. This allows the heart to have a significantly higher stroke volume (pumping more blood per beat). Because of this increased stroke volume, the heart needs to beat fewer times per minute to maintain the required cardiac output, resulting in a lower resting heart rate. During recovery, the increased stroke volume facilitates faster delivery of oxygen to cardiac and skeletal muscles, accelerating the clearance of metabolic wastes and recovery of oxygen debt.

(a) [6 marks]
- Athlete's heart rate is lower than Control's at all time points (1)
- Athlete's heart rate drops faster/more steeply (1)
- Athlete's heart rate reaches and plateaus at resting rate (60 bpm) earlier / by minute 4 (1)
- Explaining: Athlete has a higher stroke volume / stronger cardiac muscle (1)
- Explaining: More rapid delivery of oxygen to clear oxygen debt (1)
- Explaining: Faster clearance of lactic acid (1)

(b) (i) [1.5 marks]
- Correct working: ((145 - 65) / 145) * 100 (0.5)
- Correct answer: 55.2% / 55.17% (1)
(ii) [2 marks]
- Correct working: ((160 - 95) / 160) * 100 (1)
- Correct answer: 40.6% / 40.63% / 41% (1)

(c) [5 marks]
- Ventricular cardiac hypertrophy / thicker muscular walls of the heart (1)
- Increased volume of ventricle chambers (1)
- Increased stroke volume (volume of blood ejected per beat) (1)
- Higher cardiac output at peak performance (1)
- Bradycardia / lower resting heart rate due to higher stroke efficiency (1)
- Enhanced blood flow / coronary capillary density to cardiac muscle (1)

Part (a) (i) The formula is: \( \text{Magnification} = \frac{\text{Image size}}{\text{Actual size}} \). (ii) First, convert 18 mm to micrometres (\mu m): \( 18 \text{ mm} \times 1000 = 18,000 \text{ \mu m} \). Then apply the formula: \( \text{Magnification} = \frac{18,000 \text{ \mu m}}{0.3 \text{ \mu m}} = 60,000 \). The magnification is \( \times 60,000 \). (iii) The main function of the rough endoplasmic reticulum is protein synthesis and transport of synthesized proteins. Part (b) (i) Parental phenotypes: Carrier parent \( \times \) Carrier parent. Parental genotypes: \( Ff \times Ff \). Gametes: \( F \) and \( f \) from both parents. Punnett square / offspring genotypes: \( FF \), \( Ff \), \( Ff \), \( ff \). Offspring phenotypes: \( FF \) and \( Ff \) are healthy/unaffected (3/4 or 75%), \( ff \) has cystic fibrosis (1/4 or 25%). (ii) The probability that their child will be a carrier (genotype \( Ff \)) is \( \frac{2}{4} \) or \( 0.5 \) (50%). (iii) Cystic fibrosis leads to the production of thick, sticky mucus. This mucus clogs the bronchioles and alveoli, reducing the surface area available for gas exchange and making ventilation difficult, while also trapping bacteria and leading to infections.

Part (a) (i) Magnification = Image size / Actual size [1]. (ii) Converts mm to \mu m correctly: 18 mm = 18,000 \mu m [1]; Divides image size by actual size: 18,000 / 0.3 [1]; Correct final answer: (\times) 60,000 [1]. (iii) Synthesizes / translates proteins [1]. Part (b) (i) Correct parental genotypes: Ff and Ff [1]; Correct gametes: F and f from both parents [1]; Correct offspring genotypes: FF, Ff, Ff, ff [1]; Correct offspring phenotypes matched to genotypes [1]; Phenotypic ratio correctly identified (3 unaffected : 1 cystic fibrosis) [1]. (ii) Correct probability of carrier: 0.5 / 50% / 1/2 [1]. (iii) Produces thick/sticky mucus [1]; blocks airways / bronchioles / reduces gas exchange / increases infections [1].

Part (a) (i) First, convert the image diameter from mm to \mu m: \( 17.5 \text{ mm} \times 1000 = 17,500 \text{ \mu m} \). Next, use the actual size formula: \( \text{Actual size} = \frac{\text{Image size}}{\text{Magnification}} \). Apply values: \( \text{Actual size} = \frac{17,500 \text{ \mu m}}{2500} = 7 \text{ \mu m} \). (ii) Mature red blood cells lack a nucleus to create more room within the cell. This allows them to contain more haemoglobin molecules, which increases the volume of oxygen they can bind and transport. Part (b) (i) Parental Phenotypes: Normal-vision female (carrier) \( \times \) Normal-vision male. Parental Genotypes: \( X^B X^b \times X^B Y \) (Note: The mother must be a carrier, \( X^B X^b \), because she inherited her father's only X chromosome, which carried the colour-blind allele \( X^b \)). Gametes: \( X^B \) and \( X^b \) from the mother; \( X^B \) and \( Y \) from the father. Offspring Genotypes: \( X^B X^B \) (normal female), \( X^B X^b \) (carrier female), \( X^B Y \) (normal male), \( X^b Y \) (colour-blind male). (ii) Out of all possible sons (who can be \( X^B Y \) or \( X^b Y \)), the probability that a son is colour-blind is \( 0.5 \) (or 50%). (iii) Colour blindness is sex-linked and carried on the X chromosome. Males have only one X chromosome (XY). Therefore, if a male inherits a single recessive allele (\( X^b \)), he will express the condition. Females have two X chromosomes (XX) and require two copies of the recessive allele to be colour-blind, which is statistically much less likely.

Part (a) (i) Correct unit conversion: 17.5 mm = 17,500 \mu m [1]; Division step: 17,500 / 2500 [1]; Correct final answer: 7 \mu m [1]. (ii) Lacking a nucleus increases space inside the cell [1]; to allow more haemoglobin to be packed / to transport more oxygen [1]. Part (b) (i) Correct parental genotypes: X^B X^b and X^B Y [1]; Correct gametes: X^B and X^b from mother, X^B and Y from father [1]; Correct offspring genotypes: X^B X^B, X^B X^b, X^B Y, X^b Y [1]; Correct corresponding phenotypes linked to genotypes [1]; Parental phenotypes correctly stated [1]. (ii) Correct probability for a son: 0.5 / 50% / 1/2 [1] (accept 0.25 / 25% if clearly specified as the probability out of all children). (iii) Males are XY / have only one X chromosome [1]; so they only need one copy of the recessive allele to express the disease, whereas females need two recessive alleles [1].